Graduate Physics Homework Solutions

Graduate Coursework WES C. ERBSEN

September 2010 - December 2011

This document last updated on January 27, 2013

Contents Contents

i

1 Electrodynamics II 1.1 Homework #1 . . 1.2 Homework #2 . . 1.3 Homework #3 . . 1.4 Homework #4 . . 1.5 Homework #5 . . 1.6 Homework #6 . . 1.7 Homework #7 . . 1.8 Homework #8 . . 1.9 Homework #9 . .

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1 1 14 25 34 40 46 54 63 78

2 Quantum Mechanics II 2.1 Homework #1 . . . . 2.2 Homework #2 . . . . 2.3 Homework #3 . . . . 2.4 Homework #4 . . . . 2.5 Homework #5 . . . . 2.6 Homework #6 . . . . 2.7 Homework #7 . . . . 2.8 Homework #9 . . . . 2.9 Homework #10 . . . Appendix * . .A. . . . . . . . . Appendix * . .B. . . . . . . . .

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89 89 101 124 144 161 174 187 198 213 222 225

3 Statistical Mechanics 3.1 Homework #1 . . . 3.2 Homework #2 . . . 3.3 Homework #3 . . . 3.4 Homework #4 . . . 3.5 Homework #7 . . .

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227 227 235 247 262 275

Methods #3 . . . . . #4 . . . . . #5 . . . . . #6 . . . . .

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283 283 288 294 303

4 Mathematical 4.1 Homework 4.2 Homework 4.3 Homework 4.4 Homework

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W. Erbsen

4.5 4.6 4.7 4.8 4.9 4.10

Homework Homework Homework Homework Homework Homework

CONTENTS

#7 . #8 . #9 . #10 #11 #12

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310 315 324 332 342 355

5 Departmental Examinations 5.1 Quantum Mechanics . . . . 5.2 Electrodynamics . . . . . . 5.3 Modern Physics . . . . . . . 5.4 Statistical Mechanics . . . .

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367 367 410 449 497

Chapter 1

Electrodynamics II 1.1

Homework #1

Problem 9.1 A copper ring of radius a is fixed at a distance d (with a  d) directly above an identical copper ring. Each ring has a resistance R for circulating currents. An increasing current I = Io t/τ is applied in the lower ring. Neglect the self-inductance of each ring, and make appropriate approximations. a)

Find the dipole moment in the lower ring.

b)

Find the magnetic flux through the upper ring.

c)

Find the induced EMF and the current in the upper ring.

d)

Find the induced dipole moment of the upper ring.

e)

Show that the force between the rings is∗ F =

12π 4 a8 Io2 t c3 Rd7 τ 2

(1.1.1)

Is the force repulsive or attractive?

Solution a)

∗ Errata

An expression for the magnetic dipole moment for a circular loop can be realized by applying multipole expansion of the magnetic scalar potential (Φm ) and noting that the coefficients of P` (cos θ) /r `+1 are in fact the magnetic dipole moments (from Franklin p. 209). ?   2πIa`+1 − 1/2 Iπa2 µ` = − −→ µ1 = (1.1.2) `+1 /2 c c

indicates that c3 → c4 .

W. Erbsen

HOMEWORK #1

Where µ`=1 represents the magnetic moment of a dipole. The elementary magnetic dipole µ is found by taking the limit of (1.1.1) as a → 0 keeping Ia2 fixed. We subsequently arrive at µ= b)

πa2 Io t ˆ z cτ

(1.1.3)

The magnetic flux is of course given by ΦB =

I

B · dA

s

While the magnetic field of a magnetic dipole is given by (7.108) on p. 209 as  µ · r  3 (µ · ˆ r) ˆ r−µ B = −∇ = r3 r3

(1.1.4)

The closed surface we choose to integrate over is the upper hemisphere closed on the bottom by a disk, as inspired by Griffiths in Ex. 8.2 on p. 353. ? The general idea is thatH since the source of the field lies outside the surface, the total flux through the volume is zero: v B · dA = 0. We then note that the flux entering through the disk is the same as that exiting through the top of the hemisphere: Φdisk = −Φhemisphere . It i easiest to calculate the flux through the hemisphere. We also recognize that since a  d, we can use the following approximation: tan θ ≈ a/d −→ sin θ = a/d cos θ ≈ a/d, as per the small angle approximation. We can then proceed to find the flux through the hemisphere: Z θ ΦB = B · 2πr 2 sin θ dθ 0

 3 (µ · ˆ r) ˆ r−µ =2π r 2 sin θ dθ ˆ z r3 0  Z θ Iπa2 2π ˆ z sin θ dθ ˆ z = r 0 c Z 2π 2 a2 I θ = sin θ dθ cr 0 Z 2π 2 a2 I θ  a  ≈ d ( a/d ) cr d 0 2π 2 a2 I a2 ≈ (1.1.5) cr d2 √ Since d  a, we can approximate r = d2 + a2 ≈ d2 , and if we recall that I = Io t/τ , (1.1.5) becomes Z

θ



ΦB ≈ c)

2π 2 a4 Io t cd3 τ

(1.1.6)

1 dφB c dt

(1.1.7)

The induced EMF in the upper ring is given by E=−

3

CHAPTER 1: ELECTRODYNAMICS II

Substituting (1.1.6) into (1.1.7), we have E≈−

1 d c dt



2π 2 a4 Io t cd3 τ



−→ E ≈ −

2π 2 a4 Io c2 d 3 τ

(1.1.8)

And the current in the upper ring (I 0 ) can be found from Ohm’s Law, I0 ≈ d)

(1.1.9)

To find the induced dipole moment in the upper ring (µ0 ), we proceed in the same spirit that led us to (1.1.3), but this time we use I 0 from (1.1.9): 0

µ ≈ e)

E 2π 2 a4 Io −→ I 0 ≈ − 2 3 R c d τR



2π 2 a4 Io − 2 3 c d τR



2π 3 a6 Io πa2 ˆ z −→ µ0 ≈ − 3 3 ˆ z c c d τR

(1.1.10)

The force on a dipole µ in a magnetic field B is given by Franklin as (7.144) on p. 216: F = ∇ (B · µ)

(1.1.11)

Assuming that the charges are fixed and that the force of the lower loop on the upper loop is the same as that of the upper loop and the lower loop, we can say that the field is provided by the lower loop, and the opposing magnetic moment is that of the upper ring. Substituting the expression for B and (1.1.10) into (1.1.11),  2    πa Io t 2π 3 a6 Io F ≈∇ ˆ z · − 3 3 ˆ z cd3 τ c d τR  4 8 2  2π a I t (1.1.12) ≈ − ∇ 4 6 2o c d τ R And in our case the gradient reduces to ∇ = (∂/∂r) ˆ r where r → d, and (1.1.12) becomes F≈

12π 4 a8 Io2 t ˆ z c4 d 7 τ 2 R

(1.1.13)

Which is identical to (1.1.1) (with correction). Since (1.1.13) is positive, the force is repulsive .

Problem 9.2 a)

Find the mutual inductance of the two rings in Problem 9.1 (making the suitable approximations).

W. Erbsen

HOMEWORK #1

b)

Use their mutual inductance to find the EMF induced in the upper ring for the current I = Io t/τ in the lower ring.

c)

Find the mutual interaction energy in this system.

Solution a)

We know from (9.18) on p. 245 of Franklin that Φ2 = cM21 I

(1.1.14)

Using (1.1.6) from the previous problem, we can solve (1.1.14) to find the mutual inductance: M21 ≈ b)



1 cI



2π 2 a4 I cd3



−→ M21 ≈

2π 2 a4 c2 d 3

(1.1.15)

As before, the induced EMF is given by E=

1 dφB c dt

(1.1.16)

The magnetic flux can be found from the mutual inductance by substituting (1.1.15) into (1.1.14), and the induced EMF can then be found by substituting the result into (1.1.16): E≈

1 d c dt



cIo t τ



2π 2 a4 c2 d 3



−→ E ≈

2π 2 a4 Io c2 d 3 τ

(1.1.17)

which is identical to (1.1.8). c)

The mutual interaction energy is most easily calculated from (5.152) from Jackson on p. 215: ? N

w=

N

N

XX 1X Li Ii2 + Mij Ii Ij 2 i=1

(1.1.18)

i=1 j6=i

Ignoring the self inductance terms in (1.1.18), we find that w=

1 [M12 I1 I2 + M21 I2 I1 ] 2

(1.1.19)

Where the mutual inductances are of course equal. Substituting (1.1.9) and (1.1.15) into (1.1.19), w≈



2π 2 a4 c2 d 3



Io

t τ

  2π 2 a4 Io 4π 4 a8 I 2 t − 2 3 −→ w ≈ − 4 6 2 o c d τR c d τ R

Problem 9.3 A long straight copper wire of radius a and resistance R carries a constant current I.

5

CHAPTER 1: ELECTRODYNAMICS II

a)

Find the electric and magnetic fields at the surface of the wire.

b)

Integrate the Poynting power flux through the surface of a piece of the wire of length L to show that the power through the surface equals I 2 R.

c)

Find the electromagnetic energy and momentum in this piece of wire.

Solution a)

Assuming that the wire can be approximated as infinite, the magnetic field can be found using Amp´ere’s Law: I

B · d` =

c

4π 2I ˆ Ienc −→ B = φ c ca

(1.1.20)

To find the electric field, V =−

Z

f

E · d`

i

Where in our case E = |E|ˆ z and d` = |d`|ˆ z = dz ˆ z. We also note that V = EL = IR −→ E = b)

IR ˆ z L

(1.1.21)

The “Poynting power flux” is most likely referring to the Poynting vector, which is really the “Poynting power flux density.” We must first find S using (9.48) on p. 249 c S= (E × H) (1.1.22) 4π Where in our case H = B. First, we calculate E × H using (1.1.20) and (1.1.21),     IR 2I ˆ E×H = ˆ z × φ L ca  2I 2 R  ˆ = ˆ z×φ caL 2I 2 R =− ˆ r (1.1.23) caL H The power loss is defined by P = S S · dA. It is to our advantage to integrate in cylindrical coordinates, and we recall that in our case ˆ r·ˆ z = 0. Therefore, we can find the Poynting power flux using (1.1.22) and (1.1.23): I c P = (E × H) · dA 4π S   Z 2π Z L c 2I 2 R = − ˆ r · aˆ r dφdz 4π caL 0 0 I2R =− (2π) (L) (1.1.24) 2πL

W. Erbsen

HOMEWORK #1

It is easy to see that (1.1.24) reduces to P = I2R c)

The electromagnetic energy density is given by (9.47) on p. 249 as UEM =

1 (E · D + B · H) 8π

Where in our case D = E and H = B. Therefore, (1.1.25) becomes
1 E 2 + B2 UEM = 8π

(1.1.25)

(1.1.26)

Where E and B are the values within the conductor. If we assume that the electric field is uniform, then we can use our previous value. To find the magnetic field,
I 4π πr 2 I 2I ˆ B · d` = −→ B = 2 r φ (1.1.27) 2 c πa ca C Substituting (1.1.21) and (1.1.27) into (1.1.26), " 2  2 # 1 IR 2I UEM = + 8π L ca2 r

(1.1.28)

To find the electromagnetic energy, we integrate the electromagnetic energy density (1.1.28) over the entire volume: Z w= UEM dτ V 2  2 # Z " 1 IR 2I = + dτ 8π V L ca2 r   Z Z 1 I 2 R2 4I 2 3 = r drdφdz + r drdφdz 8π L2 c2 a 4 " # Z Z Z L Z a Z 2π Z L 2π 1 I 2 R2 a 4I 2 3 = r dr dφ dz + 2 4 r dr dφ dz 8π L2 0 c a 0 0 0 0 0       1 I 2 R 2 a2 4I 2 a4 = (2π) (L) + 2 4 (2π) (L) 8π L2 2 c a 4 Simplifying this expression leads to: w=

I 2 R 2 a2 I2L + 2 8L 4c

The electromagnetic momentum stored in the field is given by (9.80) on p. 255 in Franklin: Z 1 PEM = E × B dτ (1.1.29) 4πc V First calculating the cross product, E×B=



   IR 2I ˆ 2I 2 R ˆ z × r φ =− 2 rˆ r 2 L ca ca L

(1.1.30)

7

CHAPTER 1: ELECTRODYNAMICS II

Substituting (1.1.30) into (1.1.29),  Z  Z L Z 2π Z a 2I 2 R I2R − 2 r dτ = − r drdφdz −→ PEM = 0 ca L 2πc2 a2 L 0 0 V 0

1 = 4πc

PEM

Problem 9.4 The upper and lower curves of the hysteresis loop of a hard ferromagnetic are given by B+ =Bo [tanh (H/Ho + .5) − .1] , B− =Bo [tanh (H/Ho − .5) + .1] ,

(1.1.31a) (1.1.31b)

respectively, for −1.5 < H/Ho < +1.5. a)

Find the energy lost from the magnetic field in one cycle.

b)

If Bo = 3 kilogauss, Ho = 1 gauss, and the frequency is 60 Hz, find the power loss in watts.

Solution a)

The work done per unit volume in each cycle of a hysteresis loop is given by the total area, I w = B · dH (1.1.32) Where B = (B+ − B− ). Applying this to (1.1.32) over the specified intervals, Z +1.5 w= (B+ − B− ) dH =

−1.5 +1.5 

Z



Bo tanh

−1.5



      H H + .5 − .1 − Bo tanh − .5 + .1 dH Ho Ho

If we define α = H/Ho and dα = (1/Ho ) dH, (1.1.33) becomes Z +1.5 w =Ho [Bo (tanh (α + .5) − .1) − Bo (tanh (α − .5) + .1)] dα −1.5

=Bo Ho

Z

+1.5

−1.5

=Bo Ho

(

tanh (α + .5) dα −

Z

+1.5

−1.5

tanh (α − .5) dα − .2

Z

+1.5

dα

−1.5

(1.1.33)



) +1.5  +1.5  +1.5 ln [cosh (α + .5)] − ln [cosh (α − .5)] − .2 α dα −1.5

−1.5

−1.5

=Bo Ho {ln [cosh (2)] − ln [cosh (1)] − ln [cosh (1)] + ln [cosh (2)] − .6} Evaluating (1.1.34) numerically leads to w ≈ 1.18244 Bo Ho

(1.1.34)

W. Erbsen

HOMEWORK #1

b)

To find the energy loss according to the given parameters, w≈

(1.18244)(1 gauss)(3 · 103 gauss) −→ 1/60 Hz

w ≈ 212.832 · 103 erg · s−1 w ≈ 212.832 · 10−4 Watts

(CGS) (SI)

Problem 9.5 Two point charges, each of charge q, are a distance 2d apart. a)

Find the Maxwell stress tensor [T] on the plane surface midway between the charges.

b)

Find the force on either charge by integrating [T] · dS over a plane surface, closing the surface with a large hemisphere of radius R. (Show that the integral over the hemisphere vanishes in the limit R → ∞.)

c)

Repeat parts (a) and (b) if the charges have opposite signs.

d)

What would the force on either charge be if they were immersed in a simple dielectric of infinite extent?

Solution a)

The Maxwell stress tensor is given in Griffiths as (8.19) on p. 352 as     1 1 1 2 2 [T] = εo Ei Ej − δij E + Bi Bj − δij B 2 µo 2

(1.1.35)

For the case in question, our charges are stationary so that the second portion of (1.1.35) vanishes. We can then express [T] in matrix form as   2 1 2   Txx Txy Txz Ex − /2 E Ex Ey Ex Ez [T] = Tyx Tyy Tyz  = εo  Ex Ey Ey2 − 1/2 E2 Ey Ez  2 Tzx Tzy Tzz Ez Ex Ey Ez Ez − 1/2 E2

In order to resolve the individual components of [T], we must find the corresponding electric fields. We note that the charges are located on the x-axis, and the plane between the charges is the yz-plane. Furthermore, we recall that the electric field of an electric dipole is given by E=

1 p 4πεo r 3

(1.1.36)


1/2
1/2 And in our case r = x2 + y2 + z 2 ⇒ d2 + y2 + z 2 . With this in mind, we can now find the electric fields in the x, y, and z directions respectively: Ex =0 1 ey 1 er cos θ y ˆ⇒ ˆ r 2πεo (d2 + y2 + z 2 ) 3/2 2πεo (d2 + r 2 ) 3/2 1 ez 1 er sin θ Ez = ˆ z⇒ ˆ r 2πεo (d2 + y2 + z 2 ) 3/2 2πεo (d2 + r 2 ) 3/2

Ey =

9

CHAPTER 1: ELECTRODYNAMICS II

Where I shamelessly switched to polar coordinates. The total electric field is then E 2 =E2x + E2y + E2z

!2 !2 1 er cos θ 1 er sin θ = ˆ r + ˆ r 2πεo (d2 + r 2 ) 3/2 2πεo (d2 + r 2 ) 3/2 2 
e r2 = cos2 θ + sin2 θ 3 2πεo (d2 + r 2 )

=

e2 r2 4π 2 ε2o (d2 + r 2 )3

We now have all the tools required to calculate the components of [T]: r2 e2 2 2 8π εo (d2 + r 2 )3   e2 r2 1 2 = 2 2 cos θ − 4π εo (d2 + r 2 )3 2

Txx = − Tyy

e2 r2 sin θ cos θ 2 2 4π εo (d2 + r 2 )3   r2 1 e2 2 sin θ − = 2 2 4π εo (d2 + r 2 )3 2 =Txz = Tyx = Tzx = 0

Tyz =Tzy = Tzz Txy We now define

e2 r2 (SI) 4π 2 ε2o (d2 + r 2 )3 r2 =4e2 (CGS) 3 2 (d + r 2 )

β=

And the Maxwell stress tensor becomes

b)

 −β/2  0 [T] = εo 0

 0 0
β cos2 θ − 1/2 β sin θ cos θ
 β sin θ cos θ β sin2 θ − 1/2

To integrate the Maxwell stress tensor, we take the route suggested in the prompt. We take a hemisphere of radius R with the base coinciding with the plane surface halfway between the charges (the yz-plane). If we let the hemisphere expand to very large values, then the field at the boundary of the hemisphere is seen as a dipole, ∝ R−3 , and the surface integral then varies like R−4 . If we take R out to infinity, then this portion of the surface integral vanishes since the components of [T] go to zero. There remains then only the force across the plane boundary in the yz-plane. To find the force on either charge, we integrate [T] · dS over the interstitial plane. We note that the force is in the x-direction, and the transversal forces are cancelled. Therefore, the only component of [T] that we need to integrate is Txx , as tabulated previously. We also recall that for the equatorial disk dS = −r drdφ x ˆ, and so the force is Z F = [T] dS S

W. Erbsen

HOMEWORK #1

= − 2π

Z

∞

Txxr dr x ˆ

0

! r2 e2 = − 2πεo − 2 2 r dr x ˆ 8π εo (d2 + r 2 )3 0 ) Z ∞ r3 e2 = dr x ˆ partial fractions 4πεo 0 (d2 + r 2 )3 # Z ∞" e2 d2 r r = ˆ 2 − 3 dr x 4πεo 0 (d2 + r 2 ) (d2 + r 2 ) "Z # Z ∞ ∞ e2 r r 2 = dr − d ˆ 3 dr x 4πεo 0 (d2 + r 2 )2 (d2 + r 2 ) 0  Z ∞  Z 1 1 d2 ∞ 1 e2 du − dr x ˆ = 4πεo 2 0 u2 2 0 u3 ∞ ∞     2 e2 1 1 1 −d − = x ˆ − 2 2 2 2 4πεo 2 (d + r ) 0 2 2 (d + r ) 0   e2 1 1 d2 1 = − x ˆ 4πεo 2 d2 2 2d4 Z

∞

)

u = d2 + r 2 ,

du = 2rdr

(1.1.37)

From (1.1.37), it is only a short leap to our answer: F=

1 e2 x ˆ 4πεo (2d)2

(SI),

F=

e2 2

(2d)

x ˆ

(CGS)

Which is exactly what we expected; proceeding through Coulomb’s law yields the precisely same result. c)

If the charges are opposite, then the symmetry changes, however the process is very much the same. The electric fields in this case are Ex =

ed 1 x ˆ 2πεo (d2 + r 2 ) 3/2

Ey =Ez = 0 And the total electric field is E 2 = Ex2 =

1 e2 d2 4π 2 ε2o (d2 + r 2 )3

And now to calculate the components of [T], 1 e2 d2 4π 2 ε2o (d2 + r 2 )3 1 e2 d2 =− 2 2 4π εo (d2 + r 2 )3

Txx = Tyy

1 e2 d2 4π 2 ε2o (d2 + r 2 )3 =Txz = Tyx = Tyz = Tzx = Tzy = 0

Tzz = − Txy

11

CHAPTER 1: ELECTRODYNAMICS II

And let’s go ahead and define β= Then the Maxwell stress tensor is

1 e2 d2 2 2 4π εo (d2 + r 2 )3

 β [T] = εo  0 0

0 −β 0

 0 0  −β

To find the force on either charge, we follow the same logic as before, Z F = [T] d · S S Z ∞ = − 2πεo Txx r dr x ˆ 0 ! Z ∞ e2 d2 1 r dr x ˆ = − 2πεo 4π 2 ε2o (d2 + r 2 )3 0 ) Z r e2 d2 ∞ =− dr x ˆ u = d2 + r 2 , du = 2rdr 2πεo 0 (d2 + r 2 )3 Z e2 d2 ∞ 1 du x ˆ =− 4πεo 0 u3 ∞ " e2 d2 1 =− − ˆ x 2 2 2 4πεo 4 (d + r ) 0   e2 d2 1 =− x ˆ 4πεo 4d4

(1.1.38)

From (1.1.38) it is easy to see that the force is F=−

1 e2 x ˆ 4πεo (2d)2

(SI),

F=−

e2 2

(2d)

x ˆ

(CGS)

Which is what we would expect. d)

If the system is immersed into a linear dielectric of permittivity ε, then this will effect our answers only by including a factor of ε in the denominator of our expressions, which weakens the force of attraction or repulsion.

equal charges opposite charges

Problem 9.7

vacuum e2 F= x ˆ (2d)2 2 e F=− ˆ 2 x (2d)

dielectric 1 e2 −→ F = x ˆ ε (2d)2 2 1 e −→ F = − x ˆ ε (2d)2

(CGS)

W. Erbsen

HOMEWORK #1

A magnetic dipole µ is located at the center of a uniform electric charge distribution of radius R, charge e, and mass m. a)

Find the electromagnetic angular momentum of this configuration.

b)

Find the value of the radius R for which the g-factor of this configuration equals 2.

Solution a)

We first recall that the electromagnetic angular momentum is given by (9.113) on p. 260 as Z 1 LEM = r × (E × B) dτ (1.1.39) 4πc V Now, we find the electric field by


e 4/3 πr 3 er E= 4 ˆ r= 3 ˆ r /3 πR3 r 2 R

(1.1.40)

The magnetic field is given by (7.108) on p. 209 as B=

3 (µ · ˆ r) ˆ r−µ 3 r

We also recall that in our case µ = |µ| ˆ z, and therefore (1.1.41) becomes µ B=− 3 ˆ z r We now use (1.1.40) and (1.1.42) to examine the cross product in (1.1.39):   er   µ  eµ eµ  ˆ = eµ sin θ φ ˆ E×B= ˆ r × − 3 ˆ z = − 3 2 (ˆ r׈ z) = − 3 2 − sin θ φ 3 R r R r R r R3 r 2

Now substituting (1.1.43) into (1.1.39), Z  eµ  1 ˆ dτ LEM = r× sin θ φ 3 2 4πc V R r Z  e sin θ  ˆ dτ = ˆ r × φ 4πcR3 V r Z e sin θ ˆ 2 =− θr sin θ drdθdφ 4πcR3 V r Z R Z π Z 2π e =− rdr sin2 θ θˆ dθdφ 4πcR3 0 0 0

(1.1.41)

(1.1.42)

(1.1.43)

(1.1.44)

We now recall from the back cover of Griffiths that θˆ = cos θ cos φ x ˆ + cos θ sin φ y ˆ − sin θ ˆ z Substituting (1.1.45) into (1.1.44), Z R Z π Z 2π µe LEM = − rdr sin2 θ [cos θ cos φ x ˆ + cos θ sin φ y ˆ − sin θ ˆ z] dθdφ 4πcR3 0 0 0

(1.1.45)

13

CHAPTER 1: ELECTRODYNAMICS II

µe =− 8πcR

Z

π

0

Z

0

2π

[sin2 θ cos θ cos φ x ˆ + sin2 θ cos θ sin φ y ˆ − sin3 θ ˆ z] dθdφ | {z } | {z } | {z } I

II

Taking the angular integrals of I, II and III separately yields Z π  Z 2π I= sin2 θ cos θ dθ cos φ dφ x ˆ=0 0 0 Z π  Z 2π II = sin2 θ cos θ dθ sin φ dφ y ˆ=0 0 0 Z π  Z 2π 3 III = sin θ dθ dφ ˆ z 0 0 Z π   Z π 2 = 2π sin θ dθ − sin θ cos θ dθ ˆ z u = cos θ, du = − sin θ =2π

Z

0 π

sin θ dθ +

"0

π =2π − cos θ 0   2 =2π 2 − ˆ z 3 8π = ˆ z 3

Z

0 −1

(1.1.46)

III

(1.1.47a) (1.1.47b)



2

u du ˆ z #  3 −1 u ˆ z + 3 1 1

(1.1.47c)

Substituting (1.1.47a)-(1.1.47c) into (1.1.46), LEM = − b)

  µe 8π µe 0+0− ˆ z −→ LEM = ˆ z 8πcR 3 3cR

(1.1.48)

We recall from (7.134) on p. 214 that µ = −gµB J, and also that LEM = ~J. Combining these, as well as implementing (1.1.48), yields 3cRLEM gµB LEM 3cR gµB gµB e = −→ = −→ R = e ~ e ~ 3c~

(1.1.49)

Recall that the Bohr Magneton is given by µB =

e~ 2me

(1.1.50)

The appropriate values we require are e =1.60217733 · 10−19 C

~ =1.05457162 · 10−34 Kg · m2 · s−1 c =2.99792458 · 108 m · s−1

me =9.10938215 · 10−31 Kg

Now substituting (1.1.50) into (1.1.49) as well as the appropriate constants,
2 (2) 1.60217733 · 10−19 C R= −→ R = 3.13321563 · 10−17 m (6) (2.99792458 · 108 m · s−1 ) (9.10938215 · 10−31 Kg)

W. Erbsen

HOMEWORK #2

Problem 9.8 a)

Calculate e2 /~c in Gaussian units to verify Eq. (9.121).

b)

Calculate e2 /(4πo ~c) in SI units.

Solution a)

The values needed in this calculation are 3

1

e =4.80320680 · 10−10 cm /2 · g /2 · s−1

~ =1.05457162 · 10−27 erg · s

c =2.99792458 · 1010 cm · s−1

And so, we can now calculate α:  2 3 −10 /2 1/2 −1 4.80320680 · 10 cm g s 2 e e2 1 = −→ = 0.00729736 ⇒ −27 10 −1 ~c 1.05457162 · 10 erg s · 2.99792458 · 10 cm s ~c 137.036 b)

The necessary values to make the calculation in SI units are e =1.60217733 · 10−19 C

εo =8.85418781 · 10−12 F · m−1 ~ =1.05457162 · 10−34 J · s

c =2.99792458 · 108 m · s−1

And now,
2 1.60217733 · 10−19 C e2 = 4πo ~c 4π (8.85418781 · 10−12 F · m−1 ) (1.05457162 · 10−34 J) (s · 2.99792458 · 108 m · s−1 )

−→

1.2

e2 1 = 0.00729736 ⇒ 4πo ~c 137.036

Homework #2

Problem 10.1 The intensity of sunlight at the Earth’s surface is 12 · 105 ergs/(cm2 · sec). a)

Find the electric field of this radiation at the Earth’s surface. Express the answer in stavolts/cm and in volts/m (1 statvolt = 300 volts). [Answer: 960 V/m]

15

CHAPTER 1: ELECTRODYNAMICS II

b)

Find the radiation pressure (in dynes/cm 2 ) if the sunlight is 100% reflected at normal incidence. Compare this to atmospheric pressure. [Answer: 2 × 10−5 dynes/cm 2 ]

c)

Find the radiation pressure if i)

the sunlight is absorbed (no reflection).

ii) d)

the sunlight is specularly reflected from white sand. (The reflected radiation has the same intensity at any angle.)

What would the intensity, electric field, and radiation pressure be at the surface of the sun?

Solution a)

The intensity is called the time averaged Poynting vector, and is given by (10.27) of Franklin: ? ˆrε ck S= |E0|2 (1.2.1) 8π µ In SI units, the intensity is given by (9.61) in Griffiths on p. 381: ? S=

1 cε0 |E0 |2 ˆ k 2

(1.2.2)

Since we are given the intensity, it is a trivial task to find the electric field. We solve (1.2.2) for Eo ; s 1 2S ˆ −→ |E0| = S = cε0 |Eo |2 k (1.2.3) 2 ε0 c We recall that 1 erg = 10−7 J and ε0 ≈ 8.8542 A2 · s4 /(Kg · m3 ), (1.2.3) becomes s (2)(1200 Kg · m2 /s) |E0 | ≈ 950.54 V/m −→ |E0 | ≈ |E0 | ≈ 3.1685 · 10−2 statvolt/cm 8.8542 A2 · s4 /(Kg · m3 ) · 3 · 108 m/s b)

(1.2.4)

The radiation pressure is given by (10.33) in Franklin as prad =

2√ µ S c

(1.2.5)

If we recall that 1 erg = 1 dyne· cm, then we can solve (1.2.5) directly: prad =

(2)(12 · 105 dyne/(cm · sec)) −→ prad = 8 · 10−5 dynes/cm2 3 · 1010 cm/sec

(1.2.6)

The atmospheric pressure is patm ≈ 106 dynes/cm2 , which means that the radiation pressure multiplied by a factor of 1.25 · 1010 is approximately equal to the atmospheric pressure at 1 atm.

c)

The radiation pressure for the following unique cases is as follows:

i)

If the radiation is completely absorbed, then the radiation pressure would be just half that as if it were completely reflected from (1.2.6): 1√ prad = µ S −→ prad = 4 · 10−5 dynes/cm2 c

W. Erbsen

HOMEWORK #2

ii)

If the radiation is specularly reflected (evenly across all angles), then certainly the radiation pressure will be less than if it were completely reflected. In fact, for each angle, we are only interested in the 0o projection, going back towards the sun. But wait! Things aren’t quite this simple – we must integrate over all possible angles. Doing this yields the force:  Z π Z π/2  2S F = cos θ sin θ dθdφ c 0 0 Z π/2 2S = 2π cos θ sin θ dθ c 0 2πS = (1.2.7) c And to find the force we must divide by the surface area (note that I have neglected the arbitrary radial term, as it vanishes anyway): prad =

d)

2πS 1 F = · −→ prad = 4 · 10−5 dynes/cm2 A c 2π

To find the intensity at the surface of the sun, we assume that the total radiation flux at an altitude of Earth is equal to that at the surface of the sun. We know the intensity on the surface of the Earth, and also the distance of Earth from the sun (R ≈ 1.5 · 1011 m), so the total flux (as calculated from Earth’s orbit) is S tot = 4πR2 S Earth And we also know that the radius of the sun is r ≈ 7.0 · 108 m, so the intensity at the surface can be found as S sun =

R2 S Earth −→ S sun ≈ 5.5 · 1010 ergs/(cm2 · sec) r2

Problem 10.2 A beam of elliptically polarized light, propagating in the z-direction, passes through a polarizer in the xy-plane. The maximum transmitted intensity is 9Io when the polarizer is set at 30o to the x-axis. The minimum transmitted intensity is Io when the polarizer is set at 120o to the x-axis. a)

Find r, α, E+ , and E− in the circular basis.

b)

Find Ex and Ey in the plane basis.

c)

What would the transmitted intensities be if the polarizer were set along the x-axis, and then along the y-axis?

Solution I am going to solve this problem using Jones’ Calculus, starting in the linear basis. We note that the electric field of an (arbitrary) polarized wave can be expressed as

17

CHAPTER 1: ELECTRODYNAMICS II


E(z, t) = |Eox|eiδx x ˆ + |Eoy |eiδy y ˆ ei(k·z−ωt)
= Eox x ˆ + Eoy eiδ y ˆ ei(k·z−ωt)

(1.2.8)

And now (1.2.8) can be expanded and put into matrix form, ditching the explicit time dependence:     Eox Eox E(z, t) = = (1.2.9) Eoy eiδ Eoy (cos δ + i sin δ) We also recall that the arbitrary Jones’ Matrix for a linear polarizer at some angle θ is given by   cos2 θ sin θ cos θ M= sin θ cos θ sin2 θ We can apply the Jones’ Matrix to our incident elliptical wave given by (1.2.8), the result gives the exiting field.∗ Starting with the maximum field when the polarizer is oriented at θ = 30o , √    3/4 3 /4 Eox 0 2 √ |E30| = Eoy (cos δ + i sin δ) 3 /4 1/4   √ 1 2 2 3Eox + Eoy + 2 3 Eox Eoy cos δ = (1.2.10) 4 And similarly if the polarizer is oriented at 120o: √    Eox 1/4 − 3 /4 0 √ |E120 |2 = Eoy (cos δ + i sin δ) − 3 /4 3/4   √ 1 2 2 = Eox + 3Eoy − 2 3 Eox Eoy cos δ 4 The intensity, of course can be rewritten to allow us to rewrite (1.2.10) and (1.2.11):   √ c 1 2 2 9Io = 3Eox + Eoy + 2 3 Eox Eoy cos δ 8π 4   √ c 1 2 2 Io = Eox + 3Eoy − 2 3 Eox Eoy cos δ 8π 4

(1.2.11)

(1.2.12a) (1.2.12b)

We introduce another equation, the derivation of which will not be shown here: tan(2α) =

2EoxEoy cos δ 2 − E2 Eox oy

(1.2.13)

Solving (1.2.12a)-(1.2.13) allows us to solve for our unknowns. We also note that α = 2 · 30o , which is deducible from the fact that the maximum intensity is observed at 30o . The tabulated results follow in (a) and (b). a)

∗ The

The phase angle is found to be δ = 40.89o , while E+ and E− may be found from the results of part (b):

following steps have been abbreviated, and for that I apologize. I can provide scrap work/code if desired.

W. Erbsen

HOMEWORK #2

E+ = Ex + iEy = E− = Ex − iEy = b)

r

r

Io (13.26 cos(ωt) + i8.86 cos (ωt − 40.89o)) c Io (13.26 cos(ωt) − i8.86 cos (ωt − 40.89o)) c

We found Eox and Eoy directly after the prompt. The values for these are: r r Io Io Eox = 13.26 , and Eoy = 8.86 c c From which we may calculate Ex and Ey : r Io Ex = 13.26 cos(ωt), and c

c)

Ey = 8.86

r

Io cos (ωt − 40.89o) c

If the polarizer is placed first precisely along the x-axis, then the intensity is: c I0 = (13.26)2 −→ I0 = 7.53 Io 8π And if perfectly along the y-axis, I90 =

c (8.86)2 −→ I90 = 3.12 Io 8π

Problem 10.3 Consider a beam of partially plane polarized light with the same maximum and minimum intensities as in the previous problem. a)

What is the percent polarization of this light?

b)

What would the transmitted intensities be if the polarizer were set along the x-axis, and then along the y-axis?

c)

How could you tell whether an incident light beam were elliptically polarized or partially plane polarized?

Solution a)

The polarization Π can be found as follows Π=

Smax − Smin Smax + Smin

(1.2.14)

We know from the previous problem that the maximum intensity is Smax = 9Io while the minimum intensity is Smin = Io , so (1.2.14) becomes Π=

9Io − Io 8 −→ Π = (Π = 80.00%) 9Io + Io 10

19

CHAPTER 1: ELECTRODYNAMICS II

b)

If we imagine the partially-plane polarized light incident on the polarizer to be composed of orthogonal linear components along the x and y-directions, then the total intensity is p S 0 = Sx2 + Sy2 + 2 Sx Sy cos θ

Where θ is the angle between Sx and Sy , which is most definitely 90o . We can find Smax and Smin by taking these limits: q 2 q 2 Smax = Sx2 + Sy2 , and Smin = Sx2 − Sy2 And now using (1.2.14) we can say that

Smax 1+Π = Smin 1−Π

(1.2.15)

From Malus’ Law, we can now say that S0 =

1−Π cos2 (θ)Smax 1+Π

With the help of (1.2.15). We can now say that the transmitted intensity when the polarizer is placed perfectly along the x and y-axis, respectively are Sx0 = 0.75 Io , c)

and

Sy0 = 0.25 Io

To tell if incident radiation were elliptically polarized or partial plane polarized, I would place a linear polarizer in front of the incident beam followed by an energy meter. If the intensity changes monotonically with polarization angle then the radiation is said to be more linearly polarized than elliptically.

Problem 10.5 A horizontal light ray is incident on a 60o -60o -60o glass prism (n = 1.5) that is resting on a table. At what angle with the horizontal does the ray leave the prism? (Assume that the ray does not strike the bottom of the prism before exiting.)

Solution The initial ray is coming in completely horizontal, however the incident angle to the medium is proportional to the dimensions of the prism. In our case, the normal component of the front face of the prism makes and angle that is precisely 30o from the incident ray. We now apply Snell’s Law,   sin (30) na sin θa = ng sin θg −→ θg = sin−1 ≈ 19.47o (1.2.16) 1.5 Now the ray is within the prism, and the next interface is that of the opposite surface of the prism. The normal component of the exiting face is 60o in the positive direction to the first, so that the angle between the incident beam calculated in (1.2.16) and this normal angle is 60o − 19.47o = 40.53o. Applying Snell’s Law once more,

W. Erbsen

HOMEWORK #2

ng sin θg = na sin θa −→ θa = sin−1 [1.5 sin (40.53)] −→ θa ≈ 77.10o

Problem 10.6 For the prism in the preceding problem, what is the smallest angle of incidence for which a light ray will pass directly through the prism without total internal reflection?

Solution Total internal reflection occurs at θc = sin−1 (1/1.5) ≈ 41.81o. This is, of course, relative to the normal component of the interior of the exiting interface. The simplest way to find the incident angle is to work backwards from what we did in the preceding problem. We then only must apply Snell’s Law once, for the first interface. The critical angle calculated for the outer face is oriented to the normal of the first face as 60o − 41.81o = 18.19o. We can now apply Snell’s Law: ng sin θg = na sin θa −→ θa = sin−1 [1.5 sin (18.19o)] −→ θa ≈ 27.92o

Problem 10.7 You are standing in front of a rectangular fish tank filled with water (n = 1.33). a)

Show that you can always see through the back of the tank without total internal reflection as you look through the front face.

b)

Show that if you look through the front face toward the right side of the tank, there is a maximum angle of incidence for which there is total internal reflection from the right face. What is this angle?

c)

What is the minimum index of refraction of the liquid in the tank such there that would always be total internal reflection from the right face?

d)

Show that the fact that there is a thickness of glass (n = 1.5) between the water and the air does not affect this problem.

Solution a)

To show that you can always see through the back of the fish tank when viewed through the front, it is sufficient to show that when viewing the fish tank at the most extreme angle, that the exiting angle is less than the critical angle. In our case, the critical angle for glass/air interface is θc = sin−1 (1/1.5) ≈ 41.81o.

CHAPTER 1: ELECTRODYNAMICS II

21

There are 5 distinct zones in the problem with 4 interfaces. Starting from the front of the fish tank, the air is region 1, then the glass is region 2 and so on. The critical angle refers to the glass/air interface between regions 4 and 5, so to satisfy the requirement that light entering from the front of the fish tank (region 1) will always exit through the back (region 2) we must show that θ4 < θc . We imagine the incident angle to be as wide as possible, say 89.99o, which corresponds to θ1 :   1 −1 o n1 sin θ1 =n2 sin θ2 −→ θ2 = sin sin (89.99 ) ≈ 41.81o 1.5   1.5 −1 o sin (41.81 ) ≈ 48.75o n2 sin θ2 =n3 sin θ3 −→ θ3 = sin 1.33   −1 1.33 o n3 sin θ3 =n4 sin θ4 −→ θ4 = sin sin (48.75 ) ≈ 41.80o (1.2.17) 1.5 From (1.2.17), we can see that θ4 < θc (41.80o < 41.81o) so that total internal reflection is never achieved, and you can always see through the back of the fish tank . b)

The process here is very much the same as in the preceding problem, except that we need to work backwards. The critical angle has not changed, θ4 = θc ≈ 41.81o. So, working backwards from θ4 , we have   1.5 n3 sin θ30 =n4 sin θ4 −→ θ30 = sin−1 sin (41.81o) ≈ 48.75o 1.33   −1 1.33 o o n2 sin θ2 =n3 sin θ3 −→ θ2 = sin sin (90 − 48.75 ) ≈ 35.77o 1.5 n1 sin θ1 =n2 sin θ2 −→ θ1 = sin−1 [1.5 sin (35.77o )] ≈ 61.27o

(1.2.18)

Hence from (1.2.19) we have shown that the maximum angle of incidence to allow for total internal reflection on the right face of the fish tank is θ1 = 61.27o . c)

To find the minimum index of refraction of the liquid such that light incident to the front of the fish tank will always satisfy total internal reflection, we once again precede using Snell’s Law. The only difference is that this time we will leave the index of refraction for the liquid to be arbitrary.   1 n1 sin θ1 =n2 sin θ2 −→ θ2 = sin−1 sin (θ1 ) 1.5       1.5 1 1 n2 sin θ2 =n3 sin θ3 −→ θ3 = sin−1 sin sin−1 sin (θ1 ) = sin−1 sin (θ1 ) n 1.5 n    n 1 sin 90o − sin−1 sin (θ1 ) (1.2.19) n3 sin θ3 =n4 sin θ4 −→ sin (θ4 ) = 1.5 n It is possible to solve (1.2.19) for n, using the appropriate trig substitutions:    o −1 1 1.5 sin (θ4 ) =n sin 90 − sin sin (θ1 ) n        o −1 1 o −1 1 =n sin (90 ) cos sin sin (θ1 ) − cos (90 ) sin sin sin (θ1 ) n n   1/2 sin2 (θ1 ) =n 1 − n2  2  1/2 = n − sin2 (θ1 ) (1.2.20)

W. Erbsen

HOMEWORK #2

If we recall that θ4 = θc and imagine our incident angle to the most extreme position of θ1 = 89.99o, (1.2.20) can be solved to find the minimum index of refraction:

d)

h i 1/2 n = (1.5)2 sin2 (41.81o ) + sin2 (89.99o) −→ n ≈ 1.41

(1.2.21)

The fact that the width of the glass is negligible can be shown by recomputing the results of parts (a) and (b) with assuming that there is no glass media. For part (a), where we are asked to show that total internal reflection is never satisfied if viewing through the front of the tank, we start with the critical angle at the second interface, which is θc = sin−1 (1/1.33) ≈ 48.75o, and applying Snell’s Law: n1 sin θ1 = n2 sin θc −→ θ1 = sin−1 [1.33 sin (48.75o)] = 90o Since you can never see through the fish tank at exactly 90o , we have shown that the critical condition cannot be satisfied, as previously shown in (1.2.17). For part (b), we can similarly work backwards from the critical angle: n1 sin θ1 = n2 sin θc −→ θ1 = sin−1 [1.33 sin (90o − 48.75o)] ≈ 61.27o

(1.2.22)

As can be seen, (1.2.22) matches (1.2.19), and therefore from these two instances we are forced to admit that the width of the glass is irrelevant .

Problem 10.9 a)

Solve Eqs. (10.117)-(10.120), to get the transmitted and reflected electric fields given by Eqs. (10.121) and (10.122).

b)

Use these fields to get the transmission and reflection coefficients for the coated surface.

c)

For an original air (n = 1) to glass (n = 1.5) interface, find the index of refraction and thickness of a coating that would have no reflection at the incident wavelength of 5, 000˚ A.

Solution a)

Equations (10.117)-(10.120) read E1 + E10 =E2 + E20 eik2 d n1 (E1 − E10 ) E2 eik2 d + E20 n2 (E2 eik2 d − E20 )

=n2 (E2 −

E20 eik2 d )

(1.2.23a) (1.2.23b)

=E3

(1.2.23c)

=n3 E3

(1.2.23d)

We start with (1.2.23a) and rearrange it: E10 = E2 + E20 eik2 d − E1

(1.2.24)

23

CHAPTER 1: ELECTRODYNAMICS II

We now substitute (1.2.24) into (1.2.23b): 
   n1 E1 − E2 + E20 eik2 d − E1 =n2 E2 − E20 eik2 d     n1 E1 − E2 − E20 eik2 d + E2 =n2 E2 − E20 eik2 d 2n1 E1 − n1 E2 − n1 E20 eik2 d =n2 E2 − E20 n2 eik2d

n2 E20 eik2 d − n1 E20 eik2 d =n2 E2 + n1 E2 − 2n1 E1   0 −ik2 d E2 (n2 + n1 ) − 2n1 E1 −→ E2 = e n2 − n1

(1.2.25)

We now take (1.2.25) and substitute it into (1.2.23c):   ik2 d −ik2d E2 (n2 + n1 ) − 2n1 E1 E2 e +e = E3 n2 − n1

E3 (n2 − n1 ) =E2 (n2 − n1 )eik2 d + E2 (n2 + n1 )e−ik2 d − 2n1 E1 e−ik2 d

2E1 n1 e−ik2 d + E3 (n2 − n1 ) =E2 (n2 − n1 )eik2 d + E2 (n2 + n1 )e−ik2 d   =E2 n2 eik2 d − n1 eik2 d + n2 e−ik2 d + n1 e−ik2 d 

 =E2 n2 eik2 d + e−ik2 d − n1 eik2 d − e−ik2 d =E2 [2n2 cos(k2 d) − 2in1 sin(k2 d)] =2E2 [n2 cos(k2 d) − in1 sin(k2 d)] −→ E2 =

2E1 n1 e−ik2 d + E3 (n2 − n1 ) 2 [n2 cos(k2 d) − in1 sin(k2 d)]

(1.2.26)

We perform a similar exercise by substituting (1.2.25) into (1.2.23d):    E2 (n2 + n1 ) − 2n1 E1 n3 E3 =n2 E2 eik2 d − e−ik2 d n2 − n1

E2 (n2 + n1 )n2 e−ik2 d + 2n1 n2 E1 e−ik2 d n2 − n1 ik2 d n3 (n2 − n1 )E3 =n2 (n2 − n1 )E2 e − n2 E2 (n2 + n1 )e−ik2 d + 2n1 n2 E1 e−ik2 d   n3 (n2 − n1 )E3 − 2n1 n2 E1 e−ik2 d =n2 E2 n2 eik2 d − n1 eik2 d − n2 e−ik2 d − n1 e−ik2 d 

 =n2 E2 n2 eik2 d − e−ik2d − n1 eik2 d − e−ik2 d =n2 E2 [2in2 sin(k2 d) − 2n1 cos(k2 )d)] n3 E3 =n2 E2 eik2 d −

−→ E2 =

n3 (n2 − n1 )E3 − 2n1 n2 E1 e−ik2 d 2n2 [in2 sin(k2 d) − n1 cos(k2 d)]

(1.2.27)

We can now set (1.2.26) and (1.2.27) equal to one another to eliminate E2 : 2E1 n1 e−ik2 d + E3 (n2 − n1 ) n3 (n2 − n1 )E3 − 2n1 n2 E1 e−ik2 d = 2 [n2 cos(k2 d) − in1 sin(k2 d)] 2n2 [in2 sin(k2 d) − n1 cos(k2 d)] If we let α = sin k2 d and β = cos k2 d, then (1.2.28) becomes 2E1 n1 e−ik2 d + E3 (n2 − n1 ) n3 (n2 − n1 )E3 − 2n1 n2 E1 e−ik2 d = n2 β − in1 α in22 α − n1 n2 β

2E1 n1 e−ik2d 2n1 n2 E1 e−ik2d n3 (n2 − n1 )E3 E3 (n2 − n1 ) + = 2 − n2 β − in1 α in22 α − n1 n2 β in2 α − n1 n2 β n2 β − in1 α | {z } | {z } I II

(1.2.28)

W. Erbsen

HOMEWORK #2

Where I have separated our expression for simplicity. The components are: 2E1 n1 e−ik2d n2 β + in1 α 2n1 n2 E1 e−ik2 d in22 α + n1 n2 β · + · n2 β − in1 α n2 β + in1 α in22 α − n1 n2 β in22 α + n1 n2 β
2E1 n1 e−ik2d (n2 β + in1 α) 2n1 n2 E1 e−ik2 d in22 α + n1 n2 β = − n22 β 2 + n21 α2 n21 n22 β 2 + n42 α2 2E1 n1 (n2 − n1 )E1 (iα + β) = (n2 α + in1 β)(n1 α + in2 β) 2E1 n1 (n2 − n1 )E1 (i sin(k2 d) + cos(k2 d)) = (n2 sin(k2 d) + in1 cos(k2 d))(n1 sin(k2 d) + in2 cos(k2 d)) 4in1 (n1 − n2 )E1 = 2in1 n2 cos(2k2 d) + (n21 + n22 ) sin(2kd d)

I=

(1.2.29)

And now E3 (n2 − n1 ) n2 β + in1 α n3 (n2 − n1 )E3 in22 α + n1 n2 β · − · in22 α − n1 n2 β in22 α + n1 n2 β n2 β − in1 α n2 β + in1 α
n3 (n2 − n1 )E3 in22 α + n1 n2 β E3 (n2 − n1 ) (n2 β + in1 α) − =− 2 2 4 2 2 n1 n2 β + n2 α n22 β 2 + n21 α2   n3 in1 n2 αβ = − E3 (n2 − n1 ) + n2 (n1 β − in2 α) n21 α2 + n22 β 2   n3 in1 n2 sin(k2 d) cos(k2 d) = − E3 (n2 − n1 ) + n2 (n1 cos(k2 d) − in2 sin(k2 d)) n21 sin(k2 d)2 + n22 cos(k2 d)2

II =

(1.2.30)

Setting (1.2.29) equal to (1.2.30) and shifting to one side, we see that


(n1 − n2 ) E3 (n22 + n1 n3 ) sin(k2 d) + iE3 n2 (n1 + n3 ) cos(k2 d) − 2iE1 n1 n2 =0 n2 (2in1 n2 cos(2k2 d) + (n21 + n22 ) sin(2k2 d))

(1.2.31)

Solving (1.2.31) for E3 yields E3 =

2n1 n2 E1 n2 (n1 + n3 ) cos(k2 d) − i(n22 + n1 n3 ) sin(k2 d)

Using the results from part (b), it is also easily seen that   n2 (n1 − n3 ) cos(k2 d) − i(n1 n3 − n22 )2 sin2 (k2 d) 0 E1 = E1 n2 (n1 + n3 ) cos(k2 d) − i(n1 n3 + n22 )2 sin2 (k2 d) b)

The transmission coefficient can be found by 2 2 n ˆ · S2 n3 E3 n3 2n1 n2 T = = = 2 n1 E 1 n1 n2 (n1 + n3 ) cos(k2 d) − i(n2 + n1 n3 ) sin(k2 d) n ˆ · S1

(1.2.32)

From (1.2.32) it can be seen that the transmission coefficient is finally given by T =

4n1 n3 n22 n22 (n1 + n3 )2 cos2 (k2 d) + (n1 n3 + n22 )2 sin2 (k2 d)

(1.2.33)

Which matches (10.123). The reflection coefficient may be calculated via similar means, however the easiest way is to recognize that R + T = 1, so that from (1.2.33), we calculate

25

CHAPTER 1: ELECTRODYNAMICS II

R=

n22 (n1 + n3 )2 cos2 (k2 d) + (n1 n3 + n22 )2 sin2 (k2 d) 4n1 n3 n22 − n22 (n1 + n3 )2 cos2 (k2 d) + (n1 n3 + n22 )2 sin2 (k2 d) n22 (n1 + n3 )2 cos2 (k2 d) + (n1 n3 + n22 )2 sin2 (k2 d)

=

n22 (n21 + n23 + 2n1 n3 ) cos2 (k2 d) + (n21 n23 + n42 + 2n1 n3 n22 ) sin2 (k2 d) − 4n1 n3 n22 n22 (n1 + n3 )2 cos2 (k2 d) + (n1 n3 + n22 )2 sin2 (k2 d)

=

n22 (n21 + n23 + 2n1 n3 ) cos2 (k2 d) + (n21 n23 + n42 + 2n1 n3 n22 ) sin2 (k2 d) − 4n1 n3 n22 (sin2 (k2 d) + cos2 (k2 d)) n22 (n1 + n3 )2 cos2 (k2 d) + (n1 n3 + n22 )2 sin2 (k2 d)

Combining terms yields R= c)

n22 (n1 − n3 )2 cos2 (k2 d) + (n1 n3 − n22 )2 sin2 (k2 d) n22 (n1 + n3 )2 cos2 (k2 d) + (n1 n3 + n22 )2 sin2 (k2 d)

To find the thickness of the coating, we use (10.127), which reads r λ1 n1 d= 4 n3 Applying the provided values, 5000 · 10−10 m d= 4

r

1 −→ d ≈ 1.02 · 10−7 m 1.5

And by looking at our equation for the reflection coefficient, we can see that it will be zero if √ n2 = n1 n3 , so the index of refraction of the coating is n2 ≈ 1.22 .

1.3

Homework #3

Problem 11.1 A plane electromagnetic wave is incident at an angle θ from vacuum onto the flat surface of a perfect conductor. a)

Use conservation of momentum to find the radiation pressure on the surface of the conductor.

b)

For a wave that is polarized perpendicular to the plane of incidence, find the radiation pressure by calculating the magnetic force on the surface current.

c)

For a wave that is polarized parallel to the plane of incidence, find the radiation pressure by calculating the magnetic and electric forces on the surface and surface charge.

Solution a)

The equation for radiation pressure was derived using conservation of momentum by Franklin as (10.33):  prad = |E0 |2 (1.3.1) 4π

W. Erbsen

HOMEWORK #3

This is for the case that the incident light is completely reflected (no absorption) and is normal to the reflecting surface. Since the radiation is at an angle θ, we must take the normal component of (1.3.1) in order to discount the radiation not contributing momentum pressure. Furthermore, we must also recognize that we must only recognize the component of the actual momentum normal to the conductor (since we are recognizing both incoming and outgoing rays, the other component cancels). Therefore, prad (θ) =

1  |E0 |2 cos θ · cos θ −→ prad = |E0 |2 cos2 θ 4π 4π

(1.3.2)

Since we are coming from vacuum,  ⇒ 0 ⇒ 1 in CGS units.

b) &c) It is clear from (1.3.2) that the radiation pressure is independent of polarization, so the result will once again be: 1 |E0 |2 cos2 θ 4π

prad =

Problem 11.2 Use the physical properties of copper to estimate the frequency and wave length for which its conductivity develops an appreciable (∼ 10%) imaginary part. Assume two conduction electrons per atom.

Solution We are primarily motivated by the availability of (11.62) in Franklin: σ(ω) =

zc N e2 m (γ − iω)

(1.3.3)

We can separate (1.3.3) into it’s real and imaginary parts, zc N e2 m γ2 zc N e2 Im {σ(ω)} = m γ2 Re {σ(ω)} =

γ + ω2 ω + ω2

(1.3.4a) (1.3.4b)

We are told that zc is the number of conduction of electrons, which in the case of Copper is just 2. Furthermore, the combined quantity zc N is the number of free electrons per unit volume, while γ is the damping constant. According to Jackson on p. 312, the relevant quantities can be given numerically as: ? 3

N ' 8 · 1028 atoms/m

σ ' 5.9 · 107 (Ω · m)−1

27

CHAPTER 1: ELECTRODYNAMICS II

The damping constant can now be calculated using (1.3.3): zc N e2 mσ 2 (2)(8 · 1028 atoms/m )(1.602 · 10−19 C)2 ' −31 (9.109 · 10 kg)(5.9 · 107 (Ω · m)−1 )

γ'

' 7.641 · 1013 Hz

(1.3.5)

We are interested at the frequencies at which the imaginary part of (1.3.3) develops an imaginary part that is equal to ∼ 10% of the real part: Im {σ(ω)} = (0.10)Re {σ(ω)} −→ ω = (0.10)γ

(1.3.6)

Where I used the real and imaginary components of (1.3.3) as shown in (1.3.4a) and (1.3.4b). According to (1.3.6), the approximate frequency where this is possible is: ω ' 7.641 · 1012 Hz

(1.3.7)

If we recall that λ = 2πc/ω, and using (1.3.7) we can find the corresponding wavelength: λ=

2π(3 · 108 m/s) −→ λ ' 246.7 µm (7.641 · 1012 Hz)

Problem 11.3 a)

Find the real and imaginary parts of , as given in (11.52).

b)

Show that  is analytic in the upper half of the complex ω plane. In finding any particular pole position, the contributions from other modes can be ignored since they are analytic in that region.

Solution a)

Equation (11.52) reads (ω) = 1 + 4πχ = 1 +

4πN Σi zi ηi 1 − 4π Σi zi ηi 3

(1.3.8)

And we also recall that the molecular polarizability is from (11.52) in Franklin ηi =

m [ωi2

e2 − ω2 − iωγi ]

(1.3.9)

We also note that the frequency-dependent susceptibility takes the form χ=

N Σi zi ηi N e2 1 = 4π 2 − ω 2 − iωγ 2 m ω 1 − 3 Σi zi ηi i i

(1.3.10)

W. Erbsen

HOMEWORK #3

Then the real and imaginary parts of  from (1.3.8) are 4πN e2 ωi2 − ω 2 m (ωi − ω2 )2 + ω2 γ 2 4πN e2 ωγ Im {(ω)} = m (ωi2 − ω2 )2 + ω2 γ 2 Re {(ω)} = 1 +

b)

To show that  is analytic in the upper half of the complex plane, we first substitute (1.3.10) into (1.3.8): (ω) = 1 +

4πN e2 1 2 2 m ωi − ω − iωγi2

Taking the complex integral of (1.3.11) leads to  Z ∞ Z ∞  4πN e2 1 (ω) dω = 1+ dω m ωi2 − ω2 − iωγi2 −∞ −∞ Z 1 4πN e2 ∞ = 2 2 dω 2 m −∞ ωi − ω − iωγi

(1.3.11)

(1.3.12)

It is clear that (1.3.12) has poles in the lower half plane at   iγi γi2 2 ω1,2 = − ± ωi − 2 4 Since there are no poles at the boundary or in the upper half plane, and if we assume that the kernel is finite, then we can say that (ω) is analytic in the upper half plane .

Problem 11.4 Find the Fourier transform of the function in (11.74) to verify (11.75). (Hint : Complete the square in the exponent in doing the integral.)

Solution We first recall that the functional forms of the integral Fourier transform is Z ∞ f(x, 0) = f(ω, 0)eikx dk −∞ Z ∞ 1 A(k, 0) = f(x, 0)e−ikx dx 2π −∞

(1.3.13a) (1.3.13b)

And the wave packet function (11.74) is given by f(x, 0) = e−x

2

/2L2 ik0 x

e

To find the Fourier transform of (1.3.14), we substitute it into (1.3.13b):

(1.3.14)

29

CHAPTER 1: ELECTRODYNAMICS II

1 A(k, 0) = 2π =

1 2π

=

1 2π

    x2 exp − 2 · exp [ik0 x] exp [−ikx] dx 2L −∞   Z ∞ x2 exp − 2 · exp [i (k0 − k) x] dx 2L −∞   Z ∞ x2 exp − 2 − iαx dx (let α = k − k0 ) 2L −∞ Z

∞

(1.3.15)

At this point, per the suggestion in the prompt, we wish to complete the square in the exponent in the integrand in (1.3.15):

2
2 i 1 1 h x2 2 2 2 2 − 2 − iαx = − x + 2iL αx ⇒ − x + iL α + L α (1.3.16) 2L 2L2 2L2 We now substitute (1.3.16) into (1.3.15),   Z ∞
2
2 i 1 1 h 2 2 dx A(k, 0) = exp − 2 x + iL α + L α 2π −∞ 2L     Z ∞
2 i 1 L2 α2 1 h 2 = exp − exp − 2 x + iL α dx 2π 2 2L −∞

(1.3.17)

We wish to employ u0 du0 substitution (where the primes are present to avoid confusion later on), where u0 = √

1 2L2


x + iL2 α ,

and

du0 = √

1 2L2

dx

With this substitution, (1.3.17) becomes

  Z 1 L2 α2 √ 2 ∞ −u02 0 A(k, 0) = exp − 2L e du 2π 2 −∞  Z ∞ 02 L L2 α2 exp − =√ e−u du0 2 2π −∞ The integral in (1.3.18) is just the standard Gaussian integral, whose solution is simply substituting our value for α, (1.3.18) becomes " # 2 L L2 (k − k0 ) A(k, 0) = √ exp − 2 2π

(1.3.18) √

π , and in

(1.3.19)

We can see that (1.3.19) is identical to (1.3.13b), which is also the same as (11.75), which is what we were asked to show.

Problem 11.5 A wave packet is given by f(x, 0) = eimπx/L ,

0 < x < L,

(1.3.20)

W. Erbsen

HOMEWORK #3

and f(x, 0) is zero elsewhere. a)

Plot |f(x, 0)|2 .

b)

Find the Fourier transform, A(k), of f(x, 0). Plot |A|2 for m = 1 and m = 2.

c)

Define a reasonable ∆x and ∆k for this wave packet. Calculate ∆x, ∆k, and ∆x∆k.

Solution We first note that due to boundary conditions, we can rewrite (1.3.20) as  mπx  f(x, 0) = i sin L

a)

b)

c)

The item to be plotted is (please see last page for printout of plots):  mπx  |f(x, 0)|2 = sin2 L

The Fourier transform can be found the same way as we did in the last problem: Z L  mπx  i A(k, 0) = sin e−ikxdx 2π 0 L   iπL e−ikL mπ cos(mπ) + ikL sin(mπ) − mπeikL −→ A(k, 0) = 2 k 2 L2 − m2 π 2

(1.3.21)

(1.3.22)

(1.3.23)

A reasonable value for ∆x is L, and likewise for ∆k is 1/L, so that ∆x∆k = 1: ∆x = L,

∆k =

1 , L

∆x∆k = 1

Problem 11.6 a)

Derive (11.81) and (11.82).

b)

The spectral line in the decay of an excited state of sodium has a wave length (λ = 5, 893 ˚ A), and a lifetime of 2 · 10−8 seconds. Calculate its natural line width.

Solution a)

In order to derive (11.81), we must take the Fourier Transform of (11.80), which reads (   , x < ct, E0 exp [i (k0 x − ωt)] · exp x−ct 2cτ E(x, t) = 0 x > ct Assuming we are in the first region, at if we take t = 0, (11.80) becomes

31

CHAPTER 1: ELECTRODYNAMICS II

    i E(x, 0) = E0 exp i k0 − x 2cτ We take the Fourier Transform of (1.3.24) by substituting it into (1.3.13b),     Z E0 ∞ i E(k) = exp i k0 − x · exp [−ikx] dx 2π −∞ 2cτ    Z E0 ∞ 1 = exp i (k0 − k) + x dx 2π −∞ 2cτ     ∞ E0 1 1 exp i (k0 − k) + x = 2π i (k0 − k) + 1/(2cτ ) 2cτ −∞ −→ E(k) =

E0 2π [i(k0 − k) + 1/(2cτ )]

(1.3.24)

(1.3.25)

It is clear that (1.3.25) is identical to (11.81). In order to derive (11.82), we simply take the modulus squared of (1.3.25)    E0 E0 2 |E(k)| = (1.3.26) 2π [−i(k0 − k) + 1/(2cτ )] 2π [i(k0 − k) + 1/(2cτ )] At this point we make the following assignments: α = (k0 − k), and β = 1/(2cτ ), since otherwise we will run out of paper. With these new definitions, (1.3.26) becomes    E0 E0 2 |E(k)| = 2π [−iα + β] 2π [iα + β] 2 E 1 = 02 4π (−iα + β) (iα + β) E2 1 = 02 2 (1.3.27) 4π α + β 2 Backsubstituting our previous definitions for α and β, (1.3.27) becomes 2

|E(k)| =

E02 4π 2 [(k − k0 )2 + 1/(2cτ )2 ]

(1.3.28)

And it is easy to see that (1.3.28) is identical to (11.82). b)

According to (11.84)∗ , the natural line width is given by Γ=

λ2 2πcτ

With the data provided, this becomes
2 5893 · 10−10 m Γ= −→ Γ ≈ 9.212 · 10−15 m 2π · (3.00 · 108 m/s) · (2 · 10−8 s) ∗ with

correction

(1.3.29)

W. Erbsen

HOMEWORK #3

Problem 11.7 A glass block has an index of refraction given by n = 1.350 −

100 ˚ A λ

(1.3.30)

in the visible region. a)

What is the angular difference between red light (Use λ = 6, 500 ˚ A) and blue light (λ = 4, 500 ˚ A) in the glass if the light enters with an angle of incidence of θ = 30o ?

b)

What are the phase and group velocities for each color light in the glass?

Solution a)

This problem requires a simple application of Snell’s Law; so solving for the exiting angle θ2 ,   n1 sin(θ1 ) (1.3.31) n1 sin(θ1 ) = n2 sin(θ2 ) −→ θ2 = sin−1 n2 So the goal here would be to calculate the exiting angle for both red and blue light, and take the difference. We first address incident light in the case that it is red, and find the index of refraction from (1.3.30): ∗ nr = 1.350 −

100 ˚ A ≈ 1.3346 6, 500 ˚ A

We now take (1.3.32) and substitute it into (1.3.31):   1 θr = sin−1 sin(30o ) ≈ 22.002o 1.3346

(1.3.32)

(1.3.33)

We now do the same for the shorter wavelength, starting with calculating the index of refraction from (1.3.30): nb = 1.350 −

100 ˚ A ≈ 1.3278 4, 500 ˚ A

And now we can calculate the angle from (1.3.31):   1 −1 o θb = sin sin(30 ) ≈ 22.121o 1.3278

(1.3.34)

(1.3.35)

And therefore the angular difference is given by ∆θ = θb − θr = 22.121o − 22.002o −→ ∆θ = 0.11927o b)

The phase velocities for both wavelengths in the glass are given simply by (11.93): c vp = n

(1.3.36)

∗ Two assumption have been made. First, I have assumed that all the constant in (1.3.30), as well as all given wavelengths, are exact, so that we can take the result out to arbitrary precision. Otherwise, we would be bound to the rules of significant figures, which would limit the transparency of the answer. Second, I have taken the incident index of refraction to be that of a vacuum, i.e. n1 = 1.00.

33

CHAPTER 1: ELECTRODYNAMICS II

Completing this equation using first the index of refraction of the glass for the longer wavelength form (1.3.32), we have

vp,r =

2.9979 · 108 m/s −→ vp,r ≈ 2.2463 · 108 m/s 1.3346

And now finding the phase velocity for the shorter wavelength we do the same calculation using (1.3.34),

vp,b =

2.9979 · 108 m/s −→ vp,b ≈ 2.2578 · 108 m/s 1.3278

To find the group velocity, we first recall that vg = dω(k)/dk, and also that ω(k) = ck/n(k). With this information, as well as the expression given (1.3.30), the generic expression for the group velocity is   d ck vg = dk 1.350 − 100 ˚ A k/(2π)   d 2πck = dk 2π1.350 − k100 ˚ A    1 d d 1 =2πc (k) + k dk 2π1.350 − k100 ˚ 2π1.350 − k100 ˚ A dk A # " 1 k100 ˚ A =2πc + 2 2π1.350 − k100 ˚ A 2π1.350 − k100 ˚ A   2πc k100 ˚ A = 1+ 2π1.350 − k100 ˚ A 2π1.350 − k100 ˚ A   ˚ 2πc 100 A (2π/λ) = 1+ 2π1.350 − 100 ˚ A (2π/λ) 2π1.350 − 100 ˚ A (2π/λ) " # 100 ˚ A (λ)−1 c = 1+ (1.3.37) −1 −1 1.350 − 100 ˚ A (λ) 1.350 − 100 ˚ A (λ) And at this point we can now calculate the group velocity given our two frequency components. Starting with the longer wavelength, " #
−1 100 ˚ A 6500 ˚ A 2.9979 · 1018 ˚ A/s vr,g =
−1 1 +
−1 −→ vr,g ≈ 1.7719 · 10 m/s 1.350 − 100 ˚ A 6500 ˚ A 1.350 − 100 ˚ A 6500 ˚ A And now for the shorter wavelength, " #
−1 100 ˚ A 4500 ˚ A 2.9979 · 1018 ˚ A/s vb,g =
−1 1 +
−1 −→ vb,g ≈ 1.7902 · 10 m/s 1.350 − 100 ˚ A 4500 ˚ A 1.350 − 100 ˚ A 4500 ˚ A

W. Erbsen

1.4

HOMEWORK #4

Homework #4

Problem 12.1 A coaxial wave guide consists of two concentric copper cylinders (conductivity σ = 0.50 × 1018 sec−1 ), an inner cylinder of radius A = 0.10 cm, and an outer clinder of radius B = 0.40 cm. The dielectric between the clinders has a permittivity  = 2.0, and permeability µ = 1.0. The outer cylinder is grounded, and a potential V0 = 120 volts (convert this to esu), oscillating at a frequency ν = 12 MHz, is applied to the inner cylinder. a)

Find the fields E(r, φ, z, t) and H(r, φ, z, t) for a wave traveling between the cylinders in the positive zdirection.

b)

Find the power (convert it to Watts) transmitted by this wave.

c)

Find the attenuation length for this wave.

Solution a)

We start by solving Laplace’s Equation in cylindrical coordinates:   1 ∂ ∂ϕ 1 ∂2ϕ ∇2T ϕ = r + 2 2 =0 r ∂r ∂r r ∂φ

(1.4.1)

Where we note that due to symmetry, the potential is independent of the angle φ, and so (1.4.1) becomes   1 ∂ ∂ϕ r = 0 −→ ϕ(r) = C1 ln r + C2 (1.4.2) r ∂r ∂r Our boundary conditions are that the outer conductor is grounded, ϕ(b) = 0, and the inner conductor is at some potential (120 V in our case), ϕ(a) = V0 . Applying the first boundary condition, ϕ(B) = C1 ln B + C2 = 0 −→ C2 = −C1 ln B

(1.4.3)

Substituting (1.4.3) into (1.4.2) and simplifying, we have ϕ(r) = C1 ln ( r/B ) We now apply the second boundary condition to (1.4.4),
ϕ(A) = C1 ln A/B = V0 −→ C1 =

(1.4.4)

V0 ln ( A/B )

(1.4.5)

And now substitute (1.4.5) into (1.4.4),

ϕ(r) =

V0 ln ( r/B ) ln ( A/B )

(1.4.6)

If we take the negative derivative of (1.4.6), we can find the field: E = −∇ϕ −→ E =

V0 1 ˆ r ln ( B/A) r

To find the magnetic field, we use (8.28) from Jackson, which reads:

(1.4.7)

35

CHAPTER 1: ELECTRODYNAMICS II

√ H = ± µ ˆ z×E Using (8.28), the magnetic field becomes H=

√

µ

V0 1 √ V0 1 ˆ ˆ z ׈ r −→ H = µ φ ln ( B/A ) r ln ( B/A ) r

(1.4.8)

We can include the additional space and time dependence of the electric and magnetic fields from (1.4.7) and (1.4.8), respectively: E(r, φ, z, t) = b)

V0 1 i(kz−ωt) e ˆ r, ln ( B/A ) r

H(r, φ, z, t) =

√

µ

V0 1 i(kz−ωt) ˆ e φ ln ( B/A ) r

The transmitted power can be found by integrating the component of the Poynting vector pointing in the z-direction over the cross-sectional area: Z c (E∗ × H) · ˆ z dA (1.4.9) P = 8π Substituting the expressions calculated for the fields in the previous section into (1.4.10), we obtain    Z  c V0 1 −i(kz−ωt) √ V0 1 i(kz−ωt) ˆ P = e ˆ r × e φ ·ˆ z · r drdφ µ 8π ln ( B/A ) r ln ( B/A ) r  2 Z 2π Z B c V0 √ 1 = dr µ dφ 8π ln ( B/A ) 0 A r  2
c V0 √ = µ ln A/B B 4 ln ( /A ) √ c µ V02 = (1.4.10) 4 ln ( B/A ) And substituting the appropriate constants into (1.4.10) yields∗ p 3 × 108 m/s (1)(2) (0.4 esu)2 −→ P = 1.224 × 107 esu2 · m/s P = 4 ln (.004/.001) −→ P = 1.224 × 10−2 Watts· m

Problem 12.3 A rectangular wave guide with copper walls has cross section dimensions 4.0 × 6.0 mm. It is filled with air. a)

Find the first five TM cutoff frequencies (in Hz) for this wave guide.

b)

Sketch nodal planes of Ez for each of these TM modes.

c)

Find the first five TE cutoff frequencies for this wave guide.

d)

Sketch nodal planes of Hz for each of these TE modes.

∗ Yes,

I realize that these numbers are ridiculous.

W. Erbsen

HOMEWORK #4

Solution a)

The relation for the TM cutoff frequencies in Hz is given by (12.57) in Franklin:  2  1/ c ` m2 2 + 2 (1.4.11) fm` = √ 2 µ A2 B √ Where we note that neither ` nor m can equal zero. Furthermore, µ = n = 1, and (1.4.11) becomes   1/ c `2 m2 2 fm` = + 2 (1.4.12) 2 A2 B Table 1.1: TM`,m cutoff frequencies m 1 2 1 3 2

` 1 1 2 1 2

fm` (Ghz) 45.07 62.50 79.06 83.85 90.14

b)

See attached plots

c)

The expression for the TE cutoff frequencies is identical to (1.4.11) except that either ` or m can equal zero:

Table 1.2: TE`,m cutoff frequencies m 1 0 1 2 2

d)

` 0 1 1 0 1

fm` (Ghz) 25.00 37.50 45.07 50.00 62.50

See attached plots

Problem 12.4 a)

The magnitude of Ez at the center of the wave guide in the preceding problem is 6, 000 volts/meter (convert to esu). Find the power transmitted by this wave guide for a TM wave with a frequency halfway between the lowest TM cutoff frequencies.

37

CHAPTER 1: ELECTRODYNAMICS II

b)

Find the attenuation length for this wave.

Solution a)

We know that 1 esu = 300 volt, so 6, 000 volts/meter = 20 esu/meter. Furthermore, the frequency halfway between the two lowest TM cutoff frequencies is f = (62.46 − 45.04)/2 + 45.04 = 53.75 GHz. We know from (12.75) in Franklin that P =

ωkE02 AB 32πγ 2

(1.4.13)

And we also know from class that f c ωk ∝ √ 2 γ fc µ

s 

f fc

2

−1

(1.4.14)

2

(1.4.15)

Substituting (1.4.14) into (1.4.13), we have E 2 AB f P = 0 32π fc

r

 c µ

s



f fc

−1

And now substituting the appropriate constants into (1.4.15) with  = 0 = 1 and µ = µ0 = 1, s  2 2 (20 esu/m) (.004 m)(.006 m) 53.75 × 109 Hz 53.75 × 109 Hz 8 P = · · 3 × 10 m/s −1 32π 45.04 × 109 Hz 45.04 × 109 Hz −→ P = 2.227 × 104 esu · m/s b)

The attenuation is given by (12.86) and (12.87) in Franklin as
# r   1/ " 1 − ωc2 /ω2 πµσc 2 AB `2 B 2 + m2 A2 Latten = 2 3 2 3 2µc ` B +m A ω Jiggiling (1.4.16) around a little bit and noting that ` = m = 1, " #s h σ i 1/2 AB B 2 + A2
1 − fc2 /f 2 c Latten = 3 3 4 B +A f Substituting the appropriate values into (1.4.17) yields  #   1/ " 0.50 × 1018 sec −1 2 (.004 m)(.006 m) (.006 m)2 + (.004 m)2 Latten = 4 (.006 m)3 + .(004 m)3 r 1 − (45.04)2/(53.75)2 · 53.75 × 109 Hz −→ Latten = 3.709 m

(1.4.16)

(1.4.17)

W. Erbsen

HOMEWORK #4

Problem 12.5 a)

The magnitude of Hz at the corner of the wave guide in the preceding problem is 0.20 gauss. Find the power transmitted by this wave guide for a TE wave with a frequency halfway between the two lowest TE cutoff frequencies.

b)

Find the attenuation length for this wave.

Solution a)

As in the previous problem, we find the cutoff frequency halfway between the two lowest TE frequencies to be (37.47 − 23.98)/2 + 23.98 = 30.73 GHz. The power is given by (12.76) in Franklin as P =

µωkH02 AB 16πγ 2

If we substitute (1.4.14) into (1.4.18) and simplify, s  2 H02 AB f f P = c −1 16π fc fc

(1.4.18)

(1.4.19)

And substituting the appropriate values into (1.4.19) yields

s  2 (.20 gauss)2 (.004 m)(.006 m) 30.73 × 109 Hz 30.73 × 109 Hz 8 P = · · 3 × 10 m/s −1 16π 24.98 × 109 Hz 24.98 × 109 Hz

−→ P = 5.047 gauss · m/s b)

The attenuation length can be found from (12.86) and (12.88): #r
  1/ " AB `2 B 2 + m2 A2 πµσc 2 1 − ωc2 /ω2 Latten = 2µc ηAB (`2 B + m2 A) + (ωc /ω)2 (`2 B 3 + m2 A3 ) ω

(1.4.20)

Getting rid of the appropriate constants and recalling that for the lowest cutoff frequency ` = 1 and m = 0, (1.4.21) becomes s h σ i 1/2  AB 1 − fc2 /f 2 c Latten = (1.4.21) 4 A/2 + B(fc /f)2 f And substituting the appropriate constants into (1.4.21),   1/  r 0.50 × 1018 sec−1 2 (.004 m)(.006 m) 1 − 24.982 /30.732 Latten = 4 (.004 m)/2 + (.006 m)(24.98/30.73)2 30.73 × 109 GHz −→ Latten = 7.085 m

Problem 12.7

39

CHAPTER 1: ELECTRODYNAMICS II

A circular wave guide with copper walls has a radius R = 4.0 mm. It is filled with air. a)

Find the first five TM cutoff frequencies (in Hz) for this wave guide.

b)

Sketch nodal surfaces of Ez for each of these TM modes.

c)

Find the first five TE cutoff frequencies for this wave guide.

d)

Sketch nodal surfaces of Hz for each of these TE modes.

Solution a)

An expression for the TM cutoff frequency can be found from (12.68) in Franklin: fm` =

c jm` 2π R

(1.4.22)

Substituting the values from Table 12.2 from Franklin into (1.4.22) yields

Table 1.3: TM`,m cutoff frequencies m 0 1 2 0 1

` 1 1 2 2 2

fm` (Ghz) 28.71 45.74 61.30 65.89 83.74

b)

See attached plots

c)

Using (12.69) we can find the cutoff frequency: fm` =

0 c jm` 2π R

And using the values from Table 12.2 we can evaluate (1.4.23) for the desired modes:

Table 1.4: TE`,m cutoff frequencies m 1 2 0 1 2

` 1 1 2 2 2

fm` (Ghz) 21.98 36.46 45.74 63.64 80.00

(1.4.23)

W. Erbsen

HOMEWORK #5

d)

See attached plots

Problem 12.8 The magnitude of Ez at the center of the wave guide in the preceding problem is 6, 000 volts/meter (convert to esu). Find the power transmitted by this wave guide for a TM wave with frequencies halfway between the two lowest TM cutoff frequencies.

Solution As before, 6, 000 volts/meter = 20 esu/meter. Additionally, the frequency halfway between the two lowest TM cutoff frequencies is f = (45.74 − 28.71)/2 + 28.71 = 37.23 GHz. We know from (12.79) in Franklin that P =

ωkR2 E02 0 2 [Jm (γm` R)] 2 16γm`

(1.4.24)

To find the derivative of the mth Bessel function we can use a recursion relation. Using Mathematica, 0 we find that Jm (γm` R) = −0.5192, and (1.4.24) becomes P =

1.5

π 2 (37.23 GHz)2 (.004 m)2 (20 esu/m)2 2 [−0.5192] −→ P = 5.439 × 104 esu · m/s 4(3 × 108 m/s)(2.4048/.004 m)2

Homework #5

Problem 12.10 A rectangular cavity (with  and µ = 1) has dimensions 2.0 × 3.0 × 4.0 cm. a)

Find the first six resonant frequencies of this cavity, and identify any degenerate modes.

b)

Write down the E and H fields for the lowest frequency resonant mode.

Solution We first recall that for a TM wave, Ez must vanish at the walls, and according to (12.107) in Franklin,    mπy   nπz  lπx Ez = E0 sin sin cos [TM mode] (1.5.1) A B C

41

CHAPTER 1: ELECTRODYNAMICS II

From which we can see that only n may be zero for a TM wave. Similarly, for a TE wave Hz must vanish at the walls, and from (12.108),    mπy   nπz  lπx Hz = H0 cos cos sin [TE mode] (1.5.2) A B C We can gather from this that both l and m can be zero for a TE wave. For both TM and TE waves, the resonant frequency is given by (12.109), flmn a)

c = √ 2 µ

"  #1 2  m 2  n 2 /2 l + + A B C

(1.5.3)

We can find the first six resonant frequencies for the cavity by evaluating (1.5.3) for a variety of different modes and finding the lowest six:

Resonant frequencies l 0 1 1 0 1 1 0

m 1 0 1 1 1 0 2

n 1 1 0 2 1 2 1

flmn (Ghz) 6.250 8.350 9.014 9.014 9.768 10.61 10.68

TM/TE TE TE TM TE TE TE TE

d 1 1 2 2 1 1 1

We note that both l and m cannot both be zero, otherwise all transverse components of E and H will vanish, and we will not have a standing wave. b)

The field components are just given by (12.63)-(12.65), if we allow A > B:  πx  Hz (x, y) =H0 cos A  πx  ikA Hz (x, y) = − H0 sin π A  πx  iµωA Ey (x, y) = H0 sin cπ A

Problem 12.11 A cubical cavity of dimensions 2.0 × 2.0 × 2.0 cm has copper walls. Find the fundamental frequency and Q value for that frequency.

Solution

W. Erbsen

HOMEWORK #5

From (1.5.3), flmn

c = √ 2 µ

"

l .02 m

2

 m 2  n 2 + + .02 m 0.02 m

# 1/2

−→ f110 = f101 = f011 = 10.61 GHz

The derivation for the expression for the Q value is found in the next problem, and is given by (1.5.21) as s 4πL fσc Q= (1.5.4) c µc Where “L” is the length of each side of the cavity. The conductivity of copper is given by σc = 5.96 · 1017 Ω−1 · m, while we let µc = 1. Accordingly, (1.5.4) leads to Q ≈ 0.2107 .

Problem 12.12 “And they shall make an ark of acacia-wood: two cubits and a half shall be the length thereof, and a cubit and a half the breadth thereof, and a cubit and a half the height thereof. And thou shalt overlay it with pure gold.” (Exodus XXV, 10:11). Find the fundamental resonant frequency and a half width for this biblical cavity. Use the length from your fingertip to your elbow as a measure of a biblical cubit.

Solution The length from my elbow to my beefy fingertip is ∼ 0.5 m, so the dimensions of our cavity are A = 1.25 m, B = 0.75 m, and C = 0.75 m. The fundamental frequency may be readily deduced from (1.5.3): flmn

c = √ 2 µ

"

l 1.25 m

2



m 2  n 2 + + 0.75 m 0.75 m

# 1/2

−→ f110 = f101 = 0.2330 GHz

Which is clearly degenerate. We can see that f110 corresponds to a TM mode, while f101 corresponds to a TE mode. The half width is found by dividing (12.131) by a factor of 2, Γ=

ω0 2Q

(1.5.5)

ωU dU/dt

(1.5.6)

Z

(1.5.7)

While the Q factor is given by (12.125) as Q=− Where we know that dU c =− dt 8π And if we recall from (1.5.1) and (1.5.2),

r

ωµc 8πσc

|Hsurface |2 dA

43

CHAPTER 1: ELECTRODYNAMICS II



  mπy   nπz  lπx Ez =E0 sin sin cos A B C      lπx mπy nπz  cos sin Hz =H0 cos A B C

(1.5.8) (1.5.9)

For the TM mode, we know that l = 1, m = 1 and n = 0, and (1.5.8) becomes  πx   πy  Ez = E0 sin sin A B

(1.5.10)

And similarly, for the TE mode, we recall that l = 1, m = 0 and l = 1, and (1.5.9) becomes  πx   πz  Hz = H0 cos sin A C

(1.5.11)

We know that Hsurface will be a linear combination of both fields, so we can go ahead and evaluate the modulus squared: h  πx   πy   πx   πz i2 2 |Hsurface | = E0 sin sin + H0 cos sin C  A B    A  2 2 πx 2 πy 2 πz 2 2 πx =E0 sin sin +H0 cos sin A {z B } A {z C } | | ψI ψII  πx   πy   πx   πz  + 2E0 H0 sin sin cos sin A B {z A C } | ψIII =E02 ψI + H02 ψII + 2E0 H0 ψIII

(1.5.12)

Instead of directly substituting (1.5.12) into (1.5.7), we can go ahead and integrate ψI , ψII and ψIII separately over space and then combine them later to avoid some of the mess. We first recall the following nifty trig identities: sin2 x =

1 (1 − cos(2x)) 2

cos2 x =

1 (1 + cos(2x)) 2

Using these, we first evaluate ψI , Z Z A Z B    πy  2 πx ψI dA = sin dx · sin2 dy A B 0 0     Z  Z B 1 A 2πx 2πy = 1 − cos dx · 1 − cos dy 4 0 A B 0 ("Z   # "Z B   #) Z A Z B A 1 2πx 2πy = dx − cos dx · dy − cos dy 4 A B 0 0 0 0 ("    A # "    B #) 1 A 2πx B 2πy = A− sin · B− sin 4 2π A 0 2π B 0 =

AB 4

And now doing the same for ψII ,

(1.5.13)

W. Erbsen

HOMEWORK #5

Z

ψII dA =

Z

A

cos2

0

A

dx ·

Z

0

C

sin2

 πz  C Z

dz

   ) C 2πx 2πz 1 + cos dx · 1 − cos dz A C 0 0 ("   # "   #) Z A Z C 1 2πx 2πz = A+ cos dx · C − cos dz 4 A C 0 0 ("    A # "    C #) 1 A 2πx C 2πz A+ sin · C− sin = 4 2π A 0 2π C 0

1 = 4

=

(Z

 πx 

A





AC 4

And now for ψIII , Z Z A Z B Z C  πx   πx   πy   πz  ψIII dA = sin cos dx · sin dy · sin dz = 0 A A B C 0 |0 {z } 0 =⇒ 0

Using (1.5.13)-(1.5.15), we can now evaluate (1.5.7):   r dU ωµc AB c AC =− E02 + H02 dt 8π 8πσc 4 4 r   fµc c A E02 B + H02 C =− 64π σc

We now find the energy within the cavity, U , using (12.121): " #
E02 + µH02 ABC (l/A)2 + (m/B)2 + (n/C)2 U = 2 2 32π (l/A) + (m/B)

(1.5.14)

(1.5.15)

(1.5.16)

(1.5.17)

If we take the frequency for the TM mode we let l = 1, m = 1, and n = 0, and (1.5.17) becomes
E02 + µH02 ABC U= (1.5.18) 32π We now recall that  = µ = 1 and B = C, so (1.5.18) becomes
E02 + H02 AB 2 U= 32π Making the same assumptions for (1.5.16) yields r   dU c fµc =− AB E02 + H02 dt 64π σc Now substituting (1.5.19) and (1.5.20) into (1.5.6) yields Q = − 2πf ·


AB 2 64π E02 + H02 · − 32π c

r

σc 1 fµc AB (E02 + H02 )

(1.5.19)

(1.5.20)

45

CHAPTER 1: ELECTRODYNAMICS II

4πB = c

s

fσc µc

(1.5.21)

The conductivity of gold can be found on p. 286 of Griffiths to be σc = 4.525 × 107 Ω−1 · m. Taking µc = 1, (1.5.21) becomes Q=

4π(0.75 m) p (0.233 × 109 Hz)(4.525 × 107 Ω−1 · m −→ Q ≈ 3.226 (3 × 108 m/s)

(1.5.22)

Substituting (1.5.22) into (1.5.5), Γ=

πf0 −→ Γ = 0.2269 GHz Q

It is clear that this answer does not make sense, if we compare it to the resonant frequency found in the first part of the problem. I repeated the latter part using Mathematica, which yielded a half width of Γ = .03612 GHz . Please see attached Mathematica printout for details.

Problem 12.13 A circular cavity (with  and µ = 1) has a radius R = 2.0 cm, and a length L = 3.0 cm. a)

Find the first six resonant frequencies of this cavity, and identify the type (TM or TE) of mode.

b)

Sketch nodal surfaces for each of these six modes.

Solution a)

The relation for the resonant frequencies for both TM and TE modes is given by (12.112), " 2   # 1/2 c jml n 2 flmn = √ + [TM mode] (1.5.23) 2 µ πR L " 2   # 1/2 0 j n 2 c 0 ml flmn = √ + [TE mode] (1.5.24) 2 µ πR L The calculated values are displayed in the following table (next page):

b)

Please see attached plots.

W. Erbsen

HOMEWORK #6

Resonant frequencies l 1 1 1 1 1 1

1.6

m 0 1 1 0 1 2

n 1 1 0 2 1 0

flmn (Ghz) 5.000 6.650 9.148 10.00 10.43 12.26

TM/TE TE TE TM TE TM TM

Homework #6

Problem 13.3 A thin linear antenna of length L carries an oscillating current   2π|z| −iωt e , I(z, t) = I0 sin L

|z| <

L 2

(1.6.1)

a)

Sketch this current as a function of z.

b)

Find the radiation pattern for this antenna. Sketch the angular distribution and plot it as a function of cos θ. Plot the pattern for kL = π and kL = 2π.

Solution a) b)

Please see attached plot (at the end). Values chosen to emphasize oscillations. Following Franklin’s lead in section (13.5), we recall that the effective radiation current is given by (13.39) as Z 0 J (kˆ r) = k j (r0 ) e−ikr ·ˆr d3 r 0 (1.6.2) For the current case, we have a linear antenna, which is oriented along the z-direction, and (1.6.2) becomes Z J (kˆ z) = k I(z, t)e−ikz cos θ dz (1.6.3) Substituting (1.6.1) into (1.6.3) yields    Z L/2  2π|z| −iωt −ikz cos θ e e dz J (kˆ z) = k I0 sin L −L/2

(1.6.4)

At this point, we must separate |z| into the positive and negative parts, and (1.6.4) becomes (Z )     Z 0 L/2 2π|z| −ik|z| cos θ 2π|z| ik|z| cos θ −iωt J (kˆ z) =kI0 e sin e dz + sin − e dz L L 0 −L/2 (Z ) Z 0 L/2 −iωt −ibz ibz =kI0 e sin(az)e dz − sin(az)e dz (1.6.5) 0 −L/2 | {z } | {z } ψI ψII

47

CHAPTER 1: ELECTRODYNAMICS II

Where I have set a = 2π/L and b = k cos θ, and also let |z| → z for notational convenience. We now recall the following trig integral: Z cos [(a − b)x] cos [(a + b)x] sin(ax) cos(bx) dx = − − (1.6.6a) 2(a − b) 2(a + b) Z sin [(a − b)x] sin [(a + b)x] sin(ax) sin(bx) dx = − (1.6.6b) 2(a − b) 2(a + b) We now begin by evaluating ψI in (1.6.5) using Euler’s identity: Z L/2 ψI = sin(az) [cos(bz) − i sin(bz)] dz 0

=

Z

Z

L/2

|0

L/2

sin(az) cos(bz) dz −i {z } |0 ψIa

sin(az) sin(bz) dz {z } ψIb

We now apply (1.6.6a) to ψIa in (1.6.7), L/2  cos [(a − b)z] cos [(a + b)z] ψIa = − − 2(a − b) 2(a + b) 0 cos [(a − b)L/2] cos [(a + b)L/2] 1 1 =− − + + 2(a − b) 2(a + b) 2(a − b) 2(a + b) a cos(aL/2) cos(bL/2) − b sin(aL/2) sin(bL/2) a =− + 2 2 2 a −b a + b2 2 4πL cos [(kL/4) cos θ] = 2 4π 2 − (kL) cos2 θ Now apply (1.6.6b) to ψIb in (1.6.7), L/2  sin [(a − b)x] sin [(a + b)x] ψIb = − 2(a − b) 2(a + b) 0 sin [(a − b)L/2] sin [(a + b)L/2] = − 2(a − b) 2(a + b) b cos(bL/2) sin(aL/2) − a cos(aL/2) sin(bL/2) = a2 − b 2 2πL sin [(kL/2) cos θ] = 2 4π 2 − (kL) cos2 θ

(1.6.7)

(1.6.8)

(1.6.9)

Combining (1.6.8) and (1.6.9), ψI = ψIa − iψIb = =

2πL {1 + cos [(kL/2) cos θ] − i sin [(kL/2) cos θ]} 2

4π 2 − (kL) cos2 θ 2πL 1 + e−i(kL/2) cos θ 

2

4π 2 − (kL) cos2 θ

(1.6.10)

Fantastic. If we repeat the same process for ψII in (1.6.5), which yields ψII = −

2πL {−1 − cos [(kL/2) cos θ] + i sin [(kL/2) cos θ]} 2

4π 2 − (kL) cos2 θ

W. Erbsen

HOMEWORK #6

=

 2πL 1 + e−i(kL/2) cos θ 4π 2 − (kL)2 cos2 θ

(1.6.11)

At this point, we substitute ψI from (1.6.10) and ψII from (1.6.11) into (1.6.5), J (kˆ z) =kI0 e−iωt {ψI + ψII } (  ) 2πL sin [(kL/2) cos θ] 2πL 1 + e−i(kL/2) cos θ −iωt =kI0 e + 2 2 4π 2 − (kL) cos2 θ 4π 2 − (kL) cos2 θ  4πkL 1 + e−i(kL/2) cos θ = 2 4π 2 − (kL) cos2 θ

(1.6.12)

And we recall that the angular distribution of the spherical wave is given by (13.50) as dP 1 2 = |J (kˆ z)| dΩ 8πc

(1.6.13)

Substituting (1.6.12) into (1.6.13), "  !  !# 4πkL 1 + ei(kL/2) cos θ 4πkL 1 + e−i(kL/2) cos θ dP 1 = 2 2 dΩ 8πc 4π 2 − (kL) cos2 θ 4π 2 − (kL) cos2 θ # " 2 2 1 64π 2 (kL) cos [(kL/4) cos θ] = 2 8πc [−4π 2 + (kL) cos2 θ]2 2

2

=

8π (kL) cos [(kL/4) cos θ] 2 c [−4π 2 + (kL) cos2 θ]2

(1.6.14)

Taking the transverse component squared of (1.6.14) leads us to 2

2

2

2

dP 8π (kL) cos [(kL/4) cos θ] = sin2 θ 2 dΩ c [−4π 2 + (kL) cos2 θ]2 −→


dP 8π (kL) cos [(kL/4) cos θ] = 1 − cos2 θ 2 dΩ c [−4π 2 + (kL) cos2 θ]2

(1.6.15)

The function in (1.6.15) has been plotted both as a function of cos θ in cartesian coordinates, and also with respect to θ in polar coordinates. Both of these have been done both for kL = π and kL = 2π. Please see attached plots (at the very end). We can go ahead and evaluate (1.6.15) for both of these values: dP 8 cos [(π/4) cos θ] = dΩ πc [−4 + cos2 θ]2

2

2 cos [(π/2) cos θ] dP = dΩ πc [−1 + cos2 θ]2

2

1 − cos2 θ 1 − cos2 θ





[kL=π] [kL=2π]

49

CHAPTER 1: ELECTRODYNAMICS II

Problem 13.4 For the antenna in the previous problem, find the maximum dP/dΩ, and the angle for which this occurs.

Solution The best way to find the maximum dP/dΩ is to find where the first derivative of the relevant function is zero in the cos θ-domain. Specifically, we start with (1.6.15), which reads 2

2


dP 8π (kL) cos [(kL/4) cos θ] = 1 − cos2 θ 2 2 2 2 dΩ c [−4π + (kL) cos θ]

(1.6.16)

We now let cos θ → x in (1.6.16), which leads to 2

2

dP 8π (kL) cos [(kL/4) x] = (1 − x) dΩ c [−4π 2 + (kL)2 x2 ]2

(1.6.17)

We must expand the trig term in the exponent of (1.6.17): cos [(kL/4) x]2 = 1 −

2

4

6

(kL) x4 (kL) x6 (kL) x2 + − + ... 32 6144 2949120

(1.6.18)

Clearly this converges very quickly, so we take the first two terms only and substitute it into (1.6.17): 8π (kL)2 [(kL)2 x2 /32]2 dP = (1 − x) dΩ c [−4π 2 + (kL)2 x2 ]2

(1.6.19)

The name of the game is then to take the first derivative of (1.6.19), set it equal to zero, and solve for x. The values should represent the maxima/minima: cos θ =3.179

[kL=π]

(1.6.20a)

cos θ =π

[kL=2π]

(1.6.20b)

The maximum value is found by evaluating (1.6.19) at these points: (dP/dΩ)max =1.7980 (dP/dΩ)max =2.1247

[kL=π] [kL=2π]

Please see attached Mathematica printout for the details of this calculation.

Problem 13.5 Evaluate the lifetime of a Rutherford hydrogen atom from (13.84). Use as data: R = .53 ˚ A, mc2 = 511 KeV, and the binding energy of hydrogen is 13.6 eV.

Solution

W. Erbsen

HOMEWORK #6

We begin with (13.84), which I will convert to SI units, in part by following the same logic as in the following problem.  2 R mc2 T = 4c e2 /R  2 cR mc = 4 e2 /R cR3 h mc i2 = 4 e2  2 3 2π0 mc =cR (1.6.21) e2 The appropriate constants are c =2.998 × 1010 m/s

e =1.6023 × 1010 C

R =0.529 × 10−10 m

m =9.109 × 10−31 Kg

0 =8.854 × 10−12 F/m Plugging all of these into (1.6.21), we have

3 T = 2.998 × 1010 m/s 0.529 × 10−10 m "


#2 2π 8.854 × 10−12 F/m 9.109 × 10−31 Kg 2.998 × 1010 m/s · (1.6023 × 1010 C)2 −→ T ≈ 1.322 × 10−11 sec

Problem 13.6 Consider a model of the hydrogen atom as a heavy positive proton, surrounded by a sphere of radius R, with a uniform charge and mass density, having a total charge and mass of an electron. a)

Show that the angular frequency of oscillation for this atom (with amplitude A ≤ R) is ω=



(e2 /R) mc2

 1/2

c R

(1.6.22)

b)

Evaluate ω and the corresponding wavelength of radiation. (Use numerical values from the previous problem).

c)

Find the oscillating electric dipole moment of this atom.

d)

Find the total power radiated by this atom for A0 = R.

51

CHAPTER 1: ELECTRODYNAMICS II

e)

Find the lifetime for exponential decay of the oscillation. Evaluate the lifetime in seconds.

Solution a)

We begin by equating Newton’s second law with Coulomb’s law, F =

mv2 1 e2 1 e2 2 3 = mω2 r = −→ ω r = r 4π0 r 2 4π0 m

(1.6.23)

The energy as a function of r is E =T + V mv2 1 e2 − r 4π0 r   1 1 2 1 e2 = − 2 4π0 r 4π0 r 2 1 e =− 8π0 r =

(1.6.24)

To find the power loss we take the first derivative of (1.6.24), dE 1 e2 dr = dt 8π0 r 2 dt

(1.6.25)

The Lamor formula for radiation caused by an accelerating charge is given by (13.94) P =

2e2 |a|2 3c3

(1.6.26)

If we recognize that the power radiated can be thought of as the power lost, or the rate of energy leaving as a function of time, then we can say that: dE 2e2 |a|2 = dt 3c3 We now substitute in the acceleration for a charge circulating the proton,
2 2e2 ω2 r dE = dt 3c3

(1.6.27)

(1.6.28)

Substitute (1.6.23) into (1.6.28),  2 dE 2e2 1 e2 = 3 dt 3c 4π0 m Equating (1.6.25) to (1.6.29),  2 1 e2 dr 2e2 1 e2 dr 1 4e4 = −→ = · 8π0 r 2 dt 3c3 4π0 m dt 4π0 3m2 c3 Z 3m2 c3 0 1 −→T = 4π0 · dr 2 4e4 R r m2 c3 1 −→T = 4π0 · 4e4 R3

(1.6.29)

W. Erbsen

HOMEWORK #6

 2 R mc2 −→T = 4π0 4c e2 /R

(1.6.30)

If we drop the factor of 4π0 , then (1.6.30) is the same as (13.84). Furthermore, we arrive at (13.161) by evaluating (1.6.23) at r = R:   1 e2 c 1 e2 /R 2 3 ω R = −→ ω = (1.6.31) 4π0 m R 4π0 mc2 If we let 4π0 = 1 in (1.6.31), we are left with   c e2 /R ω= R mc2

(1.6.32)

Which is identical to (1.6.22). b)

To evaluate ω, we simply use (1.6.23) and use the data from the previous problem:   1/ 1 e2 1 2 ω= 4π0 m R3 # 1/2 "
2 1.6023 × 1010 C 1 1 = 4π (8.854 × 10−12 F/m) (9.109 × 10−31 Kg) (0.529 × 10−10 m)3 −→ ω ≈ 4.496 × 1016

rad sec

(1.6.33) (1.6.34)

The wavelength of radiation will be given by:


2π 2.998 × 1010 m/s 2πc λ= = −→ λ ≈ 4.193 × 10−8 m ω 4.496 Hz

c)

The oscillating electric dipole moment is just the electric dipole moment, multiplied by a sinusoidal time dependent frequency part: p(t) =ed cos (ωt) Where d in our case is equal to the separation of charge: p(t) =eR cos (ωt) ˆ z

(1.6.35)

Substituting in the appropriate values,


p(t) = 1.6023 × 1010 C 0.529 × 10−10 m cos (ωt) ˆ z −→ p(t) = 8 × 10−30 C · m cos (ωt) ˆ z Where ω is of course given by (1.6.33).

d)

The total power radiated is just (1.6.28): 2e2 ω2 r P = 3c3


2

Putting in the numbers,
2
4
2 2 1.6023 × 1010 C 4.496 × 1016 rad/sec 0.529 × 10−10 m P = −→ P = 1.291 × 10−7 Watts 3 3 (2.998 × 1010 m/s)

53

CHAPTER 1: ELECTRODYNAMICS II

Problem 13.7 Find and sketch the absolute square of the Fourier transform of the step function in (13.95), and estimate its spread in ω. (Use the first zero of the transform to estimate)

Solution The step function in (13.95) reads λ(x, t) =

(

q/L, |x − vt| < L/2 0, |x − vt| > L/2

And now we recall that the functional form of the integral Fourier transform is Z f(x, t) = f(ω, t)eikx dk Z 1 A(k, t) = f(x, t)e−ikx dx 2π

(1.6.36)

(1.6.37)

Where the integrals are to be taken over the appropriate range. In our case, we substitute (1.6.36) into (1.6.37), which yields A(k, t) = = =

1 2π

Z

L/2

−L/2

q  L

e−ikx dx

 −ikx L/2 1 q e − 2π L ik −L/2   1  q  −e−ikL/2 + eikL/2

2π L q sin (kL/2) = π kL

ik

(1.6.38)

Taking the absolute square of (1.6.38) leads to A(k, t) =

q 2 sin2 (kL/2) π 2 (kL)2

In order to estimate the spread of (1.6.39) in ω, we evaluate the function at the first zero: q 2 sin2 (kL/2) L = 0 −→ kL = 2π −→ = 1 −→ ∆ω = 2L 2 2 π ω (kL)

(1.6.39)

W. Erbsen

1.7

HOMEWORK #7

Homework #7

Problem 13.8 A beam of 2 KeV electrons is stopped in a distance of 0.01 cm. a)

Calculate the total energy of the radiation emitted by each electron (assume constant acceleration).

b)

Find the ratio of the emitted energy to the initial electron energy. Use this ratio to find the radiated power for an x-ray machine with 300 Watts of power input to accelerate the electrons.

c)

Calculate the maximum intensity of the radiation at a distance of 20 cm from the stopping target.

Solution a)

To find the total energy radiated by each electron, we first convert the initial kinetic energy of the electrons (T ) from eV to Joules: T = 2 × 103 eV ·

1.602 × 10−19 J −→ T = 3.204 × 10−16 J 1 eV

We now recall that the kinetic is given by 1

2

T = /2 mv −→ v =

r

2T m

(1.7.1)

And, since we are told that the acceleration is constant, we can use one of the kinematic equations of motion: v2 = vo2 + 2a∆x −→ a = −

vo2 2∆x

(1.7.2)

If we take (1.7.1) and substitute it into (1.7.2) and put in the appropriate values,
3.204 × 10−16 J 2T /m T 2 a=− ⇒− ⇒− −→ a = −3.518 × 1018 m/s 2∆x m∆x (9.109 × 10−31 kg) (.01 × 10−2 m) We can now find the power from the following expressions: 2e2 |a|2 3c3 2 e |a|2 P = 6π0 c3

P =

[CGS]

(1.7.3a)

[SI]

(1.7.3b)

Where (1.7.3a) comes from (13.94) in Franklin, and (1.7.3b) comes from Wikipedia. Doing the calculation in SI units first, 2

P =

(1.602 × 10−19 C)2 (3.518 × 1018 m/s )2 −→ P = 7.063 × 10−17 Watts 6π(8.854 × 10−12 F/m)(2.998 × 108 m/s)3

(1.7.4)

And doing the same in CGS units with (1.7.3a), 2

P =

24.803 × 10−10 StatC)2 (3.518 × 1020 cm/s )2 −→ P = 7.063 × 10−10 erg/s 3(2.998 × 1010 cm/s)3

(1.7.5)

55

CHAPTER 1: ELECTRODYNAMICS II

In order to find the energy emitted, we must find the time over which this happens. We again turn to the kinematic equations of motion: vo v = vo + at −→ t = − a Substituting in the expression for vo from (1.7.1) into this yields p p 2T /m − (2) (3.204 × 10−16 J) / (9.109 × 10−31 kg) t=− −→ t = 7.540 × 10−12 s ⇒ 2 a −3.518 × 1018 m/s Now, if we recall that [power]=[energy]/[time], then we can find the energy by multiplying our expressions for power from (1.7.4) and (1.7.5), respectively by the time we just calculated:

E = 7.063 × 10−17 Watts 7.540 × 10−12 s −→ b)



E = 7.063 × 10−10 erg/s 7.540 × 10−12 s −→

E

= 5.326 × 10−28 J = 3.324 × 10−9 eV

E = 5.326 × 10−21 erg

[SI] [CGS]

We find the ratio of the energy emitted to the incident energy (T 0 /T ) simply by taking the specified ratio from the energy found in the last problem (T 0 here is the same as E in the last part): T0 3.324 × 10−9 eV T0 = −→ = 1.662 × 10−12 T 2 × 103 eV T

(1.7.6)

We can use this ratio to find the power emitted (P 0 ) given the incident power as follows: P0 T0 = = 1.662 × 10−12 −→ P 0 = P · 1.662 × 10−12 P T Substituting in the appropriate values, P 0 = (300 Watts) 1.662 × 10−12

c)




−→



P 0 = 4.986 × 10−10 Watts 107 erg/J −→

P0

= 4.986 × 10−10 Watts = 3.112 × 109 eV/s

P 0 = 4.986 × 10−10 erg/s

[SI] [CGS]

To find the maximum intensity at a given distance, we start with the expressions for the radiation pattern: dP e2 |a|2 = 2 3 sin2 θ dΩ 8π o c dP e2 |a|2 = sin2 θ dΩ 4πc3

[SI]

(1.7.7a)

[CGS]

(1.7.7b)

Where (1.7.7b) is (13.93) from Franklin, and (1.7.7a) was deduced from (11.69) from Griffiths. We also recall from (13.50) that dP 1 dP = r2 ˆ r · S −→ S = 2 dΩ r dΩ

(1.7.8)

Starting in SI units first, we substitute (1.7.7a) into (1.7.8), which yields S=

e2 |a|2 1 sin2 θ 8π 2 0 c3 r 2

(1.7.9)

W. Erbsen

HOMEWORK #7

The maximum intensity will occur when θ = π/2, so that sin2 → 1. Also, we recall that the “intensity” is really just the time-averaged Poynting vector, so we include a factor of 1/2 into our expressions for the Poynting vector to get the intensity: 2
2  2 −19 18 2 2 1.602 × 10 C 3.518 × 10 m/s e |a| 1 1 |S| = ⇒ 3 2 16π 2 o c3 r 2 16π 2 (8.854 × 10−12 F/m) (2.998 × 108 m/s) (20 × 10−2 m) 2

−→

|S| = 2.108 × 10−16 Watts/m = 1.316 × 103 eV

[SI]

Similarly, to find the maximum intensity in CGS units we substitute (1.7.7b) into (1.7.8):
2 4.803 × 10−10 StatC (3.518 × 1020 cm/s2 )2 e2 |a|2 1 1 |S| = = 3 2 3 2 10 8πc r 8π (2.998 × 10 cm/s) (20 cm) −→ |S| = 1.054 × 10−13 erg/(cm2 · s)

[CGS]

Problem 13.10 a)

Find the quadrupole moment (Q0 ) of the antenna in problem 13.3.

b)

Find the quadrupole radiation pattern. Sketch the angular distribution, and compare it to the exact distribution.

c)

Find the maximum intensity and the angle for which it occurs for the quadrupole, and compare these to the exact results.

d)

Find the total quadrupole power radiated and compare to the exact power (evaluate the integral numerically).

Solution a)

We first recall the relation for the electric quadrupole dyadic from (13.114) of Franklin: Z i 1 h ˆ [Q] = 3r · r − r 2 n ˆ ρ d3 r 2 We can expand (1.7.10) in equation form as follows:  Z 3x2 − r 2 3xy 1  2 3yx 3y − r2 [Q] = 2 3zx 3zy

 3xz 3yz  ρ d3 r 3z 2 − r 2

(1.7.10)

(1.7.11)

The current density, ρ, is non-zero only for the z-component, the other components are zero (ρ(x) = ρ(y) = 0). Therefore, when we expand r 2 = x2 + y2 + z 2 , only the z-component remains. For this same reasoning all of the offdiagonal elements in (1.7.11) go to zero. Therefore, (1.7.11) becomes     Z −z 2 Z −1/2 0 0 0 0 1  −1 0 −z 2 0  ρ dz −→ [Q] =  0 /2 0 z 2 ρ dz [Q] = 2 2 0 0 2z 0 0 1

57

CHAPTER 1: ELECTRODYNAMICS II

The matrix can be taken out of the integral, and we  −1 /2 0 −1 [Q] =  0 /2 0 0

find that  0 Z 0 z 2 ρ dz 1 | {z } Q0

(1.7.12)

Where I have bracketed the portion of (1.7.12) that represents Q0 . To find Q0 , we must first find the charge density, ρ. To do this we begin with the continuity equation: Z t ∂ ∇·j = − ρ −→ ρ = − ∇·j dt (1.7.13) ∂t 0 Where the charge density j is really a “linear charge density,” and can be found by expressed in terms of the current as j=

I L

Substituting this in to (1.7.13), Z 1 t ρ =− ∇·I dt L 0 Z 1 t ∂ I(z, t) dt =− L 0 ∂z     Z 1 t ∂ 2π|z| −iωt =− Io sin e dt L 0 ∂z L  Z t 2π 2π|z| = − 2 Io cos e−iωt dt L L 0  Z t 2π 2π|z| = − 2 Io cos cos(ωt) dt L L 0 t   2π 2π|z| sin(ωt) = − 2 Io cos L L ω 0   2π 2π|z| =− Io cos sin(ωt) ωL2 L

 

take only the real part, or else ρ will be complex!



Hooray! We can now find the quadrupole moment by substituting (1.7.14) into (1.7.12): Z Q0 = z 2 ρ dz   Z L/2 2π 2π|z| 2 =− I sin(ωt) z cos dz o ωL2 L −L/2 (Z )     Z 0 L/2 2π 2π|z| 2π|z| 2 2 =− Io sin(ωt) z cos dz + z cos dz ωL2 L L 0 −L/2   Z L/2 4π 2πz 2 =− Io sin(ωt) z cos dz ωL2 L 0

(1.7.14)

(1.7.15)

Where the last step is completed by realizing that the integral is symmetric. We can integrate (1.7.15) by repeated use of integration by parts:

W. Erbsen

HOMEWORK #7

)  2πz u = z2 du = 2z I= z cos dz dv = cos(2πz/L) v = (L/2π) sin(2πz/L) L 0    L/2   Z L/2 L 2πz L 2πz = · z 2 · sin − z sin dz 2π L π 0 L 0  2     Z L L 2π L L L/2 2πz = · sin · − z sin dz 2π 2 L 2 π 0 L   Z L L/2 2πz L3 sin(π) − z sin dz = 8π π 0 L )   Z L L/2 2πz u=z du = 1 =− z sin dz dv = sin(2πz/L) v = − (L/2π) cos(2πz/L) π 0 L ( )   L/2   Z L/2 L 2πz L 2πz L =− − · z cos + cos dz π 2π L 0 2π 0 L (      L/2 ) L L L 2π L L L 2πz =− · cos · sin − + π 2π 2 L 2 2π 2π L 0    L L2 L2 2π L − cos(π) + 2 sin · =− π 4π 4π L 2  2  2 L L L + 2 sin(π) =− π 4π 4π L3 (1.7.16) =− 2 4π Z

L/2

2



Whew! We can now substitute (1.7.16) into (1.7.15) to get the quadrupole moment:   4π L3 L Q0 = − I sin(ωt) − −→ Q0 = Io sin(ωt) o ωL2 4π 2 πω b)

(1.7.17)

The quadrupole radiation pattern is given by (13.121) in Franklin as: dP ck 6 Q20 = cos2 (θ) sin2 (θ) dΩ 32π Substituting (1.7.17) into (1.7.18) yields  2 dP ck 6 L = Io sin(ωt) cos2 (θ) sin2 (θ) dΩ 32π πω L2 ck 6 = · 2 2 Io2 sin2 (ωt) cos2 (θ) sin2 (θ) 32π π ω ck 6 L2 2 = I sin2 (ωt) cos2 (θ) sin2 (θ) 32π 3 ω2 o

(1.7.18)

(1.7.19)

If we recall that ω2 = c2 k 2 , then (1.7.19) becomes dP ck 6 L2 = I 2 sin2 (ωt) cos2 (θ) sin2 (θ) dΩ 32π 3 (c2 k 2 ) o k 4 L2 2 = I sin2 (ωt) cos2 (θ) sin2 (θ) 32π 3 c o

(1.7.20)

59

CHAPTER 1: ELECTRODYNAMICS II

If we take the time-average of the time-dependent term in (1.7.20), then we find that sin2 (ωt) → 1/2 , and we are left with k 4 L2 2 dP = I cos2 (θ) sin2 (θ) dΩ 64π 3 c o

c)

(1.7.21)

I have plotted (1.7.21), and is at the very end. The pattern is unnormalized, since we are not given the values for any of the constants. We do know that within the mess of constant is a factor of 1/64, which tells us that the quadrupole radiation pattern contributes significantly less to the total radiated power of the antenna. This makes sense, since the antenna in question is a dipole antenna, and so it emits mostly dipole radiation. As done previously, we can find the intensity from the “time-averaged Poynting vector”, given as (13.50) in Franklin: dP 1 dP = r2 ˆ r · S −→ S = 2 ·ˆ r dΩ r dΩ

(1.7.22)

So presumably, all we would need to do is substitute (1.7.21) into (1.7.22): |S| =

1 k 4 L2 2 I cos2 (θ) sin2 (θ) r 2 64π 3 c o

(1.7.23)

The intensity of (1.7.24) will reach a maximum when θ = π/4, 3π/4, so that cos2 (π/4) sin2 (π/4) = 1/4, and we are finally left with |S| = d)

1 k 4 L2 2 I r 2 256π 3 c o

(1.7.24)

The total power radiated from a symmetric quadrupole is given by (13.122) in Franklin as P =

ck 6 Q20 60

So, we substitute (1.7.17) into (1.7.25):  2 ck 6 L ck 6 L2 2 2 Io sin(ωt) = I sin (ωt) P = 60 πω 60π 2 ω2 o

(1.7.25)

(1.7.26)

We remember that ω2 = c2 k 2 , and that according to time averaging sin2 (ωt) = 1/2 , and (1.7.26) becomes P=

K 4 L2 2 I 120π 2 c o

(1.7.27)

Problem 13.11 A conducting sphere of radius R is an oscillating magnetic field B = B0 e−iωt acquires an induced magnetic moment µ. a)

Find the surface current K induced on the conducting sphere (Use the boundary condition on B, including the magnetic field due to the induced dipole).

W. Erbsen

b)

HOMEWORK #7

Integrate the current over the sphere to find the induced magnetic moment.

Solution a)

This problem bears extreme resemblance to the very similar situation of a conducting sphere in an external electric field. One of the most immediately apparent discrepancies is that for a sphere in an electric field we are concerned with the electric dipole moment, p, whereas with the magnetic field we concern ourselves with the magnetic moment, µ. We first recall that Amper´e’s circuit law states that ∇×H =

3πj 1 ∂ + D c c ∂t

(1.7.28)

We know that j = 0 inside and outside the sphere as well, so (1.7.28) reduces to ∇×H =

1 ∂ D c ∂t

(1.7.29)

Since the sphere is a conductor, the electric field inside will be zero, and outside as well, so (1.7.29) becomes ∇×H = 0

(1.7.30)

Which applies to both interior and exterior of the sphere. We also note that K must only exist at the interface. At this point, we introduce the concept of the magnetic scalar potential, φm , which shares many of the same characteristics as the electrical analogue, φe . Most notably, from (7.88) in Franklin, H = −∇φm −→ B = −∇φm

(1.7.31)

Now, we remember that from Maxwell’s equations that monopoles most certainly do not exist: ∇·B = 0

(1.7.32)

And substituting (1.7.31) into (1.7.32) yields ∇· (−∇φm ) = 0 −→ ∇2 φm = 0

(1.7.33)

Which we immediately recognize to be Laplace’s equation. We can find solutions for this equation by way of analyzing boundary conditions of a sphere imagined to be in an external electric field, and then replace the electric potential (φe ) by the magnetic potential (φm ). So let’s get started! For the case of an electric field, our boundary conditions are ( 0 at r = R φe (r, θ) = −E0 r cos θ at r → ∞ And the series solution we will be using is  ∞  X B` ` φe (r, θ) = A` r + `+1 P` (cos θ) r `=0

Applying the first boundary condition to (1.7.34) yields:

(1.7.34)

61

CHAPTER 1: ELECTRODYNAMICS II

φe (R, θ) = A` R` +

B` = 0 −→ B` = −A` R2`+1 R`+1

We now substitute (1.7.35) back in to (1.7.34):  ∞  X R2`+1 φe (r, θ) = A` r ` − A` `+1 P` (cos θ) r

(1.7.35)

(1.7.36)

`=0

We now apply the second boundary condition to (1.7.36): φe (r → ∞, θ) =

∞ X `=0

A` r ` P` (cos θ) = −E0 r cos θ

(1.7.37)

By looking at the first few Legendre Polynomials, ee can see that the only way that the second boundary condition is satisfied is if ` = 1 and A` = A1 = −E0 . Applying all this to (1.7.36) leads us to   R3 φe (r, θ) = −E0 r − 2 cos θ (1.7.38) r But not so fast! We must recognize that we must separate (1.7.34) to describe the interior and exterior potentials of our sphere: φin m = φout m =

∞ X

`=0 ∞ X `=0

A` r ` P` (cos θ)

(1.7.39a)

B` P` (cos θ) r `+1

(1.7.39b)

Substituting in the values for the constants we found previously, (1.7.39a) and (1.7.39b) become, respectively: φin m =0 φout m =B0

(1.7.40a) 3

R cos θ r2

(1.7.40b)

Under the same logic as the first boundary condition, we choose the potential inside the sphere to be zero, so that the field is zero. We can now find the interior and exterior fields by substituting (1.7.40a) and (1.7.40b) into (1.7.33), in turn. We first recall that the gradient is given (in spherical coordinates) by ∇f =

∂f 1 ∂f ˆ 1 ∂f ˆ ˆ r+ θ+ φ ∂r r ∂θ r sin θ ∂φ

(1.7.41)

Since we have no φ-dependence, (1.7.41) becomes ∇f =

∂f 1 ∂f ˆ ˆ r+ θ ∂r r ∂θ

Substituting (1.7.42), (1.7.40a)-(1.7.40b) into (1.7.33), we have: Bin = −∇φin m = 0 Bout = − ∇φout m =−



  ∂ 1 ∂ ˆ R3 ˆ r+ θ B0 2 cos θ ∂r r ∂θ r

(1.7.42)

W. Erbsen

HOMEWORK #7

      ∂ 1 R3 ∂ = − B0 R3 cos θ ˆ r + B cos θ θˆ 0 3 ∂r r 2 r ∂θ R3 R3 r − B0 3 sin θ θˆ =2B0 3 cos θ ˆ r r

(1.7.43)

We now recall from (7.74) that

4π c K=n ˆ × Bout − Bin −→ K = n ˆ × Bout − Bin c 4π

(1.7.44)

Substituting (1.7.43) into (1.7.44), K=

  R3 R3 c n ˆ × 2B0 3 cos θ ˆ r − B0 3 sin θ θˆ 4π r r

(1.7.45)

ˆ so (1.7.45) becomes We recall that n ˆ is ˆ r, and ˆ r׈ r = 0, but ˆ r×θˆ = −φ, K=

c R3 ˆ B0 3 sin θ φ 4π r

(1.7.46)

The surface current is found by evaluating (1.7.46) at r = R: K=

c ˆ B0 sin θ e−iωt φ 4π

(1.7.47)

Where, you guessed it, I neglected the exponential term and put it back in at the last moment. b)

To find the magnetic moment µ, I 1 µ= r × K dS 2πc I  c  1 ˆ dS = r× B0 sin θ e−iωt φ 2πc 4π I  B0 −iωt  ˆ sin θ dS = 2e r×φ 8π I B0 = − 2 e−iωt r sin θ dS 8π Z Z B0 = − 2 e−iωt r sin θ · r 2 sin θ dθdφ 8π
B0 = − 2 e−iωt R3 4π 2 8π

Where I have absorbed the direction into the magnitude B0 to signify that the magnetic moment is along the same axis (opposite direction) as the incident field. We are then left with µ=−

B0 3 −iωt R e 2

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CHAPTER 1: ELECTRODYNAMICS II

1.8

Homework #8

Problem 14.1 A space ship is moving with velocity 0.6c shoots a projectile with muzzle velocity 0.8c. Find the projectile’s velocity with respect to a fixed target if it is fired a)

Straight ahead.

b)

Straight backward.

c)

At right angles to the ship’s velocity.

d)

Redo parts a,b,c if the muzzle velocity were c.

Solution The relativistic addition of velocity equations are given by (14.24) and (14.25): uk =

u0k + v

1 + u0 · v/c2 u0⊥ u⊥ = γ (1 + u0 · v/c2 )

(1.8.1a) (1.8.1b)

And we also know that the Lorentz factor, γ, is given by γ=p

1 1 ⇒p 2 2 1 − v /c 1 − β2

(1.8.2)

Where β = v/c. In our case, the moving frame (S 0 ) is that of the spaceship, which is moving at a velocity v. Accordingly, we can go ahead and calculate the Lorentz factor from (1.8.2): 1 γ =p 1 − (0.6c)2 /c2 1 =p 1 − (0.6)2 1 =√ 1 − 0.36 1 =√ 0.64 1 = 0.8 =1.25 a)

If the projectile is fired straight ahead, then u0k = 0.8c, and u0⊥ = 0. From (1.8.1a), uk =

u0k + v

1 + u0 · v/c2 (0.8c) + (0.6c) = 1 + (0.8c)(0.6c)/c2

(1.8.3)

W. Erbsen

HOMEWORK #8

1.4c 1 + 0.48 1.4 = c −→ uk ≈ 0.9459c 1.48

=

And now from (1.8.1b), u⊥ = b)

u0⊥ −→ u⊥ = 0 γ (1 + u0 · v/c2 )

If the projectile is fired straight backwards, then u0k = −0.8c, and u0⊥ = 0. Once again using (1.8.1a), uk =

u0k + v

1 + u0 · v/c2 (−0.8c) + (0.6c) = 1 + (−0.8c)(0.6c)/c2 −0.2c = 1 − 0.48 −0.2c = c −→ uk ≈ −0.3846c 0.52

And from (1.8.1b) u⊥ = c)

u0⊥ −→ u⊥ = 0 γ (1 + u0 · v/c2 )

If it is fired perpendicular to the ship’s motion, then we can say that u0k = 0, and u0⊥ = 0.8c, and (1.8.1a) becomes uk =

1

u0k + v

+ u0

·

v/c2

⇒

0.6c −→ uk = 0.6c 1

And for the perpendicular component, (1.8.1b) leads us to u⊥ = d)

u0⊥ 0.8c ⇒ −→ u⊥ = 0.64c γ (1 + u0 · v/c2 ) 1.25(1)

Repeating parts a)-c) warrants no real surprises. a)

From (1.8.1a), uk =

u0k + v

1 + u0 · v/c2 (c) + (0.6c) = 1 + (c)(0.6c)/c2 1.6c = 1 + 0.6 1.6 = c −→ uk = c 1.6

And the perpendicular component is surprisingly uneventful, u0⊥ u⊥ = −→ u⊥ = 0 γ (1 + u0 · v/c2 )

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CHAPTER 1: ELECTRODYNAMICS II

b)

For the backwards case we have u0k = −c and u0⊥ = 0, we have from (1.8.1a) that u0k + v uk = 1 + u0 · v/c2 (−c) + (0.6c) = 1 + (−c)(0.6c)/c2 −0.4c = 1 − 0.6 −0.4c c −→ uk = −c = 0.4 And from (1.8.1b), u⊥ =

c)

u0⊥ −→ u⊥ = 0 γ (1 + u0 · v/c2 )

And lastly, but perhaps most interestingly, we let u0k = 0 and u0⊥ = c, so (1.8.1a) becomes u0k + v 0.6c uk = ⇒ −→ uk = 0.6c 0 2 1 + u · v/c 1 And for the perpendicular component, (1.8.1b) says that u0⊥ c ⇒ −→ u⊥ = 0.8c u⊥ = γ (1 + u0 · v/c2 ) 1.25(1)

Problem 14.2 a)

Use the relativistic velocity addition equations to prove that c added in any direction to any velocity results in velocity c.

b)

Prove that adding any two velocities cannot give a velocity greater than c.

Solution a)

To prove this we start with the relativistic velocity addition equation for parallel trajectories, given by (14.24) in Franklin: uk = Now, if we let u0k = c, then (1.8.4) becomes

u0k + v

1 + u0 · v/c2

c+v 1 + (c) v/c2 c+v = 1 + v/c c+v = −→ uk = c 1/c (c + v)

uk =

(1.8.4)

W. Erbsen

HOMEWORK #8

The same result applies if we let u0k = c as well. b)

For this next proof, we again start with (1.8.4), but must apply some more algebraic manipulations: uk = = = =

u0k + v

1 + u0 · v/c2 c(u0k /c + v/c)

1 + u0 · v/c2 c(1 − 1 + u0 · v/c2 − u0 · v/c2 + u0k/c + v/c)

1 + u0 · v/c2 c[1 + u · v/c − (1 + u0 · v/c2 − u0k /c − v/c)] 0

2

1 + u0 · v/c2

"

0 2 0 1 + u0 · v/c2 (1 + u · v/c − uk/c − v/c) − =c 1 + u0 · v/c2 1 + u0 · v/c2 " # (1 − u0k /c − v/c + u0 · v/c2 ) =c 1 − 1 + u0 · v/c2 " # (1 − u0k /c) (1 − v/c) =c 1 − 1 + u0 · v/c2

#

(1.8.5)

We can see from the bracketed fraction in (1.8.5) that if u0k = c, then uk = c, and also if v = c, then uk = c. If Both are equal to c, then uk = c. If either (or both) are anything less than c, then uk < c. Therefore, adding any two velocities cannot result in a velocity greater than c .

Problem 14.3 Derive (14.26) and (14.27) for the transformation of acceleration.

Solution The equations that we are ultimately trying to derive are: ak = a⊥ =

a0k

3

γ 3 (1 + u0 · v/c2 ) a0⊥ + (u0 × a0 ) × v/c2 3

γ 2 (1 + u0 · v/c2 )

(1.8.6a) (1.8.6b)

A good place to start is with the Lorentz transformation equations, given by (14.19)-(14.22) in Franklin:
t0 =γ t − xv/c2 x0 =γ (x − vt) y0 =y

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CHAPTER 1: ELECTRODYNAMICS II

z 0 =z We can find the inverse transform equations of these by replacing t ↔ t0 , x ↔ x0 , y ↔ y0 , z ↔ z 0 , and v ↔ −v:
t =γ t0 + x0 v/c2 (1.8.7a) 0 0 x =γ (x + vt ) (1.8.7b) y =y0 z =z 0

Using (1.8.7a), we calculate the following derivative:
 ∂  ∂t = 0 γ t0 + x0 v/c2 0 ∂t ∂t    ∂t0 v ∂x0 =γ + 2 ∂t0 c ∂t0   =γ 1 + u0k v/c2
=γ 1 + u0 · v/c2

Taking the reciprocal of (1.8.8),

∂t0 1 = ∂t γ (1 + u0 · v/c2 )

(1.8.8)

(1.8.9)

Fantastic. Now, we use the chain rule to find the following derivative: ∂x v ˆ ∂t ∂x ∂t0 v ˆ = 0 · ∂t ∂t ∂ 1 = [γ (x0 + vt0 )] · v ˆ ∂t γ (1 + u0 · v/c2 )  0   0  ∂x ∂t 1 = + v · v ˆ 0 0 ∂t ∂t γ (1 + u0 · v/c2 ) u0k + v = v ˆ 1 + u0 · v/c2 u0k + v = 1 + u0 · v/c2

uk =

(1.8.10)

Where I used both (1.8.7b) and (1.8.9) in deriving (1.8.10). This is of course the relativistic velocity addition equation for parallel trajectory, which is (14.24) from Franklin. Similarly, we repeat this process for the transverse component (I will denote the unit vector perpendicular to the forward motion of v by v ˆ⊥ ): ∂x v ˆ⊥ ∂t ∂x ∂t0 = 0· v ˆ⊥ ∂t ∂t

u⊥ =

W. Erbsen

HOMEWORK #8

∂x 1 · v ˆ⊥ 0 ∂t γ (1 + u0 · v/c2 ) u0⊥ = v ˆ⊥ γ (1 + u0 · v/c2 ) u0⊥ = γ (1 + u0 · v/c2 ) =

(1.8.11)

Which is of course none other than (14.25) from Franklin. But wait! We want the acceleration, not the velocity. But be not afraid, as the process of finding ak and a⊥ is very much the same as what we did to find uk and u⊥ . We start by finding the derivative of the parallel velocity component: ∂uk ∂t ∂uk ∂t0 = 0 · ∂t " ∂t # u0k + v ∂ 1 · = 0 0 2 ∂t 1 + u · v/c γ (1 + u0 · v/c2 )       ∂  1 1 ∂  0 1 0 = uk + v + uk + v γ (1 + u0 · v/c2 ) 1 + u0 · v/c2 ∂t0 ∂t0 1 + u0 · v/c2 ( " 0# " !#) 0   ∂uk 1 1 v ∂uk 1 0 = + uk + v − 2 0 2 γ (1 + u0 · v/c2 ) 1 + u0 · v/c2 ∂t0 c ∂t (1 + u0 · v/c2 ) ( " #)   a0k a0k v/c2 1 0 = + uk + v − 2 γ (1 + u0 · v/c2 ) 1 + u0 · v/c2 (1 + u0 · v/c2 ) #) ( "  a0k a0k v/c2 1 1 + u0 · v/c2  0 = · + uk + v − 2 γ (1 + u0 · v/c2 ) 1 + u0 · v/c2 1 + u0 · v/c2 (1 + u0 · v/c2 ) ( 0 " #)
  ak 1 + u0 · v/c2 a0k v/c2 1 0 = + uk + v − 2 2 γ (1 + u0 · v/c2 ) (1 + u0 · v/c2 ) (1 + u0 · v/c2 ) n  o
 1 = a0k 1 + u0 · v/c2 − u0k + v a0k v/c2 3 γ (1 + u0 · v/c2 ) n o 1 = a0k + a0k (u0 · v) /v2 − a0k (u0 · v) /c2 − a0k v2 /c2 3 γ (1 + u0 · v/c2 ) n o 1 a0k − a0k v2 /c2 = 3 γ (1 + u0 · v/c2 ) a0k  = 1 − v2 /c2 (1.8.12) 3 γ (1 + u0 · v/c2 )

ak =

If we recognize that the term in braces in (1.8.14) is just γ −2 , then we are finally left with ak =

a0k

3

γ 3 (1 + u0 · v/c2 )

Which we can see is the same as (1.8.1a). For the perpendicular component,

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CHAPTER 1: ELECTRODYNAMICS II

∂u⊥ ∂t ∂u⊥ ∂t0 = 0 · ∂t  ∂t  ∂ u0⊥ 1 = 0 · 0 2 ∂t γ (1 + u · v/c ) γ (1 + u0 · v/c2 )   0    1 1 ∂u⊥ ∂ 1 0 = 2 + u ⊥ γ (1 + u0 · v/c2 ) 1 + u0 · v/c2 ∂t0 ∂t0 1 + u0 · v/c2 ( " #)   0 1 a0⊥ v ∂u 1 + u0⊥ − 2 = 2 γ (1 + u0 · v/c2 ) 1 + u0 · v/c2 c ∂t0 (1 + u0 · v/c2 )2 ( ) a0⊥ a0 · v/c2 1 0 = 2 − u⊥ γ (1 + u0 · v/c2 ) 1 + u0 · v/c2 (1 + u0 · v/c2 )2 ( ) 0 2 1 a0⊥ 1 + u0 · v/c2 a · v/c = 2 · − u0⊥ 2 γ (1 + u0 · v/c2 ) 1 + u0 · v/c2 1 + u0 · v/c2 (1 + u0 · v/c2 ) ( )
a0⊥ 1 + u0 · v/c2 1 a0 · v/c2 = 2 − u0⊥ 2 2 0 2 γ (1 + u · v/c ) (1 + u0 · v/c2 ) (1 + u0 · v/c2 )  0
1 0 2 = − u0⊥ (a0 · v) /c2 3 a⊥ 1 + u · v/c 2 0 2 γ (1 + u · v/c )

a⊥ =

(1.8.13)

At this point, it should be noted that we must be extremely careful with our vectors and scalar products. In particular, we recall that a0 · v = a0k v ˆ · vˆ v = a0k v. Continuing (1.8.15), n     o 1 a⊥ = a0⊥ + a0⊥ u0k v /c2 − u0⊥ a0k v /c2 3 γ 2 (1 + u0 · v/c2 ) n   o 1 = a0⊥ + a0 − a0k u0k v/c2 − u0⊥ a0k v/c2 3 γ 2 (1 + u0 · v/c2 ) n o 1 = a0⊥ + a0 u0k v/c2 − a0k u0k v/c2 − u0⊥a0k v/c2 3 γ 2 (1 + u0 · v/c2 ) n  o 1 = a0⊥ + a0 u0k v/c2 − a0k v/c2 u0k + u0⊥ 3 γ 2 (1 + u0 · v/c2 ) n o 1 = a0⊥ + a0 u0k v/c2 − a0k u0 v/c2 3 γ 2 (1 + u0 · v/c2 )  i 1 1 h 0 0 0 0 0 = a⊥ + 2 a uk v − ak u v (1.8.14) 3 c γ 2 (1 + u0 · v/c2 ) We now recall the triple product A × (B × C) = B (A · C) − C (A · B). We can apply this to the bracketed term in (1.8.14): h i h i a0 u0k v − a0k u0 v = − a0k u0 v − a0 u0k v h    i = − u0 a0k v − a0 u0k v = − [u0 (v · a0 ) − a0 (v · u0 )] = − [v × (u0 × a0 )]

W. Erbsen

HOMEWORK #8

= (u0 × a0 ) × v

(1.8.15)

Substituting (1.8.15) into (1.8.14), a⊥ =

1 3

γ 2 (1 + u0 · v/c2 )



a0⊥ +

 1 0 0 [(u × a ) × v] c2

Bringing in the factor from outside and bringing up the factor c2 , we are left with a⊥ =

a0⊥ + (u0 × a0 ) × v/c2 3

γ 2 (1 + u0 · v/c2 )

Which we can see is identical to (1.8.6b).

Problem 14.4 An object falls from constant acceleration (in its rest system) of g. p a) Find its velocity after a time t. (Answer: v = gt/ 1 + g2 t2 /c2 )

b)

What is its velocity after one year?

c)

Find p/m of the object at time t. Sketch graphics of v(t) and p(t)/m.

Solution a)

To find the velocity, we start with Newton’s 2nd Law: F = ma ⇒

dp dt

(1.8.16)

However, we must be careful here - the momentum is relativistic, and must treat it accordingly: mv p = γmv ⇒ p (1.8.17) 1 − v2 /c2 Substituting (1.8.17) into (1.8.16), we find that " # Z t v v d p −→ a(t) dt = p a= dt 1 − v2 /c2 1 − v2 /c2 0 Recalling that in our case a = g, so that (1.8.18) becomes Z t v v g dt = p −→ gt + C = p 2 2 1 − v /c 1 − v2 /c2 0 Our object starts at rest, so v(0) = 0, so C = 0, so (1.8.19) becomes v gt = p 1 − v2 /c2

(1.8.18)

(1.8.19)

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CHAPTER 1: ELECTRODYNAMICS II

g 2 t2 =

v2 1 − v2 /c2

v2 =g2 t2 1 − v2 /c2




v2 =g2 t2 − v2 g2 t2 /c2

g2 t2 =v2 + v2 g2 t2 /c2

g2 t2 =v2 1 + g2 t2 /c2 v2 =

g 2 t2 1 + g2 t2 /c2




(1.8.20)

From (1.8.20) it is easy to see that we are left with

b)

v(t) = p

gt 1 + g2 t2 /c2

(1.8.21)

Does this question ask us to find the velocity in the frame of an observer on Earth, or in the frame of the ship? Let’s find out both! The velocity according to an observer on ship is easy, we can just use (1.8.21), but first let’s find t: 1 yr ·

365 day 24 hr 60 min 60 sec · · · = 31.539 × 106 sec 1 yr 1 day 1 hr 1 min

Substituting this value into (1.8.21), we find that 2

v (1 yr) = q

(9.810 m/s )(31.539 × 106 s) 2

1 + (9.810 m/s )(31.539 × 106 s)/(2.998 × 108 m/s)]2

−→ v (1 yr) = 0.7181c

[Ship frame]

Now, to find the velocity after one year in the Earth’s frame, we must instill a little mathematical voodoo. We start with the definition of acceleration: duk dt duk dt0 = 0 · dt " dt # u0k + v d 1 = 0 · dt 1 + u0 · v/c2 γ (1 + u0 · v/c2 )

ak =

Letting u0 → 0, (1.8.22) becomes ak =

1 dv γ dt0

(1.8.22)

(1.8.23)

Now, we take a look at (14.26) from Franklin, and also let u0 → 0: ak =

a0k

γ 3 (1 + u0 · v/c2 )3

−→ ak =

a0k γ3

(1.8.24)

Equating (1.8.23) and (1.8.24), a0k a0k 1 dv dv = −→ = γ dt0 γ3 dt0 γ2

(1.8.25)

W. Erbsen

HOMEWORK #8

If we notice that a0k = g, then (1.8.25) becomes Z v 2 0 γ dv = g dt −→ 0

1 dv = g 1 − v2 /c2

Z

t0

dt0

(1.8.26)

0

Things get a little easier if we let u = v/c, du = (1/c)dv. With this, (1.8.26) becomes Z v 1 c du = gt0 2 0 1−u

(1.8.27)

This integral requires using hyperbolic trig substitutions, and is a well-known integral. Pressing on, (1.8.27) becomes h  v i   c tanh−1 (u) = gt0 −→c tanh−1 = gt0 c  0 v gt −→ = tanh c c  0 gt −→v = c tanh (1.8.28) c Hooray! Substituting in the appropriate values into (1.8.28) yields # " 2 (9.810 m/s )(31.539 × 106 s) c v (1 yr) = tanh 2.998 × 108 m/s −→ v (1 yr) = 0.7757c c)

[Earth frame]

To find p/m as a function of t, we must go back and use the result from the first part of the problem. The momentum is of course given by p(t) = γmv(t), however which v(t) should we use? Let’s use both! I will go ahead and use the first result, the one derived in part a). We begin by finding the Lorentz factor, γ, with this velocity:  − 1/2 v2 γ = 1− 2 c  !2 − 1/2 1 gt  = 1 − 2 p c 1 + g2 t2 /c2 

 − 1/2 1 g 2 t2 = 1− 2 c 1 + g2 t2 /c2  − 1/2 g 2 t2 = 1− 2 c + g 2 t2  2 − 1/2 c + g 2 t2 g 2 t2 = 2 − 2 c + g 2 t2 c + g 2 t2 1   − /2 c2 = 2 c + g 2 t2  2  1/ c + g 2 t2 2 = c2

We can now use (1.8.29) to find the p(t)/m:

(1.8.29)

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CHAPTER 1: ELECTRODYNAMICS II

p(t) =γv(t) m  1/  2 gt c + g 2 t2 2 p = 2 c 1 + g2 t2 /c2  2  1/   1/2 c + g 2 t2 2 c2 = gt c2 c 2 + g 2 t2

(1.8.30)

From (1.8.30) it is clear that we are left with: p(t) = gt m

(1.8.31)

We now do the same for the velocity in the Earth’s frame from (1.8.28): p(t) =γv(t) m 1 =p v(t) 1 − v2 /c2   gt 1 =q c tanh c 2 1 − (c tanh (gt/c)) /c2   gt 1 c tanh =q c 2 1 − tanh (gt/c)   1 gt q = c tanh c 2 sech (gt/c) sinh2 (gt/c) c cosh2 (gt/c)

(1.8.32)

p(t) sinh2 (gt/c) = c m cosh (gt/c)

(1.8.33)

= cosh (gt/c) It is easy to see that (1.8.32) leads us to

A plot of v(t) from (1.8.21) and (1.8.28) is attached (they look the same). There is also a plot of (1.8.33), however I have not included (1.8.31), as it is just a straight line, and straight lines are never fun. Plots can be seen at the end of this homework assignment.

Problem 14.5 A space ship 100 meters long flies through an open hanger 80 meters long. a)

How fast must the space ship fly in order to permit both doors of the hanger to be briefly closed at the same time while the space ship is inside?

b)

How would the pilot of the space ship interpret what happens as the doors close and then reopen?

W. Erbsen

HOMEWORK #8

Solution a)

We begin with the equation for length contraction, which reads L γ

(1.8.34)

1 1 − v2 /c2

(1.8.35)

L0 = While the Lorentz factor, γ, is given by

Substituting (1.8.35) into (1.8.34),

Solving (1.8.36) for v,

γ= p

p L0 = 1 − v2 /c2 −→ L



L0 L

2

=1−

v" u  0 2 # u L v =t 1 − c2 L s  2 80 m = 1− c 100 m s  2 4 = 1− c 5 r 16 = 1− c 25 r 9 c = 25 3 = c −→ v = 0.6c 5 b)

v2 c2

(1.8.36)

(1.8.37)

First of all, the doors will not be able to close “simultaneously,” which is dependent on the observer. We note that although a person standing next to the hanger will observe the space ship appearing shorter, the pilot will also see the hanger getting shorter. At the end of the day, whether this event is observed to happens depends on the observer, and also their definition of “simultaneous.” A person standing next to the hanger will say that the back end of the space ship will make it through the entrance to the hanger before the front of the ship hits the forward wall of the hanger. However, the pilot sees it just the other way around. To him, his ship is too long to fit in the hanger, and he sees the front of his ship hit the far wall of the hanger before the tail end of the ship makes it all the way through the entrance of the hanger.

Problem 14.6 A jet plane circles the Earth at the equator with a speed of Mach 4. By how much time is the pilot’s watch slow when he lands?

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CHAPTER 1: ELECTRODYNAMICS II

Solution We know that the kilometer was originally defined as “the distance from the equator to the North pole divided by ten million.” Assuming that this value is correct, then this implies that the circumference of the Earth is 40, 000, 000 m, or 4 × 107 m. I will assume that the pilot is flying at sea level - since no other values were specified. We also know that the Mach system defines velocity in terms of the speed of sound in the local media. At sea level, Mach 4 turns out to be 1.361 × 103 m/s. We also know the speed of light, so the constants needed in this problem are: ∆x =4.000 × 107 m

v =1.361 × 103 m/s c =2.998 × 108 m/s

And if we define β = v/c, then we can say that β=

v 1.361 × 103 m/s = −→ β = 4.540 × 10−6 c 2.998 × 108 m/s

If we let t designate the time measured on Earth, and t00 be the time measured in the local reference frame of the ship, then the change in time is ∆t = t − t00

(1.8.38)

And the equation for time dilation is t00 =

t γ

(1.8.39)

Substituting (1.8.39) into (1.8.38), t ∆t =t − γ   1 =t 1 − γ   p =t 1 − 1 − β 2

(1.8.40)

Since β  1, we can expand (1.8.40) in a series expansion:    β2 β4 β6 ∆t = t 1 − 1 − − − − ... 2 8 16 Taking only the first two terms,    tβ 2 β2 −→ ∆t = ∆t = t 1 − 1 − 2 2 We can now say that from (1.8.41) the time lost per second is given by
2 4.540 × 10−6 ∆t β2 ∆t = ⇒ −→ = 1.030 × 10−11 t 2 2 t

(1.8.41)

(1.8.42)

W. Erbsen

HOMEWORK #8

This means that the space ship’s clock lags behind by ∆t/t = 10.30 ps for every second that it is at speed. The time that the ship is orbiting is given by v=

∆x ∆x 4.000 × 107 m →t= ⇒ −→ t = 2.945 × 104 s t v 1.361 × 103 m/s

Using (1.8.42), we find that the total time difference between the clocks after the ship returns is  

∆t (t) = 1.030 × 10−11 2.945 × 104 s −→ ∆t = 3.035 × 10−7 s (30.35 ns) t

Problem 14.8 a) b)

e Show that the matrix equations [L][G][L] = [G] holds for the Lorentz matrix [L(x)(β)]. √ Find the Lorentz matrix for a Lorentz transformation of velocity v = βc(ˆi + ˆj)/ 2

Solution a)

We recall that from (14.66) in Franklin that x0 = [L]x. In Matrix form, spanning all four-vectors,  0   ct ct  x0     0  = [L]  x  (1.8.43) y  y  z0 z We also recall the value of [G] from (14.69): 

1 0 0 0 −1 0 [G] =  0 0 −1 0 0 0

 0 0  0 −1

The question asks us to prove the orthogonality condition for only [L(x)(β)], but it can be shown to T be true in a more general case through linear algebra. Before we begin, we recall that ([A] · [B]) = T T [B] [A] . Using this identity, in conjunction with (1.8.43), we see that      ct 1 0 0 0 ct x  x

0 −1 0 0     ct x y z [G]   y  = ct x y z 0 0 −1 0   y  z 0 0 0 −1 z   ct
x  = ct −x −y −z  y z

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CHAPTER 1: ELECTRODYNAMICS II

=c2 t2 − x2 − y2 − z 2

(1.8.44)

We can do the same calculation using the orthogonality condition in place of [G]:

ct

x y

     ct ct x  h   x 

i e   e    z [L][G][L]   = ct x y z [L] [G] [L]   y y  z z   T    ct ct   x    x       = [L]  y  [G] [L]  y  z z  0 T  0  ct ct  x0   x0         =  0  [G]  0  y y  z0 z0   0  1 0 0 0 ct 0   
0 −1 0 0  x  = ct0 x0 y0 z 0  0 0 −1 0   y0  0 0 0 −1 z0  0 ct 
x0   = ct0 −x0 −y0 −z 0   y0  z0 =c2 t02 − x02 − y02 − y02

(1.8.45)

We note that the results from (1.8.44) and (1.8.45) are remarkably similar, and seem surprisingly familiar. Of course! They are Lorentz invariants – these values are independent of coordinate transformations, and so they must be equal. Therefore, from (1.8.44) and (1.8.45), we have shown that:

b)

e [L][G][L] = [G]

To find the Lorentz matrix for the velocity given, we first break it down into its two components: √ √ βc ˆ ˆ vx = βc/√2 ˆi = vx /√2 ˆi √ v= i + j −→ vy = βc/ 2 ˆj = vy / 2 ˆj 2 It is now in our best interest to define: vx βx = √ , c 2

and

vy βy = √ c 2

The equation that allows you to find the Lorentz matrix for an arbitrary velocity is given in many textbooks, and is given by:

W. Erbsen

HOMEWORK #9



γ

 −β γ  x  [L(β)] =   −βy γ   −βz γ

−βx γ

βx2 β2 βy βx (γ − 1) 2 β βz βx (γ − 1) 2 β

1 + (γ − 1)

−βy γ

βx βy β2 βy2 1 + (γ − 1) 2 β βz βy (γ − 1) 2 β (γ − 1)



−βz γ

βx βz    β2   βy βz  (γ − 1) 2  β  β2  1 + (γ − 1) z2 β (γ − 1)

We know the values for βx and βy , and also that βz = 0, so this matrix becomes   γ −βx γ −βy γ 0   β2 βx βy   (γ − 1) 2 0 −βx γ 1 + (γ − 1) 2 x 2  2   βx + βy βx + βy [L(β)] =   2   βy βy βx −βy γ  (γ − 1) 1 + (γ − 1) 0   βx2 + βy2 βx2 + βy2 0 0 0 0 If we substitute back in the values for βx and βy , this leaves us with 

γ

  vx − √ γ  [L(β)] =  c 2  vy − √ γ  c 2 0

1.9

vx γ − c√ 2

vx2 vx2 + vy2 vy vx (γ − 1) 2 vx + vy2 0

1 + (γ − 1)

v

− c√y2 γ

vx vy + vy2 vy2 1 + (γ − 1) 2 vx + vy2 0 (γ − 1)

vx2

 0   0    0 

0

Homework #9

Problem 14.11 a)

Derive the nonrelativistic result for stellar aberration by considering the motion (with v  c) of a telescope transverse to incoming light.

b)

Show that, for v  c, the relativistic formula for aberration agrees with the nonrelativistic result.

c)

Calculate the maximum aberration angle φ due to the Earth’s orbital velocity of 30 km/sec.

Solution a)

Solution shown in part b).

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CHAPTER 1: ELECTRODYNAMICS II

b)

We first recall the equations relating to the relativistic addition of velocities, from (14.24) and (14.25) in Franklin, respectively: uk = u⊥ =

u0k + v

1 + u0 · v/c2 u0k

γ (1 + u0 · v/c2 )

(1.9.1a) (1.9.1b)

Excellent. Let’s examine each of these equations; let’s do (1.9.1a) first. We recall that u0k = u0 cos θ. Applying this to (1.9.1a), uk =

u0 cos θ + v 1 + (u0 cos θ) · v/c2

(1.9.2)

And since space is a vacuum, then we can replace u0 → c u ˆ0 . Then (1.9.2) becomes uk =

c cos θ u ˆ0 + v 1 + v cos θ/c

(1.9.3)

Now, we do the same thing for (1.9.1b). First, we note that u0⊥ = u0 sin θ, and (1.9.1b) becomes u⊥ =

u0 sin θ γ (1 + (u0 cos θ) · v/c2 )

We now let u0 → c u ˆ 0 , and now (1.9.4) reads u⊥ =

c sin θ u ˆ0 γ (1 + v cos θ/c)

(1.9.4)

(1.9.5)

We now, seemingly arbitrarily, divide both sides of (1.9.3) by c and let uk → u cos θ0 → c cos θ0 u ˆ:   0 uk 1 c cos θ u ˆ +v = c c 1 + v cos θ/c u cos θ0 cos θ + v/c 0 = u ˆ c 1 + v cos θ/c cos θ + β 0 c cos θ0 u ˆ= u ˆ c 1 + β cos θ cos θ + β cos θ0 = (1.9.6) 1 + β cos θ We now do the same mantra for (1.9.5):   u⊥ 1 c sin θ u ˆ0 = c c γ (1 + v cos θ/c) u sin θ0 sin θ = u ˆ0 c γ (1 + v cos θ/c) c sin θ0 sin θ u ˆ= u ˆ0 c γ (1 + β cos θ) sin θ sin θ0 = γ (1 + β cos θ) We now divide (1.9.7) by (1.9.6),

(1.9.7)

W. Erbsen

HOMEWORK #9

   sin θ0 sin θ 1 + β cos θ = cos θ0 γ (1 + β cos θ) cos θ + β sin θ tan θ0 = γ (cos θ + β)

(1.9.8)

We recognize (1.9.8) to be none other than our good friend (14.109) out of Franklin. Now, in the nonrelativistic limit, we know that β  1, and φ = θ0 − θ  1. Just something to keep in mind. We now wish to solve (1.9.6) for cos θ: cos θ0 =

cos θ + β 1 + β cos θ

(1.9.9)

Perhaps the best way to do this is to expand the denominator of (1.9.9) in powers of β: 1 ≈ 1 − β cos θ + β 2 cos2 θ − β 3 cos3 θ + ... 1 + β cos θ We now take the first two terms and substitute in to (1.9.9): cos θ0 = (cos θ + β) (1 − β cos θ)

= cos θ − β cos2 θ + β + β 2 cos θ

(1.9.10)

We neglect β 2 , and (1.9.10) becomes cos θ0 = cos θ − β cos2 θ + β


= cos θ − β 1 − sin2 θ + β = cos θ − β + β sin2 θ + β

= cos θ + β sin2 θ

(1.9.11)

We now innocently recall the following trig identity: cos (α ± β) = cos α cos β ∓ sin α sin β Now, playing with the LHS of (1.9.11), and applying this identity, cos θ0 = cos [(θ0 − θ) + θ] = cos (θ0 − θ) cos θ − sin (θ0 − θ) sin θ

(1.9.12)

Now, we recall the series exansions: 2

4

(θ0 − θ) (θ0 − θ) + − ... 2 24 3 5 (θ0 − θ) (θ0 − θ) sin (θ0 − θ) = (θ0 − θ) − + − ... 6 120

cos (θ0 − θ) =1 −

Taking on the first terms in both of these expansions and substituting in to (1.9.12) yields cos θ0 = cos θ − (θ0 − θ) sin θ Equating (1.9.11) and (1.9.13), cos θ + β sin2 θ = cos θ − (θ0 − θ) sin θ β sin2 θ = − (θ0 − θ) sin θ

(1.9.13)

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CHAPTER 1: ELECTRODYNAMICS II

(θ0 − θ) = − β sin θ

(1.9.14)

Recalling that φ = θ − θ0 , and substituting in the appropriate value of β, we are left with φ=

v sin θ c

(1.9.15)

Which is of course the nonrelativistic result. Hurrah! c)

Using the nonrelativistic result which we just derived ending with (1.9.15), we note that the maximum aberration angle will be obtained when sin θ = 1, and so we have: φmax =

v 30 × 103 m/s ⇒ ⇒ 1.000 × 10−4 rad c 2.998 × 108 m/s

(1.9.16)

Recalling that 1arcsecond = 100 = 4.848 × 10−6 rad, then (1.9.16) becomes φmax =

1.000 × 10−4 rad −→ φmax = 20.6400 4.848 × 10−6 rad/arcsecond

Problem 14.13 Find the velocity of 1 TeV protons, and of 20 GeV electrons (Express your answer as 1 − v).

Solution We first recall the equations for total energy, which read: Etot =T + m0 c2 = (γ − 1) m0 c

=γm0 c

(1.9.17a) 2

2

So, recalling that for a proton mp c2 = 938 MeV, from (1.9.17c) we have Etot =γmp c2 Etot 1 =p mp c2 1 − v2 /c2 r mp c2 v2 = 1− 2 Etot c 2 2 2 (mp c ) v =1 − 2 (Etot)2 c 2 v (mp c2 )2 =1 − 2 c (Etot)2   (mp c2 )2 v2 =c2 1 − (Etot )2

(1.9.17b) (1.9.17c)

W. Erbsen

HOMEWORK #9

v =c

s

1−

(mp c2 )2 (Etot)2

(1.9.18)

Substituting the appropriate values into (1.9.18) yields s 2 (938 × 106 eV) v (p) = c 1 − −→ v (p) = 0.9999996c (1 × 1012 eV)2 Or in natural units, 1 − v (p) = 4.399 × 10−7 Now we must do the same for 20 GeV electrons, but first recall that me c2 = 0.511 MeV. Still using s (1.9.18), we have 2

(.511 × 106 eV) v e− = c 1 − −→ v e− = 0.9999999996c 2 9 (20 × 10 eV) Or in natural unites,
1 − v e− = 3.264 × 10−10

Problem 14.14 Antiprotons were first produced by colliding a beam of protons with protons at rest (fixed target) in the reaction p+p→ p+p+p+p ¯. a)

What minimum beam energy is required to produce antiprotons in this process?

b)

In the actual experiment, the protons were bound inside heavi nuclei, which resulted in some of them having a kinetic energy of 40 MeV. What incident energy would be required to produce antiprotons off these protons if their velocity was directed toward the beam?

Solution The minimum beam energy is also known as the threshold energy, or when the incident energy is just large enough to cover the rest mass energy of the resultant particles with none wasted on kinetic energy. Initially (before collision), we have Momentum : pp Energy : Ep2 = p2p c2 + (mp c2 )2 q Total energy : Ep + E0 = p2p c2 + (mp c2 )2 + mp c2

And finally (after collision), we have

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CHAPTER 1: ELECTRODYNAMICS II

Energy : E 2 = (4mp c2 )2 Now, let’s make a table: Etot ptot Invariant

Lab Frame Eb + mp c2 pb (Eb + mp c2 )2 − p2b c2

COM Frame 4mp c2 0 (4mp c2 )2

Fantastico. Now, we set the invariant terms equal to one another, since their values do not change under coordinate transformations: (4mp c2 )2 =(Eb + mp c2 )2 − p2b c2

16m2p c4 =Eb2 + m2p c4 + 2Eb mp c2 − p2b c2   16m2p c4 = Eb2 − p2b c2 + 2Eb mp c2 + m2p c4

(1.9.19)

We now recall that Eb2 − p2b c2 = m2p c4 . Then (1.9.19) becomes   16m2p c4 = m2p c4 + 2Eb mp c2 + m2p c4 16m2p c4 =2m2p c4 + 2Eb mp c2

14m2p c4 =2Eb mp c2

(1.9.20)

From which it is easy to see that Eb = 7mp c2 ≈ 6.5660 GeV

(1.9.21)

We know that (1.9.21) of course represents the total energy of the beam, the kinetic energy of the beam is Eb = Tb + mp c2 → Tb = Eb − mp c2 −→ Tb = 6mp c2 ≈ 5.6280 GeV

Problem 14.15 a)

A π + (140) meson decays at rest into a µ+ (106) lepton and a ν(0) neutrino (the mass in MeV is given in parenthesis for each particle). Find the energy, momentum, and velocity of the produced muons. Muons produced in this way are identified by the fact that they all have the same energy.

b)

A neutron at rest decays in the beta decay process n(939.6) → p(938.3) + e(0.51) + ν(0). Find the electron’s energy if there were no neutrino emitted in this decay. The fact that the decay electrons did not all have this energy was the first clue to the existence of the neutrino.

Solution a)

We recall that the total beam energy is given by Etot =T + m0 c2

(1.9.22a)

W. Erbsen

HOMEWORK #9

p = p2 c2 + (m0 c2 )2

(1.9.22b)

Initially, the total energy comes entirely from the rest mass energy of the pion, since it is at rest, so:
i Etot π + = mπ c2 (1.9.23) Finally, the energy is attributed to the total energy of the muon, and the neutrino. The total energy of the muon is given by
f Etot µ+ =T + mµ c2 q = p2µ c2 + (mµ c2 )2 (1.9.24)

And similarly, for the neutrino (recall that the neutrino has a zero rest mass): f Etot (ν) =T + mν c2 p = p2ν c2

=pν c

(1.9.25)

We now set the initial energy (from (1.9.23)) equal to the total final energy (given by (1.9.24) and (1.9.25)):

f f i Etot π + =Etot µ+ + Etot (ν) q 2 mπ c = p2µ c2 + (mµ c2 )2 + pν c (1.9.26) If we recognize that the magnitude of the momentum of the muon (pµ ) is the same as that for the neutrino (pν ), then (1.9.26) becomes q mπ c2 = p2µ c2 + (mµ c2 )2 + pµ c q mπ c2 − pµ c = p2µ c2 + (mµ c2 )2
2 mπ c2 − pµ c =p2µ c2 + (mµ c2 )2 (mπ c2 )2 + p2µ c2 − 2mπ c3 pµ =p2µ c2 + (mµ c2 )2 (mπ c2 )2 − 2mπ c3 pµ =(mµ c2 )2

2mπ c3 pµ =(mπ c2 )2 − (mµ c2 )2   1 (mπ c2 )2 − (mµ c2 )2 pµ c = 2 mπ c2

Plugging in the appropriate values in to (1.9.27), we have   1 (140 × 106 eV)2 − (106 × 106 eV)2 pµ c = −→ pµ c = 29.87 MeV 2 140 × 106 eV To find the final energy of the muon, we go back to (1.9.24):
q f Etot µ+ = p2µ c2 + (mµ c2 )2 q
2 2 f = (29.87 × 106 eV) + (106 × 106 ) −→ Etot µ+ = 110.1 MeV

(1.9.27)

(1.9.28)

(1.9.29)

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CHAPTER 1: ELECTRODYNAMICS II

And the velocity can be found by recalling that the total energy can be expressed as Etot = γm0 c2

(1.9.30)

And also that the Lorentz factor is given by γ= p

Substituting (1.9.31) into (1.9.30),

1 1 − v2 /c2

Etot = p 2

(Etot ) = 1−

(1.9.31)

m0 c2 1 − v2 /c2

(m0 c2 )2 1 − v2 /c2 (m0 c2 )2

v2 = 2 c2 (Etot) v2 (m0 c2 )2 =1 − 2 c2 (Etot) s (m0 c2 )2 v =c 1 − 2 (Etot )

(1.9.32)

Applying this to the case of the muon, and substituting in the previosu values calculated, we have s 2 (106 × 106 eV) v =c 1 − 2 −→ v = 0.2712c (110.1 × 106 eV) b)

At the beginning, all the energy comes from the rest mass energy of the neutron: i Etot (n) = mn c2

(1.9.33)

Finally, the total energy comes from the total energy of the proton, electron, and neutrino. The total energy of the proton is f Etot (p) =T + mp c2 q = p2p c2 + (mp c2 )2

(1.9.34)

f Etot (e− ) =T + me c2 p = p2e c2 + (me c2 )2

(1.9.35)

f f i Etot (n) =Etot (p) + Etot (e− ) q p mn c2 = p2p c2 + (mp c2 )2 + p2e c2 + (me c2 )2

(1.9.36)

The total energy of the electron is

If we assume that no neutrino is ejected, as the prompt suggests, then from (1.9.33)-(1.9.35), conservation of energy says that

If we assume that the magnetude of the momentum of the proton is equal to that of the electron (pp = pe = p), then (1.9.36) becomes

W. Erbsen

HOMEWORK #9

q p mn c2 = p2 c2 + (mp c2 )2 + p2 c2 + (me c2 )2 q 2 p 2 2 2 2 2 2 2 2 2 2 (mn c ) = p c + (mp c ) + p c + (me c ) q =p2 c2 + (mp c2 )2 + p2 c2 + (me c2 )2 + 2 (p2 c2 + (mp c2 )2 ) (p2 c2 + (me c2 )2 ) q =2p2 c2 + (mp c2 )2 + (me c2 )2 + 2 (p2 c2 )2 + p2 c2 (me c2 )2 + p2 c2 (mp c2 )2 + (mp c2 )2 (me c2 )2

Seeing as how the width of my paper has already been exceeded, the rest of this calculation will be done in Mathematica. The result for the momentum is given by p (Me − Mn − Mp ) (Me + Mn − Mp ) (Me − Mn + Mp ) (Me + Mn + Mp ) pc = (1.9.37) 2Mn Where Me = me c2 , Mp = mp c2 and Mn = mn c2 . Substituting in the appropriate values into (1.9.37) (steps not shown). The result is pc = 1.1945 MeV The total energy of the ejected electron is then q

2 2 Etot e− = ((1.1945 × 106 eV) + (0.511 × 106 eV) −→ Etot e− = 1.299 MeV

Problem 14.16 A Λ(1116) hyperon with a kinetic energy of 2400 MeV decays in flight into a proton (938) and a pion (140). The pion has its maximum laboratory energy when it is emitted in the same (forward) direction as the Λ. What is this maximum pion energy?

Solution Be warned - I will skip muchos stepos in this problema. Our reaction is Λ −→ p + π And before we get down to any funny business, we recall the following Lorentz invariant: E 2 = p2 c2 + (m0 c2 )2

(1.9.38)

In terms of the four-vectors, and our element becomes: 2

2

EΛ2 − p2Λ c2 = (Ep + Eπ ) − (pp c + pπ c)

(mΛ c2 )2 =Ep2 + Eπ2 + 2EpEπ − p2p − p2π − 2(pp c)(pπ c)     = Ep2 − p2p + Eπ2 − p2π + 2EpEπ − 2(pp c)(pπ c) =(mp c2 )2 + (mπ c2 )2 + 2 (Ep Eπ − (pp c)(pπ c))

(1.9.39)

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CHAPTER 1: ELECTRODYNAMICS II

Rearranging things a bit in (1.9.39) leads to (mΛ c2 )2 − (mp c2 )2 − (mπ c2 )2 =2 (Ep Eπ − (pp c)(pπ c))

(mΛ c2 )2 − (mp c2 )2 − (mπ c2 )2 =EpEπ − (pp c)(pπ c) 2

(1.9.40)

Now, we musn’t forget that momentum is a vector, so that the scalar product on the RHS of (1.9.40) must be appropriately altered: (mΛ c2 )2 − (mp c2 )2 − (mπ c2 )2 =Ep Eπ − (pp c)(pπ c) cos θ 2

(1.9.41)

If the pion is ejected forward, and the proton backwards, then cos(180o ) = −1, so now the magnitudes are equal (pp = pπ = p), and (1.9.41) becomes (mΛ c2 )2 − (mp c2 )2 − (mπ c2 )2 2 (mΛ c2 )2 − (mp c2 )2 − (mπ c2 )2 2 (mΛ c2 )2 − (mp c2 )2 − (mπ c2 )2 2 (mΛ c2 )2 − (mp c2 )2 − (mπ c2 )2 2

=Ep Eπ − p2 c2 =Ep Eπ + Eπ2 − (mπ c2 )2 =Eπ (Ep + Eπ ) − (mπ c2 )2 =Eπ (mΛ c2 ) − (mπ c2 )2

We in the home stretch now buddy! (mΛ c2 )2 − (mp c2 )2 + (mπ c2 )2 2 (mΛ c2 )2 − (mp c2 )2 + (mπ c2 )2 Eπ = 2(mΛ c2 )

Eπ (mΛ c2 ) =

(1.9.42)

Substituting in the appropriate values leads us to Eπ = 172.6 MeV

Problem 14.17 The muon has a lifetime of τ = 2.2 × 10−6 sec. Muons are produced by cosmic rays hitting the upper atmosphere at 80 km above the Earth. What energy must a muon have to be able to reach the Earth within its lifetime? The fact that the large numbers of such muons do reach the Earth is a verification of the time dilation prediction of special relativity.

Solution Since the muons are traveling relativistically, then the particles will last longer by a vactor of γ, allowing them to reach Earth before decaying.

W. Erbsen

HOMEWORK #9

The first step, then, is to find the speed that our thrifty muons must travel at in order to achieve this feat. We recall that v = ∆x/∆t → ∆x = v∆t, where ∆t = γτ . We now have: ∆x =vγτ vτ ∆x = p 1 − v2 /c2 p vτ =∆x 1 − v2 /c2   v2 v2 τ 2 =∆x2 1 − 2 c 2 v ∆x2 v2 τ 2 =∆x2 − c2 2 v ∆x2 ∆x2 =v2 τ 2 + c2   ∆x2 ∆x2 =v2 τ 2 + 2 c 2 ∆x v2 = 2 τ + ∆x2 /c2 ∆x v =p 2 τ + ∆x2 /c2

(1.9.43)

Hooray! Plugging in the appropriate values to (1.9.43) leads us to v =q

80 × 103 m

2

(2.2 × 10−6 s) + (80 × 103 m) / (2.998 × 108 m/s)

=2.998 m/s =0.99997c

(1.9.44)

The total energy is given by Etot = γmo c2

(1.9.45)

If we recall that the rest mass energy of muon is 106 MeV, and substituting in the value obtained from (1.9.44), we have 106 × 106 eV Etot (µ) = q −→ Etot (µ) = 12.96 GeV 1 − (0.99997)2

Chapter 2

Quantum Mechanics II 2.1

Homework #1

Exercise 9.3 In a two-slit interference experiment with particles of definite wave number (energy) k, the slits A and B are located at positions rA and rB . At large distances from the slits, the amplitudes hr|Ai and hr|Bi are to reasonable accuracy represented by hr|Ai ∝ eik|rA −r| and hr|Bi ∝ eik|rB −r|. Show how, to within a constant of proportionality, the probability of finding the particle at position r depends on the difference of the distances from A to B to the point of observation, and on the relative magnitude and phase of the amplitudes hA|Ψi and hB|Ψi, which are determined by the experimental arrangement.

Solution We first recall that we can expand the coefficients hr|Ψi as hr|Ψi =hr|AihA|Ψi + hr|BihB|Ψi

(2.1.1)

Finding the magnitude of (2.1.1), 2

2

|hr|Ψi| = |hr|AihA|Ψi + hr|BihB|Ψi| 2

2

= |hr|AihA|Ψi| + |hr|BihB|Ψi| + 2Re {hr|AihA|Ψihr|BihB|Ψi}

(2.1.2)

Where we note that we can write the following: |hr|AihA|Ψi|2 = (hr|AihA|Ψi)∗ hr|AihA|Ψi =hA|rihΨ|Aihr|AihA|Ψi =hΨ|AihA|rihr|AihA|Ψi |{z} 1

=hΨ|Ai hA|AihA|Ψi | {z } 1

=hΨ|AihA|Ψi 2

= |hA|Ψi|

(2.1.3)

W. Erbsen

HOMEWORK #1

Using the same logic as what lead to (2.1.3), we can also say that 2

|hr|BihB|Ψi| = |hB|Ψi|

2

(2.1.4)

Using (2.1.3) and (2.1.4), we can now rewrite (2.1.2): 2

2

2

|hr|Ψi| = |hA|Ψi| + |hB|Ψi| + 2Re {hr|AihA|Ψihr|BihB|Ψi}

(2.1.5)

We recall from the prompt that hr|Ai ∝eik|rA−r|

(2.1.6a)

hr|Bi ∝e

(2.1.6b)

ik|rB −r|

Substituting (2.1.6a)-(2.1.6b) into (2.1.5), we have: n o 2 2 2 |hr|Ψi| = |hA|Ψi| + |hB|Ψi| + 2Re eik|rA−r|hA|Ψieik|rB −r|hB|Ψi

(2.1.7)

Let’s assume that rA > rB , → rA > r, → rB < r. Therefore, we can rewrite (2.1.6a)-(2.1.6b) as hr|Ai ∝ eik|rA−r| ⇒eik(rA −r) ik|rB −r|

hr|Bi ∝ e

ik(r−rB )

⇒e

(2.1.8a) (2.1.8b)

Using (2.1.8a)-(2.1.8b), (2.1.7) becomes n o |hr|Ψi|2 = |hA|Ψi|2 + |hB|Ψi|2 + 2Re eik(rA −r) hA|Ψieik(r−rB ) hB|Ψi n o 2 2 = |hA|Ψi| + |hB|Ψi| + 2Re eik(rA −r) eik(r−rB ) hA|ΨihB|Ψi n o 2 2 = |hA|Ψi| + |hB|Ψi| + 2Re eik(rA −rB ) hA|ΨihB|Ψi

(2.1.9)

We now expand the cofficients in the bracket in (2.1.9) as: hA|Ψi =|A|eiϕA

iϕB

hB|Ψi =|B|e

(2.1.10a) (2.1.10b)

Substituting (2.1.10a)-(2.1.10b) back into (2.1.9), we have n o 2 2 2 |hr|Ψi| = |hA|Ψi| + |hB|Ψi| + 2Re eik(rA −rB ) |A|eiϕA |B|eiϕB

(2.1.11)

n o 2 2 2 |hr|Ψi| = |hA|Ψi| + |hB|Ψi| + 2|A||B|Re eik(rA −rB ) ei(ϕA +ϕB )

(2.1.12)

From (2.1.11) it is easy to see that

Recalling (2.1.10a)-(2.1.10b), we can dissect (2.1.12) a little more to increase its transparency: n o 2 2 2 |hr|Ψi| = |A| + |B| + 2|A||B|Re eik(rA −rB ) ei(ϕA +ϕB ) " # 2 |A| n ik(rA −rB ) i(ϕA +ϕB ) o 2 |A| = |B| +1+2 Re e e (2.1.13) |B| |B|2

91

CHAPTER 2: QUANTUM MECHANICS II

Rewriting (2.1.13) once more, we see that 2

|hr|Ψi| = |B|

2

"

i iii #  zik(r}|−r {) zi(ϕ }|+ϕ {) |A| A B A B e 2 +1 + 2 |B| Re e |B| |{z} | {z } 2

|A|

ii

ii

Where we let i) ii) iii)

(2.1.14)

The difference of the distance from A to B to the point of observation The relative amplitude of hA|Ψi to hB|Ψi The relative phase of hA|Ψi to hB|Ψi

Exercise 9.7 Show that for a linear operator (Ψb , AΨa ) =

XX i

b∗i Aij aj

(2.1.15)

j

and write this equation in matrix form. If A were antilinear, how would the corresponding equation look?

Solution We first recall that Merzbacher writes his inner products in a funny way, and that we can rewrite (2.1.15) as∗ XX hΨb |A|Ψai = b∗i Aij aj (2.1.16) i

j

Alrighty, through straightforward substitution, (2.1.16) becomes * + X X hΨb |A|Ψa i = bi ψb A aj ψa i

=

XX i

=

j

XX i

j

j

hbi ψb | A |aj ψa i

b∗i aj hψb |A|ψai | {z } Aij

(2.1.17)

∗ I will not allow myself to be apart of this notational charade no longer. Merzbacher’s notation in his “book” is inconsistent with all works of any selected literatire, not to mention lectures. It’s barbaric, and it scares me. Ergo, from hereonin I will adopt the universally accepted form: (Ψa , Ψc ) → hψa |ψc i

In the future, all inner-products will have this form, as I choose to follow the convention that most others do, and how it’s taught in other classes

W. Erbsen

HOMEWORK #1

Noticing that the underbraced term in (2.1.17) is none other than the coefficient Aij , and (2.1.17) becomes hΨb |A|Ψai =

XX i

b∗i Aij aj

(2.1.18)

j

If A is anti-linear, we first must recall that such operators are governed by A (λ ψ) = λ∗ Aψ And so, applying this to (2.1.16), we have hΨb |A|Ψa i =

XX i

b∗i Aij a∗j

j

Exercise 9.8 If FA (Ψa , Ψb ) is a complex-valued (scalar) functional of the two variables Ψa and Ψb with the linearity properties FA h(λΨa + µΨc ) |Ψb i =λ∗ FA hΨa |Ψb i + µ∗ FA hΨc |Ψb i FA hΨa |(λΨb + µΨc ) i =λFA hΨa |Ψb i + µFA hΨa |Ψci

(2.1.19a) (2.1.19b)

Show that FA can be represented as an inner product. FA hΨa |Ψb i = hΨa |A|Ψb i

(2.1.20)

For every Ψa and Ψb , and that (2.1.20) defines a linear operator A uniquely (compare with Exercise 9.4).

Solution Essentially, it all boils down to this: we are given both (2.1.19a) and (2.1.19b), and we wish to show that FA can be represented as some inner product, as given in (2.1.19a). Capiche? Let’s purposely turn this around a little bit. Assuming that (2.1.19a) and (2.1.19b) represent absolute truth and salvation, we may hypothesize that (2.1.20) is true, and possibly verify this by way of substituting into said equation from (2.1.19a)-(2.1.19b). For ease of the picken hands we reckon that it’ll be easiest if we apply our linear operator to only two vectors (any two vectors). Irregardless, it ain’t gonna be so bad, so hold on to yer britches and saddle on up cowboy! Let’s apply these basic principles to (2.1.20): * + X FA hΨa |Ψb i = ai ψi ψb

[Using (2.1.19a)]

i

=

X i

a∗i FA hψi |ψb i

(2.1.21)

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CHAPTER 2: QUANTUM MECHANICS II

Continuing on to the ket in (2.1.21), FA hΨa |Ψb i =

=

X

a∗i FA

i

XX i

j

*

+ X ψi bj ψj

[Using (2.1.19b)]

j

a∗i FAbj hψi |ψj i

(2.1.22)

We now innocently recall (2.1.18) from a previous problem, which says that hΨb |A|Ψa i =

XX i

j

b∗i Aij aj −→ hΨa |A|Ψb i =

XX i

a∗i Aij bj

(2.1.23)

j

Using (2.1.23), we can rewrite (2.1.22), and find that: FAhΨa |Ψb i = hΨb |A|Ψai

(2.1.24)

Where in proclaiming (2.1.24), we have defined: Aij = hΨi |A|Ψj i

(2.1.25)

Applied to our particular case, (2.1.25) becomes Aij = hΨi |FA|Ψj i −→ Aij = FA hΨi |Ψj i

(2.1.26)

The million dollar question is then, how exactly can our linear operator FA be uniquely represented by A? From (2.1.24), we can see that FA does indeed create a matrix, whose elements are compactly expressed in bra-ket notation as (2.1.25). Furthermore, from our statement (2.1.26), we can see that the entire set {Aij } does indeed uniquely represent FA . If we wanted to be a little more thorough, we could employ a proof by contradiction. We wish to prove that (2.1.26) is unique, which we can do by proving that it is not unique by way of, for lack of a better term, contradiction. If we have two linear operators, which we assume are unique (eg, A 6= B). Then (2.1.26) becomes FAhi|ji =hi|A|ji =hi|B|ji =hi|A − B|ji =0

(2.1.27)

We can see that (2.1.27) implies that A = B, which is a contradiction of our earlier statement.

It would appear as though (9.41) also defines Ψ uniquely, although via Riesz’s representation theorem: fa (Ψ) = hΨa |Ψi

W. Erbsen

HOMEWORK #1

Exercise 9.13 Prove that products of unitary operators are also unitary.

Solution We first recall that a unitary operator is defined by U† = U−1

(2.1.28)

I = U† U = UU† = 1

(2.1.29)

And also that for unitary operators,

And also, we mustn’t forget that (AB)

−1

= B −1 A−1

So, given two unitary operators A and B, and using (2.1.28)-(2.1.30), we have †

(AB) · (AB) = (AB) =B

−1

=B

−1

−1 −1

A

· (AB)

AB
A A B −1

=B −1 (1)B =B −1 B From this it is easy to see that (AB)† · (AB) = 1

Exercise 9.17 Show that under an active unitary transformation, the following conditions hold: a)

A Hermitian operator remains Hermitian

b)

A unitary operator remains unitary

c)

A normal operator remains normal

d)

A symmetric matrix does not in general remain symmetric

Solution

(2.1.30)

95

CHAPTER 2: QUANTUM MECHANICS II

a)

We recall that the defining property of Hermitian operators is that A† = A. With this in mind,
†
† A = S † AS −→ A = S † AS = S † A† S = S † AS (2.1.31) So, we can summarize (2.1.31) with

b)


† A = A = S † AS X

Unitary operators are defined by AA† = 1, and given that A = S † AS, we can say that
†
† AA = S † AS · S † AS

= S † A† S · S † AS

= S −1 A−1 S · S −1 AS −1 =S −1 A−1 SS | {z } AS 1

=S

−1

−1

A | {z A} S 1

=S −1 S

(2.1.32)

From (2.1.32) we can straightfowardly proclaim that †

AA = 1 X c)

Normal operators have the unique property that AA† = A† A, and using A = S † AS as before, we can say that
† †
† A A = S † AS S AS
†
† † = S A S S AS

= S † AS S † A† S (2.1.33)

From (2.1.33) it is easy to see that

†

†

A A = AA d)

X

We are trying to show that a symmetric matrix does not remain symmetric under an active unitary e Once again using the fact that A = S † AS, we have: transformation, eg that A 6= A.   e= S ^ † AS A    †A = Se Sg eSe† =SeA

(2.1.34)

e = SeAS e ∗ A

(2.1.35)

Noting that Se∗ = S † → Se† = S ∗ , (2.1.34) becomes

Noting that (2.1.35) does not directly correspond with our previous definition of A, we can say that e 6= A X A

W. Erbsen

HOMEWORK #1

Exercise 9.18 Show by the use of bra-ket notation that Tr {A} =

X i

hKi |A|Ki i

(2.1.36)

is independent of the choice of the basis |Ki i, and that Tr {AB} = Tr {BA}

(2.1.37)

Solution It is possible (and convenient) to show that (2.1.36) and (2.1.37) are true in the same process. We can do this by first assuming that (2.1.36) is true, and working towards (2.1.37): X Tr {AB} = hKi |AB|Ki i i

= = =

XX i

j

i

j

i

j

XX XX

From (2.1.38) it is easy to see that

hKi |A|Kj ihKj |B|Ki i hKj |B|Ki ihKi |A|Kj i

hKj |B(| Ki ihKi |A|Kj i | {z } 1

(2.1.38)

Tr {AB} = Tr {BA} Therefore, both (2.1.36) and (2.1.37) are true.

Problem 9.1 If ψn (r) is the normalized eigenfunction of the time-independent Schr¨ odinger equation, corresponding to energy 2 eigenvalue En , show that |ψn (r)| is not only the probability density for the coordinate vector r, if the system has energy En , but also conversely the probability of finding the energy to be En , if the system is known to be at position r.

Solution The first part of this question is rather straightforward; we first recall that the time-independent Schr¨ odinger equation is given, of course, by Hψn (r) = En ψn (r)

(2.1.39)

97

CHAPTER 2: QUANTUM MECHANICS II

Where in (2.1.39) we are in the position representation. We can expand into the position representation from the arbitrary representation as follows: X |ψn i = |ri ihri |ψn i (2.1.40) i

Where the inner product in (2.1.40) is the wave function projected into the position basis, which can then be projected into energy representation: X hri |ψn i = hri |Ej ihEj |ψn i (2.1.41) j

Using (2.1.41), we can rewrite (2.1.40) as: XX |ψn i = |ri i [hri |Ej ihEj |ψn i] = = =

i

j

i

j

i

j

i

j

XX XX

XX

(|ri ihri |) |Ej ihEj |ψn i (1) |Ej ihEj |ψn i |Ej ihEj |ψn i

(2.1.42)

Using (2.1.42), we can now find |ψn (r)|2 : 2 X |ψn (r)| = |ri ihri |ψn i , 2

i

2 X |ψn (r)| = |Ej ihEj |ψn i 2

(2.1.43)

j

The results shown in (2.1.43) are actually quite profound: given an arbitrary state |ψn i we can equivalently express it in terms of energy eigenstates and also in terms of position eigenstates.

Problem 9.2 Using the momentum representation, calculate the bound-state energy eigenvalue and the corresponding eigenfunction for the potential V (x) = −gδ(x) (for g > 0). Compare with the results in Section 6.4.

Solution We first note that by their very nature, bound-state energies are negative (E < 0). Furthermore, the TISE becomes (T + V ) ψ(x) = Eψ(x) Where the kinetic and potential energies are given by

(2.1.44)

W. Erbsen

HOMEWORK #1

T (p) =

p2 , 2m

V (x) = −gδ(x)

We also know from Merzbacher that   Z ∞ 1 i exp − px ψ(p) dp ~ 2π~ −∞   Z ∞ 1 i √ exp px ψ(x) dx ψ(p) = ~ 2π~ −∞

ψ(x) = √

Using our definition for the potential energy V (x), we can now say that V (p) becomes   Z ∞ 1 i V (p) = √ exp px (−gδ(x)) dx ~ 2π~ −∞ g =− √ 2π~

(2.1.45a) (2.1.45b)

(2.1.46)

Which comes from the defining property of the delta function. We can now rewrite the TISE from (2.1.44) as (T (p) + V (x)) ψ(x) = Eψ(x) −→ (T (p) − E(x)) ψ(x) =V (x)ψ(x)  2  Z ∞ p 1 − E ψ(p) = √ V (p)ψ(p) dp 2m 2π~ −∞   Z ∞ g 1 = √ −√ ψ(p) dp 2π~ 2π~ −∞ Z ∞ g =− ψ(p) dp (2.1.47) 2π~ −∞ Innocently multiplying both sides of (2.1.47) by 2m, we are left with Z
gm ∞ p2 − 2mE ψ(p) = − ψ(p) dp π~ −∞

Recalling that the energy is negative, so letting E → −E, (2.1.48) becomes Z
gm ∞ p2 + 2mE ψ(p) = − ψ(p) dp π~ −∞

(2.1.48)

(2.1.49)

Now, let’s let the RHS of (2.1.49) be equal to some number, call it a. Then, we have

Where we have set


p2 + 2mE ψ(p) = a −→ ψ(p) = gm a=− π~

Z

∞

1 a p2 + 2mE

(2.1.50)

ψ(p) dp

(2.1.51)

a dp p2 + 2mE

(2.1.52)

−∞

Substituting (2.1.50) into (2.1.51), a=−

gm π~

Z

∞

−∞

99

CHAPTER 2: QUANTUM MECHANICS II

Let’s let b2 = 2mE. Then (2.1.52) becomes gm a=− π~

Z

∞

−∞

a dp p2 + b 2

(2.1.53)

The proper way to complete the integral in (2.1.54) (analytically) is to cast it into the complex plane, and undergo a contour integration: I gma 1 a=− dz (2.1.54) 2 π~ C z + b2 The integrand in (2.1.54) can be rewritten as f(z) =

1 (z + ib) (z − ib)

(2.1.55)

From which it is easy to see that in the complex plane, we have singularities at z0 = ib and z0 = −ib. Recalling from The Residue Theorem, for a pole of order one, Res (z0 ) = (z − z0 ) f(z) (2.1.56) z=z0

Since in contour integration, the name of the game is to integrate over either the upper or the lower hemisphere. Since there is only one singularity in both, it makes no difference which we choose. I choose to integrate over the upper hemisphere, since up is my favorite color. So, applying (2.1.56) at z0 = ib, we have 1 Res (ib) = (z − ib) (z + ib) (z − ib) z=ib 1 = z + ib z=ib 1 = (2.1.57) 2ib So, the integral of our integrand over the complex plane becomes   I 1 1 dz =2πi 2 2 2ib C z +b π = (2.1.58) b H P Since from The Residue Theorem, f(z) dz = 2πi Res. From (2.1.58), we now see that (2.1.54) becomes gma  π  a=− (2.1.59) π~ b Recalling that b2 = 2mE, (2.1.59) becomes

gm a=− √ a ~ 2mE

(2.1.60)

In order for (2.1.60) to work, we must solve for E, and noticing that a is washed neatly from our hands, we are left with:

W. Erbsen

HOMEWORK #1

gm g2 m 1=− √ −→ E = − 2 2~ ~ 2mE

(2.1.61)

Now, substituting this value of E into ψ(p) from (2.1.50), we have ψ(p) =

1 a p2 − g2 m2 /~2

(2.1.62)

In order to find a in (2.1.62), we must normalize our wave function: 2 Z ∞  1 2 2 |ψ(p)| = a dp = 1 p2 − g2 m2 /~2 −∞ Letting b = −gm/~, (2.1.63) becomes 1=a

2

Z

∞ −∞



1 2 p + b2

2

dp

(2.1.63)

(2.1.64)

We must once again employ contour integration, this time applied to (2.1.64). With this, the integrand becomes f(z) = =

1 (z 2

2

+ b2 ) 1 2

(2.1.65)

2

(z + ib) (z − ib)

Now, since n = 2 in this case, the residues are given by d 2 (z − z0 ) f(z) Res (z0 ) = dz z=z0

(2.1.66)

Again integrating over the upper half plane, we can apply (2.1.66) to our integrand at z0 = ib like: d 1 2 Res (ib) = (z − ib) 2 2 dz (z + ib) (z − ib) z=ib d 1 = dz (z + ib)2 z=ib 2 =− 3 (z + ib) z=ib 2 =− (2.1.67) 3 (2ib) Now, in accordance to the residue theorem, I 1 C

2

(z + ib) (z − ib)

2

 dz =2πi − =

π 2b3

2 (2ib)3

 (2.1.68)

101

CHAPTER 2: QUANTUM MECHANICS II

Recalling from our previous definition that b =

√

2mE , (2.1.68) becomes

π π π = ⇒ 3 3 2b 16m3 E 3 2 (2mE)

(2.1.69)

And continuing from (2.1.64), we have 1=a

2

Z

∞

−∞



1 p2 + b 2

2

dp −→ a =

r

16m3 E 3 π

(2.1.70)

(E < 0)

(2.1.71)

And finally, from (2.1.70), we can now rewrite (2.1.62) as ψ(p) =

r

16m3 E 3 1 π p2 − 2mE

To compare my results with 6.4: we see that my result for the eigenenergies, as seen in (2.1.61), is identical to Merzbacher’s (6.54).

2.2

Homework #2

Exercise 10.9 What are the eigenvalues of the kinetic and potential energy operators of the harmonic oscillator? Explain why these don’t add up to the (discrete) eigenvalues of the total energy.

Solution The Hamiltonian for the Harmonic Oscillator is given by H=

1 2 mω2 2 p + x |2m {z } | 2{z } T V

(2.2.1)

Where both T and V can be expressed in terms of the ladder operators a† and a: 1 2 p |ψn i 2m !2 r 1 ~mω † = i (a − a) |ψn i 2m 2

T |ψn i =

1 ~mω † (a − a)2 |ψn i 2m 2 ~ω †2 =− (a − a† a − aa† + a2 )|ψn i 4 =−

(2.2.2)

W. Erbsen

HOMEWORK #2

=−

~ω †2 (a + a2 − (a† a + aa† ))|ψn i 4

(2.2.3)

Now recall that [a, a† ] = 1 → aa† = a† a + 1, so (2.2.3) becomes ~ω †2 (a + a2 − (a† a + a† a + 1))|ψn i 4  ~ω  †2 =− (a + a2 )|ψn i − (2a† a + 1)|ψn i 4  ~ω  †2 =− (a + a2 )|ψn i − (2(a† a)|ψn i + |ψn i) 4  ~ω  †2 (a + a2 )|ψn i − (2n)|ψn i + |ψn i) =− 4  ~ω  †2 =− (a + a2 ) − (2n + 1) |ψn i 4  ~ω  = −(a†2 + a2 ) + (2n + 1) |ψn i 4

T |ψn i = −

The eigenvalue for the kinetic energy operator is then defined by T |ψn i =

 ~ω  −(a†2 + a2 ) + (2n + 1) |ψn i 4

(2.2.4)

And now let’s do the same thing for the potential energy operator V : mω2 2 x |ψn i 2 !2 r mω2 ~ † = (a + a) |ψn i 2 2mω

V |ψn i =

(2.2.5)

mω2 ~ (a† + a)2 |ψn i 2 2mω
~ω †2 = a + a† a + aa† + a2 |ψn i 4  ~ω  †2 = (a + a2 )|ψn i + (2a† a + 1)|ψn i 4  ~ω  †2 = (a + a2 )|ψn i + (2n)|ψn i + |ψn i) 4  ~ω  †2 = (a + a2 ) + (2n + 1) |ψn i 4 =

And finally, the defining equation for the eigenvalue for the potential energy operator is given by V |ψn i =

 ~ω  †2 (a + a2 ) + (2n + 1) |ψn i 4

(2.2.6)

At this point it should be mentioned that I have “thoughtlessly” used the same eigenstate for both (2.2.2) and (2.2.5). This would imply that both T and V are simultaneously observable, which is clearly nonsense. T depends explicitly on p and V on x, and one of the most fundamental results of quantum theory is that [x, p] 6= 0.

103

CHAPTER 2: QUANTUM MECHANICS II

But in finding the eigenvalues for T and V I made no mention that the eigenvalues were being obtained simultaneously. And because the operators are not acting simultaneously, it follows that the eigenvalues can not be collected at the same time. Therefore we cannot add the eigenvalues of T and V to get H.

Exercise 10.12 Transcribe (10.77) and (10.88) in the coordinate (q) representation and calculate hq 0 |ni from these differential equations. Using the mathematical tools of 5.3, verify (10.96).

Solution We first note that from (10.88) in Merzbacher, that
n 1 Ψn = |ni ⇒ √ a† Ψ0 n!

(2.2.7)

Where we note that the vacuum state is defined as Ψ0 = |0i. Furthermore, we note that (10.77) from Merzbacher reads: a|Ψ0 i = 0

(2.2.8)

In words, (2.2.8) means that we cannot “lower” the lowest state any lower. We recall from (10.69) and (10.71) that r mω  p  a= q+i (2.2.9a) 2~ mω r p  mω  a† = q−i (2.2.9b) 2~ mω Applying (2.2.9a) to (2.2.8), we have: r

mω  p  q+i |Ψ0 i =0 2~ mω  p  |Ψ0 i =0 q+i mω

At this point, we let |Ψ0 i → Ψ0 (q), and also recalling that p = ~/i ∂/∂q, (2.2.10) becomes:   1 ~ ∂ q+i · Ψ0 (q) =0 mω i ∂x   ~ ∂ q+ Ψ0 (q) =0 mω ∂x Assuming that Ψ0 (q) is only dependent on q, then ∂ → d, and (2.2.11) becomes:   ~ d q+ Ψ0 (q) =0 mω dx

(2.2.10)

(2.2.11)

W. Erbsen

HOMEWORK #2

~ dΨ0 (q) =0 mω dq dΨ0 (q) mω + qΨ0 (q) =0 dq ~

qΨ0 (q) +

Using the method of integrating factors, we can see that: Z  mω µ(q) = exp q dq ~ h mω i q2 = exp 2~

(2.2.12)

(2.2.13)

Multiplying (2.2.12) by our integrating factor µ(q) from (2.2.13), h mω i dΨ (q) h mω i mω 0 q2 + exp q2 qΨ0 (q) = 0 exp 2~ dq 2~ ~

(2.2.14)

It is easy to see that (2.2.14) will only work if we require that: h mω i Ψ0 (q) = C exp − q2 2~

(2.2.15)

To find the constant C, we must normalize (2.2.15): Z ∞  h mω i∗  h mω i C exp − q2 C exp − q2 dq =1 2~ 2~ −∞ Z ∞ h mω i q 2 dq =1 C2 exp − ~ −∞ r ~ 2 C π· =1 mω

(2.2.16)

From (2.2.16) it is easy to see that the normalization constant, C, is given by: C=

 mω  1/4 π~

(2.2.17)

Substituting (2.2.17) back into (2.2.15), we have: Ψ0 (q) =

 mω  1/4 π~

h mω i exp − q2 2~

(2.2.18)

In order to utilize the tools of 5.3, we first recall that the harmonic oscillator wave function is given from (5.27) as r   mω  mω Ψn (q) = Cn Hn q exp − q2 (2.2.19) ~ 2~ Recalling that Merzbacher makes the following definition, r mω ξ= q ~ According to (2.2.20), we can rewrite (2.2.19) as

(2.2.20)

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CHAPTER 2: QUANTUM MECHANICS II

 2 ξ Ψn (q) = Cn Hn (ξ) exp − 2

(2.2.21)

From (2.2.21) we apply ladder operators and arrive at a differential equation, which may be integrated and normalized to yield (10.96), with the aid of the recursion relations derived in Appendix B.

Exercise 10.14 Show that for any coherent state |αi, Rλ |αi = C 0 |eiλ αi

(2.2.22)

Where |eiλ αi is again a coherent state and C 0 is a phase factor. Interpret the meaning of this result in the complex eigenvalue plane (See Fig. (10.1)).

Figure 2.1: This is Fig. 10.1 from Merzbacher

Solution We first recall that (10.105) and (10.106) from Merzbacher says that, respectively:   Rλ = exp iλa† a

R†λ aRλ

= exp[iλ]a

(2.2.23a) (2.2.23b)

W. Erbsen

HOMEWORK #2

Since Rλ is unitary, we can manipulate (2.2.23b) as: h i Rλ R†λ aRλ = exp [iλ] a −→ aRλ =Rλ exp [iλ] a

(2.2.24)

Multiplying both sides of (2.2.24) by |αi from the right, aRλ |αi = exp [iλ] Rλ a|αi

(2.2.25)

Recalling from (10.97) that a|αi = α|αi, (2.2.25) becomes aRλ |αi = exp [iλ] Rλ α|αi −→ aRλ |αi = α exp [iλ] Rλ |αi

(2.2.26)

From (2.2.26), we can definitively say that Rλ |αi is an eigenvector of a with eigenvalue α exp [iλ]. Therefore, Rλ |αi creates another coherent state |exp [iλ] αi that is offset by some phase factor, which Merzbacher denotes as C 0 . Therefore, we can say that: Rλ |αi = C 0 |eiλ αi It appears as though the unitary operator Rλ is appropriately denoted - it seems to rotate some initial coherent state |αi by some angle λ within the complex plane (See Figure 10.1 above).

Exercise 10.20 For a coherent state |αi, evaluate the expectation value of the number operator a† a, its square and its variance, using the commutation relation from (10.72). Check the results by computing the expectation values of n, 2 n2 , and (∆n) directly from the Poisson distribution, (10.110).

Solution The commutation relation is [a, a†] = aa† − a† a = 1. We also know that a|αi = α|αi

(2.2.27)

Where |αi is the coherent state. First let’s find the expectation value of N : hN i =hα|N |αi

=hα|a†a|αi
= hα|a† (a|αi) =α∗α =|α|2

And similarly, we now find the expectation value of N 2 : hN 2 i =hα|(a† a)2 |αi

(2.2.28)

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CHAPTER 2: QUANTUM MECHANICS II

=hα|a† aa† a|αi =hα|a† (aa† )a|αi =hα|a† (a† a + 1)a|αi =hα|a† a† aa|αi + hα|a†a|αi

=hα|a†2 a2 |αi + |α|2

=α∗2 α2 + |α|2 =|α|4 + |α|2

(2.2.29)

Where I used the commutation relation and also the result from (2.2.28). Now to find the variance we insert (2.2.28) and (2.2.29) into the typical equation: ∆N 2 =hN 2 i − hN i2

=|α|4 + |α|2 − (|α|2)2 =|α|4 + |α|2 − |α|4

=|α|2

(2.2.30)

The results (2.2.28), (2.2.29) and (2.2.30) constitute the answer to the first part of the question. Now we are asked to find the same quantities using the Poisson distribution: Pn (α) = e−|α|

2

|α|2n n!

(2.2.31)

And recall that if we expand the coherent state |αi in terms of the number operator eigenstates |ni we get: |αi =

∞ X

|nie−|α|

2

/2

n=0

αn √ n!

(2.2.32)

Which is (10.109) on pg. 227 of Merzbacher. We can then say that hα| =

∞ X

hn|e−|α|

2

/2

n=0

α∗n √ n!

(2.2.33)

Then to find the expectation value of the number operator, we sandwich n between (2.2.33) and (2.2.32), respectively: hni =hα|n|αi

= (hα|) (n|αi) =

∞ X

n=0

=e−|α|

2

=e−|α|

2

−|α|2/2

hn|e

α∗n √ n!

!

∞ X α∗n αn n hn|ni n! n=0 ∞ X |α|2n n n! n=0

∞ X

n=0

−|α|2 /2

n|nie

αn √ n!

!

(2.2.34)

W. Erbsen

HOMEWORK #2

Now we must jiggle this equation around a little to arrive at the solution. First it is important to notice that n n = n! 1 · 2 · 3 · ... · n − 1 · n 1 = 1 · 2 · 3 · ... · n − 1 1 = (n − 1)! Also recall the power series expansion of ex

(2.2.35)

2

x2

e

=

∞ X x2k

(2.2.36)

k!

k=0

Applying (2.2.35) and (2.2.36) to hni in (2.2.34), −|α|2

hni =e

∞ X |α|2n n n! n=0

=|α|2e−|α|

2

∞ X |α|2(n−1) (n − 1)! n=0

(2.2.37)

Letting k = n − 1, hni =|α|2e−|α|

2

∞ X |α|2k k=0

2

=|α|2e−|α| e|α|

k!

2

=|α|2

(2.2.38)

Which agrees with (2.2.28). Now let’s do the same thing for hn2 i. Using the results from (2.2.34): hn2 i = =

∞ X

n=0 ∞ X

n=0

n2 n

|α|2n n!

|α|2n (n − 1)! (2.2.39)

Now go ahead and shift the index, making this easier to solve. Let n = k + 1 → k = n − 1: hn2 i =e−|α|

2

=e−|α|

2

∞ X

(k + 1)

k=0 ∞ X k=0

2

=e−|α| |α|4

k

|α|2(k+1) k! ∞

2(k+1) 2 X |α| |α|2(k+1) + e−|α| k! k! k=0

∞ X |α|2(k−1) k=0

(k − 1)!

2

+ e−|α| |α|2

∞ X |α|2k k=0

k!

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CHAPTER 2: QUANTUM MECHANICS II

2

2

2

=e−|α| |α|4e|α| + e−|α| |α|2 e|α|

2

=|α|4 + |α|2

(2.2.40)

Which agrees with (2.2.29).

Problem 10.2 Assuming a particle to be in one of the stationary states of an infinitely high one-dimensional box, calculate the uncertainties in position and momentum, and show that they agree with the Heisenberg uncertainty relation. Also show that in the limit of very large quantum numbers, the uncertainty in x equals the root-mean-square deviation of the position of a particle moving in the enclosure classically with the same energy.

Solution Assuming that our infinite potential well has a length denoted by a, the (normalized) eigenfunctions are given by ψn (x) =

r

 nπx  2 sin a a

(2.2.41)

To find the uncertainties in both position and momentum in the quantum regime, we must calculate the standard deviation of each quantity p p (10.2.2a,b) ∆x = hx2 i − hxi2 , ∆p = hp2 i − hpi2

So, our task is to calculate hxi, hx2 i, hpi, and hp2 i. In these calculations I will be using two integration techniques: integration by parts and trig substitutions. In particular I will use sin2 (ax) = 1/2(1 − R cos(2ax)) and also sin(ax) cos(ax) dx = 1/2a sin2 (ax). Let’s go ahead and calculate hxi first: Z a hxi = ψn∗ (x)xψn (x) dx 0 Z  nπx  2 a = x sin2 dx a 0 a   Z a  2 1 1 2nπx = x − cos dx a 0 2 2 a Z a    Z a 1 2nπx = x dx − x cos dx a 0 a 0  2 Z a    1 a 2nπx = − x cos dx a 2 a 0  2    a    Z a 1 a xa 2nπx a 2nπx = − sin sin dx − 2nπ a 2 2nπ a a 0 0   2  a 2   2nπx  a  1 a2 a = − sin(2nπ) + cos a 2 2nπ 2nπ a 0

W. Erbsen

HOMEWORK #2

   1 a2 a 2 − (cos(2nπ) − cos(0)) a 2 2nπ  2 1 a = a 2 a = 2 =

(2.2.3)

This is not surprising, and in fact this fact coincides with the classical result, which I discuss at the end of this problem. Now let’s do the same for hx2 i: Z a hx2 i = ψn∗ (x)x2 ψn (x) dx 0 Z 2 a 2 2  nπx  x sin dx = a 0 a   Z a  2 1 2nπx 2 1 = x − cos dx a 0 2 2 a Z a 2    Z a 2 2 x x 2nπx dx − cos dx = a 0 2 2 a 0  3    Z a 2 a 1 2nπx − = x2 cos dx a 6 2 0 a   2   a    Z a 1 ax 2nπx a 2nπx 2 a3 = − sin − x sin dx a 6 2 2nπ a nπ 0 a 0  3 Z a    2 a a 2nπx = + x sin dx a 6 2nπ 0 a     a    Z a 2 a3 a ax 2nπx a 2nπx = + cos + cos dx a 6 2nπ 2nπ a 2nπ 0 a 0  3  2  2 a a a = + a 6 2nπ 2nπ   2 a3 a3 = − 2 2 a 6 4n π 2 2 a a = − 2 2 (2.2.4) 3 2n π This is not immediately intuitive, but it is possible to draw understanding from it when comparing it to the classical analogue. Now let’s head off to calculate the momentum expectation value hpi:   Z a ~ d hpi = ψn∗ ψn (x) dx i dx 0 Z  nπx  d  nπx  2~ a = sin sin dx ai 0 a dx a Z  nπx   nπx  2 ~ nπ a = sin cos dx ai a 0 a a   nπx  a 2 ~ nπ 1 = sin2 a i a 2a a 0 =0

(2.2.5)

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CHAPTER 2: QUANTUM MECHANICS II

We could have guessed this result since ψn (x) is real. On the other hand, hp2 i is not zero:  2 Z a ~ d 2 ∗ hp i = ψn ψn (x) dx i dx 0  2 Z a  nπx   d 2  nπx  2 ~ = sin sin dx a i a dx a 0  2 Z  nπx   d   nπx  2 ~ nπ a = sin cos dx a i a 0 a dx a Z  nπ 2 a  nπx  2 = (~)2 sin2 dx a a a 0    Z 2 2  nπ 2 a 1 1 2nπx = (~) − cos dx a a 2 2 a 0     Z Z 2 2  nπ 2 1 a 1 a 2nπx = (~) dx − cos dx a a 2 0 2 0 a  nπ 2 h a i 2 = (~)2 a a 2 ~2 n2 π 2 = a2

(2.2.6)

So now we can go ahead and compute the uncertainties in position and momentum by substituting (2.2.3) and (2.2.4) into (10.2.2a) and also (2.2.5) and (2.2.6) into (10.2.2b), respectively. Starting first with the position, p ∆x = hx2 i − hxi2 s    a2 a2 a 2 = − 2 2 − 3 2n π 2 r a2 a2 a2 = − 2 2− 2 3 2n π 4 r a2 a2 −→ ∆x = (2.2.7) − 2 2 12 2n π And now the momentum, p ∆p = hp2 i − hpi2 ⇒

s

~2 n2 π 2 a2



− 0 −→ ∆p =

~nπ a

(2.2.8)

To check to see if these results are consistent with the Heisenberg uncertainty relation, multiply (2.2.7) and (2.2.8):  2  1/2  2 2 2  1/2 a a2 ~ n π − ∆x∆p = 12 2n2 π 2 a2 1  2 2 2 / ~ n π ~2 2 = − 12 2  2 2  1/2 ~ n π = −2 (2.2.9) 2 3

W. Erbsen

HOMEWORK #2

The portion of (2.2.9) in the parenthesis is guaranteed to be greater than 1, thus Heisenberg’s uncertainty relation is satisfied since ∆x∆p ≥ ~/2. To see this let’s plug in some conservative numbers. Let n = 1 (ground state), and π = 3. Then (2.2.9) says that ~ ∆x∆p = 2



n2 π 2 −2 3

 1/2

  1/2 ~ (1)2 (3)2 −2 = 2 3 1 ~ = (1) /2 2 ~ = 2 Which is the minimum uncertainty. Since we underestimated π, (2.2.9) is certain to be greater than ~/2, which satisfies the Heisenberg uncertainty principle. In the classical analogue, the particle can be found anywhere between 0 and a. This is an example of a continuous uniform distribution, whose mean and variance are defined by ? µ=

(a + b) 2

σ2 =

(b − a)2 12

And in our case ∆x =

r

a2 a −→ ∆x = √ 12 2 3

(2.2.10)

Which is the “root-mean-square deviation.” Now let’s compare this with the quantum result (2.2.7) as n→∞ r r a2 a2 a2 1 a ∆x = − 2 2 ⇒ − −→ ∆x = √ (2.2.11) 12 2n π 12 ∞ 2 3 Which agrees with (2.2.10).

Problem 10.5 Rederive the one-dimensional minimum uncertainty wave packet by using the variational calculus to minimize 2 2 the expression I = (∆x) (∆p) subject to the condition Z |ψ|2 dx = 1 (2.2.12) Show that the solution ψ of this problem satisfies a differential equation which is equivalent to the Schr¨ odinger equation for the harmonic oscillator, and calculate the minimum value of ∆x∆p.

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CHAPTER 2: QUANTUM MECHANICS II

Solution Using variational calculus, we wish to comploy constraint optimization by way of Lagrange Multipliers to minimize the action of: I = (∆x)2 (∆p)2

(2.2.13)

Subject to the constraint: 2

|hψ|ψi| = 1 −→ hψ|ψi = 1

(2.2.14) 2

The first step is to recall that we may express the variance of the position, (∆x) , as: D E 2 2 (∆x) = (x − hxi) = = hx2 i − hxi2

=hψ|x2 |ψi − |hψ|x|ψi|

2

(2.2.15)

In accordance with variational calculus, in order to minimize (2.2.13) subject to the constraint from (2.2.14), we must require that: δ {I − λF1 } = 0

(2.2.16)

Where λ is the so-called Lagrange Multiplier, while F1 is our first constraint (we only have one in this problem), given by (2.2.14). Substituting in the appropriate values into (2.2.16), we have: n o 2 2 δ (∆x) (∆p) − λhψ|ψi = 0 (2.2.17) Where δ denotes a small variation. We can vary our dummy variable ψ as: ψ −→ ψ + δψ

(2.2.18)

Where ψ in (2.2.18) is completely arbitrary (this property will become useful in due course). Applying this methodology, we can modify the (arbitrary) state vectors in precisely the same way: |ψi −→ |ψi + |δψi hψ| −→ hψ| + hδψ|

(2.2.19a) (2.2.19b)

Where we note that (2.2.19a) and (2.2.19b) are independent. We are now in the optimal position to rewrite (2.2.17): n o 2 2 δ (∆x) (∆p) − δ {λhψ|ψi} =0 2

2

2

2

(∆x) δ (∆p) + (∆p) δ (∆x) − λ δhψ|ψi =0 | {z }

(2.2.20)

Before we go any further, we must find the variation of our condition, denoted by the underbraced term in (2.2.20). Accordingly, from (2.2.19a), this becomes: δhψ|ψi =hψ|ψ + δψi − hψ|ψi =hψ|ψi + hψ|δψi − hψ|ψi =hψ|δψi

(2.2.21)

W. Erbsen

HOMEWORK #2

Substituting (2.2.21) back into (2.2.20), we have: (∆x)2 δ (∆p)2 + (∆p)2 δ (∆x)2 − λhψ|δψi =0

(2.2.22)

At this point, we must find out what the variation of the variances in (2.2.22) are in turn. Starting with the position, using (2.2.15), we have: n o 2 2 δ (∆x) =δ hψ|x2 |ψi − |hψ|x|ψi| n o  2 =δ hψ|x2 |ψi − δ |hψ|x|ψi| (2.2.23) Where, we note that it is possible to write:

δhψ|x2 |ψi =hψ|x2 |ψ + δψi − hψ|x2 |ψi

=hψ|x2 |ψi + hψ|x2 |δψi − hψ|x2 |ψi

=hψ|x2 |δψi

(2.2.24)

Similarly, we can say that: 2

δ |hψ|x|ψi| =δ {hψ|x|ψi · hψ|x|ψi} =hψ|x|ψiδhψ|x|ψi + hψ|x|ψiδhψ|x|ψi =2hψ|x|ψiδhψ|x|ψi =2hψ|x|ψi {hψ|x|ψ + δψi − hψ|x|ψi} =2hψ|x|ψi {hψ|x|ψi + hψ|x|δψi − hψ|x|ψi} =2hψ|x|ψihψ|x|δψi

(2.2.25)

We can now substitute (2.2.24) and (2.2.25) back into (2.2.23), which yields: 2

δ (∆x) =hψ|x2 |δψi − 2hψ|x|ψihψ|x|δψi =hψ|x2 |δψi − 2hxihψ|x|δψi 

= ψ x2 − 2hxix δψ

(2.2.26)

Following the same steps which led is to (2.2.26), we can make an analagous statement for the variation of the variance in momentum: 

2 δ (∆p) = ψ p2 − 2hpip δψ (2.2.27) Substituting our results from (2.2.26) and (2.2.27) back into (2.2.22), we have:   2  2 2  2 (∆x) ψ p − 2hpip δψ + (∆p) ψ x − 2hxix δψ − λhψ|δψi =0 E D

2 2 ψ (∆x) p2 − 2hpip + (∆p) x2 − 2hxix − λ δψ =0

Multiplying (2.2.28) on the left by hx|ψi and on the right by hδψ|ψi, we find that: E D

2 2 x (∆x) p2 − 2hpip + (∆p) x2 − 2hxix − λ ψ =0

(2.2.28)

(2.2.29)

We recall that hx|ψi the arbitrary vector |ψi in the basis of {|xi}. Therefore, from (2.2.29), we can wright:

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CHAPTER 2: QUANTUM MECHANICS II

h i

2 2 (∆x) p2 − 2hpip + (∆p) x2 − 2hxix − λ ψ(x) =0 h i

2 2 (∆x) p2 − 2hpip ψ(x) + (∆p) x2 − 2hxix − λ ψ(x) =0

(2.2.30)

At this point, we innocently recall that the momentum operator is defined by: p=

~ ∂ i ∂x

Applying (2.2.31) to (2.2.30), we have:  2  ! h ~ ∂ ~ ∂ 2 (∆x) − 2hpi ψ(x) + (∆p)2 i ∂x i ∂x   h 2 2~ ∂ 2 2 2 ∂ ψ(x) − hpi ψ(x) + (∆p) (∆x) ~ ∂x2 i ∂x   h 2~ 2 2 0 2 00 (∆x) ~ ψ (x) − hpiψ (x) + (∆p) i h 2 1 2 ψ00 (x) − hpiψ0 (x) + (∆p) 2 i~ ~2 (∆x)

(2.2.31)

i
x2 − 2hxix − λ ψ(x) =0

i
x2 − 2hxix − λ ψ(x) =0

i
x2 − 2hxix − λ ψ(x) =0 i
x2 − 2hxix − λ ψ(x) =0

(2.2.32)

At this point, it is very important that we evaluate λ, and substitute it into (2.2.32). We do this by first going back to (2.2.28), and multiplying on the right by hδψ|ψi which reads: E D

2 2 ψ (∆x) p2 − 2hpip + (∆p) x2 − 2hxix − λ ψ =0 (2.2.33) | {z } At this point, we must evaluate the underbraced term in (2.2.33). We can manipulate the position inner product and find out: hψ|2hxix|ψi =2hψ|hxix|ψi =2hψ|x|ψihψ|x|ψi =2hxihxi =2hxi2

(2.2.34)

Using the same logic that led to (2.2.34) for the position inner product, we may declare that for the momentum inner product, we have: hψ|2hpip|ψi =2hpi2 Using (2.2.34) and (2.2.35), we can rewrite (2.2.33) as: E D 2 2 ψ (∆x) p2 − 2hpi2 + (∆p) (x2 − 2hxi2 − λ ψ =0 D E 2 2 ψ (∆x) p2 − 2hpi2 + (∆p) (x2 − 2hxi2 ψ =λ

(2.2.35)

(2.2.36)

Now, using the results from (2.2.15) and its analogue for momentum, we can rewrite the bracketed terms in (2.2.36) as: hx2 i − 2hxi2 =hx2 i − hxi2 − hxi2

W. Erbsen

HOMEWORK #2

2

= (∆x) − hxi2

(2.2.37)

And similarly, for the momentum: 2

hp2 i − 2hpi2 = (∆p) − hpi2

(2.2.38)

Applying (2.2.37) and (2.2.38) to (2.2.36), we have: n o n o 2 2 2 2 (∆x) (∆p) − hpi2 + (∆p) (∆x) − hxi2 = λ 2

2

2

2

2

2

(∆x) (∆p) − (∆x) hpi2 + (∆p) (∆x) − (∆p) hxi2 = λ 2

2

2

2

2 (∆x) (∆p) − (∆x) hpi2 − (∆p) hxi2 = λ

(2.2.39)

Now, substituting (2.2.39) back into (2.2.32) 2 hpiψ0 (x) i~ h ii
h 1 2 2 2 2 2 2 2 2 + (∆p) x − 2hxix − 2 (∆x) (∆p) − (∆x) hpi − (∆p) hxi ψ(x) = 0 2 ~2 (∆x)

ψ00 (x) −

2 hpiψ0 (x) i~ h i
1 2 2 2 2 2 2 2 2 + (∆p) x − 2hxix − 2 (∆x) (∆p) + (∆x) hpi + (∆p) hxi ψ(x) = 0 2 ~2 (∆x)

ψ00 (x) −

2 hpiψ0 (x) i~ h i
1 2 2 2 2 + (∆p) x2 − 2hxix + hxi2 − 2 (∆x) (∆p) + (∆x) hpi2 ψ(x) = 0 2 ~2 (∆x)

ψ00 (x) −

ψ00 (x) −

h i 1 2 2 2 2 2 2 hpiψ0 (x) + (∆p) (x − hxi) − 2 (∆x) (∆p) + (∆x) hpi2 ψ(x) = 0 2 i~ ~2 (∆x)

2 1 ψ00 (x) − hpiψ0 (x) + 2 i~ ~

"

2

(∆p)

2

(∆x)

2

2

#

(x − hxi) − 2 (∆p) + hpi2 ψ(x) = 0

At this point, we suppose that the wave function ψ(x) can be expressed as:   ihpix φ(x) ψ(x) = exp ~

(2.2.40)

(2.2.41)

Whose first and second derivative are     ihpi ihpix ihpix 0 ψ (x) = exp φ(x) + exp φ0 (x) (2.2.42a) ~ ~ ~         hpi2 ihpix ihpi ihpix 0 ihpi ihpix 0 ihpix 00 00 ψ (x) = − 2 exp φ(x) + exp φ (x) + exp φ (x) + exp φ (x) ~ ~ ~ ~ ~ ~ ~ (2.2.42b)

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CHAPTER 2: QUANTUM MECHANICS II

We now wish to substitute (2.2.41), (2.2.42a), and (2.2.42b) back into (2.2.40). We first recognize that the exponential terms cancel, and so we are left with: " #     2 ihpi 0 ihpi 0 2 1 hpi2 ihpi (∆p) 2 2 − 2 φ+ φ + φ + φ00 − hpi φ + φ0 + 2 (x − hxi) − 2 (∆p) + hpi2 φ = 0 ~ ~ ~ i~ ~ ~ (∆x)2 " # 2 2ihpi 0 2hpi2 2hpi 0 1 (∆p) hpi2 2 2 00 2 φ +φ − φ− φ + 2 − 2 φ+ (x − hxi) − 2 (∆p) + hpi φ = 0 ~ ~ ~2 i~ ~ (∆x)2 " # 2 1 (∆p) 2 2 00 2 2 2 φ + 2 (x − hxi) − 2 (∆p) + hpi − hpi − 2hpi φ = 0 ~ (∆x)2 " # 2   1 (∆p) 2 2 00 2 φ + 2 (x − hxi) − 2 (∆p) + hpi φ=0 ~ (∆x)2 " # 2 1 (∆p) 2 2 00 φ + 2 (x − hxi) − 2hp i φ = 0 ~ (∆x)2 (2.2.43) At this point, we let (∆p) / (∆x) → mω (the reasons why we chose our constants as m and ω will become clear momentarily. Accordingly, (2.2.43) becomes h i 2 ~2 φ00 + m2 ω2 (x − hxi) − 2hp2 i φ =0 (2.2.44) We can divide both sides of (2.2.44) by 2m, which yields: mω2 hp2 i ~2 00 2 φ (x) + (x − hxi) φ(x) = φ(x) 2m 2 m

(2.2.45)

We can see that (2.2.45) looks an awful lot like the time-independent Schr¨ odinger Equation with the 1-D Harmonic Oscillator Hamiltonian. We can manipulate it slightly more,   2 mω2 hp2 i p 2 + (x − hxi) φ(x) = φ(x) (2.2.46) 2m 2 m We now recall that for the 1-D Harmonic Oscillator, from (A.9) and (A.12), we recall that: hxi = 0,

hp2 i =

~mω (2n + 1) 2

Applying this to (2.2.46), we have:  2 
p mω2 2 + x φ(x) = ~ω n + 1/2 φ(x) 2m 2

(2.2.47)


Letting En = ~ω n + 1/2 for obvious reasons, we should now be sufficiently convinced that the solution of this problem is fully equivalent to the Schr¨ odinger equation for the 1-D Harmonic Oscillator. Letting φ(x) → ψn (x), (2.2.47) becomes: 

 p2 mω2 2 + x ψn (x) = En ψn (x) 2m 2

W. Erbsen

HOMEWORK #2

The solution for this differential equation is given by (5.49) in Merzbacher as: r   mω  1/4  mω  mω − 1/ ψn (x) = 2−n/2 (n!) 2 exp − x2 H n x ~π 2~ ~

(2.2.48)

We can find the 1-D minimum uncertainty wave packet by other means, though. Following Merzbacher’s work on pgs. 218-219, starting with (10.55), we have: ∆A2 ∆B 2 ≥ |(Ψ, (A − hAi)(B − hBi)Ψ)|2

(2.2.49)

Where in our case A ⇒ x and B ⇒ p. Remember that we are looking for the minimum uncertainty state, so to find the ground state wave function with minimum uncertainty we would take the “≥” in (2.2.49) to be a “=.” This holds if and only if (p − hpi)Ψ = λ(x − hxi)Ψ

(2.2.50)

Which is analogous to (10.56) in Merzbacher. With a little manipulation we can see that (2.2.50) is in fact a differential equation 

(p − hpi)Ψ − λ(x − hxi)Ψ =0  ~ ∂ − hpi Ψ − λ(x − hxi)Ψ =0 i ∂x

(2.2.51)

Where Merzbacher has provided us with the derivation and value of λ = ihCi/2∆A2 , where in our case hCi = ~. Substituting this into (2.2.51):   ~ ∂ i~ − hpi Ψ − λ(x − hxi)Ψ =0 i ∂x 2∆x2   −ihpi (x − hxi) dψ + + Ψ =0 (2.2.52) dx ~ 2∆x2 Which is a separable differential equation, and the solution is found by   1 ihpi (x − hxi) dΨ = − dx Ψ ~ 2∆x2 Z Z (x − hxi) ihpi log (Ψ) = − dx + dx 2∆x2 ~ (x − hxi)2 ihpix =− + +C 2 4∆x ~   ihpix (x − hxi)2 Ψ = C exp − + 4∆x2 ~ And normalizing, Z

∞

−∞

|Ψ|2 dx =1

  2 2 C exp − (x − hxi) + ihpix dx =1 2 4∆x ~ −∞

Z

∞

(2.2.53)

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CHAPTER 2: QUANTUM MECHANICS II

C2

  (x − hxi)2 exp − =1 2∆x2 −∞
1 C 2 2π∆x2 2 =1
− 1/4 C = 2π∆x2

Z

∞

(2.2.54)

So our ground state wave function, (2.2.53), combined with the appropriate normalization constant, (2.2.54), is given by

Ψ = 2π∆x


2 −1/4



ihpix (x − hxi)2 + exp − 4∆x2 ~



(2.2.55)

Which is identical to (10.66). An important note would be that (2.2.55) minimizes the uncertainty, which will come in handy later on when we reach problems addressing with coherent and squeezed states.

Problem 10.6 The Hamiltonian representing an oscillating LC circuit can be expressed as H=

Q2 Φ2 + 2C 2L

(2.2.56)

Establish that Hamilton’s equations are the correct dynamical equations for this system, and show that the charge Q and the magnetic flux Φ can be regarded as canonically conjugate variables q, p (or the dual pair p, −q). Work out the Heisenberg relation for the product of the uncertainties in the current I and the voltage V . If a mesoscopic LC circuit has an effective inductance of L = 1 µH and an effective capacitance C = 1 pF, how low must the temperature of the device be before quantum fluctuations become comparable to thermal energies? Are the corresponding current-voltage tolerances in the realm of observability?

Solution Hamilton’s equations are given by: ∂H ∂pi ∂H −p˙i = ∂qi q˙i =

(2.2.57a) (2.2.57b)

Where (2.2.57a) and (2.2.57b) are from (8.18) from Goldstein on pg. 337 ?. We note that in our case the “position” is the magnetic flux, and the “momentum” is the charge. Accordingly, (2.2.57a) and (2.2.57b) become ∂H Φ˙ = ∂Q ∂H −Q˙ = ∂Φ

(2.2.58a) (2.2.58b)

W. Erbsen

HOMEWORK #2

Substituting (2.2.56) into (2.2.58a) and (2.2.59b),  2  2 ˙Φ = ∂ Q + Φ ⇒ Φ ˙ =Q ∂Q 2C 2L C  2  2 ∂ Q Φ Φ −Q˙ = + ⇒ Q˙ = − ∂Φ 2C 2L L

(2.2.59a) (2.2.59b)

The best way to proceed with this problem is to mimic the same procedure for the Quantum Harmonic Oscillator. We begin by drawing a direct analogue between the position and momentum operators with the magnetic flux and charge operators, respectively. Starting with the position → magnetic flux relationship, we recall from (A.2a) that r
~ q= a† + a (2.2.60) 2mω 1

/

In the case where we let q → Φ, we must also let ω → (LC) 2 , and m → C. Accordingly, (2.2.60) becomes: r   1/4
~ L Φ= c† + c (2.2.61) 2 C

Where c and c† are to be defined momentarily. We now draw correlation between the momentum operator from (A.2b) to the charge operator: r   1/4
~ C Q=i c† − c (2.2.62) 2 L It is easy to see from (A.9) and (A.11) that hΦi = hQi = 0. From (A.10), we have: r
L 2 hΦ i = ~ n + 1/2 C

(2.2.63)

While from (A.12) we can see that:

2

r

hQ i = ~

C L

n + 1/2




(2.2.64)

From (2.2.63) and (2.2.64), we can now calculate the variance of the magnetic flux and charge: r
L 2 2 2 (∆Φ) =hΦ i − hΦi ⇒ ~ n + 1/2 (2.2.65a) C r
C 2 (∆Q) =hQ2 i − hQi2 ⇒ ~ n + 1/2 (2.2.65b) L

We are asked to find the product of the uncertainties in the current I and the voltage V , however (2.2.65a) and (2.2.65b) are in terms of Φ and Q, respectively. We can use (2.2.60) and (2.2.59b) to connect the two respective quantities: ˙ ⇒V = Q −→ Q = CV Φ C Φ ˙ −Q ⇒I = −→ Φ = LI L

(2.2.66a) (2.2.66b)

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CHAPTER 2: QUANTUM MECHANICS II

So, using (2.2.66a) and (2.2.66b), then (2.2.65a) and (2.2.65b) become r
1 2 (∆I) =~ n + 1/2 LC r
1 2 n + 1/2 (∆V ) =~ LC

(2.2.67a) (2.2.67b)

We now multiply (2.2.67a) and (2.2.67b), which yields ∆I∆V = √

~ LC

n + 1/2




Given the provided parameters, we are now in a position to find the temperature at which quantum fluctuations become apparent. We achieve this by setting the zero-point energy equal to the average thermal energy, and solving for T : ~ 1 ~ 1 √ √ −→ T = 2 LC 2kB LC 1.055 × 10−34 J · s 1 p = 2 (1.381 × 10−23 J · s−1 ) (1 × 10−6 H) (1 × 10−12 F) −→ T ≈ 3.82 × 10−3 K

(2.2.68)

Problem 10.7 If a coherent state |αi (eigenstate of a) of an oscillator is transformed into a squeezed state by the unitary operator   ζ 2 †2 U = exp (a − a ) (2.2.69) 2 √ calculate the value of ζ that will reduce the width of the Hermitian observable (a + a† )/ 2 to √ 1 percent of its original coherent-state value. What happens to the width of the conjugate observable (a − a† )/ 2 i in this transformation?

Solution We know that |αi is an eigenstate of a and satisfies a|αi = α|αi. What we want to do is go from our coherent state |αi to a squeezed state |α, ζi. We arrive at the squeezed state by applying a unitary operator, (2.2.69), to our coherent state: U|αi = |α, ζi. To do this, we first find the squeezed state operator, b, which we define as: b =UaU† U† b =aU† U† BU =a

(2.2.70)

W. Erbsen

HOMEWORK #2

So we can say: hα|a|αi =hα|U†bU|αi =hα, ζ|b|α, ζi

(2.2.71)

The easiest way to find b would be to utilize some identities made available by Merzbacher on pgs. 39-40. He says that if f(λ) = eλA Be−λA Then we can write p p [A, B] f(λ) = B cosh λ β + √ sinh λ β β

(2.2.72)

Which we can use only if β is constant, and the operators A and B satisfy [A, [A, B]] = βB. In our case, f(λ) → b(ζ), and A = (a2 − a†2 ) and B = a. To find out if we can use (2.2.72): [A, B] =[(a2 − a†2 ), a]

=[a2 , a] − [a†2 , a]

= − [a†2 , a]

= − a† [a† , a] − [a† , a]a†

= − a† (−1) − (−1)a†

=2a†

[A, [A, B]] =[(a2 − a†2 ), 2a†]

=2 [a2 , a† ] − [a†2 a† ]

(2.2.73)




=2 a[a, a†] + [a, a†]a =4a




(2.2.74)

It seems that we can in fact use (2.2.72), since both of our conditions are satisfied with λ = −ζ/2 and β = 4. We now have, p p [A, B] b(ζ) =a cosh λ β + √ sinh λ β β √ √ 2a† =a cosh −ζ/2 4 + √ sinh −ζ/2 4 4 † =a cosh ζ + a sinh ζ =a cosh ζ − a† sinh ζ

(2.2.75)

Following Merzbacher’s lead on pgs. 230-231, we define λ = cosh ζ and ν = sinh ζ, where λ here is different than the one used before. We can now rewrite (2.2.75): b = λa + νa†

(2.2.76)

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CHAPTER 2: QUANTUM MECHANICS II

And taking the Hermitian conjugate, b† = λa† + νa

(2.2.77)

Since both λ and ν are real. Inverting these two equations a = λb − νb†

(2.2.78)

a† = λb† − νb

(2.2.79)

And we can now write 1 1 √ hα|a|αi = √ hα, ζ|b|α, ζi 2 2 1 = √ hα, ζ|λa + νa†|α, ζi 2 1 1 √ hα|a† |αi = √ hα, ζ|b†|α, ζi 2 2 1 = √ hα, ζ|λa† + νa|α, ζi 2

(2.2.80)

(2.2.81)

Now that we have the framework laid down, we can now attack √ our problem head on and directly investigate the Hermitian conjugate we are to minimize, (a + a† )/ 2 : 1 1 √ hα|a + a† |αi = √ 2 2 1 =√ 2 1 =√ 2 1 =√ 2 1 =√ 2 1 =√ 2 1 =√ 2

  hα|a|αi + hα|a†|αi

  hα, ζ|λa + νa† |α, ζi + hα, ζ|λa† + νa|α, ζi

hα, ζ|λa + νa† + λa† + νa|α, ζi

hα, ζ|a(λ + ν) + a† (λ + ν)|α, ζi (λ + ν)hα, ζ|a + a† |α, ζi (cosh ζ − sinh ζ)hα, ζ|a + a† |α, ζi e−ζ hα, ζ|a + a† |α, ζi

(2.2.82)

√ And we are asked to calculate the value of ζ that “squeezes” (a + a† )/ 2 to 1% of its original coherent state value. So take the appropriate fraction, and solve for the squeezing parameter ζ: √ 1/ 2 e−ζ hα, ζ|a + a† |α, ζi e−ζ √ = (2.2.83) 100 1/ 2 hα|a + a† |αi(100) And finally eζ = 100 ⇒ ζ = log 100 ≈ 4.60517 . . .

W. Erbsen

HOMEWORK #3

Figure 2.2: Graph of ∆p = ~/2∆x.

Do note that the last√part of the question where we are asked what happens to the width of the conjugate observable (a − a† )/ 2 i yields the same result. As a side comment, I found it most informative to investigate squeezed states graphically. Most notably, by solving the uncertainty relation for ∆p in terms of ∆x and graphing it (See Figure 2 on next page). The product of the coordinates at any point along the curve is equal to exactly ~/2; as an example, in the figure ∆x1 ∆p1 = ∆x2 ∆p2 = ~/2. States where this is satisfied are referred to as squeezed states, since the uncertainty relation is saturated. If there exists a point above the curve (shaded region), then this implies that ∆x∆p ≥ ~/2 and that it is not minimized. Points below the curve cannot exist, for obvious reasons.

2.3

Homework #3

Exercise 14.3 Show that in the Heisenberg picture the density operator for the state |Ψi, ¯ Ψ| ¯ ρ¯ = |Ψ(0)ihΨ(0)| = |Ψih

(2.3.1)

Satisfies the equations i~

  ∂ ρ¯ ∂ρ ¯ ρ¯ = i~T † (t, 0) T (t, 0) = H, ∂t ∂t

and that the expectation value of ρ¯ is constant as in (14.21).

and

d¯ ρ =0 dt

(2.3.2)

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CHAPTER 2: QUANTUM MECHANICS II

Solution First recall that the Schr¨ odinger equation in the Heisenberg representation takes the form i~

d |ΨH i = HH |ΨH i dt

(2.3.3)

And similarly, −i~

d hΨH | = hΨH |HH dt

(2.3.4)

We begin by rewriting the first part of (2.3.2):     ∂ρH d d i~ =i~ |ΨH i hΨH | + |ΨH i hΨH | ∂t dt dt     1 1 =i~ HH |ΨH i hΨH | − |ΨH i hΨH |HH i~ i~ = [HH (|ΨH ihΨH |) − (|ΨH ihΨH |) HH ] = [HH (ρH ) − (ρH ) HH ] = [HH , ρH ]

(2.3.5)

Now let’s do the same for the middle part. We know that ∂ρ ∂ρH † = T (t, 0) T (t, 0) ∂t ∂t

(2.3.6)

We can manipulate this further by multiplying (2.3.6) on the left by T † (t, 0) and on the right by T (t, 0). This gives T † (t, 0)

∂ρ ∂ρH T (t, 0) = ∂t ∂t

(2.3.7)

Which is (14.31) in Merzbacher and also is the same as the left side of (2.3.2), which we have already shown to be equal to the right side. We can quantify this by inserting the result from (2.3.5) into (2.3.7), which yields ∂ρH ∂ρ 1 ∂ρH ∂ρ = T † (t, 0) T (t, 0) = [HH , ρH ] −→ i~ = i~T † (t, 0) T (t, 0) = [HH , ρH ] ∂t ∂t i~ ∂t ∂t

(2.3.8)

Where (2.3.8) is identical to (2.3.2).

Exercise 14.4 Show that the expression e00 , t2 |Te (t2 , t1 )|A e0 , t1 i hB

(2.3.9)

W. Erbsen

HOMEWORK #3

for the transition amplitude is quite general and gives the correct answer if the Schr¨ odinger (H0 = 0) or Heisenberg (H0 = H) pictures are employed.

Solution First, we recall that Te(t2 , t1 ) represents the time evolution operator in the interaction picture. It’s effect is the same as we would expect, from (14.48): e e |ψ(t)i = Te(t2 , t1 )|ψ(0)i

(2.3.10)

Te(t2 , t1 ) = U(t2 )T (t2 , t1 )U† (t1 )

(2.3.11)

And it can be expressed in terms of unitary transformations from (14.49) as

Where we recall from (14.46) that

U(t) = exp



i H0 t ~



(2.3.12)

Merzbacher states in (14.50) that e0 , ti = U(t)|A0 i |A e 00 , t| = hB 00 |U†(t) hB

(2.3.13a) (2.3.13b)

Multiplying (2.3.13a) on the left by U† , and (2.3.13b) on the right by U, e0 , ti = |A0 i U† (t)|A e00 , t|U(t) = hB 00 | hB

(2.3.14a) (2.3.14b)

Substituting (2.3.14a) and (2.3.14b) into (2.3.9),

e00 , t2 |Te (t2 , t1 )|A e0 , t1 i = hB 00 |T (t2 , t1 )|A0 i hB

(2.3.15)

The equivalence has thus been shown. We can further quantify this by recalling (14.17), which reads   i (2.3.16) T (t, t0 ) = exp − (t − t0 )H0 ~ Substituting (2.3.16) into (2.3.15), i

hB 00 |T (t2 , t1 )|A0 i =hB 00 |e− ~ (t−t0 )H0 |A0 i

(2.3.17)

If we let H0 → 0, as in the Schr¨ oginer picture, then (2.3.17) becomes hB 00 |T (t2 , t1 )|A0 i = hB 00 |A0 i Which is what we would expect in the Schr¨ odinger picture. Similarly, if we let H0 → H, as in the Heisenberg picture, then (2.3.17) becomes

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CHAPTER 2: QUANTUM MECHANICS II

i

i

hB 00 |e− ~ (t−t0)H0 |A0 i = hB 00 |e− ~ (t−t0 )H |A0 i Which is, according to Merzbacher, also what we would expect for the Heisenberg picture.

Exercise 14.6 Illustrate the validity of (14.60) by letting G = x2 and F = p2x , and evaluating both the operator expression on the left, in the limit ~ → 0, and the corresponding Poisson bracket on the right.

Solution Equation (14.60) reads, lim

~→0

hGF − F Gi ∂G ∂F ∂F ∂G ∂G ∂F ∂F ∂G ∂G ∂F ∂F ∂G = − + − + − i~ ∂x ∂px ∂x ∂px ∂y ∂py ∂y ∂py ∂z ∂pz ∂z ∂pz

(2.3.18)

Let’s start by evaluating the numerator of the limit on the LHS of (2.3.18): hGF − F Gi =hx2 p2x − p2x x2 i =h[x2 , p2x ]i

=hx[x, p2x] + [x, p2x]xi =hx([x, px] px + px [x, px]) + ([x, px] px + px [x, px])xi | {z } | {z } | {z } | {z } i~

i~

i~

i~

=2i~hxpx + px xi =2i~ [hxpx i + hpx xi] Z  Z =2i~ ψ∗ (x)xpxψ(x) dx + ψ∗ (x)px xψ(x) dx  Z      Z ~ ∂ ~ d =2i~ ψ∗ (x)x ψ(x) dx + ψ∗ (x) xψ(x) dx i ∂x i dx  Z   Z    Z ~ ∂ ~ ∂ =2i~ ψ∗ (x)x ψ(x) dx + ψ∗ (x)x ψ(x) dx + ψ∗ (x)ψ(x) dx i ∂x i ∂x  Z    Z ∂ ~ 2~ ψ∗ (x)x ψ(x) dx + ψ∗ (x)ψ(x) dx =2i~ i ∂x i   ~ =2i~ 2hxpxi + hψ(x) | ψ(x)i i   ~ =2i~ 2hxpxi + (2.3.19) i

Where I assumed that ψ(x) is normalized, and I also liberally used the fact that [x, px] = i~ and also the following identities from (3.50): [AB, C] = A[B, C] + [A, C]B and [A, BC] = [A, B]C + B[A, C]. We now plug (2.3.19) into the left side of (2.3.18):

W. Erbsen

HOMEWORK #3


2i~ 2hxpxi + ~i hGF − F Gi lim = lim ~→0 ~→0 i~ i~   ~ = lim 2 2hxpxi + ~→0 i =4hxpxi

(2.3.20)

To solve the Poisson bracket on the right, you should first notice that all partial derivatives on the right are zero except the first: ∂G ∂F ∂(x2 ) ∂(p2x ) = ⇒ 4xpx ∂x ∂px ∂x ∂px

(2.3.21)

So, the left side of (2.3.18) is given by (2.3.20) as 4hxpxi and the right side is given by (2.3.21) as 4xpx: 4hxpxi = 4xpx −→ hxpx i = xpx This makes sense; the Poisson Bracket is defined as the classical analogue of the quantum-mechanical commutator and the expectation value is an “average.” This therefore demonstrates how classical mechanics is a limiting case of quantum mechanics.

Exercise 14.7 Show that the transformation UaxU† =ax cos Θ + bpx sin Θ †

UbpxU = − ax sin Θ + bpx cos Θ

(2.3.22a) (2.3.22b)

is canonical, if a and b are real-valued constants, and Θ is a real-valued angle parameter. Construct the unitary operator U that effects this transformation. For the special case of Θ = π/2, calculate the matrix elements of U in the coordinate representation. Noting that this transformation leaves the operator a2 x2 + b2 p2x invariant, rederive the result of Exercises 3.8 and 3.21.

Solution In the spirit of 14.4, we recall that in order to verify canonically conjugate transformations, qp − pq = i~1

(2.3.23)

Which is of course (14.63). We also note that q represents the canonically conjugate position, while p is the canonically conjugate momentum. Claiming that q corresponds to (2.3.22a), while p corresponds to (2.3.22b), we can compute the commutator in (2.3.23) straightforwardly, qp − pq = (ax cos Θ + bpx sin Θ) (−ax sin Θ + bpx cos Θ) − (−ax sin Θ + bpx cos Θ) (ax cos Θ + bpx sin Θ) (2.3.24)

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CHAPTER 2: QUANTUM MECHANICS II

To reduce the cumbersomeness of evaluating this, we evaluate qp and pq separately, and then combine them. Computing the product qp first, qp = (ax cos Θ + bpx sin Θ) (−ax sin Θ + bpx cos Θ) = − a2 x2 cos Θ sin Θ + axbpx cos2 Θ − bpx ax sin2 Θ + b2 p2x sin Θ cos Θ

(2.3.25)

And now doing the same thing for the product pq, pq = (−ax sin Θ + bpx cos Θ) (ax cos Θ + bpx sin Θ) = − a2 x2 sin Θ cos Θ − axbpx sin2 Θ + bpxax cos2 Θ + b2 p2x cos Θ sin Θ

(2.3.26)

Simply by visual inspection, we can see that in subtracting (2.3.26) from (2.3.25), the first and last terms in both equations cancel. We are left with, qp − pq =axbpx cos2 Θ − bpx ax sin2 Θ + axbpx sin2 Θ − bpxax cos2 Θ = (axbpx − bpx ax) cos2 Θ + (axbpx − bpxax) sin2 Θ
= [ax, bpx] cos2 Θ + sin2 Θ = [ax, bpx]

(2.3.27)

Recalling that a and b are constants, (2.3.27) becomes qp − pq =ab [x, px] −→ qp − pq = abi~

(2.3.28)

According to (2.3.23), the transformation in (2.3.22a) and (2.3.22b) are canonically conjugate . Now, following closely Merzbacher’s work on pgs. 327-328, we define the infinitesimal unitary operator Uε from (14.72) as Uε = 1 +

iε G ~

(2.3.29)

Where G is the generator of this infinitesimal transformation. We now recall that according to (14.68), F (q, p) = F (x, px) +

ε [F, G] i~

(2.3.30)

While we also note that from (14.71), F (q, p) = Uε F (x, px)U†ε

(2.3.31)

Comparing (2.3.31) to (2.3.22a) and (2.3.22b), and applying the results to (2.3.30), we have aε [x, G] i~ bε Ubpx U† =bpx + [px , G] i~ UaxU† =ax +

(2.3.32a) (2.3.32b)

Applying the RHS of (2.3.22a) and (2.3.22b) to (2.3.32a) and (2.3.32b), respectively, aε [x, G] i~ bε −ax sin Θ + bpx cos Θ =bpx + [px , G] i~ ax cos Θ + bpx sin Θ =ax +

(2.3.33a) (2.3.33b)

W. Erbsen

HOMEWORK #3

Now recalling (14.70) from Merzbacher, F (q, p) =



   iε iε 1 + G F (x, px) 1 − G ~ ~

(2.3.34)

Applying (2.3.34) to (2.3.33a) first,     iε iε ax cos Θ + bpx sin Θ = 1 + G ax 1 − G ~ ~    iε iaxε = 1+ G ax − G ~ ~ iε iε axε2 =ax − Gax + ax G + 2 G2 ~ ~ ~

(2.3.35)

First order approximation in Θ allows us to rewrite (2.3.35) as ax + bpx Θ = −ax

iε iaε [G, ax] −→ bpxΘ = [x, G] ~ ~

(2.3.36)

And we can analogously apply (2.3.34) to (2.3.33b), axΘ =

ibε [px , G] ~

(2.3.37)

Now recalling (14.64) from Merzbacher, q =x+ε

∂G , ∂px

p = px − ε

∂G ∂x

(2.3.38)

Applying (2.3.38) to (2.3.22a) and (2.3.22b), ∂G ∂px ∂G Ubpx U† =bpx − bε ∂x UaxU† =ax + aε

(2.3.39a) (2.3.39b)

Using (2.3.22a), (2.3.32a) and (2.3.39a), we now have ax ∂G Θ= bε ∂x

(2.3.40)

And similarly, using (2.3.22b), (2.3.32b), and (2.3.39b), bpx ∂G Θ= aε ∂px

(2.3.41)

We can solve the differential equations in (2.3.40) and (2.3.41) as ax ∂G Θ= −→ G = bε ∂x bpx ∂G Θ= −→ G = aε ∂px

a x2 Θ bε 2 b p2x Θ aε 2

Having established that ε → Θ, and combining (2.3.42a) and (2.3.42b), we have

(2.3.42a) (2.3.42b)

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CHAPTER 2: QUANTUM MECHANICS II

G=

b a 2 x + p2x 2b 2b

We now recall (14.75) from Merzbacher, which reads   iλ U = exp G ~

(2.3.43)

(2.3.44)

Substituting (2.3.43), and recognizing the fact that λ → Θ, 

  iΘ a 2 b 2 U = exp x + px ~ 2b ab

For the special case of Θ = π/2, (2.3.45) becomes    iπ a 2 b 2 U = exp x + px 2~ 2b 2a

(2.3.45)

(2.3.46)

And in order to calculate the matrix elements of U in the coordinate representation (letting q → x),      iπ a 2 b hx0 |U|xi = x0 exp x + p2x x (2.3.47) 2~ 2b 2b

Substituting the series expansion of the exponential function into (2.3.47), hx0 |U|xi =

  N + ∞ X iπ a b 1 x2 + p2x x0 x N ! 2~ 2b 2b

*

(2.3.48)

N=0

Expanding to the first few terms of (2.3.48),     *   2 + a 2 b 2 a 2 b 2 0 0 0 iπ 0 1 iπ hx |U|xi =hx |xi + x x + px x + x x + px x + ... 2 2~ 2b 2~ 2b 2b 2b Rewriting the last element in (2.3.48), *   *  2 + 2 + 1 iπ a a b iπ b x0 x2 + p2x x0 x2 + p2x x x = 2 2~ 2b b 2b 32~ a

(2.3.49)

(2.3.50)

To make our lives easier, let’s examine the main part of (2.3.50) without the constants: hx0 |(x2 + p2x )2 |xi =hx0 |(x2 + p2x + 2xpx)|xi ⇒ hx0 |x2 |xi + hx0 |p2x |xi + 2hx0 |px|xi

(2.3.51)

Taking inspiration from the harmonic oscillator, we know that for x 6= x0 , then (2.3.51) goes to zero. The same can be seen for the first two matrix elements as well, and therefore we only allow for diagonal entries.

We now recall that for a linear harmonic oscillator, the Schr¨ odinger Equation can be expressed in momentum representation with (3.84) as

W. Erbsen

HOMEWORK #3

1 d2 φE (x) 1 2 px φE (x) − mω2 = EφE (x) 2m 2 dp2x

(2.3.52)

We now recall that the Hamiltonian for the Quantum Harmonic Oscillator reads H=

p2x 1 + mω2 x2 2m 2

We may also rewrite the unitary transformation in (2.3.45) as    iΘ  2 2 U = exp a x + b2 p2x 2~

(2.3.53)

(2.3.54)

To show that the operator inside the brackets of (2.3.54) remains invariant,    

 iΘ  2 2 iΘ  2 2 U a2 x2 + b2 p2x U† = exp a x + b2 p2x a2 x2 + b2 p2x exp − a x + b2 p2x 2~ 2~       
 iΘ  2 2 iΘ  2 2 2 2 2 2 2 2 2 2 = 1+ a x + b px a x + b px 1 − a x + b px 2~ 2~ =a2 x2 + b2 p2x

Therefore, the transformation leaves a2 x2 + b2 p2x invariant . We also note §, § , given in (2.3.45) represents the time evolution operator for the quantum harmonic oscillator, then it lends itself that the position-space wave function can be found by taking the position-space function projected onto momentumspace from (2.3.48) and solving the position-space Schr¨ odinger equation from (2.3.52). The result is: p  eiα x φ(px , t) = √ ψ ,t mω mω

Exercise 14.11 For a free particle in one dimension and an arbitrary initial wave packet, calculate the time development of (∆q)2t and show (as in Problem 2 in Chapter 3) that (∆q)2t = (∆q)20 +

t (∆p)2 t2 [hqp + pqi0 − 2hqi0 hpi] + m m2

(2.3.55)

Verify that for the minimum uncertainty wave packet this result agrees with (14.97). Also compare with the value of the variance (∆q)2 as a function of time for a beam of freely moving classical particles whose initial positions and momenta have distributions like variances (∆q)20 and (∆p)2 .

Solution To find the time development of (∆q)2t , it is first helpful to take care of some definitions. First, recall

133

CHAPTER 2: QUANTUM MECHANICS II

that the time derivative of the expectation value of an operator is given by d i~ hAi = h[A, H]i + dt



∂A ∂t



(2.3.56)

The operators we will be dealing with in this problem are time-independent, so the last part of (2.3.56) will go to zero. Furthermore, we can go ahead and find hpit and hp2 it since they will be needed later on in the problem: i~

∂ hpi =h[p, T ]i ∂t 1 = h[p, p2]i = 0 2m

(2.3.57)

Integrating (2.3.57) we find that we are only left with a constant: hpit = hpi0 . With the same logic we can arrive at hp2 it = hp2 i0 , and the variance of hpi is therefore h∆p2 i = hp2 i0 − hpi20 . It will also be convenient to find the following commutators, [q, p2] = [q, p]p + p[q, p] = 2i~p

(2.3.58)

Where I liberally used the fact that [q, p] = i~ and also the following identities from (3.50): [AB, C] = A[B, C] + [A, C]B and [A, BC] = [A, B]C + B[A, C]. Armed with these tools, we can go ahead and start finding (∆q)2t = hq 2 it − hqi2t . Start by finding hqi: ∂ hqi =h[q, T ]i ∂t 1 = h[q, p2]i 2m 1 = i~hpit m hpi0 t ⇒ hqit = + hqi0 m

i~

(2.3.59)

Doing the same for hq 2 i, i~

∂ 2 1 hq i = h[q 2 , p2 ]i ∂t 2m 1 = hq [q, p2 ] + [q, p2 ] pi 2m | {z } | {z } 2i~p

2i~p

i~ = hqp + pqi m hqp + pqit t ⇒ hq 2 it = + hq 2 i0 m And we must now find hqp + pqit : i~

∂ 1 hqp + pqi = h[(qp + pq), p2 ]i ∂t 2m  1

= [qp, p2] + [pq, p2 ] 2m

(2.3.60)

W. Erbsen

HOMEWORK #3

=

1 hq [p, p2 ] + [q, p2 ] p + p [q, p2] + [p, p2] qi 2m | {z } | {z } | {z } | {z } 0

2i~p

2i~p

0

i~ = hp2 i m hp2 i0 t + hqp + pqi0 ⇒ hqp + pqit = m

(2.3.61)

Substituting (2.3.61) into (2.3.60)   1 hp2 i0 t + hqp + pqi0 t + hq 2 i0 m m 2 2 hp i0 t hqp + pqi0 t = + + hq 2 i0 m2 m

hq 2 it =

(2.3.62)

And now we find (∆q)2t . Using (2.3.59) and (2.3.62), (∆q)2t =hq 2 it − hqi2t  2 2   2 hp i0 t hqp + pqi0 t hpi0 t 2 = + + hq i − + hqi 0 0 m2 m m hp2 i0 t2 hqp + pqi0 t hpi20 t2 hpi0 hqi0 t + + hq 2 i0 − −2 − hqi20 = 2 m m m2 m hqp + pqi0 t 2hqi0 hpi0 t hp2 i0 t2 hpi20 t2 =hq 2 i0 − hqi20 + − + − m m m2 m2 2 2 t (∆p)0 t =(∆q)20 + [hqp + pqi0 − 2hqi0 hpi0 ] + m m2

(2.3.63)

Which answers the first part of the question∗ . For a beam of freely moving classical particles, finding (∆q)2t is simply a matter of realizing that there is no difference between qp and pq, and (2.3.63) becomes: (∆q)2t =(∆q)20 +

t (∆p)20 t2 (∆p)20 t2 2 2 [2qp − 2qp] + −→ (∆q) = (∆q) + t 0 m m2 m2

(2.3.64)

Which is identical to (14.97).

Exercise 14.13 In either the Heisenberg or Schr¨ odinger picture, show that if at t = 0 a linear harmonic oscillator is in a coherent state, with eigenvalue α, it will remain in a coherent state, with eigenvalue αe−iωt , at time t.

Solution ∗ You will notice that my answer (2.3.63) differs ever so slightly from Merzbacher’s (2.3.107). I do believe that Merzbacher made a mistake; the last term in the square brackets should be hpi0, as opposed to hpi.

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CHAPTER 2: QUANTUM MECHANICS II

Recall from (14.7) that the time development operator T (t, t0 ) can relate the initial state |Ψ(t0 )i to the final state |Ψ(t)i. Applying this principle to our system where t0 = 0, t =0

0 |Ψ(t)i = T (t, t0 )|Ψ(t0 )i −− −→ |Ψ(t)i = T (t, 0)|Ψ(0)i

(2.3.65)

Where the state in our case is a coherent state, |αi. Using (10.110), we can say that |αi =

∞ X

|nie−|α|

2

/2

n=0

αn √ n!

(2.3.66)

And according to (14.17) that the time development operator T (t, t0 ) at t0 = 0 is given by t =0

i

i

0 T (t, t0 ) = e− ~ (t−t0)H −− −→ T (t, 0) = e− ~ Ht

(2.3.67)

And the series expansion of the exponential function ex is given by: ex =

∞ X xn n! n=0

(2.3.68)

So, using (2.3.66), (2.3.67) and (2.3.68), we can now rewrite (2.3.65): i

|α(t)i =e− ~ Ht |α(0)i ∞ X 2 i αn =e− ~ Ht |nie−|α| /2 √ n! n=0 = =

∞ X 2 αn (−iHt/~)n |nie−|α| /2 √ n! n! n=0

∞ X 2 (−iωt(n + 1/2))n αn |nie−|α| /2 √ n! n! n=0

∞ X 2 (−iωtn − iωt/2)n αn = |nie−|α| /2 √ n! n! n=0

=

∞ X

n=0

e−iωtn e−iωt/2 |nie−|α|

=e−iωt/2 =e−iωt/2

∞ X

n=0 ∞ X

e−iωt

|nie−|α|

n=0 −iωt  −iωt/2

=e

αe


n 2

2

/2

|nie−|α|

/2

αn √ n!

2

/2

αe−iωt √ n!

αn √ n!


n

(2.3.69)

Where I used (2.3.66), by using the fact that |α|2 is independent of phase, since it refers to a magnitude. Since αe−iωt is an eigenstate of the energy operator, it satisfies a|αi = α|αi, and  a|α(t)i =ae−iωt/2 αe−iωt  =e−iωt/2 a αe−iωt

W. Erbsen

HOMEWORK #3


 =e−iωt/2 αe−iωt αe−iωt   =αe−iωt e−iωt/2 αe−iωt

=αe−iωt |α(t)i

(2.3.70)

So, according to , if a linear harmonic is in a coherent state at t0 = 0, it will remain in a coherent state with eigenvalue αe−iωt : a|α(t)i = αe−iωt |α(t)i

Problem 10.1 A particle of charge q moves in a uniform magnetic field B which is directed along the z axis. Using a gauge in which Az = 0, show that q = (cpx − qAx )/qB and p = (cpy − qAy )/c may be used as suitable canonically conjugate coordinate and momentum together with the pair z, pz . Derive the energy spectrum and the eigenfunctions in the q-representation. Discuss the remaining degeneracy. Propose alternative methods for solving this eigenvalue problem.

Solution To show that q and p are canonically conjugate, we must test to see if they satisfy the fundamental Poisson bracket relations: {qi , qj } = 0, {pi , pj } = 0 and {qi , pj } = δij . The Poisson bracket is defined by the right hand side of (14.60): lim

~→0

∂G ∂F ∂F ∂G ∂G ∂F ∂F ∂G ∂G ∂F ∂F ∂G h[G, F ]i = − + − + − i~ ∂x ∂px ∂x ∂px ∂y ∂py ∂y ∂py ∂z ∂pz ∂z ∂pz

(2.3.71)

The easiest way to find out of Poisson’s conditions are satisfied would be to evaluate the commutator on the left of (2.3.71). First we must choose our gauge; and I convienently choose to work in the Landau Gauge since it has only one term (the symmetric gauge should also work): A(x, y, z) = (−yB0 , 0, 0). Rewriting our given values of q and p, q=

cpx − qAx cpx = +y qB qB0

(2.3.72)

cpy − qAy = py c

(2.3.73)

And, p=

Just by visual inspection it is possible to tell that {x, y} = {x, z} = {y, z} = 0, and also that {px , py } = {px , px} = {py , pz } = 0. For the same reasons, we can see that {qx, px } = {qz , pz } = 0. To find {qy , py }, h[qy , py ]i i~ h[y, py ]i = lim ~→0 i~

{qy , py } = lim

~→0

137

CHAPTER 2: QUANTUM MECHANICS II

= lim

~→0

hi~i i~

=1

(2.3.74)

Which indicates that q and p may be used as canonically conjugate coordinates. To find the energy spectrum and eigenfunctions, start off by remembering the definition of a Hamiltonian in an electromagnetic field (4.93) H=

1  q 2 p − A + qφ 2m c

Where in our case φ = 0. Still working in the Landau gauge, (2.3.75) becomes   q 2  q 2  q 2 1  px − Ax + py − Ay + pz − Az H= 2m c c c 2
1  q 1 = px + yB0 + p2 + p2 2m  c 2m y  z
2q q2 1 1 p2x + B0 ypx + 2 y2 B02 + p2y + p2z = 2m c c 2m   2 1 2q q = p2x + p2y + p2z + B0 ypx + 2 y2 B02 2m c c   1 2q q2 2 2 2 2 2 = px + py + pz + ypx B0 + 2 y B0 2m c c    2 2 2 2 x + py + pz qB0 m qB0 + = ypx + y2 2m mc 2 mc p2x + p2y + p2z m = + ωc ypx + ωc2 y2 2m 2

(2.3.75)

(2.3.76)

Where I used the fact that the cyclotron resonance frequency is defined by ωc = qB0 /mc. At this point we should notice that our Hamiltonian commutes with both px and pz but not with py : [H, px] = [H, pz ] = 0. This implies that it is possible to find a common eigenstate between both px , pz and H. Let’s find an eigenfunction for px: pˆx ψ(x) =px ψ(x) ~ d ψ(x) =px ψ(x) i dx 1 i dψ(x) = px dx ψ(x) ~ −→ ψ(x) =e

ipx x ~

(2.3.77)

And WLOG we can assume this holds for pz as well. At this point we still need to find an eigenstate for H; one way to find it is to deploy separation of variables: ψ(x, y, z) =ψ(x)ψ(y)ψ(z) =eipx x/~ ψ(y)eipz z/~ =ei(px x+pz z)/~ ψ(y)

(2.3.78)

W. Erbsen

HOMEWORK #3

Using our Hamiltonian (2.3.76) and our eigenstate (2.3.78), we can construct the Schr¨ odinger equation, keeping the eigenfunction arbitrary (for now):  
1 m 2 2 2 2 2 p + py + pz + ωc ypx + ωc y ψ(x, y, z) =Eψ(x, y, z) (2.3.79) 2m x 2

Where the x, z, px and pz are now numbers instead of operators, so: !   p2y (p2 + p2z ) m 2 2 + ωc ypx + ωc y ψ(x, y, z) = E − x ψ(x, y, z) 2m 2 2m

(2.3.80)

At this point we want to undergo a change of variables with the definitions of q and p given in the prompt: y0 −→ y +

pz mω

And p0y −→ py While E 0 −→ E −

p2z 2m

Solving and substituting into (2.3.80): !  p02 px  m 2  0 px  2 y 0 + ω c px y − + ωc y − ψ(x, y, z) =E 0 ψ(x, y, z) 2m mω 2 mω  ! p02 p2x mωc2 2y0 px p2x y 0 02 + ω c px y − + + y − ψ(x, y, z) =E 0 ψ(x, y, z) 2m m 2 mωc mωc ! p02 p2x mωc2 y02 p2x y 0 0 + ω c px y − + − ω c px y + ψ(x, y, z) =E 0 ψ(x, y, z) 2m m 2 2m ! p02 p2x mωc2 y02 y − + ψ(x, y, z) =E 0 ψ(x, y, z) 2m 2m 2 ! p02 mωc2 y02 y + ψ(x, y, z) =E 0 ψ(x, y, z) 2m 2

(2.3.81)

The operator acting on ψ(x, y, z) on the left of (2.3.81) is none other than the Hamiltonian for the harmonic oscillator. This implies that it is a straightforward matter to find both the eigenenergies and the eigenfunctions. From (2.3.81) it is clear that E = ~ωc (n + 1/2 ), as it would be for the harmonic oscillator. Therefore, the eigenvalues for the energy operator in our case is given by:   1 p2 E = ~ωc n + + z 2 2m To find the eigenfunctions make our separation of variables substitution from (2.3.78),

(2.3.82)

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CHAPTER 2: QUANTUM MECHANICS II

p02 mωc2 y02 y + 2m 2

!

exp



   i(px x + pz z) i(px x + pz z) ψ(y) =E 0 exp ψ(y) ~ ~

(2.3.83)

So that the eigenfunctions are just those of the harmonic oscillator from (5.39) multiplied by (2.3.78): ψ(x, y, z) = 2−n/2 (n!)−1/2

 mω 1/4 c

~π

  r  mωc mωc i(px x + pz z) 2 exp − + x Hn x 2~ ~ ~

(2.3.84)

These results agree with the literature; the energy spectrum is referred to as “Landau Levels.” In fact, the whole of the problem alludes to “Landau Quantization,” or the quantization cyclotron orbits of charged particles subjected to an external magnetic field. The fact that the system boils down to the harmonic oscillator problem is no surprise; when we think of a classical particle in a static magnetic field, the particle follows a helical path, moving in the direction of the field and oscillating in a plane normal to the direction of propagation. The quantum case, however is more interesting. The momentum components in the x and y plane do not commute: [px, py ] 6= 0. Therefore it is impossible for us to precisely define the momentum in these two directions, which causes degeneracy. This is made more obvious when we notice that neither of these terms are present in (2.3.82). We also note that the degeneracy is directly proportional to the strength of the applied magnetic field B0 . ?

Problem 10.2 A linear harmonic oscillator is subjected to a spatially uniform external force F (t) = Cη(t)e−λt where λ is a positive constant and η(t) the Heaviside step function (A.23). If the oscillator is in the ground state at t < 0, calculate the probability of finding it at time t in an oscillator eigenstate with quantum number n. Assuming 1 C = (~mλ3 ) /2 , examine the variation of the transition probabilities with n and with the ratio λ/ω, ω being the natural frequency of the harmonic oscillator.

Solution This is an instance of a forced linear harmonic oscillator, whose Hamiltonian is given by (14.105): H=

p2 1 + mω2 x2 − qQ(t) − pP (t) 2m 2

(2.3.85)

Where Q(t) is a real-valued function of t and corresponds to an external time-dependent force (F (t) in our case). P (t) is a velocity dependent term and needn’t be considered for this problem. The Hamiltonian can also be written in terms of the raising and lowering operators:   1 H =~ω a† a + + f(t)a + f ∗ (t)a† (2.3.86) 2 Which is (14.106). The function f(t) is given by r r ~ ~mω f(t) = − Q(t) + i P (t) 2mω 2

W. Erbsen

HOMEWORK #3

r

~ Cη(t)e−λt 2mω r ~ √ ~mλ3 η(t)e−λt =− 2mω r ~2 λ3 =− η(t)e−λt 2ω =−

(2.3.87)

Since the oscillator is initially in the ground state at t < 0 and we are asked to find the probability at some time t, we should use the time evolution operator, |Ψ(t)i = T (t, 0)|Ψ(0)i, where T (t, 0) can be found from (14.148), letting t2 = t, t1 = 0 and H0 = ~ω(a† a + 1/2 ) from (14.121):  † T (t, 0) =eiβ(t,0) exp −ζ ∗ (t, 0)aeiωt + ζ(t, 0)a† e−iωt e−iω(a a+1/2)t (2.3.88) And to find the probability at some later time t of finding the oscillator in an eigenstate with quantum number n would be given by Pn (t) = |hn |Ψ(t)i |

2 2

= |hn|T (t, 0) | Ψ(0)i| 2  † = hn| eiβ(t,0) exp −ζ ∗ (t, 0)aeiωt + ζ(t, 0)a† e−iωt e−iω(a a+1/2)t |Ψ(0)i

(2.3.89)

At this point we should note that −iω(a† a+1/2)t

e

∞ X −iω(a† a + 1/2)t |Ψ(0)i = k!

=

k=0 ∞ X

k=0

|Ψ(0)i

(−iωa† a − iωt/2)k |Ψ(0)i k!

−iωt/2

=e

=e−iωt/2

∞ X (−iωa† a)k k=0 ∞ X k=0

−iωt/2

=e


k

k!

|Ψ(0)i

(−iω)k † k (a a) |Ψ(0)i k!

"

|Ψ(0)i +

=e−iωt/2 |Ψ(0)i

∞ X (−iω)k k=1

k!

†

k

(a a) |Ψ(0)i

# (2.3.90)

Because the number operator acting on the ground state (any number of times) has an eigenvalue of zero since you can’t go any lower than the ground state: a† a|Ψ(0)i −→ a† (a|Ψ(0)i) = 0. We also know that β(t, 0) is given by (14.140): Z t Z t n o 0 00 0 00 i 0 β(t, 0) = 2 dt dt00 f(t0 )f ∗ (t00 )e−iω(t −t ) − f ∗ (t0 )f(t00 )eiω(t −t ) (2.3.91) 2~ 0 0 Luckily, we don’t have to evaluate (2.3.91) since β is real, such that eiβ(t,0) is itself a phase and whose magnitude is 1, and it therefore doesn’t matter. The same argument can be applied to (2.3.90). Therefore, (2.3.89) becomes

141

CHAPTER 2: QUANTUM MECHANICS II

  2 Pn (t) = n exp −ζ ∗ (t, 0)aeiωt + ζ(t, 0)a† e−iωt Ψ(0)

(2.3.92)

Where ζ(t, 0) can be found from (14.138):

Z i t iωt0 ∗ 0 e f (t ) dt0 ζ(t, 0) = − ~ 0 r Z t 0 0 λ3 =i eiωt η(t0 )e−λt dt0 2ω 0 And recall that from (A.23), the Heaviside step function is  Z x 1 x>0 η(x) = δ(u) du = 0 x 0, then η(t0 ) = 1 in (2.3.93), according to (2.3.94). So, ζ(t, 0) becomes: r Z t 0 λ3 ζ(t, 0) =i ei(ω+iλ)t dt0 2ω 0 r 0 h 0 t =t λ3 1 =i ei(ω+iλ)t 2ω i(ω + iλ) t0 =0 r λ3 1 = ei(ω+iλ)t 2ω ω + iλ

(2.3.93)

(2.3.94)

(2.3.95)

Using this, we can now rewrite (2.3.92): * ("r # "r # ) + 2 λ3 1 λ3 1 i(ω+iλ)t † −iωt −i(ω+iλ)t iωt Pn (t) = n exp e a e − e ae Ψ(0) 2ω ω + iλ 2ω ω − iλ * ("r # "r # ) + 2 λ3 1 λ3 1 e−λt a† − eiλ a Ψ(0) (2.3.96) = n exp 2ω ω + iλ 2ω ω − iλ Define a new complex number α(t):

α(t) =

r

λ3 1 e−λt 2ω ω + iλ

For reasons to become clear momentarily. Then (2.3.96) becomes   2 Pn (t) = n exp α(t)a† − α∗ (t)a Ψ(0)

(2.3.97)

(2.3.98)

Which reminds us of the displacement operator (10.98): †

Dα =eαa Which has the property

−α∗ a

(2.3.99)

|αi =Dα |0i

(2.3.100)

Which is (10.101). Using (2.3.99) and (2.3.100) we can now rewrite (2.3.98): 2

Pn (t) = |hn |Dα | Ψ(0)i|

W. Erbsen

HOMEWORK #3

2

= |hn|αi|

(2.3.101)

Recall that we can define |αi in terms of eigenstates of the number operator with the Poisson distribution, (10.109), such that (2.3.101) becomes (10.110) Pn (t) =e|α|

2

|α|2n n!

(2.3.102)

Where we can find α from (2.3.97): r 2 λ3 1 2 −λt |α| = e 2ω ω + iλ ! r ! r 3 λ3 1 λ 1 = e−λt e−λt 2ω ω − iλ 2ω ω + iλ λ3 −2λt 1 e 2ω (ω − iλ)(ω + iλ) λ3 = e−2λt 2ω(ω2 + λ2 ) =

(2.3.103)

So that (2.3.102) becomes n  λ3 λ3 −2λt 1 −2λt e e 2 2 2 2 2ω(ω + λ ) n! 2ω(ω + λ )      n λ3 λ3 −2λt 1 −2nλt = exp e e 2ω(ω2 + λ2 ) n! 2ω(ω2 + λ2 )

Pn (t) = exp



(2.3.104)

We see that (2.3.104) can be easily rewritten as 

λ3 Pn (t) = exp 2ω(ω2 + λ2 )



λ3 2ω(ω2 + λ2 )

n

1 −2λt(n+1) e n!

(2.3.105)

Which answers the first part of the question. To see how the transition probability Pn (t) varies with n and λ/ω, it will be most instructive to create a graph. For the first graph we seek to see how Pn (t) varies with n, so we set ω = λ = 1 in (2.3.105) and construct a 3-D graph (Fig. (1)). Similarly, to see how Pn (t) varies with γ (which I have defined as γ = λ/ω), simply plot (2.3.105) with γ for various values of n and see how it looks (Figs. (2)-(4)). From Fig. (1) it is clear that as n increases the probability that the system will remain in the lowest state subjected to the time-dependent force F (t) decreases with higher n. Similarly, from Figs. (2)-(4) we can see that as γ increases the probability that the system will remain in the ground state decreases with respect to lower γ.

Problem 10.3

143

CHAPTER 2: QUANTUM MECHANICS II

Figure 2.3: Pn (t) vs γ

Figure 2.4: Pn (t) vs n for n = 0

Figure 2.5: Pn (t) vs n for n = 1

Figure 2.6: Pn (t) vs n for n = 2

If the term V (t) in the Hamiltonian changes suddenly (“impulsively”) between time t and t + ∆t, in a time ∆t short compared with all relevant periods, and assuming only that [V (t0 ), V (t00 )] = 0 during the impulse, show that the time development operator is given by " # Z i t+∆t 0 0 T (t + ∆t, t) = exp − V (t ) dt (2.3.106) ~ t Note especially that the state vector remains unchanged during a sudden change of V by a finite amount.

Solution Recall that (14.11) reads i~

dT (t, t0 ) = H(t)T (t, t0 ) dt

(2.3.107)

Where in our case we recognize that t0 → t and t → t + ∆t. We also note that H(t) = H0 + V (t). So, let’s solve (2.3.107) for T (t + ∆t, t): 1 1 dt(t + ∆t, t) = H(t0 ) dt0 T (t + ∆t, t) i~

W. Erbsen

HOMEWORK #4

"

i T (t + ∆t, t) = exp − ~ " i = exp − ~

Z

t+∆t

0

H0 + V (t ) dt

t

H0

Z

t+∆t

t

dt0 +

Z

0

#

t0 +∆t

V (t0 ) dt0

t

!#

(2.3.108)

Where we recognize that since the Hamiltonian changes suddenly, the first integral is approximately zero: R t+∆t 0 dt ≈ 0, and (2.3.108) becomes t " # Z i t+∆t T (t + ∆t, t) = exp − (2.3.109) V (t0 ) dt0 ~ t Which agrees with (2.3.106). To see that the state vector remains unchanged during a sudden change of V by a finite amount, recall that |ψ(x, t)i =T (t, 0)|ψ(x, 0)i −→ |ψ(x, t + ∆t)i = T (t + ∆t, t)|ψ(x, t)i

(2.3.110)

And with a sudden change ∆t ≈ 0, such that the right side of (2.3.110) becomes T (t, t)|ψ(x, t)i = |ψ(x, t)i

2.4

Homework #4

Problem 14.4 A linear harmonic oscillator in its ground state is exposed to a spatially constant force which at t = 0 is suddenly removed. Compute the transition probabilities to the excited states of the oscillator. Use the generating function for Hermite polynomials to obtain a general formula. How much energy is transferred?

Solution The Hamiltonian for a forced harmonic oscillator is given by (15.65) in Merzbacher as H = H0 + H 0 ⇒

p2 1 + mω2 x2 − xF (t) 2m 2

(2.4.1)

Where the perturbing term, H 0 = −xF (t) in this case. We now wish to employ perturbation theory to see how the energy changes when this force is added. Due to parity, the first order shift is zero: En(1) = hψn(0) |H 0 |ψn(0)i ⇒ 0 The reason for this is because the perturbation is odd and we are integrating over the entire universe. The second order shift is not zero, though. The easiest way to solve this problem is to follow the work done by Merzbacher on Pgs. 84-88. From (5.42),

145

CHAPTER 2: QUANTUM MECHANICS II

Z

∞

ψn∗ (x) · x · ψk (x) dx "r # r r ~ n n+1 = δk,n−1 + δk,n+1 mω 2 2

hn|x|ki =

−∞

(2.4.2)

And all the other terms vanish. So, we have En(2) =

X |hn|H 0|ki|2 (0)

n6=k

=

(0)

Ek − En

X |hn|F x|ki|2 (0)

(0)

En − Ek √ √ F 2 ~ | n δn,k−1 + n + 1 δn,k+1 |2 = (0) (0) 2mω Ek − En ( √ ) √ F 2 ~ | n δn,k−1 |2 | n + 1 δn,k+1 |2 = + (0) (0) (0) (0) 2mω Ek − En Ek − En √   √ F 2~ | n δn,k−1|2 | n + 1 δn,k+1 |2 + = 2mω (n + 3/2 ) ~ω − (n + 1/2 ) ~ω (n − 1/2 ) ~ω − (n + 1/2 ) ~ω   F 2~ n n+1 − = 2mω ~ω ~ω   F 2~ 1 − = 2mω ~ω F2 =− 2mω2 n6=k

(2.4.3)

We know that (2.4.3) is, of course, approximate. We can find the exact value in a few different ways. The method I am choosing to employ entails substituting the perturbing Hamiltonian into the TISE. We start by defining the following quantity: x0 = x −

F mω2

(2.4.4)

And we recall that the TISE is   ~2 ∂ 2 + V (x) ψ(x) = Eψ(x) − 2m ∂x2

(2.4.5)

We now recall from (2.4.1), that our potential in this problem is given by V (x) =

1 mω2 x2 − F x 2

Substituting (2.4.4) into (2.4.48),  2   1 F F 0 V (x) = mω2 x0 + − F x + 2 mω2 mω2     1 F2 2F x0 F 0 = mω2 x02 + 2 4 + − F x + 2 m ω mω2 mω2

(2.4.6)

W. Erbsen

HOMEWORK #4

1 = mω2 x02 + 2 1 = mω2 x02 + 2 1 = mω2 x02 − 2

mω2 F2 mω2 2F x0 F2 · 2 2+ · − F x0 − 2 2 m ω 2 mω mω2 2 2 F F + F x0 − F x0 − 2 2mω mω2 2 F 2mω2

Substituting (2.4.48) back into the TISE from (2.4.5),   ~2 ∂ 2 1 F2 2 02 − + mω x − ψ(x0 ) = Eψ(x0 ) 2m ∂x02 2 2mω2     ~2 ∂ 2 1 F2 2 02 0 − + mω x ψ(x ) = E + ψ(x0 ) 2m ∂x02 2 2mω2

(2.4.7)

(2.4.8)

If we allow for the following substitution, we see that our old harmonic oscillator differential equation and can jump to the solution: E0 = E +

F2 2mω2

(2.4.9)

Then we can rewrite (2.4.8) as   ~2 ∂ 2 1 2 02 − + mω x ψ(x0 ) = E 0 ψ(x0 ) 2m ∂x02 2

(2.4.10)

The solution of (2.4.10) is as we would expect,
En0 = n + 1/2 ~ω

Substituting (2.4.11) back into (2.4.9),



F2 F2 n + 1/2 ~ω = En + −→ En = n + 1/2 ~ω − 2 2mω 2mω2

(2.4.11)

(2.4.12)

The solution for the eigenfunctios of (2.4.10) have been solved exhaustively, and the solution is given by (5.39) in Merzbacher (with a few modifications) as ψn (x0 ) = √

 0  1/4  02  ξ ξ exp − Hn (ξ 0 ) 2 2n n! π 1

Where ξ 0 may be deduced from the original definition and our definition of x0 from (2.4.4), r mω 0 ξ0 = x ~ r   mω F = x− ~ mω2 r r mω mω F = x− ~ ~ mω2 F =ξ − √ ~mω3

(2.4.13)

147

CHAPTER 2: QUANTUM MECHANICS II

=ξ − a

(2.4.14)

Where a is constant. From (2.4.13), we may deduce that our ground state wave function is  0  1/4  02  ξ ξ ψ0 (x ) = exp − π 2 0

(2.4.15)

Our new goal is to find the transition probability, which is defined as Pn = |hψn (x)|ψ0 (x)i|2 . Using our previous definitions for ψn (x) and ψ0 (x0 ), we see that Pn =|hψn (x)|ψ0 (x)i|2 Z ∞ = ψn∗ (x)ψ0 (x) dx −∞

=

Z

∞

−∞

1 √ 2n n!

!  1   1/4  2  02 ! / ξ ξ ξ0 4 ξ exp − Hn (ξ) exp − dξ π 2 π 2

Substituting in our value for ξ 0 from (2.4.14) into (2.4.16), Z !  !! 2   1/4  2  1/ ∞ 1 ξ ξ (ξ − a) 4 (ξ − a)2 √ Pn = exp − Hn (ξ) exp − dξ −∞ 2 π 2 2n n! π

(2.4.16)

(2.4.17)

We now note that (2.4.17) strikes an uncanny resemblance to integral which led to the result obtained in Problem 6 in Chapter 5, where we are asked to find the transitional probability of a forced harmonic oscillator. In this problem, we found that   α2n 1 2 Pn = n exp − α (2.4.18) 2 n! 2 Where we defined α to be α=

r

mω a ~

Where, uncoincidentally, I have chosen a in my problem to coincide with a from Problem 5.6. Therefore, using our previous definition of a, we can use (2.4.18) to say Pn =

 n   1 F2 1 F2 · exp − 2n n! ~mω3 2 ~mω3

Problem 14.5 In the nuclear beta decay of a tritium atom (3 H) in its ground state, an electron is emitted and the nucleus changes into an 3 He nucleus. Assume that the change is sudden, and compute the probability that the atom is found in the ground state of the helium atom after the emission. Compute the probability of atomic excitation to the 2S and 2P states of the helium ion. How probable is excitation to higher levels, including the continuum?

W. Erbsen

HOMEWORK #4

Solution In β decay, we know that a neutron decays into a proton, electron, and electron anti-neutrino. For Tritium, 3 H →3 He + e− + ν¯, where ν¯ is an anti-neutrino. To find the probability that the atom is in He 2 the ground state of helium after the emission, we must find P = |hΨH 1,0,0(r, θ, φ)|Ψ1,0,0(r, θ, φ)i| , which is the overlap integral between the two states particular to the transition in question. The superscript H signifies the state of Tritium where Z = 1, and similarly He is the Helium ion with Z = 2. Since both 3 H and 3 He+ have only one electron, they are “Hydrogenic,” and we can therefore use (12.98) from Merzbacher: Ψ1,0,0 (r, θ, φ) =



Z3 πa3

 1/2

  Zr exp − a

(2.4.19)

Additionally, we must know the functional dependence of the radius (a) on Z, which is given by (4.72) in Griffiths: ? a=

4π0 ~2 me2

(2.4.20)

And m in our case will translate to the reduced mass µ, which will most certainly be different. The reduced mass of Tritium is µ(3 H) =

me · 3mp ≈ me me + 3mp

(2.4.21)

And similarly for µ(3 He+ ), which just says that a(3 H) = a(3 He+ ) = a0 . Now we must determine (2.4.19) for both Tritium and the 3 He nucleus. For Tritium Z = 1, so that (2.4.19) is just ΨH 1,0,0(r, θ, φ)

=



13 πa3

 1/2

h ri exp − a

(2.4.22)

  2r exp − a

(2.4.23)

And for 3 He+ Z = 2, so: ΨHe 1,0,0 (r, θ, φ) =



8 πa3

 1/2

Taking the square of the overlap integral between (2.4.22) and (2.4.23), He 2 P =|hΨH 1,0,0(r, θ, φ)|Ψ1,0,0(r, θ, φ)i| *  !  1/2  ! + 2 h r i  8  1/2 1 2r = exp − exp − πa3 a πa3 a "√ Z # 2   Z π Z ∞ 8 2π 3r 2 = dφ sin θ dθ r exp − dr πa3 0 a 0 0  Z ∞   2 8 3r 2 = 2 6 4π r exp − dr π a a 0 ) Z ∞ 2 8 = 2 6 · 16π 2 r 2 exp [−αr] dr α = 3/a π a 0

CHAPTER 2: QUANTUM MECHANICS II

 2 128 2 = 6 a α3  3 2 128 2a = 6 a 27 128 4a6 = 6 · a 729 512 = −→ P ≈ 0.702332 729

149

(2.4.24)

To find the probability of excitation into the 2S and 2P states, the process is very much the same. However instead of blindly implementing (12.98) we must manually find the excited eigenfunctions. This is not so bad, since we know that Ψn,l,m (r, θ, φ) = Rn,l (r)Ylm (θ, φ). The respective radial wave functions and spherical harmonics can be looked up in any QM textbook, such as chapter 4 in Griffiths. The wave function for 2S ends up being, 0 ΨHe 2,0,0 (r, θ, φ) =R2,0 (r)Y0 (θ, φ)   1/2   h r i 1r1 8 2r 1 = √ 2− exp − · a a 2 π 2 2 a3   1/2   h i 1 r r = 1− exp − 3 πa a a

(2.4.25)

Taking the inner product with (2.4.19) and squaring the result, He 2 P =|hΨH 1,0,0 (r, θ, φ)|Ψ2,0,0(r, θ, φ)i| *  ! +  1/2 h r i  1  1/2  h r i 2 1 r = exp − 1− exp − πa3 πa3 a a a D h i   h i E 2 1 r r r = 2 6 exp − exp − 1− π a a a a Z 2π   2 Z π Z ∞  1 r 2r 2 = 2 6 dφ sin θ dθ r 1− exp − dr π a a a 0 0 0 Z ∞     2 Z 16 2r 1 ∞ 3 2r r 2 exp − dr − r exp − dr = 6 a a a 0 a 0 Z ∞ 2 ) Z 16 1 ∞ 3 2 = 6 r exp [−αr] dr − r exp [−αr] dr α = 2/a a a 0 0  2 1 6 16 2 = 6 − a α3 a α4  2 16 2 3 3 3 = 6 a − a a 8 8 16 = −→ P = 0.25 64

(2.4.26)

To find the wave function corresponding to the 2P state we repeat the same process that we did in (2.4.25):

W. Erbsen

HOMEWORK #4

0 ΨHe 2,1,0 (r, θ, φ) =R2,1(r)Y1 (θ, φ)   1/2 h r i 1r 3 r 4 = exp − · cos θ 3a3 a a 2 π   1/2 h ri 1 1 = r exp − cos θ πa3 a a

(2.4.27)

And now, He 2 P =|hΨH 1,0,0(r, θ, φ)|Ψ2,1,0(r, θ, φ)i| *  ! + 2  1/2 h r i  1  1/2 1 h ri 1 exp − r exp − cos θ = πa3 πa3 a a a h r i h ri E 2 1 D = 4 exp − r exp − cos θ πa a a     Z 2π Z π Z ∞ 1 2r 3 = dφ sin θ cos θ dθ r exp − dr πa4 0 a 0 0

−→ P = 0

(2.4.28)

Because of the θ integral. The probability of excitation to higher states is just the probability that is left over from the previous states, eg 1.00 − 0.70 − 0.25 = 0.05. Sadly Tritium is not available to the public, and this is unfortunate for restless high-schoolers who wish to make a low-power inertial confinement fusion device, as Tritium is the ideal source ?, ?.

Problem 14.7 At t < 0 a system is in a coherent state |αi (eigenstate of a) of an oscillator and subjected to an impulsive interaction
i~ V (t) = ζ a2 − a†2 δ(t) (2.4.29) 2

Where ζ is a real-valued parameter. Show that the sudden change generates a squeezed state. If the oscillator frequency is ω, derive the time dependence of the variances   2   2 a + a† a − a† √ √ ∆ , and ∆ (2.4.30) 2 2

Solution We know that |αi is an eigenstate of a and satisfies a|αi = α|αi. What we want to do is go from our coherent state |αi to a squeezed state |α, ζi. We see that we arrive at a squeezed state by first finding the squeezed state operator, b, where we can say hα|a|αi =hα|U†bU|αi

151

CHAPTER 2: QUANTUM MECHANICS II

=hα, ζ|b|α, ζi

(2.4.31)

The easiest way to find b would be to utilize some identities made available by Merzbacher on pgs. 39-40. He says that if f(λ) = eλA Be−λA Then we can write p p [A, B] f(λ) = B cosh λ β + √ sinh λ β β

(2.4.32)

Which we can use only if β is constant, and the operators A and B satisfy [A, [A, B]] = βB. In our case, f(λ) → b(ζ), and A = (a2 − a†2 ) and B = a. To find out if we can use (2.4.32): [A, B] =[(a2 − a†2 ), a]

=[a2 , a] − [a†2 , a]

= − [a†2 , a]

= − a† [a† , a] − [a† , a]a†

= − a† (−1) − (−1)a†

=2a†

[A, [A, B]] =[(a2 − a†2 ), 2a†]

=2 [a2 , a† ] − [a†2 a† ]

(2.4.33)




=2 a[a, a†] + [a, a†]a =4a




(2.4.34)

It seems that we can in fact use (2.4.32), since both of our conditions are satisfied with λ = −ζ/2 and β = 4. We now have, p p [A, B] sinh λ β b(ζ) =a cosh λ β + √ β √ √ 2a† =a cosh −ζ/2 4 + √ sinh −ζ/2 4 4 † =a cosh ζ + a sinh ζ =a cosh ζ − a† sinh ζ

(2.4.35)

Following Merzbacher’s lead on pgs. 230-231, we define λ = cosh ζ and ν = sinh ζ, where λ here is different than the one used before. We can now rewrite (2.4.35): b = λa + νa†

(2.4.36)

b† = λa† + νa

(2.4.37)

And taking the Hermitian conjugate,

W. Erbsen

HOMEWORK #4

Since both λ and ν are real. Inverting these two equations a = λb − νb†

(2.4.38)

a† = λb† − νb

(2.4.39)

And we can now write 1 1 √ hα|a|αi = √ hα, ζ|b|α, ζi 2 2 1 = √ hα, ζ|λa + νa†|α, ζi 2 1 1 √ hα|a† |αi = √ hα, ζ|b†|α, ζi 2 2 1 = √ hα, ζ|λa† + νa|α, ζi 2

(2.4.40)

(2.4.41)

So, we can say that the impulsive action of (2.4.29) generates a squeezed state . We can find the variances from (2.4.30) by recalling that we can write   2 i
2
a + a† 1h √ ∆ = hα, ζ| a + a† |α, ζi − hα, ζ| a + a† |α, ζi2 (2.4.42a) 2 2 2   i
2
1h a − a† √ ∆ = hα, ζ| a − a† |α, ζi − hα, ζ| a − a† |α, ζi2 (2.4.42b) 2 2 We can evaluate (2.4.42a) and (2.4.42b) according to the previous formulas laid out in this problem. Unfortunately, I did not have time as I have been preparing a presentation for WildCorn. The framework has been laid out, the rest of this problem is very straightforward.

Exercise 15.5 From the definition (15.31) of the propagator and the Hermitian character of the Hamiltonian, show that K(r0 , r; t0, t) = K ∗ (r, r0 ; t, t0 )

(2.4.43)

linking the transition amplitude that reverses the dynamical development between spacetime points (r, t) and (r0 , t0 ) to the original propagator.

Solution We first recall (15.31), which states K(r, r0 ; t, t0 ) = hr|T (t, t0 )|r0 i

(2.4.44)

153

CHAPTER 2: QUANTUM MECHANICS II

Where the time evolution operator is given by (14.41) as   i 0 0 T (t, t ) = exp − H(t − t ) ~

(2.4.45)

Substituting (2.4.45) into (2.4.44) (letting ~ → 1 for convenience),    0 0 K(r, r0 ; t, t0) =hr|e−iH(t−t ) |r0 i ⇒ hr|e−iHt eiHt |r0 i

(2.4.46)

We now take the complex conjugate of both sides of (2.4.46), h  i∗  ∗  ∗ 0 0 K ∗ (r, r0; t, t0 ) = hr|e−iHt eiHt |r0 i ⇒ eiHt |r0 i hr|e−iHt

(2.4.47)

Recalling the Hermitian property of the Hamiltonian H ∗ = H † , we can rewrite (2.4.47) as    0 0 K ∗ (r, r0 ; t, t0) = hr0 |e−iHt eiHt |ri ⇒ hr0 |e−iH(t −t) |ri

(2.4.48)

Going back to (2.4.46), if we interchange r and r0 and also t and t0 , then (2.4.46) becomes    0 0 K(r0 , r; t0, t) = hr0 |e−iHt eiHt |ri ⇒ hr0 |e−iH(t −t)|ri

(2.4.49)

We now see that (2.4.49) is equal to (2.4.48), and we can now say K(r0 , r; t0, t) = K ∗ (r, r0 ; t, t0)

Exercise 15.8 Verify that (15.48) solves the time-dependent Schr¨ odinger equation and agrees with the initial contition (15.32). If the initial (t0 = 0) state of a free particle is represented by ψ(x, 0) = eik0 x , verify that (15.35) produces the usual plane wave.

Solution Equation (15.48) reads K(x; x0 ; t − t0 ) =

r

m exp 2πi~(t − t0 )



m(x − x0 )2 2i~(t − t0 )



We want to use this in the time-dependent Schr¨ odinger equation (15.33): " #  2 ∂ ~2 iq 0 0 i~ K(r, r ; t, t ) = − ∇ − A + V K(r, r 0 ; t, t0 ) ∂t 2m ~c The LHS of (2.4.51) is

(2.4.50)

(2.4.51)

W. Erbsen

HOMEWORK #4

 m(x − x0 )2 2i~(t − t0 )   1/2   −m(x − x0 )2 + i(t − t0 )~ im im(x − x0 )2 = · − · exp − 2(t − t0 )2 2π~t − 2πt0 ~ 2(t − t0 )~

∂ ∂ i~ K(r, r 0 ; t, t0) =i~ ∂t ∂t

r

m exp 2πi~(t − t0 )



(2.4.52)

While the RHS of (2.4.51) is # " #r "  2  2   iq ~2 iq ~2 m m(x − x0 )2 0 0 ∇ − A + V K(r, r ; t, t ) = − ∇− A +V − exp 2m ~c 2m ~c 2πi~(t − t0 ) 2i~(t − t0 )   r ~2 2 m(x − x0 )2 m =− ∇ exp 2m 2πi~(t − t0 ) 2i~(t − t0 )   1/2 im −m(x − x0 )2 + i(t − t0 )~ · − = 2(t − t0 )2 2π~t − 2πt0 ~   im(x − x0 )2 · exp − (2.4.53) 2(t − t0 )~ Since we have a free particle and A = 0 and V = 0. All derivatives were computed with Mathematica. We do notice, however that the left side of (2.4.51), (2.4.52) is equal to the right side, (2.4.53), therefore (2.4.50) solves the time-dependent Schr¨ odinger equation. We are now asked to verify that this agrees with the initial condition given by (15.32), which says: K(r, r 0 ; t, t) = δ(r − r 0 )

(2.4.54)

Which clearly satisfies the TDSE for t−t0 −→ t−t = 0 because of the completeness of the eigenfunctions: K(r, r 0 ; t, t0) =

∞ X

ψn∗ (r 0 )ψn (r)e−iEnt/~

(2.4.55)

ψn∗ (r 0 )ψn (r) = δ(r − r 0 )

(2.4.56)

n=0

And in our case, K(r, r 0 ; t, t) =

∞ X

n=0

For the last part of the question, (15.35) reads Z ψ(r, t) = K(r, r 0 ; t, t0 )ψ(r 0 , t0 )d3 r 0

(2.4.57)

Where in our case t0 = 0, ψ(x, 0) = eik0 x and K(r, r 0 ; t, t0) is given by (2.4.50). So, (2.4.57) becomes: Z ∞ ψ(r, t) = K(r, r 0 ; t, 0)ψ(r 0 , 0)d3r 0 −∞   Z ∞ r 0 m m(x − x0 )2 = exp · eik0 x dx0 2πi~t 2i~t −∞   Z ∞ r m m(x − x0 )2 = exp + ik0 x0 dx0 2πi~t 2i~t −∞

155

CHAPTER 2: QUANTUM MECHANICS II

r



 m(x2 − 2xx0 − x02 ) 0 exp = + ik0 x dx0 2i~t −∞ r   Z ∞ mxx0 mx02 m mx2 0 exp − − + ik0 x dx0 = 2πi~t −∞ 2i~t i~t 2i~t r   Z ∞ m mx02 mxx0 mx2 exp − = − + + ik0 x0 dx0 2πi~t −∞ 2i~t i~t 2i~t r    Z ∞   m mx  0 mx2 m 02 = exp − x + ik0 − x + dx0 2πi~t −∞ 2i~t i~t 2i~t r Z ∞  m = exp ax02 + bx0 + c dx0 2πi~t −∞ r  2  r −b m π = exp +c 2πi~t −a 4a m 2πi~t

Z

∞

(2.4.58)

Where I have set m 2i~t mx b =ik0 − i~t mx2 c= 2i~t

(2.4.59a)

a=−

(2.4.59b) (2.4.59c)

We now wish to evaluate (2.4.58) one bit at a time. The first factor is r

π = −a

r

π·

2i~t ⇒ m

r

2πi~t m

(2.4.60)

The exponential term becomes " #
2  − ik0 − mx −b2 mx2 i~t
+ exp + c = exp m 4a 2i~t 4 − 2i~t     1 2i~t 2mxk0 m2 x2 mx2 = exp · −k02 − + 2 2 + 4 m ~t ~ t 2i~t   2 2 2 ik ~t imx imx = exp − 0 − ixk0 + − 2m 2~t 2~t   ik0 (k0 ~t + 2mx) = exp − 2m 

(2.4.61)

Substituting (2.4.60) and (2.4.61) back into (2.4.58), r r     m 2πi~t ik0 ik0 ψ(r, t) = · exp − (k0 ~t + 2mx) −→ ψ(r, t) = exp − (k0 ~t + 2mx) 2πi~t m 2m 2m

(2.4.62)

Which is indeed a plane wave.

W. Erbsen

HOMEWORK #4

Exercise 15.9 Calculate |ψ(x, t)|2 from (15.50) and show that the wave packet moves uniformly and at the same time spreads so that " # ~2 t2 2 2 (∆x)t = (∆x)0 1 + (2.4.63) 4 4m2 (∆x)0

Solution Equation (15.50) reads: 1

ψ(x, t) = [2π(∆x)20 ]− /4



i~t · 1+ 2m(∆x)20

− 1/2



· exp 

2

x 2 ~t − 4(∆x) 2 + ik0 x − ik0 2m 0

1+

i~t 2m(∆x)20

 

(2.4.64)

To make calculating |ψ(x, t)|2 from (2.4.64) easier recall that if we have (a + ib)/(c + id), then a + ib c − id (a + ib)(c − id) · ⇒ c + id c − id c2 + d 2

(2.4.65)

Furthermore we remember that if f(x) = exp (e + if) then |f(x)| = exp (e − if) · exp (e + if) = exp (2e), so in further calculations I will recognize that the imaginary phase cancels out ahead of time to simplify the calculations. So, we have:  2   − 1/2 x2 2 ~t − 4(∆x) 2 + ik0 x − ik0 2m 1 i~t 2 2 − /4 0   |ψ(x, t)| = [2π(∆x)0 ] · 1+ · exp 2 i~t 2m(∆x)0 1 + 2m(∆x)2 0  2   − 1/2 2 x2 2 ~t + ik x − ik − 2 0 1 0 2m i~t 4(∆x)0   = [2π(∆x)20 ]− /4 · 1 + · exp (2.4.66) i~t 2m(∆x)20 1 + 2m(∆x) 2 0 | {z } | {z } 2 2 |ψI | |ψII | Such that

 − 1/2 2  − 1/2 1 ~2 t2 i~t 2 2 − /4 2 − 1/2 · 1+ |ψI | = [2π(∆x)0 ] · 1+ ⇒ [2π(∆x)0] 2m(∆x)20 4m2 (∆x)40

And also

  2 x2 2 ~t − 4(∆x) 2 + ik0 x − ik0 2m 2 0   |ψII | = exp i~t 1 + 2m(∆x)2 0   x2 x2 2 ~t 2 ~t − 4(∆x) − 4(∆x) 2 − ik0 x + ik0 2m 2 + ik0 x − ik0 2m 0 0  = exp  + i~t i~t 1 − 2m(∆x) 1 + 2m(∆x) 2 2 0

0

(2.4.67)

157

CHAPTER 2: QUANTUM MECHANICS II



x 2 ~t − 4(∆x) 2 − ik0 x + ik0 2m



x − 4(∆x) 2 +



x − 2(∆x) 2 +

= exp  = exp  = exp 

2

0

1−

i~t 2m(∆x)20

2

k0 ~xt 2m(∆x)20

0

−

·

1+

k02~2 t2 4m2 (∆x)20

1+ 2

0

1



i~t 2m(∆x)20 i~t 2m(∆x)20

1+

k 2 ~2 t2 k0 ~xt − 2m02 (∆x)2 m(∆x)20 0 ~2 t2 + 4m(∆x) 4 0

 2 k 2~2 t2 ~xt − 02m2 − x2 + k0m   = exp  ~2 t2 (∆x)20 1 + 4m(∆x) 4

2

+

x2 4(∆x)20 ~2 t2 4m(∆x)40

−

x 2 ~t − 4(∆x) 2 + ik0 x − ik0 2m

+

0

1+ k0~xt 2m(∆x)20



i~t 2m(∆x)20

−

k02 ~2 t2 4m2 (∆x)20

·



1− 1−

i~t 2m(∆x)20 i~t 2m(∆x)20

 





0

(2.4.68)

Combining (2.4.67) and (2.4.68), |ψ(x, t)|2 =|ψI |2 |ψII |2 1

=[2π(∆x)20]− /2



~2 t2 · 1+ 4m2 (∆x)40

− 1/2





 0 ~t − x − k2m + x2    · exp  ~2 t2 2 (∆x)0 1 + 4m(∆x)4 k0 ~t m

(2.4.69)

0

Which is the answer. By graphing this we can see that this does indeed move uniformly and spreads in time. If we substitute (2.4.63) into (2.4.69), 2

|ψ(x, t)| =

1 [2π(∆x)2t ]− /2



(k0 ~t/m − x) (−k0 ~t/(2m) + x/2) · exp (∆x)2t



(2.4.70)

With the definition of (∆x)t provided by (2.4.63), (2.4.70) is very clearly Gaussian (not that it didn’t look that way before). Please see attached Mathematica printout for plots.

Exercise 15.13 For the harmonic oscillator, derive ψ(x, t) directly from ψ(x, 0) by expanding the initial wave function, which represents a displaced ground state as in (15.60), in terms of stationary states. Use the generating function (5.33) to obtain the expansion coefficients and again to sum the expansion. Rederive (15.61).

Solution Using the definition of the propagator from (15.59),

K(x, x0 ; t) =

z 

α }| {  1/2   mω mω exp − (x2 cos(ωt) − 2xx0 + x02 cos(ωt)) 2πi~ sin(ωt) 2i~ sin(ωt)

W. Erbsen

HOMEWORK #4

 =α exp −

 mω (x2 cos(ωt) − 2xx0 + x02 cos(ωt)) 2i~ sin(ωt)

(2.4.71)

We can find ψ(x, t) using Z ∞ ψ(x, t) = K(x, x0 ; t)ψ(x0 , 0) dx0 −∞   Z ∞ n mω o mω =αN exp − (x2 cos(ωt) − 2xx0 + x02 cos(ωt)) · exp − (x0 − x0 )2 dx0 2i~ sin(ωt) 2~ −∞   Z ∞ mω mω 0 (x2 cos(ωt) − 2xx0 + x02 cos(ωt)) − (x − x0 )2 dx0 =αN exp − 2i~ sin(ωt) 2~ −∞   2  Z ∞ 0 02 mω x cos(ωt) 2xx x cos(ωt) 0 2 =αN exp + − − (x − x0 ) dx0 − 2~ i sin(ωt) i sin(ωt) i sin(ωt) −∞  2   Z ∞ x cos(ωt) 2xx0 x02 cos(ωt) mω 02 0 2 + − − x + 2x x0 − x0 dx0 =αN exp − 2~ i sin(ωt) i sin(ωt) i sin(ωt) −∞  Z ∞ mωx2 cos(ωt) 2mωxx0 mωx02 cos(ωt) =αN exp − + − 2i~ sin(ωt) 2i~ sin(ωt) 2i~ sin(ωt) −∞  mωx02 2mωx0 x0 mωx20 − + − dx0 2~ 2~ 2~  Z ∞ mωx02 mωx02 cos(ωt) mωxx0 mωx0 x0 =αN exp − − + + 2~ 2i~ sin(ωt) i~ sin(ωt) ~ −∞  mωx2 cos(ωt) mωx20 − − dx0 2i~ sin(ωt) 2~     Z ∞ mω mω cos(ωt) mωx mωx0 =αN exp − − x02 + + x0 2~ 2i~ sin(ωt) i~ sin(ωt) ~ −∞   mωx2 cos(ωt) mωx20 + − − dx0 (2.4.72) 2i~ sin(ωt) 2~ Where we set   mω mω cos(ωt) mω cos(ωt) − =− 1+ 2~ 2i~ sin(ωt) 2~ i sin(ωt)   mωx mωx0 mω x b= + = + x0 i~ sin(ωt) ~ ~ i sin(ωt)   mωx2 cos(ωt) mωx20 mω x2 cos(ωt) c =− − =− + x20 2i~ sin(ωt) 2~ 2~ i sin(ωt) a=−

(2.4.73) (2.4.74) (2.4.75)

Such that (2.4.72) becomes Z

∞

 exp ax02 + bx0 + c dx0 −∞ r  2  π b =αN exp − + c −a 4a

ψ(x, t) =αN

Where we have

(2.4.76)

159

CHAPTER 2: QUANTUM MECHANICS II

r

"   −1 # 1/2 π mω cos(ωt) 1+ = π −a 2~ i sin(ωt) "   −1 # 1/2 mω i sin(ωt) + cos(ωt) = π 2~ i sin(ωt)    1/2 2π~ i sin(ωt) = mω i sin(ωt) + cos(ωt)

(2.4.77)

And also, we can say that (   2   −1  )  2 mω x 1 mω cos(ωt) mω x2 cos(ωt) b 2 + c = exp + x0 · 1+ − + x0 exp − 4a ~ i sin(ωt) 4 2~ i sin(ωt) 2~ i sin(ωt)

We can simplify this further, (

 2   −1  ) x 1 mω i sin(ωt) + cos(ωt) mω x2 cos(ωt) m2 ω2 2 + x0 · − + x0 exp {%} = exp ~2 i sin(ωt) 4 2~ i sin(ωt) 2~ i sin(ωt)       mω x2 2xx0 i sin(ωt) mω x2 cos(ωt) 2 2 = exp − + + x − + x 0 0 2~ sin(2 ωt) i sin(ωt) i sin(ωt) + cos(ωt) 2~ i sin(ωt)        x2 2xx0 i sin(ωt) x2 cos(ωt) mω − + + x20 − + x20 = exp 2 2~ sin( ωt) i sin(ωt) i sin(ωt) + cos(ωt) i sin(ωt)      2  mω ix2 1 x cos(ωt) = exp − + 2xx0 + ix20 sin(ωt) − + x20 2~ sin(ωt) i sin(ωt) + cos(ωt) i sin(ωt)    2     mω − ix + 2xx0 + ix2 sin(ωt) (cos(ωt) − i sin(ωt)) 2 x cos(ωt) 0  sin(ωt) = exp · − + x20    2~ cos(ωt) + i sin(ωt) (cos(ωt) − i sin(ωt)) i sin(ωt)   2  mω − ix cos(ωt) + 2xx0 cos(ωt) + ix20 sin(ωt) cos(ωt) − x2 − 2ixx0 sin(ωt) + x20 sin(2 ωt) sin(ωt)  = exp −  2~ cos(2 ωt) + sin(2 ωt)  x2 cos(ωt) − − x20 i sin(ωt)   2 mω ix cos(ωt) = exp − + 2xx0 cos(ωt) + ix20 sin(ωt) cos(ωt) − x2 − 2ixx0 sin(ωt) + x20 sin(2 ωt)− 2~ sin(ωt)  x2 cos(ωt) 2 − − x0 i sin(ωt) n mω  o = exp 2xx0 cos(ωt) + ix20 sin(ωt) cos(ωt) − x2 − 2ixx0 sin(ωt) + x20 sin(2 ωt) − x20  2~  
mω ix2 sin(2ωt) = exp 2xx0 cos(ωt) + 0 − x2 − 2ixx0 sin(ωt) + x20 1 − cos(2 ωt) − x20 2~ 2    2 mω ix sin(2ωt) = exp 2xx0 cos(ωt) + 0 − x2 − 2ixx0 sin(ωt) − x20 cos(2 ωt) 2~ 2

W. Erbsen

HOMEWORK #4



  mω ix2 sin(2ωt) −x2 + 2xx0 cos(ωt) − x20 cos(2 ωt) − 2ixx0 sin(ωt) + 0 2~ 2    mωixx0 sin(ωt) imωx20 sin(2ωt) mω  2 2 2 x − 2xx0 cos(ωt) + x0 cos( ωt) − + = exp − 2~ ~ 4~   2 mω mωixx0 sin(ωt) imωx0 sin(2ωt) 2 (x − x0 cos(ωt)) − + = exp − 2~ ~ 4~    mω mωxx0 sin(ωt) mωx20 sin(2ωt) 2 = exp − (x − x0 cos(ωt)) − i − (2.4.78) 2~ ~ 4~ = exp

Going back to (2.4.76), implementing (2.4.77) (keeping (2.4.78) arbitrary for the time being), and substituting back in our value for α,  2  π b exp − + c −a 4a  1/2    1/2  2   2π~ i sin(ωt) b mω =N · exp − + c 2πi~ sin(ωt) mω cos(ωt) + i sin(ωt) 4a    1/2  2  mω 2π~ i sin(ωt) b =N · exp − + c 2πi~ sin(ωt) mω cos(ωt) + i sin(ωt) 4a    1/2  2  1 i sin(ωt) b =N exp − + c i sin(ωt) cos(ωt) + i sin(ωt) 4a   1/2  2  1 b =N exp − +c cos(ωt) + i sin(ωt) 4a    2    1/2 1 b iωt) iωt) =N +i e − e )/(2i) exp − + c 4a (eiωt) + eiωt))/2   1/2  2  1 b =N +c exp − 4a eiωt) /2 + e−iωt)/2 + eiωt)/2 − e−iωt)/2  1/2  2   b 1 =N exp − +c 4a eiωt)  2  b −iωt)/2 =N e exp − + c 4a

ψ(x, t) =N



mω 2πi~ sin(ωt)

 1/2 r

Now substituting (2.4.78) into (2.4.79), n mω  mω o iωt) mω 2 2 ψ(x, t) =N e− 2 exp − (x − x0 cos(ωt)) − i xx0 sin(ωt) − x0 sin(2ωt) 2~ ~ 4~   mω  iω  mω mω 2 2 (x − x0 cos(ωt)) − i xx0 sin(ωt) − x sin(2ωt) − t =N exp − 2~ ~ 4~ 0 2 n mω ω o mω mω 2 2 =N exp − (x − x0 cos(ωt)) − i t+ xx0 sin(ωt) − x0 sin(2ωt) 2~ 2 ~ 4~ Such that we finally arrive at n mω ω o mω mω 2 2 ψ(x, t) = N exp − (x − x0 cos(ωt)) − i t+ xx0 sin(ωt) − x0 sin(2ωt) 2~ 2 ~ 4~

(2.4.79)

(2.4.80)

161

CHAPTER 2: QUANTUM MECHANICS II

Which is identical to (15.61). ΨHx,tL 1.0

0.8

0.6

0.4

0.2

x -20

-10

10

20

Figure 2.7: Graph of |ψ(x, t)| vs. x for various values of t. We can also show that |ψ(x, t)| oscillates without any change in shape, we proceed with the usual mantra of complex analysis, recognizing that for a complex number z = |z|eiφ, and in our case: n mω o 2 |ψ(x, t)| =N exp − (x − x0 cos(ωt)) (2.4.81) 2~ As can be seen in Fig. (2.4), plotting |ψ(x, t)| for various times (30 in this case) clearly shows that |ψ(x, t)| does not change shape, eg it does not broaden, or otherwise distort. We also note that |ψ(x, t)| is normalized.

2.5

Homework #5

Part I Problem 1 Consider the “canonical ensemble,” defined by the statistical operator h  i−1 ρ = Tr e−H/kB T e−H/kB T


for the 1-D harmonic oscillator (with n-degenerate eigenvalues En = n + 1/2 ~ω with n = 0, 1, 2, ...)
a) Show that Tr ρ2 < 1. What happens for T → 0?

b)

Calculate the occupation probability of level hni.

(2.5.1)

W. Erbsen

HOMEWORK #5

c)

Calculate the averaged energy hEi and specific heat ∂/∂T hEi.

d)

Evaluate and discuss the limits of very large and very small temperatures in part c).

Solution As is the case most of the time in world of statistical mechanics, we must first find the partition function (the single-particle variety, in our case) Q1 =

∞ X

n=0


 exp −β~ω n + 1/2

∞  X = exp − 1/2 β~ω exp [−nβ~ω]

  = exp − 1/2 β~ω 

= exp − 1/2 β~ω

n=0 ∞ X

n=0

∞ X

n=0

n

{exp [−β~ω]}

1 1 − exp [−β~ω]

1 = exp [ 1/2 β~ω] − exp [− 1/2 β~ω] 1 = 2 sinh [ 1/2 β~ω] a)

(2.5.2)

We first note that the density operator, ρ, is given in the coordinate representation by    1/2    mω ~ω mωq 2 ~ω tanh tanh ρˆ = hq|ρˆ|qi = exp − (2.5.3) π~ 2kBT ~ 2kBT   Which is (5.4.25) from Pathria. To show that Tr ρˆ2 < 1, we evaluate ρˆ2 by squaring (2.5.3):      mω ~ω 2mωq 2 ~ω 2 ρˆ = tanh exp − tanh (2.5.4) π~ 2kB T ~ 2kB T Taking the trace of (2.5.4), we find that Z ∞  2 Tr ρˆ = ρˆ2 dq −∞      Z ∞  ~ω 2mωq 2 ~ω mω tanh exp − tanh dq = π~ 2kBT ~ 2kBT −∞  Z ∞    mω ~ω 2mωq 2 ~ω = tanh exp − tanh dq π~ 2kB T ~ 2kB T −∞ Let’s define:

Then (2.5.5) becomes

  mω ~ω α= tanh ~ 2kBT

(2.5.5)

163

CHAPTER 2: QUANTUM MECHANICS II

  α Tr ρˆ2 = π

Z

∞

−∞

  exp −2αq 2 dq

r α π = π 2α r α = 2π s   1 mω ~ω = · tanh (2.5.6) 2π ~ 2kB T  2 We can qualitatively show that Tr ρˆ < 1 by plotting (2.5.6) as a function of temperature (see  2 Fig. (1)). We see that Tr ρˆ approaches 0.4, which by most accounts is less than 1. Tr@Ρ^2D 0.5

0.4

0.3

0.2

0.1

0.0

T 0

5

10

15

20

25

  Figure 2.8: Graph of Tr ρˆ2 vs T for Problem 1. b)

In general, the way to find the average number of things in statistical mechanics is ∞ 1 X hni = n · exp [−nβEn ] Q1 n=0

(2.5.7)

In our case, (2.5.7) can be written hni =

∞ 
 1 X n · exp −β~ω n + 1/2 Q1 n=0

∞   1 X n · exp [−nβ~ω] exp − 1/2 β~ω Q1 n=0   ∞ exp − 1/2 β~ω X = n · exp [−nβ~ω] Q1 n=0

=

(2.5.8)

We now, seemingly arbitrarily, take the following derivative: ∂ {exp [−nβ~ω]} = − n~ω exp [−nβ~ω] ∂β Manipulating this a bit more, (2.5.9) becomes

(2.5.9)

W. Erbsen

HOMEWORK #5

n exp [−nβ~ω] = −

1 ∂ {exp [−nβ~ω]} ~ω ∂β

Substituting (2.5.10) back into (2.5.8),   ∞   exp − 1/2 β~ω X 1 ∂ hni = − {exp [−nβ~ω]} Q1 ~ω ∂β n=0 (∞ )   1 exp − 1/2 β~ω ∂ X =− exp [−nβ~ω] ~ω Q1 ∂β n=0     1 exp − 1/2 β~ω ∂ 1 =− ~ω Q1 ∂β 1 − exp [−β~ω] )  1 ( 1 exp − /2 β~ω exp [−β~ω] =− −~ω 2 ~ω Q1 (1 − exp [−β~ω]) ( )   exp − 3/2 β~ω 1 = Q1 (1 − exp [−β~ω])2

(2.5.10)

(2.5.11)

Substituting in to (2.5.11) our partition function from (2.5.2), we are left with (   ) 1  exp − 3/2 β~ω hni = 2 sinh /2 β~ω 2 (1 − exp [−β~ω]) c)

The average energy, hEi, can be found from the partition function from (2.5.2) as ∂ log [Q1 ] ∂β   1 ∂ =− log ∂β 2 sinh [ 1/2 β~ω]    ∂ log 2 sinh 1/2 β~ω = ∂β

hEi = −

Taking the simple derivative in (2.5.12), we arrive at for the average energy   hEi = 1/2 ~ω coth 1/2 β~ω

(2.5.12)

(2.5.13)

The specific heat is defined as

CV =

∂hEi ∂T

Substituting (2.5.13) into (2.5.14),      ∂ ~ω ~ω ~ω ~ω ~ω 2 CV = coth ⇒ · csch ∂T 2 2kBT 2 2kBT 2 2kBT

(2.5.14)

(2.5.15)

We see that (2.5.15) may be readily reduced to CV =

  ~2 ω2 ~ω 2 csch 4kB T 2 2kBT

(2.5.16)

165

CHAPTER 2: QUANTUM MECHANICS II

d)

By plotting (2.5.16) as a function of T (where we let ~ = ω = kB = 1, we can clearly see the trend at high and low temperatures (see Fig. (2)). As T → 0, CV → 0 , and as T → ∞, CV → 1 . For the low-temperature limit, we approach the quantum zero-point energy of E = 1/2 ~ω, which is independent of temperature. Therefore, in the low-temperature limit, the specific heat at constant volume approaches zero. For the high-temperature limit, CV approaches 1, which is what we would expect classically. CV 1.0

0.8

0.6

0.4

0.2

0.0 0.0

T 0.5

1.0

1.5

2.0

2.5

Figure 2.9: Graph of CV vs T for Problem 1.

Problem 2 An electron is in the spin state |↑iz with respect to the z-axis and you measure the electron’s spin with respect to a new quantization axis, given by the vector e = (0.6, 0, 0.8). Find the probability for measuring the electron’s spin in the +e direction.

Solution We first recall that S = ~/2 σ, where the pauli spin matrices are given by       0 1 0 −i 1 0 σx = , σy = , σz = 1 0 i 0 0 −1 Our electron is originally spin up in the z-direction, and our goal is to find the probability that it is spin up in the +e direction, where e = (0.6, 0, 0.8). To find this probability, we must calculate:  (e)† (z) 2 χ+ (2.5.17) P = χ+ (z)

Where it is known that χ+ is given simply by

(z)

χ+ =

  1 0

(2.5.18)

W. Erbsen

HOMEWORK #5

(e)

In order to find χ+ , we must diagonalize the spin matrix Se . We do that by taking the normal component with S: Se =ˆ e·S =0.6Sx + 0.8Sz      ~ 0 1 1 0 = 0.6 + 0.8 1 0 0 −1 2     ~ 0 0.6 0.8 0 = + 0.6 0 0 −0.8 2   ~ 0.8 0.6 = 2 0.6 −0.8

(2.5.19)

To find the eigenvalues and eigenvectors of (2.5.19), we take the determinate of the matrix and set it to zero: ~ 0.8 − λ 0.6 = 0 −→ (0.8 − λ) (−0.8 − λ) − (0.6) (0.6) = 0 −0.8 − λ 2 0.6 −→ − 0.64 − 0.8λ + 0.8λ + λ2 − 0.36 = 0 −→λ = ±1

To find the eigenvector corresponding to λ = +1,     ~ 0.8 − 1 ~ −0.2 0.6 0.6 ⇒ 0.6 −0.8 − 1 2 2 0.6 −1.8

(2.5.20)

To find the eigenvectors, ~ 2



    −0.2 0.6 α 0 = 0.6 −1.8 β 0

−→

−0.2α + 0.6β = 0 0.6α − 1.8β = 0

For the first line, we can let α = −0.6, and β = 0.2. Accordingly, our eigenspinor becomes   −0.3 (e) χ+ = 0.1 We can now substitute (2.5.21) and (2.5.18) back into (2.5.17) to find the probability,  †   2  
1 2 1 −0.3 ⇒ |−0.3|2 −→ P = 0.09 P = ⇒ −0.3 0.1 0.1 0 0

Part II

(2.5.21)

(2.5.22)

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CHAPTER 2: QUANTUM MECHANICS II

Exercise 15.25 If an ensemble E consists of an equal-probability mixture of two nonorthogonal (but normalized) states |Ψ1 i and |Ψ2 i with overlap C = hΨ1 |Ψ2 i, evaluate the Shannon mixing entropy H(E ) and the von Neumann entropy, S(ρ). Compare the latter with the former as |C| varies between 0 and 1. What happens as C → 0?

Solution The Shannon mixing entropy is given by (15.126): H(E ) = −

N X

pi log [pi ]

(2.5.23)

i=1

And the density operator is given by ρ=

N X i=1

pi ρi ⇒

N X i=1

pi |Ψi ihΨi |

And since the states have equal probabilities, p1 = p2 =

1 2

Substituting this into (2.5.23) we have H(E ) = −

2 X i=1

pi log [pi ] = −

  2 X log 1/2 = log [2] = 0.693147... 2 i=1

(2.5.24)

The Von Neumann entropy is defined by (15.128): S(ρ) = −

n X i=1

pi log [pi ] ⇒ −

n X i=1

hpi |ρ log [ρ] |pi i ⇒ −Tr(ρ log [ρ])

(2.5.25)

Where p1,2 is given in Exercise 15.24. So, p

|hΨ1 |Ψ2 i|2 1±C ⇒ 2 2 1+C 1−C −→ p1 = , p2 = 2 2 p1,2 =

1±

Putting this back into (2.5.25), we have         1+C 1+C 1−C 1−C S(ρ) = − log − log 2 2 2 2

(2.5.26)

For the functional dependence of S(ρ) on C please see Fig. (3). As we can see, when C = 0 then S(ρ) = log [2] = H(E ), however as C is increased S(ρ) decreases. Explicitly, as C → 0, we have S(ρ) = log [2]

W. Erbsen

HOMEWORK #5

SHΡL 0.7

0.6

0.5

0.4

0.3

0.2

0.1

C -1.0

0.5

-0.5

1.0

Figure 2.10: Graph of S(ρ) for Problem 15.25.

Problem 15.1 For a system that is characterized by the coordinate r and the conjugate momentum p, show that the expectation value of an operator F can be expressed in terms of the Wigner distribution W (r0 , p0 ) as ZZ hF i = hΨ|F |Ψi = FW (r0 , p0)W (r0 , p0 ) d3 r0 d3 p0 (2.5.27) where FW (r0 , p0) =

Z

0

eip ·r

00

/~

hr0 − r00 /2|F |r0 + r00 /2i d3 r00

(2.5.28)

and where the function W (r0 , p0 ) is defined in Problem 5 in Chapter 3. Show that for the special cases F = f(r) and F = g(p) these formulas reduce to those obtained in Problem 5 and 6 in Chapter 3, that is, FW (r0 ) = f(r0 ) and FW (p0 ) = g(p0 ).

Solution The Wigner distribution function defined in Problem 5 in Chapter 3 is:     Z 1 r00 r00 0 0 −ip0 ·r00 /~ ∗ 0 0 W (r , p ) = e ψ r − ψ r + d3 r00 (2π~)3 2 2 So, putting (2.5.28) and (2.5.29) into (2.5.27), ZZ hF i = FW (r0 , p0)W (r0 , p0) d3 r0 d3 p0  Z Z Z 0 00 = eip ·r /~ hr0 − r00 /2|F |r0 + r00 /2i d3 r00   Z 1 −ip0 ·r000 /~ ∗ 0 000 0 000 3 000 · e ψ (r − r /2) ψ (r + r /2) d r d3 r0 d3 p0 (2π~)3

(2.5.29)

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CHAPTER 2: QUANTUM MECHANICS II

=

1 (2π~)3

ZZ Z Z

eip ·(r 0

00

−r000 )/~

hr0 − r00 /2|F |r0 + r00 /2i · ψ∗ (r0 − r000/2) ψ (r0 + r000/2) d3 r000 d3 r00



d3 r0 d3 p0

(2.5.30)

We recognize that r000 integral collapses when looking at the exponent, which reduces to the delta function δ (r00 − r000). So, (2.5.30) becomes ZZ 1 hF i = hr0 − r00 /2|F |r0 + r00 /2iψ∗ (r0 − r00/2) ψ (r0 + r00 2) d3 r00d3 r0 (2.5.31) (2π~)3 We also recall from Chapter 17 that the Wigner Exckart Theorem holds if the determinate of J = 1, where J is      ∂ r00 ∂ r00 0 0 ! r −  ∂r0 r − 2 ∂r00 2  1 − 1/2   J=     ⇒ 1  ∂ 1 /2 r00 ∂ r00  0 0 r + r + 0 00 ∂r 2 ∂r 2 Taking the determinate of J, 1 − 1/2 Det {J} = ⇒ (1)( 1/2 ) − (− 1/2 )(1) ⇒ 1 X 1 1 /2

(2.5.32)

We can define two quantities r1 and r2 as

r00 2 r00 0 r2 =r + 2 r1 =r0 −

Using these, we can then rewrite (2.5.31) as ZZ hF i = hr1 |F |r2iψ∗ (r1 ) ψ (r2 ) d3 r1 d3 r2 ZZ = hr1 |F |r2ihr1 |r2 i d3 r1 d3 r2 ZZ = hr1 |F |r2iδ (r1 − r2 ) d3 r1 d3 r2 =hF i

(2.5.33)

Using the results from (2.5.32) and also the results from (2.5.33), we can therefore safely make the conclusion that hF i can be expressed in terms of W (r0 , p0 ) as in (2.5.27). The next parts of this problem ask us to show that FW (r0 ) = f(r0 ) and FW (p0 ) = g(p0 ). Starting with FW (r0 ), we rewrite (2.5.28) as Z 0 00 FW (r0 , p0 ) = eip ·r /~ hr0 − r00 /2|f(r)|r0 + r00 /2i d3 r00 Z 0 00 = eip ·r /~ f(r)hr0 − r00 /2|r0 + r00 /2i d3 r00

W. Erbsen

HOMEWORK #5

= =

Z

Z

0

00

/~

f(r)δ (r0 − r00/2 − (r0 + r00 /2)) d3 r00

0

00

/~

f(r)δ (r00) d3 r00

eip ·r eip ·r

(2.5.34)

From (2.5.34) we can safely conclude that FW (r0 , p0 ) = f(r0 ). To do the same for FW (p0 ), we proceed in very much the same way, except we expand g(p0 ) in p eigenstates: Z  Z 0 00 FW (r0 , p0 ) = eip ·r /~ hr0 − r00 /2|g(p0) |pihp| d3 p |r0 + r00 /2i d3 r00 ZZ 0 00 = eip ·r /~ g(p0 )hr0 − r00 /2|pihp|r0 + r00 /2i d3 r00 d3 p (2.5.35) Recalling from a previous homework set that hr|pi =

1 (2π~)

3/ 2

eir·p/~

(2.5.36)

Rewriting (2.5.35), and applying (2.5.36), ZZ 0 00 ∗ 0 0 FW (r , p ) = eip ·r /~ g(p0 )hr0 − r00 /2|pi (hr0 + r00 /2|pi) d3 r00 d3 p ( )( ) ZZ 1 1 i(r0−r00 /2)·p/~ −i(r0+r00 /2)·p/~ ip0 ·r00 /~ 0 = e g(p ) e e d3 r00 d3 p 3 3 / / (2π~) 2 (2π~) 2 )( ) ( ZZ 1 1 i(r0−r00 /2)·p/~ −i(r0+r00 /2)·p/~ ip0 ·r00 /~ 0 = e g(p ) e e d3 r00 d3 p 3 3 /2 /2 (2π~) (2π~) ZZ 0 00 0 00 0 00 1 = eip ·r /~ g(p0 )ei(r −r /2)·p/~ e−i(r +r /2)·p/~ d3 r00 d3 p 3 (2π~) ZZ 0 00 1 = g(p0 )ei(p −p)·r /~ d3 r00 d3 p (2.5.37) 3 (2π~) It is easy to see that the exponential term in (2.5.37) collapses to a δ-function, and we are left with nothing other than FW (r0 , p0) = g(p0 ).

Exercise 16.8 Prove that the only matrix which commutes with all three Pauli matrices is a multiple of the identity. Also show that no matrix exists which anti-commutes with all three Pauli matrices.

Solution Imagining some matrix A:

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CHAPTER 2: QUANTUM MECHANICS II



a b A= c d



To answer the first part of the question, we look to evaluate: [A, σ] = [A, σx] + [A, σy ] + [A, σz ] = 0

(2.5.38)

The first commutator is given by [A, σx] =Aσx − σxA       a b 0 1 a b 0 1 = − c d 1 0 c d 1 0     b a c d = − d c a b

(2.5.39)

While the second is [A, σy ] =i



   b −a −c −d −i d −c a b

(2.5.40)

And finally, the third commutator is [A, σz ] =

 a c

  −b a − −d −c

b −d



Plugging (2.5.39)-(2.5.41) back into (2.5.38),          b a c d b −a −c −d a [A, σ] = − +i −i + d c a b d −c a b c          b a b −a c d −c −d a = +i − +i + d c d −c a b a b c   b + ib + c − ic a − d − 2b = d + id − (a + ia) + 2c c − ic − (b + ib)

(2.5.41)

   −b a b − −d −c −d    −b a b − −d −c −d (2.5.42)

This leaves us with four equations and four unknowns: b(1 + i) + c(1 − i) = 0 a − 2b − d = 0 d(1 + i) − a(1 + i) + 2c = 0 c(1 − i) − b(1 + i) = 0

(2.5.43a) (2.5.43b) (2.5.43c) (2.5.43d)

Solving (2.5.43a) for b(1 + i), our four equations become b(1 + i) = −c(1 − i)

(1 − i) a + 2c −d = 0 (1 + i) d(1 + i) − a(1 + i) + 2c = 0

(2.5.44a) (2.5.44b) (2.5.44c)

W. Erbsen

HOMEWORK #5

c(1 − i) + c(1 − i) = 0 −→ c = 0

(2.5.44d)

Now that we know that we require that c = 0, (2.5.44a)-(2.5.44c) become b(1 + i) = 0 −→ b = 0 a − d = 0 −→ a = d d(1 + i) − a(1 + i) = 0

(2.5.45a) (2.5.45b) (2.5.45c)

From (2.5.44d), (2.5.45a) and (2.5.45b), it is clear that b = c = 0 and a = d. Therefore, we can now say that the only matrix which commutes with all three Pauli matrices is a multiple of the identity. We note that multiplying the commutator by a constant will not change this. The next part of the question asks us to show that no matrix exists which anti-commutes with all three Pauli matrices. To show this, we proceed in the same way we did in the last problem:     b a c d [A, σx] = − 6= 0 (2.5.46) d c a b     b −a −c −d [A, σy ] =i −i 6= 0 (2.5.47) d −c a b     a −b a b [A, σz ] = − 6= 0 (2.5.48) c −d −c −d Following the same logic as before, we can see that the anti-commutation condition will only exist if a = b = c = d = 0, so therefore no matrix exists which anti-commutes with all three Pauli matrices.

Exercise 16.15 In many applications, conservation laws and selection rules cause a decaying two-level system to be prepared
in an eigenstate of σz , say 10 , and governed by the simple normal Hamiltonian matrix H = a1 + bσx

(2.5.49)

where a and b are generally complex constants. In terms of the energy difference ∆E = E02 − E01 and the decay rates Γ1 and Γ2 , calculate the probabilities of finding the system at time t in state α or state β, respectively.

Solution
We first recall that from the prompt, we have α = 10 , and according to (2.5.49),     1 0 0 1 H =a +b 0 1 1 0     a 0 0 b = + 0 a b 0   a b = b a

(2.5.50)

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CHAPTER 2: QUANTUM MECHANICS II

We also note that from the prompt that ∆E = E02 − E01, and we recall from (16.81) from Merzbacher, Γ1 2 Γ2 E2 =E02 − i 2

E1 =E01 − i

(2.5.51a) (2.5.51b)

Taking the difference of (2.5.51a) and (2.5.51b), Γ2 Γ1 − E01 − i 2 2 ∆Γ =∆E − i 2

E2 − E1 =E02 − i

Where ∆Γ = Γ2 − Γ1 . We also recall from (16.83),      hβ|H|αi iE2 t iE1 t hβ|T (t, 0)|αi = exp − − exp − E2 − E1 ~ ~ We already know what hβ|H|αi is:

 b α a   
a b 1 = 0 1 b a 0

(2.5.52)

(2.5.53)

 a hβ|H|αi =β b †

=b

Substituting (2.5.52) and (2.5.54) into (2.5.53), ( " "
#
#) i E02 − i/2 Γ2 t i E01 − i/2 Γ1 t b hβ|T (t, 0)|αi = exp − − exp − ∆E − i/2 ∆Γ ~ ~

(2.5.54)

(2.5.55)

Taking the modulus squared of (2.5.56), 2

Pα→β = |hβ|T (t, 0)|αi|

  " "
#
# 2   i i i E02 − /2 Γ2 t i E01 − /2 Γ1 t |b| = − exp − exp −  |∆E − i/2 ∆Γ|2  ~ ~   " "
#
# 2   i i 2 i E02 − /2 Γ2 t i E01 − /2 Γ1 t |b| = − exp − exp −  ~ ~ (∆E)2 − 1/4 (∆Γ)2  2

(2.5.56)

The bracketed term in (2.5.56) can be reduced to " " "
#
#!
# i E02 + i/2 Γ2 t i E01 + i/2 Γ1 t i E02 − i/2 Γ2 t {...} = exp − exp exp − ~ ~ ~ "
#! i E01 − i/2 Γ1 t − exp − ~

W. Erbsen

HOMEWORK #6

        Γ2 (2i∆E − (Γ1 + Γ2 )) t (2i∆E − (Γ1 + Γ2 )) t Γ1 = exp − − exp − − exp + exp − ~ 2~ 2~ ~

(2.5.57)

Substituting (2.5.57) into (2.5.56),      |b|2 (2i∆E − (Γ1 + Γ2 )) t Γ2 − exp − Pα→β = exp − 2 2 ~ 2~ (∆E) − 1/4 (∆Γ)     (2i∆E − (Γ1 + Γ2 )) t Γ1 − exp + exp − 2~ ~ To find Pα→α , the only difference is that hα|H|αi = a, and so we have      |a|2 Γ2 (2i∆E − (Γ1 + Γ2 )) t Pα→α = exp − − exp − 2 2 ~ 2~ (∆E) − 1/4 (∆Γ)     (2i∆E − (Γ1 + Γ2 )) t Γ1 + exp − − exp 2~ ~

2.6

Homework #6

Exercise 16.23 If a constant magnetic field B0 , pointing along the z-axis, and a field B1 , rotating with angular velocity ω in the xy-plane, act in concert on a spin system (gyromagnetic ratio γ), calculate the polarization vector P as a function of time. Assume P to point in the z-direction at t = 0. Calculate the Rabi oscillations in the rotating frame, and plot the average probability that the particle has “spin down” as a function of ω/ω0 for a value of B1 /B0 = 0.1. Show that a resonance occurs when ω = −γB0 . (This arrangement is a model for all magnetic resonance experiments.)

Solution The processing magnetic field, B1 , can be expressed in terms of its angular frequency ω and the time t as B1 = B1 cos (ωt) ˆi ∓ B1 sin (ωt) ˆj

(2.6.1)

The sign in (2.6.1) determines the direction of rotation. Let’s let it be rotating clockwise, so that ∓ → −. Furthermore, we are told that the static field points purely in the z-direction, so that B0 = (0, 0, B0). The total magnetic field is of course ˆ B = B1 + B0 ⇒ B = B1 cos (ωt) ˆi − B1 sin (ωt) ˆj + B0 k

(2.6.2)

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CHAPTER 2: QUANTUM MECHANICS II

We now set our minds to finding the Hamiltonian matrix, H, which describes our system. We now recall that in general, the Hamiltonian for a particle with magnetic moment µ in some magnetic field B is given by H = −µ · B, where H is a matrix. For the case of a spin in a magnetic field, the magnetic moment becomes µ = −γ~σ/2, where σ denotes the respective Pauli spin matrix. Accordingly, the Hamiltonian matrix becomes i γ~ h ˆ H =− B1 cos (ωt) σx ˆi − B1 sin (ωt) σy ˆj + B0 σz k (2.6.3) 2 We can rewrite (2.6.3) by replacing our trig functions with exponentials:  

γ~ B1 iωt B1 iωt −iωt −iωt ˆ ˆ ˆ H =− e +e σx i − e −e σy j + B0 σz k 2 2 2i    

γ~ B1 1 iωt iωt −iωt −iωt ˆ ˆ ˆ e −e σy j + B0 σz k =− e +e σx i − 2 2 i We also recall that the Pauli spin matrices are     0 1 0 −1 , σy = i , σx = 1 0 1 0

Substituting in σx , σy and σz into (2.6.4),   
0 γ~ B1 iωt −iωt H=− e +e 1 2 2   
0 γ~ B1 =− eiωt + e−iωt 1 2 2

σz =



1 0 0 −1

(2.6.4)



    
0 −1 1 iωt 1 1 0 −iωt e −e i + B0 − 0 1 0 0 −1 i     
0 −1 1 1 0 iωt −iωt − e −e + B0 0 1 0 0 −1

We can combine the first two terms in (2.6.5) as    0 eiωt + e−iωt 0 − iωt eiωt + e−iωt 0 e − e−iωt

− eiωt − e−iωt 0








=2



0

e−iωt

eiωt 0

(2.6.5)



Using this, (2.6.5) becomes H =−

  γ~ 0 B1 −iωt e 2

eiωt 0

+ B0



1 0 0 −1

(2.6.6)

We now define the following quantities ω0 =

γB0 , 2

ω1 =

γB1 2

Accordingly, (2.6.6) becomes H = − ~ω1



0 e−iωt

   eiωt 1 0 − ~ω0 0 0 −1

At this point, we define the following time-dependent wave function   c (t) χ(t) = 1 c2 (t)

(2.6.7)

(2.6.8)

W. Erbsen

HOMEWORK #6

Whose first derivative with respect to time is of course   d c˙ 1 (t) χ(t) = c˙ 2 (t) dt

(2.6.9)

We now require that our Hamiltonian and associated wave functions satisfy the TDSE as i~

d χ(t) = Hχ(t) dt

Substituting (2.6.7)-(2.6.9) into (2.6.10) yields          c˙1 (t) 0 eiωt 1 0 c1 (t) i~ = −~ω1 −iωt − ~ω0 · c˙2 (t) e 0 0 −1 c2 (t)       c˙ (t) c2 (t)eiωt c1 (t) i 1 = − ω1 − ω0 c˙2 (t) c1 (t)e−iωt −c2 (t)       c˙1 (t) c2 (t)eiωt c1 (t) =iω1 + iω 0 c˙2 (t) c1 (t)e−iωt −c2 (t)

(2.6.10)

(2.6.11)

It is easy to see that (2.6.11) yields two coupled differential equations, c˙ 1 (t) =iω1 c2 (t)eiωt + iω0 c1 (t) −iωt

c˙ 2 (t) =iω1 c1 (t)e

− iω0 c2 (t)

(2.6.12a) (2.6.12b)

The two equations (2.6.12a) and (2.6.12b) constitute a set of coupled differential equations. The most straightforward way to solve these coupled equations is to evaluate the derivative of one of the equations and then plug in the other one. This was done via Mathematica, including the initial conditions. The result is 
 c1 (t) = cos (ω0 t/2) + i/ω0 ω0 − 1/2 ω sin (ω0 t/2) eiωt/2 (2.6.13a)   q 2 2 c2 (t) = i/ω0 (ω0 ) − ( 1/2 ω − ω0 ) sin (ω0 t/2) e−iωt/2 (2.6.13b)

q 2 Where I have defined ω0 = ( 1/2 ω − ω0 ) + ω12 . We can further simplify (2.6.13a) and (2.6.13b) by letting q 2 2 (ω0 ) − ( 1/2 ω − ω0 ) ω0 t ω0 − ω ωt α= , β = , γ= , δ= (2.6.14) 0 0 ω 2 ω 2 Using α, β and γ, we can now rewrite (2.6.13a) and (2.6.13b) as c1 (t) = [cos (β) + iγ sin (β)] eiδ −iδ

c2 (t) =iα sin (β) e

(2.6.15a) (2.6.15b)

We recall that the Pauli spin matrices from before, and to find the polarization vector P as a function of time, we must find each component by taking the inner product of the respective spin matrix. For instance, we can find Px as follows: Px =hχ(t)|σx |χ(t)i

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CHAPTER 2: QUANTUM MECHANICS II

=χ† (t)σx χ(t) =

  
0 1 c1 (t) 1 0 c2 (t)  
c1 (t) c∗1 (t) c2 (t)

c∗1 (t)

c∗2 (t)

= c∗2 (t)

=c∗2 (t)c1 (t) + c∗1 (t)c2 (t)

(2.6.16)

Substituting in to (2.6.16) the results from (2.6.15a) and (2.6.15b),
∗    ∗
Px = iα sin (β) e−iδ (cos (β) + iγ sin (β)) eiδ + (cos (β) + iγ sin (β)) eiδ iα sin (β) e−iδ = − iα sin (β) eiδ (cos (β) + iγ sin (β)) eiδ + iα (cos (β) − iγ sin (β)) e−iδ sin (β) e−iδ  =iα − sin (β) cos (β) e2iδ − iγ sin2 (β) e2iδ + sin (β) cos (β) e−2iδ − iγ sin2 (β) e−2iδ      =iα − sin (β) cos (β) e2iδ − e−2iδ − iγ sin2 (β) e2iδ + e−2iδ  =iα − sin (β) cos (β) [2i sin (2δ)] − iγ sin2 (β) [2 cos (2δ)] =2α sin (β) cos (β) sin (2δ) + 2αγ sin2 (β) cos (2δ) =2α sin (β) [cos (β) sin (2δ) + sin (β) cos (2δ)]

(2.6.17)

Similarly, we can find Py in the following way: Py =hχ(t)|σy |χ(t)i =χ† (t)σy χ(t)

  
0 −i c1 (t) c∗2 (t) i 0 c2 (t)  
c1 (t) = ic∗2 (t) −ic∗1 (t) c2 (t) = c∗1 (t)

=ic∗2 (t)c1 (t) − ic∗1 (t)c2 (t)

(2.6.18)

Substituting in the appropriate values,
∗ 
  ∗
Py =i iα sin (β) e−iδ cos (β) + iγ sin (β) eiδ − i (cos (β) + iγ sin (β)) eiδ iα sin (β) e−iδ = − α sin (β) eiδ (cos (β) + iγ sin (β)) eiδ + α (cos (β) − iδ sin (β)) e−iδ sin (β) e−iδ  =α sin (β) (cos (β) + iγ sin (β)) e2iδ + (cos (β) − iγ sin (β)) e−2iδ  =α sin (β) cos (β) e2iδ + iγ sin (β) e2iδ + cos (β) e−2iδ − iγ sin (β) e−2iδ      =α sin (β) cos (β) e2iδ + e−2iδ + iγ sin (β) e2iδ − e−2iδ =α sin (β) {cos (β) [2 cos (2δ)] + iδ sin (β) [2i sin (2δ)]} =2α sin (β) {cos (β) cos (2δ) − γ sin (β) sin (2δ)}

And finally, for Pz we have Pz =hχ(t)|σz |χ(t)i =χ† (t)σz χ(t) = c∗1 (t)

  
1 0 c1 (t) c∗2 (t) 0 −1 c2 (t)

(2.6.19)

W. Erbsen

HOMEWORK #6

= c∗1 (t)

−c∗2 (t)

 
c1 (t) c2 (t)

=c∗1 (t)c1 (t) − c∗2 (t)c2 (t)

(2.6.20)

With the coefficients, (2.6.20) turns in to  ∗  
∗
Pz = (cos (β) + iγ sin (β)) eiδ (cos (β) + iγ sin (β)) eiδ − iα sin (β) e−iδ iα sin (β) e−iδ = (cos (β) − iγ sin (β)) (cos (β) + iγ sin (β)) + α2 sin2 (β) = cos2 (β) + γ 2 sin2 (β) + α2 sin2 (β)
= cos2 (β) + sin2 (β) α2 + γ 2

(2.6.21)

So, the polarization vector, P, is given by (2.6.17), (2.6.19) and (2.6.21) as Px = 2α sin (β) [cos (β) sin (2δ) + sin (β) cos (2δ)] Py = 2α sin (β) {cos (β) cos (2δ) −
γ sin (β) sin (2δ)} Pz = cos2 (β) + sin2 (β) α2 + γ 2

Where α, β, γ and δ are defined by (2.6.14). To find the probability that the spin polarization is up as a function of time, we evaluate   2
c1 (t) 2 (z) ⇒ |c1 (t)|2 ⇒ c∗ (t)c1 (t) P↑ (t) = hχ+ |χ(t)i ⇒ 1 0 (2.6.22) 1 c2 (t)

And now we have

∗

P↑ (t) = [cos (β) + iγ sin (β)] [cos (β) + iγ sin (β)] −→ P↑ (t) = cos2 (β) + γ 2 sin2 (β)

(2.6.23)

And similarly,

And now

  2
c1 (t) 2 (z) ⇒ |c2 (t)|2 ⇒ c∗2 (t)c2 (t) P↓ (t) = hχ− |χ(t)i ⇒ 0 1 c2 (t) ∗

P↓ (t) = (iα sin (β)) (iα sin (β)) −→ P↓ (t) = α2 sin2 (β)

(2.6.24)

(2.6.25)

Resonance occurs when the probability from (2.6.25) equals 1. To evaluate the condition, we rewrite (2.6.25) by substituting back in the values of α and β from (2.6.14): q  1/ ω − ω )2 + (ω )2 2 ( 2 0 1 (ω1 ) P↓ (t) = sin2  t (2.6.26) 2 ( 1/2 ω − ω0 )2 + (ω1 )2 The maximum value of the sin2 (...) term is 1, so we are left with: 2

(ω1 ) ( 1/2 ω

2

2

− ω0 ) + (ω1 )

= 1 −→ 1/2 ω − ω0

Using (2.6.27) and our previous definition of ω0 , we find that


2

= 0 −→ ω = 2ω0

(2.6.27)

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CHAPTER 2: QUANTUM MECHANICS II

ω=2



γB0 2



−→ ω = γB0

We can rewrite (2.6.26) once again by expressing it in terms of ω/ω0 as suggested in the prompt, 2

P↓ (t) =

(ω1 /ω0 )

2

2

2

( 1/2 ω/ω0 − 1) + (ω1 /ω0 )

sin

"

1

/2

r

ω02

h

2 ( 1/2 ω/ω0 )

2

+ (ω1 /ω0 )

i

t

#

(2.6.28)

The time-averaged version of (2.6.28) is (recalling that hsin2 (...)i = 1/2 ) 2

hP↓i =

(ω1 /ω0 ) h i 2 2 2 ( 1/2 ω/ω0 − 1) + (ω1 /ω0 )

(2.6.29)

Please see Fig. (2.11) for a plot of (2.6.29). We can clearly see that resonance occurs at ω/ω0 = 2. 0.5

0.4

0.3

0.2

0.1

Ω -1

0

1

2

3

4 Ω0

Figure 2.11: Resonance plot of hP↓ i for Exercise 16.23.

Exercise 16.26 Show that if the incident spin state is a pure transverse polarization state, the scattering amplitudes for the initial polarizations P0 = ±ˆ n are g = ±ih and the scattering leaves the polarization unchanged, P = P0 .

Solution

W. Erbsen

HOMEWORK #6

We first recall from (16.130) from Merzbacher that
|g|2 − |h|2 P0 + i (g∗ h − gh∗ ) n ˆ + 2|h|2 P0 · n ˆ·n ˆ + (g∗ h + gh∗ ) P0 × n ˆ P= 2 2 ∗ ∗ |g| + |h| + i (g h − gh ) P0 · n ˆ

(2.6.30)

If the initial beam has transverse polarization with the scattering plane perpendicular to P0 where P0 = P0 n ˆ, (2.6.30) becomes
|g|2 + |h|2 P0 + i (g∗ h − gh∗ ) P= n ˆ (2.6.31) |g|2 + |h|2 + i (g∗ h − gh∗ ) P0 Which is of course (16.131) from Merzbacher. We now recall (16.129), which states   Tr Sρinc S† · σ P= dσ/dΩ

(2.6.32)

We now recall (16.127) and (16.125) from the book, which are (respectively): ρinc = 1/2 (1 + P0 · σ) ,

  dσ = Tr ρinc S† · S dΩ

Substituting these equations from (2.6.33) into (2.6.32), we have   Tr S · 1/2 (1 + P0 · σ) · S† · σ P= Tr [ 1/2 (1 + P0 · σ) · S† · S] With the scattering matrix S defined by (16.120) as  S11 S= S21

S12 S22

(2.6.33)

(2.6.34)



(2.6.35)

While the Pauli spin matrices are ˆ σ = σx ˆi + σy ˆj + σz k Where the individual spin matrices that constitute (2.6.36) are      0 1 0 −i 1 σx = , σy = , σz = 1 0 i 0 0

(2.6.36)

0 −1



(2.6.37)

We now have the tools to finish the problem, unfortunately I have run out of time. Using (2.6.31)-(2.6.37), the scattering amplitudes turn out to be g = ±i~ and the polarization vector unchanged, P = P0 .

Exercise 16.28 Assuming that P0 is perpendicular to the scattering plane, evaluate the asymmetry parameter A, defined as a measure of the right-left asymmetry by (dσ/dΩ)+ − (dσ/dΩ)− A= (2.6.38) (dσ/dΩ)+ + (dσ/dΩ)−

181

CHAPTER 2: QUANTUM MECHANICS II

where the subscripts + and − refer to the sign of the product P0 · n ˆ. Show that if P0 = ±ˆ n, the asymmetry A equals the degree of polarization P defined in (16.132). In particle polarization experiments, this quantity is referred to as the analyzing power.

Solution We first note that (16.132) reads P=P ·n ˆ=i

g∗ h − gh∗ n ˆ |g|2 + |h|2

(2.6.39)

We now recall that (16.128) from Merzbacher is dσ = |g|2 + |h|2 + i(g∗ h − gh∗ )P0 · n ˆ dΩ

(2.6.40)

Using (2.6.40) we can rewrite (2.6.38) as

|g|2 + |h|2 + i(g∗ h − gh∗ ) − |g|2 + |h|2 − i(g∗ h − gh∗ ) A= (|g|2 + |h|2 + i(g∗ h − gh∗ )) + (|g|2 + |h|2 − i(g∗ h − gh∗ ))

(2.6.41)

We note that (2.6.41) may be trivially reduced to A=i

g∗ h − gh∗ n ˆ |g|2 + |h|2

(2.6.42)

Where we see that (2.6.42) is identical to (2.6.39).

Exercise 16.33 For a mixed state given by the density matrix ρ=

1 75



 41 7 + 7i 7 − 7i 34

(2.6.43)

Check the inequalities (15.120), and calculate the eigenvalues and eigenstates. Evaluate the von Neumann entropy, and compare this with the outcome entropy for a measurement of σz .

Solution We first wish to evaluate the eigenvalues and eigenvectors of ρ. We find the eigenvectors in the usual way,    41/75 − λ (7 + 7i) /75 34 1 = 0 −→ 41 − λ − λ − 2 (7 + 7i) (7 − 7i) = 0 (7 − 7i) /75 34/75 − λ 75 75 75 (41)(34) 41 34 (49 − 49i + 49i + 49) −→ − λ − λ + λ2 − =0 2 75 75 75 752

W. Erbsen

HOMEWORK #6



2

−→λ − λ −

36 75

2

=0

(2.6.44)

Evaluating the quadratic equation from (2.6.44), q 2 1 ± 1 − 4 (36/75) 16 λ1,2 = −→ λ1 = , 1 25

λ2 =

9 25

Let’s find the eigenvectors. Starting with λ1 first,      41/75 − 16/25 (7 + 7i) /75 α 0 = (7 − 7i) /75 34/75 − 16/25 β 0      −7/75 (7 + 7i) /75 α 0 = (7 − 7i) /75 −14/75 β 0      −1 1 + i α 0 = 1 − i −2 β 0

(2.6.45)

(2.6.46)

The corresponding eigenvector for (2.6.46) is: 

χ1 =

 1+i 1

Normalizing according to the usual prescription N = [hχ1 |χ1 i]   1 1+i χ1 = √ 1 3

(2.6.47) −1

/2

, (2.6.47) becomes

Using λ2 from (2.6.45) and following the same prescription, we find that   1+i χ2 = 1 The normalization factor is, of course, the same. Therefore, our eigenvectors are   1 1+i χ = χ1 = χ2 = √ 1 3

(2.6.48)

(2.6.49)

(2.6.50)

The inequalities from (15.120) read  2 0 ≤ Tr ρ2 ≤ [Tr {ρ}] = 1 2

ρii ρjj ≥ |ρij |

(2.6.51a) (2.6.51b)

Let’s tackle showing (2.6.51a) first. We first note that from (2.6.43), we can easily find the trace: Tr {ρ} =

1 (41 + 34) ⇒ 1 75

Now, we need to find ρ2 : ρ2 =

1 (75)2



†   41 7 + 7i 41 7 + 7i 7 − 7i 34 7 − 7i 34

(2.6.52)

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CHAPTER 2: QUANTUM MECHANICS II

  41 7 + 7i 41 7 + 7i 7 − 7i 34 7 − 7i 34   2 2 1 (41) + (7 − 7i) 41 (7 + 7i) + 34 (7 − 7i) = 2 2 (75)2 41 (7 + 7i) + 34 (7 − 7i) (7 + 7i) + (34) =

1 (75)2



(2.6.53)

The trace of (2.6.53) is  2 2 2 2 Tr ρ2 = (41) + (7 − 7i) + (7 + 7i) + (34) ≈ 0.50

(2.6.54)

Therefore, from (2.6.52) and (2.6.54), we can now say that

 2 0 ≤ Tr ρ2 ≤ [Tr {ρ}] = 1

(2.6.55)

Furthermore, from (2.6.43) we can now say that

ρii ρjj = (41) (34) ⇒ 1394

(2.6.56)

While we also see that 2

|ρij | = (7 + 7i) (7 − 7i) ⇒ 98

(2.6.57)

Therefore, from (2.6.56) and (2.6.57) we can say that ρii ρjj ≥ |ρij |

2

The von Neumann entropy is defined by (15.129) as S(ρ) = −

N X

pi log [pi ]

(2.6.58)

i=1

Substituting in our eigenvalues to (2.6.58), we find that     16 16 9 9 S(ρ) = − log − log −→ S(ρ) = 0.653 25 25 25 25 The outcome entropy is defined by (15.131) in Merzbacher as H(K) = −

n X

p(Kj ) log [p(Kj )]

(2.6.59)

j=1

But we mustn’t forget (15.132), which reads p(Kj ) = Tr {|Kj ihKj |ρ} Which can be applied to our previous work as Tr {|K1 ihK1 |ρ} =

41 , 75

Therefore, applying (2.6.60) to (2.6.59), we find that

Tr {|K2 ihK2 |ρ} =

34 75

(2.6.60)

W. Erbsen

HOMEWORK #6

    41 41 34 34 H(K) = − log − log −→ H(K) = 0.688 75 75 75 75

(2.6.61)

(z)

To find the probability of measuring χ+ , we first must evaluate   
1 0 1 1+i (z) (z) hχ+ |σz |χ+ i = 1 − i 1 0 −1 1 3    1 1−i 1+i = 1 3 −1 1 = (1 − i) (1 + i) − 1 3 1 = 3 We see here that the outcome entropy is larger than the result from a measurement of σz . Realizing now that I may have misinterpreted this last part of the problem, we can also say that H(K) > S(ρ)

Problem 16.1a The spin-zero neutral kaon is a system with two basis states, the eigenstates of σz , representing a particle K 0 ¯ 0 : The operator σx = CP represents the combined parity (P ) and charge conjugation (C), and its antiparticle K ¯ 0 i. The dynamics is governed by the or particle-anti-particle, transformation and takes α = |K 0 i into β = |K Hamiltonian matrix Γ (2.6.62) H = M−i 2 Where M and Γ are Hermitian 2 × 2 matrices, representing the mass-energy and decay properties of the system, respectively. The matrix Γ is positive definite. A fundamental symmetry (under combined CP and time reversal transformations) requires that σxM∗ = Mσx and σx Γ∗ = Γσx . Show that in the expansion of H in terms of the Pauli matrices, the matrix σz is absent. Derive the eigenvalues and eigenstates of H in terms of the matrix elements of M and Γ. Are the eigenstates orthogonal?∗

Solution ¯ 0 i. Furthermore, we recall that the four matrices 1, Charge-parity reversal requires that σx |K 0 i = 1|K σx , σy and σz are linearally dependent, and any 2 × 2 matrix can be represented as A = λ0 1 + λ1 σx + λ2 σy + λ3 σz Which is (16.57) from Merzbacher. Applying this principle to our matrix M, ∗ What’s

your favorite color?

185

CHAPTER 2: QUANTUM MECHANICS II

M =a1 1 + b1 σx + c1 σy + d1 σz        1 0 0 1 0 −i 1 =a1 + b1 + c1 + d1 0 1 1 0 i 0 0   a1 + d1 b1 − ic1 = b1 + ic1 a1 − d1

0 −1

 (2.6.63)

We recall from the prompt that σxM∗ = Mσx , which is a requirement. Applying the LHS of this statement to our matrix M from (2.6.63),   ∗    0 1 a1 + d1 b1 − ic1 0 1 a1 + d1 b1 + ic1 σx M∗ = ⇒ (2.6.64) 1 0 b1 + ic1 a1 − d1 1 0 b1 − ic1 a1 − d1 We now compute the RHS of our statement:     a1 + d1 b1 − ic1 0 1 b − ic1 Mσx = ⇒ 1 b1 + ic1 a1 − d1 1 0 a1 − d 1

a1 + d 1 b1 + ic1



(2.6.65)

So, equating (2.6.64) and (2.6.65),  b − ic1 σx M = Mσx −→ 1 a1 + d 1 ∗

a1 − d 1 b1 + ic1





b − ic1 = 1 a1 − d 1

a1 + d 1 b1 + ic1



(2.6.66)

This yields two equations, b1 − ic1 + a1 − d1 = b1 − ic1 + a1 + d1 −→ −d1 = d1

a1 + d1 + b1 + ic1 = a1 − d1 + b1 + ic1 −→ d1 = −d1

(2.6.67a) (2.6.67b)

We see that in order for M to meet the requirement that σxM∗ = Mσx , we must have d1 = 0. We can rewrite M from (2.6.63) with this change,   a1 b1 − ic1 M= (2.6.68) b1 + ic1 a1 We can equally well apply this principle to the matrix Γ:   a2 + d2 b2 − ic2 Γ= b2 + ic2 a2 − d2

(2.6.69)

Following the logic which led us to (2.6.68), we may use (2.6.69) to deduce that d2 = 0 according to the requirement that σx Γ∗ = Γσx , and we can now write   a2 b2 − ic2 Γ= (2.6.70) b2 + ic2 a2 We now substitute (2.6.68) and (2.6.70) into (2.6.62),     i a1 b1 − ic1 a2 b2 − ic2 H= − b1 + ic1 a1 a2 2 b2 + ic2    i  i a1 b1 − ic1 / ·a /2 (b2 − ic2 ) = − i 2 2 i b1 + ic1 a1 /2 (b2 + ic2 ) /2 · a2

W. Erbsen

HOMEWORK #6

   a1 b1 − ic1 ia2 /2 (ib2 + c2 ) /2 − b1 + ic1 a1 (ib2 − c2 ) /2 ia2 /2   a1 − ia2 /2 b1 − ic1 − (ib2 + c2 ) /2 = b1 + ic1 − (ib2 − c2 ) /2 a1 − ia2 /2 =



(2.6.71)

We must now find the eigenvalues of H, in the usual way a1 − ia2 /2 − λ b1 − ic1 − (ib2 + c2 ) /2 b1 + ic1 − (ib2 − c2 ) /2 =0 a1 − ia2 /2 − λ
2

a1 − i/2 a2 − λ − b1 − ic1 − 1/2 (ib2 + c2 ) b1 + ic1 − 1/2 (ib2 − c2 ) =0

(2.6.72)

We can just go ahead and solve for λ, p λ = a1 − i/2 a2 ± (b1 − ic1 − 1/2 (ib2 + c2 )) (b1 + ic1 − 1/2 (ib2 − c2 ))

(2.6.73)

We can simplify (2.6.73) ever so slightly, giving eigenvalues λ1,2 = a1 − i/2 a2 −

q

2

2

(2b1 − ib2 ) + (2c1 − ic2 )

(2.6.74)

To find the eigenvectors, we substitute λ1 and λ2 from (2.6.74) into our Hamiltonian from (2.6.72) as      a1 − ia2 /2 − λ1,2 b1 − ic1 − (ib2 + c2 ) /2 c1 0 = (2.6.75) b1 + ic1 − (ib2 − c2 ) /2 a1 − ia2 /2 − λ1,2 c2 0 The eigenvectors were found using Mathematica. The result is ! √ (2b1−ib2 )2 +(2c1 −ic2 )2 ± 2b1−ib2+2ic1 +c2 χ1,2 = N 1

(2.6.76)

The normalization in (2.6.76) is obtained as we usually do:  
∗ c1 1 2 † 2 hχ1 |χ1 i = 1 −→ N χ1 · χ1 = 1 −→ N c1 c2 = 1 −→ N = p ∗ c2 c1 c1 + c∗2 c2

(2.6.77)

This normalization constant was determined from (2.6.77) using Mathematica, the result is q q  −1/2 (2b1 + ib2 )2 + (2c1 + ic2 )2 (2b1 − ib2 )2 + (2c1 − ic2 )2  N = + 2b1 + ib2 − 2ic1 + c2 2b1 − ib2 + 2ic1 + c2

(2.6.78)

Combining (2.6.76) and (2.6.78), the normalized eigenvectors are q

χ1,2 = 

2

2

(2b1 + ib2 ) + (2c1 + ic2 ) 2b1 + ib2 − 2ic1 + c2

+

q

2

2

(2b1 − ib2 ) + (2c1 − ic2 ) 2b1 − ib2 + 2ic1 + c2

 −1/2 

±

√

(2b1−ib2 )2 +(2c1−ic2 )2 2b1−ib2 +2ic1 +c2

To see if the eigenstates from (2.6.79) are orthogonal, we apply the following test:

1

!

(2.6.79)

187

CHAPTER 2: QUANTUM MECHANICS II

hχ1 |χ2 i = 0

(if orthogonal)

(2.6.80)

Using Mathematica, it is easy to show that (2.6.80) is not satisfied, and therefore χ1 and χ2 are not orthogonal .

2.7

Homework #7

Exercise 17.12 Derive the recursion relation (17.54), as indicated.

Solution We first recall three three relations defining various quantities specifying the behavior of the total angular momentum: J =J1 + J2 Jz =Jz1 + Jz2 J+ =J1+ + J2+

(2.7.1a)

We now recall the three equations from (17.27) of Merzbacher: Jz |j mi =m~|j mi p J+ |j mi = (j − m)(j + m + 1) ~ |j1 j2 j m + 1i p J− |j mi = (j + m)(j − m + 1) ~ |j1 j2 j m − 1i

(2.7.2a) (2.7.2b) (2.7.2c)

We also recall (17.52), which reads

|j1 j2 j mi =

X

m1 ,m2

|j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi

(2.7.3)

Which is the link between the direct product basis and the total angular momentum basis. According to Merzbacher, in order to obtain the recursion relations from (17.54), all we have to do is apply J+ and J− to (2.7.3). The behavior of J+ and J− is defined by (2.7.2b) and (2.7.2c), which may be more compactly redefined as p J± |j mi = (j ∓ m)(j ± m + 1) ~ |j1 j2 j m ± 1i (2.7.4) Similarly, we may generalize (2.7.1a) as

J± = J1± + J2± We can apply (2.7.5) to (2.7.3), which yields

(2.7.5)

W. Erbsen

HOMEWORK #7

J± {|j1 j2 j mi} = J±

(

X

m1 ,m2

)

|j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi

(2.7.6)

We must now evaluate the LHS and the RHS of (2.7.6) in turn. Starting with the LHS, we have p J± |j1 j2 j mi = (j ∓ m)(j ± m + 1) |j1 j2 j m ± 1i (2.7.7)

Which comes as no surprise. Evaluating the RHS of (2.7.6) poses a more arduous problem, which we shall see momentarily: X X J± |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi = (J1± + J2±) |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi m1 ,m2

=J1±

X

m1 ,m2

+J2±

X

m1 ,m2

m1 ,m2

|j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi (2.7.8)

We express the first term on the RHS of (2.7.8) as I and the second as II, and we can rewrite it as X J± |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi = I + II (2.7.9) m1 ,m2

So, now we must evaluate I and II in turn. In doing this, we must recognize that J1± and J2± only work on particle 1 and 2, respectively. Accordingly, X I =J1± |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi m1 ,m2

X p = (j1 ∓ m1 )(j1 ± m1 + 1) |j1 j2 m1 ± 1 m2 ihj1 j2 m1 m2 |j1 j2 j mi

(2.7.10)

m1 ,m2

And similarly, we can also evaluate II as X II =J2± |j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 j mi m1 ,m2

X p = (j1 ∓ m1 )(j1 ± m1 + 1) |j1 j2 m1 m2 ± 1ihj1 j2 m1 m2 |j1 j2 j mi

(2.7.11)

m1 ,m2

Substituting (2.7.10) and (2.7.11) into (2.7.9) and combining the result with (2.7.7), p

p (j ± m)(j ∓ m + 1) hj1 j2 m1 m2 |j m ∓ 1i = p(j1 ± m1 )(j1 ± m1 + 1) |j1 j2 m1 ± 1 m2 ihj1 j2 m1 m1 |j mi + (j2 ∓ m2 )(j1 ± m2 + 1) |j1 j2 m1 m2 ± 1ihj1 j2 m1 m2 |j mi

Where I have suppressed the presence of j1 j2 in the ket to save room. We note that this is identical to (12.54), and is what we were trying to show.

Exercise 17.13

189

CHAPTER 2: QUANTUM MECHANICS II

Show that a symmetry exists along the three quantum numbers j1 , j2 and j, and that in addition to (17.57) they satisfy the equivalent relations |j − j2 | ≤ j1 ≤ j + j2

and

|j − j1 | ≤ j2 ≤ j + j1

(2.7.12)

Solution We first recall (17.57) from Merzbacher, which reads |j1 − j2 | ≤ j ≤ j1 + j2

(2.7.13)

We note that the magnitude on the LHS of (2.7.18) can in fact be two different things: |j1 − j2 | = j1 − j2

or

|j1 − j2 | = j2 − j1

With these possibilities, in addition to the RHS of (2.7.18), we have three inequalities to play with: j ≤j1 + j2

(2.7.14a)

j1 ≤j + j2 j2 ≤j + j1

(2.7.14b) (2.7.14c)



(2.7.15)

Using (2.7.19) and (2.7.21), we can write j1 ≥ j − j2 j1 ≥ j2 − j

−→ j1 ≥ |j − j2 |

Using (2.7.22) and (2.7.20), we can now say |j − j2 | ≤ j1 ≤ j + j2 Similarly, using (2.7.19) and (2.7.20), we can declare that  j2 ≥ j − j1 −→ j2 ≥ |j − j1 | j2 ≥ j1 − j

(2.7.16)

And now combining (2.7.16) with (2.7.21), |j − j1 | ≤ j2 ≤ j + j1

Exercise 17.15 Verify the values of two special C-G coefficients: hj 1 j 0|j 1 j ji =

s

j j +1

(2.7.17a)

W. Erbsen

HOMEWORK #7

hj 2 j 0|j 2 j ji =

s

j(2j − 1) (j + 1)(2j + 3)

(2.7.17b)

Determine the value of the trivial C-G coefficient hj 0 m 0|j 0 j mi. How does it depend on the value of m?

Solution As was done in class, we start our journey by calculating the C-G coefficients by starting with the largest magnetic quantum number m = j1 + j2 and repeatedly applying the lowering operator J− = J1+ + J2−. So, ~ |j1 , j2 , m1 = j1 , m2 = j2 i 2(j1 + j2 )

(2.7.18)

J− |j1 , j2 , j = j1 + j2 , m = j1 + j2 i =|j1 , j2 , j = j1 + j2 , m = j1 + j2 − 1i

(2.7.19)

|j1 , j2 , j = j1 + j2 , m = j1 + j2 i = p

Applying J− to the LHS of (2.7.18) first,

And now applying J− to the RHS of (2.7.18), J− {N |j1 , j2 , m1 = j1 , m2 = j2 i} = (J1+ + J2−) {N |j1 , j2 , m1 = j1 , m2 = j2 i} =N {|j1 , j2 , m1 = j1 − 1, m2 = j2 i + |j1 , j2 , m1 = j1 , m2 = j2 − 1i} (2.7.20) Using these results, and substituting back in the normalization factor N , we find that s s j1 j2 |j = j1 + j2 , m = j1 + j2 − 1i = |m1 = j1 − 1, m2 = j2 i + |m1 = j1 , m2 = j2 − 1i j1 + j2 j1 + j2 (2.7.21) In our special case, (2.7.21) becomes hm1 = j, m2 = 0|j, m = ji =

s

j j+1

We continue along the same lines, and using Mathematica to take care of the matrix algebra we find that s j2 (2j2 − 1) hm1 = j1 − 2, m2 = j2 |j = j1 + j2 − 2, m = j1 + j2 − 2i = (2.7.22) (j1 + j2 − 1) (2j1 + 2j2 − 1) Applying the appropriate constants, (2.7.22) becomes hm1 = j, m2 = 2|j = j, m = ji =

s

j (2j − 1) (j + 1) (2j + 3)

We evaluate the last part of the question by simply using (2.7.18). The result is

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CHAPTER 2: QUANTUM MECHANICS II

1 hm1 = m, m2 = 0|j = j, m = mi = p ~ 2 (j1 + j2 )

1 −→ hm1 = m, m2 = 0|j = j, m = mi = √ ~ 2j

Exercise 17.16 Work out the results (17.63) from the recursion relations for C-G coefficients and the normalization and standard phase conditions.

Solution The results from (17.63) read





 j1 , 1/2 , m − 1/2 , 1/2 | j1 , 1/2 , j1 ± 1/2 , m = ±

s

j1 ± m + 1/2 2j1 + 1

s  j1 ∓ m + 1/2 j1 , 1/2 , m + 1/2 , − 1/2 | j1 , 1/2 , j1 ± 1/2 , m = 2j1 + 1

(2.7.23a) (2.7.23b)

While the recursion relation from (17.54) reads p p (j ± m)(j ∓ m + 1) hj1 j2 m1 m2 |j m ∓ 1i = p(j1 ± m1 )(j1 ± m1 + 1) |j1 j2 m1 ± 1 m2 ihj1 j2 m1 m1 |j mi + (j2 ∓ m2 )(j1 ± m2 + 1) |j1 j2 m1 m2 ± 1ihj1 j2 m1 m2 |j mi (2.7.24) Where it is important to recognize that the j1 j2 terms in the ket are still suppressed. We also notice that (2.7.23a) and (2.7.23b) are C-G coefficients, and must take the form of hj1 , j2 , m1 , m2 |j1 , j2 , j, mi. Therefore, comparing this to the LHS of (2.7.23a), we may deduce that j2 = 1/2 ,

m1 = m − 1/2 ,

m2 = 1/2 ,

j = j1 ± 1/2

(2.7.25)

We substitute these values into the recursion relation from (2.7.24). We denote the LHS of (2.7.24) as LHS, while the first and second terms of the RHS are RHS1 and RHS2 , respectively. Accordingly, have: p LHS = (j ± m)(j ∓ m + 1) hj1 , j2 , m1 , m2 |j1 , j2 , j, m ∓ 1i p = (j1 ± 1/2 ± m)(j1 ± 1/2 ∓ m + 1) hj1 , 1/2 , m − 1/2 , 1/2 |j1 , 1/2 , j1 ± 1/2 , m ∓ 1i p RHS1 = (j1 ± (m − 1/2 ))(j1 ± (m − 1/2 ) + 1) |j1 , 1/2 , m − 1/2 ± 1, 1/2 ihj1 , 1/2 , m − 1/2 , 1/2 |j1 , 1/2 , j1 ± 1/2 , mi p = (j1 ± m ∓ 1/2 )(j1 ± m ∓ 1/2 + 1) |j1 , 1/2 , m − 1/2 ± 1, 1/2 ihj1 , 1/2 , m − 1/2 , 1/2 |j1 , 1/2 , j1 ± 1/2 , mi p RHS2 = (j2 ∓ m2 )(j1 ± m2 + 1) |j1 , j2 , m1 , m2 ± 1ihj1 , j2 , m1 , m2 |j1 , j2 , j, mi p = ( 1/2 ∓ 1/2 )(j1 ± 1/2 + 1) |j1 , 1/2 , m − 1/2 , 1/2 ± 1ihj1 , 1/2 , m − 1/2 , 1/2 |j1 , 1/2 , j1 ± 1/2 , mi

W. Erbsen

HOMEWORK #7

Seemingly arbitrarily, we choose the top sign in the above equations: p LHS = (j1 + m + 1/2 ) (j1 − m + 3/2 ) hj1 , 1/2 , m − 1/2 , 1/2 |j1 + 1/2 , m1 i p RHS1 = (j1 + m − 1/2 ) (j1 + m + 1/2 ) |j1 , 1/2 , m + 1/2 , 1/2 ihj1 , 1/2 , m − 1/2 , 1/2 |j1 + 1/2 , mi RHS2 =0 Combining these equations and rearranging, s hm − 1/2 , 1/2 |j1 + 1/2 , mi =

j1 + m + 1/2 hm + 1/2 , 1/2 |j1 + 1/2 , m + 1i j1 + m + 3/2

Shifting the index according to m → m + 1, we now have s j1 + m + 3/2 hm + 3/2 , 1/2 |j1 + 1/2 , m + 2i hm + 1/2 , 1/2 |j1 + 1/2 , m + 1i = j1 + m + 5/2 Going back to (2.7.23a) and applying m = j1 + 1/2 to (2.7.26) and (2.7.27), we find that s j1 + m + 1/2 1 1 1 1 1 hj1 , /2 , m − /2 , /2 |j1 , /2 , j1 − /2 , mi = ± 2j1 + 1

(2.7.26)

(2.7.27)

(2.7.28)

Substituting back in the appropriate ± signs into (2.7.28),

s

 j1 ± m + 1/2 1 1 1 1 1 j1 , /2 , m − /2 , /2 | j1 , /2 , j1 ± /2 , m = ± 2j1 + 1

(2.7.29)

Which we note is identical to (2.7.23a). To derive (2.7.23b), we repeat the same process. Evaluating the LHS and RHS of our recursion relation yields p LHS = (j1 ± 1/2 ± m)(j1 ± 1/2 ∓ m + 1) hj1 , 1/2 , m + 1/2 , − 1/2 |j1 ± 1/2 , m ∓ 1i p RHS1 = (j1 ± (m + 1/2 ))(j1 ± (m + 1/2 ) + 1) |j1 , 1/2 , m + 1/2 ± 1, − 1/2 ihj1 , 1/2 , m + 1/2 , − 1/2 |j1 ± 1/2 , mi p RHS2 = ( 1/2 ∓ (− 1/2 ))(j1 ± (− 1/2 ) + 1) |j1 , 1/2 , m + 1/2 , − 1/2 ± 1ihj1 , 1/2 , m + 1/2 , − 1/2 |j1 ± 1/2 , mi The steps taken to arrive at (2.7.23b) are from henceforth identical to that which was done to derive (2.7.23a), and accordingly they will not be shown. From the above equations, we choose an arbitrary sign (which will be stuck back in at the end) in order to simplify the equation. We then shift the index of m, finagle the equations and evaluating the coefficient in terms of j1 and m, we get

 j1 , 1/2 , m + 1/2 , − 1/2 | j1 , 1/2 , j1 ± 1/2 , m = Which is of course (2.7.23b).

Exercise 17.18

s

j1 ∓ m + 1/2 2j1 + 1

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CHAPTER 2: QUANTUM MECHANICS II

Starting with the uncoupled basis (17.67), work out the 4 × 4 matrices Sz and S2 , and show by explicit diagonalzation that the singlet and triplet states are the eigenvectors with the appropriate eigenvalues.

Solution We recall that (17.67) from Merzbacher reads α1 ⊗ α2 = α1 α2 ,

α1 ⊗ β2 = α1 β2 ,

β1 ⊗ α2 = β1 α2 ,

β1 ⊗ β2 = β1 β2

(2.7.30)

We introduce the following shorthand notation, as indicated in lecture, which constitutes the product basis: {|++i, |+−i, |−+i, |−−i} Where the first term in each ket acts on particle 1, while the second acts on particle 2. We also recall that we can write the total spin operator S in terms of the single particle spin operators S1 and S2 as S = S1 ⊗ 1(2) + 1(1) ⊗ S2

(2.7.31)

We also note that when referring to the individual components of the spin operators, it will be most convenient to work in terms of the raising and lowering spin operators (S+ and S− , respectively). The raising and lowering operators are described in terms of the cartesian components by S± = Sz ± iSy So that S (Sx , Sy , Sz ) → S (S± , Sz ). We now recognize that the S acts on the individual spin kets according to Sz |+i = ~/2 |+i, Sz |−i = − ~/2 |−i S+ |+i = 0, S+ |−i = ~|+i S− |+i = ~|−i, S− |−i = 0 We also musn’t forget how the total components act on each particle, following the convention of (2.7.31), Sz =S1z ⊗ 1(2) + 1(1) ⊗ S2z

S+ =S1+ ⊗ 1(2) + 1(1) ⊗ S2+

S− =S1− ⊗ 1(2) + 1(1) ⊗ S2−

Accordingly, we can now write Sz |++i = ~/2 |++i + ~/2 |++i ⇒ ~|++i

Sz |+−i =0 Sz |−+i =0

Sz |−−i = − ~/2 |−−i − ~/2 |−−i ⇒ −~|−−i We can express (2.7.32a)-(2.7.32d) in matrix form as

(2.7.32a) (2.7.32b) (2.7.32c) (2.7.32d)

W. Erbsen

HOMEWORK #7

(Sz )m1 ,m2



~ 0 = 0 0

0 0 0 0

 0 0 0 0   0 0  0 −~

(2.7.33)

The value of the eigenvalues from (2.7.34) are transparent, λ1 = ~,

λ2 = −~

We now wish to find S2 , first representing it in the product basis: S2 =S21 + S22 + 2S1z S2z + S1+ S2− + S1− S2+

(2.7.34)

In order to evaluate all the matrix elements for S2 , let’s make a table: |++i |+−i |−+i |−−i

S21 /4 ~2 |++i 3 /4 ~2 |+−i 3 /4 ~2 |−+i 3 /4 ~2 |−−i 3

S22 /4 ~2 |++i 3 /4 ~2 |+−i 3 /4 ~2 |−+i 3 /4 ~2 |−−i 3

S1z /2 ~|++i 1 /2 ~|+−i − 1/2 ~|−+i − 1/2 ~|−−i 1

S2z /2 ~|++i − 1/2 ~|+−i 1 /2 ~|−+i − 1/2 ~|−−i 1

S1+ 0 0 ~|+i ~|+i

S2− ~|+i 0 ~|−i 0

S1− ~|+i ~|−i 0 0

S2+ 0 ~|+i 0 ~|−i

Substituting the values of the tabulated matrix elements into (2.7.34), we find that:

S2 |++i = 3/4 ~2 |++i + 3/4 ~2 |++i + 2 1/2 ~|++i 1/2 ~|++i ⇒ 2~2 |++i

S2 |+−i = 3/4 ~2 |+−i + 3/4 ~2 |+−i + 2 1/2 ~|+−i − 1/2 ~|+−i + ~2 |−+i ⇒ ~2 (|+−i + |−+)

S2 |−+i = 3/4 ~2 |−+i + 3/4 ~2 |−+i + 2 − 1/2 ~|−+i 1/2 ~|−+i + ~2 |+−i ⇒ ~2 (|−+i + |+−i)

S2 |−−i = 3/4 ~2 |−−i + 3/4 ~2 |−−i + 2 − 1/2 ~|−−i − 1/2 ~|−−i ⇒ 2~2 |−−i

(2.7.35a) (2.7.35b) (2.7.35c) (2.7.35d)

We can find the expectation values of S2 (i.e. the matrix elements) easily, and the non-zero elements are h+ + |S2 |++i =2~2

h+ − |S2 |+−i =~2

h+ − |S2 |−+i =~2

h− + |S2 |+−i =~2 h− + |S2 |−+i =~2

h+ + |S2 |++i =2~2 While all the other elements are of course zero. We are now in a place to write down the matrix S2 :

S2




m1 ,m2



2  2 0 =~  0 0

0 1 1 0

0 1 1 0

 0 0  0 2

(2.7.36)

Which is not diagonal. We can remedy this by diagonalizing the central matrix, allowing us to find the eigenvalues and eigenvectors of S2 .

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CHAPTER 2: QUANTUM MECHANICS II



~2 (S ) = 2 ~ 0 2

~2 ~2



Finding the eigenvalues, 2 ~ − λ 2 ~



~2 = 0 −→ ~2 − λ ~2 − λ − ~4 = 0 2 ~ −λ
−→λ λ − 2~2 = 0 −→ λ1 = 0, λ2 = 2~2

Whose corresponding eigenvectors are 1 V1 = √ (|+−i − |−+i) , 2

1 V2 = √ (|+−i + |−+i) 2

The eigenvectors which we did not include here are found from the first and fourth column of the matrix S2 . Combining the results, we arrive at a series of states whose total angular momentum is either 0 or 1, i.e. either singlet or triplet. The states are 1 |0 0i = √ (|+−i − |−+i) 2 |1 mi = |++i 1 |1 mi = √ (|+−i + |−+i) 2 |1 mi = |−−i

Problem 17.1 In the notation of (17.64) the state of a spin one-half particle with sharp total angular momentum j, m is jm jm aYj− 1/ + bYj+ 1/ 2 2

(2.7.37)

Assume this state to be an eigenstate of the Hamiltonian with no degeneracy other than that demanded by rotational invariance. a)

If H conserves parity, how are the coefficients a and b restricted?

b)

If H is invariant under time reversal, show that a/b must be imaginary.

c)

Verify explicitly that the expectation value of the electric dipole moment −e r vanishes if either parity is conserved or time reversal invariance holds (or both).

Solution

W. Erbsen

HOMEWORK #7

We first note that (17.64) reads Y`jm

=

`± 1/ , m Y` 2

1 = √ 2` + 1

p ± ` ± m + 1/2 p ` ∓ m + 1/2

m− 1/2

Y`

m+ 1/2

Y`

!

(2.7.38)

Where Y`jm denotes the common eigenstates of Jz and J2 . Furthermore, if parity is conserved, then we require that h i j− 1/ j+ 1/2 jm jm jm jm Pˆ aYj− = (−1) 2 aYj− bYj+ (2.7.39) 1/ + bY1+ 1/ 1/ + (−1) 1/ 2 2 2 2

Problem 17.4 The Hamiltonain of the positronium atom in the 1S state in a magnetic field B along the z-axis is to good approximation eB0 H = AS1 · S2 + (S1z − S2z ) (2.7.40) mc if all higher energy states are neglected. The electron is labeled as particle 1 and the positron as particle 2. Using 2 the coupled representation in which S2 = (S1 + S2 ) and Sz = S1z +S2z are diagonal, obtain the energy eigenvalues and eigenvectors and classify them according to the quantum numbers associated with constants of the motion. Empirically, it is known that for B = 0 the frequency of the 13 S → 11 S transition is 2.0338 × 105 MHz and that the mean lifetime for annihilation are 10−10 s for the singlet state (two photon decay) and 10−7 s for the triplet state (three-photon decay). Estimate the magnetic field strength B0 which will cause the lifetime of the longer lived m = 0 state to be reduced (“quenched”) to 10−8 s.

Solution We start by finding the constants of motion of the problem, which is found by evaluating the commutator of the Hamiltonian from (2.7.40) with various quantities and noting which equals zero. To do this, we recall the following commutation relations for S2 :  2  S , Sz =0  2   2  S , Sz1 6=0, S , Sz2 6= 0  2    S , S1 = S2 , S2 = 0  2 2  2 2  S , S1 = S , S2 = 0 And similarly, some relations for Sz are

[Sz , Sz1 ] = [Sz , Sz2 ] = 0     Sz , S21 = Sz , S22 = 0

From the prompt, we are told to use the coupled representation, where S2 = (S1 + S2 )2 and Sz = Sz1 +Sz2 are diagonal. We can evaluate the dot product in (2.7.40) by expanding the given definition of S2 :

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CHAPTER 2: QUANTUM MECHANICS II

2

S2 = (S1 + S2 ) ⇒ S21 + S22 + 2S1 · S2 −→ S1 · S2 = 1/2 S2 − S21 − S22 Substituting (2.7.41) into (2.7.40), the Hamiltonian becomes





eB0 H = 1/2 A S2 − S21 − S22 + (S1z − S2z ) mc

(2.7.42)

Let’s not evaluate some commutators to find the constants of motion:   
 H, S2 = 1/2 A S2 − S21 − S22 + ω (Sz1 − Sz2 ) , S2           = 1/2 A S2 , S2 − S21 , S2 − S22 , S2 + ω Sz1 , S2 − Sz2 , S2 6= 0 | {z } | {z } | {z } | {z } | {z } =0

=0

=0

6=0

=0

=0

=0

(2.7.43)

6=0


 [H, Sz ] = 1/2 A S2 − S21 − S22 + ω (Sz1 − Sz2 ) , Sz        = 1/2 A S2 , Sz − S21 , Sz − S22 , Sz + ω [Sz1 , Sz ] − [Sz2 , Sz ] = 0 | {z } | {z } | {z } | {z } | {z } =0 =0 =0 =0 =0  1
 [H, S1] = /2 A S2 − S21 − S22 + ω (Sz1 − Sz2 ) , S1        = 1/2 A S2 , S1 − S21 , S1 − S22 , S1 + ω [Sz1 , S1 ] − [Sz2 , S1 ] = 0 | {z } | {z } | {z } | {z } | {z } =0 =0 =0 =0 =0  1
 2 2 2 [H, S2] = /2 A S − S1 − S2 + ω (Sz1 − Sz2 ) , S2        = 1/2 A S2 , S2 − S21 , S2 − S22 , S2 + ω [Sz1 , S2 ] − [Sz2 , S2 ] = 0 | {z } | {z } | {z } | {z } | {z } =0

(2.7.41)

(2.7.44)

(2.7.45)

(2.7.46)

=0

So, S2 is not a constant of motion according to (2.7.43), while Sz , S1 , and S2 are constants of motion from (2.7.44)-(2.7.46). We now wish to switch to the following basis notation, {|++i, |+−i, |−+i, |−−i} We now wish to evaluate the diagonal matrix elements for H, first recalling that
H = 1/2 A S2 − S21 − S22 + ω (S1z − S2z ) |++i |+−i |−+i |−−i

S2 2~2 |++i 0 0 2~2 |−−i

S21 /4 ~2 |++i 3 /4 ~2 |+−i 3 /4 ~2 |−+i 3 /4 ~2 |−−i 3

S21 /4 ~2 |++i 3 /4 ~2 |+−i 3 /4 ~2 |−+i 3 /4 ~2 |−−i 3

S1z /2 ~|++i 1 /2 ~|+−i 1 − /2 ~|−+i − 1/2 ~|−−i 1

S2z /2 ~|++i − 1/2 ~|+−i 1 /2 ~|−+i − 1/2 ~|−−i 1

Substituting the tabulated elements into our Hamiltonian,
H|++i = 1/2 A 2~2 − 3/4 ~2 − 3/4 ~2 |++i ⇒ 1/4 A~2 |++i


H|+−i = 1/2 A − 3/4 ~2 − 3/4 ~2 |+−i + ω 1/2 ~ + 1/2 ~ |+−i ⇒ − 3/4 A~2 + ω~ |+−i


H|−+i = 1/2 A − 3/4 ~2 − 3/4 ~2 |−+i + ω − 1/2 ~ − 1/2 ~ |−+i ⇒ − 3/4 A~2 − ω~ |−+i
H|−−i = 1/2 A 2~2 − 3/4 ~2 − 3/4 ~2 |−−i ⇒ 1/4 A~2 |−−i

These calculations were clearly done in the direct product basis, whereas it could just as easily be done in the total angular momentum basis.

W. Erbsen

2.8

HOMEWORK #9

Homework #9

Problem 17.6 The magnetic moment operator for a nucleon of mass mn is µ = e (g` L + gS S) /2mn c, where g` = 1 and gS = 5.587 for a proton, g` = 0 and gS = −3.826 for a neutron. In a central field with an additional spin-orbit interaction, the nucleons move in shells characterized by the quantum numbers ` and j = ` ± 1/2 . Calculate the magnetic moment of a single nucleon as a function of j for the two kinds of nucleons, distinguishing the two cases j = ` + 1/2 and j = `− 1/2 . Plot j times the effective gyromagnetic ratio versus j, connected in each case the points by straight-line segments (Schmidt lines).

Solution We start by rewriting the magnetic moment operator µ: µ=

e (g`L + gs S) 2mn c

(2.8.1)

Where mn is the mass of the nucleon. If we choose the quantization axis of the interaction to be along the ˆ z-direction, the expectation value of the magnetic moment operator is given by hµz i = hα`sjmj |µz |α`sjmj i

(2.8.2)

We can rewrite (2.8.1) according to our chosen quantization axis as µz =

e (g`Lz + gs Sz ) 2mn c

(2.8.3)

Substituting (2.8.3) into (2.8.2), we have hµz i =

e hα`sjmj |(g` Lz + gs Sz )|α`sjmj i 2mn c

(2.8.4)

At this point, we recall that Jz = Lz + Sz → Lz = Jz − Sz . Substituting this into (2.8.4), e hα`sjmj |(g` (Jz − Sz ) + gs Sz )|α`sjmj i 2mn c e = hα`sjmj |(g`Jz + (gs − g` ) Sz )|α`sjmj i 2mn c e = {g`hα`sjmj |Jz |α`sjmj i + (gs − g` ) hα`sjmj |Sz |α`sjmj i} 2mn c

hµz i =

(2.8.5)

Our task is now to evaluate the expectation values hJz i and hSz i on the RHS of (2.8.5). The tool that allows us to do this is the Wigner Eckart Theorem (W.E.T.). Applying this to hSz i first, hα`sjmj |Sz |α`sjmj i = hj1mj 0|j1jmj ihαjkSkαji

(2.8.6)

We can apply the W.E.T. along the same lines to the expectation value of hJz i: hα`sjmj |Jz |α`sjmj i = hj1mj 0|j1jmj ihαjkJkαji

(2.8.7)

We now notice that the conveniently uncalculated C-G coefficients in (2.8.6) and (2.8.7) are in fact equal. We can therefore take the ratio of these two equations, thus ridding them from our lives:

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CHAPTER 2: QUANTUM MECHANICS II

hα`sjmj |Sz |α`sjmj i hj1mj 0|j1jmj ihαjkSkαji = hα`sjmj |Jz |α`sjmj i hj1mj 0|j1jmj ihαjkJkαji hα`sjmj |Sz |α`sjmj i hαjkSkαji = hα`sjmj |Jz |α`sjmj i hαjkJkαji hαjkSkαji −→ hα`sjmj |Sz |α`sjmj i = hα`sjmj |Jz |α`sjmj i hαjkJkαji

(2.8.8)

At this point, we make a small deviation to evaluate the reduced matrix elements. In order to do this, we must evaluate hJ · Si and hJ · Ji: hα`sjmj |J · S|α`sjmj i =hj1mj 0|j1jmj ihαjkJkαjihαjkSkαji hα`sjmj |J · J|α`sjmj i =hj1mj 0|j1jmj ihαjkJkαjihαjkJkαji

(2.8.9a) (2.8.9b)

Not surprisingly, we notice once again that the C-G coefficients are again the same in (2.8.9a) and (2.8.9b). Taking the ratio of these two equations, hα`sjmj |J · S|α`sjmj i hj1mj 0|j1jmj ihαjkJkαjihαjkSkαji = hα`sjmj |J · J|α`sjmj i hj1mj 0|j1jmj ihαjkJkαjihαjkJkαji hα`sjmj |J · S|α`sjmj i hαjkSkαji = hα`sjmj |J · J|α`sjmj i hαjkJkαji

(2.8.10)

Switching the LHS and RHS in (2.8.10), and substituting in the appropriate eigenvalues, (2.8.10) becomes   hαjkSkαji 1/2 j(j + 1)~2 + s(s + 1)~2 − `(` + 1)~2 = hαjkJkαji j(j + 1)~2 j(j + 1) + s(s + 1) − `(` + 1) = (2.8.11) 2j(j + 1) Substituting (2.8.11) back into (2.8.4), hα`sjmj |Sz |α`sjmj i =

j(j + 1) + s(s + 1) − `(` + 1) hα`sjmj |Jz |α`sjmj i 2j(j + 1)

(2.8.12)

And finally, substituting (2.8.12) into (2.8.5):     e j(j + 1) + s(s + 1) − `(` + 1) hµz i = g` hα`sjmj |Jz |α`sjmj i + (gs − g` ) hα`sjmj |Jz |α`sjmj i 2mn c 2j(j + 1) (2.8.13) Substituting in the appropriate eigenvalues for hJz i, (2.8.13) simplifies to    ej~ j(j + 1) + s(s + 1) − `(` + 1) hµz i = g` + (gs − g`) 2mn c 2j(j + 1) Now, in our particular case, each of the nucleons has s = 1/2 , and so (2.8.14) becomes    ej~ j(j + 1) − `(` + 1) + 3/4 hµz i = g` + (gs − g` ) 2mn c 2j(j + 1)

(2.8.14)

(2.8.15)

Now, we mustn’t forget that we are interested when we do not have an empty sub-shell, since a filled sub-shell must have a zero (total) angular momentum. Because j is always equal to an integer plus ± 1/2 ,

W. Erbsen

HOMEWORK #9

the occupancy of the sub-shell is always even. Since the nuclear spin is non-zero in our case, we have a contribution from the nuclei towards a magnetic dipole moment. This is justified since both the neutrons and the protons have an intrinsic magnetic moment. Additionally, we must consider that the proton is charged, meaning that it can produce an additional magnetic moment due to its orbital motion. We therefore look at two unique cases of the total angular momentum: when j = ` + 1/2 , and also when j = ` − 1/2 . Let’s look at j = ` + 1/2 first. To do this, we substitute the value back into (2.8.15):    e(` + 1/2 )~ (` + 1/2 )((` + 1/2 ) + 1) − `(` + 1) + 3/4 g` + (gs − g`) hµz i = 2mn c 2(` + 1/2 )((` + 1/2 ) + 1)    e(` + 1/2 )~ (` + 1/2 )(` + 3/2 ) − `(` + 1) + 3/4 g` + (gs − g`) = 2mn c 2(` + 1/2 )(` + 3/2 ))    e(` + 1/2 )~ `2 + 3/2 ` + 1/2 ` + 3/4 − `2 − ` + 3/4 = g` + (gs − g`) 2mn c 2(` + 1/2 )(` + 3/2 ))    ` + 3/2 e(` + 1/2 )~ g` + (gs − g`) = 2mn c 2(` + 1/2 )(` + 3/2 ))   e(` + 1/2 )~ 1 = g` + (gs − g`) 2mn c 2(` + 1/2 ) h e~ gs g` i g`(` + 1/2 ) + − = 2mn c 2 2 e~ h gs i = `g` + (2.8.16) 2mn c 2 We now wish to do the same thing for j = ` − 1/2 :    (` − 1/2 )((` − 1/2 ) + 1) − `(` + 1) + 3/4 e(` − 1/2 )~ hµz i = g` + (gs − g`) 2mn c 2(` − 1/2 )((` − 1/2 ) + 1)    e(` − 1/2 )~ (` − 1/2 )(` + 1/2 ) − `(` + 1) + 3/4 = g` + (gs − g`) 2mn c 2(` − 1/2 )(` + 1/2 )    e(` − 1/2 )~ 1 `(` + 1) + 3/4 = g` + (gs − g`) − 2mn c 2 2(` − 1/2 )(` + 1/2 )   e~ g` − 2gs (` + 1) `g` + = 2mn c 2(2` + 1)      e~ 1 `+1 = g` ` + − gs 2mn c 2(2` + 1) 2` + 1   i 1 h e~ ` − /2 gs = (` + 1)g` − 2mn c ` + 1/2 2 So, in summary, we have from (2.8.16) and (2.8.17) that  e~ n gs o   j = ` + 1/2  2m c `g` + 2 , n   hµz i =  e~ ` − 1/2 h gs i   (` + 1)g` − , j = ` − 1/2 2mn c ` + 1/2 2 We can rewrite (2.8.18) slightly as

(2.8.17)

(2.8.18)

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CHAPTER 2: QUANTUM MECHANICS II

 e~ n gs o   j = ` + 1/2  2m c `g` + 2 , n   hµz i = e~  j h gs i   (` + 1)g` − , j = ` − 1/2 2mn c j + 1 2

A plot of jg` is plotted vs. j on attached printout.

Problem 17.9 A system that is invariant under rotation is perturbed by a quadrupole interaction V =

2 X

Cq T2q

(2.8.19)

q=−2

where the Cq are constant coefficients and T2q are the components of an irreducible spherical tensor operator, defined by one of its components: 2

T22 = (Jx + iJy )

(2.8.20)

a)

Deduce the conditions for the coefficients Cq if V is to be Hermitian.

b)

Consider the effect of the quandrupole perturbation on the manifold of a degenerate energy eigenstate of the unperturbed system with angular momentum quantum j, neglecting all other unperturbed energy eigenstates. What is the effect of the perturbation on the manifold of an unperturbed j = 1/2 state?

c)

If C±2 = C0 and C±1 = 0, calculate the perturbed energies for a j = 1 state, and plot the energy splittings as a function of the interaction strength C0 . Derive the corresponding unperturbed energy eigenstates.

Solution a)

In order for V to be Hermitian, we require that it be self-adjoint. From (4.5) in Merzbacher, V† =V Where self-adjoint implies that the matrix is its own conjugate transpose, e.g. V

∗

=V

(2.8.21)

We must also recall that applying the self-adjoint condition to an operator with two elements, we have:
∗ ∗ ∗ ∗ (A · B) = B · A ⇒ B · A In order to evaluate the specific conditions we must require for each of the components, we must expand (2.8.19) and apply the Hermitian requirement from (2.8.21): V =C−2 T2−2 + C−1 T2−1 + C0 T20 + C1 T21 + C2 T22

(2.8.22)

W. Erbsen

HOMEWORK #9

Applying the self-adjoint condition from (2.8.21) to (2.8.22),
∗
∗ ∗ ∗ ∗ ∗ V = C−2 T2−2 + C−1 T2−1 + (C0 T20 ) + (C1 T21 ) + (C2 T22 ) −2 ∗

=T 2

∗

−1 ∗

C −2 + T 2

∗

0∗

∗

1∗

∗

2∗

∗

C −1 + T 2 C 0 + T 2 C 1 + T 2 C 2

(2.8.23)

The coefficients Cq in (2.8.23) are non-conjugatable, as coefficients tend not to be. Therefore, (2.8.23) can be rewritten as V

∗

−2 ∗

=T 2

−1 ∗

∗ C−2 + T2

0∗

1∗

2∗

∗ C−1 + T 2 C0∗ + T 2 C1∗ + T 2 C2∗

∗ ∗ =T2−2 † C−2 + T2−1 † C−1 + T20 † C0∗ + T21 † C1∗ + T22 † C2∗

(2.8.24)

q∗

Where I have rewritten the tensor operators as T k = Tkq † , for reasons that will become clear momentarily. In the last homework, we showed that Tkq † = (−1)q Tkq ∗ , which applied to (2.8.24) yields
∗ ∗
∗ ∗
∗
∗
∗ ∗ V = T2−2 C−2 + −T2−1 C−1 + T20 C0∗ + −T21 C1∗ + T22 C2∗ ∗ ∗ =T2−2 ∗ C−2 − T2−1 ∗ C−1 + T20 ∗ C0∗ − T21 ∗ C1∗ + T22 ∗ C2∗

(2.8.25)

Equating (2.8.22) and (2.8.25), ∗ ∗ C−2 T2−2 + C−1 T2−1 + C0 T20 + C1 T21 + C2 T22 =T2−2 ∗ C−2 − T2−1 ∗ C−1 + T20 ∗ C0∗ − T21 ∗ C1∗ + T22 ∗ C2∗ (2.8.26)

We note that (2.8.26) can only be satisfied if: C0 ∈ R ∗ C1 = C−1 ∗ C2 = C−2 b)

In order to study the effect of the quadrupole perturbation on the manifold of a degenerate energy eigenstate of the unperturbed system with generic total angular momentum quantum number j, we must investigate the selection rules for an arbitrary irreducible 2nd rank tensor operator. We first recall the two main stipulations on q and k, defined by (17.88) and (17.89), respectively as: q = m0 − m 0

|j − j | ≤ k ≤ j + j

(2.8.27a) 0

(2.8.27b)

We also recollect the Wigner Eckart Theorem from (17.87): hα0 j 0 m0 |Tkq |αjmi = hjkmq|jkj 0 m0 ihα0 j 0 kTk kαji

(2.8.28)

Where we require that the bounds on Tkq are limited by ±q. For a second rank irreducible tensor operator, we have q = 2, and therefore: q = −2, −1, 0, 1, 2 Now, applying the possible values of q to (2.8.27a) first, we find that ∆m = − 2, −1, 0, 1, 2 −→ ∆m = 0, ±1, ±2 And similarly for (2.8.27b),

(2.8.29)

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CHAPTER 2: QUANTUM MECHANICS II

|∆j| ≤ −2, −1, 0, 1, 2 ≤ ∆j −→ ∆j = 0, ±1, ±2

(2.8.30)

It is easy to see that (2.8.29) and (2.8.30) express the relevant selection rules for an irreducible second-rank tensor operator Tq2 . Another important result worth noting is that when both of these conditions are satisfied, we require that hα0 j 0 m0 |Tkq |αjmi = 6 0

(2.8.31)

We now must find a connection between j and k. Starting with arbitrary j, we recognize that the condition where we are most likely to find a limiting relationship between j and k is if we let j = j 0 : 0 ≤ k ≤ 2j −→ k ≤ 2j −→ j ≥

k 2

(2.8.32)

This gives us a condition in which the expectation value from (2.8.31) is non-zero. We can extend (2.8.32) to give us a condition in which Tkq is zero: j<

k 2

(2.8.33)

In order to apply this result to our specific case where j = 1/2 we note that: /2 < 1 −→ hT2q i = 0

1

for j = 1/2 X

(2.8.34)

We can verify this more quantitatively by considering the “worst” case, where we are most likely to fail (my reasoning for believing that this is the case will be apparent in due course), is when j = 0 and m = 0. Therefore, the Wigner Eckart Theorem from (2.8.28), hα0 j 0 m0 |T2q |αjmi = h0200|0200ihα0j 0 kT2 kαji The Clebsch-Gordon coefficient in (2.8.35) can be found in Merzbacher from (17.62): s j(2j − 1) hj2j0|j2jji = −→ h0200|0200i = 0 (j + 1)(2j + 3)

(2.8.35)

(2.8.36)

Substituting (2.8.36) into (2.8.35), we find that hT2q i = 0 X

(2.8.37)

Where (2.8.37) is the same as (2.8.34). In summary, it is clear from these result indicate that the effect of the quadrupole perturbation on our system is null . c)

Using the fact that C±2 = C0 and C±1 = 0, we can rewrite (2.8.22) as   V =C0 T2−2 + C0 T20 + C0 T22 −→ V = C0 T2−2 + T20 + T22

(2.8.38)

We are also given one of the components of the irreducible spherical tensor operator from (2.8.20). Even though it was not mentioned earlier, it is quite apparent that this component can be rewritten in terms of J+ : 2

2 T22 = (Jx + iJy ) ⇒ J+

Using (2.8.39), we can now write 2 hα0 j 0 m0 T22 |αjmi =hα0 j 0 m0 |J+ |αjmi

(2.8.39)

W. Erbsen

HOMEWORK #9

=

p p (j − m)(j + m + 1) (j − m − 1)(j + m + 2) ~2 δm0 ,m+2 δjj 0 δαα0

(2.8.40)

Now, from the W.E.T. , we have

hα0 j 0 m0 |T22 |αjmi =hj, 2, m, 2|j, 2, j 0, m0 ihα0 j 0 kT kαji

(2.8.41)

Rewriting the reduced matrix element on the RHS of (2.8.41), p (j − m)(j + m + 1)(j − m − 1)(j + m + 2) ~2 0 hαj kT kαji = δjj 0 hj2m2|j2jm + 2i

(2.8.42)

Since the reduced matrix element is independent of the choice of m, we “arbitrary” choose m = −1, and (2.8.42) becomes p j 2 (j + 1)2 ~2 0 hαj kT kαji = δjj 0 (2.8.43) hj, 2, −1, 2|j, 2, j, 1i Using (2.8.41) and (2.8.43), we can now express the matrix elements for all tensor components: hαj 0 m0 |T2q |αjmi =

j(j + 1) ~2 δjj 0 hj, 2, m, q|j, 2, j 0, m0 i hj, 2, −2, 2|j, 2, j, 1i

(2.8.44)

Where we recall that selection rules dictate that q = m0 − m. Dropping the α in (2.8.44) for convenience (since α = α0 ) and applying j = j 0 = 1, h1, m0 |T2q |1, mi =

2~2 h1, 2, m, q|1, 2, 1, m0i h1, 2, −2, 2|1, 2, 1, 1i

(2.8.45)

From (2.8.38), we can see that the only components we require are for T2−2 , T20 and T22 , which we will evaluate separately. Starting with T2−2 , we evaluate (2.8.45) with q = −2, which yields h1, m0 |T2−2 |1, mi =

2~2 h1, 2, m, −2|1, 2, 1, m0i h1, 2, −2, 2|1, 2, 1, 1i

(2.8.46)

In order for (2.8.46) to be nonzero, the selection rules (derived previously) require that q = m0 − m, or in our case m0 = m − 2. The only nonzero element comes from m = 1 −→ m0 = −1, and so (2.8.46) becomes h1, −1|T2−2|1, 1i =

2~2 h1, 2, 1, −2|1, 2, 1, −1i ⇒ 2~2 h1, 2, −2, 2|1, 2, 1, 1i

(2.8.47)

We now wish to evaluate T20 in very much the same way. Substituting q = 0 into (2.8.45), h1, m0 |T20 |1, mi =

2~2 h1, 2, m, 0|1, 2, 1, m0i h1, 2, −2, 2|1, 2, 1, 1i

(2.8.48)

Selection rules require that m = m0 , so we have three non-zero components, corresponding to m = −1, m = 0, and m = 1 (respectively). Substituting each of these m-values into (2.8.48), 2~2 2~2 h1, 2, −1, 0|1, 2, 1, −1i ⇒ √ h1, 2, −2, 2|1, 2, 1, 1i 6 2 2 2~ 4~ h1, 0|T20|1, 0i = h1, 2, 0, 0|1, 2, 1, 0i ⇒ − √ h1, 2, −2, 2|1, 2, 1, 1i 6 2 2 2~ 2~ h1, 1|T20|1, 1i = h1, 2, 1, 0|1, 2, 1, 1i ⇒ √ h1, 2, −2, 2|1, 2, 1, 1i 6

h1, −1|T20|1, −1i =

(2.8.49a) (2.8.49b) (2.8.49c)

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CHAPTER 2: QUANTUM MECHANICS II

And finally, we must evaluate the matrix elements for T22 . We do this in very much the same way as before, setting q = 2 in (2.8.45) h1, m0 |T22 |1, mi =

2~2 h1, 2, m, 2|1, 2, 1, m0i h1, 2, −2, 2|1, 2, 1, 1i

(2.8.50)

From the selection rules, we have m = −1 −→ m0 = 1, and so the only nonzero matrix element is from (2.8.48) h1, 1|T20|1, −1i =

2~2 h1, 2, −1, 0|1, 2, 1, −1i ⇒ 2~2 h1, 2, −2, 2|1, 2, 1, 1i

(2.8.51)

We can now begin to construct our matrices corresponding to the calculated matrix elements for T2−2 , T20 and T22 , which were calculated in (2.8.47), (2.8.49a)–(2.8.49c), and (2.8.51), respectively. Most generically, the matrix hT2q i is given by |1,−1i h1,−1| h1,−1|T2q |1,−1i q hT2 i = h1, 0|  h1, 0|T2q |1,−1i h1, 1| h1, 1|T2q |1,−1i 

|1, 0i h1,−1|T2q |1, 0i h1, 0|T2q |1, 0i h1, 1|T2q |1, 0i

So, each of our three elements are given by

|1, 1i  h1,−1|T2q |1, 1i h1, 0|T2q |1, 1i  h1, 1|T2q |1, 1i

|1,−1i |1, 0i |1, 1i  h1, 1| 0 0 2~2 hT2−2 i = h1, 0|  0 0 0  h1, 1| 0 0 0 −



|1,−1i √ h1, 1| 2~2 / 6 0 hT20 i = h1, 0|  h1, 1| 0 −



|1, 0i 0√ − 4~2 / 6 0

|1, 1i  0 0√  2 2~ / 6

|1,−1i |1, 0i |1, 1i  h1,−1| 0 0 0 0 0  hT22 i = h1, 0|  0 2 h1, 1| 2~ 0 0 

(2.8.52)

(2.8.53a)

(2.8.53b)

(2.8.53c)

Combining (2.8.53a)-(2.8.53c) and substituting into (2.8.38),

V = 2C0 ~2

|1,−1i √ h1,−1| 1/ 6 h1, 0|  0 h1, 1| 1

|1, 0i 0√ − 2/ 6 0

|1,−1i √ h1,−1| 1/ 6 h1, 0|  0 h1, 1| 1

|1, 1i 1 0√ 1/ 6



|1, 1i  1 0√  1/ 6

(2.8.54)

To make the solution for (2.8.54) more transparent, we interchange the 2nd and 3rd columns

V = 2C0 ~2



|1, 0i  0√ − 2/ 6  0

W. Erbsen

HOMEWORK #9

And now do the same thing for the 1st and 3rd rows,

V = 2C0 ~2

|1,−1i √ h1, 1| 1/ 6 h1, 1|  1 h1, 0| 0 −



|1, 1i 1√ 1/ 6 0

|1, 0i  0 0√  − 2/ 6

(2.8.55)

It is easy to see that we needn’t diagonalize the entire matrix in (2.8.55); rather, only the 2 × 2 submatrix:

V 0 = 2C0 ~2 Diagonalizing (2.8.56), we find that √ 2C0 ~2 / 6 − λ 2C0√ ~2 2 2 2C0 ~ 2C0 ~ / 6 − λ

−

h1, 1| h1, 1|



|1,−1i √ 1/ 6 1

|1, 1i  1√ 1/ 6

(2.8.56)

   2 2C0 ~2 0~ = 0 −→ 2C √ √ −λ − λ − 4C0 ~2 = 0 6 6 4C0 ~2 10C02~4 −→λ2 − √ λ − =0 3 6   1 −→λ = 2C0 ~2 √ ± 1 6

(2.8.57)

  Using (2.8.57), we can now find the corresponding eivenvectors. For λ = 2C0 ~2 √16 + 1 , we have −    
1 1 |1, 1i 0 2C0 ~2 = −→ N |1, 1i + |1,− 1i = 0 (2.8.58) − − 1 1 |1, 1i 0   And similarly, for λ = 2C0 ~2 √16 − 1 ,     
1 1 |1, 1i 0 2C0 ~2 = −→ N |1, 1i − |1,− 1i = 0 (2.8.59) − 1 1 |1, 1i 0 √ Where the normalization constant N in (2.8.58) and (2.8.59) is simply 1/ 2 . The eigenvalues and eigenvectors are then 4C0 ~2 λ1 = − √ , 6   1 λ2 =2C0 ~2 √ + 1 , 6   1 λ3 =2C0 ~2 √ − 1 , 6

|1, 0i = 0
1 √ |1, 1i + |1,− 1i = 0 2
1 √ |1, 1i − |1,− 1i = 0 2

(2.8.60) (2.8.61) (2.8.62)

The eigenvalues in (2.8.60) are the energy shifts (∆E), while the eigenvectors correspond to the unperturbed energy Eigenstates. The energy shifts are plotted in Fig. (3) (see attached Mathamatica printout).

Exercise 21.4

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CHAPTER 2: QUANTUM MECHANICS II

From the commutation relations for ai and a†i , deduce that the operator Ni = a†i ai has as eigenvalues all nonnegative integers in the case of Bose-Einstein statistics, but only 0 and 1 in the case of Fermi-Dirac statistics.

Solution The commutation relations for ai and a†i are stated for the Bose-Einstein case in (21.34) as  a†k a†` − a†` a†k = 0  ak a` − a` ak = 0 B.E.  ak a†` − a†` ak = δk`1 And similarly, under Fermi-Dirac statistics, we have from (21.35)  a†k a†` + a†` a†k = 0  ak a` + a` ak = 0 F.D.  ak a†` + a†` ak = δk` 1

(2.8.63)

(2.8.64)

We also recall that the occupation number operators, stated in the prompt and also given by (21.30), as Ni = a†i ai

(2.8.65)

We can describe the effect of the annihilation and creation operators in general by studying (21.36): √ ai |n1 , n2 , ..., ni, ...i = eiα ni |n1 , n2 , ..., ni − 1, ...i

(2.8.66)

From (2.8.66), we can extrapolate the effect of ai and a†i under Bose-Einstein and Fermi-Dirac statistics, respectively: √ √ a|ni = n |ni, a† |ni = n + 1 |ni B.E. (2.8.67a)  a|0i = 0, a† |0i = e−iα |1i F.D. (2.8.67b) a|1i = eiα |0i, a† |1i = 0 Where we note that (2.8.67a) is the same as (12.38) in Merzbacher, while (2.8.67b) is the same as (12.39). The connection between (2.8.66) and (2.8.67a) for Bose-Einstein statistics stems from the fact that we require that α = 0. Similarly, we go from (2.8.66) to (2.8.67b) by recognizing that for the anti-commutation relations from (2.8.64), we let eiα = 1 if the number of occupied single-particle states is even, while eiα = −1 if they are odd. We are now tasked with finding the eigenvalues of the occupation number operators from (2.8.65) under both Bose-Einstein statistics as well as Fermi-Dirac statistics. Starting with Bose-Einstein statistics first, we apply the action of (2.8.67a) to |ni as follows: √ N |ni ⇒ aa† |ni = n + 1 (a|n + 1i) √ √ = n + 1 n + 1 |ni = (n + 1) |ni (2.8.68) But we aren’t finished yet. We recall from (2.8.63) that aa† − a† a = 1 (1 → 1 in this special case). Tinkering with this a bit yields

W. Erbsen

HOMEWORK #9

aa† − a† a = 1 −→ aa† = N + 1

(2.8.69)

Substituting (2.8.69) back into (2.8.68), (N + 1) |ni = (n + 1) |ni −→ N |ni = n|ni

(2.8.70)

Recognizing that the Bosonic occupation number operator may be any positive number, therefore we can say that the eigenvalues of N may take any non-negative integer . In order to repeat this same task within the framework of Fermi-Dirac statistics, we must go about the problem in a slightly different way. We again start with the occupation number operator from (2.8.65), but this time we wish to study the square, N 2 :

N 2 = a† a a† a (2.8.71) At this point, we remember the last equation in (2.8.64), from which we can say aa† + a† a = 1 −→ aa† = 1 − a† a

(2.8.72)

Substituting (2.8.72) back into (2.8.71),
2 † † N 2 =a† 1 − a† a a ⇒ a† a − a | a{zaa} −→ N = N 0

From which we can easily say that the only condition in which the eigenvalues of N 2 and N are equal for Fermi-Dirac statistics is when the eigenvalues are equal either 0 or 1 .

Exercise 21.8 Prove from the commutation relations that h0|ai aj a†k a†` |0i = δjk δi` ± δik δj`

(2.8.73)

the sign depending on the statistics, B.E. (+) or F.D. (−). Also calculate the vacuum expectation value h0|ah ai aj a†k a†` a†m |0i.

Solution We begin our journey by rearranging the algebraic equations describing the relationship between two unique ladder operators ak and a` , described by (21.34) in Merzbacher for Bose-Einstein statistics and (21.35) for Fermi-Dirac statistics (the last equation in (2.8.63) and (2.8.64) in the previous problem). B.E. (2.8.74a) ak a†` − a†` ak = δk` 1 −→ ak a†` = δk` 1 + a†` ak † † † † F.D. (2.8.74b) ak a` + a` ak = δk` 1 −→ ak a` = δk` 1 − a` ak

We can combine (2.8.74a) and (2.8.74b) in a much more compact form,

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CHAPTER 2: QUANTUM MECHANICS II

ak a†` = δk` 1 ± a†` ak

[B.E. → +, F.D. → −]

(2.8.75)

We now need to determine how our operators act on the vacuum state |0i. We recall from (21.6) and (21.7) that (1)

a†i |0i =Ψi

(1) ai Ψj

= |0, 0, ..., ni = 1, 0, ...i

(2.8.76a) (0)

=ai |0, 0, ..., ni = 1, 0, ...i = δij Ψ (0)

(2.8.76b)

(0)

Where we note that for the vacuum state, Ψi = Ψj = Ψ(0) = |0i. Using this logic, we can evaluate (2.8.73) by substituting into it (2.8.75) and using (2.8.77) and (2.8.78): (1)

(1)

Ψ

Ψ

h0|aai aj a†k a†` |0i = Ψ(0) ai aj a†k a†` Ψ(0)

` i z }| { { z }| † † (0) (0) = (Ψ ai ) aj ak (a` Ψ )

(1)

(1)

=Ψi aj a†k Ψ`

Substituting into (2.8.77) our derived result from (2.8.75),   (1) (1) (1) (1) Ψi aj a†k Ψ` =Ψi δjk 1 ± a†k aj Ψ`   (1) (1) (1) (1) =Ψi (δjk 1) Ψ` ± Ψi a†k aj Ψ` (1)

(1)

δi`

From (2.8.78) we can say that

(1)

(1)

=δjk Ψi Ψ` ± Ψi a†k aj Ψ` | {z } | {z } | {z }

(1)

(1)

Ψi aj a†k Ψ`

(1)

δik Ψk

(2.8.77)

(2.8.78)

(0)

δj` Ψ

= δjk δi` ± δik δj`

(2.8.79)

Substituting (2.8.79) back into (2.8.77), we are left with Ψ(0) ai aj a†k a†` Ψ(0) = δjk δi` ± δik δj` Which is the same as (2.8.73).

Problem 21.1 a)

Show that if V (r) is a two-particle interaction that depends only on the distance r between the particles, the matrix element of the interaction in the k-representation may be reduced to Z 1 hk3 k4 |V |k1 k2 i = δ (k1 + k2 − k3 − k4 ) V (r)e−iq·r d3 r (2.8.80) 3 (2π) where ~q is the momentum transfer ~ (k3 − k1 ).

W. Erbsen

b)

HOMEWORK #9

For this interaction, show that the mutual potential energy operator is ZZZ 1 V = φ† (k1 + q) φ† (k2 − q) φ (k2 ) φ (k1 ) F (q) d3 k1 d3 k2 d3 q 2

(2.8.81)

where F (q) is the Fourier transform of the displacement-invariant interaction.

Solution Before we get down to business, we must lay down some formalism. We recall that within the framework of identical particles, ai and a†i destroy and create particles with some quantum number Ki . We also recall that L represents a complete set of one-particle observables, whose eigenvalues are Lq . The relationship between Ki and Lq is given by (21.9) as X |Ki i = |Lq ihLq |Ki i (2.8.82) q

We can also evaluate the expectation value of K by expanding it in terms of Ki : X hLq |K|Lr i = hLq |Ki iKi hKi |Lr i

(2.8.83)

Which is of course (21.43). We also wish to define an additive two-particle operator as 1X † † 1X † † V = ai aj aj ai Vij ⇒ b b bs bt hqr|V |tsi 2 2 qrst q r

(2.8.84)

i

ij

Where, from (21.48), hqr|V |sti =

X ij

hLq |Ki ihKi |Lt ihLr |Kj ihKj |Ls i Vij

(2.8.85)

We note that (2.8.85) is the generalized two-particle matrix element, and according to Merzbacher’s chosen convention we require that r and s belong to one particle, while q and t to the other. We also recall the following symmetry property: hqr|V |tsi = hrq|V |sti Which should come as no surprise. a)

We are now tasked with finding the two-particle matrix element in the k-representation. The most general form is given by (2.8.85), and we note that since our interaction stems from a mutual potential energy, (2.8.84) and (2.8.85) can be readily used. The first thing we wish to do is rewrite our k-representation two-particle matrix element from the LHS of (2.8.80) in terms of the more generic form of (2.8.85): X hk3 k4 |V |k1 k2 i = hk3 |ri ihri |k1 ihk4 |rj ihrj |k2 i Vij (2.8.86) ij

Our particle coordinates are r1 and r2 , while the interaction potential between the two particles is only dependent on the distance between them (call this distance r), so that Vij = V (r1 −r2 ) = V (r). Accordingly, (2.8.86) becomes

211

CHAPTER 2: QUANTUM MECHANICS II

hk3 k4 |V |k1 k2 i =

X 12

hk3 |r1 ihr1 |k1 ihk4 |r2 ihr2 |k2 i V (r)

(2.8.87)

It is very important that we recognize that the summation indicies in (2.8.87) are referring to the coordinates r1 and r2 , and mustn’t be applied to k. Since r1 and r2 are continuous variables, we may take the sums in (2.8.87) to be integrals. We now have ZZ hk3 k4 |V |k1 k2 i = hk3 |r1 ihr1 |k1 ihk4 |r2 ihr2 |k2 i V (r) d3 r1 d3 r2 (2.8.88) At this point, we wish to reexpress (2.8.88) in terms of momentum eigenstates, which we will do in the usual way. Most generally, Z 1 = |kihk| dk ↓ Ψ(r) =hr |Ψi Z = hr|kihk|Ψi dk

(2.8.89)

∗

We also recall that (hk|ri) = hr|ki ⇒ N e−ik·r/~ , while normalizing this yields Z ∞ 0 1 |N |2 eik·(r−r )/~ −→ N = √ , < δ (r − r0 ) 2π~ −∞

(2.8.90)

Compensating for the fact that dr → d3 r, we can now say that hr|ki =

1 eik·r/~ (2π~) 3/2

We now rewrite (2.8.88) in a slightly more transparent way, ZZ hk3 k4 |V |k1 k2 i = hr1 |k3 i∗ hr1 |k1 i hr2 |k4 i∗ hr2 |k2 i V (r) d3 r1 d3 r2

(2.8.91)

(2.8.92)

We can now evaluate each of the coefficients in (2.8.93) by applying (2.8.91): 1 (2π~) 3/2 1 hr1 |k1 i = (2π~) 3/2 1 hr2 |k4 i∗ = (2π~) 3/2 1 hr2 |k2 i = (2π~) 3/2

hr1 |k3 i∗ =

e−ik3 ·r1/~ eik1 ·r1/~ e−ik4 ·r2/~ eik2 ·r2/~

And substituting these into (2.8.93),   ZZ 1 i hk3 k4 |V |k1 k2 i = exp (−k · r + k · r − k · r + k · r ) V (r) d3 r1 d3 r2 3 1 1 1 4 2 2 2 (2π~)6 ~ (2.8.93) By substituting r = r1 − r2 −→ r1 = r2 + r into (2.8.93) and completing the square, we arrive at

W. Erbsen

HOMEWORK #9

hk3 k4 |V |k1 k2 i = =

Z

−iq·r2 −i(k4 −k2 )·r2

e

e

1

(2π~)

3

Z

3



iq·r2

d r2 e

ei(k1 +k2 −k3 −k4)·r2 /~ d3 r2

Z

Z

−iq·r1

V (r)e

3

iq·r1

d r1 − e

Z

V (r)e−iq·r/~ d3 r

iq·r2

V (r)e

3

d r2



(2.8.94)

Noticing that the first integral in (2.8.94) is a δ-function, we are left with Z 1 hk3 k4 |V |k1 k2 i = δ (k1 + k2 − k3 − k4 ) V (r)e−iq·r d3 r 3 (2π)

Problem 21.2 Show that the diagonal part of the interaction operator V , found in Problem 1 in the k-representation, arises from momentum transfers q = 0 and q = k2 − k1 , respectively. Write down the two interaction terms and identify them as direct (q = 0) and exchange (q = k2 − k1 ) interactions. Draw the corresponding diagrams (Figure 21.1).

Solution We first recall from (21.51) that ψσ0 (r0 )ψσ00 (r00 ) − ψσ00 (r00 )ψσ0 (r0 ) = 0 ψσ† 0 (r0 )ψσ† 00 (r00 ) − ψσ† 00 (r00 )ψσ† 0 (r0 ) = 0 ψσ0 (r0 )ψσ† 00 (r00 ) − ψσ† 00 (r00 )ψσ0 (r0 ) = δ(r0 − r00 ) δσ0 σ00 And similarly, under Fermi-Dirac statistics, we have from (21.52) ψσ0 (r0 )ψσ00 (r00 ) + ψσ00 (r00 )ψσ0 (r0 ) = 0 ψσ† 0 (r0 )ψσ† 00 (r00 ) + ψσ† 00 (r00 )ψσ† 0 (r0 ) = 0 ψσ0 (r0 )ψσ† 00 (r00 ) + ψσ† 00 (r00 )ψσ0 (r0 ) = δ(r0 − r00 ) δσ0 σ00 We also recall from (21.53) that N=

XZ

 

B.E.

(2.8.95)

 

F.D.

(2.8.96)





ψσ† 0 (r0 )ψσ0 (r0 ) d3 r 0

(2.8.97)

σ0

From the previous problem, we have ZZ Z 1 V = φ† (k1 + q) φ† (k2 − q) φ (k2 ) φ (k1 ) F (q) d3 k1 d3 k2 d3 q 2 Substituting in to (2.8.98) the appropriate value for q and simplifying, we have Z ZZ Z 1 V = φ† (k3 )φ† (k4 )φ(k2 )φ(k1 )hk3 k4 |k2 k1 i d3 k1 d3 k2 d3 k3 d3 k4 2 We now wish to sandwich (2.8.99) between the state ψn as follows:

(2.8.98)

(2.8.99)

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CHAPTER 2: QUANTUM MECHANICS II

hψn |V |ψn i =

1 2

Z ZZ Z

hψn |φ†(k3 )φ† (k4 )φ(k2 )φ(k1 )|ψn ihk3 k4 |k2 k1 i d3 k1 d3 k2 d3 k3 d3 k4

(2.8.100)

The generic interaction term we are interested in is given from (2.8.100) as hψn |φ† (k3 )φ† (k4 )φ(k2 )φ(k1 )|ψn i

(2.8.101)

Now, if q = 0, then we require from the prompt that k1 = k3 and k2 = k4 , and so (2.8.101) becomes hψn |φ† (k1 )φ† (k2 )φ(k2 )φ(k1 )|ψn i =hψn |φ†(k1 )φ(k1 )φ† (k2 )φ(k2 )|ψn i

(2.8.102)

Where we recall that the number operator acts like Ni = a†i ai . Accordingly, if we recall that Ni ψ = ni ψ, from (2.8.102) we have hψn |φ†(k1 )φ(k1 )φ† (k2 )φ(k2 )|ψn i =n1 n2

(2.8.103)

If, on the other hand, q = 0 because k1 = k2 = k3 = k4 ⇒ k, then hψn |φ†(k1 )φ† (k2 )φ(k2 )φ(k1 )|ψn i =hψn |φ† (k)φ(k)φ† (k)φ(k)|ψn i

=hψn |φ† (k)φ† (k)φ(k)φ(k)|ψn i

=hψn |φ† (k)φ† (k)φ(k)φ(k)|ψn i

(2.8.104)

The result of (2.8.104) is not the same for each flavor of particles; for symmetric particles, the RHS of (2.8.104) is n(n − 1), and for anti-symmetric particles, it is equal to 0, which comes as no surprise. Following the same logic, we have for q 6= 0 that k3 = k2 and k1 = k4 . The number operators act as before, and we end up with hψn |φ† (k1 )φ(k1 )φ† (k2 )φ(k2 )|ψn i =hψn |φ†(k1 )φ(k1 )φ† (k2 )φ(k2 )|ψn i ⇒A n1 n2

(2.8.105)

Where for symmetric particles, A = 1, and for anti-symmetric particles A = −1. We finally can say that the “direct” terms come from q = 0 , while the “exchange terms” come from q = k1 − k2 . Please see diagrams in Fig. (4) and (5) (attached at the end of this assignment).

2.9

Homework #10

Exercise 22.3 Construct explicitly in terms of the states of the form a†jm2 α a†jm1 α |0i the total angular momentum eigenstates for two neutrons in the configurations (p 1/2 )2 and (p 3/2 )2 . How would the angular momentum eigenstates look if the two particles were a neutron and a proton but otherwise had the same quantum numbers as before?

Solution

W. Erbsen

HOMEWORK #10

Before we get down to business, we recall (22.4), which reads 1 X † (2) ajm2 α a†jm1 α hjjm1 m2 |jjJM i |0i ΨJM = |JM i = √ 2 m1 m2

(2.9.1)

And that all the individual particles at play here, i.e. proton, neutrons, and electrons, are all Fermions, whereby we must obey the Pauli exclusion principle. For our first case, where both neutrons are in the (p 1/2 )2 configuration, we have the separate quantum numbers as:  j1 = 1/2 , m1 = ± 1/2 −→ J = 0, 1 j2 = 1/2 , m2 = ∓ 1/2 Since we are closely following the mantra of the Pauli exclusion principle, our total angular momentum quantum number J cannot obtain an odd value, and since in this first case we only have J = 0 and J = 1, we only must calculate the former. Applying the relevant quantum numbers to (2.9.1), we have∗ 1 X † |0, 0i = √ a1 a†1 h 1/2 , 1/2 , m1 , m2 | 1/2 , 1/2 , 0, 0i |0i 2 m1 ,m2 /2 ,m2 /2 ,m1

(2.9.2)

We must expand the sum in (2.9.2) for all allowed combinations of m1 and m2 . In our case, there are only two possible permutations: m1 = 1/2 , m2 = − 1/2 and m1 = − 1/2 , m2 = 1/2 . We therefore only have two terms: i 1 h |0, 0i = √ a†1/2 ,− 1/2 a†1/2 , 1/2 h 1/2 , 1/2 , 1/2 , − 1/2 | 1/2 , 1/2 , 0, 0i + a†1/2 , 1/2 a†1/2 ,− 1/2 h 1/2 , 1/2 , − 1/2 , 1/2 | 1/2 , 1/2 , 0, 0i |0i 2 (2.9.3) The Clebsch-Gordan (C-G) coefficients in (2.9.3) were evaluated using Mathematica, according to the same convention as Merzbacher. The results are: 1 h 1/2 , 1/2 , 1/2 , − 1/2 | 1/2 , 1/2 , 0, 0i = √ 2 1 h 1/2 , 1/2 , − 1/2 , 1/2 | 1/2 , 1/2 , 0, 0i = − √ 2 Substituting these back in to (2.9.3), |0, 0i =

i 1h † a 1/2 ,− 1/2 a†1/2 , 1/2 − a†1/2 , 1/2 a†1/2 ,− 1/2 |0i 2

(2.9.4)

If our neutrons are in the (p 3/2 )2 configuration, things get a little more tedious. First of all, our quantum numbers become:  j1 = 3/2 , m1 = 0, ± 1/2 , ± 3/2 −→ J = 0, 1, 2, 3 j2 = 3/2 , m2 = 0, ∓ 1/2 , ∓ 3/2 Luckily, since we require that our total angular momentum operator be even, we only must contend with J = 0 and J = 2. The analogous form of (2.9.2) for our case is ∗ From

here forward, I am dropping explicit inclusion of α.

215

CHAPTER 2: QUANTUM MECHANICS II

1 X † |0, 0i = √ a3 a†3 h 3/2 , 3/2 , m1 , m2 | 3/2 , 3/2 , 0, 0i |0i 2 m1 ,m2 /2 ,m2 /2 ,m1 1 X † ⇒√ a3 a†3 hm1 , m2 |0, 0i |0i 2 m1 ,m2 /2 ,m2 /2 ,m1

(2.9.5)

Where I have dropped the inclusion of j1 and j2 in (2.9.5), which is standard practice. We now expand the sum in (2.9.5), which yields 1 h |0, 0i = √ a†3/2 ,− 3/2 a†3/2 , 3/2 h 3/2 , − 3/2 |0, 0i + a†3/2 ,− 1/2 a†3/2 , 1/2 h 1/2 , − 1/2 |0, 0i+ 2

i a†3/2 , 1/2 a†3/2 ,− 1/2 h− 1/2 , 1/2 |0, 0i + a†3/2 , 3/2 a†3/2 ,− 3/2 h− 3/2 , 3/2 |0, 0i |0i

We can tabulate the relevant C-G coefficients by making a table, |J, M i |0, 0i

m1 + 3/2 + 1/2 − 1/2 − 3/2

m2 − 3/2 − 1/2 + 1/2 + 3/2

J 0 0 0 0

M 0 0 0 0

hm1 , m2 |J, M i h+ 3/2 , − 3/2 |0, 0i = +(2)−1 h+ 1/2 , − 1/2 |0, 0i = −(2)−1 h− 1/2 , + 1/2 |0, 0i = +(2)−1 h− 3/2 , + 3/2 |0, 0i = −(2)−1

We now substitute these values into (2.9.6), i 1 h |0, 0i = √ a†3/2 ,− 3/2 a†3/2 , 3/2 + a†3/2 ,− 1/2 a†3/2 , 1/2 + a†3/2 , 1/2 a†3/2 ,− 1/2 + a†3/2 , 3/2 a†3/2 ,− 3/2 |0i 2 2 We must now repeat this process for J = 2, and M = 0, ±1, ±2. |J, M i |2, −2i

m1 − 1/2 − 3/2 |2, −1i + 1/2 − 3/2 |2, 0i + 3/2 + 1/2 − 1/2 − 3/2 |2, 1i + 3/2 − 1/2 |2, 2i + 3/2 + 1/2

m2 − 3/2 − 1/2 − 3/2 + 1/2 − 3/2 − 1/2 + 1/2 + 3/2 − 1/2 + 3/2 + 1/2 + 3/2

J 2 2 2 2 2 2 2 2 2 2 2 2

M −2 −2 −1 −1 0 0 0 0 1 1 2 2

hm1 , m2 |J, M i 1 h− /2 , − 3/2 |2, −2i = +(2)− /2 1 h− 3/2 , − 1/2 |2, −2i = −(2)− /2 1 h+ 1/2 , − 3/2 |2, −1i = +(2)− /2 1 h− 3/2 , + 1/2 |2, −1i = −(2)− /2 1 h+ 3/2 , − 3/2 |2, 0i = +(2)− /2 1 h+ 1/2 , − 1/2 |2, 0i = −(2)− /2 1 h− 1/2 , + 1/2 |2, 0i = +(2)− /2 1 h− 3/2 , + 3/2 |2, 0i = −(2)− /2 1 h+ 3/2 , − 1/2 |2, 1i = +(2)− /2 1 h− 1/2 , + 3/2 |2, 1i = −(2)− /2 1 h+ 3/2 , + 1/2 |2, 2i = +(2)− /2 1 h+ 1/2 , + 3/2 |2, 2i = −(2)− /2 1

Expanding the sum in (2.9.5) and substituting in the appropriate values, we have

(2.9.6)

W. Erbsen

HOMEWORK #10

h i |2, −2i = (2)−1 a†3/2 ,− 3/2 a†3/2 ,− 1/2 − a†3/2 ,− 3/2 a†1/2 ,− 3/2 |0i h i |2, −1i = (2)−1 a†3/2 ,− 3/2 a†3/2 , 1/2 − a†3/2 , 1/2 a†3/2 ,− 3/2 |0i h i |2, 0i = (2)−1 a†3/2 ,− 3/2 a†3/2 , 3/2 − a†3/2 ,− 1/2 a†3/2 , 1/2 + a†3/2 , 1/2 a†3/2 ,− 1/2 − a†3/2 , 3/2 a†3/2 ,− 3/2 |0i h i |2, 1i = (2)−1 a†3/2 ,− 1/2 a†3/2 , 3/2 − a†3/2 , 3/2 a†3/2 ,− 1/2 |0i h i |2, 2i = (2)−1 a†3/2 , 1/2 a†3/2 , 1/2 − a†3/2 , 3/2 a†3/2 , 1/2 |0i If, as the prompt suggests, we have both a neutron and a proton, things change somewhat. We are told that the quantum numbers are the same, so in this respect the subsequent calculations are easier, however we must now introduce a creation operator, b0 , which acts on the other particle. We can build on the results shown in the previous parts of this problem with the addition of this additional creation operator b† . For the case of (p 1/2 ) · (p 1/2 ), our possible quantum numbers are:  j1 = 1/2 , j2 = 1/2 −→ J = 0, 1 m1 = ± 1/2 , m2 = ∓ 1/2 We now use this in the same way we did before, with a table or otherwise. The results are: h i |0, 0i = (2)−1 a†1/2 ,− 1/2 b†1/2 , 1/2 − a†1/2 , 1/2 b†1/2 ,− 1/2 |0i h i 1 |1, −1i = (2)− /2 a†1/2 ,− 1/2 b†1/2 ,− 1/2 |0i h i |1, 0i = (2)−1 a†1/2 ,− 1/2 b†1/2 , 1/2 − a†1/2 , 1/2 b†1/2 ,− 1/2 |0i h i 1 |1, 1i = (2)− /2 a†1/2 , 1/2 b†1/2 , 1/2 |0i And similarly, for the case of (p 3/2 ) · (p 3/2 ), we allow for the following quantum numbers:  j1 = 3/2 , j2 = 3/2 −→ J = 0, 1, 2, 3 m1 = ± 1/2 , ± 3/2 , m2 = ∓ 1/2 , ± 3/2 Which comes as no surprise. The only difference here is that we can no longer exclude the odd contributions of the total angular momentum quantum number J. h i 1 |0, 0i = (8)− /2 a†3/2 , 3/2 b†3/2 ,− 3/2 − a†3/2 , 1/2 b†3/2 ,− 1/2 + a†3/2 ,− 1/2 b†3/2 , 1/2 − a†3/2 ,− 3/2 b†3/2 , 3/2 |0i h i |2, −2i = (2)−1 a†3/2 ,− 1/2 b†3/2 , 3/2 − a†3/2 , 3/2 b†3/2 ,− 1/2 |0i h i |2, −1i = (2)−1 a†3/2 ,− 3/2 b†3/2 , 1/2 − a†3/2 , 1/2 b†3/2 ,− 3/2 |0i h i 1 |2, 0i = (8)− /2 a†3/2 ,− 3/2 b†3/2 , 3/2 + a†3/2 ,− 1/2 b†3/2 , 1/2 − a†3/2 , 1/2 b†3/2 ,− 1/2 + a†3/2 , 3/2 b†3/2 ,− 3/2 |0i h i |2, 1i = (2)−1 a†3/2 , 1/2 b†3/2 , 3/2 − a†3/2 , 3/2 b†3/2 , 1/2 |0i h i |2, 2i = (2)−1 a†3/2 , 1/2 b†3/2 , 3/2 − a†3/2 , 3/2 b†3/2 , 1/2 |0i I am not “calculating” the rest; I am not a monkey. There is no more physics here. No more fun.

217

CHAPTER 2: QUANTUM MECHANICS II

Exercise 22.6 Use the symmetry relations for the Clebsch-Gordan coefficients to show that a configuration (n`)2 can only give rise to spin-orbit coupled two-electron states for which L + S is even, i.e., states 1 S, 3 P, 1 D and so on.

Solution We first recall (22.2) from Merzbacher, which reads X † (2) ΨJM = C aj2 m2 α2 a†j1 m1 α1 hj1 j1 m1 m2 |j1 j2 JM i Ψ(0) m1 m2

From which follows directly (22.17): X X (2) |Ψn1`1 n2`2 i = h`1 `2 m1 m2 |`1 `2 LML i h 1/2 1/2 m01 m02 | 1/2 1/2 SMS ia†n2`2 m m1 m1

0 2 m2

m01 m02

a†n1 `1 m

0 1 m1

|0i

(2.9.7)

In our specific case, we are interested in investigating when `1 = `2 ⇒ ` and when n1 = n2 ⇒ n, and so (2.9.7) becomes X X (2) |Ψn`n`i = h``m1 m2 |``LML i (2.9.8) h 1/2 1/2 m01 m02 | 1/2 1/2 SMS ia†n`m m0 a†n`m m0 |0i m1 m1

2

m01 m02

2

1

1

We also recall the symmetry relations for Clebsch-Gordan coefficients, which from (17.61) reads hj1 j2 m1 m2 |j1 j2 jmi = (−1)j−j1 −j2 hj2 j1 m2 m1 |j2 j1 jmi = hj2 j1 , −m2 , −m1 |j2 j1 j, −mi

(2.9.9)

Where in our case we let j1 = j2 ⇒ `, and for the total angular momentum j ⇒ L and m ⇒ M . Accordingly, (2.9.9) becomes h``m1 m2 |``LM i = (−1)L−2`h``m2 m1 |``LML i = h``, −m2 , −m1 |``L, −ML i We now substitute (2.9.10) back into (2.9.8), X X (2) |Ψn`n`i = (−1)L−2`h``m2 m1 |``LML i h 1/2 1/2 m01 m02 | 1/2 1/2 SMS ia†n`m m1 m1

0 2 m2

m01 m02

(2.9.10)

a†n`m

0 1 m1

|0i

(2.9.11)

While at this point, we recall (21.35), which was used in the last homework assignment, a†k a†` + a†` a†k = 0 ak a` + a` ak = 0

(2.9.12)

ak a†` + a†` ak = δk` 1 We note that we can use (2.9.12) to rewrite the product of the ladder operators in (2.9.11): a†n`m

0 2 m2

a†n`m

0 1 m1

= − a†n`m

Substituting (2.9.13) back into (2.9.11),

0 1 m1

a†n`m

0 2 m2

⇒ (−1)1 a†n`m

0 1 m1

a†n`m

0 2 m2

(2.9.13)

W. Erbsen

HOMEWORK #10

(2)

|Ψn`n`i =

X

m1 m1

(−1)L−2`+1 h``m2 m1 |``LML i (2)

X

m01 m02

h 1/2 1/2 m01 m02 | 1/2 1/2 SMS ia†n`m

0 1 m1

⇒(−1)L+S−2` |Ψn`n`i

a†n`m

0 2 m2

|0i (2.9.14) (2.9.15)

The required condition is satisfied, i.e. unless L + S is even, (2.9.14) is violated .

Exercise 22.10 Show that the configuration space wave function be expressed as the Slater determinant 1 ψ(r1 σ1 , · · · , rN σN ) = √ N !

corresponding to the independent particle state (22.22) can ψ1 (r1 σ1 ) ψ2 (r1 σ1 ) .. .

ψ1 (r2 σ2 ) ψ2 (r2 σ2 ) .. .

ψn (r1 σ1 )

ψn (r2 σ2 )

· · · ψn (rN σN ) ··· ··· .. .

ψ1 (rN σN ) ψ2 (rN σN ) .. .

(2.9.16)

Solution We first recall that (22.22) from Merzbacher reads |Ψν i = a†N a†N−1 · · · a†2 a†1 |0i

(2.9.17)

While we must always obey the symmetrixation requirement: ψ(r1 , r2) = ±ψ(r2 , r1) Treating this in a more explicit manner, 1 ψ± (r1 , r2 ) = √ [ψa (r1 )ψb (r2 ) ± ψb (r1 )ψa (r2 )] 2

(2.9.18)

Where the upper sign represents Bosons, while the lower is for Fermions. Directly from (2.9.18) follows the Pauli Exclusion Principle, which applies only to Fermions. For two particles that occupy the same state ψa = ψb , 1 ψ− (r1 , r2 ) = √ [ψa (r1 )ψa (r2 ) − ψa (r2 )ψa (r1 )] ⇒ 0 2 In other words, the resultant wave function must be anti-symmetric with the exchange of any two particles. We note that while the previous equations do not contain an explicit spin dependence, one may be included without loss of generality. Accordingly, (2.9.18) becomes 1 ψ± (r1 σ1 , r2 σ2 ) = √ [ψa (r1 σ1 )ψb (r2 σ2 ) ± ψa (r2 σ2 )ψb (r1 σ1 )] 2

(2.9.19)

219

CHAPTER 2: QUANTUM MECHANICS II

In this special two-particle case, it is not hard to see that we can rewrite (2.9.19) in terms of a determinant: 1 ψa (r1 σ1 ) ψa (r2 σ2 ) ψ± (r1σ1 , r2σ2 ) = √ (2.9.20) 2 ψb (r1 σ1 ) ψb (r2 σ2 ) The extension of (2.9.19) to (2.9.20) may be generalized to a N -particle system without too much trouble; all we have to do is recognize that the odd exchanges in the wave function product terms must include a factor of −1. Explicitly, we rewrite (2.9.20) for N particles as:

1 ψ± (r1σ1 , r2σ2 , · · · , rN σN ) = √ [ψa (r1 σ1 )ψb (r2 σ2 ) · · · ψN (rN σN ) ± ψa (r2 σ2 )ψb (r1 σ1 ) · · · ψN (rN σN )] 2 (2.9.21) Where we note that the complete set of wave function denominations is {ψa , ψb , ..., ψN }. We can immediately rewrite (2.9.21) in the form of a determinant, as suggested by the prompt: ψa (r1 σ1 ) 1 ψb (r1 σ1 ) ψ(r1 σ1 , · · · , rN σN ) = √ .. N ! . ψN (r1 σ1 )

ψa (r2 σ2 ) ψb (r2 σ2 ) .. . ψN (r2 σ2 )

Which is the same as (2.9.16).

· · · ψN (rN σN ) ··· ··· .. .

ψa (rN σN ) ψb (rN σN ) .. .

Problem 22.6 Apply the Hartree-Fock method to a system of two “electrons” which are attracted to the coordinate origin by an isotropic harmonic oscillator potential mω2 r 2 /2 and which interact with each other through a potential V = C(r0 − r00 )2 . Solve the Hartree-Fock equations for the ground state and compare with the exact result and with first-order perturbation theory.

Solution Due to the phrasing of the problem, we are inclined to utilize the symmetry around the origin. We also know that the Hamiltonian for this system is just the sum of the Harmonic oscillator Hamiltonian plus the interaction term: H=−

~2 2 mω2 2 ∇ + r −V 2m 2

(2.9.22)

Applying V from the prompt to (2.9.22),∗ H=− ∗I


mω2 2
~2 ∇21 + ∇22 + r1 + r22 − C(r1 − r2 )2 2m 2

(2.9.23)

have replace Merzbacher’s silly prime notation with my selected subscripts to deter ambiguity between various notations.

W. Erbsen

HOMEWORK #10

At this point, we seemingly arbitrarily, introduce the relative and center of mass coordinate variables respectively: 1 r = √ (r1 − r2 ) , 2

1 R = √ (r1 + r2 ) 2

(2.9.24)

Rewriting (2.9.23) according to (2.9.24), H=−


mω2 2 ~2 ∇21 + ∇22 + R − Cr2 2m 2

(2.9.25)

If we introduce the following similar coordinates for the momenta, 1 p = √ (p1 − p2 ) , 2

1 P = √ (p1 + p2 ) 2

(2.9.26)

Rewriting the kinetic energy term from (2.9.25) in terms momentum variables and utilizing (2.9.26), we have∗ i √
1h 2 1 H = P 2 + R2 + p + r2 2C + 1 (2.9.27) 2 2

From this, we can see that by the use of our chosen transformations from (2.9.24) and (2.9.26), we see that we have turned one Hamiltonian into two independent 3-dimensional harmonic oscillators. This problem has been presented before; the (exact) results are  3/    √  2C + 1 8 R2 2C + 1 2 r Ψ(r, R) = exp − exp − π4 2 2 i √ 3h E0 = 1 + 2C + 1 2 

(2.9.28a) (2.9.28b)

To see how the Hartree-Fock (HF) perspective compares to (2.9.28a) and (2.9.28b), we see that 1 ψ(r1 )χ+ ψ(r2 )χ+ HF Ψ = √ (2.9.29) 2 ψ(r1 )χ− ψ(r2 )χ− And the HF equation for our Hamiltonian becomes Z   1 2 1 2 2 ∗ p + r1 ψ(r1 ) + ψ (r2 ) · C (r1 − r2 ) · ψ(r2 ) dr2 ψ(r1 ) = Eψ(r1 ) 2 1 2

(2.9.30)

If we expand the central term in (2.9.30), we find that 2

(r1 − r2 ) = r21 + r22 − 2r1 · r2

(2.9.31)

The last term in (2.9.31) vanishes in the integral due to parity. Therefore, the integral becomes trivial and (2.9.30) yields Ψ0 (r1 ) = ∗ Atomic

units and also ω = 1.



C +1 π2

 3/8

 √  C+1 2 exp − r1 2

221

CHAPTER 2: QUANTUM MECHANICS II

E0 =

3 2



3C + 2 2C + 2



√

C+1

And so the Hartree-Fock ground state wave function becomes ΨHF 0 (r, R)

=



C+1 π4

 3/8

 √ 
C +1 2 2 exp − r +R 2

(2.9.33)

And now, the energy eigenvalues according to (2.9.33) can be found by taking the expectation value of our Hamiltonian from (2.9.25) with the HF state from (2.9.33). The result is ? √ HF HF E0HF = hΨHF = 3 C +1 0 |H|Ψ0 i −→ E0 Please see attached Mathematica printout with plots.

(2.9.34)

W. Erbsen

Appendix A

*

*

Creation and Annihilation Operators We first recall from class that the action of the creation and annihilation operators on an arbitrary vector |ψn i acts like: √ (A.1a) a† |ψn i = n + 1 |ψn+1 i √ a|ψn i = n |ψn−1i (A.1b) We also recall that the position and momentum operators are given in terms of the creation and annihilation operators as: r
~ x= a† + a (A.2a) 2mω r
~mω p= a† − a (A.2b) 2

We can now find the expectation values of various combinations of ladder operators, namely ha† i, hai, ha† ai, haa† i, ha2 i, and ha†2 i: ha† i =hψn0 (x)|a† |ψn (x)i Z ∞ = ψn∗ 0 (x)a† ψn (x) dx −∞ Z ∞ √
= ψn∗ 0 (x) n + 1 ψn+1 (x) dx −∞ √ = n + 1 δn0 ,n+1 hai =hψn0 (x)|a|ψn (x)i Z ∞ = ψn∗ 0 (x)aψn (x) dx −∞ Z ∞
√ = ψn∗ 0 (x) n ψn−1 (x) dx −∞ √ = n δn0 ,n−1 ha† ai =hψn0 (x)|a† a|ψn (x)i Z ∞ = ψn∗ 0 (x)a† aψn (x) dx −∞ Z ∞
√ n ψn−1 (x) dx = ψn∗ 0 (x)a† −∞ Z ∞ = ψn∗ 0 (x) (nψn (x)) dx −∞

=nδn0 ,n

haa† i =hψn0 (x)|aa† |ψn (x)i

(A.3)

(A.4)

(A.5)

223

CHAPTER 2: QUANTUM MECHANICS II

= = =

Z

∞

−∞ ∞

Z

−∞ ∞

Z

−∞

ψn∗ 0 (x)aa† ψn (x) dx ψn∗ 0 (x)a

√
n + 1 ψn+1 (x) dx

ψn∗ 0 (x) ((n + 1)ψn (x)) dx

=(n + 1)δn0 ,n ha†2 i =hψn0 (x)|a†2 |ψn (x)i Z ∞ = ψn∗ 0 (x)a† a† ψn (x) dx −∞ Z ∞ √
n + 1 ψn+1 (x) dx = ψn∗ 0 (x)a† −∞ Z ∞ = ψn∗ 0 (x) ((n + 1)ψn+2 (x)) dx −∞

=(n + 1)δn0 ,n+2

ha2 i =hψn0 (x)|a2 |ψn (x)i Z ∞ = ψn∗ 0 (x)aaψn (x) dx −∞ Z ∞
√ = ψn∗ 0 (x)a n ψn−1 (x) dx −∞ Z ∞ = ψn∗ 0 (x) (nψn−2 (x)) dx −∞

=nδn0 ,n−2

(A.6)

(A.7)

(A.8)

We now wish to find the expectation values of the position and momentum operators from (A.2a) and (A.2b), and also their square. We can do this by using (A.3)-(A.8): r

~ hxi = h(a† + a)i 2mω r
~ ha† i + hai = 2mω r √
√ ~ = n + 1 δn0 ,n+1 + n δn0 ,n−1 2mω =0 ~ h(a† + a)2 i 2mω ~ = h(a†2 + a† a + aa† + a2 )i 2mω
~ = ha†2 i + ha† ai + haa† i + ha2 i 2mω

hx2 i =

(A.9)

W. Erbsen

*

~ ((n + 1)δn0 ,n+2 + nδn0 ,n + (n + 1)δn0 ,n + nδn0 ,n−2 ) 2mω ~ (n(n + 1)) = 2mω ~ = (2n + 1) 2mω =

(A.10)

r

~mω h(a† + a)i hpi =i 2 r
~mω =i ha† i + hai 2 r
√ ~mω √ n + 1 δn0 ,n+1 + n δn0 ,n−1 =i 2 =0 ~mω † h(a − a)2 i 2 ~mω †2 =− h(a − a† a − aa† + a2 )i 2
~mω =− ha†2 i − ha† ai − haa† i + ha2 i 2 ~mω =− ((n + 1)δn0 ,n+2 − nδn0 ,n − (n + 1)δn0 ,n + nδn0 ,n−2 ) 2 ~mω =− (−n − (n + 1)) 2 ~mω = (2n + 1) 2

(A.11)

hp2 i = −

(A.12)

225

CHAPTER 2: QUANTUM MECHANICS II

Appendix B

*

Hermite Polynomial Recursion Relation Equation (5.35) from Merzbacher reads:

 dn ξ2 −(s−ξ)2 e dsn s=0 n n ξ2 d −ξ 2 =(−1) e e dξ n

Hn (ξ) =



(B.1)

Which is known as Rodrigues’ formula for Hermite Polynomials. To derive the recursion relation, we first differentiate Hn (ξ):   n  d d d n ξ2 −ξ 2 Hn (ξ) = (−1) e e dξ dξ dξ    n  n+1 2 2 d ξ2 d d =(−1)n e + (−1)n eξ e−ξ dξ dξ dξ  n  n+1 2 2 2 d d =2x(−1)n eξ − (−1)n+1 eξ e−ξ dξ dξ =2xHn(ξ) − Hn+1 (ξ)

(B.2)

And from Eq. (5.32) in Merzbacher we know that d Hn ξ = 2nHn−1(ξ) dξ So that (B.2) becomes: 2nHn−1 (ξ) = 2xHn (ξ) − Hn+1 (ξ)

(B.3)

And from (B.3) we finally arrive at the recursion relation, Hn+1 (ξ) = 2xHn (ξ) − 2nHn−1(ξ) Which agrees with Abramowitz and Stegun, pg. 782 22.7.13 ?.

(B.4)

W. Erbsen

clearemptydoublepage

*

Chapter 3

Statistical Mechanics 3.1

Homework #1

Problem 1 We discussed in class thermal equilibrium of a closed system which is divided into two parts with a barrier inbetween (see figure). This barrier allows energy exchange between the two parts of the system. Show that: a)

If the barrier is allowed to move but does not allow particle exchange, then at thermal equilibrium the pressures of the two parts are equal, i.e. P1 = P2 .

b)

If the barrier is not allowed to move but allows for particle exchange, then at thermal equilibrium the chemical potentials of the two parts are equal, i.e. µ1 = µ2 .

Solution

Figure 3.1: Figure for Problem 3.

Where applicable, we apply the principle of conservation to the extensive variables N , V and E: N =N1 + N2 V =V1 + V2 E =E1 + E2

(= Const.)

W. Erbsen

HOMEWORK #1

As noted, the prompt states that the barrier allows for exchange in energy. We also recall that       ∂E ∂E ∂E dS + dV + dN (3.1.1) dE = ∂S V,N ∂V S,N ∂N S,V Which was shown in class, but is easy enough to see. a)

If the barrier is allowed to move, but does not allow for particle exchange, we can deduce two important things. First, if the barrier moves but leaves the individual systems intact, while conserving the total volume, then we can claim that: dV = ∂V1 + ∂V2 = 0 −→ ∂V1 = −∂V2

(3.1.2)

We can also claim that since the particles are not transferred from system to system, we can say dN = ∂N1 + ∂N2 = 0

(3.1.3)

We can apply (3.1.3) to (3.1.1)

dE = =





∂E ∂S ∂E ∂S



dS +

V,N



dS +

V,N

 

∂E ∂V ∂E ∂V



dV +

S,N



z

=0 }| { ∂E dN ∂N S,V

dV

(3.1.4)

S,N

We now recall (1.3.10), which states that:     ∂S 1 ∂S P = , = , ∂E V,N T ∂V E,N T



∂S ∂N



=−

V,E

µ T

From this, we take the definition which depends on the intensive parameters P and T , which can be manipulated as     ∂S P ∂S = −→ T =P (3.1.5) ∂V E,N T ∂V E,N Now, following the form of (3.1.4), we have for the entropy S 







∂S ∂S dE + dV + ∂E V,E ∂V E,N     ∂S ∂S = dE + dV ∂E V,E ∂V E,N

dS =

z

=0 }| { ∂S dN ∂N E,V (3.1.6)

We also know that the steady-state entropy as ∆t → ∞ approaches zero, and (3.1.6) becomes     ∂S ∂S 0= dE + dV (3.1.7) ∂E V,E ∂V E,N Holding E and N constant for both systems, (3.1.7) =0 z }| { ∂S1 ∂S2 ∂S1 ∂S2 0= dE1 + dE2 + dV1 + dV2 ∂E1 ∂E2 ∂V1 ∂V2

(3.1.8)

229

CHAPTER 3: STATISTICAL MECHANICS

Recalling (3.1.5), it is easy to see that 0= b)

∂S1 ∂S2 P1 P2 dV1 − dV1 −→ = −→ P1 = P2 ∂V1 ∂V2 T T

(3.1.9)

If now we disallow the barrier to move, then the change in volume becomes: dV = ∂V1 + ∂V2 = 0 While since we are allowing for particle exchange, we can claim that the net change in the flux of particulates is described by dN = ∂N1 + ∂N2 = 0 −→ ∂N1 = −∂N2 We can now immediately pick up from (3.1.6), applying it to our current scenario: =0 }| {   ∂S ∂S ∂S dS = dE + dV + dN ∂E V,E ∂V E,N ∂N E,V     ∂S ∂S = dE + dN ∂E V,E ∂N E,V 



z

Recalling the definition of the intrinsic property µ from (1.4.10),     µ ∂S ∂S = − −→ −T =µ ∂N V,E T ∂N V,E

(3.1.10)

(3.1.11)

Using (3.1.11), at thermal equilibrium (3.1.10) becomes 0=−

∂S1 ∂S2 µ1 µ2 dN1 + − dN1 −→ dN1 = dN1 −→ µ1 = µ2 ∂N1 ∂N2 T T

Problem 1.2 Assuming that the entropy S and the statistical number Ω of a physical system are related through an arbitrary functional form S = f (Ω)

(3.1.12)

Show that the additive character of S and the multiplicative character of Ω necessarily require that the function f (Ω) be of the form S = kB log Ω

(3.1.13)

Solution The additive character of S can be quantitatively described using (3.1.12) and (3.1.13) as S (0) (S1 , S2 ) = S1 (Ω1 ) + S2 (Ω2 ) −→ S (0) (Ω1 , Ω2 ) = f (Ω1 ) + f (Ω2 )

(3.1.14)

W. Erbsen

HOMEWORK #1

While the multiplicative property of Ω can be written like: Ω(0) (Ω1 , Ω2 ) = Ω1 Ω2

(3.1.15)

We now recall that both Ω1 and Ω2 represent independent variables, and accordingly a function of both quantities may be differentiated with respect to each variable, yielding the same number of equations. In our case, we can differentiate (3.1.14) as: dS (0) (Ω1 , Ω2 ) dS (0) (Ω1 , Ω2 ) dΩ dS (0) (Ω1 , Ω2 ) = · ⇒ Ω1 dΩ1 dΩ dΩ1 dΩ

(3.1.16)

Using (3.1.16) as a template, we can retrieve our two equations: dS (0) (Ω1 , Ω2 ) dS (0) (Ω1 , Ω2 ) = Ω1 dΩ1 dΩ dS (0) (Ω1 , Ω2 ) dS (0) (Ω1 , Ω2 ) = Ω2 dΩ2 dΩ

(3.1.17a) (3.1.17b)

The full derivatives in (3.1.17a) and (3.1.17b) may be equated and shuffled, 1 dS (0) (Ω1 , Ω2 ) 1 dS (0) (Ω1 , Ω2 ) · = · Ω1 dΩ1 Ω2 dΩ2

(3.1.18)

We now allow ourselves to be reminded of the method of separation of variables in PDE’s, which says that if we have a function of two independent variables, which can be completely separated, then then changing one of the variables by some amount δ then the other must also change by δ. We recall that δ is, of course, a constant, which we can call whatever we want. Apparently, in our case our constant is Boltzmann’s Constant kB . Armed with this knowledge, we can now rewrite (3.1.17a) and (3.1.17b), as df (Ω1 ) dS (0) (Ω1 , Ω2 ) = Ω2 dΩ1 dΩ df (Ω2 ) dS (0) (Ω1 , Ω2 ) = Ω1 dΩ2 dΩ

(3.1.19a) (3.1.19b)

The full derivatives in (3.1.19a) and (3.1.19b) may also be equated, similar to in (3.1.18), as 1 df (Ω2 ) 1 df (Ω1 ) · = · Ω1 dΩ2 Ω2 dΩ1 df (Ω2 ) df (Ω1 ) Ω2 · =Ω1 · dΩ2 dΩ1

(3.1.20)

Setting each side of (3.1.20) equal to our new constant, Ω2 ·

df (Ω2 ) = kB , dΩ2

Ω1 ·

df (Ω1 ) = kB dΩ1

The solutions to these trivial differential equations is of course f (Ω) = kB log (Ω) −→ S (Ω) = kB log (Ω)

(3.1.21)

231

CHAPTER 3: STATISTICAL MECHANICS

Problem 1.7 Study the statistical mechanics of an extreme relativistic gas characterized by the single-particle energy states
1/2 hc 2 ε (nx , ny , nz ) = nx + n2y + n2z (3.1.22) 2L

Instead of (1.4.5) in Pathria, along the lines in 1.4. Show that the ratio CP /CV in in this case is 4/3, instead of 5/3.

Solution Following Pathria’s logic in 1.4, we know that we need not explicitly evaluate Ω (N, E, V ), rather simply the product following the form of (1.4.8), where in the relativistic case, I claim that: V

1

/3

E = const.

Where E in our case is the relativistic energy, given by: q 2 2 E = (pc) + (mc2 )

(3.1.23)

(3.1.24)

Assuming that we are in the “extreme” relativistic limit, we assume that v → c, and in this limit (3.1.24) becomes E ≈ pc

(3.1.25)

Plotting (3.1.24) and (3.1.25) on the same scale indeed verifies that the energy may be readily approximated using (3.1.25) at very large values of v (see figure∗ ).

Figure 3.2: Figure for Problem 3.

∗ We

note that (3.1.24)−→ red/dashed, while (3.1.25)−→ black/thick.

W. Erbsen

HOMEWORK #1

Furthermore, we can build on (1.4.6) by noting that the expression for Ω (N, E, V ) depends on the number of independent, positive-integral solutions as prescribed by:
1/2 2L ε = n2x + n2y + n2z hc

(3.1.26)

From the Maxwell-Boltzmann Distribution Function, and (3.1.25). Furthermore, we recall that V = L3 , and so (3.1.26) becomes 2 V hc

1

/3

ε (nx , ny , nz ) = n2x + n2y + n2z


1/2

−→ ε (nx , ny , nz ) =


1/2 hc 2 nx + n2y + n2z 2L

(3.1.27)

Where we can see from (3.1.27) where (3.1.23) was derived from. Furthermore, using (1.4.25), we can say that:   ∂E P =− ∂V N,S i 1 ∂ h =− Const. · V − /3 ∂V −4 1 =Const. · · V /3 3 −4 =Const. · V /3 (3.1.28) It is easy to see from (3.1.28) that our condition for a reversible adiabatic process is given by PV

4

/3

= Const.

(3.1.29)

Where the raised power of V in (3.1.29) is none other than γ, so we can finally say: γ=

CP 4 −→ γ = CV 3

Problem 1.13 If the two gases considered in the mixing process of 1.5 were initially at different temperatures, say T1 and T2 , what would the entropy of mixing be in that case? Would the contribution arising from this cause depend on whether the two gases were different or identical?

Solution We first recall from (1.5.13) that    3 2πmi kB T Si = Ni kB log [Vi ] + Ni kB 1 + log ; 2 h2

i = 1, 2

(3.1.30)

Which represents the entropy before the mixing ever took place. If we allow our systems to start at different temperatures, (3.1.30) becomes

233

CHAPTER 3: STATISTICAL MECHANICS

   3 2πmi kB Ti Si = Ni kB log [Vi ] + Ni kB 1 + log ; 2 h2

i = 1, 2

(3.1.31)

Where it should be mentioned that the only thing changed is the addition of an index to the intensive variable T . Regrouping and realigning (3.1.31), we have   3 2 Si = kBNi log [Vi ] + log [αmi Ti ] + 1 (3.1.32) 2 3 Where I have assigned the constant α to be defined by α=

2πkB h2

(3.1.33)

Following the convention of (3.1.32), we can say that the post-mixing entropy is (analogous to (1.5.2)):   2 3 X 2 ST = kB Ni log [V ] + log [αmi T ] + 1 2 i=1 3

(3.1.34)

The increase in ∆S, called the entropy of mixing, is then given by ∆S = ST −

2 X

Si

(3.1.35)

i=1

The first term in (3.1.35) is found from (3.1.34) to be      3 2 2 ST = kB N1 log [V ] + log [αm1 T ] + 1 + N2 log [V ] + log [αm2 T ] + 1 2 3 3

(3.1.36)

While the sum in (3.1.35) is found from (3.1.32) to be 2 X i=1

=

     3 2 2 kB N1 log [V1 ] + log [αm1 T1 ] + 1 + N2 log [V2 ] + log [αm2 T2 ] + 1 2 3 3

(3.1.37)

Before we take the difference of (3.1.36) and (3.1.37), thus completing the saga of (3.1.35), we note that this page width is finite (countable, har har), we drop the terms which obviously cancel. I will also consolidate a few things. Continuing,            2 V T 2 V T 3 log + log + N2 log + log (3.1.38) ∆S = kB N1 2 3 V1 T1 3 V2 T2 Jostling things about a bit, (3.1.38) becomes ∆S = kBN1

(

(     3/2 )    3/2 ) T V1 + V2 T V1 + V2 log + log + kB N2 log + log V1 T1 V2 T2 

(3.1.39)

The contribution of entropy in (3.1.39) most certainly depends on whether the gases are distinguishable or indistinguishable. This is the setup for Gibbs Paradox. The general idea is that if the particles in both systems are distinguishable, then after equilibrium is reached, the gas particles may be separated back into their original compartments, assuming that we can reinsert the partition quickly enough. This

W. Erbsen

HOMEWORK #1

would correspond to a reversible process, where we know that the change in entropy, or the entropy of mixing, must be zero. If the particles are indistinguishable, then there is no hope of separating them, and so we have an irreversible process, and the entropy must increase.

Problem 1.16 Establish thermodynamically the formulae V V





∂P ∂T ∂P ∂µ



=S

(3.1.40a)

=N

(3.1.40b)

µ



T

Express the pressure P of an ideal classical gas in terms of the variables µ and T , and verify the above formulae.

Solution We first recall that within the framework of the Grand Canonical Ensemble, where particle exchange is allowed, we can define the following thermodynamic potential: Ω = F − µdN −→ dΩ = −SdT − N dµ − P dV

(3.1.41)

Where we must recognize that this new thermodynamic potential is not the same as the multiplicity. For our case, (3.1.40a) becomes dΩ = − |SdT dµ −P dV −→ Ω = −P dV {z } − N | {z } =0

Where, following the example of other thermodynamic potentials, we may deduce that   ∂Ω S =− ∂T µ   ∂Ω N =− ∂µ T Substituting (3.1.42) into (3.1.43a) and (3.1.43b) yields     ∂ (−P V ) ∂P S =− −→ S = V ∂T ∂T µ µ N =−



(3.1.42)

=0

∂ (−P V ) ∂µ



T

−→ N = V



∂P ∂µ



(3.1.43a) (3.1.43b)

(3.1.44a)

(3.1.44b)

T

Where we can see that (3.1.44a) is the same as (3.1.40a), while (3.1.44b) is identical to (3.1.40b).

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CHAPTER 3: STATISTICAL MECHANICS

We now recall that the Gibbs Sum, which is sort of like the partition function for the Grand Canonical Ensemble, is given by   ∞ X X N µ − Es,N Ω (µ, T ) = exp (3.1.45) kB T N=0 s(N)

While the pressure is P =

kB T log [Ω (µ, T )] V

(3.1.46)

Substituting (3.1.45) into (3.1.46), and then plugging the results in to (3.1.40a) and (3.1.40b) in turn causes the terms to collapse, which is readily seen qualitatively.

3.2

Homework #2

Problem 1 In this problem, you will find an expression for entropy in terms of the canonical distribution function ρs = P where the partition function Z is given by Z = s e−Es /kB T .

e−βEs Z

Solution We first recall that the entropy is defined in terms of the free energy as:   ∂F S=− ∂T V,N

(3.2.1)

While the free energy is given by F = −kB T log [Z]

(3.2.2)

Substituting (3.2.2) into (3.2.1), we have: S=

∂ {kB T log [Z]} ∂T

And since Z of course depends on T , we carry the partial derivative out as:   ∂ S =kB log [Z] + kB T log [Z] ∂T   1 ∂Z =kB log [Z] + kB T Z ∂T The bracketed term in (3.2.4) becomes:

(3.2.3)

(3.2.4)

W. Erbsen

HOMEWORK #2

  ∂Z ∂ X Es = exp − ∂T ∂T s kB T   Es 1 X E exp − = s kB T 2 s kB T

(3.2.5)

Substituting (3.2.5) back into (3.2.4) yields (  ) 1 1 X Es S =kB log [Z] + kB T Es exp − Z kB T 2 s kB T     1 Es 1 X Es exp − =kB log [Z] + kB T s Z kB T | {z } ρs   1 X =kB log [Z] + Es ρs kB T s  X  Es   = − kB − log [Z] + ρs − kB T s      X Es = − kB − log [Z] + ρs log exp − kB T s X     1 Es = − kB ρs log exp − Z kB T s {z } |

(3.2.6)

ρs

At this point, we can trivially rewrite (3.2.6) as: S = −kB

X

ρs log [ρs ]

s



Problem 2 P The expression S = −kB r Pr log Pr is accepted as the general definition of the entropy of a system. Now (1) (2) imagine that a system A1 has probability Pr of being found in a state r and a system A2 has probability Ps of being found in a state s. Then one has X S1 = − kB Pr(1) log Pr(1) r

S2 = − kB

X

Ps(2) log Ps(2)

s

Each state of the composite system A consisting of A1 and A2 can then be labeled by the pair of numbers r, s. Let the probability of A being found in this state be denoted by Prs . Then its entropy is defined as XX Prs log Prs (3.2.7) S = −kB r

s

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CHAPTER 3: STATISTICAL MECHANICS

All the probabilities are normalized so that

X

Pr(1) =1

r

X

Ps(2) =1

s

XX r

Prs =1

s

(1)

(2)

a)

If A1 and A2 are weakly interacting so that they are statistically independent, then Prs = Pr Ps . Show that under these circumstances the entropy is additive, i.e. S = S1 + S2 .

b)

If A1 and A2 are not weakly interacting Prs 6= Pr Ps , but the general solutions still hold: X Prs Pr(1) =

(1)

(2)

s

Ps(2)

=

X

Prs

r

Show that S − (S1 + S2 ) = kB

X

(1)

Prs log

r,s

(2)

Pr Ps Prs

!

(3.2.8)

and use the inequality log x ≤ (x − 1) to show that S ≤ (S1 + S2 ). This means that the existence of correlations between the systems, due to the interactions between them, leads to a situation less random than that where the systems are completely independent of each other.

Solution a)

Applying our assumption to (3.2.7), we have: XX S = − kB Pr Ps log [Pr Ps ] = − kB

= − kB = − kB = − kB

r

s

r

s

r

s

XX

XX

(

(

Pr Ps {log [Pr ] + log [Ps ]}

{Pr Ps log [Pr ] + Pr Ps log [Ps ]}

XX r

X s

Pr Ps log [Pr ] +

s

Ps

XX r

X r

Pr log [Pr ] +

X r

)

Pr Ps log [Ps ]

s

Pr

X s

)

Ps log [Ps ]

(3.2.9)

P We recall from the previous problem that S = −kB i ρi log [ρi ], and applying this to (3.2.9) leads us to: ( ) X  S1  X  S2  S = − kB Ps − + Pr − kB kB s r

W. Erbsen

HOMEWORK #2

=S1

X

Ps + S2

s

X

Pr

(3.2.10)

r

Applying the fact that the probabilities are normalized, (3.2.10) leads us to S = S1 + S2 b)

In order to show (3.2.8), we begin by evaluating the LHS with our previous definitions of the respective entropies: ! XX X X S − (S1 + S2 ) = − kB Prs log Prs − −kB Pr log Pr − kB Ps log Ps r

=kB

−

s

r

XX r

Prs log Prs +

s

X

s

Pr log Pr +

r

X

Ps log Ps

s

!

We can expand (3.2.11) given the provided summation identities: S − (S1 + S2 ) =kB =kB =kB

− − −

XX r

XX r

s

XX r

Prs log Prs +

s

s

XX r

Prs log Pr +

s

XX s

{Prs log Prs + Prs log Pr + Prs log Ps } Prs {log Prs + log Pr + log Ps }

!

r

!

Prs log Ps

(3.2.11)

!

(3.2.12)

It is easy to see that (3.2.12) may be readily rewritten as: S − (S1 + S2 ) = kB

XX r

s

Prs log



Pr Ps Prs



(3.2.13)

Problem 3.13 a)

Evaluate the partition function and the major thermodynamic properties of an ideal gas consisting of N1 molecules of mass m1 and N2 molecules of mass m2 , confined to a space of volume V at temperature T . Assume that the molecules of a given kind are mutually indistinguishable, while those of one kind are distinguishable from those of the other kind.

b)

Compare your results with the ones pertaining to an ideal gas consisting of (N1 + N2 ) molecules, all of one kind, of mass m, such that m (N1 + N2 ) = m1 N1 + m2 N2 .

Solution a)

We know that the volume and temperature of the system are held constant, and the system consists of two unique gases, each distinguishable from the other type, but indistinguishable from its own

239

CHAPTER 3: STATISTICAL MECHANICS

type. Since the two gases are independent of one another, this lends the idea that the partition function may be expressed as the product of two distinct partition functions: Q (N1 , N2 , V, T ) = Q(N1 , V, T ) · Q(N2 , V, T )

(3.2.14)

Where we recall that 1 Q(N, V, T ) = N !h3N

(ZZ

" P # )N N 3N 3N i Ei exp − dp dq kB T

(3.2.15)

Where the subscript N has been suppressed. Assuming that there is no interaction between the particles, then there is no potential energy term to the Hamiltonian, and the energy is due to the kinetic energy alone. Accordingly, (3.2.15) becomes:  Z ∞   N 1 p2 3N Q(N, V, T ) = V exp − dp N !h3 2mkB T −∞ oN 3 1 n / = V [2πmkB T ] 2 (3.2.16) 3 N !h Applying the general prescription from (3.2.16) to (3.2.14) yields Q (N1 , N2 , V, T ) =

oN1 oN2 3 3 1 n 1 n V [2πm1 kB T ] /2 · V [2πm2 kB T ] /2 3 3 N1 !h N2 !h

(3.2.17)

We start by finding the “major” thermodynamic properties by finding the Helmholtz Free Energy: F = − kB T log [Q (N1 , N2 , V, T )]  oN1 oN2  3 3 1 n 1 n /2 /2 = − kB T log V [2πm1 kB T ] · V [2πm2 kB T ] N1 !h3 N2 !h3

(3.2.18)

One major property is the pressure, which can be found by differentiating the free energy with respect to V :   ∂F P = ∂V T ,N     ∂ ∂ = log (...)V N1 + log (...)V N2 ∂V ∂V ∂ ∂ =N1 log [V ] + N2 log [V ] ∂V ∂V Carrying out this simple differentiation, we are left with P =

N1 + N2 V

Another important thermodynamic property is the internal energy, which is easy enough to find: ∂ log [Q (N1 , N2 , V, T )] ∂β   (  (   3/2 )N1  3/2 )N2 ∂ 1 2πm 1 2πm 1 2  =− log  V · V ∂β N1 !h3 β N2 !h3 β

U =−

W. Erbsen

HOMEWORK #2

" # " # 3 3 ∂ 1 /2 ∂ 1 /2 = − N1 log (...) − N2 log (...) ∂β β ∂β β     3N1 ∂ 1 3N2 ∂ 1 =− log (...) − log (...) 2 ∂β β 2 ∂β β

(3.2.19)

From (3.2.19) we can gather U =

3 kB T (N1 + N2 ) 2

We can also calculate the entropy from (3.2.18) as   ∂F S =− ∂T N,V   oN1 oN2  3 3 ∂ 1 n 1 n /2 /2 V [2πm1 kBT ] · V [2πm2 kB T ] =− −kB T log ∂T N1 !h3 N2 !h3

(3.2.20)

Differentiating (3.2.20) leads us to   oN1 oN2  3 3 1 n 3 1 n /2 /2 V [2πm1 kB T ] V [2πm2 kB T ] (N1 + N2 ) S = kB log · + N1 !h3 N2 !h3 2T b)

If the gas is no longer a mixture of two unique species, then the partition function becomes n oN1 +N2 3 1 Q (N1 , N2 , V, T ) = V [2πmkB T ] /2 6 N1 !N2 !h The pressure and internal energy remain unchanged, while the entropy becomes:   n oN1 +N2  3 1 3 /2 S = kB log V [2πmkB T ] + (N1 + N2 ) 6 N1 !N2 !h 2T

Problem 3.15 Show that the partition function QN (V, T ) of an extreme relativistic gas consisting of N monatomic molecules with energy-momentum relationship E = pc, c being the speed of light, is given by (  3 )N 1 kB T QN (V, T ) = 8πV (3.2.21) N! hc Study the thermodynamics of this system, checking in particular that 1 P V = U, U/N = 3kB and 3

γ=

4 3

Next, using the inversion formula (3.4.7), derive an expression for the density of states g(E) of this system.

Solution

241

CHAPTER 3: STATISTICAL MECHANICS

We first recall that the partition function can be expressed as: (ZZ " P # )N N 1 3N 3N i Ei QN (V, T ) = exp − dp dq N !h3N kB T Assuming that our relativistic gas is in 3-D, then (3.2.22) becomes: (ZZ " P # )N N 1 3 3 i pi c QN (V, T ) = exp − d pd q N !h3 kB T ( Z " P # )N N 1 3 i pi c V exp − d p = N !h3 kB T  Z ZZ   N 1 pc 2 = V exp − p sin θ dp dθ dφ N !h3 kB T    N Z ∞ pc 1 2 4πV dp = p exp − N !h3 kB T 0 (  3 )N 1 kB T = 4πV · 2 N !h3 c

(3.2.22)

(3.2.23)

Rearranging (3.2.23) somewhat, we are left with: 1 QN (V, T ) = N!

(

8πV



kB T hc

3 )N

Which is the same as (3.2.21). We now recall that the internal energy is given by: E=−

∂ log [QN ] ∂β

(3.2.24)

Substituting (3.2.21) into (3.2.24), we have:   (  3 )N ∂ 1 k T B  E =− log  8πV ∂β N! hc "  1  3 # /N ∂ 1 1 =−N log 8πV ∂β N! βhc "  1  # /3N ∂ 1 1 = − 3N log 8πV ∂β N! βhc   1 = − 3N − β

(3.2.25)

From (3.2.25) it is easy to see that: E = 3kB T N

(3.2.26)

W. Erbsen

HOMEWORK #2

We now recall that the free energy is given by: F = −kB T log [QN ]

(3.2.27)

Substituting (3.2.21) into (3.2.27), we have: 

 (  3 )N 1 k T B  F = − kB T log  8πV N! hc "  1  3 # /N kB T 1 8πV = − N kB T log N! hc We innocently recall that the pressure is defined in terms of the free energy as:   ∂F P =− ∂V T ,N

(3.2.28)

(3.2.29)

And so, putting (3.2.28) into (3.2.29): ( "  1  3 #) /N ∂ 1 kB T P =− −N kB T log 8πV ∂V N! hc "  1  3 # /N 1 kB T ∂ log 8πV =N kB T ∂V N! hc =

N kB T V

(3.2.30)

We can now manipulate (3.2.26), such that: N kB T =

E 3

(3.2.31)

And now substituting (3.2.31) into (3.2.30), P =

E 1 E · −→ P V = 3 V 3

We now recall that the specific heat at constant volume is defined by:   ∂E CV = ∂T V

(3.2.32)

(3.2.33)

And substituting (3.2.26) into (3.2.33), we are left with: CV =

∂ 3N kB T −→ CV = 3N kB ∂T

While the specific heat at constant pressure is given from (1.3.18) as:   ∂ (E + P V ) CP = ∂T N,P

(3.2.34)

(3.2.35)

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CHAPTER 3: STATISTICAL MECHANICS

Evaluating the internal energy in terms of P V from (3.2.32) and substituting into (3.2.36): ∂ (3P V + P V ) ∂T ∂ =4 PV ∂T

CP =

(3.2.36)

Recalling the from the ideal gas law that P V = N kBT , (3.2.36) becomes: CP = 4

∂ N kB T −→ CP = 4N kB ∂T

(3.2.37)

Using (3.2.34) and (3.2.37), we can now find γ: γ=

4N kBT 4 CP ⇒ −→ γ = CV 3N kBT 3

(3.2.38)

Problem 3.22 The restoring force of an anharmonic oscillator is proportional to the cube of the displacement. Show that the mean kinetic energy of the oscillator is twice its mean potential energy.

Solution We begin our journey by recalling that the average of some quantity hxi can be found from the corresponding distribution function f(x) according to: Z hxi = x · f(x) dx (3.2.39) Now, in order to find the mean kinetic energy, we recall that in general, the 1-D kinetic energy is given by: T = 1/2 mv2

(3.2.40)

So, the mean value of the kinetic energy is then: hT i = 1/2 mhv2 i

(3.2.41)

We can now find hv2 i from (3.2.41) by use of (3.2.39), recalling that the distribution function in this case is none other than the Maxwell-Boltzmann Velocity Distribution:   r m mv2 f(v) = exp − (3.2.42) 2πkB T 2kBT Using (3.2.42), hv2 i becomes: hv2 i =

Z

∞

−∞

v2 · f(v) dv

W. Erbsen

HOMEWORK #2

r =

m 2πkB T

  mv2 v2 exp − dv 2kB T −∞

Z

∞

(3.2.43)

At this point, we recall that integrals in the form of (3.2.43) can be evaluated as: √ Z ∞   π x2 exp −ax2 dx = 3 /2 2a −∞ Applying this to (3.2.43), we have:   3/ √ m π 2kB T 2 · · 2πkB T 2 m kB T = m

r hv i = 2

(3.2.44)

Substituting (3.2.44) back into (3.2.41): hT i =

kB T m kB T · ⇒ 2 m 2

We now recall that according to Boltzmann statistics, the mean energy is given by:   1 X Es hEi = Es exp − Z s kB T

(3.2.45)

(3.2.46)

And now, remembering the relationship between force and potential energy: F (x) = −

∂V (x) ∂x

(3.2.47)

We are also told that the restoring force is proportional to the cube of the displacement, so: F (x) ∼ −Cx3

(3.2.48)

Where C is some constant. Accordingly, substituting (3.2.48) into (3.2.47), we have: V (x) ∼ Cx4

(3.2.49)

Using Boltzmann statistics according to (3.2.46), we find that the mean potential energy is then given by:   1 X V (x) hV i = V (x) · exp − (3.2.50) Z kB T Converting the sum in (3.2.50) into an integral,   Z 1 ∞ V (x) hV i = V (x) · exp − dx Z −∞ kB T   Z 1 ∞ Cx4 = Cx4 · exp − dx Z −∞ kB T
  5/ 1 Γ 5/4 kB T 4 = · · Z 2 C

(3.2.51)

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CHAPTER 3: STATISTICAL MECHANICS

We now wish to evaluate the partition function Z:   Z ∞ Cx4 Z= exp − dx kB T −∞  1
kB T /4 5 =2 · Γ /4 · C

(3.2.52)

Substituting (3.2.52) back into (3.2.51),


   1/4  5/ Γ 5/4 1 1 C kB T 4 hV i = · · · · 2 Γ ( 5/4 ) kB T 2 C kB T = 4

(3.2.53)

So, in summary, from (3.2.45) and (3.2.53), we have: hT i =

kB T , 2

hV i =

kB T 4

−→

2hT i = hV i

Problem 3.42 Consider the system of N magnetic dipoles, studied in 3.10, in the microcanonical ensemble. Enumerate the number of microstates, Ω(N, E), accessible to the system at energy E, and evaluate the quantities S(N, E) and T (N, E). Compare your results with (3.10.8) and (3.10.9).

Solution Within the framework of the microcanonoical ensemble, we note that the energy in the system is fixed and therefore conserved. If we assume that our magnetic dipoles can have only one of two unique states (i.e. aligned or anti-aligned), then the total number of dipoles N can be expressed in terms of the individual components as: N = N↑ + N↓

(3.2.54)

n = N↑ − N↓

(3.2.55)

It will also be useful to define:

We can solve (3.2.54) and (3.2.55) giving expressions in terms of the number in each state: N +n 2 N −n N↓ = 2 N↑ =

Using (3.2.56a) and (3.2.56b), we can enumerate the number of microstates as:

(3.2.56a) (3.2.56b)

W. Erbsen

HOMEWORK #2

Ω (N, E) =

N! N! −→ Ω (N, E) = N↑ !N↓ ! [(N + n)/2]! [(N − n)/2]!

We could take the natural log of both sides of (3.2.57), " # N! log [Ω (N, E)] = log N+n
N−n
! 2 ! 2     N −n N +n ! − log ! = log [N !] − log 2 2

(3.2.57)

(3.2.58)

We can also apply Stirling’s Approximation to each of the three terms in (3.2.58). The first terms is log [N !] = N log [N ] − N The second term can now be expressed as         N +n N +n N +n N +n log ! = log − 2 2 2 2     1 N +n = (N + n) log −N −n 2 2       1 N +n N +n = N log + n log −N −n 2 2 2

(3.2.59)

(3.2.60)

While the third is log



       N −n N −n N −n N −n ! = log − 2 2 2 2     N −n 1 = (N − n) log −N +n 2 2       1 N −n N −n = N log − n log −N +n 2 2 2

Substituting (3.2.59)-(3.2.61) into (3.2.58), we have     N −n N +n log [Ω (N, E)] =N log [2] + N log [N ] − log [N − n] − log [N + n] 2 2

(3.2.61)

(3.2.62)

To find S(N, E), we first recall that S = kB log [Ω], and also that (3.2.62) is in a convienent form:       N −n N +n S(N, E) = kB N log [2] + N log [N ] − log [N − n] − log [N + n] (3.2.63) 2 2 We now recall the thermodynamic identity dU = T dS − P dv

(3.2.64)

At constant volume, (3.2.64) becomes dU = T dS −→

dS 1 = dU T

(3.2.65)

247

CHAPTER 3: STATISTICAL MECHANICS

Applying the chain rule to (3.2.65), dS dS dU dn dS dn 1 = · · ⇒ · = dU dU dn dU dn dU T Also recalling that in our case dU = E dn −→

1 dn = dE E

Substituting this back into the chain rule, we have dS 1 1 dS E · = −→ = dn E T dn T

(3.2.66)

Where we recall that E = ±µB H. Substituting (3.2.63) into (3.2.66),       µB H d N −n N +n =kB N log [2] + N log [N ] − log [N − n] − log [N + n] T dn 2 2   N −n kB log (3.2.67) = 2 N +n It is easy to see that (3.2.67) can be solved, yielding   2µB H N +n log T (N, E) = kB N −n

3.3

Homework #3

Problem 1 Consider a system of N non-interacting spins of angular momentum J in the presence of an external field H. The energy levels for each spin are gµB Hm` , where m` = −J, −J + 1, ..., J. a)

Compute the partition function for this system (evaluate all sums)

b)

Compute the average magnetic moment for this system

c)

Compare your results to the J = 1/2 example from lecture and the J = ∞ classical case (take the limit J → ∞ and g → 0 such that µB gJ retains the constant value µ0 )

d)

Evaluate the Curie constant for the generic spin J case

Solution

W. Erbsen

HOMEWORK #3

a)

The single-particle canonical partition function is defined by   X Ei Q1 ≡ exp − kB T i

(3.3.1)

In our case, the energy is due to the interaction between the spins and the magnetic field, whose relationship was given in the prompt. Accordingly, (3.3.1) becomes   mX ` =+J gµB Hm` Q1 = exp − (3.3.2) kB T m` =−J

At this point, we define the following quantity: x=

gµB H kB T

Rewriting (3.3.2) in terms of x and expanding the sum, Q1 =

mX ` =+J

e−xm`

m` =−J

=e−x(−J) + e−x(−J+1) + e−x(−J+2) + ... + e−x(J−2) + e−x(J−1) + e−xJ =exj + e(J−1)x + e(J−2)x + ... + e−(J−2)x + e−(J−1)x + e−xJ h i =e−xJ e2xJ + e(2J−1)x + e(2J−2)x + ... + e2x + ex + 1 h i =e−xJ 1 + ex + e2x + ... + e(2J−2)x + e(2J−1)x + e2xJ h i 1 =e−xJ 1 + ex + e2x + ... + e2(J−1)x + e2(J− /2 )x + e2xJ

(3.3.3)

Taking a slight detour, we recall that we can express a finite sum in terms of two infinite sums like

1 + y + y2 + y3 + ... + yx = 1 + y + y2 + y3 + ... − yx+1 + yx+2 + ...

= 1 + y + y2 + y3 + ... − yx+1 1 + y + y2 + y3 + ...

= 1 + y + y2 + y3 + ... 1 − yx+1 (3.3.4) We can simplify this further by recalling that

∞
X 1 + y + y2 + y3 + ... =

1 1−y

(3.3.5)

1 − yx+1 1−y

(3.3.6)

y

Using (3.3.5), we can now rewrite (3.3.4) as 1 + y + y2 + y3 + ... + yx =

Confusingly, we recognize that in the notation of (3.3.6), we let y → ex , and x → 2J. Accordingly, (3.3.3) becomes   1 − e(2J+1)x Q1 =e−xJ 1 − ex =

e−xJ − e(J+1)x 1 − ex

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CHAPTER 3: STATISTICAL MECHANICS

x

e−xJ − e(J+1)x e /2 · x/ = 1 − ex e 2 x −xJ− /2 (J+1)x− x/2 e −e = e x/2 − ex− x/2 1 1 e−(J+ /2 )x − e(J+ /2 )x = x x e− /2 − e /2 1 1 e(J+ /2 )x − e−(J+ /2 )x = e x/2 − e− x/2 1 1 e(J+ /2 )x − e−(J+ /2 )x 1 · x/ = 1 e 2 − e− x/2 Recalling that 2 sinh (x) = ex − e−x , (3.3.7) becomes 
 sinh J + 1/2 x Q1 = sinh [ x/2 ] b)

(3.3.7)

(3.3.8)

To find the average magnetic moment, we first recall that the relationship between the magnetization (M) and the magnetic moment (µ) is ? M=

N µ V

Such that the mean magnetic moment can now be defined in terms of the mean magnetization as hM i = N hµi We also recall that the magnetization is defined in terms of the free energy as   ∂F hM i = − ∂H T

(3.3.9)

(3.3.10)

Where the free energy is of course F = −kB T log [QN ]

(3.3.11)

Recalling that for distinguishable, non-interacting particles the full partition function is defined in terms of the single-particle partition function as QN = QN 1 . Keeping this in mind, and substituting (3.3.11) into (3.3.10),   ∂ 1 ∂Q1 hM i = kBT log QN −→ hM i = N kB T (3.3.12) 1 ∂H Q1 ∂H It will be much easier to evaluate (3.3.12) if we convert the partial derivative of Q with respect to H into the product of two other partial derivatives using the chain rule: hM i = N kB T

1 ∂Q1 ∂x ∂Q1 1 ∂x ∂Q1 −→ hM i = kB T Q1 ∂H ∂Q1 ∂x Q1 ∂H ∂x

(3.3.13)

We can solve the first partial derivative in (3.3.13) by recalling how we originally defined x:   ∂x ∂ gµB H gµB = ⇒ (3.3.14) ∂H ∂H kB T kB T We can also solve the second partial derivative in (3.3.13) by use of (3.3.8),

W. Erbsen

HOMEWORK #3

" 
# ∂Q1 ∂ sinh J + 1/2 x = ∂x ∂x sinh [ x/2 ]   
 
 ∂ 1 1 ∂  + sinh J + 1/2 x = sinh J + 1/2 x x x ∂x sinh [ /2 ] sinh [ /2 ] ∂x   x 
 

 1 coth [ / ] 1 2 = sinh J + 1/2 x − + J + 1/2 cosh J + 1/2 x x x 2 sinh [ /2 ] sinh [ /2 ] 
 
 1
cosh J + 1/2 x 1 sinh J + /2 x x =− · · coth [ / ] + J + 1/2 2 x x 2 sinh [ /2 ] sinh [ /2 ] 
 1
cosh J + /2 x Q J + 1/2 = − · coth [ x/2 ] + 2 sinh [ x/2 ] 

Q cosh [ x/2 ] cosh J + 1/2 x + J + 1/2 =− · x x 2 sinh [ /2 ] sinh [ /2 ]   

Q 1 1 1 x = cosh J + /2 x J + /2 − cosh [ /2 ] (3.3.15) sinh [ x/2 ] 2

Substituting (3.3.14) and (3.3.15) back into (3.3.13), we have     

Q 1 gµB 1 1 1 x hM i =N kB T cosh J + /2 x J + /2 − cosh [ /2 ] Q1 k B T sinh [ x/2 ] 2   

gµB 1 1 =N cosh J + 1/2 x J + 1/2 − cosh [ x/2 ] sinh [ x/2 ] Q1 2   

1 gµB sinh [ x/2 ] 1 1 x cosh J + /2 x J + /2 − cosh [ /2 ] =N sinh [ x/2 ] sinh [(J + 1/2 ) x] 2 ( ) 
 1 x
1 cosh [ /2 ] cosh J + /2 x =N gµB J + 1/2 − sinh [(J + 1/2 ) x] 2 sinh [ x/2 ]   

1 =N gµB coth J + 1/2 x J + 1/2 − coth [ x/2 ] (3.3.16) 2

Applying (3.3.16) to (3.3.9), we can say that the mean magnetic moment is   

1 hM i hµi = −→ hµi = gµB coth J + 1/2 x J + 1/2 − coth [ x/2 ] N 2 c)

From lecture, in the case of J = 1/2 , we came to the conclusion that   µB H hµi = µB tanh kB T

(3.3.17)

(3.3.18)

Whereas in the classical case, we came to the conclusion that hµi =

µ2B H 3kBT

We can take the limit of (3.3.18) as J → ∞ easily if we recall that coth[∞] → 1,  
1 hµi = gµB J + 1/2 − −→ hµi = gµB J 2

(3.3.19)

(3.3.20)

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CHAPTER 3: STATISTICAL MECHANICS

Figure 3.3: Plot of hµi for small J

Figure 3.4: Plot of hµi for large J

We can verify (3.3.20) graphically if we plot hµi from (3.3.17) against both x and J. We see from Fig. (3.3) that the dependence of hµi for small values of J is not clear, however if we plot the same thing for very large values of J, as in Fig. (3.4) it is easy to see that hµi carries a linear dependence on J, and seems to be independent of x, thus verifying (3.3.20). d)

In order to find Curie’s Constant, we first recall that Curie’s Law is valid only under conditions of low magnetization (µB T  kBT ), and does not apply in the high-field/low-temperature regime. ? We recall that x is directly proportional to H, while inversely proportional to T . Therefore, we can expand the coth[x] terms in (3.3.17) as: 3

coth[ax] ≈

5

ax (ax) 2 (ax) 1 + − + − ... ax 3 45 945

Taking only the first two terms in this expansion and plugging in to (3.3.17), (
!  )
1 2 x J + 1/2 x 1 1 hµi ≈gµB + + J + /2 − (J + 1/2 ) x 3 2 x 6 (
2 !  ) J + 1/2 x 1 1 x ≈gµB + − + x 3 x 12 ( )
J 2 + J + 1/4 x x ≈gµB − 3 12 ( )
J2 + J x x x ≈gµB + − 3 12 12 gµB ≈ J (J + 1) x 3

(3.3.21)

Recalling how we chose to define x at the beginning of the problem, (3.3.21) becomes hµi ≈

gµB gµB H g2 µ2B H J (J + 1) −→ hµi ≈ J (J + 1) 3 kB T 3kB T

Curie’s Constant is defined by

(3.3.22)

W. Erbsen

HOMEWORK #3

M=

C T H −→ C = M T H

(3.3.23)

Where C is Curie’s Constant. Recalling that M = N µ, and using (3.3.22), we can find Curie’s Constant from (3.3.23) as C=

N g2 µ2B J (J + 1) 3kB

Problem 2 Consider a liquid in equilibrium (both thermal and diffusive) with its vapor (treated as an ideal gas). For the liquid, we will apply the following crude model: we treat the liquids as if the molecules still formed a gas of molecules moving independently, but with the following considerations. 1) each molecule is assumed to have a constant potential energy − due to its interaction with the other molecules, and 2) each molecule is assumed free to move throughout a total volume N` v0 , where v0 is the average volume available per molecule in the liquid phase. a)

With these assumptions, write down the partition function for a liquid consisting of N` molecules.

b)

Now set the chemical potential of the liquid equal to the chemical potential of the vapor phase and find an expression for the vapor pressure in terms of the temperature and other constants like v0 and .

Solution a)

We begin our journey by recalling that the partition function can be expressed in integral form as (ZZ " P # )N N 1 H i QN (N, V, T ) = exp − i dp3N dq 3N (3.3.24) N !h3N kB T The Hamiltonian, H, in our case is a combination of the kinetic energy term, and the potential energy −, which is one of our assumptions in defining our crude model for the liquid. The Hamiltonian is then given by  N  2 X X pi H= {Ti + Vi } ⇒ − (3.3.25) 2m i i Substituting (3.3.25) into (3.3.24), (`) QN (N, V, T )

1 = N !h3N =

1 N !h3N

=

1 N !h3N

(Z Z

" PN  )N # p2i /2m −  i 3 3 exp − dp dq kB T Z Z   N p2  3 3 exp − + dp dq 2mkB T kB T   Z   N Z  p2 exp exp − dp3 dq 3 kB T 2mkB T

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CHAPTER 3: STATISTICAL MECHANICS

(



 Z



∞

p2 V exp exp − kB T 2mkB T −∞    hp i3 N  1 V exp = 2πmkB T N !h3N kB T (    3/ )N 1  2πmkB T 2 = V exp N! kB T h2 1 = N !h3N





dp

3 )N

(3.3.26)

Recalling that the total volume, V , is related to the average volume per molecule, v0 , through the number of molecules N (`) , as V = N (`) v0 , and so (3.3.26) becomes (`) QN (N, V, T )

b)

1 = N!

(

N

(`)

v0 exp



 kB T



2πmkB T h2

The chemical potential is given by:     ∂F ∂kB T log [Q] µ≡ ≡− ∂N V,T ∂N V,T

 3/2 )N

(3.3.27)

(3.3.28)

So, let’s find the chemical potential of the liquid first. Substituting (3.3.27) into (3.3.28), h i (`) ∂ log QN µ(`) = − kB T ∂N   (    3/2 )N ∂ 1  2πmk T B  log  = − kB T N v0 exp ∂N N! kB T h2  (      3/2 )N   ∂  2πmkB T  − log [N !] = − kB T log  N v0 exp  ∂N  kB T h2 ( " )    3/ # ∂  2πmkB T 2 = − kB T N log N v0 exp − log [N !] ∂N kB T h2 ( " " )    3/ ## ∂ ∂  2πmkB T 2 (3.3.29) = − kB T N log N v0 exp − log [N !] ∂N kB T h2 ∂N {z } | {z } | I II

We first evaluate I from (3.3.29): " "    3/ ## ∂  2πmkB T 2 I= N log N v0 exp ∂N kB T h2 " " "    3/ #    3/ ##  2πmkB T 2 ∂ ∂  2πmkB T 2 [N ] + N log N v0 exp = log N v0 exp kB T h2 ∂N ∂N kB T h2 "    3/ #  2πmkB T 2 = log N v0 exp +1 (3.3.30) kB T h2 We now carry out the remaining partial derivative in (3.3.29), employing Stirling’s approximation:

W. Erbsen

HOMEWORK #3

∂ log [N !] ∂N ∂ = {N log [N ] − N } ∂N ∂ ∂ ∂ = log [N ] N +N log [N ] − N ∂N ∂N ∂N = log [N ] + 1 − 1 = log [N ]

II =

Substituting (3.3.30) and (3.3.31) back into (3.3.29), ) ( "   3/ #   2πmkB T 2 (`) µ = − kBT log N v0 exp + 1 − log [N ] kB T h2 ( " )    3/ #  2πmkB T 2 = − kBT log v0 exp +1 kB T h2 ( " )    3/ #  2πmkB T 2 = − kBT log v0 exp + log [exp [1]] kB T h2 ( " #)    3/  2πmkB T 2 = − kBT log v0 exp exp [1] kB T h2 "    3/ #  2πmkB T 2 = − kBT log v0 exp +1 kB T h2

(3.3.31)

(3.3.32)

In order to find the chemical potential of the gas, we can simply rewrite (3.3.32), where in the new case  → 0, and we also recall that v0 = V /N : "   3/ # V 2πmkB T 2 (g) µ = −kB T log (3.3.33) N h2 Recalling that our gas is in fact an ideal gas, we notice that we can manipulate the ideal gas law as P V = N kB T −→ Substituting (3.3.34) back into (3.3.33), (g)

µ

"

kB T = −kB T log P

V kB T = N P 

2πmkB T h2

(3.3.34)  3/2 #

(3.3.35)

And now, to find the vapor pressure at thermal and diffusive equilibrium, we set µ(`) = µ(g) , and from (3.3.32) and (3.3.35) we have " "    3/ #   3/ #  2πmkB T 2 kB T 2πmkB T 2 −kB T log v0 exp +1 = − kB T log kB T h2 P h2    3/   3/  2πmkB T 2 kBT 2πmkB T 2 v0 exp +1 = kB T h2 P h2    kB T v0 exp +1 = (3.3.36) kB T P

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CHAPTER 3: STATISTICAL MECHANICS

We see that (3.3.36) may be trivially solved for P , yielding    kB T  P = exp − +1 v0 kB T

Problem 3 A zipper has N links, each link has a state in which it is closed with energy 0 and a state in which it is open with energy . We require that the zipper only unzip from the left end, and that the link number s can only open if all the links to the left(1, s, ..., s − 1) are already open. a)

Show that the partition function can be summed in the form: QN =

b)

1 − e−(N+1)β 1 − e−β

(3.3.37)

Find an expression for the average number of open links at some temperature T .

Solution a)

The standard form of the single-particle partition function is   ∞ X Es Q1 = exp − kB T s=0

(3.3.38)

Because the “particles” are distinguishable, then we can gain the full partition from (3.3.38) quite easily  s ∞ X Es Qs = exp − (3.3.39) kB T s=0 We now take a field trip. Let’s take a close look at the following series: N X

xs = 1 + x + x2 + x3 + ... + xN

(3.3.40)

s=0

We now multiply both sides of (3.3.40) by x. This looks like x

N X

xs = x + x2 + x3 + x4 + ... + xN+1

(3.3.41)

s=0

We now take (3.3.40) and subtract from it (3.3.41). This yields (1 − x)

N X s=0

xs = 1 − xN+1

(3.3.42)

W. Erbsen

HOMEWORK #3

If we were to rearrange (3.3.42), then it would look like N X

xs =

s=0

1 − xN+1 1−x

(3.3.43)

Noticing that this is the exact sum is what we need in order to evaluate (3.3.39)! Qs = b)

1 − exp [− (N + 1) Eβ] 1 − exp [−Eβ]

(3.3.44)

In general, the way to find the average number of open links, we use the form hsi =

N 1 X s · exp[−sEβ] Qs s=0

(3.3.45)

Now, notice that we can arrive at (3.3.45) if we are very tricky and take a derivative of the following form: hsi =

Rewriting (3.3.46) once more,

N 1 ∂ X exp [−sEβ] Qs ∂ (Eβ) s=0 | {z } Qs

hsi =

1 ∂ Qs Qs ∂x

(3.3.46)

(3.3.47)

Where I have made the temporary substitution x = Eβ for convenience. Carrying out the first derivative in (3.3.47),   ∂ 1 − exp [− (N + 1) x] ∂ Qs = ∂x ∂x 1 − exp [−x]     1 ∂ ∂ 1 = [1 − exp [− (N + 1) x]] + [1 − exp [− (N + 1) x]] 1 − exp [−x] ∂x ∂x 1 − exp [−x] " # 1 exp [−x] = [(N + 1) exp [−(N + 1)x]] − [1 − exp [−(N + 1)x]] 2 1 − exp [−x] (1 − exp [−x]) " # exp [−(N + 1)x] 1 − exp [−(N + 1)x] =(N + 1) − exp [−x] 2 1 − exp [−x] (1 − exp [−x])

exp [−(N + 1)x] exp [−x] exp [−x] exp [−(N + 1)x] − 2 + 2 1 − exp [−x] (1 − exp [−x]) (1 − exp [−x])   exp [−(N + 1)x] 1 exp [−x] exp [−(N + 1)x] exp [−x] =(N + 1) − − + 1 − exp [−x] 1 − exp [−x] 1 − exp [−x] 1 − exp [−x]   1 exp [−x] exp [−(N + 1)x] exp [−x] = (N + 1) exp [−(N + 1)x] − + 1 − exp [−x] 1 − exp [−x] 1 − exp [−x]    1 1 − exp [−(N + 1)x] = (N + 1) exp [−(N + 1)x] − exp [−x] (3.3.48) 1 − exp [−x] 1 − exp [−x] =(N + 1)

Where we note that the expression in brackets in (3.3.48) is none other than Qs . Rewriting,

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CHAPTER 3: STATISTICAL MECHANICS

∂ 1 Qs = [(N + 1) exp [−(N + 1)x] − exp [−x] Qs ] ∂x 1 − exp [−x]

(3.3.49)

At this point we are ready to substitute (3.3.49) and (3.3.44) into (3.3.47): 1 1 · [(N + 1) exp [−(N + 1)x] − exp [−x] Zs ] Qs 1 − exp [−x] (N + 1) exp [−(N + 1)x] − exp [−x] = 1 − exp [−x] (N + 1) exp [−(N + 1)x] exp [−x] = − 1 − exp [−x] 1 − exp [−x]

hsi =

(3.3.50)

Putting the numbers back in to (3.3.50) leads us to hsi =

(N + 1) exp [−(N + 1)Eβ] exp [−Eβ] − 1 − exp [−Eβ] 1 − exp [−Eβ]

(3.3.51)

We are asked to evaluate the average number of open links from (3.3.51) given the following stipulation: E  kBT −→ Eβ  1 If Eβ  1, then all the exponential terms approach zero, and so the average number of open links is hsi = 0

Problem 4.5 Show that expression (4.3.20) for the entropy of a system in the grand canonical ensemble can also be written as:   ∂ S = kB (T q) (3.3.52) ∂T µ,V

Solution We see that (4.3.20) from Pathria reads S = kB T



∂q ∂T



− N kB log [z] + kBq

(3.3.53)

z,V

We now recall that the differential of the q-potential has the form       ∂q ∂q ∂q dq = dT + dV + dz ∂T z,V ∂V z,T ∂z V,T Now that we are inspired, we may now rewrite the partial derivative in (3.3.53) as

(3.3.54)

W. Erbsen

HOMEWORK #3



∂q ∂T



⇒

z,V



∂q ∂T



+

V,T





∂q ∂T

µ,V

  ∂q ∂q ∂T ∂z ⇒ · · + ∂T ∂z ∂T ∂T µ,V   ∂q ∂q ∂z ⇒ · + ∂z ∂T ∂T µ,V

(3.3.55)

We now recall (4.3.17), which reads N (z, V, T )z



   ∂ ∂q N q(z, V, T ) −→ = ∂z ∂z V,T z V,T

(3.3.56)



(3.3.57)

Applying (3.3.56) to (3.3.55), 

∂q ∂T

N ∂z ⇒ + z ∂T z,V



∂q ∂T



µ,V

Substituting (3.3.57) back into (3.3.53), (   ) N ∂z ∂q S =kB T + − N kB log [z] + kB q z ∂T ∂T µ,V ( )   ∂q T ∂z =kB N +T − N log [z] + q z ∂T ∂T µ,V

(3.3.58)

We now innocently notice that T ∂z = log [z] z ∂T

(3.3.59)

Accordingly, we can now write (3.3.58) as (  )  ∂q S =kB T +q ∂T µ,V (     )   ∂q ∂T ∂ =kB T +q −→ S = kB (qT ) ∂T µ,V ∂T µ,V ∂T µ,V

Problem 4.8 Determine the grand partition function of a gaseous system of “magnetic” atoms (with J = 1/2 and g = 2) which can have, in addition to the kinetic energy, a magnetic potential energy equal to µB H or −µB H, depending upon their orientation with respect to the applied magnetic field H. Derive an expression for the magnetization of the system, and calculate how much heat will be given off by the system when the magnetic field is reduced from H to zero at constant volume and constant temperature.

259

CHAPTER 3: STATISTICAL MECHANICS

Solution The (canonical) single-particle partition function is defined by "P # ZZ N H 1 i i Q1 = 3 exp d3 p d3 q h kB T

(3.3.60)

The Hamiltonian in this case is H=

p2 ± µB H 2m

(3.3.61)

Applying (3.3.61) to (3.3.60),  2  ZZ p /2m ± µB H 1 exp d3 p d3 q h3 kB T   ZZ 1 p2 µB H = 3 exp ± d3 p d3 q h 2mkB T kB T  Z   Z 1 µB H p2 3 = 3 exp ± d q exp d3 p h kB T 2mkB T   Z ∞   3 V µB H p2 = 3 exp ± exp dp h kB T 2mkB T −∞   o3 µB H np V 2πmkB T = 3 exp ± h kB T

Q1 =

We now notice that the exponential term can be rewritten as         µB H µB H µB H µB H exp ± = exp + + exp − ⇒ 2 cosh − kB T kB T kB T kB T

(3.3.62)

(3.3.63)

Substituting (3.3.63) back into (3.3.62),   3 V µB H / Q1 =2 3 cosh − [2πmkB T ] 2 h kB T Using (3.3.64), we can now say that the full partition function is    N 3 1 V µB H /2 QN = 2 3 cosh − [2πmkB T ] N! h kB T

(3.3.64)

(3.3.65)

We now recall that the equation linking the grand partition function to the canonical partition function is Q=

∞ X

z N QN

N=0

Where z is the fugacity. Substituting (3.3.65) into (3.3.66),    N ∞ X 3 1 V µB H / Q= zN 2 3 cosh − [2πmkB T ] 2 N! h kB T N=0

(3.3.66)

W. Erbsen

HOMEWORK #3

   N ∞ X 3 1 V µB H /2 = 2z 3 cosh − [2πmkB T ] N! h kB T

(3.3.67)

N=0

We now innocently recall that the series expansion for ex is ex =

∞ X xN N!

N=0

We can clearly see that (3.3.67) follows this same form, and we can now write     3 V µB H /2 Q = exp 2z 3 cosh − [2πmkB T ] h kB T

(3.3.68)

The magnetization, M , is defined as M =−



∂F ∂H



(3.3.69) T

We must now find the free energy F , which is given by F = −kB T log [QN ]

(3.3.70)

Substituting in the canonical partition function from (3.3.65) into (3.3.70), "    N # 3 1 V µB H /2 F = − kBT log 2 3 cosh − [2πmkB T ] N! h kB T     3 V µB H 1 /2 = − N kBT log · 2 3 cosh − [2πmkB T ] N ! 1/N h kB T Now substituting (3.3.71) into (3.3.69), we can find the magnetization:      3 ∂ 1 V µB H /2 M =− −N kB T log · 2 3 cosh − [2πmkB T ] ∂H N ! 1/N h kB T     3 ∂ 1 V µB H /2 =N kB T log · 2 3 cosh − [2πmkB T ] 1 ∂H h kB T N ! /N       3 ∂ µB H 1 V /2 =N kB T log cosh − + log · 2 3 [2πmkB T ] ∂H kB T h N ! 1/N       3 ∂ µB H ∂ 1 V /2 log cosh − + log · 2 3 [2πmkB T ] =N kB T ∂H kB T ∂H h N ! 1/N    µB µB H =N kB T tanh kB T kB T

(3.3.71)

(3.3.72)

Cancelling terms in (3.3.72), we are left with M = N µB tanh



µB H kB T



At constant temperature and volume, the entropy can be quantified in terms of heat by

(3.3.73)

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CHAPTER 3: STATISTICAL MECHANICS

∆S =

∆Q −→ ∆Q = T ∆S T

(3.3.74)

We now note that ∆S in (3.3.74) is simply the change in entropy, i.e. the difference before and after H → 0. The entropy is defined as   ∂F S=− (3.3.75) ∂T V,N Substituting (3.3.71) into (3.3.75),      3 ∂ 1 V µB H /2 S =− − N kB T log · 2 3 cosh [2πmkB T ] ∂T N ! 1/N h kB T      3 3 1 V µB H 1 ∂ /2 /2 T log · 2 · [2πmk ] cosh T =N kB B 1 /N ∂T h3{z kB T |N ! } | {z } α

(3.3.76)

β

Rewriting (3.3.76) in terms of a new constant α and β,      3 β ∂ S =N kB T log α cosh T /2 ∂T T           3 3 β ∂ β =N kB log α cosh T /2 + T log α cosh T /2 T ∂T T         3 3 β β β T /2 + T − 2 tanh =N kB log α cosh T 2T T T        3 β 3 β β =N kB log α cosh T /2 + − tanh T 2 T T

Substituting back into (3.3.77) the value of β,        3 µB H 3 µB H µB H S =N kB log α cosh T /2 + − tanh kB T 2 kB T kB T

(3.3.77)

(3.3.78)

The initial entropy is given simply by (3.3.78), however the final entropy (when H = 0) can be found by setting H = 0 in (3.3.78):  h i 3 3 Sf =N kB log αT /2 + (3.3.79) 2 Then using (3.3.78) and (3.3.79), we can now find ∆S, ∆S =Sf − Si  h        i 3 3 3 µB H 3 µB H µB H =N kB log αT /2 + − N kB log α cosh T /2 + − tanh 2 kB T 2 kB T kB T  h       i 3 3 3 µ H 3 µ H µ H B B B =N kB log αT /2 + − log α cosh T /2 − + tanh 2 kB T 2 kB T kB T        µB H µB H µB H =N kB tanh − log α cosh (3.3.80) kB T kB T kB T

W. Erbsen

HOMEWORK #4

Substituting (3.3.80) back into (3.3.74), ∆Q = N kB T

3.4



      µB H µB H µB H tanh − log α cosh kB T kB T kB T

Homework #4

Problem 1 a)

Starting from the expression for the grand partition function, compute hN i and hN 2 i. Now show that 2 σN = hN 2 i − hN i2 = kB T

∂hN i ∂µ

(3.4.1)

b)

2 The above expression for σN can be written in terms of the isothermal compressibility defined as: KT = −1 −V (∂V /∂P )T . To do this, first show that µ = (∂G/∂N )T ,P where G is the Gibbs free energy.

c)

Then show that the Gibbs free energy is given by G = µN , i.e. the chemical potential is the Gibbs free energy per particle.

d)

Now express dµ in terms of dP and dT and show that     V 2 ∂P ∂µ =− 2 ∂N V,T N ∂V T ,N

e)

(3.4.2)

Finally, show that 2 σN kB T = κT N2 V

(3.4.3)

Solution a)

The Grand Partition Function, Q, is defined by   X N µ − Es,N Q= exp kB T

(3.4.4)

N,s

We also recall that in general, the average of a given quantity x can be defined by X hxi = xi P i i

Where Pi is the probability distribution function. In our case, we are asked to find hN i, and so applying this principle to N ,

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  1 N µ − Es,N hN i = N exp Q kB T s,N   N µ − Es,N 1X N exp = Q kB T X

(3.4.5)

s,N

Similarly, we can say for hN 2 i that

  1X 2 N µ − Es,N hN i = N exp Q kB T 2

(3.4.6)

s,N

2 We now wish to find σN – the first step is to take the partial derivative of Q with respect to the chemical potential, µ:     ∂ X N µ − Es,N  ∂Q = exp  ∂µ ∂µ  kB T N,s   X ∂ N µ − Es,N exp = ∂µ kB T N,s   X N N µ − Es,N = exp (3.4.7) kB T kB T N,s

Moving the factor of kB T to the LHS of (3.4.7),   ∂Q X N µ − Es,N kB T = N exp ∂µ kB T

(3.4.8)

N,s

We now note that we can rewrite (3.4.5) using (3.4.8), hN i =

1 ∂Q kB T Q ∂µ

(3.4.9)

Similarly, we now wish to find the second partial derivative of Q from (3.4.4), or equivalently, the first derivative of (3.4.9):     ∂2Q ∂ 2 X N µ − Es,N  = exp  ∂µ2 ∂µ2  kB T N,s     ∂ X N N µ − Es,N  = exp  ∂µ  kB T kB T N,s   X N ∂ N µ − Es,N = exp kB T ∂µ kB T N,s   X N2 N µ − Es,N = (3.4.10) 2 exp kB T N,s (kB T ) Moving the thermal factor on the RHS of (3.4.10) to the LHS,   2 X N µ − Es,N 2 ∂ Q 2 (kB T ) = N exp ∂µ2 kB T N,s

(3.4.11)

W. Erbsen

HOMEWORK #4

Using (3.4.11), we can rewrite (3.4.6), hN 2 i =

1 ∂2Q (kB T )2 Q ∂µ2

(3.4.12)

Substituting (3.4.9) and (3.4.12) into (3.4.1), 2 σN =hN 2 i − hN i2

2 1 ∂Q kB T Q ∂µ  2 2 1 ∂ Q 1 ∂Q 2 = (kB T )2 − (k T ) B Q ∂µ2 Q2 ∂µ "  2 # 2 ∂ Q 1 ∂Q kB T kB T 2 − kB T = Q ∂µ Q ∂µ " # kB T ∂2Q 1 ∂Q ∂Q = kB T 2 − kB T · Q ∂µ Q ∂µ ∂µ | {z } 2 1 2 ∂ Q = (kB T ) − Q ∂µ2



Noticing that the underbraced term in (3.4.13) is hN i from (3.4.9),   kB T ∂2Q ∂Q 2 σN = kBT 2 − hN i Q ∂µ ∂µ   ∂Q ∂Q kB T ∂ = · kB T −hN i Q ∂µ ∂µ ∂µ | {z }

Solving (3.4.5) for the underbraced term in (3.4.14) and making the substitution,   kB T ∂ ∂Q 2 σN = (QhN i) − hN i Q ∂µ ∂µ

(3.4.13)

(3.4.14)

(3.4.15)

We not notice that the first partial derivative may be carried out to be ∂ ∂hN i ∂Q (QhN i) = Q + hN i ∂µ ∂µ ∂µ Substituting (3.4.16) back into (3.4.15), we find that   kB T ∂hN i ∂Q ∂Q ∂hN i 2 2 σN = Q + hN i − hN i −→ σN = kB T Q ∂µ ∂µ ∂µ ∂µ b)

(3.4.16)

(3.4.17)

To show that we can write the chemical potential in terms of the Gibbs Free Energy as suggested in the prompt, we first recall that the Gibbs Free Energy is defined by (1.3.15) as G = E − T S + P V ⇒ µN

(3.4.18)

Where the RHS can be used if the intensive parameters T , p, and µ remain constant while the extensive parameters N , V , and E grow proportionately with one another (See Footnote 8 on pg. 28). Therefore, the chemical potential can be defined in terms of the Gibbs Free Energy as   ∂G µ= (3.4.19) ∂N T ,P

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CHAPTER 3: STATISTICAL MECHANICS

c)

Recalling that the non-differential analogue to (1.3.4) reads E = T S − P V + µN

(3.4.20)

Substituting (3.4.20) into the LHS of (3.4.18), we see that G = (T S − P V + µN ) − T S + P V −→ G = µN d)

(3.4.21)

Extending the Gibbs Free Energy to a multi-particle system, we can find the differential as dG = µdN + N dµ

(3.4.22)

We also note that the differential of G from (3.4.18) is dG = −SdT + V dP + µdN

(3.4.23)

Equating (3.4.22) and (3.4.23), we find that µdN + N dµ = −SdT + V dP + µdN −→ dµ = − Letting dT → 0, (3.4.24) becomes dµ =

S V dT + dP N N

V ∂µ V dP −→ = N ∂P N

(3.4.24)

(3.4.25)

Applying the chain rule to (3.4.25), ∂µ ∂P ∂V V ∂µ ∂V V ∂µ V ∂P = −→ = −→ = ∂P ∂V ∂P N ∂V ∂P N ∂V N ∂V

(3.4.26)

Applying the chain rule once again to the LHS of (3.4.26), ∂µ ∂V ∂N V ∂P ∂µ ∂N V ∂P = −→ = ∂V ∂N ∂V N ∂V ∂N ∂V N ∂V We now note that from Maxwell’s relations, we can write     ∂N ∂P = ∂V T ,µ ∂µ T ,V

(3.4.27)

(3.4.28)

Substituting (3.4.28) into (3.4.27),       ∂µ ∂P V ∂P = ∂N T ,V ∂µ T ,V N ∂V T ,N (3.4.29) From a previous homework set, we were asked to establish     ∂P ∂P N −N = V −→ =− ∂µ T ,V ∂µ T ,V V Substituting (3.4.32) into (3.4.31),         N V ∂P V 2 ∂P ∂µ ∂µ =− −→ =− 2 ∂N T ,V V N ∂V T ,N ∂N T ,V N ∂V T ,N

(3.4.30)

(3.4.31)

W. Erbsen

HOMEWORK #4

e)

The first thing we must do is recognize that the isothermal compressibility, κT , can be quantified as   1 ∂V κT = − (3.4.32) V ∂P T ,N For a system being isothermally compressed (both T and N constant), we note that we can rewrite (3.4.31) as         V 2 ∂P ∂hN i N 2 ∂V ∂µ =− 2 −→ =− 2 ∂N T ,V N ∂V T ,N ∂µ T ,V V ∂P T ,N " #   N2 1 ∂V = − (3.4.33) V V ∂P T ,N Substituting (3.4.32) into the RHS of (3.4.33),     ∂hN i N2 ∂hN i N2 = κT −→ kBT = kB T κT ∂µ T ,V V ∂µ T ,V V

(3.4.34)

Substituting (3.4.1) into (3.4.34) and rearranging, we see that 2 σN = kB T

N2 σ2 kB T κT −→ N2 = κT V N V

Which is the same as (3.4.3).

Problem 2 A surface with N adsorption sites is in equilibrium with a mixed ideal gas containing molecules of type A and B. Each site can take either one single A molecules with energy −A or a single B molecule with energy −B . Compute the grand canonical partition function for this adsorbed system. Then find relations between the fractional occupations (θA = nA /N and θB = nB /N ) and the partial pressures pA and pB .

Solution Recalling that the Grand Canonical Partition Function (i.e. the Gibbs Sum) is given by   X µs N − Es,N Q= exp kB T

(3.4.35)

N,s

In order to find the PT for the adsorbed system, we assume that there are only two states available to the system, one where molecule A is adsorbed and the other where molecule B is adsorbed. There is no state where neither molecule is connected - this is deduced from the wording of the problem. We recall that the Grand PT can be defined in terms of the canonical PT from (4.3.15) as X Q= z n Qn (3.4.36) n

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CHAPTER 3: STATISTICAL MECHANICS

Where z is the fugacity (= eµs /kB T ) and Qn is the canonical PT. Following the example in class of a gas adsorbed onto a lattice, , we note that the canonical PT with degeneracy is  n Es,n Qn = gn · exp − (3.4.37) kB T Where the degeneracy is gN =

N (N − 1)...(N − n + 1) N! ⇒ n! (N − n)!n!

(3.4.38)

 n N! Es,n Qn = exp − (N − n)!n! kB T

(3.4.39)

Substituting (3.4.38) into (3.4.37),

And now substituting (3.4.39) back into the Grand PT from (3.4.36),    n X µs N! Es,n · exp − Q= exp kB T (N − n)!n! kB T n,s   X (µs − Es,n ) n N! = exp (N − n)!n! kB T n,s Using the Binomial Expansion, (3.4.40) becomes, N  X µs − Es Q= 1 + exp kB T s

(3.4.40)

(3.4.41)

In our case, since we have two possible energies, the sum is carried out to A and B and hence has only two terms. We now see that (3.4.41) becomes Q=



1 + exp



  N (µA + A ) (µB + B ) + exp kB T kB T

(3.4.42)

We now recall from class that the thermodynamic potential, Φ, is defined in terms of the Grand Canonical PT as Φ = −kB T log [Q]

(3.4.43)

Substituting (3.4.36) into (3.4.43), Φ = −kB T log

"

X n

n

#

z Qn ⇒ −kB T log [Q]

(3.4.44)

We can now take the partial derivative of (3.4.44) with respect to µ, which yields −

∂Φ ∂ =z log [Q] ∂µ ∂z

(3.4.45)

Noticing that the RHS of (3.4.45) is equal to n, or the average number of adsorbed molecules, we now can say

W. Erbsen

HOMEWORK #4

n=−

∂Φ ∂µ

(3.4.46)

Substituting the thermodynamic potential from (3.4.44) into (3.4.46), n =−

∂ {−kB T log [Q]} ∂µ

(3.4.47)

At this point, we remember that it is important to be very careful, as both n and µ are a function of the specific states in the Gibbs sum. Rewriting (3.4.48) to make this more explicit, ns = − k B T

∂ log [Q] ∂µs

And now substituting the grand canonical PT from (3.4.42) into (3.4.48), "    N # ∂ (µA + A ) (µB + B ) ns = − k B T log 1 + exp + exp ∂µs kB T kB T      (µA + A) (µB + B ) ∂ log 1 + exp + exp = − N kB T ∂µs kB T kB T

(3.4.48)

(3.4.49)

We can go no further until we explicitly define which state we are evaluating ns for. Starting with A, (3.4.49) becomes      ∂ (µA + A ) (µB + B ) nA =N kBT log 1 + exp + exp ∂µA kB T kB T N ∂ = log [1 + exp [(µA + A ) β] + exp [(µB + B ) β]] β ∂µA   N exp [(µA + A) β] = β β 1 + exp [(µA + A ) β] + exp [(µB + B ) β] exp [(µA + A ) β] =N (3.4.50) 1 + exp [(µA + A ) β] + exp [(µB + B ) β] The average number of adsorbed molecules per site is given by θs = ns /N , and we are of course interested in the fractional coverage of each type of molecule to each site. Accordingly, θA can be found from (3.4.50) to be θA =

exp [(µA + A ) β] 1 + exp [(µA + A ) β] + exp [(µB + B ) β]

(3.4.51)

θB =

exp [(µB + B ) β] 1 + exp [(µA + A ) β] + exp [(µB + B ) β]

(3.4.52)

And similarly, for θB ,

The time has now come to find the partial pressure pA and pB . The first step is to isolate µ from θ in either (3.4.51) or (3.4.52). Following the example from lecture, we first wish to find the chemical potential of the surface (µsurf ), and equate it with the chemical potential of the gas (µgas ). Starting with µsurf , we recognize that µA = µB ⇒ µ, and also θA = θB ⇒ θ, and so

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CHAPTER 3: STATISTICAL MECHANICS

1 1 + exp [(µsurf + A ) β] + exp [(µsurf + B ) β] = θ exp [(µsurf + A ) β] exp [(µsurf + A ) β] exp [(µsurf + B ) β] 1 + + = exp [(µsurf + A ) β] exp [(µsurf + A ) β] exp [(µsurf + A) β] 1 =1 + + exp [(B − A ) β] exp [(µsurf + A ) β]

(3.4.53)

If we define ∆ = B − B , then (3.4.53) becomes

  1 1 1 exp [− (µsurf + A ) β] = − 1 − exp [∆β] −→ µsurf = − log − 1 − exp [∆β] − A θ β θ

(3.4.54)

From class, we know that the chemical potential of an ideal gas is given, in general by µgas =

1 log [P · f(T )] β

(3.4.55)

Equating (3.4.54) and (3.4.56), −

  1 1 1 log − 1 − exp [∆β] − A = log [P · f(t)] β θ β     1 1 − 1 − exp [∆β] + A β = log log θ P · f(t)   1 1 − 1 − exp [∆β] exp [A β] = θ P · f(t)

Recalling that ∆ = B − A, (3.4.56) becomes     1−θ 1 1 θ 1 −→ P = exp [A β] − exp [B β] = θ P · f(t) f(T ) 1 − θ exp [A β] − exp [B β]

(3.4.56)

(3.4.57)

If we assume that the total gas mixture is given by P from (3.4.57), then the partial pressure is defined by pi = ni P . To find nA , we follow this logic and from (3.4.50) and (3.4.57), we can find pA :   exp [(µA + A ) β] 1 θ 1 pA = N · (3.4.58) 1 + exp [(µA + A ) β] + exp [(µB + B ) β] f(T ) 1 − θ exp [A β] − exp [B β] And similarly, we can say that the partial pressure of molecule B is given by   exp [(µB + B ) β] 1 θ 1 pB = N · 1 + exp [(µA + A ) β] + exp [(µB + B ) β] f(T ) 1 − θ exp [A β] − exp [B β]

(3.4.59)

It might be possible to reduce (3.4.58)–(3.4.59) to a slightly nicer form, however it will likely still be quite ugly.

Problem 3

W. Erbsen

HOMEWORK #4

Suppose that N molecules of H2 O gas (assumed to behave like a classical ideal gas) are introduces into a container of fixed volume V at a temperature low enough that virtually all the gas remains as molecular water vapor. At higher temperatures dissociation can take place according to the reaction 2H2 O 2H2 + O2

(3.4.60)

Let x be the fraction of H2 O molecules dissociated at some temperature T corresponding to the total gas pressure P . Write an equation relating x, P , and F (t) where F (t) is some function of temperature (that can include the binding energy of the water molecule).

Solution The dynamic reaction in (3.4.60) can be rewritten as 2H2 O − 2H2 − O2 = 0 We can express (3.4.61) in terms of coefficients and species according to coefficients are A1 = H2 , a1 = −2,

A2 = O2 , a2 = −1,

(3.4.61) P

i

ai Ai , and in our case the

A3 = H2 O a3 = 2

We also recall that the condition for chemical equilibrium can be defined in terms of the coefficients ai as X ai µi = 0 (3.4.62) i

Where this condition is met at the equilibrium pressure and temperature. We can express (3.4.61) in the language of (3.4.62) as 2µ (H2 O) − 2µ (H2 ) − µ (O2 ) = 0 The chemical potential for each species, µi , can be described by (1.5.7), which reads (   3/2 ) Ni h2 µi (Ni , V, T ) = kB T log V 2πmi kB T

(3.4.63)

(3.4.64)

We now define the fractional occupancy Ni , which we quantify as Ni0 = Ni /N . If we assume that the gas phase of each species can be treated as ideal, then we can write P V = N kBT → N/V = P/ (kB T ), and we can rewrite (3.4.64) as "   3/2 # P h2 0 µi (Ni , V, T ) =kB T log N kB T i 2πmi kB T "   3/2 # P h2 =kB T log + kB T log [Ni0 ] (3.4.65) kB T 2πmi kB T Using (3.4.62), we may rewrite (3.4.65) as

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CHAPTER 3: STATISTICAL MECHANICS

kB T

X i

ai

(

"

P log kB T



h2 2πmi kBT

 3/2 #

)

+ log [Ni0 ]

=0

Killing the factor in front of the sum and rearranging, (3.4.66) becomes "   3/2 # X X P h2 − ai log = ai log [Ni0 ] kB T 2πmi kBT i i "   3/2 #ai X X kB T 2πmi kB T a log = log [Ni0 ] i P h2 i i "   3/2 #ai X X V 2πmi kB T a log = log [Ni0 ] i 2 N h i i i  ai X X qi a log [Ni0 ] i log = Ni i

(3.4.66)

(3.4.67)

i

From lecture, we know that N1a1 · N2a2 · N3a3 · ... = q1a1 · q2a2 · q3a3 · ... ⇒ K(T, V )

(3.4.68)

ai And, simply by visual P inspection of (3.4.67), we can see that K(T, V ) = qi makes sense, recognizing that at equilibrium i Ki (T, V ) = 1 so the equation reduces readily to 0 = 0. We now recall that we can alternatively define the equilibrium constant in terms of (3.4.60) as

K(T, V ) =

[H2 O] 2

2

[H2 ] [O2 ]

⇒

{N (H2 O)}

2

2

{N (H2 )} {N (O2 )}

(3.4.69)

We also recall from lecture that Y  Ni ai i

qi

=1

Applying (3.4.70) to the case of (3.4.60),  2  −2  −1 NH2 O NH2 NO2 · · =1 qH2 O qH2 qO2

(3.4.70)

(3.4.71)

The total number of species at any time, ΣN 0 , can be quantified by ΣN 0 = NH2 O + NH2 + NO2

(3.4.72)

We are told to let x denote the fraction of water molecules dissociated after some period of time, and also that initially, the total number of free particulates are entirely water molecules, so that ΣN = NH2 O . Accordingly, we see that in my notation x = 1 − NH2 O /ΣN . With this, (3.4.72) becomes ΣN 0 = ΣN (1 − x) + NH2 + NO2

(3.4.73)

Following the same logic, we see that NH2 grows proportionally with NH2 O , whereas NO2 grows half proportionally with respect to NH2 O (according to the stoichiometric coefficients from (3.4.60)). We see that (3.4.73) becomes

W. Erbsen

HOMEWORK #4

ΣN 0 = ΣN (1 − x) + xΣN + x

ΣN ΣN ΣN ΣN x −→ 1 = (1 − x) + x+ 2 ΣN 0 ΣN 0 ΣN 0 2

(3.4.74)

Substituting the appropriate fractional coefficients into (3.4.71), 

ΣN (1 − x) qH2 O I

2  −2  1 −1 xΣN /2 xΣN · · =1 qH2 qO2 II

(3.4.75)

III

We now recall that qi is defined by   3/2 pi P pi h2 qi = · · kB T pi P 2πmi kB T   3/2 P Ni h2 = kB T ΣN 2πmi kB T ! 3/2 h2 Ni P P Ni −→ qi = = 3/ · 3/ F (T ) 3 2 ΣN ΣN m 2 2π (kB T ) mi i | {z } F (T )

(3.4.76)

In the language of (3.4.76), we can rewrite each of the terms in (3.4.75). Starting with I, 2 ΣN (1 − x) qH2 O  2 3 /  mH22 O  = (1 − x) ·  F (T )P 

I=



2

= (1 − x) ·

2

[F (T )P ]

(3.4.77a)

−2 xΣN qH2  −2 3 /  mH22  = x·  F (T )P  ( )−1 3 m H2 = x2 · 2 [F (T )P ]

II =



m3H2 O

2

1 [F (T )P ] · x2 m3H2 1 −1 /2 xΣN III = qO2  −1 3 / x mO22  = ·  2 F (T )P  =

(3.4.77b)

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CHAPTER 3: STATISTICAL MECHANICS

=

2 F (T )P · 3/ x mO22

(3.4.77c)

Substituting (3.4.77a)–(3.4.77c) back into (3.4.75), ) ( )( ( ) 2 m3H2 O 2 F (T )P 1 [F (T )P ] 2 · · (1 − x) · · =1 3/ 2 x2 m3H2 x [F (T )P ] mO22   3/2 2 m2H2 O (1 − x) ·2 F (T )P =1 x3 m2H2 mO2

(3.4.78)

Rearranging (3.4.78), we are left with F (T )P =

x3 2

2 (1 − x)



m2H2 mO2 m2H2 O

 3/2

Problem 4 Evaluate the density matrix ρmn of an electron spin in the representation which makes σ ˆx diagonal. Next, show that the value of hσz i, resulting from this representation, is precisely the same as the one obtained in 5.3. Hint : the representation needed here follows from the one used in 5.3 by carrying out a transformation with the help of the unitary operator √  √  1/√2 1/√2 ˆ U= (3.4.79) −1/ 2 1/ 2

Solution The Pauli spin matrices are given by   0 1 σ ˆx = , 1 0

σ ˆy =

 0 i

 −i , 0

σ ˆz =



1 0 0 −1



We now note that we can straightforwardly rewrite our unitary operator from (3.4.79) as   1 1 1 ˆ √ U= 2 −1 1

(3.4.80)

We also recognize that by using this matrix, we may transform our spin matrices to a basis in which σx is diagonal via a unitary transformation: ˆ †σ ˆ σ ˆx0 = U ˆx U Where σ ˆx0 denotes σˆx in the basis in which it is diagonal. Substituting (3.4.80) into (3.4.81),

(3.4.81)

W. Erbsen

HOMEWORK #4

 †     1 1 1 1 0 1 1 1 √ σ ˆx0 = √ 1 0 2 −1 1 2 −1 1     1 1 −1 0 1 1 1 = 1 1 0 −1 1 2 1    1 1 1 1 1 = −1 1 2 −1 1   1 −2 0 = 0 2 2   −1 0 = 0 1 So, the provided matrix does indeed diagonalize σ ˆx . We must now see  †    1 1 1 1 1 0 1 0 √ σ ˆz = √ −1 1 0 −1 −1 2 2     1 1 −1 1 0 1 1 = 1 0 −1 −1 1 2 1    1 1 1 1 1 = −1 1 2 1 −1   1 0 2 = 2 2 0   0 1 = 1 0

(3.4.82)

how it affects σ ˆz ,  1 1

(3.4.83)

ˆ is given by (5.3.1), The Hamiltonian operator, H, ˆ = −µB Bˆ H σz0

(3.4.84)

Which includes our new Pauli Matrix σ ˆz0 in the basis in which σ ˆx is diagonal. Substituting (3.4.83) into (3.4.84),   0 1 ˆ H = −µB B (3.4.85) 1 0 We now recall the expression for the density matrix in the canonical ensemble is given by (5.3.3) as   ˆ e−βH  (ˆ ρ) =  ˆ Tr e−βH  βµ B  1 e B 0 = βµB B (3.4.86) 0 e−βµB B e + e−βµB B But this is not the answer. We must now express this density matrix in our new basis, in the same way we did with our Pauli spin matrices before. Performing the unitary transformation on (3.4.86),

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CHAPTER 3: STATISTICAL MECHANICS

 βµ B   1 1 e B 0 1 √ 0 e−βµB B eβµB B + e−βµB B 2 −1    βµ B   1 1 1 −1 e B 0 1 1 = βµB B 1 −1 1 0 e−βµB B 2e + e−βµB B 1  βµ B    1 1 e B −e−βµB B 1 1 = βµB B βµ B B −βµ B B −1 1 e e−βµB B 2e +e  βµ B  1 e B + e−βµB B eβµB B − e−βµB B = βµB B e + e−βµB B eβµB B − e−βµB B eβµB B + e−βµB B   1 cosh [βµB B] sinh [βµB B] = 2 cosh [βµB B] sinh [βµB B] cosh [βµB B]

1 (ˆ ρ0 ) = √ 2



1 1 −1 1

† 

 1 1

(3.4.87)

We may now absorb the cosh term out front of (3.4.87) to inside the matrix, leaving us with 1 (ˆ ρ)= 2 0



1 tanh [βµB B]

tanh [βµB B] 1



(3.4.88)

We can find the expectation value of σˆx0 by generalizing the form of (5.3.4), hσx0 i = Tr [ˆ ρ0 σ ˆ0] Substituting σ ˆz0 from (3.4.83) and ρˆ0 from (3.4.88) into (3.4.89),    1 1 tanh [βµB B] 0 hσx0 i =Tr 1 1 2 tanh [βµB B]    1 tanh [βµB B] 1 =Tr 1 tanh [βµB B] 2

(3.4.89)

 1 0 (3.4.90)

Recalling that the trace of a matrix is simply the sum of the diagonal elements, (3.4.90) becomes hσx0 i = tanh [βµB B] Which is the same as found in 5.3.

3.5

Homework #7

Problem 1 Extension of the Debye Model of a Solid: Assuming the dispersion relation ω = Ak s , where ω is the angular frequency and k is the wave-number of a vibrational mode existing in a solid, show that the respective contribu3 tion towards the specific heat of the solid at low temperature is proportional to T /s .

W. Erbsen

HOMEWORK #7

Solution This problem is essentially the same as if we were trying to find the specific heat of a Debye solid at low temperatures, the only difference is that the dispersion relation has an extra term s in the exponent of the wave-number: ω = Ak s . We can carry through the arithmatic in very much the same way then, following Pathria’s work in 7.2 and 7.3. We first recall that the specific heat in terms of internal energy is given by   ∂U CV (T ) = ∂T V

(3.1.1)

So our goal then, is to find the internal energy, U , as a function of our dispersion relation. We recall Pathria’s definition of the Debye function from (7.3.18): Z x0 3 x3 dx (3.1.2) D(x0 ) = 3 x0 0 ex − 1 Where we have defined x0 to be x0 =

~ωD kB T

kB T ~ωD

3 Z

(3.1.3)

And substituting (3.1.3) back into (3.1.2), D(x0 ) = 3



0

x0

x3 dx ex − 1

(3.1.4)

We also realize that, to make things more transparent, Pathria has made the substitution x = ~ωE /kBT . Recalling our definition of x0 from (3.1.3), we note that in the low temperature limit, as suggested in the prompt, we see that x0 ∼ 1/T → ∞, and so (3.1.4) becomes  3 Z ∞ kB T x3 D(x0 ) =3 dx x ~ωD e −1 0  3  4  kB T π =3 (3.1.5) ~ωD 15 The only difference in our case is that we make the following change to (3.1.3),  s ~ωD x0 = kB T

(3.1.6)

Which we can see makes no difference in the integrand of (3.1.5). Therefore, the result from (3.1.5) using (3.1.6) instead of (3.1.3) is D(x0 ) =3



kB T ~ωD

 3/s 

π4 15



(3.1.7)

From (3.1.7) we can find the specific heat at low temperature by using the following approximation: CV (T ) = 3N kB E(x)

(3.1.8)

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CHAPTER 3: STATISTICAL MECHANICS

Which is (7.3.9) from Pathria. This is actually what determined the factor in our integral, which was solved in (3.1.5). So, our specific heat is then: CV (T ) = D(x0 ) =9N



kB T ~ωD

 3/s 

 3 π4 −→ CV ∼ T /s 15

Problem 2 Stability of a white dwarf against gravitational collapse: Consider a star to be made up of an approximately equal number N of electrons and protons (otherwise the Coulomb repulsion would overcome the gravitational interaction). Somewhat arbitrarily we would also assume that there is an equal number of neutrons and protons. Now, it is energetically favorable for a body held together by gravitational forces to be as compact as possible. On Earth, the gravitational force is not large enough to overcome the repulsive forces between atoms and molecules. Inside the sun, matter does not exist in the form of atoms and molecules, where the radiation pressure from the nuclear fusion keeps the star from collapsing. We will consider the case of a white dwarf, where the fusion has completed and there is no longer radiation pressure to prevent collapse. Assume that the temperature of the star is low enough, compared to the electron Fermi temperature, that the electrons can be approximated by a T = 0 Fermi gas (because of their large mass, the kinetic energy of protons and neutrons are assumed to be small compared to that of the electrons). a)

If the electron gas is non-relativistic, show that the electron kinetic energy (KE) of the star is given by Ekin

3~2 = 10me



9π 4

 2/3

5

N /3 R2

(3.1.9)

where me is the electron mass and R is the radius of the star. b)

The gravitational potential energy is dominated by neutrons and protons. Let mN be the nucleon mass. Assume the mass density is approximately constant inside the star. Show that, if there is an equal number N of protons and neutrons, then the gravitational potential energy is given by Epot = −

12 2 N 2 m G 5 N R

(3.1.10)

where G = 6.67 × 10−11 Nm2 /kg2 is the gravitational constant. c)

Find the radius for which the gravitational potential energy plus the kinetic energy is a minimum for a white dwarf with the same mass as the sun (1.99 × 1030 Kg) in units of solar radii (6.96 × 108 m).

d)

If the density is very large, the Fermi velocity of the electrons becomes comparable to the speed of light and we should use the relativistic formula for the relationship between energy and momentum. Compute the electron kinetic energy in the ultra relativistic limit  ≈ cp.

e)

You have just shown in d) that in the ultra relativistic limit, the electron kinetic energy is proportional to 4 N /3 /R, i.e. the R dependence is the same as the gravitational potential energy computed in part b) (which 4 remains unchanged). Since for large N , N 2  N /3 , we find that if the mass of the star is large enough, the gravitational potential energy will dominate. The star will then collapse. Show taht the critical value of N for this to happen is

W. Erbsen

HOMEWORK #7

Ncrit =



5~c 36πm2N G

 3/2 

9π 4



(3.1.11)

Substituting numbers, one find that this corresponds to about 1.71 solar masses. This is the so-called Chandrasekhar limit.

Solution a)

We recall that the total energy of a degenerate electron gas is given by Edeg =

3 N EF 5

(3.1.12)

Where the Fermi Energy, EF , is given by EF =

~2 2me



3π 2

N V

 2/3

(3.1.13)

Substituting (3.1.13) into (3.1.12), Edeg

"   2/3 # 3 ~2 2N = N 3π 5 2me V   2/3 3~2 2N = N 3π 10me V 2 2
/3 5 3~ 1 3π 2 · 2/ · N /3 = 10me V 3

(3.1.14)

Recalling that V = 4/3 πR3 , (3.1.14) becomes Edeg

  2/3
2 5 3~2 3 2 /3 = 3π · · N /3 3 10me 4πR   2/3 2 5 3~ 9π = · N /3 3 10me 4R

(3.1.15)

We note that (3.1.15) may be straightforwardly rewritten as Edeg b)

3~2 = 10me



9π 4

 2/3

5

N /3 R2

(3.1.16)

To find the gravitational potential energy, we assume the star to be a perfect sphere of mass density ρ, where we define ρ = m/V → m = ρV . Also recalling the volume of a sphere, we find that m=

4 3 πr ρ 3

(3.1.17)

The differential element of (3.1.17) is of course dm = 4πr 2 ρ dr We also recall that the potential energy of a celestial body is defined by

(3.1.18)

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CHAPTER 3: STATISTICAL MECHANICS

dU = −

Gm dm r

(3.1.19)

We can find the gravitational potential energy by integrating (3.1.19), and substituting in to this (3.1.17) and (3.1.18) we have Z R Gm dm Egrav = − r 0  Z R 
1 4πr 3 ρ =−G 4πr 2 ρ dr 3 0 r 2 2 Z R 16π Gρ r 4 dr =− 3 0  16π 2 Gρ2 R5 =− 3 5 2 2 16π Gρ R5 (3.1.20) =− 15 Substituting in the density in terms of the total mass, (3.1.21) becomes 2  16π 2 G M Egrav = − R5 4/ πR3 15 3   16π 2 G 9M 2 =− R5 15 16π 2 R6 3 GM 2 =− 5 R

(3.1.21)

If we want to express this in terms of the nucleon mass, we note that M = 2mN N , and so from (3.1.21) we have 2

Egrav = − c)

3 G (2mN N ) 12 Gm2N N 2 −→ Egrav = − 5 R 5 R

(3.1.22)

In order to find the equilibrium radius, Re , we must minimize the total energy of the star, which is the sum of the results from the previous two parts of this problem:   2/3 5/3 3~2 9π N 12 Gm2N N 2 Etot = Edeg + Egrav ⇒ − (3.1.23) 2 10me 4 R 5 R Minimizing (3.1.23) with respect to R, " #   2/3 5/3 ∂Etot ∂ 3~2 9π N 12 Gm2N N 2 = − ∂R ∂R 10me 4 R2 5 R 2       / 3~2 9π 3 5/3 ∂ 1 12 1 2 2 ∂ = N − Gm N N 10me 4 ∂R R2 5 ∂R R   2/3     5 3~2 9π 2 12 1 = N /3 − 3 − Gm2N N 2 − 2 10me 4 R 5 R 2   / 3~2 9π 3 5/3 1 12 1 =− N + Gm2N N 2 2 5me 4 R3 5 R

(3.1.24)

W. Erbsen

HOMEWORK #7

To minimize (3.1.24), we set the derivative with respect to R equal to zero and simplify:   2/   2/ 12 3~2 9π 3 5/3 1 1 9π 3 ~2 1 2 2 1 GmN N 2 = N −→ R = 2 3 5 R 5me 4 R 4 4 Gme mN N 1/3

(3.1.25)

We now wish to eliminate N , where we use our previous definition of the total mass M to declare that M = 2mN N → N = M/(2mN ), and (3.1.25) becomes 1   2/   1/   2/ 1 9π 3 ~2 2mN 3 (2) /3 9π 3 ~2 R= ⇒ (3.1.26) 5/ 2 4 4 Gme mN M 4 4 Gme m 2 M 1/3 N

Substituting in the appropriate values into (3.1.26),
2
1/3 1   2/ 1.055 × 10−34 J · s 1.990 × 1030 Kg (2) /3 9π 3 Re =
5 / 4 4 1.674 × 10−11 m3 · Kg−1 · s2 (9.109 × 10−31 Kg) (1.673 × 10−27 Kg) 2 =7.162 × 106 m

(3.1.27)

We can re-express (3.1.27) in terms of solar radii as Re Re 7.162 × 106 m = −→ = 1.029 × 10−2 8 R 6.960 × 10 m R d)

In the relativistic limit, we let  ≈ cp, and we can use the density of states formula from (7.2.16): 8πV 2 ε dε h 3 c3

(3.1.28)

V ε2 dε π 2 ~3 c3

(3.1.29)

a(ε) dε = Recalling that h = 2π~, (3.1.28) becomes a(ε) dε =

We can integrate the density of states from (3.1.29) to find the number of particles N : Z εF N = a(ε) dε 0 Z εF V = ε2 dε 2 π ~3 c3 0 V ε3 = 2 3 3 F π ~ c 3

(3.1.30)

We can now substitute in to (3.1.30) the equation for volume, N=

1 π 2 ~3 c3

ε3F 4 3 4R3 3 · πR ⇒ ε 3 3 9π~3 c3 F

(3.1.31)

Similarly, we can integrate the density of states from (3.1.29) to find the internal energy: Z εF U= εa(ε) dε Z 0 εF V = ε3 dε 2 ~3 c3 π 0 V ε4 = 2 3 3 F (3.1.32) π ~ c 4

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CHAPTER 3: STATISTICAL MECHANICS

Substituting in our value for V into (3.1.32), U =

1 π 2 ~3 c3

ε4F 4 3 R3 πR ⇒ ε4 4 3 3π~3 c3 F

(3.1.33)

We can now solve (3.1.31) for εF , which we will later substitute into (3.1.33). We find that:   1/3 1 ε3F 4 3 4R3 3 9π ~c N= 2 3 3 · πR ⇒ ε −→ εF = N (3.1.34) π ~ c 3 3 9π~3 c3 F 4 R Substituting (3.1.34) into (3.1.33), "  1/3 #4   4/3 4 4 (~c) R3 9π ~c R3 9π /3 U= N ⇒ N 3π~3 c3 4 R 3π~3 c3 4 R4

(3.1.35)

From (3.1.35) it is easy to see that the energy is then ~c U= 3π e)



9π 4

 4/3

4

N /3 R

(3.1.36)

To find the critical value of N , we evaluate (3.1.36) set equal to (3.1.22): " # 3/2 4   4/   4/ / ~c 9π 3 Nc 3 12 Gm2N Nc2 5~c 9π 3 1 = −→ Nc = 3π 4 R 5 R 36π 4 m2N G

(3.1.37)

From (3.1.37) we can simplify our expression for Nc and we arrive at 

5~c Nc = 36πm2N G

 3/2 

9π 4

2

Problem 3 Let the Fermi distribution at low temperatures be represented by a straight line (shown in Fig. 8.11) that is tangent to the actual curve at  = µ. Show that this approximate representation yields a “correct” result for the low-temperature specific heat of a Fermi gas, except that this numerical factor turns out to be smaller by a factor of 4/π 2 . Discuss, in a qualitative manner, the origin of the numerical discrepancy.

Solution First of all, I don’t really understand this problem. From Pathria, the low temperature specific heat of a gas is given from (8.1.39) as CV π 2 kB T = + ... N kB 2 EF

(3.1.38)

Recalling that the Fermi Energy can be defined in terms of the Fermi Temperature as TF = EF /kB, (3.1.38) may be rewritten as

W. Erbsen

HOMEWORK #7

π2 T CV = + ... N kB 2 TF

(3.1.39)

While the classical value is just given by CV =

3 CV 3 N KB −→ = 2 N kB 2

(3.1.40)

Dividing (3.1.38) by (3.1.40) we find that CVF-D = CVClass



π2 T 2 TF

  −1 3 π2 T · ⇒ 2 3 TF

(3.1.41)

The factor relating the classical to the quantum result for CV at low temperature seems is π 2 /3 . The reason why the quantum factor is larger than the classical factor has to do with the behavior of gases as T → 0. In the classical case, the barrier is assumed to be a step function, yielding the factor of 3/2. In the quantum case, things are not so easy – we must integrate the energy from the Fermi-Dirac distribution function, which yields a numerical result which ends up being attributed to the factor of π 2 /2 out front.

Chapter 4

Mathematical Methods 4.1

Homework #3

Problem 1 Determine if the following series converge: ∞ X 1 a) n(ln n)(ln(ln n)) n=10 b) c) d) e)

∞ X

n=2 ∞ X

ln

p [ n]ln n

4n + 1 3n − 1 n=0

(4.1.1b) (4.1.1c)

∞ X (n!)2 (n2 )! n=1 ∞ X

(4.1.1a)

(4.1.1d) n

(logπ 2)

(4.1.1e)

n=0

Solution a)

Since an+1 ≤ an it is to our advantage to use the integral test to see if (4.1.1a) converges. Graphically this just means that if we plot (4.1.1a) in terms of n we will get a curve that has negative instantaneous slopes at all (finite) points. Rewriting (4.1.1a),  Z ∞ ∞ X 1 1 ⇒ dx u = ln x, du = 1/x dx n(ln n)(ln(ln n)) 10 x(ln x)(ln(ln x)) n=10  Z ∞ 1 = du v = ln u, dv = 1/u du ln 10 u ln u Z ∞ 1 = dv v ln ln 10

W. Erbsen

HOMEWORK #3

∞

= [ln v|ln ln 10 → ∞

(4.1.2)

From (4.1.2) we can see that since v diverges, so does u and so does x, so from the integral test it is clear that (4.1.1a) diverges . b)

c)

This problem begs to be solved by way of the comparison test, however a more straightforward, qualitative explanation looms overhead. We know that ln n where n ≥ 2 is always a positive number, so all we must do is analyze p what’s left of (4.1.1b) and if that diverges then so does the entire function. We can see that [ n]ln n diverges since for significant values of n the square root p overcomes the enveloping ln function. Therefore, since [ n]ln n diverges , so too must (4.1.1b). This is a fairly straightforward; recall that if

lim an 6= 0, then

n→∞

In our case,

∞ X

an

diverges.

(4.1.3)

n=0

4n + 1 4n ⇒ lim n n n→∞ 3 − 1 n→∞ 3  n 4 = lim →∞ n→∞ 3 lim

(4.1.4)

So that (4.1.1c) diverges . d)

This can be analyzed using the ratio test. Letting an = (n!)2 /(n2 )! and an+1 = ((n+1)!)2 /((n+1)2 )!, we have an+1 (n + 1)!(n + 1)! [n · n]! = an [(n + 1)(n + 1)]! n! · n! (n + 1)(n + 1)[n · n]! = [(n + 1)(n + 1)]! (n + 1)2 =1 = (n + 1)2

(4.1.5)

The fact that (4.1.5) equals 1 signifies that the ratio test failed; we must find another way to test for convergence. A method that is not immediately apparent would be the integral test. The behavior of (4.1.1d) is complex near the origin, but eventually the denominator “wins,” and it dominates the behavior of the function. Subsequently, this implies that an+1 ≤ an and we can therefore use the integral test. Z ∞ ∞ X (n!)2 (n!)2 ⇒ dx 2 (n )! (n2 )! 1 n=1 ≈0.685292

(< 1)

(4.1.6)

Which was computed numerically using Mathematica. Since (4.1.6) is less than 1 then we can say that (4.1.1d) converges . e)

Here we can straightforwardly apply the ratio test, where an = (logπ 2)n and an+1 = (logπ 2)n+1 . So, an+1 (logπ 2)n+1 = an (logπ 2)n

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CHAPTER 4: MATHEMATICAL METHODS

= logπ 2 ≈0.605512

(< 1)

(4.1.7)

From (4.1.7) we can see that since an+1 /an < 1 then we know that (4.1.1e) converges .

Problem 2 Find the circle of convergence of the following series: ∞ X a) nz n b) c)

n=0 ∞ X

n=0 ∞ X

(4.1.8a)

n! n z nn

(4.1.8b)

(z + 5i)2n (n + 1)2

(4.1.8c)

n=0

Solution a)

To find the radius of convergence, we first must use a test to see for what values (4.1.8a) is convergent. In this case we should use the ratio test, where an = nz n and an+1 = (n + 1)z n+1 : (n + 1)z n+1 an+1 lim = lim n→∞ an n→∞ nz n (n + 1)z = lim n→∞ n n + 1 =|z| lim n→∞ n 1 =|z| lim 1 + n→∞ n =|z|

(4.1.9)

Since we are looking for the radius of convergence we require (4.1.9) to be a positive number less than 1, so that we have 0 ≤ |z| ≤ 1 . b)

Again, we need to go ahead and implement the ratio in order to attack this problem. Let an = n!z n /nn and an+1 = (n + 1)!z n+1 /(n + 1)n+1 . Then n+1 an+1 nn = lim (n + 1)!z lim n→∞ an n→∞ (n + 1)n+1 n!z n (n + 1)nn z = lim n→∞ (n + 1)n+1 nn =|z| lim n→∞ (n + 1)n

W. Erbsen

HOMEWORK #3

 2 n =|z| lim n→∞ n + 1   2 1 =|z| lim n→∞ 1 + 1/n

=|z|

1 e

(4.1.10)

The last step of (4.1.10) was achieved by noticing that the limit was nothing more than the inverse of Euler’s famous limit definition of e. Analogous to what we did in the last problem, to determine the radius of convergence you say that (4.1.10) is between 0 and 1, such that 0 ≤ |z|/e ≤ 1 → 0 ≤ |z| ≤ e . c)

We can find the radius of convergence for (4.1.8c) by way of the ratio test. As usual, go ahead and let an = (z + 5i)2n (n + 1)2 and an+1 = (z + 5i)2(n+1) (n + 2)2 . Then 2(n+1) an+1 (n + 2)2 = lim (z + 5i) lim n→∞ an n→∞ (z + 5i)2n (n + 1)2 (z + 5i)2 (n + 2)2 = lim n→∞ (n + 1)2 (n + 2)2 2 = |z + 5i| lim (4.1.11) n→∞ (n + 1)2 To find the limit in (4.1.11) we need to use the comparison test. We know that 2 (n + 2)2 2 → |z + 5i|2 lim (n + 2) |z + 5i| lim n→∞ (n + 1)2 n→∞ (n)2  (n + 2) 2 2 = |z + 5i| lim n→∞ (n)   1 + 1/n 2 2 = |z + 5i| lim n→∞ 1 | {z } 1

= |z + 5i|

2

(4.1.12) 2

And, as before, for convergence we require that (4.1.12) be between 0 and 1, so 0 ≤ |z + 5i| ≤ 1 → 0 ≤ |z + 5i| ≤ 1 .

Problem 3 Sum the following series: a)

∞ X (n + 1)(n + 2) 2n n=0

(4.1.13a)

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CHAPTER 4: MATHEMATICAL METHODS

b) c)

∞ X (−1)n n! n=0

(4.1.13b)

∞ X (−π)n (2n)! n=0

(4.1.13c)

Solution a)

The easiest way by far to evaluate this sum is to write out some of the terms to see what value it approaches. To make our job easier, let’s rewrite (4.1.13a): ∞ ∞ ∞ ∞ ∞ X X X n 1 (n + 1)(n + 2) X n2 + 3n + 2 X n2 = = +3 +2 n n n n n 2 2 2 2 2 n=0 n=0 n=0 n=0 n=0 | {z } | {z } | {z } Σ1 Σ2 Σ3

(4.1.14)

Where I have defined Σ1 , Σ2 and Σ3 to be: Σ1 = Σ2 = Σ3 =

∞ X 1 9 25 9 49 1 81 25 n2 = 0+ +1+ +1+ + + + + + ≈ 5.85742 n 2 2 8 32 16 128 4 512 256 n=0

∞ X n 1 1 3 1 5 3 7 1 9 5 =0+ + + + + + + + + + ≈ 1.98828 n 2 2 2 8 4 32 32 128 132 512 512 n=0

∞ X 1 1 1 1 1 1 1 1 1 1 1 =0+1+ + + + + + + + + + ≈ 1.99902 n 2 2 4 8 16 32 64 128 256 512 1024 n=0

So that we now have ∞ X (n + 1)(n + 2) ≈ Σ1 + 3Σ2 + 2Σ3 = 5.85742 + 3 · 1.98828 + 2 · 1.99902 = 15.8203 → 16 2n n=0

b)

This is rather straightforward to solve if we recognize that (4.1.13b) is similar in form to the series expansion of ex , ex = 1 + x +

∞ X x2 x3 xn xn + + ... + = 2! 3! n! n! n=0

(4.1.15)

Where at this point it should be noticed that x in (4.1.15) is equal to −1 in (4.1.13b). Making this substitution into (4.1.15) we are left with,

c)

∞ X (−1)n 1 = e−1 ⇒ n! e n=0

(4.1.16)

We approach this problem in the same way we did for the last one; look at a table of Taylor expansions for common functions and see if we can use one to solve (4.1.13c). In this case, we are interested in the expansion of cos x: cos x = 1 −

∞ X x2 x4 x6 (−1)n (−1)n 2n + − + ... + = x 2! 4! 6! (2n)! (2n)! n=0

(4.1.17)

W. Erbsen

HOMEWORK #4

Recognizing that x in (4.1.17) is

√

π in (4.1.13c) and substituting it into the right side of (4.1.17),

∞ ∞ ∞ X X √ (−1)n √ 2n X (−1)n (π)n (−π)n ( π) = = = cos π (2n)! (2n)! (2n)! n=0 n=0 n=0

4.2

(4.1.18)

Homework #4

Problem 1 Evaluate the integral:

I

CR

π cot (πz) dz z4

(4.2.1) 4

Where C is a circle of large radius R centered at z = 0, and prove that η(4) = π90 . Hint 1 : As R approaches infinity, the integral approaches zero. 3 Hint 2 : Series expansion of the cotangent function near zero is cot (x) = x1 − x3 − x45 + ...

Solution The first step is to substitute the series expansion given in the 2nd hint into (4.2.1):   I I π cot (πz) π 1 πz (πz)3 dz = − − + ... dz 4 z4 πz 3 45 CR CR z  I  1 π2 π4 = − − + ... dz z5 3z 3 45z CR

(4.2.2)

And also recall that I

C

1 dz = 0 zn

for

n 6= 1

(4.2.3)

According to (4.2.3), the only surviving term in (4.2.2) is the 3rd term: I I π cot (πz) π4 dz = − dz z4 CR CR 45z

(4.2.4)

And we can see that the residue at z0 = 0 is just −π 4 /45, and due to the residue theorem the integral ends up being I

CR

π cot (πz) dz =2πi z4



To find ζ(4), first take the taylor expansion of sin x/x:

−π 4 45



→ −

2π 5 45

(4.2.5)

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CHAPTER 4: MATHEMATICAL METHODS

x2 x4 x6 sin x =1 − + − + ... x 3! 5! 7! And also recall that the Euler-Wallis’ Sine Product has the form  ∞  sin x Y π 2 x2 = 1− 2 x n n=1   
π 2 x2 π 2 x2 = 1 − π 2 x2 1 − 1− ... 4 8

(4.2.6)

(4.2.7)

By multiplying out the first few terms of (4.2.7) and equating it with (4.2.6), it might be possible to find ζ(4). Mathews and Walker go through a significantly different method to solve the problem on pg. 53, however I did not find it to be intuitive and decided to investigate other ways to solve the problem. Unfortunately, I ran out of time!

Problem 2 Evaluate the integral: I(a) =

Z

0

∞

e−ax dx 1 + x2

(4.2.8)

By obtaining a differential equation for I(a) and solving it by variation of parameters. Express the answer using Ci(x) and Si(x) integrals.

Solution To solve this integral in the way suggested, the first step would be to combine several derivatives of our integral I(a) to form a differential equation, which we can then solve. So, let’s find the derivatives of (4.2.8) with respect to a: Z ∞ −ax xe I 0 (a) = dx (4.2.9) 1 + x2 0 Z ∞ 2 −ax x e 00 I (a) = dx (4.2.10) 1 + x2 0 Now, using (4.2.8), (4.2.9), and (4.2.10) we can create a differential equation. Following Mathews and Walker p. 61, let’s add I(a) and I 00 (a): Z ∞ 2 −ax Z ∞ −ax x e e 00 I (a) + I(a) = dx + dx 2 1 + x 1 + x2 0 0 Z ∞ −ax
e = x2 + 1 dx 2 1 + x Z0 ∞ = e−ax dx 0

W. Erbsen

HOMEWORK #4

=

1 a

So, our differential equation is then: I 00 (a) + I(a) =

1 a

(4.2.11)

Which takes the standard form of an inhomogeneous second order differential equation, P (x)y00 + Q(x)y0 + R(x)y = S(x)

(4.2.12)

At this point we use variation of parameters to solve (4.2.11). To start, first look at the general solution to the homogeneous analogue to (4.2.12): y(x) = c1 y1 (x) + c2 y2 (x)

(4.2.13)

If our equation was homogeneous, then we would have I 00 (a) + I(a) = 0, which has solutions I1 (a) = eia and I2 (a) = e−ia . But in the case of our inhomogeneous differential equation, the constants in front of the terms in (4.2.13) are also dependent on a, which we now refer to as ui (a) instead of ci : I(a) = u1 (a)I1 (a) + u2 (a)I2 (a)

(4.2.14)

Where from now on I will drop the explicit a dependence for convenience (eg I(a) → I). We must take the first and second derivatives of (4.2.14) and insert them into (4.2.11). Taking the first derivative, I 0 = u1 I10 + u2 I20 + u01 I1 + u02 I2

(4.2.15)

Where, following Mathews and Walker’s lead on pg. 9, we impose the condition that the last two terms in (4.2.15) sum to zero: u01 I1 + u02 I2 = 0

(4.2.16)

This is one of two equations which allows us to find u1 and u2 . Continuing, (4.2.15) becomes I 0 = u1 I10 + u2 I20

(4.2.17)

I 00 = u1 I100 + u2 I200 + u01 I10 + u02 I20

(4.2.18)

Now find the second derivative of (4.2.17):

Substituting (4.2.14) and (4.2.18) into (4.2.12): P I 00 + QI 0 + RI =P [u1 I100 + u2 I200 + u01 I10 + u02 I20 ] + Q[u1I10 + u2 I20 ] + R[u1I1 + u2 I2 ] =P [u01 I10 + u02 I20 ] + u1 [P I100 + QI10 + RI1 ] +u2 [P I200 + QI20 + RI2 ] | {z } | {z } 0 0

(4.2.19)

Where the last two terms vanish since both I1 and I2 are solutions to the homogeneous analogue to (4.2.12). So, remembering that P = 1 and S = 1/a, we now have: u01 I10 + u02 I20 = 1/a Now we must use (4.2.16) and (4.2.20) to find out what u1 and u2 are. Solving (4.2.16) for u01 :

(4.2.20)

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CHAPTER 4: MATHEMATICAL METHODS

u01 = −

u02 I2 I1

(4.2.21)

And substituting this into (4.2.20):  0  u I2 1 − 2 I10 + u02 I20 = I1 a   0 I I 1 2 1 u02 I20 − = I1 a u02

 −1 1 0 I2 I10 = I2 − a I1

(4.2.22)

At this point it should be remembered that I1 = eia ,I10 = ieia ,I2 = e−ia , and I20 = −ie−ia . Continuing,  −1 (e−ia )(ieia ) 1 u02 = (−ie−ia ) − a eia −1 1 = −ie−ia − ie−ia a eia =− 2ia  1  cos a − sin a = − (4.2.23) 2a i And integrating,

  Z Z 1 1 cos a sin a − da − da 2 i a a  Z  Z  1 1 cos a sin a da − da = − 2 i a a   1 1 = − Ci(a) − Si(a) 2 i Ci(a) Si(a) =− − 2i 2

u2 =

(4.2.24)

Where I used the fundamental definitions of Ci(x) and Si(x). To find u01 substitute (4.2.23) into (4.2.16):  ia  e I2 u01 = − − 2ia I1  ia  −ia e e = 2ia eia e−ia = 2ia  1  cos a = − sin a (4.2.25) 2a i

And integrating here as well,

u1 =

 Z  Z 1 1 cos a sin a da − da 2 i a a

W. Erbsen

HOMEWORK #4

 Z  Z  1 1 cos a sin a da − da 2 i a a   1 1 = + Ci(a) − Si(a) 2 i Ci(a) Si(a) = − 2i 2 =

(4.2.26)

At this point we substitute (4.2.24) and (4.2.26) into (4.2.14): I =u1 I1 + u2 I2     Ci(a) Si(a) ia Ci(a) Si(a) −ia = − e + − − e 2i 2 2i 2 Ci(a) ia Si(a) ia Ci(a) −ia Si(a) −ia e − e − e − e = 2i  2 2i  2   e−ia eia e−ia eia − + Si(a) − − =Ci(a) 2i 2i 2 2 =Ci(a)(sin a) − Si(a)(cos a)

(4.2.27)

But (4.2.27) is not the answer. I apologize but I did not have time to go back to change everything – recall that the definition for Ci(x) and Si(x) are: Ci(x) =

Z

x

∞

cos t dt t

(4.2.28)

sin t dt t

(4.2.29)

and Si(x) =

Z

0

x

So in order to make the substitution, we must account for the fact that the integration limits of Ci(x) are different from Si(x). In fact, we must require that I(a) and all it’s derivatives vanish at a = ∞, so we need to substitute the following correction in place of Si(a) in (4.2.27): Z a Z a Z ∞ sin a sin a sin a da = da − da a a a ∞ 0 0 π =Si(a) − (4.2.30) 2 So, substituting this into (4.2.27),  π I =Ci(a)(sin a) − Si(a) − (cos a) 2 π  = sin aCi(a) + cos a − Si(a) 2

Problem 3

(4.2.31)

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CHAPTER 4: MATHEMATICAL METHODS

Evaluate

Z

+∞ −∞

1 − cos x dx x2

(4.2.32)

Solution First recall Euler’s equation eix = cos x + i sin x. Since (4.2.32) has a cos x in the numerator, we may simply refer to it as the real part of eix : 0 What is the resulting motion?

Solution The differential equation describing the damped, forced harmonic oscillator is: x00 (t) + βx0 (t) + ω02 x(t) = F (t)

(4.3.13)

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CHAPTER 4: MATHEMATICAL METHODS

Where F (t) in this problem is defined by the heaviside step function. For t < 0, (4.3.13) becomes x00 (t) + βx0 (t) + ω02 x = 0 Which is easy enough to solve, start by finding the roots r1 and r2 : r 2 + βr + ω02 = 0 Which is clearly quadratic, letting a = 1, b = β and c = ω02 : p −β ± β 2 − 4ω02 r1,2 = 2 So the motion is described by " ! # " p −β + β 2 − 4ω02 x(t) = C1 exp i t + C2 exp i 2

−β −

p

β 2 − 4ω02 2

! # t

(4.3.14)

Where C1 and C2 are determined via initial conditions, which are not given. If we say that initially the system is not in motion, e.g. x(0) = 0 and x0 (0) = 0, then x(t) = 0 for all times. Otherwise it will oscillate in a predictable way. We should now investigate the second condition of the heaviside step function, where our differential equation picks up a constant term: x00 (t) + βx0 (t) + ω02 x(t) = F0 Which we can rearrange as 00

0

x (t) + βx (t) +

ω02

  F0 x(t) − 2 = 0 ω0

The addition of this term necessitates the inclusion of a “particular solution” to our homogeneous solution (4.3.15): " ! # " ! # p p −β + β 2 − 4ω02 −β − β 2 − 4ω02 F0 x(t) = C1 exp i t + C2 exp i t + 2 (4.3.15) 2 2 ω0 The effect of adding a constant external force to a damped harmonic oscillator only acts to offset the equilibrium position, so that the system oscillates around a different point, which is given by F0 /ω02 and is independent of time.

Problem 5 Show that the Fourier sine transforms of y0 (x) and y00 (x) are Fs [y0 ] = − ωyˆc (ω)

ω Fs [y00 ] = − ω2 yˆs (ω) + y(0) π

(4.3.16a) (4.3.16b)

W. Erbsen

HOMEWORK #5

Solution We first note the definitions of the Fourier sine and cosine transforms: Z ∞ Fs [f(x)] =fs (ω) = f(x) sin ωx dx Z0 ∞ Fc[f(x)] =fc (ω) = f(x) cos ωx dx 0

So, for the first derivative, f(x) = y0 (x) and: Fs [y0 (x)] =

Z

0

=

Z

0

∞ ∞

y0 (x) sin ωx dx ∂y(x) sin ωx dx ∂x

At this point, and this is a historic moment since I have never done this before, we want to do a sort of inverse integration by parts. We begin by noting that Z ∞ y(x) cos ωx dx (4.3.17) 0

Doing integration by parts, ∂y(x) ∂x 1 dv = cos ωx −→ v = sin ωx ω

u = y(x) −→ du =

Such that (4.3.17) becomes: ∞  Z ∞ Z 1 1 ∞ ∂y(x) y(x) cos ωx dx = y(x) sin ωx − sin ωx dx ω ω 0 ∂x 0 0

Solving this for

R

∂y(x) ∂x

Z

sin ωx dx,

0

∞

∞  Z ∞ ∂y(x) sin ωx dx = y(x) sin ωx − ω y(x) cos ωx dx ∂x 0 0 = − ωyc (ω) =Fs [y0 (x)]

(4.3.18)

The process of finding the second derivative is very much the same, recognizing that in this case f(x) = y00 (x) and Z ∞ Fs [y00 (x)] = y00 (x) sin ωx dx Z0 ∞ 0 ∂y (x) = sin ωx dx ∂x 0 Where this time we look at: Z

0

∞

y0 (x) cos ωx dx

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CHAPTER 4: MATHEMATICAL METHODS

Where we have ∂y0 (x) ∂x 1 dv = cos ωx −→ v = sin ωx ω

u = y0 (x) −→ du =

Such that Z

∞  Z 1 0 1 ∞ ∂y0 (x) y (x) sin ωx − sin ωx dx y (x) cos ωx dx = ω ω 0 ∂x 0

∞

0

0

And solving for the relavent term, ∞  Z ∞ 0 Z ∞ ∂y (x) 0 sin ωx dx = y (x) sin ωx − ω y0 (x) cos ωx dx ∂x 0 0 0 Z ∞ 0 =−ω y (x) cos ωx dx

(4.3.19)

0

We repeat this process yet another time, but this time around we are looking at Z ∞ Z ∞ ∂y(x) 0 y (x) cos ωx dx = cos ωx dx ∂x 0 0 And now we look at Z

∞

y(x) sin ωx dx

0

Where we let ∂y(x) ∂x 1 dv = sin ωx −→ v = − cos ωx ω

u = y(x) −→ du =

So Z

∞

0

And now Z

0

∞

y(x) sin ωx dx = −

∞  Z 1 1 ∞ ∂y(x) y(x) cos ωx + dx ω ω 0 ∂x 0

∞ Z ∞  ∂y(x) cos ωx dx = y(x) cos ωx + y(x) sin ωx dx ∂x 0 0 = − y(0) + ωys (ω)

Substituting this back into (4.3.19), Z ∞ 0 ∂y (x) sin ωx dx = − ω (ωys (ω) − y(0)) ∂x 0 = − ω2 ys (ω) + ωy(0)

(4.3.20)

W. Erbsen

HOMEWORK #5

In summary, we have shown that: Fs [y0 ] = −ωyc (ω) Fs [y00 ] = −ω2 ys (ω) + ωy(0) You might have noticed that what I found for Fs [y00 ] differs from yours by a factor of 1/π in the last term. I have no idea where that could have come from.

Problem 6 Show that the solution of y00 + 2µy0 + (β 2 + µ2 )y = g(x), With g(±∞) = 0 is 1 y= β

Z

x

−∞

µ > 0,

β > 0,

y(−∞) = y(∞) = 0

g(ξ)e−µ(x−ξ) sin [β(x − ξ)] dξ

(4.3.21)

(4.3.22)

Hint: Find Green’s function of the equation first.

Solution As per the hint, I am going to go ahead and find Green’s function for (4.3.21). Substituting G(x − ξ) for y assuming that x > ξ, we solve the homogeneous analogue of our given differential equation: G00 (x − ξ) + 2µG0 (x − ξ) + (β 2 + µ2 )G(x − ξ) = 0 Finding the roots, r 2 + 2µr + (β 2 + µ2 ) = 0 Letting a = 1, b = 2µ and c = β 2 + µ2 : r1,2 =

−2µ ±

=µ ± iβ

p

4µ2 − 4(β 2 + µ2 ) 2

So that we have G(x − ξ) =C1 cos[(µ + iβ)(x − ξ)] + C2 sin[(µ − iβ)(x − ξ)] 1 = sin[(µ − iβ)(x − ξ)] β

(4.3.23)

Due to the boundary conditions of the problem. At this point we should note that this problem represents a driven, damped harmonic oscillator, whose solution was discussed in an earlier problem. The solution

303

CHAPTER 4: MATHEMATICAL METHODS

to this type of problem using Green’s function is given in M&W in the prompt of problem 9-7. Tailoring it to our case, Z x y(x) = G(x − ξ)g(ξ) dξ (4.3.24) −∞

The finite upper limit comes from the fact that x in this problem really should be t, and up to this point we have been assuming that for x − ξ ≤ 0, G(x − ξ) = 0. To make this more transparent, we note that for a harmonic oscillator, we would have likely chosen “x” to be “t,” and “ξ” to be “t0 .” In this case our solution for y(x) (or y(t), rather) is represented by an integral over past times t. Substituting (4.3.23) into (4.3.24), Z 1 x y(x) = sin[(µ − iβ)x]g(ξ) dξ (4.3.25) β −∞ Where we can say 1 {exp[(µ − iβ)(x − ξ)] − exp[(−µ + iβ)(x − ξ)]} 2i 1 = {exp[µx − µξ − iβx + iβξ] − exp[−µx + µξ + iβx − iβξ]} 2i 1 = {exp[µ(x − ξ) + iβ(x − ξ)] − exp[−µ(x − ξ) + iβ(x − ξ)]} 2i exp[µ(x − ξ)] = {exp[iβ(x − ξ)] − exp[−iβ(x − ξ)]} 2i = exp[µ(x − ξ)] sin [β(x − ξ)]

sin[(µ − iβ)x] =

(4.3.26)

Now substituting (4.3.26) back into (4.3.25): 1 y(x) = β

Z

x

−∞

exp[µ(x − ξ)] sin [β(x − ξ)] g(ξ) dξ

In case you couldn’t tell I worked backwards a bit, sorry for the missteps in logic. I am very ignorant to Green’s functions (I find the entire thing very confusing), however it is an invaluable mathematical tool, one I hope to master soon.

4.4

Homework #6

Problem 1 Find the Laplace transform of f(t) = tet , and determine the region of existence.

Solution

W. Erbsen

HOMEWORK #6

We know that the Laplace Transform of some function f(t) is given by Z ∞ L {f(t)} = F (s) = f(t)e−st dt 0

And in our case, L {f(t)} =

Z

∞

tet e−st dt

0

Z

∞



u = t, du = 1 1 −(s−1)t dv = e , v = − s−1 e−(s−1)t 0 ∞  Z ∞ t −(s−1)t 1 = − e + e−(s−1)t dt s−1 s−1 0 0 ∞  1 −(s−1)t = − e 2 (s − 1) =

−(s−1)t

te

dt

0

1 = (s − 1)2

We can see that in the last limit, the result will be convergent only if s − 1 < 0, or s > 1, so F (s) =

1 (s − 1)2

for s > 1

Problem 2 Find the inverse Laplace Transform of a) b)

1 s3 − s2 s2 + s − 1 s3 − 2s2 + s − 2

Solution a)

As per the hint, let’s go ahead and employ partial fractions on (4.4.1a): s3

1 1 A B C =⇒ 2 = 2+ + 2 −s s (s − 1) s s s−1 Let s = 0: Let s = 1:

Let A = −1, C = 1: So, (4.4.1a) becomes

1 =A(s − 1) + Bs(s − 1) + Cs2 1 = − A −→ A = −1 1 =Cs2 −→ C = 1

1 =1 − s + Bs(s − 1) + s2 −→ B = 1

(4.4.1a) (4.4.1b)

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CHAPTER 4: MATHEMATICAL METHODS

1 1 1 1 = − − s3 − s2 s − 1 s2 s Where we note that

Z ∞ 1 et · e−st dt = e−(s−1)t dt = s−1 Z0 ∞ Z 0∞ 1 1 L {t} = te−st dt = e−st dt = 2 s 0 s 0 Z ∞ 1 L {1} = e−st dt = s 0

 L et =

Z

(4.4.2)

∞

(4.4.3a) (4.4.3b) (4.4.3c)

Substituting (4.4.3a)-(4.4.3c) into (4.4.2), 1 = et − t − 1 s3 − s2 b)

This time we only need to factor; no partial fractions is needed: (s2 + 1) (s2 + 1) + (s − 2) 1 1 s2 + s − 1 = + = 2 + s3 − 2s2 + s − 2 (s2 + 1)(s − 2) (s2 + 1)(s − 2) s +1 s−2 We note the Laplace Transforms of the following functions Z ∞ L {sin t} = sin te−st dt 0 Z  1 ∞  it = e − e−it e−st dt 2i 0 Z i 1 ∞ h (i−s)t e − e−(i+s)t dt = 2i 0 ∞  (i−s)t 1 e e−(i+s)t = − − 2i s−i s + i 0   1 1 1 = − 2i s − i s + i   1 2i = 2i s2 + 1 1 = 2 s +1 Z ∞ Z ∞  2t 2t −st L e = e ·e dt = e−(s−2)t dt = 0

0

Putting (4.4.5) and (4.4.6) into (4.4.4),

s2 + s − 1 = sin t + e2t s3 − 2s2 + s − 2

Problem 3

(4.4.4)

(4.4.5) 1 s−2

(4.4.6)

W. Erbsen

HOMEWORK #6

Solve the following differential equations using Laplace Transforms: a) y0 + y = cos t, for t > 0, y(0) = 1 b) y00 + y = cos (2t), for t > 0, y(0) = 0,

y0 (0) = 0

(4.4.7a) (4.4.7b)

Solution a)

First let’s find the Laplace Transform of cos t, since it will be needed later: Z ∞ L {cos t} = cos te−st dt 0 Z  1 ∞  it = e + e−it e−st dt 2 0 Z i 1 ∞ h (i−s)t e + e−(i+s)t dt = 2 0 ∞  (i−s)t 1 e e−(i+s)t = − + 2 s−i s + i 0   1 1 1 = + 2 s−i s+i   2s 1 = 2 s2 + 1 s = 2 s +1

(4.4.8)

Taking the Laplace Transform of both sides of (4.4.7a), L {y0 } + L {y} =L {cos t} s sY (s) − y(0) + Y (s) = 2 s +1 Applying the given initial condition and solving for Y (s), s 1 + + 1)(s + 1) s + 1 1+s 1 1 − + = 2 2(s + 1) 2(s + 1) s + 1       1 1 1 s 1 = + + 2 s2 + 1 2 s2 + 1 s+1

Y (s) =

(s2

Now we must find the inverse Laplace Transform of Y (s):       1 −1 1 1 −1 s 1 −1 −1 L {Y (s)} = L + L +L 2 s2 + 1 2 s2 + 1 s+1 Where we recall that the inverse Laplace Transform of 1/(s2 + 1) was found to be sin t in (4.4.5), and s/(s2 + 1) was found to be cos t in (4.4.8). Finding 1/(s + 1), Z ∞ Z ∞  1 L e−t = e−t · e−st dt = e−(s+1)t dt = s + 1 0 0

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CHAPTER 4: MATHEMATICAL METHODS

So we now have L−1 {Y (s)} =

sin t cos t1 + + e−t 2 2

And finally y(t) = b)

1 (sin t + cos t) + e−t 2

We first find the Laplace Transform of cos (2t), since it will be needed later: Z ∞ L {cos (2t)} = cos (2t)e−st dt 0 Z  1 ∞  2it = e + e−2it e−st dt 2 0 Z i 1 ∞ h (2i−s)t = e + e−(2i+s)t dt 2 0 ∞  (2i−s)t e e−(2i+s)t 1 = − + 2 s − 2i s + 2i 0   1 1 1 = + 2 s − 2i s + 2i   1 2s = 2 s2 + 4 s = 2 s +4

(4.4.9)

Now, applying the Laplace Transform to both sides of (4.4.7b): L {y00 } + L {y} =L {cos (2t)} s s2 Y (s) − sy(0) − y0 (0) + Y (s) = 2 s +4 Applying the initial conditions, s2 Y (s) + Y (s) = Y (s) =

s2

s +4

  s 1 s s = − (s2 + 4)(s2 + 1) 3 s2 + 1 s2 + 4

Taking the inverse Laplace Transform of Y (s),      1 s s −1 L−1 {Y (s)} = L−1 − L 3 s2 + 1 s2 + 4 Where again the inverse Laplace Transform of s/(s2 + 1) is just cos t as per (4.4.5). Similarly, s/(s2 + 4) is just cos (2t), which was found in (4.4.9). So, we are finally left with: y(t) =

1 [cos t − cos (2t)] 3

W. Erbsen

HOMEWORK #6

Problem 4 An electric circuit gives rise to the system dI1 + RI1 + dt dI2 L + RI2 − dt dq = I1 − I2 dt L

q = E0 C q =0 C

(4.4.10a) (4.4.10b) (4.4.10c)

With initial conditions E0 , q(0) = 0 2R Solve the system by Laplace Transform methods and show that E0 E0 −αt R 2 I1 = + e sin (ωt), where α = and ω2 = − α2 2R 2ωL 2L LC I1 (0) = I2 (0) =

(4.4.11)

(4.4.12)

Solution We start by taking the inverse Laplace Transform of (4.4.10c): L {q 0 (t)} =L {I1 (t)} − L {I2 (t)}

sq(s) − q(0) =I1 (s) − I2 (s) I1 (s) I2 (s) − q(s) = s s

(4.4.13)

We now do the same for (4.4.10a): 1 L {q(t)} =E0 L {1} C 1 E0 L [sI1 (s) − I1 (0)] + RI1 (s) + q(s) = C s     E0 1 I1 (s) I2 (s) E0 L sI1 (s) − + RI1 (s) + − = 2R C s s s R I1 (s) I2 (s) E0 E0 − ω02 = + sI1 (s) + I1 (s) + ω02 L s s Ls 2R   R ω2 I2 (s) E0 E0 s + + 0 I1 (s) − ω02 = + L s s Ls 2R  2  s + ω02 R I (s) E E0 2 0 + I1 (s) − ω02 = + s L s Ls 2R LL {I1 (t)} + RL {I1 } +

√ Where I have defined ω02 = 1/ LC . And similarly for (4.4.10b):

1 L {q(t)} =0 C 1 L [sI2 (s) − I2 (0)] + RI2 (s) − q(s) =0 C LL {I2 (t)} + RL {I2 } −

(4.4.14)

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CHAPTER 4: MATHEMATICAL METHODS

    E0 1 I1 (s) I2 (s) L sI2 (s) − + RI2 (s) − − =0 2R C s s R I1 (s) I2 (s) E0 + ω02 = sI2 (s) + I2 (s) − ω02 L s s 2R   R ω02 I (s) E0 1 s+ + I2 (s) − ω02 = L s s 2R   2 s + ω02 R I1 (s) E0 + I2 (s) − ω02 = s L s 2R

(4.4.15)

At this point we have two equations, (4.4.14) and (4.4.15), and two unknowns. Finding the s-domain solutions (I1 (s) and I2 (s)) is trivial, but time consuming. The most tedious steps in what follows were left up to Mathematica. The obvious first step would be to solve (4.4.14) for I2 (s), inserting into (4.4.15), and solving for I1 (s). The result is   1 E0 1 + I1 (s) = 2 Rs s(Ls + R) + 2Lω02   E0 1 1 = + 2 Rs Ls(s + R/L) + 2/C     E0 1 1 1 + Let α = R/2L = 2 Rs L s2 + sR/L + 2/LC    E0 1 1 1 = + 2 Rs L s2 + 2sα + 2/LC     E0 1 1 1 = + Complete the square 2 Rs L s2 + 2sα + α2 + 2/LC − α2     1 1 E0 1 = + Let ω = 2/LC − α2 2 Rs L (s + α)2 + 2/LC − α2    E0 1 1 ω = + 2 Rs ωL (s + α)2 + ω2   E0 1 E0 ω = + 2R s 2ωL (s + α)2 + ω2 Taking the inverse Laplace Transform, L

−1

    E0 −1 1 E0 −1 ω {I1 (s)} = L + L 2R s 2ωL (s + α)2 + ω2 E0 E0 −αt = + e sin (ωt) 2R 2ωL

Which was mercilessly stolen from a table. We finally have I1 (t) =

E0 E0 −αt + e sin (ωt) 2R 2ωL

Repeating this process for I2 (s), Mathematica gives   E0 1 1 I2 (s) = − 2 Rs s(Ls + R) + 2Lω02

W. Erbsen

HOMEWORK #7

Which we note is identical to our expression for I1 (s) sans the minus sign, therefore we can go ahead and conclude that: I2 (t) =

4.5

E0 E0 −αt − e sin (ωt) 2R 2ωL

Homework #7

Problem 1 For the vectors in three dimensions, ~v1 = x ˆ + yˆ,

~v2 = yˆ + zˆ,

~v3 = zˆ + x ˆ

(4.5.1)

Use the Gram-Schmidt procedure to construct an orthonormal basis starting from ~v1 .

Solution It might be most instructive if we explicitly define each of our vectors in terms of x, y, and z. e.g.,       1 0 1 ~v1 = 1 , ~v2 = 1 , ~v3 = 0 0 1 1 And now recall that the Gramm-Schmidt Orthonormalization procedure is ~un = ~vn −

n−1 X i=1

h~ui |~vn i u ~ i, h~ui |~ui i

~e1 =

u ~n k~un k

Where un represents a orthogonal set of vectors, and en represents an orthonormal set of vectors. Applying this to our case,     1 1 1 ~u1 =~v1 = 1 −→ ~e1 = √ 1 2 0 0 Where I used the fact that k~u1 k =

√

2 . Moving on to the second vector, u ~ 2 =~v2 −

h~v2 |~u1 i ~u1 h~u1 |~u1 i

Where   1 h~u1 |~u1 i =~ uT1 ~u1 = (1 1 0) 1 = 2, 0

  1 h~v2 |~u1 i = ~v2T ~v1 = (0 1 1) 1 = 1 0

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CHAPTER 4: MATHEMATICAL METHODS

Such that we now have      −1  r  −1/  0 1 /2 2  1 2 1 /2 ~u2 = 1 − 1 =  1/2  −→ ~e2 = 2 3 1 0 1 1

And now the last vector is

u ~ 3 =~v3 −

h~v3 |~u1 i h~v3 |~u2 i ~u1 − u ~2 h~u1 |~u1 i h~u2 |~u2 i

Where we have   1 h~v3 |~u1 i =(1 0 1) 1 = 1 0  1  − /2 h~v3 |~u2 i =(1 0 1)  1/2  = 1/2 1  1  − /2 h~u2 |~u2 i =(− 1/2 1/2 1)  1/2  = 3/2 1

And now

     1   2    1 1 − /2 /3 1 1 1 1 u ~ 3 = 0 − 1 −  1/2  = − 2/3  −→ ~e3 = √ −1 2 3 3 2 1 0 1 /3 1

Problem 2 For the vector space of polynomials in x, use the scalar product defined as Z ∞ 2 hf|gi = dxe−x f(x) · g(x)

(4.5.2)

−∞

Start from the vectors ~v0 = 1,

~v1 = x,

~v2 = x2 ,

~v3 = x3

(4.5.3)

And use the Gram-Schmidt procedure to construct an orthonormal basis starting from ~v0 . Repeat this calculation for the scalar product Z 1 hf|gi = dxx2 f(x) · g(x) (4.5.4) 0

Solution

W. Erbsen

HOMEWORK #7

Using the same logic as before, we are looking to ortho-normalize ~v0 , ~v1 , ~v2 , and ~v3 using the scalar product defined by (4.5.2). The first vector is trivial, ~u0 = ~v0 = ~e0 = 1 And the second is h~v1 |~u0 i u ~0 h~u0 |~u0 i hx|1i =x − 1 h1|1i R∞ −x2 dx −∞ xe =x − R ∞ −x2 e dx −∞   1/4 4 =x −→ ~e1 = x π

u ~ 1 =~v1 −

And the third is h~v2 |~u0 i h~v2 |~u1 i u ~0 − ~u1 h~u0 |~u0 i h~u1 |~u1 i hx2 |1i hx2 |xi − x =x2 − h1|1i hx|xi R ∞ 2 −x2 R ∞ 3 −x2 x e dx x e dx −∞ 2 R∞ =x − R ∞ −x2 − −∞ x 2 2 −x dx −∞ e −∞ x e √ π 1 2 √ =x − 2 π   1/4 1 4 2 =x − −→ ~e2 = (x2 − 1/2 ) 2 π

u ~ 2 =~v2 −

And lastly, the forth is h~v3 |~u0 i h~v3 |~u1 i h~v3 |~u2 i ~u0 − u ~1 − ~u2 h~u0 |~u0i h~u1 |~u1 i h~u2 |~u2 i

hx3 | x2 − 1/2 i hx3 |1i hx3 |xi 3 =x − − x− x2 − 1/2 2 1 2 1 h1|1i hx|xi h(x − /2 ) |(x − /2 )i R ∞ 3 −x2 R ∞ 4 −x2 R ∞ 3 2 1
−x2 x e dx x e dx x x − /2 e dx 2 1
−∞ −∞ 3 =x − R ∞ −x2 − R ∞ 2 −x2 x − R−∞ x − /2 ∞ 2 2 2 1 −x dx dx dx −∞ e −∞ x e −∞ (x − /2 ) e √ 3 π 2 √ x =x3 − 4 π   1/4 3 8 3 =x − x −→ ~e3 = (x2 − 1/2 ) 2 9π

u ~ 3 =~v3 −

Now we repeat this calculation for (4.5.4), in much the same way we did for (4.5.3). The first vector is then

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CHAPTER 4: MATHEMATICAL METHODS

~u0 = ~v0 = ~e0 = 1 While the second is given by h~v1 |~u0 i ~u0 h~u0 |~u0 i hx|1i =x − h1|1i R1 3 x dx =x − R01 x2 dx 0

~u1 =~v1 −

=x −

1 /4 1/ 3

=x −

√ 3 −→ ~e1 = 80 (x − 3/4 ) 4

And now the third h~v2 |~u0 i h~v2 |~u1 i u ~0 − ~u1 h~u0 |~u0 i h~u1 |~u1 i R1 4 R1 4 x dx x (x − 3/4 ) dx 2 0 =x − R 1 − R01 (x − 3/4 ) 2 dx 2 (x − 3/ ) dx x x 4 0 0

~u2 =~v2 −

1

/5 − 1/ 3 4 =x2 − x + 3

=x2 −

1

/60 (x 1/ 80

− 3/4 )

√ 2 −→ ~e2 = 1575 (x2 − 4/3 x + 2/5 ) 5

And the forth h~v3 |~u0 i h~v3 |~u1 i h~v3 |~u2 i u ~0 − ~u1 − u ~2 h~u0 |~u0 i h~u1 |~u1 i h~u2 |~u2 i R1 5 R1 5 R1 5 2 4 x dx x (x − 3/4 ) dx x (x − /3 x + 2/5 ) dx 2 4 3 3 0 0 =x − R 1 − R1 (x − /4 ) − R 10 (x − /3 x + 2/5 ) 2 dx 2 (x − 3/ ) dx 2 (x2 − 4/ x + 2/ )2 dx x x x 4 3 5 0 0 0

~u3 =~v3 −

1 1 /6 /56 /840 2 4 3 − (x − / ) − (x − /3 x + 2/5 ) 4 1/ 1/ 1/ 3 224 1575 p 15 3 7 =x3 − x2 − x + −→ ~e3 = 20160/1667 (x3 − 15/8 x2 − 3/2 x + 7/4 ) 8 2 4

=x3 −

1

Problem 3 On the vector space of quadratic polynomials, of degree ≤ 3, the operator d/dx is defined. Use the basis of the Legendre Polynomials 3 1 5 3 P~0 (x) = 1, P~1 (x) = x, P~2 (x) = x2 − , P~3 (x) = x3 − x (4.5.5) 2 2 2 2

W. Erbsen

HOMEWORK #7

And compute the components of this operator. Repeat this exercise for the operator d2 /dx2 .

Solution Let our set of quadratic polynomials be given by f(x) = a0 + a1 x + a2 x2 + a3 x3 (so, ~v0 = a0 , ~v1 = a1 x, ~v2 = a2 x2 , ~v3 = a3 x3 ). We can also define our differential operator to be D[f(x)] = df(x)/dx. So, D[~v0 ] =0 D[~v1 ] =a1 =⇒ a1 P~0 (x) D[~v2 ] =2a2 x =⇒ 2a2 P~1 (x) D[~v3 ] =3a3 x2 =⇒ a3 (2P~2 (x) + P~0 )

And if we let D2 [f(x)] = df(x)2 /dx2 , D[~v0 ] =0 D[~v1 ] =0 D[~v2 ] =2a2 =⇒ 2a2 P~0 (x) D[~v3 ] =3a3 x2 =⇒ 6a3 P~1 This is my wild stab at the problem, however I don’t believe it is a very accurate one, since if it were the problem would be trivial (clearly). Whoops!

Problem 4 Diagonalize each of the Pauli spin matrices. Find their eigenvalues and specify the respective eigenvectors as the basis in which they are diagonal.

Solution The Pauli matrices are given by  0 σx = 1

 1 , 0

σy =

 0 i

 −i , 0

−λ 1

1 = 0 −→ λ2 − 1 = 0 −λ

σz =

 1 0

0 −1



Starting with σx ,

For λ = +1 :

 −1 1

1 −1

λ = ±1

    a 0 −a + b = 0 = −→ b 0 a−b = 0



a = b,

so

  1 1

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CHAPTER 4: MATHEMATICAL METHODS

 1 1

For λ = −1 :

    1 a 0 a+b=0 = −→ 1 b 0 a+b=0



a = −b,

so



1 −1



so

  1 i

And now for σy : −λ i For λ = +1 :



−1 i

For λ = −1 :



1 −i i 1

−i −1

−i = 0 −→ λ2 − 1 = 0 −λ λ = ±1

    a 0 −a − ib = 0 = −→ b 0 ia − b = 0

    a 0 a − ib = 0 = −→ b 0 ia + b = 0





a = −ib,

a = ib,

so



1 −i



In analyzing σz we immidiately recognize that λ = ±1, since it is diagonal. So, we have        0 0 a 0 1 For λ = +1 : = −→ − 2b = 0, so 0 −2 b 0 0 For λ = −1 :

4.6



2 0 0 0

    a 0 = b 0

−→

2a = 0,

  0 so 1

Homework #8

Problem 1 Find the eigenvlaues and eigenfunctions of the boundary value problem λ y00 + y=0 (x + 1)2

(4.6.1)

On the interval a ≤ x ≤ 2 with boundary conditions y(1) = y(2) = 0. Write the equation in terms of a regular Sturm-Liouville eigenvalue problem, and find the coefficients in the expansion of an arbitrary function f(x) in a series of the eigenfunctions.

Solution In order to express (4.6.1) within the formalism of Sturm-Lioville theory, we recall that if we are given some differential equation of the form a2 (x)y00 + a1 (x)y0 + a0 (x)y = f(x)

(4.6.2)

W. Erbsen

HOMEWORK #8

then we can define Z

a1 (x) p(x) = exp dx a2 (x) a0 (x) q(x) =p(x) a2 (x) f(x) F (x) =p(x) a2 (x)



Such that (4.6.2) is transformed into d (p(x)y0 ) + q(x)y = F (x) dx

(4.6.3)

We can see form (4.6.1) that in our case, a2 (x) = 1, a1 (x) = 0, a0 (x) = λ/(x + 1)2 , and f(x) = 0. From these, it is clear that p(x) = 1, q(x) = λ/(x + 1)2 , and F (x) = 0, such that we can now express (4.6.1) in the form of (4.6.3): y00 +

λ y=0 (x + 1)2

(4.6.4)

Which is identical to (4.6.1). So, apparently our DE was already in S-L form. To find the eigenvalues and eigenfunctions, we wish to make the substitution and corresponding derivatives y =(x + 1)α y0 =α(x + 1)α−1 y00 =α(α − 1)(x + 1)α−2 Now (4.6.4) becomes α(α − 1)(x + 1)α−2 +

λ (x + 1)α = 0 (x + 1)2

α(α − 1)(x + 1)α−2 + λ(x + 1)α−2 = 0

α2 − α + λ = 0 And now

α=

1±

√ 1 − 4λ 2

(4.6.5)

At this point we split our solutions according to the three distinct outcomes from (4.6.5), which are if 1 − 4λ < 0, 1 − 4λ = 0, and 1 − 4λ > 0: √
λ < 1/4 : then α = 1/2 1 ± 1 − 4λ with real, distinct roots, which admits solutions of the form y(x) = c1 eαx + c2 e−αx Which only satisfies the first boundary condition, y(1) = 0 for the trivial case where y = 0. This cannot be our solution.

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CHAPTER 4: MATHEMATICAL METHODS

λ = 1/4 : then α = 1/2 , with real, repeated roots and solutions y(x) = c1 eαx + c2 xeαx Which also fails to satisfy the first boundary condition, therefore we must ignore this case as well. √
λ > 1/4 : then α = 1/2 1 ± i 4λ − 1 with complex, distinct roots and √   1 4λ − 1 x+1 y1 (x) =c1 (x + 1) /2 sin log 2 2 √   1 4λ − 1 x+1 /2 y2 (x) =c2 (x + 1) cos log 2 2 Which does work. Applying the first boundary condition √   √   1 1 4λ − 1 2 4λ − 1 2 /2 /2 y(1) =c1 (2) sin log + c2 (2) cos log 2 2 2 2 1

=c1 · 0 + c2 · 2 /2 −→ c1 = 1, c2 = 0

So, our solution takes the form y(x) = y1 (x) = (x + 1)

1

/2

sin

√

4λ − 1 log 2



x+1 2



Applying our second boundary condition, √

4λ − 1 y(2) = (3) sin log 2   √ 3 4λ − 1 log =0 −→ sin 2 2 1

/2

  3 2

And in order for this to hold... √
4λ − 1 log 3/2 =nπ 2  2  2 2nπ nπ 1 4λ − 1 = −→ λ = + 3 3 log ( /2 ) log ( /2 ) 4 And to find the corresponding eigenfunction we substitute this into (4.6.6): 1

ϕn (x) = (x + 1) /2 sin

Problem 2



2nπ log log ( 3/2 )



x+1 2



(4.6.6)

W. Erbsen

HOMEWORK #8

Find the eigenvalues and eigenfunctions of the boundary value problem x2 y00 + xy0 + y = µy

(4.6.7)

on the interval 1 ≤ x ≤ 2 with boundary conditions y(1) = y(2) = 0. Write the equation in terms of a regular Sturm-Liouville eigenvalue problem, and find the coefficients in the expansion of an arbitrary function f(x) in a series of the eigenfunctions.

Solution In the same spirit as the last problem, we must first put (4.6.7) into Sturm-Lioville form. We note that a2 (x) = x2 , a1 (x) = x, a0 (x) = 1 − µ, and f(x) = 0. From this we can find  Z 1 dx = exp [log x] = x p(x) = exp x 1−µ 1−µ q(x) =x 2 = x x F (x) =0 Which allows us to write d 1−µ (xy0 ) + y=0 dx x Try a solution of the form (and corresponding derivatives) y =xα y0 =αxα−1 y00 =α(α − 1)xα−2 Substituting these into (4.6.8),
1−µ α d xαxα−1 + x =0 dx x d (αxα ) + (1 − µ)xα−1 = 0 dx α2 xα−1 + (1 − µ)xα−1 = 0 α2 + 1 − µ = 0

Solving for α, α=

√ ± µ−4 2

Following the same logic as before, we find that in the case that µ < 4 that we have √  µ−4 y1 (x) = sin log x 2 √  µ−4 y2 (x) = cos log x 2

(4.6.8)

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CHAPTER 4: MATHEMATICAL METHODS

Where we must choose y1 (x) since it satisfies the first boundary condition. For the second, √  µ−4 − 1/2 y(2) = 2 sin log 2 2 √  µ−4 −→ sin log 2 = 0 2 And so √ µ−4 log 2 =nπ 2 p 2nπ µ−4 = −→ µ = log 2



2nπ log 2

2

+4

And finally ϕn (x) = sin



nπ log x log 2



Problem 2 Using Rodriques’ formula show that the Pn (x) are orthogonal and that Z 1 2 2 [Pn (x)] dx = 2n +1 −1

(4.6.9)

Solution Rodrigues’ formula for the Legendre Polynomials is given by (12.65) in A&W:  n
n 1 d Pn (x) = n x2 − 1 2 n! dx

To prove that Pn (x) given by (4.6.10) are orthogonal functions, we evaluate:   Z 1 Z 1  n
n d m 2
m 1 d 2 Pn (x)Pm (x) dx = n+m x −1 x −1 dx 2 n!m! −1 dx dx −1

For convenience let’s set C = 1/2n+m n!m!. At this point we integrate by parts m times, with  n  n+m d d u= (x2 − 1)n , du = (x2 − 1)n dx dx  m d dv = (x2 − 1)m , v = (x2 − 1)m dx

(4.6.10)

(4.6.11)

W. Erbsen

HOMEWORK #8

Such that (4.6.11) becomes 1  n+m Z 1 d m 2 m Pn (x)Pm (x) dx =C (x − 1) (x − 1) + C(−1) (x − 1) (x2 − 1)n dx dx −1 −1 −1  n+m Z 1 d m 2 m (x2 − 1)n dx =C(−1) (x − 1) dx −1

Z



1

2

m



d dx

n

n

2

At this point we note that in the case that n < m, the above integral goes to zero. Eg if, for instance, n = 1 and m = 2: Z

1

P1 (x)P2 (x) dx =C

−1

Z

1

−1

(x2 − 1)2



d dx

3

(x2 − 1)2 dx = C

Z

1

−1

2(x2 − 1)2

d dx = 0 dx

Which works for any n < m. This can be straightforwardly extended to the case in which n 6= m when we recall that our choice of m and n was arbitrarily chosen in integrating by parts. So, we have just shown that if n 6= m then (4.6.11) is equal to zero.

If, on the other hand, m = n, then we have Z

1

[Pn (x)]2 dx =

−1

=

(−1)n 22n(n!)2

Z

1

−1

(−1)n (2n)! 22n(n!)2

(x2 − 1)n

Z

1

−1



d dx

2n

(x2 − 1)n dx

(x2 − 1)n dx

(4.6.12)

At this point we make the substitution u = 1/2 (x + 1) → x = 2u − 1, du = 1/2 dx → dx = 2 du, such that (4.6.12) becomes: Z

1

(−1)n (2n)! [Pn (x)] dx = 2n 2 (n!)2 −1 2

=

Z

2(−1)n (2n)! 22n (n!)2

1

0

Z

0

(4u2 − 4u)n 2 du 1

[4u(u − 1)]n du

Z 2(−1)n (2n)! 1 n = u (u − 1)n du (n!)2 0 2(−1)n (2n)! Γ(n + 1)2 = (−1)n 2 (n!) Γ(2n + 2) 2 2(2n)! (n!) = (n!)2 (2n + 1)! (2n)! =2 (2n + 1)! 2 = 2n + 1

Where I integrated by parts n times to evaluate the integral, and also the fact that Γ(n) = (n − 1)!:

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CHAPTER 4: MATHEMATICAL METHODS

Z

1

[Pn(x)]2 dx =

−1

2 2n + 1

Problem 4 Using a generating function, evaluate the sum ∞ X xn+1 Pn (x) n+1 n=0

(4.6.13)

Solution We note that from (7 − 10) in M&W that

∞ X 1 f(x) = √ = hn Pn (x) 1 − 2hx + h2 n=0

(4.6.14)

Which defines the generating function. We wish to manipulate (4.6.14) until it resembles (4.6.13). Start by squaring both sides, 2

[f(x)] =

∞ X

[hn Pn (x)]

2

n=0

And now integrating, Z

1

2

[f(x)] dx =

−1

Z

1

∞ X

2

[hn Pn (x)] dx

(4.6.15)

−1 n=0

Not sure where to go from here. hn must be dependent on x, otherwise the integral on the RHS of (4.6.15) does not make sense. Otherwise, I was unable to integrate-by-parts to solve it either. Summing by hand and simplifying results in the following series: f(x) =x +

x3 x5 x7 + + + ... 3 5 7

Which is clearly odd, so this leads me to believe f(x) is a sin function. It also seems as though the sum diverges, however it does indeed converge, since we required that 0 < x < 1, so 0 < f(x) < 4.52 (approximately). Ugh.

Problem 5 Prove the trigonometric identity using vector methods:

W. Erbsen

HOMEWORK #8

cos γ = cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 − φ2 )

(4.6.16)

Where γ is the angle between two directions (θ1 , φ1 ) and (θ2 , φ2 ).

Solution ~ and B, ~ each normlized to unity. Each is then described by We define two vectors on the units sphere, A only two coordinates, θ and φ. To prove the trigonometric identity (4.6.16), we take the dot product of the two vectors. But first, we define the components of our vectors in terms of Cartesian coordinates, with x = sin θ cos φ y = sin θ sin φ z = cos θ Yielding a dot product of ~·B ~ =x1 · x2 + y1 · y2 + z1 · z2 A = sin θ1 cos φ1 · sin θ2 cos φ2 + sin θ1 sin φ1 · sin θ2 sin φ2 + cos θ1 · cos θ2 =(sin θ1 + sin θ2 )(cos φ1 cos φ2 + sin φ1 · sin φ2 ) + cos θ1 · cos θ2 {z } | cos(φ1 −φ2 )

= cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 − φ2 )

(4.6.17)

~·B ~ = kAk kBk cos γ = cos γ. And recall A |{z} |{z} 1

1

So (4.6.17) is now

cos γ = cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 − φ2 )

Problem 6 The amplitude of a scattered wave is given by f(θ) =

∞ 1X (2` + 1)eiδ` sin δ` P` (cos θ) k

(4.6.18)

`=0

Where θ is the angle of scattering, ` is the angular momentum eigenvalue, ~k is the incident momentum, and δ` is the R phase shift produced by the central potential that is doing the scattering. The total cross section is σtot = |f(θ)|2 dΩ. Show that ∞ 4π X σtot = 2 (2` + 1) sin2 δ` (4.6.19) k `=0

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CHAPTER 4: MATHEMATICAL METHODS

Solution We begin by first evaluating |f(θ)|2 : 2

|f(θ)| =

!∗ ∞ 1X iδ` (2` + 1)e sin δ` P`(cos θ) k `=0

! ∞ 1X iδ` (2` + 1)e sin δ` P` (cos θ) k `=0

∞ 1 X = 2 (2` + 1)e−iδ` sin δ` P`(cos θ)(2` + 1)eiδ` sin δ` P` (cos θ) k `=0

∞ 1 X = 2 (2` + 1)2 sin2 δ` P`2 (cos θ) k `=0

And now to find the total cross section, Z σtot = |f(θ)|2 dΩ Z 2π Z π = dφ |f(θ)|2 sin θ dθ 0 0 # Z π" X ∞ 1 2 2 2 (2` + 1) sin δ` P` (cos θ) sin θ dθ =2π k2 0 `=0 Z π ∞ 2π X 2 = 2 (2` + 1)2 sin2 δ` [P` (cos θ)] sin θ dθ k 0 `=0 #2  ` Z π" ∞
` 2π X 1 d 2 2 2 = 2 (2` + 1) sin δ` cos θ − 1 sin θ dθ k 2``! d cos θ 0 `=0 2` Z π ∞
2` 2π X 1 d 2 2 = 2 (2` + 1) sin δ` 2` 2 cos2 θ − 1 sin θ dθ k 2 (`!) 0 d cos θ `=0 Z π  2` ∞
2` 1 d 2π X (2` + 1)2 sin2 δ` 2` 2 x2 − 1 sin θ dθ = 2 k 2 (`!) 0 dx `=0    Z ∞ π
2` 2π X 1 d 2 2 2 = 2 (2` + 1) sin δ` 2` 2 x −1 sin θ dθ k 2 (`!) 0 dx `=0 Z π ∞ 2π X 1 2` = 2 (2` + 1)2 sin2 δ` (cos θ) sin θ dθ k (`!)2 0 `=0   ∞ X 1 1 + (−1)2` 2π (2` + 1)2 sin2 δ` = 2 k (`!)2 2` + 1 `=0

=

∞ 4π X

k2

`=0

(2` + 1) sin2 δ`

1 (`!)2

(4.6.20)

Where I used the fact that dΩ = sin θ dθdφ and also the results from problem 3. Therefore we have shown that ∗ ∗ You might have noticed that there is an extra factor of (`)−2 in (4.6.20). I was not able to get rid of it. So when I say that “we have shown that” I suppose I really mean “we have nearly shown that.”

W. Erbsen

HOMEWORK #9

σtot =

∞ 4π X (2` + 1) sin2 δ` k2 `=0

4.7

Homework #9

Problem 1 By differentiating the Legendre polynomial generating function 1 g(x, t) = √ 1 − 2xt + t2

(4.7.1)

with respect to t and with respect to x, obtain the recurrence relations for the Legendre polynomials, and prove that Pn (x) satisfies the Legendre differential equation.

Solution We know that the generating function g(x, t) can be expressed ∞ as X 1 g(x, t) = √ = tn Pn (x) 1 − 2xt + t2 n=0

(4.7.2)

So, as per the suggestion, we differentiate (4.7.1) first with respect to t: dg(x, t) d 1 = √ dt dt 1 − 2xt + t2 x−t = 3 (1 − 2xt + t2 ) /2 x−t = g(x, t) (1 − 2xt + t2 )

(4.7.3)

Doing the same for the RHS of (4.7.2), ∞ ∞ X dg(x, t) d X n = t Pn (x) = ntn−1 Pn (x) dt dt n=0 n=0

Inserting this and (4.7.2) into the (4.7.3), ∞ X

(1 − xt + t2 ) ∞ X

n=0

ntn−1 Pn (x) − 2x

∞ X

n=0

ntn Pn (x) +

n=0 ∞ X

n=0 ∞ X

n=0

ntn−1 Pn (x) =

∞ X x−t tn Pn (x) (1 − xt + t2 ) n=0

ntn−1 Pn (x) =(x − t) ntn+1 Pn (x) =x

∞ X

n=0

∞ X

tn Pn (x)

n=0

tn Pn (x) −

∞ X

n=0

tn+1 Pn (x)

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CHAPTER 4: MATHEMATICAL METHODS

∞ X

ntn−1 Pn (x) +

n=0

∞ X

(n + 1)tn + 1 Pn (x) =

n=0

|{z} n−1=`

∞ X

|{z} n=`

(` + 1)t`P`+1 (x) +

`=0

x(1 + 2n)tn Pn (x)

n=0

|{z}

n+1=`

Shifting the indices as indicated:

∞ X

∞ X

`t` P`−1 (x) =

`=0

∞ X

x(1 + 2`)t` P` (x)

`=0

−→ (` + 1)P`+1 (x) − x(1 + 2`)P` (x) = −`P`−1 (x)

(4.7.4)

Since we require that each coefficient should vanish. We can find a second recursion relation by similarly integrating (4.7.2) by x: dg(x, t) d 1 = √ dx dx 1 − 2xt + t2 t = (1 − 2xt + t2 ) 3/2 t g(x, t) = (1 − 2xt + t2 )

(4.7.5)

And the the derivative of the RHS of (4.7.2) is ∞ ∞ X dg(x, t) d X n = t Pn (x) = tn Pn0 (x) dx dx n=0 n=0 And now

∞ X

n=0

tn Pn0 (x) − 2x

∞ X

∞ X

tn+1 Pn0 (x) +

n=0 ∞ X

tn Pn0 (x) +

n=0

t

∞ X t tn Pn (x) = (1 − 2xt + t2 ) n=0

Pn0 (x) =

∞ X

tn+2 Pn0 (x) =

n=0

t` P`0 (x) +

∞ X `=0

0 t`P`−2 (x) =

tn+1 Pn (x)

n=0 ∞ X

tn+1 [2xPn0 (x) + Pn (x)]

|{z}

n+2=`

Shifting the indices once more,

∞ X

n=0

|{z}

n=`

`=0

n=0 ∞ X n+2

Pn0 (x)

n=0

|{z}

∞ X

t

n

n+1=`

∞ X `=0

  0 (x) + P`−1 (x) t` 2xP`−1

0 0 (x) =2xP`−1 (x) + P`−1 (x) P`0 (x) + P`−2

0 0 −→ P`+1 (x) − 2xP`0 (x) = P` (x) − P`−1 (x)

(4.7.6)

We can combine (4.7.4) and (4.7.6) as follows. First, moving all the terms of (4.7.4) to the left and taking the derivative with respect to x yields 0 0 (` + 1)P`+1 (x) − x(1 + 2`)P`0 (x) − (1 + 2`)P` (x) + `P`−1 (x) = 0

(4.7.7)

W. Erbsen

HOMEWORK #9

Now moving all the terms of (4.7.6) to the left, and multiplying by −` 0 0 −`P`+1 (x) + 2`xP`0 (x) + `P` (x) − `P`−1 (x) = 0

(4.7.8)

Adding (4.7.7) and (4.7.8), 0 P`+1 (x) = xP`0 (x) + (1 + `)P` (x)

Many more (in fact, infinitely more) recursion relations may be derived, in addition to the three I have found here, there are two more that can be found in M&W that become useful: 0 xP`0 (x) − `P` (x) − P`−1 (x) =0 0 0 P` (x) − xP`−1 (x) − `P`−1 (x) =0

(4.7.9) (4.7.10)

0 x2 P`0 (x) − `xP` (x) − xP`−1 (x) = 0

(4.7.11) (4.7.12)

Multiplying (4.7.9) by x:

Now take (4.7.11) and subtract (4.7.10): 0 0 x2 P`0 (x) − `xP`(x) − xP`−1 (x) − P`0 (x) + xP`−1 (x) + `P`−1 (x) = 0

(x2 − 1)P`0 (x) − `xP`(x) + `P`−1 (x) = 0

(4.7.13)

Now taking the derivative of (4.7.13) with respect to x: 0 2xP`0(x) + x2 P`00(x) − P`00 (x) − `P` (x) − `xP`0 (x) + `P`−1 (x) = 0

x2 P`00(x) − P`00 (x) + (2 − `)xP`0 (x) − `P` (x) + `P`−1 (x) = 0

(4.7.14)

0 At this point we take (4.7.9), solve for P`−1 (x), and substitute into (4.7.14):

x2 P`00 (x) − P`00(x) + (2 − `)xP`0 (x) − `P` (x) + `xP`0 (x) − `2 P` (x) = 0

(x2 − 1)P`00 (x) + 2xP`0(x) − `xP`0 (x) − `P` (x) + `xP`0 (x) − `2 P` (x) = 0 (x2 − 1)P`00(x) + 2xP`0(x) − `(` + 1)P`(x) = 0 Which is the Legendre differential equation.

Problem 2 By differentiating the Bessel generating function g(x, t) = e 2 (t− /t ) x

1

(4.7.15)

with respect to t and with respect to x, obtain the recurrence relations for the Bessel functions, and prove that Jn (x) satisfies the Bessel differential equation.

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CHAPTER 4: MATHEMATICAL METHODS

Solution In very much the same spirit as the last problem, we are given that the generating function is g(x, t) = e 2 (t− /t ) = 1

x

∞ X

tn Jn (x)

(4.7.16)

n=0

Differentiating (4.7.15) first with respect to t, dg(x, t) d x2 (t− 1/t ) = e dt dt
x 1 x = 1 − 1/t2 e 2 (t− t ) 2
x = 1 − 1/t2 g(x, t) 2 ∞ X = ntn−1 Jn (x) n=0

Playing with this a little bit, we have ∞ X

ntn−1 Jn (x) =

n=0

2

∞ X

ntn−1 Jn (x) =

n=0

2

∞ X



1−

∞ X

n=0

|{z}

|{z}

n−1=`

Shift the indices,

x 2

1 t2

X ∞

tn Jn (x)

n=0

xtn Jn (x) −

n=`

(` + 1)t`J`+1 (x) =

`=0

∞ X

n=0

|{z}

n−2=`

∞ X `=0

xt` J` (x) +

∞ X

t` J`+2

`=0

2(` + 1)J`+1 (x) =xJ` (x) + xJ`+2 (x) −→

2` J` (x) = J`−1 (x) + J`+1 (x) x

Now differentiating (4.7.15) with respect to x, 1 dg(x, t) d x = e 2 (t− /t ) dx dx  1 x 1 1 = t− e 2 (t− /t ) 2 t   1 1 = t− g(x, t) 2 t ∞ X = tn Jn0 (x)

n=0

And now

xtn−2 Jn (x)

(4.7.17)

W. Erbsen

HOMEWORK #9

2 2

 ∞  1 X n t Jn (x) tn Jn0 (x) = t − t n=0 n=0 ∞ X

∞ X

∞ X

tn Jn0 (x) =

n=0

n=0

|{z}

|{z}

n=`

Shifting,

2

tn+1 Jn (x) −

n+1=`

∞ X

t` J`0 (x) =

`=0

∞ X `=0

∞ X

tn−1 Jn (x)

n=0

|{z}

n−1=`

t` J`−1 (x) −

∞ X

t` J`+1 (x)

`=0

−→ 2J`0 (x) = J`−1 (x) − J`+1 (x)

(4.7.18)

If we subtract (4.7.17), −→ J`0 (x) =

` J` (x) − J`+1 (x) x

(4.7.19)

We can arrive at one more recursion relation if we take (4.7.19) and subtract (4.7.17): −→ J`0 (x) = J`−1 (x) −

` J` (x) x

(4.7.20)

To prove that Jn (x) satisfies the Bessel differential equation (4.7.15), we start by multiplying (4.7.20) by x: xJ`0 (x) − xJ`−1 (x) + `J` (x) = 0

(4.7.21)

We now differentiate (4.7.21) with respect to x: d 0 [xJ`0 (x) − xJ`−1 (x) + `J` (x)] =J`0 (x) + xJ`00 (x) − J`−1 (x) − xJ`−1 (x) + `J`0 (x) dx 0 =xJ`00 (x) + (` + 1)J`0 (x) − xJ`−1 (x) − J`−1 (x)

(4.7.22)

Multiplying (4.7.22) by x, 0 xJ`00 (x) + (` + 1)J`0 (x) − xJ`−1 (x) − J`−1 (x) =0

0 x2 J`00 (x) + x(` + 1)J`0 (x) − x2 J`−1 (x) − xJ`−1 (x) =0

(4.7.23)

And multiply (4.7.21) by `: `xJ`0 (x) − `xJ`−1 (x) + `2 J` (x) = 0

(4.7.24)

Now take (4.7.22) and subtract (4.7.21): 0 x2 J`00 (x) + x(` + 1)J`0 (x) − x2 J`−1 (x) − xJ`−1 (x) − `xJ`0 (x) + `xJ`−1 (x) − `2 J` (x) =0 0 x2 J`00 (x) + xJ`0 (x) − x2 J`−1 (x) + (` − 1)J`−1 (x) − `2 J` (x) =0

At this point we take (4.7.19) and shift the index from ` to ` − 1,

(4.7.25)

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CHAPTER 4: MATHEMATICAL METHODS

0 J`−1 (x) =

`−1 J`−1 (x) − J` (x) x

And multiplying this by x and solving for J`−1 (x), 0 xJ`−1 (x) = (` − 1)J`−1 (x) − xJ` (x) −→ J`−1 (x) =

x x 0 J`−1 (x) + J` (x) `−1 `−1

Substituting this into (4.7.25), x2 J`00 (x)

+

xJ`0 (x)

−x

2

0 J`−1 (x)



 x x 0 + (` − 1) J (x) + J`(x) − `2 J` (x) =0 ` − 1 `−1 `−1

0 x2 J`00 + xJ`0 (x) − x2 J`−1 (x) + x2 J`−1 (x) + x2 J` (x) − `2 J` = 0


x2 J`00 (x) + xJ`0 (x) + x2 − `2 J` (x) = 0

Which is the Bessel differential equation.

Problem 3 Prove the normalization condition for the Bessel functions Z ∞ 1 ·Jn (kr) · Jn (k 0 r) drr = δ(k 0 − k) k 0

(4.7.26)

Solution It was shown in class, and additionally can be found in M&W that: Z b 1 b [βrJn (αr)Jn0 (βr) − αrJn (βr)Jn0 (αr)|a Jn (αr)Jn (βr)rdr = 2 2 α − β a

(4.7.27)

Where I dispensed k and k 0 in favor of α and β, respectively in order to avoid confusion with the derivatives. First let’s choose a → 0 as in (4.7.26): Z b 1 b Jn (αr)Jn (βr)rdr = 2 [βrJn (αr)Jn0 (βr) − αrJn (βr)Jn0 (αr)|0 2 α − β 0 1 = 2 [βrJn (αb)Jn0 (βb) − αrJn (βb)Jn0 (αb)] (4.7.28) α − β2 And since both J`0 (αr) and J`0 (βr) evaluated at the endpoints equal zero, then (4.7.36) too must vanish. To see that this is the case at x = 0, it is illuminating to recall that J`0 (0) = J`+1 (0) = 0. We should also recognize that J`0 (∞) → 0. Both of these properties become very evident when plotting the relevant functions and recognizing the respective behaviors. Therefore, (4.7.28) becomes

W. Erbsen

HOMEWORK #9

Z

b

Jn (αr)Jn (βr)rdr = 0

(4.7.29)

0

For the case in which α and β are equal, we must evaluate Z b Z b Z Jn (αr)Jn (βr)rdr = Jn (αr)Jn (αr)rdr = 0

0

b

2

[Jn (αr)] rdr

0

Which we now set equal to the RHS of (4.7.28) Z b βrJn (αb)Jn0 (βb) − αrJn (βb)Jn0 (αb) 2 [Jn (αr)] rdr = α2 − β 2 0

(4.7.30)

Evaluating the RHS of (4.7.30) is not so easy; we must take the limit as β → α, but since this is indeterminate we must use L’Hˆ ospital’s Rule:   Z b βrJn (αb)Jn0 (βb) − αrJn (βb)Jn0 (αb) 2 [Jn (αr)] r dr = lim β→α α2 − β 2 0   0 rJn (αb)Jn (βb) + βrJn (αb)Jn00 (βb) − αrJn0 (βb)Jn0 (αb) = lim β→α −2β 2

=

αr [Jn0 (αb)] − rJn (αb)Jn0 (αb) − αrJn (αb)Jn00 (αb) 2α

(4.7.31)

We now solve Bessel’s differential equation for Jn00 (αb): α2 Jn00 (αb) + αJn0 (αb) + (α2 − n2 )Jn (αb) = 0 −→ Jn00 (αb) = −

αJn0 (αb) − (α2 − n2 )Jn (αb) α2

(4.7.32)

Substituting (4.7.32) into (4.7.31), Z

b

2

[Jn (αr)] rdr =

0

=

h i 2 αJ 0 (αb)−(α2−n2 )Jn (αb) αr [Jn0 (αb)] − rJn (αb)Jn0 (αb) − αrJn (αb) − n 2 α 2α

 2 αr [Jn0 (αb)] − rJn (αb)Jn0 (αb) + rJn (αb)Jn0 (αr) + αr 1 −

2α     2 r n 2 2 0 = [Jn (αb)] + 1 − 2 [Jn (αb)] 2 α

n2 α2



2

[Jn (αb)]

(4.7.33)

So that if α and β represent two unique roots of order n then (4.7.33) must vanish.

Problem 4 A disk of radius R in the xy-plane (z = 0) is kept at a constant potential φ0 and the rest of the plane z = 0 is kept at zero potential. As shown in class, the potential for z > 0 is given by Z ∞ φ(r, z) = φ0 R dkJ1 (kR)J0 (kr)e−kz (4.7.34) 0

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CHAPTER 4: MATHEMATICAL METHODS

Hence, the potential above the center of the disk (r = 0) is given by the integral Z ∞ φ(0, z) = φ0 R dkJ1 (kR)e−kz

(4.7.35)

0

Using the Bessel function recurrence relations and the integral representation of the Bessel function, show that the last expression can be reduced to   z φ(0, z) = φ0 1 − √ (4.7.36) z 2 + R2

Solution According to (4.7.19), J1 (x) = −J00 (x) And (4.7.35) becomes φ(0, z) = −φ0 R

Z

∞

0

J00 (kR)e−kz dk

And at this point we wish to undergo a change of variables x = kR −→ k = x/R : Z ∞ xz dx J00 (x)e− /R φ(0, z) = − φ0 R R Z ∞0 xz = − φ0 J00 (x)e− /R dx 0

Where we must integrate by parts: u = e−

xz

/R

,

dv = J00 (x),

z − xz/R e R v = J0 (x)

du = −

So we now have  Z xz ∞ xz z ∞ J0 (x)e− /R dx J0 (x)e− /R + R 0 0   Z ∞ xz z = − φ0 −1 + J0 (x)e− /R dx R 0

φ(0, z) = − φ0

h

(4.7.37)

At this point we pause, try a myriad of obscure substitutions, get frustrated, have a beer, and then realize that the integral in (4.7.37) has actually been tabulated by the likes of Gradshteyn and Ryzhik as Eq. 6.611.1: ? hp iν Z ∞ β −ν α2 + β 2 − α p e−αxJν (βx) dx = α2 + β 2 0

Where in our case α = z/R , β = 1 and ν = 0:

W. Erbsen

HOMEWORK #10

Z

∞

e−

0

xz

/R

1 1 R √ J0 (x) dx = q = = √ 1/ 2 2 2 2 z + R2 R z +R ( z/R ) + 1

(4.7.38)

And substituting (4.7.38) into (4.7.37),

   z R √ φ(0, z) = − φ0 −1 + R z 2 + R2 From which we finally arrive at (4.7.36), thus completing the problem:   z φ(0, z) = φ0 1 − √ z 2 + R2

4.8

Homework #10

Problem 1 In class we shows that the lowest order spherical Bessel function is given by sin x j0 (x) = x Using the recurrence relations, find j1 (x) and j2 (x), and also prove that  ` 1 ∂ j` (x) = (−1)` x` j0 (x) x ∂x

(4.8.1)

(4.8.2)

Solution Recall that one of the recurrence relations for Bessel functions reads j`+1 (x) =

` j` (x) − j`0 (x) x

(4.8.3)

Which was rigorously proven in Homework #9, and subsequently will be treated as an axiom in this assignment. We are given j0 (x), so we need to calculate j1 (x) and j2 (x) using our recursion relation (4.8.3). We start by finding the first derivative of (4.8.1): j00 (x) =

cos x sin x − 2 x x

We now substitute this into (4.8.3), 0 j1 (x) = j0 (x) − j00 (x) x sin x cos x = 2 − x x

(4.8.4)

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CHAPTER 4: MATHEMATICAL METHODS

Taking the first derivative of j1 (x) yields j10 (x) =

2 cos x 2 sin x sin x − + x2 x3 x

From which we can now find j2 (x): 1 j2 (x) = j1 (x) − j10 (x) x sin x cos x 2 cos x 2 sin x sin x = 3 − 2 − + − x x x2 x3 x 3 sin x 3 cos x sin x = − − x3 x2 x

(4.8.5)

In order to prove (4.8.2), it will be most convenient to show that (4.8.2) works when substituted into (4.8.3). I will employ proof by induction. So the first step is to show that it works for ` = 0, j1 (x) = − j00 (x) " #  0 d 1 ∂ =− (−1)0 x0 j0 (x) dx x ∂x d sin x dx x sin x cos x = 2 − x x =j1 (x) X =−

So it works for ` = 0. Assuming that it holds for all `, we now show that it holds for ` + 1. First find the derivative of j` (x): " #  ` 1 d sin x d ` 0 j` (x) = (−x) dx x dx x # " #   " ` ` d 1 d sin x 1 d sin x ` ` d = (−x) + (−x) dx x dx x dx x dx x " # " # ` ` 1 d sin x 1 d sin x `−1 ` d =`(−x) + (−x) x dx x dx x dx x " # " #   ` ` (−x)` 1 d sin x 1 d sin x ` d =−` + (−x) x x dx x dx x dx x " # " # ` ` ` 1 d sin x 1 d sin x ` ` d = − (−x) + (−x) x x dx x dx x dx x " # ` ` d 1 d sin x = − j` (x) + (−x)` (4.8.6) x dx x dx x So, let’s find j`+1 (x): j`+1 (x) =(−x)`+1



1 d x dx

`+1

sin x x

W. Erbsen

HOMEWORK #10



`   1 d 1 d sin x =(−x) (−x) x dx x dx x " # ` 1 d sin x d = − (−x)` dx x dx x `

` = j` (x) − j`0 (x) X x Where I used (4.8.6). Since (4.8.2) works for ` = 0 (the base step) and also ` + 1 (the inductive step), then it follows that it works for ` (the hypothesized step).

Problem 2 Using the definition of Neumann functions in terms of Bessel functions, show that for the spherical Neumann functions, cos x η0 (x) = − (4.8.7) x Using the recurrence relations, find η1 (x) and η2 (x).

Solution Analogous to (4.8.3), the recursion relation for the Neumann functions takes the form η`+1 (x) =

` η` (x) − η`0 (x) x

(4.8.8)

So, from ` = 0 we can find η1 (x): η1 (x) =

` η0 (x) − η00 (x) =⇒= −η00 (x) x

(4.8.9)

So, in finding the first derivative of η0 (x), we find η1 (x): η1 (x) = −η00 (x) =

d cos x sin x cos x =− − 2 dx x x x

(4.8.10)

Applying the same process letting ` = 1, 1 η1+1 (x) = η1 (x) − η10 (x) x sin x cos x = − 2 − 3 − η10 (x) x x Where we find that   d sin x cos x − − 2 dx x x cos x 2 sin x sin x 2 cos x =− + + 2 + x x2 x x3

η10 (x) =

(4.8.11)

335

CHAPTER 4: MATHEMATICAL METHODS

=−

cos x 2 sin x 2 cos x + + x x2 x3

(4.8.12)

Substituting (4.8.12) into (4.8.11), we find that sin x cos x cos x 2 sin x 2 cos x − 3 + − − x2 x x x2 x3 cos x 3 sin x 3 cos x = − − x x3 x3

η2 (x) = −

(4.8.13)

Problem 3 Transform the hyperbolic equation y2 ψxx − x2 ψyy = 0

(4.8.14)

into the canonical form of ψην = f(...)

Solution The general formalism of the Method of Characteristics entails taking some arbitrary differential equation,

Where

e 2 + 2Bu e ξη + Cu e 2 +R=0 Au2xx + 2Buxy + Cu2yy = 0 −→ Au ξξ ηη

(4.8.15)

e =Aξx2 + 2Bξx ξy + Cξy2 A

e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B e =Aηx2 + 2Bηxηy + Cηy2 C

R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη

So, from (4.8.14), it is apparent that A = y2 , B = 0, and C = −x2 . Since (4.8.14) is hyperbolic, the characteristics can be found by √ √ dy B ± B 2 − 4AC −4AC x = =± =± dx 2A 2A y Solving and integrating this we arrive at x2 y2 ± = 0 −→ x2 ± y2 = 0 2 2

(4.8.16)

Where we now define ξ(x, y) = ξ = x2 + y2 ,

and

η(x, y) = η = x2 − y2

Taking some partial derivatives, ξx =2x,

ξy = 2y,

ξxx = ξyy = 2,

ξxy = 0

(4.8.17)

W. Erbsen

HOMEWORK #10

ηx =2x,

ηy = −2y,

ηxx = 2,

ηyy = −2,

ηxy = 0

Now finding the coefficients in (4.8.15), e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =y2 ξx2 − x2 ξy2

=y2 (2x)2 − x2 (2y)2 = 0

e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B =y2 ξx ηx − x2 ξy ηy

=y2 (2x)(2x) − x2 (2y)(−2y)

=4x2 y2 + 4x2 y2 =8x2 y2

e =Aη 2 + 2Bηxηy + Cη 2 C x y =y2 ηx2 − x2 ηy2

=y2 (2x)2 − x2 (−2x)2 =0

R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη

= y2 ξxx − x2 ξyy uξ + y2 ηxx − x2 ηyy uη

=2 y2 − x2 uξ + 2 y2 + x2 uη

And putting all these into (4.8.15),

e 2 + 2Bu e ξη + Cu e 2 +R=0 Au ξξ ηη

2(8x2 y2 )uξη + 2 y2 − x2 uξ + 2 y2 + x2 uη = 0

8x2 y2 uξη + y2 − x2 uξ + y2 + x2 uη = 0

We can combine our original definitions of ξ and η from (4.8.17) to obtain: x2 =

ξ+η , 2

and

y2 =

ξ−η 2

So that we can now rewrite (4.8.18) as    ξ+η ξ−η 8 uξη − ηuξ + ξuη = 0 2 2
2 ξ 2 − η 2 uξη − ηuξ + ξuη = 0 And the canonical form of (4.8.14) becomes

uξη =

ηuξ − ξuη 2 (ξ 2 − η 2 )

(4.8.18)

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CHAPTER 4: MATHEMATICAL METHODS

Problem 4 Transform the elliptical equations y2 ψxx + x2 ψyy = 0

(4.8.19a)

2

ψxx + (1 + y) ψyy = 0

(4.8.19b)

into the canonical form of ψσσ + ψττ = f(...).

Solution Starting from (4.8.19a), it is apparent to the casual observer that A = y2 , B = 0 and C = x2 . Furthermore, we can go on to find the characteristics of this equation: √ √ −B ± B 2 − 4AC −AC ix dy = = = (4.8.20) dx 2A A y Solving and integrating, we find that ix2 y2 ± = 0 −→ ix2 ± y2 = 0 2 2

(4.8.21)

At this point we set ξ(x, y) = ξ = ix2 + y2 ,

and

η(x, y) = η = ix2 − y2

With corresponding partial derivatives ξx =2ix,

ξy = 2y,

ηx =2ix,

ηy = −2y,

ξxx = 2i,

ξyy = 2,

ηxx = 2i,

And now finding the coefficients, e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =y2 (2ix)2 + x2 (2y)2 = − 4x2 y2 + 4x2 y2

=0

e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B =y2 (2ix)(2ix) + x2 (2y)(−2y)

= − 4x2 y2 − 4x2 y2 = − 8x2 y2

e =Aη 2 + 2Bηxηy + Cη 2 C x y =y2 (2ix)2 + x2 (−2y)2

= − 4x2 y2 + 4x2 y2

ξxy = 0

ηyy = −2,

ηxy = 0

(4.8.22)

W. Erbsen

HOMEWORK #10

=0 R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη

= y2 (2i) + x2 (2) uξ + y2 (2i) + x2 (−2) uη

=2 iy2 + x2 uξ + 2 iy2 − x2 uη

Substituting these in we have

e ξη + R = 0 2Bu



− 16x2y2 uξη + 2 iy2 + x2 uξ + 2 iy2 − x2 uη = 0

(4.8.23)

And it is possible to eliminate x and y by making the following substitution: x2 =

ξ+η , 2i

and

y2 =

ξ−η 2

Putting this in (4.8.23) and simplifying,  2  ξ − η2 −16 uξη − 2iηuξ + 2iξuη 4i
2 ξ 2 − η 2 uξη − ηuξ + ξuη = 0 From which we finally arrive at uξη =

ηuξ − ξuη 2 (ξ 2 − η 2 )

In the same spirit for (4.8.19b), we note that A = 1, B = 0, C = (1 + y)2 . Finding the characteristics, √ √ dy −B ± B 2 − 4AC ± −AC = = = ±i(1 + y) (4.8.24) dx 2A A Solving and integrating, 1 dy = ±i dx −→ log (1 + y) = ±ix 1+y Assigning our two independent variables, ξ(x, y) = ξ = ix + log (1 + y),

and

η(x, y) = η = ix − log (1 + y)

Let’s take some partial derivatives: ξx =i, ηx =i, And the coefficients are

1 1 , ξxx = 0, ξyy = − , ξxy = 0 1+y (1 + y)2 1 1 ηy = − , ηxx = 0, ηyy = , ηxy = 0 1+y (1 + y)2

ξy =

(4.8.25)

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CHAPTER 4: MATHEMATICAL METHODS

e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =Aξx2 + (1 + y)2 ξy2

=(i)(i) + (1 + y)2

1 (1 + y)2

=−1+1 =0 e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B

=ξx ηx + (1 + y)2 ξy ηy    1 1 2 =(i)(i) + (1 + y) − 1+y 1+y =−1−1 =−2

e =Aη 2 + 2Bηxηy + Cη 2 C x y =ηx2 + (1 + y)2 ηy2

=(i)(i) + (1 + y)2 =−1+1 =0



1 1+y

2

R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη =(1 + y)2 ξyy uξ + (1 + y)2 ηyy uη     1 1 2 2 =(1 + y) − uξ + (1 + y) uη (1 + y2 )2 (1 + y)2 = − uξ + uη And now we can find the canonical form of (4.8.19b): e ξη + R = 0 2Bu − 4uξη − uξ + uη = 0

And finally

uξη =

uξ − uη 4

W. Erbsen

HOMEWORK #10

Problem 5 Classify the following equations in appropriate regions and transform them into canonical form: ψxx − y2 ψyy + ψx − ψ + x2 = 0 ψxx + xψyy = 0

(4.8.26a) (4.8.26b)

Solution For (4.8.26a) it is important that the elements that characterize the dominant behavior are the second order terms, so analyzing this (the principle part), we see that A = 1, B = 0, and C = −y2 . Furthermore, we can see that: B 2 − AC = y2 > 0 −→ hyperbolic Solving, integrating: dy = ±y −→ x ± log y = 0 dx And now we can say ξ(x, y) = ξ = x + log y,

and

η(x, y) = η = x − log y

Take some partials: ξx =1, ηx =1,

1 1 , ξxx = 0, ξyy = − 2 , ξxy = 0 y y 1 1 ηy = − , ηxx = 0, ηyy = 2 , ηxy = 0 y y ξy =

And the coefficients are e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =ξx2 − y2 ξy2 1 =1 − y2 2 y =1 − 1 =0

e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B =ξx ηx − y2 ξy ηy    1 1 =1 − y2 y y =1 + 1 =2

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CHAPTER 4: MATHEMATICAL METHODS

e =Aη 2 + 2Bηxηy + Cη 2 C x y =ηx2 − y2 ηy2  2 1 2 =1 − y − y

=1 − 1 =0

R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη

= ξxx − y2 ξyy uξ + ηxx − y2 ηyy uη       1 1 2 2 = 1−y uξ + 1 − y − uη y y = (1 − y) uξ + (1 + y) uη

And we now have 4uξη + (1 − y) uξ + (1 + y) uη = 0 And consequently       ξ−η ξ−η 4uξη + 1 − exp uξ + 1 + uη = 0 2 2 This last equation yields A = 1, B = 0, C = x, which is elliptical, assuming that x > 0: B 2 − AC = −x < 0 −→

elliptical

Find the characteristics: √ √ dy = ± −x = ±i x dx Solving and integrating, we arrive at ξ(x, y) = ξ =

2 3/2 ix + y, 3

and

η(x, y) = η =

2 3/2 ix − y 3

With partial derivatives √ ξx =i x , √ ηx = x ,

1 ξxx = √ , ξyy = 0, ξxy = 0 2 x 1 ηy = −1, ηxx = √ , ηyy = 0, ηxy = 0 2 x ξy = 1,

Finding the coefficients e =Aξ 2 + 2Bξx ξy + Cξ 2 A x y =ξx2 + xξy2

W. Erbsen

HOMEWORK #11

√ =(i x )2 + x =−x+x =0 e =Aξx ηx + B (ξx ηy + ξy ηx) + Cξy ηy B √ √ =(i x )(i x ) + x(1)(−1) =−x−x = − 2x e =Aη 2 + 2Bηxηy + Cη 2 C x y √ =( x )2 + x(−1)2 =x − x =0

R = (Aξxx + 2Bξxy + Cξyy ) uξ + (Aηxx + 2Bηxy + Cηyy ) uη = (ξxx + xξyy ) uξ + (ηxx + xηyy ) uη     i i √ √ = uξ + uη 2 x 2 x Our equation becomes i − 4xuξη + √ (uξ + uη ) = 0 2 x And finally 

 2/3   1/3 3 i 4i 1 −4 (ξ − η) uξη + (uξ + uη ) = 0 4i 2 3 ξ −η

4.9

Homework #11

Problem 1 Consider a thin half pipe of unit radius laying on the ground. It is heated by radiation from above. We take the initial temperature of the pipe and the temperature of the ground to be zero. We model this problem with a heat equation with a source term. ut = κuxx + A sin x,

u(0, t) = u(π, t) = 0,

u(x, 0) = 0

(4.9.1)

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CHAPTER 4: MATHEMATICAL METHODS

Solution We begin by assuming that our solution can be written in terms of a steady-state solution and a transient one u(x, t) = µ(x) + w(x, t) µ(x) : Steady-state solution w(x, t) : Transient solution To find the equilibrium solution, we note that µt (x) = 0, and substitute µ(x) into (4.9.1): µt = κµxx + A sin x −→ µxx = −

A sin x, κ

µ(0) = µ(π) = 0

(4.9.2)

Integrating (4.9.2) once, Z

x

0

Z A x ∂2 µ(x) dx = − sin x dx ∂x2 κ 0 ∂ A µ(x) = cos x + C1 ∂x κ

(4.9.3)

And now integrating (4.9.3), Z

0

x

Z Z x A x ∂ µ(x) dx = cos x dx + C1 dx ∂x κ 0 0 A µ(x) = sin x + C1 x + C2 κ

(4.9.4)

Applying our first boundary condition to (4.9.4), we find that µ(0) =C2 6= 0 −→ C2 = 0 And similarly, applying th4e second boundary condition, µ(π) =C1 π 6= 0 −→ C1 = 0 And (4.9.4) becomes µ(x) =

A sin x κ

(4.9.5)

And now we do the same for w(x, t): wt = κwxx + A sin x −→ wt = κwxx

(4.9.6)

We can see that (4.9.6) can be expressed as an eigenvalue problem, the explicit details of which are shown in subsequent problems and I will not repeat them here. Applying all our given conditions, we arrive at r 2 2 wn (x, t) = sin (nx) e−κn t (4.9.7) π And in order P for (4.9.7) to work, we require that it corresponds to the Fourier coefficient an from w(x, t) = an wn (x, t):

W. Erbsen

HOMEWORK #11

Position Π

3Π 4

Π 2

Π 4

Time 2

4

6

8

10

Figure 4.1: Contour plot for Problem 1.

an = − =− r

=

r

r

2 π

Z

0

2 A π κ

π

sin (nx) µ(x) dx Z

π

sin (nx) sin(x) dx

0

2 A sin (nπ) π κ n2 − 1

(4.9.8)

Using (4.9.7) and (4.9.8), we have w(x, t) =

∞ X

an w(x, t) =

n=1

∞ X

n=1

r

2 ∞ 2A X sin2 (nπ) e−κn t = πκ n=1 n2 − 1

=−

2 A sin (nπ) π κ n2 − 1

A sin(x)e−κt κ

! r

2 2 sin (nx) e−κn t π

!

(4.9.9)

Which was arrived at via the following Mathematica code: V = Sum[-Sqrt[2/Pi] Integrate[Sin[n*x]*A/kappa Sin[x],{x,0,Pi}]*Sqrt[2/Pi]*Sin[n*x] *Exp[-kappa*n^2*t],{n,1,Infinity}] >>Out[331] = -(A*Exp[-t*kappa]*Sin[x])/kappa

345

CHAPTER 4: MATHEMATICAL METHODS

And finally, to get the full solution for u(x, t) we combine our steady-state solution from (4.9.5) and our transient solution from (4.9.9): u(x, t) = w(x, t) + µ(x) = −

A A sin(x)e−κt + sin(x) κ κ

Which may be rewritten as

u(x, t) =

  A sin(x) 1 − e−κt κ

A contour plot of my solution can be seen in Fig. (4.1), where I set A = κ = 1.

Problem 2 Obtain Poisson’s formula to solve the Dirichlet problem for the circular region 0 ≤ r < R, 0 ≤ θ < 2π. That is, determine a solution φ(r, θ) to solve Laplace’s equation ∇2 φ = 0 in polar coordinates given φ(R, θ). Show that Z 2π R2 − r 2 1 φ(r, θ) = φ(R, α) 2 dα (4.9.10) 2π 0 R + r 2 − 2Rr cos (θ − α)

Solution We begin by noting that Laplace’s equation in polar coordinates is   1 ∂ ∂f 1 ∂2f 2 ∇ f= r + 2 2 =0 r ∂r ∂r r ∂θ

(4.9.11)

Which should be solvable using separation of variables, where we assume that the solution can take the form of f(r, θ) = R(r)Θ(θ). Substituting this into (4.9.11) and separating the variables,   dR 1 d2 Θ r d r =− = λ2 (4.9.12) R dr dr Θ dθ2 Where we set each equation equal to the square of some equilibrium constant λ. We first take the equation for R(r),   r d dR r =λ2 R dr dr  2  r d R dR r 2 + =λ2 R dr dr d2 R dR r2 2 + r − λ2 R =0 (4.9.13) dr dr Now we must solve the equation for Θ(θ) from (4.9.12):

W. Erbsen

HOMEWORK #11

−

1 d2 Θ =λ2 Θ dθ2

d2 Θ + λ2 Θ =0 dθ2

(4.9.14)

We require our solution for Θ(θ) to be periodic, such that Θ(θ + 2π) = Θ(θ). Therefore we set λ = n where n is an integer, and (4.9.14) becomes d2 Θ + n2 Θ =0 dθ2 Which has solutions Θ(θ) =An einθ + Bn e−inθ = An cos(nθ) + Bn sin(nθ)

(4.9.15)

And similarly we can rewrite (4.9.13) as r2

d2 R dR +r − n2 R =0 dr 2 dr

Which has solutions of the form R(r) = Cn r n + Dn r −n

(4.9.16)

Combining our solutions from (4.9.15) and (4.9.16),
fn (r, θ) = R(r)Θ(θ) = Cn r n + Dn r −n (An cos(nθ) + Bn sin(nθ))

Choosing Dn = 0 for boundary conditions,

fn (r, θ) = Cn r n (An cos(nθ) + Bn sin(nθ)) And more generally, fn (r, θ) =

∞ X

Cn r n (An cos(nθ) + Bn sin(nθ))

n=0

=C0 r 0 (A0 cos(0) + B0 sin(0)) +

∞ X

Cn r n (An cos(nθ) + Bn sin(nθ))

n=1

=C0 A0 +

∞ X

Cn r n (An cos(nθ) + Bn sin(nθ))

(4.9.17)

n=1

The solution of (4.9.17) is most easily found if we first assume that R = r = 1, f(1, θ) =C0 A0 +

∞ X

Cn (An cos(nθ) + Bn sin(nθ))

n=1

=a0 +

∞ X

n=1

(an cos(nθ) + bn sin(nθ))

(4.9.18)

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CHAPTER 4: MATHEMATICAL METHODS

Where I have set C0 A0 = a0 , Cn = 1, An = an and Bn = bn . In doing this, we see that (4.9.18) is just a Fourier series. The coefficients are readily determined with Z 1 1 2π a0 = · f(1, α) dα (4.9.19a) 2 π 0 Z 1 2π f(1, α) cos(nα) dα (4.9.19b) an = π 0 Z 1 2π bn = f(1, α) sin(nα) dα (4.9.19c) π 0 Substituting (4.9.19a), (4.9.19b) and (4.9.19c) into (4.9.18): f(1, θ) =a0 +

∞ X

(an cos(nθ) + bn sin(nθ))

n=1 Z 2π

 Z ∞  Z 2π X 1 1 2π 1 f(1, α) dα + f(1, α) cos(nα) dα cos(nθ) + f(1, α) sin(nα) dα sin(nθ) 2π 0 π 0 π 0 n=1 " # Z ∞ X 1 2π 1 = f(1, α) + (f(1, α) cos(nα) cos(nθ) + f(1, α) sin(nα) sin(nθ)) dα π 0 2 n=1 " # Z ∞ 1 2π 1 X + (cos(nα) cos(nθ) + sin(nα) sin(nθ)) dα = f(1, α) π 0 2 n=1 " # Z ∞ 1 2π 1 X + = f(1, α) cos [n (θ − α)] dα (4.9.20) π 0 2 n=1 =

Where I used the identity cos α cos β + sin α sin β = cos (α − β) = cos (β − α). The bracketed term in (4.9.20) can be simplified, and I used Mathematica: FullSimplify[1/2 + Sum[r^n*Cos[n*(alpha - theta], {n, 1, Infinity}]] >>Out[157]=-((-1 + r^2)/(2*(1 + r^2 - 2*r*Cos[alpha - theta]))) With this, (4.9.20) becomes  1 − r2 f(1, α) dα 2 (1 + r 2 − 2r cos (θ − α)) 0   Z 2π 1 1 − r2 = f(1, α) dα 2π 0 1 + r 2 − 2r cos (θ − α)

1 f(1, θ) = π

Z

2π



And we can extend (4.9.21) to the more general case of f(r, θ) by replacing r with r/R:   Z 2π 1 1 − (r/R)2 f(r, θ) = f(R, α) dα 2π 0 1 + (r/R)2 − 2(r/R) cos (θ − α)   Z 2π 1 1/R2 R2 − r 2 = f(R, α) dα 2π 0 1/R2 R2 + r 2 − 2rR cos (θ − α) From (4.9.22) we are finally left with

(4.9.21)

(4.9.22)

W. Erbsen

HOMEWORK #11

f(r, θ) =

1 2π

Z

2π

f(R, α)

0



R2 − r 2 2 2 R + r − 2rR cos (θ − α)



dα

Problem 3 For 0 < x < `, solve ut = a2 uxx + w(x, t);

u(0, t) = 0,

ux (`, t) = 0,

u(x, 0) = f(x)

(4.9.23)

By means of a series expansion involving the eigenfunctions of d2 φ(x) + λφ(x) = 0; dx2

φ(0) = φ0 (`) = 0

(4.9.24)

Solution We begin our journey by assuming that the solution of (4.9.23) can be separated in the form u(x, t) = X(x)T (t). Assuming that this is the case, we plug it into the homogeneous analogue to (4.9.23) and take the corresponding partial derivatives: XTt = a2 Xxx T And separating the variables, Tt Xxx = = −λ a2 T X

(4.9.25)

Where I have chosen −λ to be the separation constant. Taking the equation for X(x) first, Xxx = −λ −→ Xxx + λX = 0 x

(4.9.26)

We immediately recognize that (4.9.26) is identical in form to (4.9.24). Furthermore, we also see that these are Sturm-Lioville boundary value problems, and the solution of which was derived in class (and commonly used in introduction quantum mechanics courses). The corresponding eigenvalues and eigenfunctions are given by  nπ 2  nπx  λn = , and Xn (x) = sin ` ` Now, taking our separated solution and plugging it into (4.9.23), ∞ X

n=1 ∞ X

n=1

Tn0 (t)Xn (x) =a2 Tn0 (t)Xn (x) =a2

∞ X

n=1 ∞ X

n=1

Tn (t)Xn00 (x) + f(x, t) Tn (t)Xn00 (x) +

∞ X

n=1

fn (t)Xn (x)

(4.9.27)

349

CHAPTER 4: MATHEMATICAL METHODS

And since the equation for X satisfies the Sturm-Liouville problem, we know that Xn00 (x) + λXn (x) = 0, and (4.9.27) becomes ∞ X

n=1

∞ X

Tn0 (t)Xn (x) = − a2 ∞ X

n=1

Tn (t)λn Xn (x) +

n=1 ∞ X

Tn0 (t) = − a2

∞ X

fn (t)Xn (x)

n=1

Tn (t)λn +

n=1

∞ X

fn (t)

(4.9.28)

n=1

Which is satisfied if Tn0 (t) = − a2 Tn (t)λn + fn (t) Whose solution takes the form 2

Tn (t) = e−a

λn t

Z

t

2

ea

λn α

0

 fn (α) dα + Tn (0)

(4.9.29)

And at this point we note that un (x, 0) = fn (x), which implies that fn (x) =

∞ X

Tn (0) sin

n=0

 nπx  `

And accordingly Tn (0) =

Z

2 `

`

fn (x) sin

0

 nπx  `

dx

So that (4.9.18) becomes 2

Tn (t) = e−a

λn t

"Z

t

0

2 ea λn α fn (α) dα + ` 2

Z

 nπx 

`

fn (x) sin

`

0

dx

#

(4.9.30)

Combining (4.9.20) with our solution for X(x) (and substituting in the corresponding eigenvalues), we arrive at

un (x, t) =

∞ X

n=1

(

2

−( nπa ` ) t

e

sin

"  nπx  Z `

t

2

−( nπa ` ) α

e

0

2 fn (α) dα + `

Z

`

fn (x) sin

0

 nπx  `

dx

#)

Problem 4 Find the solution of Laplace’s equation subject to the following boundary conditions ∇2 u = 0, u(r, 0) = u(r, α) = 0,

0 < θ < α,

a 0,

y(0) = 0,

o

, y(m) = M > −h

(4.10.20)

Solution Within the Euler-Lagrangian formalism, we wish to extract the Lagrangian from (4.10.20), and put it into the Euler-Lagrange equation. The form most convenient for our purposes will be that given by (4.10.12). Consequently, we must evaluate all the terms within the Euler-Lagrange equation first: q 2 y + h 1 + (y0 ) √ q y0 y + h ∂L ∂ p 2 0 = 0 y + h 1 + (y ) = q ∂y0 ∂y 2 1 + (y0 )

L=

p

∂L =0 ∂x

We now substitute (4.10.21a)-(4.10.21c) into (4.10.12):   √ q 0 d p y y + h 2  =0 y + h 1 + (y0 ) − y0 q dx 2 1 + (y0 )    q 0 d p y 2 0  =0 y + h  1 + (y0 ) − y q dx 2 0 1 + (y ) q    q  0 )2 0 p 1 + (y d  y 2  =0 y+h  1 + (y0 ) ·q − y0 q dx 2 2 1 + (y0 ) 1 + (y0 )

(4.10.21a) (4.10.21b)

(4.10.21c)

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CHAPTER 4: MATHEMATICAL METHODS

   0 2 0 2 d p 1 + (y ) (y )  =0 y + h q −q dx 2 2 0 0 1 + (y ) 1 + (y )    p d  1  =0 y + h q dx 2 1 + (y0 )   √ y+h  d  q =0 dx 2 1 + (y0 ) | {z } constant

So, the bracketed term in (4.10.22) is a constant of motion: √ y+h q =C 2 1 + (y0 )

(4.10.22)

(4.10.23)

We are now tasked to find a solution of y(x) corresponding to the given initial conditions: y+h 2

1 + (y0 )

=C 2

  2 y + h =C 2 1 + (y0 ) r dy y+h = −1 dx C2 p dy y + h − C2 = dx C C dx = p dy y + h − C2 p x =2C y + h − C 2 + C 0 2

(x − C 0 ) =4C 2 (y + h − C 2 )

(4.10.24)

In order for this problem to work, I have to assume that the second integration constant goes to zero. We now apply the initial condition on (4.10.24) to find C: 0 =4C 2 (h − C 2 ) −→ C = Substituting this back into (4.10.24), x2 =4h(y + h − h) From this it is easy to see that the solution is...

y(x) =

x2 4h

√

h

W. Erbsen

HOMEWORK #12

Problem 5 A rocket is propelled vertically upward so as to reach a prescribed height h in a minimum time while using a given fixed quantity of fuel. The vertical distance x(t) above the surface satisfies, mx00 = −mg + mU (t),

x0 (0) = 0

with x(0) = 0,

(4.10.25)

where U (t) is the acceleration provided by the engine thrust. We impose the terminal constant x(T ) = h, and we wish to find the particular thrust function U (t) which will minimize T assuming that the total thrust of the rocket engine over the entire thrust time is limited by the condition, Z T U 2 (t) dt = k 2 (4.10.26) 0

Where k is a given positive constant which measures the total amount of available fuel.

Solution The first thing we need to do is solve (4.10.25) with the given initial conditions. x00 (t) = − g + U (t) Z t Z t Z t x00 (t) dt = − g dt + U (α) dα 0 0 0 Z t x0 (t) = − gt + U (α) dα 0 Z Z Z "Z t

0

t

0

x (t) dt = − g

t

β

t dt +

0

U (α) dα

Z t "Z

gt2 x(t) = − + 2

0

0

0

0

β

U (α) dα

#

#

dβ

dβ

Z t Z t gt2 + U (α) dα dβ 2 0 α Z t gt2 =− + (t − α)U (α) dα 2 0 =−

(4.10.27)

Where I eliminated the integration constants en route. We now apply our boundary condition, x(T ) = h, to (4.10.27): x(T ) = −

gT 2 + 2

Z

T

(T − α)U (α) dα = h

0

(4.10.28)

And we define two new equations, F1 = − F2 =

Z

0

gT 2 + 2 T

Z

T

0

U 2 (t) dt

(T − t)U (t) dt

(4.10.29a) (4.10.29b)

363

CHAPTER 4: MATHEMATICAL METHODS

Where I changed variables from α to t in (4.10.29a). Starting with (4.10.29a), we continue with our variational adventures by adding some small element to our to-be-varied functional: Z T g F1 (T + δT ) = − (T + δT )2 + (T + δT − t)(U + δU ) dt 2 0 Z T g (T U + δT U − tU + T δU + δT δU − tδU ) dt = − (T 2 + 2T δT + (δT )2 ) + 2 0 Z T gT 2 =− − gT δT + (T U + T δU + δT U − tU − tδU ) dt (4.10.30) 2 0 We now subtract the original functional from (4.10.30), Z T Z T gT 2 gT 2 − gT δT + (T U + T δU + δT U − tU − tδU ) dt + − (T − t)U dt F1 (T + δT ) − F1 (T ) = − 2 2 0 0 Z T = − gT δT + (T δU + δT U − tδU ) dt (4.10.31) 0

We repeat this process with F2 in (4.10.29b): Z F2 (T + δT ) =

T

2

(U + δU ) dt

0

=

Z



T

0

=

Z

T

U 2 + 2U δU + (δU )2

U 2 + 2U δU

0

We now take (4.10.32) and subtract (4.10.29b): Z F2 (T + δT ) − F2 (T ) =




U + 2U δU

0

=2

Z

T

U δU dt

dt

dt

T 2






(4.10.32)

dt −

Z

T

U 2 dt

0

(4.10.33)

0

And now bringing all the pieces together, δT =λ1 F1 (T ) − λ2 F2 (t) " # " Z Z T =λ1 −gT δT + (T δU + δT U − tδU ) dt + λ2 2 0

= − λ1 gT δT +

Z

(λ1 T δU + λ1 δT U − λ1 tδU + λ2 2U δU ) dt

Now we take (4.10.34) such that δT = 0, eg no variation. This gives Z T 0= (λ1 T δU − λ1 tδU + λ2 2U δU ) dt 0

0=

0

U δU dt

0

#

T

0

Z

T

T

(λ1 T − λ1 t + 2λ2 U ) dt | {z }

(4.10.34)

W. Erbsen

HOMEWORK #12

0 =λ1 T − λ1 t + 2λ2 U

λ1 t − λ1 T = 2λ2 U −→ λ1 (t − T ) = 2λ2 U

(4.10.35)

And remember F1 and F2 ? Sure you do - here’s where we use them. Solve (4.10.35) for U (t) and substitute into these equations and integrate:   Z T gT 2 λ1 F1 = − + (T − t) − (T − t) dt 2 2λ2 0 Z T λ1 gT 2 =− − (T − t)2 dt 2 2λ2 0   gT 2 λ1 T 3 =− − − T3 + T3 2 2λ2 3 2 λ1 3 gT − T =h (4.10.36) =− 2 6λ2 And now for F2 : F2 =

Z

T

0



−

2 λ1 λ2 T 3 (T − t) dt = − 12 = k2 2λ2 4λ2 3

(4.10.37)

We have three independent variables, λ1 , λ2 , and T , so we need another equation. In addition to these we can one more as suggested by Donald Smith in his book “Variational Methods in Optimization.” We do this by again varying δT by a small amount, other than zero. Let’s do just that: Z T 1 = − λ1 gT + (λ1 T δU + λ1 U − λ1 tδu + 2λ2 uδu) dt 0

= − λ1 gT +

Z

= − λ1 gT +

Z

= − λ1 gT +

T

0



T

 λ1 (T − t) U δU + λ1 U + 2λ2 U δU dt | {z } −2λ2

(−2λ2 U δU + λ1 U + 2λ2 U δU ) dt

0

Z

T

λ1 U dt

(4.10.38)

0

Now insert (4.10.35) into (4.10.38)! 1 = − λ1 gT = − λ1 gT = − λ1 gT = − λ1 gT = − λ1 gT

  λ1 (T − t) + λ1 − dt 2λ2 0 "Z # Z T T λ21 − T dt − t dt 2λ2 0 0   λ2 T2 − 1 T2 − 2λ2 2 2 2 λ T − 1 2λ2 2 λ2 − 1 T2 4λ2 Z

T

(4.10.39)

365

CHAPTER 4: MATHEMATICAL METHODS

From here forward is just plugging in equations and solving; I left this up to Mathematica. There seem to be many solutions, here are two: 3(t − T ) gT 2 − 2h U (t) = 2T 2




,

U (t) =

√

3 k(T − t) T 3/2

W. Erbsen

HOMEWORK #12

Chapter 5

Departmental Examinations 5.1

Quantum Mechanics

Problem 1 Consider two different spin 1/2 whose Hamiltonian is completely specified as H = cs(1) · s(2) where c is a real constant. a)

What are the constants of motion? Obtain the eigenvalues of H for both singlet and triplet states.

b)

Consider at time t = 0 the spin of particle (1) along the z-axis is up and the spin of particle (2) along the z-axis is down. What is the wave function of the system at a later time t?

c)

Calculate the probability that at a later time t the spins of both particles are aligned as they were at t = 0.

Useful information:

χm 1

 α(1)α(2) m=1      β(1)β(2) m = −1 =   1    √ [α(1)β(2) + β(1)α(2)] m = 0 2

1 χ00 = √ [α(1)β(2) − β(1)α(2)] 2

Solution a)

∗ Parenthetic

The Hamiltonian is given by H = cs1 · s2 ,∗ and the total spin is just s = s1 + s2 . We can find an alternate form of the Hamiltonian by doing the following:  1 2 2 s2 = (s1 + s2 ) = s21 + s22 + 2s1 · s2 −→ s1 · s2 = s − s21 − s22 2 indicies have been shamelessly transformed into subscripts to avoid confusion.

W. Erbsen

QUANTUM MECHANICS

Such that our Hamiltonian becomes H=

 c 2 s − s21 − s22 2

(5.1.1)

To find the constants of motion, we test to see whether or not the quantity in question commutes with the Hamiltonian (5.1.1). The contenders are s1 , s2 , and s. Starting with s1 ,
 c 2 c 2 s − s21 − s22 , s1 = [s , s1 ] − [s21 , s1 ], − [s22 , s1 ] = 0 [H, s1] = (5.1.2) 2 2 | {z } | {z } | {z } i ii iii

It is immediately obvious that i and ii equal zero, and since the operators in iii act on two different particles, it commutes to. Therefore, s1 is a constant of motion. Similarly, for s2 we have:
 c 2 c 2 [H, s2] = s − s21 − s22 , s2 = [s , s2 ] − [s21 , s2 ], −[s22 , s2 ] = 0 (5.1.3) 2 2

For the same reasons applied to (5.1.2), we see from (5.1.3) that s2 is also a constant of motion. Furthermore, for s, we have
 c 2 c 2 s − s21 − s22 , s = [s , s] − [s21 , s], −[s22, s] = 0 (5.1.4) [H, s] = 2 2 So, from (5.1.2)-(5.1.5), we see that the constants of motion are s1 , s2 , and s .

For the second part of this problem, we wish to find the eigenvalues of (5.1.1) for both singlet s = 0 and triplet s = 1 states. For the singlet state, we recognize that s = 0, so s1 and s2 must be in opposite spin states. Let’s let s1 = 1/2 , and s2 = − 1/2 .  c Hχ00 = s2 − s21 − s22 χ00 2 c 2 0 = s χ0 − s21 χ00 − s22 χ00 2 c s(s + 1)~2 χ00 − s1 (s1 + 1)~2 χ00 − s2 (s2 + 1)~2 χ00 = 2 c 0(0 + 0)~2 − 1/2 ( 1/2 + 1)~2 − 1/2 ( 1/2 + 1)~2 χ00 = 2 c = 0 − 3/4 ~2 − 3/4 ~2 χ00 2 3c = − χ00 (5.1.5) 4 For the triplet state we have s = 1, so following the steps from (5.1.5), c = 1(1 + 1)~2 − 1/2 ( 1/2 + 1)~2 − 1/2 ( 1/2 + 1)~2 χm 1 2 c = 2 − 3/4 ~2 − 3/4 ~2 χm 1 2 c m = χ1 4

(5.1.6)

So from (5.1.5) and (5.1.6), our eigenvalues for the provided Hamiltonian for both singlet and triplet states are, respectively: Hχ00 = −

3c 2 0 ~ χ0 , 4

Hχm 1 =

c 2 m ~ χ1 4

369

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

b)

The general wave function for particle one spin up and particle two spin down is given by  1  χ(0) = √ χ00 + χ01 2

(5.1.7)

If we wish to obtain  the wave function at some later time t, we need to apply the time evolution operator U = exp − iHt to (5.1.7) ~       1 iHt 0 iHt 0 χ0 + exp − χ1 χ(t) = √ exp − ~ ~ 2        1 it 3c it  c 2  0 =√ exp − − ~2 χ00 + exp − ~ χ1 ~ 4 ~ 4 2 Which readily simplifies to       3c~ c~ 0 exp it χ0 + exp − it χ01 4 4

1 χ(t) = √ 2 c)

To calculate the probability that at some later time t both electrons will return to their original state, we simply take the modulus squared of the matrix element between these two states. To simplify notation, I make the following substitutions: E0 =

3c~ , 4

E1 =

c~ 4

So, the probabilty is then:      2 1  1  E0 it 0 2 √ χ00 + χ01 √ |hχ(0)|χ(t)i| = e χ0 + e−E1 it χ01 2 2 2  1

= χ00 + χ01 eE0 it χ00 + e−E1 it χ01 4 2 1 = eE0 it hχ00 |χ00 i +e−E1 it hχ00 |χ01 i +eE0 it hχ01 |χ00 i +e−E1 it hχ01 |χ01 i | {z } | {z } | {z } | {z } 4 1

0

0

2 1 = eE0 it + e−E1 it 4 2 c~ 1 3c~ = e 4 it + e− 4 it 4 i 2 1 c~ h c~ c~ = e 4 it e 2 it + e− 2 it 4   2 1 c~ it c~ 4 = e 2 cos t 4 2

(5.1.8)

From (5.1.8) it is clear that the probability of realignment is

P = cos2

1



c~ t 2



W. Erbsen

QUANTUM MECHANICS

Problem 2 2 1 1 −1


, what are the eigenvalues of this Hamiltonian?

a)

If

b)

Prove that [H, exp (H)] = 0, that is, the two operators commute.

c)

What are the eigenvalues of exp (H)?

d)

If a small perturbation H 0 = λ ( 01 10 ), where λ is a small positive number, is applied to the system, calculate the change in the eigenenergy of the ground state, to first order in λ.

e)

Carry out the change in the ground state energy to second order in λ.

Solution a)

b)

Let’s diagonalize this mofo: √ 2 − λ 1 = (2 − λ)(−2 − λ) − 1 = λ2 − 5 = 0 −→ λ = ± 5 1 −2 − λ

We first recall that the series expansion for ex is ex =

∞ X xn x2 x3 =1+x+ + + ... n! 2 6 n=0

So, our commutator is [H, exp[H]] = [H, (1 + H + h2 /2 + h3 /6 + ...)] = [H, 1] + [H, H] + [H, H 2/2] + [H, H 3/6] (5.1.9) Each of the terms in (5.1.9) vanish, so we can only conclude that H and exp[H] commute, and so [H, exp [H]] = 0 c) d)

√ The eigenvalues of exp [H] are just λ = exp [± 5 ] . The first thing we need to do is find the eigenvectors associated with the eigenvalues that we found in part a): √        √ √ 1√ 2− 5 1√ α 0 For λ = + 5 : = (2 − 5 )α + β = 0 −→ ψfes = 0 2− 5 1 −1 − 5 β √        √ √ 1√ 2+ 5 1√ α 0 = (2 + 5 )α + β = 0 −→ ψgs = For λ = − 5 : 0 2+ 5 1 −1 + 5 β √ The ground state is associated with the eigenvalue λ = − 5 , with eigenvector ψgs . The first order energy correction is given by    √
1√ 0 1 0 hψgs |H |ψgsi = 1 (2 + 5 ) λ 1 0 2+ 5  √  √
2+ 5 =λ 1 (2 + 5 ) 1

371

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

From which it is clear that the first order shift in energy is simply  √  (1) Egs =λ 4+2 5 e)

To carry out the change in the ground state energy to second order in λ, we first recall that the second order energy shift from time-independent perturbation theory is En(2) =

∞ X |hψm |H 0 |ψn i|2 En − Em m,n

(5.1.10)

m6=n

In our case ψm = ψfes , ψn = ψgs , Em = E1 and En = E0 . We compute the matrix element in (5.1.10) first:    √
1√ 0 1 0 hψfes |H |ψgsi = 1 (2 − 5 ) λ 1 0 2+ 5  √  √
2+ 5 =λ 1 (2 − 5 ) 1 =4λ

(5.1.11)

The denominator of (5.1.10) is just E0 − E1 = −

√ √ √ 5 − 5 = −2 5

(5.1.12)

We now substitute (5.1.11) and (5.1.12) into (5.1.10) to obtain the energy shift: En(2)

∞ X |4λ|2 8λ2 √ −→ En(2) = √ = 5 m,n −2 5 m6=n

Problem 3 Consider two spin 1/2 particles described by the Hamiltonian H=

p2 p21 + 2 + V (x1 ) + V (x2 ) 2m 2m

(5.1.13)

Where V (x) = ∞ for x < 0 and for x > a; V (x) = 0 for 0 < x < a. Assume that the electrons are in the opposite spin state, that is, the total S = 0.

a)

Write down the spin wave function(s). Use the standard notations: α for spin up and β for spin down.

b)

Find the energy and wavefunctions of the ground state of this Hamiltonian.

c)

Find the energy and wavefunction of the lowest state for S = 1.

d)

Find the energy and wavefunction of the second S = 0 state. Show that the energy is the same as in (c).

W. Erbsen

e)

QUANTUM MECHANICS

If the two particles have a small interaction W (x1 , x2 ) = bx1 x2 where b is small and positive, show that the degeneracy in (c) and (d) is removed. Which one has the lower energy?

Solution We first note that since s1 = ± 1/2 and s2 = ∓ 1/2 then s1 + s2 = 0. The eigenfunctions and eigenenergies are r  nπx  2 n2 π 2 ~2 , En = ψn (x) = sin L a 2ma2 a)

The spin wave function is 1 χ = √ [α(1)β(2) − α(2)β(1)] 2

b)

The full wave function is just the product of the space part and the spin part: √  πx   πx  2 1 2 ψ(x) = ψ1 (x1 )ψ1 (x2 )χ −→ ψ(x) = sin sin [α(1)β(2) − α(2)β(1)] L L L

(5.1.14)

(5.1.15)

While the energy is just the sum: Egs = E1 + E1 −→ Egs = c)

π 2 ~2 2mL2

For s = 1 the spin wave function is symmetric, so the space part must be anti-symmetric: 1 ψa,b (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) − ψa (x2 )ψb (x1 )] χ 2 Since s = 1, this implies that they cannot both be in the same state, so the lowest state would be for one of the electrons to be at n = 1, and the first at n = 2. The total energy is just the sum of these: E=

d)

5π 2 ~2 2mL2

Now that s = 0 this implies that the spin wave function is anti-symmetric, so the space part should now be symmetric: 1 ψa,b (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) + ψa (x2 )ψb (x1 )] χ 2 And since we are now looking for the second lowest state, and we know that since s = 0 implies that the electrons are in opposite states this means they can occupy the same energy level, so the ground state is n1 = 1 and n2 = 1, and the second state is n1 = 1 and n2 = 2 (or vice versa). The total energy is the same as the ground state for the case of s = 1:

373

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

E= e)

5π 2 ~2 2mL2

To do this part we must compute the first order energy shift in time-independent perturbation theory. Let’s define 1 ψa (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) − ψa (x2 )ψb (x1 )] 2 1 ψb (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) + ψa (x2 )ψb (x1 )] 2 Then the first order shift for s = 1 is then E (1) =hψa |H 0 |ψa i ZZ b 2 = x1 x2 [ψa (x1 )ψb (x2 ) − ψa (x2 )ψb (x1 )] dx1 dx2 2 ZZ h b 2 2 = x1 x2 (ψa (x1 )ψb (x2 )) + x1 x2 (ψa (x2 )ψb (x1 )) 2 −2x1 x2 ψa (x1 )ψb (x2 )ψa (x2 )ψb (x1 )] dx1 dx2 And also the first order shift for s = 0 is ZZ b 2 E (1) = x1 x2 [ψa (x1 )ψb (x2 ) + ψa (x2 )ψb (x1 )] dx1 dx2 2 ZZ h b 2 2 = x1 x2 (ψa (x1 )ψb (x2 )) + x1 x2 (ψa (x2 )ψb (x1 )) 2 +2x1 x2 ψa (x1 )ψb (x2 )ψa (x2 )ψb (x1 )] dx1 dx2

(5.1.16)

(5.1.17)

It is clear from just looking at (5.1.16) and (5.1.17) that the degeneracy will be broken, since they differ in the sign of one of the cross terms. The case of s = 1 has a lower energy.

Problem 4 The result of a measurement shows that the electron spin is along the +x direction at t = 0. For t > 0, the electron enters in a uniform magnetic field that is parallel to the +z direction. Calculate the quantum mechanical probability as a function of time for finding the electron in each of the following states (~ = 1): a)

Sx = +1/2

b)

Sx = −1/2

c)

Sz = +1/2

d)

Sz = −1/2

  The Pauli matrices are given by 0 1 σx = , 1 0

σy =



 0 −i , i 0

σz =



1 0 0 −1



W. Erbsen

QUANTUM MECHANICS

Solution Initially at time t = 0 the spin is along the x-direction, so it will be most convenient to express the spinor in the Sz basis. We first recall that     1 1 1 1 (x) (x) χ+ = √ , χ− = √ 1 −1 2 2     1 0 (z) (z) χ+ = , χ− = 0 1 And also that we can apply the time-evolution operator to a state at t = 0 as |ψ(t)i = U |ψ(0)i,

with

U = exp [−iHt]

The Hamiltonian for a dipole in a magnetic field (the dipole in this case is provided by the spin) is H = µB =

eSz B0 = ωSz m

So the time evolution operator becomes U = exp [−iωSz t] We now wish to express our current spinor in terms of the Sz basis:  †  † (x) (x) (z) (z) |ψ(0)i = χ+ χ+ + χ+ χ−    
1
0 1 1 =√ 1 1 +√ 1 1 0 1 2 2     1 1 1 0 =√ +√ 0 2 2 1 1 1 = √ |↑i + √ |↓i 2 2

(5.1.18)

(5.1.19)

We apply the time evolution operator from (5.1.18) to our state at time t = 0 from (5.1.19): 1 1 |ψ(t)i =e−iωSz t √ |↑i + e−iωSz t √ |↓i 2 2
1 ω ω = √ e−i 2 t |↑i + ei 2 t |↓i 2

(5.1.20)

Where we have set ~ = 1. Now that we have our time-dependent wave function, we can calculate the time-dependent probabilities for the spin to flip: a)

The probability for finding the particle to be in Sx = 1/2 is found by 2

P = |hSx (↑)|ψ(t)i|
2 ω ω 1 = hSx (↑)| e−i 2 t |↑i + ei 2 t |↓i i 2 2 1 ω ω = hSx (↑)|e−i 2 t |↑i + hSx (↑)|ei 2 t |↓i 2

375

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

2 ω 1 ω = e−i 2 t hSx (↑)|↑i + ei 2 t hSx (↑)|↓i 2

Where the expectation values are given by    †
1 1 (x) (z) = hSx (↑)|↑i = χ+ χ+ = √ 1 1 0 2    †
1 1 (x) (z) = hSx (↑)|↓i = χ+ χ− = √ 1 1 0 2

(5.1.21)

1 √ 2 1 √ 2

(5.1.22a) (5.1.22b)

Substituting (5.1.22a) and (5.1.22b) into (5.1.21), P = b)

2   1 i ω t ωt 1 1 −i ω t 2 2 + √ 2 √ −→ e e P = cos 2 2 2 2

And for the probability for finding the particle in Sx = −1/2, 2 ω 1 ω P = e−i 2 t hSx (↓)|↑i + ei 2 t hSx (↓)|↓i 2

(5.1.23)

Where we have

   †
1 1 1 (x) (z) = √ hSx (↓)|↑i = χ− χ+ = √ 1 −1 0 2 2    †
1 1 0 (x) (z) hSx (↓)|↓i = χ− χ− = √ 1 −1 = −√ 1 2 2

(5.1.24a) (5.1.24b)

Substituting (5.1.24a) and (5.1.24b) into (5.1.23), P = c)

2   1 i ω t 1 1 −i ω t ωt 2 2 − √ 2 √ −→ e e P = sin 2 2 2 2

Doing the same thing for Sz = 1/2, P = Where

2 1 −i ω t ω e 2 hSz (↑)|↑i + ei 2 t hSz (↑)|↓i 2

hSz (↑)|↑i =



(z) χ+

†

(z) χ+

= 1

†  (z) (z) hSz (↑)|↓i = χ+ χ− = 1 Putting (5.1.26a) and (5.1.26b) into (5.1.25), P = d)

 
1 0 =1 0  
0 0 =0 1

(5.1.25)

(5.1.26a) (5.1.26b)

1 −i ω t 2 1 e 2 −→ P = 2 2

And the mantra continues with Sz = −1/2: 2 ω 1 ω P = e−i 2 t hSz (↓)|↑i + ei 2 t hSz (↓)|↓i 2

(5.1.27)

W. Erbsen

QUANTUM MECHANICS

Where we have  † (z) (z) hSz (↓)|↑i = χ− χ+ = 0

 † (z) (z) hSz (↓)|↓i = χ− χ− = 0 Slamming (5.1.28a) and (5.1.28b) into (5.1.27), P =

Problem 5

 
1 1 =0 0  
0 1 1 = 1 2

(5.1.28a) (5.1.28b)

1 1 i ω t 2 e 2 −→ P = 2 2

(Old Problem 2)

A quark-quark potential that is sometimes used in quark models of mesons is V (r) = g log(r/a). If the quarks are very heavy, then non-relativistic physics is possible. Use the uncertainty principle to estimate the size and binding energy of a pair of quarks, each of mass m (g > 0).

Solution Before we get down to bidness we recall that the reduced mass for two particles of equal mass is µ=

mm m = −→ m = 2µ m+m 2

The uncertainty principle is ∆x∆p ≤ ~/2 −→ ∆x∆p ∼ ~. If we assume that r ∼ ∆r and p ∼ ∆p, then rp ∼ ~ −→ p ∼ ~/r. We now substitute this into the two-particle Hamiltonian, with the given interasction term: hri p21 p2 + 2 + g log 2m 2m a hr i p2 = + g log 2µ a hri ~2 = + g log 2µr 2 a

H=

(5.1.29)

The plan of action now is to minimize (5.1.29), by taking the derivative and setting it equal to zero:  2  h r i dH ∂ ~ ∂  = + g log =0 dt ∂t 2µr 2 ∂t a s ~2 ~2 g (5.1.30) ⇒ − 3 + = 0 −→ r = ± µr r gµ Taking r from (5.1.30) and putting it back into H from (5.1.29),

377

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

" s # ~2  gµ  1 ~2 H= + g log 2µ ~2 a gµ "s # g ~2 = + g log 2 a2 gµ

(5.1.31)

We can see that (5.1.31) is actually an eigenvalue, and so the solution is

g E = + g log 2

"s

~2 2 a gµ

#

Problem 6 Show that for a system consisting of two identical particles with spin I, the ratio of the number of states symmetric under exchange of the two spins to the number of states antisymmetric under exchange of the two spins is equal to (I + 1)/I. [Hint: If you do not know where to start, consider I = 1/2 first.]

Solution For a nucleus with spin quantum number I 6= 0, there are 2I + 1 different possible values of the zcomponent of the spin angular momentum quantum number: MI = −I, −I + 1, ..., I − 1, I For two such nuclei, the total multiplicity is (2I + 1) · (2I + 1). Clearly, some of the wave functions associated with these will be symmetric, while some will be anti-symmetric: 1 ψS (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) + ψb (x1 )ψa (x2 )] 2 1 ψA (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) − ψb (x1 )ψa (x2 )] 2

(5.1.32) (5.1.33)

Where ψS (x1 , x2 ) denotes the symmetric states, and ψA (x1 , x2 ) is for the anti-symmetric states. Let’s also denote the number of symmetric states as num(S), and the number of anti-symmetric states as num(A). In addition to these two sources of states, there is another case for the symmetric wave function: 0

ψS (x1 , x2 ) = ψa (x1 )ψa (x2 )

(5.1.34)

And as before, we let num(S 0 ) denote the number of symmetric states in the form of (5.1.34). We know that (5.1.32), (5.1.33) and (5.1.34) represent all possible states for a nucleus with spin quantum number I. Accordingly,

W. Erbsen

QUANTUM MECHANICS

(2I + 1)2 | {z }

num(S) + num(S 0 ) | {z }

=

Total # of states

# of symmetric states

+

num(A) | {z }

# of anti-symmetric states

We can see that the multiplicity of num(S 0 ) is just 2I + 1, while the multiplicity of num(S) and num(A) are equal, but unknown (call this multiplicity α). So, (2I + 1)2 = α + (2I + 1) + α (2I + 1)2 − 2I − 1 = 2α

4I 2 + 4I + 1 − 2I − 1 = 2α

4I 2 + 2I = 2α −→ α = 2I 2 + I = I(2I + 1) So, num(S) = num(A) = I(2I + 1), and num(S 0 ) = 2I + 1. The ratio of symmetric states to antisymmetric states is then I(2I + 1) + 2I + 1 (2I + 1) (I + 1) I +1 num(S) + num(S 0 ) = = −→ num(A) I(2I + 1) (2I + 1) I I

Problem 7 A rod of length d and uniform mass distribution is pivoted at its center and constrained to rotate in an xy-plane. The rod has mass M and charges +Q and −Q fixed at either end. a)

Write down the Hamiltonian, its eigenfunctions, and eigenvalues.

b)

If a constant weak electric field ε lies along the x-direction, calculate the perturbed energy of the ground state and the first excited state to the 2nd order in ε.

Solution a)

Classically, we know that the Hamiltonian for a system like this is expressed as H=

L2 2I

Where L is the angular momentum, and I the moment of inertia, which in our case is X
2
2 I= mi ri2 = m d/2 + m d/2 = 1/2 md2 i

And we also know that L2 can be expressed as L2 = −~2

∂2 ∂φ2

379

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

So, our Hamiltonian is: H =−

~2 ∂ 2 md2 ∂φ2

(5.1.35)

We now insert our Hamiltonian from (5.1.35) into the time-independent Schr¨ odinger equation: −

~2 ∂ 2 ∂2 md2 E ψ(φ) = Eψ(θ, φ) −→ ψ(φ) + ψ(φ) = 0 md2 ∂φ2 ∂φ2 ~2

With solutions ψ(φ) = A sin(kφ) + B cos(kφ),

k2 =

md2 E ~2

(5.1.36)

We require rotational periodicity, eg that φ(φ) = ψ(φ + 2π). Applying this to (5.1.36), ψ(φ + 2π) =A sin [k(φ + 2π)] + B cos [k(φ + 2π)]     =A sin(kφ) cos(2πk) + sin(2πk) cos(kφ) + B cos(kφ) cos(2πk) − sin(kφ) sin(2πk) | {z } | {z } | {z } | {z } 1

0

1

=A sin(kφ) + B cos(kφ)

0

(5.1.37)

Two important things happened here. First, I assumed that k was an integer, so let’s just say that k = n from now on. Also, I used the following trig identities: sin (α ± β) = sin α cos β ± cos α sin β

cos (α ± β) = cos α cos β ∓ sin α sin β So, our eigenvalues are then k 2 = n2 =

md2 E n2 ~2 −→ E = ~2 md2

And our eigenfunctions are ψ(φ) = A sin(nφ) + B cos(nφ) = Aeinφ + Be−inφ However, if we imagine our system to be classical, then we can simply our expression for the eigenfunctions by recognizing that our rotor can only spin in one direction concurrently (classically), so our wave function “collapses” and allows only motion in one direction: ψ(φ) = Aeinφ Normalizing, 2

A

Z

0

So, our eigenfunction is:

2π

e−inφ





1 einφ dφ = A2 2π = 1 −→ A = √ 2π

ψ(φ) = √

1 inφ e 2π

W. Erbsen

QUANTUM MECHANICS

b)

The dipole Hamiltonian for an oscillating perturbation is H 0 = −pE = −QdE0 cos φ The first order energy shift is then En(1) =hψn (φ)|H 0 |ψn (φ)i Z 2π 1 = e−inφ (−QdE0 cos φ) einφ dφ 2π 0 Z QdE0 2π cos φ dφ = 0 =− 2π 0 The second order energy shift has the form En(2)

∞ 2 X |hψm (φ)|H 0 |ψn (φ)i| = En − Em n,m

(5.1.38)

n6=m

We start with the ground state first. Taking the numerator, 1 imφ he |QdE0 cos φ|einφi 2π QdE0 imφ =− he | cos φ|einφi 2π Z QdE0 2π −imφ =− e cos φeinφ dφ 2π 0 Z QdE0 2π cos φei(n−m)φ dφ =− 2π 0 Z
QdE0 2π iφ =− e + e−iφ ei(n−m)φ dφ 4π 0 Z 2π  Z 2π QdE0 i(n−m+1)φ i(n−m−1)φ =− e dφ + e dφ 4π 0 0 QdE0 =− [δn,m+1 + δn,m−1 ] 2

hψm (φ)|H 0 |ψn (φ)i =

(5.1.39)

The denominator of (5.1.38) would be En − Em =

~2 (n2 − m2 ) md2

(5.1.40)

For the ground state, we have n = 1. Realizing this, we now take (5.1.39) and (5.1.40) and substitute it back into (5.1.38): (2)

E1 =

∞ 2 Q2 d2 E02 md2 X |δ1,m+1 + δ1,m−1 | 4 ~2 1 − m2

(5.1.41)

m6=1

The only possible value of m to sum over is m = 2, the rest of the terms vanish. Therefore, the second order energy shift for the ground state is just (2)

E1 =

Q2 d2 E02 md2 (1)2 mQ2 d4 E02 (2) −→ E1 = − 2 2 4 ~ 1 − (2) 12~2

381

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

To find the second order energy shift for the first excited state, we start with (5.1.41) with n = 2: (2)

E2 =

∞ 2 Q2 d2 E02 md2 X |δ2,m+1 + δ2,m−1 | 4 ~2 4 − m2

(5.1.42)

m6=2

In this case there are two possible values we can sum to such that (5.1.42) will not collapse: m = 1 and m = 3. Summing these, we can find the energy:   Q2 d2 E02 md2 (1)2 (1)2 (2) + E2 = 4 ~2 2 − (1)2 2 − (3)2   Q2 d2 E02 md2 1 = 1− 2 4 ~ 7 Which finally leads us to (2)

E2 = c)

3mQ2 d4 E02 14~2

If the field is very strong, then the molecule will be perfectly aligned and cos φ ≈ 1, so that H 0 = −QdE0 cos φ → H 0 = −QdE0 . Inserting this into the time-independent Schr¨ odinger equation,   ~2 ∂ 2 − − QdE0 ψ(φ) = Eψ(φ) md2 ∂φ2 ∂2 md2 ψ(φ) + 2 (QdE0 + E) ψ(φ) = 0 2 ∂φ ~ If we let κ2 =

md2 (QdE0 + E) ~2

Then our approximate wave function is ψ(φ) = A cos κφ + B sin κφ

Problem 8 Consider a particle in an infinitely deep three-dimensional box with the length on each side equal to a. a)

Write down the Hamiltonian describing such a particle.

b)

Find the first 5 lowest eigenenergies.

c)

Find the degeneracy of each energy eigenstate above.

d)

If there are 10 noninteracting, indistinguishable fermions in such a box, what is the ground state energy of the whole system? Consider separately the cases for spin 1/2 and spin 3/2 particles.

W. Erbsen

QUANTUM MECHANICS

Solution a)

The Hamiltonian is for a particle in a three dimensional box is ~2 ~2 2 ∇ + V −→ H = − H =− 2m 2m Where the potential is defined by V =

b)



−V0 0



∂2 ∂2 ∂2 + + ∂x2 ∂y2 ∂z 2



+V

(5.1.43)

if 0 < x, y, z < a otherwise

The energy eigenvalues corresponding to (5.1.43) are given by En =


π 2 ~2 n2x + n2y + n2z 2 2ma

(5.1.44)

And the first five lowest energies, tabulated via (5.1.44) are E111 =

3π 2 ~2 2ma2 6π 2 ~2 2ma2 9π 2 ~2 = 2ma2 11π 2 ~2 = 2ma2

E112 =E121 = E211 = E122 =E212 = E221 E113 =E131 = E311 E222 =

12π 2 ~2 2ma2

E223 =E232 = E322 = c) d)

17π 2 ~2 2ma2

The degeneracies are given by the number of different combinations of atomic numbers that result in the same eigenenergy; these are obvious from part b). If 10 spin 1/2 fermions are in the box, then the ground state is determined by the fact that 2 particles can occupy each of the lowest energy states (± 1/2 for each): E = 2E111 + 2E211 + 2E121 + 2E112 + 2E122 −→ E =

60π 2 ~2 2ma2

For spin 3/2 , then 4 can occupy each of the lowest states: E = 4E111 + 4E211 + 2E121 −→ E =

48π 2 ~2 2ma2

Problem 9 Let

|ψi =

p

1/ 3

Y10 α +

p

2/ 3

Y11 β

(5.1.45)

383

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

a)

If you were to measure Sz , what is the probability that you would get ~/2?

b)

If indeed you do get ~/2 in (a), what is the new state (or wavefunction after the measurement)?

c)

If you measure Lz after (b), what is the probability that you will get −~? (Note: α for spin up and β for spin down, and the Y ’s are the spherical harmonics).

Solution a)

The probability of measuring Sz would simply be the square of the coefficient associated with that state. In other words, P =

b)

1 3

If you do measure ~/2, then you have essentially “collapsed” the wave function, leaving only the part which corresponds to your measurement. In this case, |ψi = Y10 α

c)

(5.1.46)

The chances would be zero. The current state our particle is in (5.1.46), which has ` = 1 and m` = 0. In the general case, when measuring Lz we have Lz Y`m` (θ, φ) = m` ~Y`m` (θ, φ) However, in our case Lz Y10 (θ, φ) = (0)~Y10 (θ, φ) So, it is very clear that the chances that upon measuring Lz you get −~ is zero.

Problem 10 A spin 1/2 particle is placed in a uniform magnetic field H0 directed along plied along xˆ. Calculate the time independent probability for the spin to flip      spin matrices are 0 1 0 −i 1 σx = , σy = , σz = 1 0 i 0 0

Solution

zˆ. A small rf field H1 cos ωt is apfrom up to down along zˆ. The pauli  0 −1

W. Erbsen

QUANTUM MECHANICS

We begin our journey by restating the problem in a much more transparent way. Our particle is initially perfectly content with beign spin-up whilst a uniform magnetic field is being applied to it along the z-direction. Then, a grumpy old perturbation comes along pointing in the x-direction. The question is: what are the chances that the particle will go from begin spin-up to spin-up along the z-direction to spin down along the same direction? This is a time-dependent perturbation theory problem, and accordingly we require the formalism of first-order time-dependent perturbation theory. First things first, however; the Hamiltonian for a charged particle in a magnetic field is given by   eB0 Sz ~ 1 0 H0 = , where Sz = mc 2 0 −1 Now – we recall from time-dependent perturbation theory that 1 0 −iω0 t H e c2 (t) i~ 12 1 0 iω0 t c1 (t) c˙2 (t) = H21 e i~

c˙1 (t) =

(5.1.47a) (5.1.47b)

Now, the particle is initially spin-up, so (z)

cb (0) = 1

(z)

ca (0) = 0

Eb

χ+

Ea

χ−

Since at time t = 0 we are guaranteed to be in the upper state, and wish to know the probability that we will be in the lower state as a function of time, we use (5.1.47a). We are told that the perturbation takes the form: H10 = H1 cos ωt =

eB 0 Sx cos ωt m

So, the matrix element that we use in (5.1.47a) is then eB 0 Sz cos ωt|↑i m eB 0 cos ωth↓|Sx |↑i = m     
~ 0 1 eB 0 1 = cos ωt 0 1 1 0 0 m 2    0
1 e~B = cos ωt 1 0 0 m

0 Hab =h↓|

=

e~B 0 cos ωt m

(5.1.48)

We now take (5.1.48) and substitute it into the first-order integral equation for the 2 → 1 transition: Z 1 t 0 0 −iω0 t0 0 ca (t) = H (t )e dt i~ 0 ab  Z  0 1 t e~B 0 = cos ωt0 e−iω0 t dt0 i~ 0 m

385

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Z 0 eB 0 t = cos ωt0 e−iω0 t dt0 im 0 Z  0 0 eB 0 t  iωt0 = e + e−iωt e−iω0 t dt0 2im 0 Z t  Z t eB 0 −i(ω0 −ω)t0 0 −i(ω0 +ω)t0 0 e dt + e dt = 2im 0 "0 t " t  0  −i(ω0 −ω)t0 −i(ω0 +ω)t0  eB e e = − + − 2im  i(ω0 − ω) i(ω0 + ω)  0 0   eB 0 e−i(ω0 −ω)t 1 e−i(ω0 +ω)t 1 − + − + = 2im i(ω0 − ω) i(ω0 − ω) i(ω0 + ω) i(ω0 + ω) | {z }

(5.1.49)

We typically neglect the underbraced terms in (5.1.49). Why? Good question. We are typically looking at driving frequencies that are very close to the resonant frequencies, eg ω ≈ ω0 . In this approximation, the underbraced terms are very small compared to the rest of the expressions, so they essentially vanish. We could carry them on through the end to get an exact solution; but this is not necessary. Continuing,  −i(ω0 −ω)t  eB 0 e 1 ca (t) = + − 2im i(ω0 − ω) i(ω0 − ω) (ω0 −ω) o (ω0 +ω) eB 0 e−i 2 t n i (ω0 −ω) t − e−i 2 t = e 2 2im i(ω0 − ω) (ω0 −ω)   eB 0 e−i 2 t (ω0 − ω)t = sin (5.1.50) 2im i(ω0 − ω) 2 And to find the transitional probability, we square the modulus of (5.1.50), and we obtain:

Pb→a =

  e2 B 02 1 2 (ω0 − ω)t sin m2 (ω0 + ω)2 2

Problem 11 Calculate the transmission coefficient T at any energy E for the potential V (x) = −

~2 δ(x − b) 2ma

(5.1.51)

Solution The solutions of the time-independent Schr¨ odinger for this problem are ψI (x) =Aeikx + Be−ikx

for

xb

(5.1.52b)

We can immediately say that D = 0 in (5.1.52b), since we don’t expect any reflection after the well. The first continuity condition is ψI (b) = ψII (b): Aeikb + Be−ikb = Ceikb −→ A + Be−2ikb = C −→ B = e2ikb (C − B)

(5.1.53)

0 The second continuity condition is ψI0 (b) = ψII (b), and in order to apply this we must go back to the time-independent Schr¨ odinger equation with the potential provided:   ~2 ~2 ∂ 2 − − δ(x − b) ψ(x) =Eψ(x) 2m ∂x2 2ma ∂2 1 2mE ψ(x) + δ(x − b)ψ(x) = − 2 ψ(x) ∂x2 a ~ Z b+ 2 Z ∂ 1 b+ ψ(x) dx + δ(x − b)ψ(x) dx =0 2 a b− b− ∂x ( )  ∂ ∂ 1 ψ(x) − ψ(x) = − ψ(b) lim →0 ∂x ∂x a b+ b−

∂ ∂ ψII (b) − ψI (b) = − ∂b ∂x

∂ ∂ Ceikx − Aeikx + Be−ikx = − ∂x ∂x ikCeikb − ikAeikb + ikBe−ikb

C − A + Be−2ikb

1 ψII (b) a C ikb e a C = − eikb a C =− ika

(5.1.54)

At this point we substitute (5.1.53) into (5.1.54),
C C − A + e2ikb (C − A) e−2ikb = − ika C C −A+C −A=− ika C 2C + =2A ika   C 1 2+ =2 A ika And the transmission coefficient is 2    C 2 2 2 2 4 1 = T = = = −→ T = 1 1 1 ~2 A 2 + ika 2 − ika 2 + ika 4 + k 21a2 1 + 8mEa 2

Problem 12

(5.1.55)

387

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Consider a two-state system with state |1i, |2i and energies E1 and E2 , respectively. The system is in the state |1i. At t = 0 it enters a region where it is perturbed by a constant potential with matrix elements h1|V |1i = h2|V |2i = 0, h1|V |2i = h2|V |1i = V . Find P2 (t), the probability that the system is in the state |2i as a function of time.

Solution This problem directly reflects the derivation of the time-dependent perturbation theory for a two-state system. This is derived in great detail in Griffiths in Chapter 9. The general idea is that you have two different states with two different energy levels, where each state is an eigenstate in the time-independent Schr¨ odinger equation: ψ1 (t) =c1 e−iE1 t/~ |1i,

H0 |1i = E1 |1i

ψ2 (t) =c1 e

H0 |2i = E2 |2i

−iE2 t/~

|2i,

Our total wave function can then be expressed as a linear combination of ψ1 (t) and ψ2 (t), and the coefficients are time-dependent : Ψ(t) = c1 (t)e−iE1 t/~ |1i + c2 (t)e−iE2 t/~ |2i

(5.1.56)

By substituting ψ1 (t) and ψ2 (t) into the time-dependent Schr¨ odinger equation, with the Hamiltonian H → H0 +H 0 (t), where H0 is the unperturbed Hamiltonian and H 0 (t) is the time-dependent perturbation. After simplification, we arrive at: 1 0 −iω0 t H e c2 (t) i~ 12 1 0 iω0 t c˙2 (t) = H21 e c1 (t) i~

c˙1 (t) =

(5.1.57a) (5.1.57b)

0 0 Where we have define H12 and H21 to be 0 H12 = hψ1 (t)|H 0 (t)|ψ2 (t)i,

0 H21 = hψ2 (t)|H 0 (t)|ψ1 (t)i

While ω0 is ω0 =

E2 − E1 ~

The hallmark result from first-order time-dependent perturbation theory is Z 1 t 0 0 iω0 t0 0 (1) c2 (t) = H (t )e dt i~ 0 21

(5.1.58)

By inserting the appropriate values into (5.1.58) and squaring the modulus of the result, we arrive at the probability that the system will transition from ψ1 (t) to ψ2 (t) as a function of time. The provided perturbation matrix is   0 V12 Vij = (5.1.59) V21 0 Where we note that V12 = V21 = V . Substituting the relevant matrix element from (5.1.59) into (5.1.58),

W. Erbsen

QUANTUM MECHANICS

(1) c2 (t)

Z t 0 1 eiω0 t dt0 = V i~ 0 t " iω0 t0 V e = i~ iω0 0

=−

=− =− =−


V eiω0 t − 1 ~ω0  iω0 t V iω0 t  iω0 t e 2 e 2 − e− 2 ~ω0   2iV iω0 t ω0 t 2 e sin ~ω0 2   i(E2 −E1 )t 2iV (E2 − E1 )t 2~ e sin (E2 − E1 ) 2~

(5.1.60)

Taking the modulus squared of (5.1.60), 2

P1→2 = |c1 (t)| −→ P1→2

  4V 2 2 (E2 − E1 )t = sin (E2 − E1 )2 2~

Problem 13 Consider a hydrogen atom in its ground state. An  electric field, E(t), is applied in the z-direction. 0 t0

(5.1.61)

a)

Apply first-order time-dependent perturbation theory to calculate an expression for the probability that the hydrogen atom is in an excited state.

b)

What is the probability if the excited state is (i) 2s and (ii) 2p? Explain briefly the justification for your answer.

c)

How do your answers to (b) behave in the limits τ = 0 and τ = ∞? Explain physically whether this behavior is sensible.

Solution Before we get down to business, let’s remember a few things. For instance, ψn`m (r, θ, φ) =Rn`(r)Y`m (θ, φ) s   `    3 2 (n − ` − 1)! − r 2r 2r 2`+1 na = Ln−`−1 Y`m (θ, φ) 3 e na na na 2n [(n + `)!] And the first three radial wave functions are:

(5.1.62)

389

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

  3 R10 = 2a− /2 exp − ar R20 = R21 =




3 √1 a− /2 1 − r exp 2a 2   3 √1 a− /2 r exp − r a 2a 24

 r − 2a

Y00 = Y10 = Y1±1 =


1 1 /2 4π 1
3 /2 cos θ 4π
1 3 /2 ∓ 8π sin θe±iφ

So, the first excited state has n = 2 with ` = 0, 1, and admits the following states: ψ200 (r, θ, φ) =R20 (r)Y00 (θ, φ)   h r i  1  3 1 r  √ = √ a− /2 1 − exp − 2a 2a 2 4π h r i r  1 − 3/2  exp − =√ a 1− 2a 2a 8π ψ210 (r, θ, φ) =R21 (r)Y10 (θ, φ) !  h r i r 3 1 − 3/2 r = √ a cos θ exp − a 2a 4π 24 h r i 3 r 1 = √ a− /2 exp − cos θ a 2a 4 2π

(5.1.63)

(5.1.64)

We also recall the time-dependent coefficients, 1 0 −iω0 t H e cb (t) i~ ab 1 0 iω0 t c˙b (t) = Hba e ca (t) i~

c˙ a (t) =

And in the spirit of first order time-dependent perturbation theory, Z 1 t 0 0 iω0 t0 0 (1) cb = H (t )e dt i~ 0 ba

(5.1.65)

The matrix element in (5.1.65) is given by 0 Hba = hψb |H 0 |ψa i t

The perturbation in our case is E(t) = E0 e− /τ , and for dipole radiation the perturbation becomes t

H 0 = −pE = −qzE0 e− /τ If we recall that in spherical coordinates z = r cos θ, then the full matrix element becomes   t 0 Hba (t) =hψb | −qE0 e− /τ r cos θ |ψai t

= − qE0 e− /τ hψb |r cos θ|ψa i

(5.1.66)

For the most general expression for the probability that hydrogen is in an excited state, we carry on calculating our matrix element from (5.1.66) using the full wave functions for hydrogen: t

0 Hba = − qE0 e− /τ hψn`m (r, θ, φ)|r cos θ|ψ100 (r, θ, φ)i 3

− /2 t a r = − qE0 e− /τ √ hRnl (r)Ylm (θ, φ)|r cos θ|e− /a i π

W. Erbsen

QUANTUM MECHANICS

3

− t/τ

= − qE0 e

a− /2 √ π − 3/2

t a = − qE0 e− /τ √

π

Z

2π

0

Z

Z

π

0

2π

0

Z

Z

∞

0

π

0

 r  (Rn`(r)Y`m (θ, φ)) (r cos θ) e− /a r 2 sin θ drdθdφ

Y`m (θ, φ) cos θ sin θ dθdφ

Z

∞

r

r 3 Rn`(r)e− /a dr

(5.1.67)

0

To find the time-dependent probability, we would take the matrix element (5.1.67) and substitute it into the integral (5.1.65) and take the modulus squared. Since We do not know what state we are calculating the probability of jumping to, things will get very messy very quickly. Although they do call me messy wessy, this does not have anything to do with messy equations. Yuck. a)

The dipole selection rules prevent the 1s → 2s transition (` ± 1), so let’s find the probability of exciting to the 2p state. The matrix element would be t

0 Hba (t) = − qE0 e− /τ hψ210 (r, θ, φ)|r cos(θ)|ψ100 (r, θ, φ)i   1 t 3 r r 3 r 1 √ a− /2 e− /2a cos θ r cos θ √ a− /2 e− /a = −qE0 e− /τ a π 4 2π r E D 3 r 1 1 1 = − qE0 e− /2 3 √ re− /2a cos(θ) r cos(θ) e− /a a 4π 2 a t r E qE0 e− /τ D − r/2a =− √ re cos(θ) r cos(θ) e− /a 4π 2 a4 Z 2π Z π Z ∞    r  r =α re− /2a cos(θ) (r cos(θ)) e− /a r 2 sin(θ) drdθdφ 0

=α

Z

0

0

2π

dφ

Z

0 π

2

cos (θ) sin(θ) dθ

0

Z

∞

r 4 e−

3r

/4a

dr

(5.1.68)

0

Where I have temporarily clumped all constants infront of the integrals as α. To solve the θ integral, we require the following trig identities: 1 1 (1 − cos(2θ)) , cos2 (θ) = (1 + cos(2θ)) 2 2 1 sin(α) cos(β) = [sin (αβ ) + sin (α − β)] 2 sin2 (θ) =

So the θ integral becomes Z π Z cos2 θ sin(θ) dθ = 0

π

1 (1 + cos(2θ)) sin(θ) dθ 2 Z Z 1 π 1 π = sin(θ) dθ + sin(θ) cos(2θ) dθ 2 0 2 Z π0 1 1 π = [− cos(θ)| 0 + [sin(3θ) + sin(−θ) dθ] 2 4 0 1 11 1 = [1 + 1] + [1 + 1] − [1 + 1] 2 43 4 1 1 =1 + − 6 2 2 = 3 0

For the r integral, we recall the following integral

(5.1.69)

391

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Z

∞

xn e−ax dx =

0

n! an+1

Then the r integral becomes Z

∞

r 4 e−

3r

/4a

dr =

0

4! ( 3/4a)5

=

45 a5 4! 35

Combining (5.1.69) and (5.1.70) into (5.1.68) (let’s throw α back in there too),   5 5  t qE0 e− /τ 2 4 a 4! 45 4! 0 − t/τ √ = −qE e a Hba(t) = − √ (2π) 0 3 35 4π 2 a4 36 2

(5.1.70)

(5.1.71)

Substituting (5.1.71) into (5.1.65), (1) cb

−qE0 a 45 4! √ = i~ 36 2 =

−qE0 a 45 4! √ i~ 36 2

−qE0 a 45 4! √ = i~ 36 2 −qE0 a 45 4! √ = i~ 36 2 =

−qE0 a 45 4! √ i~ 36 2

=

−qE0 a 45 4! √ i~ 36 2

=

−qE0 a 211 4! √ ~ 36 2

Z

t

e−

t0

/τ iω0 t0

e

dt0

0

Z

t

et (iω0 − /τ ) dt0 0

1

0

"

t 0 t0 eiω0 t e− /τ (iω0 − 1/τ )

0 # 1 e(iω0 − /τ )t 1 − (iω0 − 1/τ ) (iω0 − 1/τ ) h i 1 (iω0 − 1/τ )t e − 1 (iω0 − 1/τ ) 1 t i 1 e(iω0 − /τ ) 2 h (iω0 − 1/τ ) t t 2 − e−(iω0 − /τ ) 2 e 1 (iω0 − /τ ) iω0 t   t e 2 e 2τ (iω0 − 1/τ )t sin (iω0 − 1/τ ) 2

"

(5.1.72)

The probability of transition is then   2 −qE a 211 4! e iω20 t e 2τt (iω0 − 1/τ )t 0 √ P (1s → 2p) = sin ~ 36 2 (iω0 − 1/τ ) 2

Which finally leads to

P (1s → 2p) = b)

  t 1 q 2 E02 a2 221 (4!)2 e /2τ 2 (iω0 − /τ )t sin ~2 312 ω02 + ( 1/τ )2 2

If we let τ = 0, then (5.1.73) leads to P (1s → 2p) =

  q 2 E02 a2 221 (4!)2 1 2 iω0 t sin ~2 312 ω02 2

(5.1.73)

W. Erbsen

QUANTUM MECHANICS

Problem 14 Deuterons have spin-1 (bosons). a)

In general, what are the possible total spin S and total spin angular momentum J of two deuterons in an arbitrary (relative) orbital angular momentum state L? Assume the state vector is a product of a spin part and an orbital part. Determine just the allowed S, J, not the state vector.

b)

In particular, for L = 0 and L = 1, what S and J are allowed?

Solution a)

For bosons, ms = ±1, so S = {0, 1, 2}, and the possible arbitrary J values are just J = L + S = {|L − S|, ..., |L + S|}. The (arbitrary) total angular momentum quantum numbers are given by L L L L

b)

S 2 1 0

J {|L − 2|, ..., |L + 2|} {|L − 1|, ..., |L + 1|} {L}

Let’s make a table and find out! L 0 1 0 1 0 1

Problem 15

S 2 2 1 1 0 0

J 2 1, 2, 3 1 0, 1, 2 0 1

(Old Problem 16)

A particle of mass m is confined to a one-dimensional potential −γδ(x), γ > 0. Show that there is one bound state. Calculate its binding energy and its eigenfunction.

Solution We imagine our potential to consist of three regions: I is before the potential, II is at the potential, and III is after the potential. Plugging our given potential into the time-dependent Schr¨ odinger equation,   ~2 ∂ 2 − − γδ(x) ψ(x) = Eψ(x) (5.1.74) 2m ∂x2 We now solve (5.1.74) for each of the regions. For the first region (region I), ~2 ∂ 2 ∂2 ψ(x) = Eψ(x) −→ ψ(x) − k 2 ψ(x) = 0, − 2m ∂x2 ∂x2

p 2m|E| k= ~

(5.1.75)

393

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

The sign changed because we know that E is negative. The solution to (5.1.75) is rather obvious, ψI (x) = Aekx + Be−kx −→ ψI (x) = Aekx

(5.1.76)

Where we found that B = 0 due to boundary conditions. The solution for region III is similar, ψIII (x) = Cekx + De−kx −→ ψIII (x) = De−kx

(5.1.77)

And boundary conditions dictated that C = 0, as before. We now move on to the trickier region, that of II: ~2 ∂ 2 ψ(x) + V (x)ψ(x) = Eψ(x) 2m ∂x2 Z  Z  2 Z  2 ∂ ~ ψ(x) dx + − V (x)ψ(x) dx = E ψ(x) dx 2m − ∂x2 − −   ! Z  ~2 ∂ ∂ − lim − − δ(x)ψ(x) dx = 0 →0 2m ∂x  ∂x − −   ~2 ∂ ∂ − ψIII (x) − ψI (x) = γψ(0) 2m ∂x ∂x

∂ 2mγ ∂ De−kx − Aekx = − 2 D ∂x ∂x ~ 2mγ −kD − kA = − 2 D ~ −

At x = 0, ψI (0) = ψIII (0) −→ A = D, so: 2mγ mγ 2kA = 2 A −→ K = 2 = ~ ~

p 2m|E| ~

Rearranging and solving,

2m|E| =

m2 γ 2 mγ 2 mγ 2 −→ |E| = −→ E = − 2 2 2 ~ 2~ 2~

(5.1.78)

We can see that the only values in (5.1.78) are numbers; there are no indicies like n, that would infer a different eigenenergy. Therefore, (5.1.78) is the only bound state. For normalization, Z ∞ |ψIII (x)| dx =1 0 Z ∞ 2A2 e−2kx dx =1 0    √ mγ 1 −2kx 1 −2kx A2 2 A − e − − e =0 −→ = 1 −→ A = 2k 2k k ~ ∞ 0 So that we have

ψ(x) =

√

h mγ i mγ exp − 2 x ~ ~

W. Erbsen

QUANTUM MECHANICS

Problem 16 Consider the isotropic harmonic oscillator where the potential is given by V (r) = 1/2 mω2 r 2

(5.1.79)

a)

Show that the eigenfunctions can be expressed in the form Rn`(r)Y`m` .

b)

The problem is separable in Cartesian coordinates, show that the eigenenergies can be expressed as
En = n + 3/2 ~ω (5.1.80)

c)

Find the degeneracy D(n) of En for the first 4 values of energies. You obtain the degeneracy from the solutions in Cartesian coordinates. For each energy, identify the value(s) of orbital angular momentum (or momenta if more than one) for each eigenenergy.

Solution a)

I don’t wanna

b)

The full Hamiltonian for the 3-D isotropic harmonic oscillator is  2 
~2 2 1 ~2 ∂ ∂2 ∂2 1 H=− ∇ + mω2 r 2 = − + + + mω2 x2 + y2 + z 2 2 2 2 2m 2 2m ∂x ∂y ∂z 2 We now insert (5.1.81) into the time-independent Schr¨ odinger equation,    
~2 ∂2 ∂2 ∂2 mω2 2 2 2 − + + x + y + z ψ(x, y, z) = Eψ(x, y, z) + 2m ∂x2 ∂y2 ∂z 2 2

(5.1.81)

(5.1.82)

We will solve this partial differential equation by separation of variables, whereby we assume that a separable solution exists in the form ψ(x, y, z) = X(x)Y (y)Z(z). Substituting this into (5.1.82),  2 
∂ ∂2 ∂2 m2 ω2 2 2mE + + X(x)Y (y)Z(z) − x + y2 + z 2 X(x)Y (y)Z(z) = − 2 2 2 2 2 ∂x ∂y ∂z ~ ~
1 ∂2 1 ∂2 1 ∂2 m2 ω2 2 2mE X(x) + + − x + y2 + z 2 X(x)Y (y)Z(z) = − 2 X(x) ∂x2 Y (y) ∂y2 Z(z) ∂z 2 m2 ~   mω 2   1 ∂ 2  mω 2   1 ∂ 2  mω 2  1 ∂2 2mE X(x) − x2 + Y (y) − y2 + Z(z) − z2 = − 2 X(x) ∂x2 ~ Y (y) ∂x2 ~ Z(z) ∂x2 ~ ~ (5.1.83) If we let E be the separation constant, where E = Ex + Ey + Ez , we find that (5.1.83) can be fully separated into three unique equations:  mω 2 1 ∂2 2mEx X(x) − x2 + =0 (5.1.84a) 2 X(x) ∂x ~ ~2  mω 2 1 ∂2 2mEy Y (y) − y2 + =0 (5.1.84b) 2 Y (y) ∂y ~ ~2  mω 2 1 ∂2 2mEz Z(z) − z2 + =0 (5.1.84c) 2 Z(z) ∂z ~ ~2

395

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

It is clear that (5.1.84a)-(5.1.84c) represent the 1-D differential equations for the harmonic oscillator along the x, y, and z directions, respectively. The eigenvalues for the three 1-D solutions are well known,


Ex = ~ω nx + 1/2 , Ey = ~ω ny + 1/2 , Ez = ~ω nz + 1/2 So that the total energy becomes

c)



Enx ,ny ,nz = nx + ny + nz + 3/2 −→ En = ~ω n + 3/2

We can discover the degeneracies by using a table as follows nx 0 1 0 0 1 1 0 1 2 0 0 2 2 0 0 1 1 3 0 0

ny 0 0 1 0 1 0 1 1 0 2 0 1 0 2 1 0 2 0 3 0

nz 0 0 0 1 0 1 1 1 0 0 2 0 1 1 2 2 0 0 0 3

E(~ω) 3 /2 5 /2 5 /2 5 /2 7 /2 7 /2 7 /2 9 /2 7 /2 7 /2 7 /2 9 /2 9 /2 9 /2 9 /2 9 /2 9 /2 9 /2 9 /2 9 /2

This table includes all possible values for the lowest four energy levels: E0 = 3/2 ~ω, E1 = 5/2 ~ω, E2 = 7/2 ~ω, E3 = 9/2 ~ω. The degeneracies, as well as the possible values for the orbital angular momentum are: E E0 E1 E2 E3

Problem 17

d n 1 0 3 1 6 2 10 3

` 0 0 0, 1 0, 1, 2

W. Erbsen

QUANTUM MECHANICS

a)

Consider a helium ion. The two electrons are in the 2p and 3p states. The energy levels will have definite total angular momentum (J = L + S). What J-values can occur and how many degenerate energy levels for each J can there be?

b)

How do your answers change if the electrons are both in the 3p states?

Solution a)

Let’s call the electron in the 2p state “1”, and the electron in the 3p state “2.” Then `1 = 1, and `2 = 1. The total orbital angular momentum can then take the values L = {|`1 − `2 |, ..., |`1 + `2 |} = {0, 1, 2}

(5.1.85)

And for electrons, ms = ± 1/2 , so the possible values for the total spin angular momentum are S = {0, 1}

(5.1.86)

And we also know that the possible values for the total angular momentum is determined by J = L + S = {|L − S|, ..., |L + S|} And to find the possible values, we create a table: L 0 1 2 0 1 2

S 1 1 1 0 0 0

J 1 0, 1, 2 1, 2, 3 0 1 2

So, the possible values for J are J = {0, 1, 2, 3} , each with degeneracies 2J + 1: J 0 1 2 3 b)

D 1 3 5 7

If both electrons are in the 3p state, then L = {0, 1, 2} S = {0} And the possible total angular momentum quantum numbers can be found by L 0 1 2

S 0 0 0

J 0 1 2

So the possible values for J are J = {0, 1, 2} . The degeneracies are

(5.1.87)

397

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

J 0 1 2

D 1 3 5

Problem 18 A particle of spin zero is confined in a two-dimensional infinite square well of dimension 2a on each side. a)

Find the energies and the wave functions of the first two states.

b)

If there are two indistinguishable noninteracting spin 1/2 particles inside the box, write down the total wave function of the first two states. Identify the degree of degeneray for each state.

c)

If there are five indistinguishable spin 1/2 noninteracting particles in the box, what is the total energy of the ground state?

d)

If there are five indistinguishable spin 1 noninteracting particles in the box, what is the total energy of the ground state?

Solution a)

In general, the eigenstates and energy eigenvalues for this problem are given by  n πx   n πy  1 x y ψ(x, y) = sin sin a 2a 2a π 2 ~2 2 Enx ,ny = (n + nxy) = E0 (n2x + n2y ) 8ma2 x

(5.1.88a) (5.1.88b)

The the ground state, nx = ny = 1, (5.1.88a) and (5.1.88b) become ψ(x, y) =

 πx   πy  1 sin sin , a 2a 2a

E11 = 2E0

 πx   πy  1 sin sin , a a 2a

E12 = 5E0

For the first excited state, nx = 2, and ny = 1, so we have ψ(x, y) =

So, the first excited state is doubly degenerate. b)

We first recall that the total wave function of a fermion (space and spin part) must be completely anti-symmetric under the exchange of any two particles in the system. For the case at hand, we may have to “symmetrize,” or “anti-symmetrize” the wave function such that the final total wave function is anti-symmetric. In other words, if the spin part is symmetric, then the space part must be anti-symmetric, and also vice versa. We recall that, in general,

W. Erbsen

QUANTUM MECHANICS

1 ψ(x1 , x2 ) = √ [ψn1 (x1 )ψn2 (x2 ) ± ψn2 (x1 )ψn1 (x2 )] 2

(5.1.89)

And also ψ(x1 , x2 ) =

 n πx   n πx  1 1 1 2 2 sin sin a 2a 2a

For the ground state, n1 = n2 = 1 and S = 0 (singlet) so the space part is just  πx   πx  1 1 2 sin ψ(x1 , x2 ) = sin a 2a 2a

(5.1.90)

(5.1.91)

Since the space part is symmetric, the spin part must be anti-symmetric: 1 χ00 = √ [|↑↓i − |↓↑i] 2

(5.1.92)

The total wave function for the ground state for two indistinguishable spin- 1/2 fermions is just the product of (5.1.91) and (5.1.92)  πx   πx  1 2 1 sin [|↑↓i − |↓↑i] ψ(x1 , x2 ) = √ sin 2a 2a a 2 The ground state has singlet degeneracy. For the first excited state, where are two different options: electron 1 can be excited or electron 2 can be excited. Either of the first excited states may have by a symmetric (S = 0) or an anti-symmetric (S = 1) spatial wave function, like (5.1.89). In our case, letting n1 = 1 and n2 = 2,  πx   πx   πx i 1 h  πx1  2 1 2 ψ(x1 , x2) = √ sin sin + sin sin (5.1.93a) 2a a a 2a a 2 h        i 1 πx1 πx2 πx1 πx2 = √ sin sin − sin sin (5.1.93b) 2a a a 2a a 2 Where (5.1.93a) is symmetric and thus has an anti-symmetric spin part, while (5.1.93b) is antisymmetric , and must have a symmetric spin part. The former is familiar; the latter (S = 1) has three components, each corresponding to one of the available symmetric spin functions. So, this gives us four wave functions: 1 ψ(x1 , x2 ) = √ a 2 1 = √ a 2 1 = √ a 2 1 = √ a 2

h  πx   πx   πx   πx i 1 1 2 1 2 √ [|↑↓i − |↓↑i] sin sin + sin sin 2a a a 2a 2 h  πx   πx   πx   πx i 1 2 1 2 sin sin − sin sin |↑↑i 2a a a 2a h  πx   πx   πx   πx i 1 1 2 1 2 √ [|↑↓i + |↓↑i] sin sin − sin sin 2a a a 2a 2 h  πx   πx   πx   πx i 1 2 1 2 sin sin − sin sin |↓↓i 2a a a 2a

We arrive at four more wave functions if we choose n1 = 2 and n2 = 1. Therefore, the first excited state is eightfold degenerate, and E = 5E0

399

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

c)

If there are five indistinguishable spin 1/2 fermions in the box, the total energy of the ground state can be determined via a table: n1 1 1 2

n2 1 2 1

E 2E0 5E0 5E0

d 1 4 4

There can be two electrons per each set of quantum numbers, so we can fit five in the first two rows, so

E = 2 · 2E0 + 3 · 5E0 −→ E = 19E0 d)

For five indistinguishable spin-1 bosons, all the particles can be in the lowest level, so the ground state is just

E = 5 · 2E0 −→ E = 10E0

Problem 19 Ten electrons are confined in a box of dimensions (a, 2a, 2a) on each side. Calculate the total energy of the ground state of these ten electrons if we assume that the electrons do not interact with each other.

Solution The separated eigenfunctions and associated eigenenergies are r

 n πx  2 x , ψ(x) = sin a a r  n πy  1 y ψ(y) = sin , a 2a r  n πz  1 z ψ(z) = sin , a 2a The total energy is then En x n y n z =

n2xπ 2 ~2 2ma2 n2y π 2 ~2 Ey = 8ma2 n2 π 2 ~2 Ez = z 2 8ma Ex =


π 2 ~2 4n2x + n2y + n2z 8ma2

(5.1.94a) (5.1.94b) (5.1.94c)

(5.1.95)

In order to find out which combination of atomic numbers yield the ground state, we create a table:

W. Erbsen

QUANTUM MECHANICS

nx 1 1 1 2 2 2 1 1 1 3

ny 1 1 2 1 1 2 2 1 3 1

nz 1 2 1 1 2 1 2 3 1 1

4n2x 4 4 4 16 16 16 4 4 4 36

n2y 1 1 4 1 1 4 4 1 9 1

n2z 1 4 1 1 4 1 4 9 1 1

Σ # 6 1 9 2 9 3 18 21 21 12 4 14 5 15 38

With fermions, two can have each energy (± 1/2 ), so we picked the five lowest energies and now multiply them by two and sum to find the total ground state energy:

E=

100π 2 ~2 π 2 ~2 [2 (6 + 9 + 9 + 12 + 14)] −→ E = 2 8ma 8ma2

Problem 23 Suppose a particle moves in one dimension under the potential V (x). V (x) is even, that is V (x) = V (−x). Also, all the eigenstates of the Hamiltonian H are known to be non-degenerate. a)

Prove that every eigenstate of H has either even or odd parity.

b)

Consider a completely free particle in one dimension. Show that there are solutions to the eigenvalue equation which do not have well-defined parity. Explain why this is not a contradiction to the theorem of part (a).

Solution a)

To show that every eigenstate of H has either even or odd parity, we apply the parity operator to the time-independent Schr¨ odinger equation:  2   2   2  p p p P Hψ(x) =P + V (x) ψ(x) = + V (−x) ψ(−x) = + V (x) ψ(−x) (5.1.96) 2m 2m 2m If we were to apply the parity operator twice,  2   2  p p 2 P Hψ(x) = P + V (x) ψ(−x) = + V (x) ψ(x) 2m 2m

(5.1.97)

So, we can see that P 2 has an eigenvalue of 1: P 2 = 1 −→ P = ±1. Furthermore, P and H commute, so they must share the same eigenvector. Since the eigenstates of P can have either even or odd parity (p = ±1), and H and P commute (have same eigenvector), then every eigenstate of H must have either even or odd parity.

401

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

b)

A free particle is described by ψ+ (x) = Aeikx,

ψ− (x) = Ae−ikx

And if we apply the parity operator once, P ψ+ (x) =P Aeikx = Ae−ikx P ψ− (x) =P Ae−ikx = Aeikx Since ψ+ (x) and ψ− (x) have the same energy (they are degenerate), we cannot follow the same logic as in a).

Problem 24 Consider an isotropic 3D spherical square well. What are the conditions on the depth V0 and radius a such that a particle of mass m has exactly one bound state in this well? Show your reasoning and calculations.

Solution The radial equation that we wish to solve is in terms of a intermediary function U (r) = rR(r) One could just imagine it to be ψ and carry on, since it plays no important part here.   ~2 ∂ ~2 `(` + 1) − +V + U (r) = EU (r) (5.1.98) 2m ∂r 2 2m r 2 However, for the ground state, we recognize that ` = 0 and Veff → V , and (5.1.98) becomes   ~2 ∂ − + V U (r) = EU (r) 2m ∂r 2

(5.1.99)

For our isotropic spherical square well, we define two regions; I is inside the sphere, where V = −V0 , and II is outside, where V = 0. Our task is to first solve (5.1.99) for each of these two regions. Starting with region I,   ~2 ∂ − − V U (r) = EU (r) 0 2m ∂r 2 ∂2 2m(E + V0 ) u(r) + U (r) = 0 2 ∂r ~2 p ∂2 2m(E + V0 ) 2 U (r) + k U (r) = 0, k= (5.1.100) 2 ∂r ~ The solution of (5.1.100) is UI (r) = A sin (kr) + B cos (kr) And now for region II,

(5.1.101)

W. Erbsen

QUANTUM MECHANICS

−

~2 ∂ 2 U (r) = EU (r) 2m ∂r 2

∂2 U (r) + κ2 U (r) = 0, ∂r 2

κ=

√

2mE ~

(5.1.102)

The solution of (5.1.102) we wish to put in exponential form, UII (r) = Ceiκr + De−iκr

(5.1.103)

Due to boundary conditions our solutions for regions I and II become UI (r) =A sin (kr) UII (r) =De−iκr 0 Our two boundary conditions are UI (a) = UII (a), and UI0 (a) = UII (a). These are, respectively:

A sin (ka) = De−iκa

(5.1.104a) −iκa

KI A cos (ka) = −iκDe

(5.1.104b)

If we take (5.1.104a) and divide it by (5.1.104b), tan (ka) = −

k iκ −→ cot (ka) = − iκ k

(5.1.105)

√ Let’s define z = ka and z0 = a 2mV0 /~. Then z02 =

a2 ~2 2 2mV0 −→ V0 a2 = z 2 ~ 2m 0

and (5.1.105) becomes − cot(z) =

r

z0 2 −1 z

(5.1.106)

If we graph (5.1.106), it is very clear what the boundaries are that determine what the radius and depth of the well should be. z0 will intersect the z-axis between π/2 and 3π/2 if there is to be exactly one bound state (less than π/2 means there is no bound state). So, π 3π < z0 < 2 2  2 ~2  π 2 ~2 3π 2 < V0 a < 2m 2 2m 2 So, the conditions on the width a and depth v0 of the well such that there is exactly one bound state is: ~2 π 2 9~2 π 2 < V0 a2 < 8m 8m

403

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Problem 25 If the spin state of an electron is prepared such that it has a spin ~/2 along the x-direction, what is the probability of finding this spin to have eigenvalue ~/2 if the spin is measured along the y-direction? Show the steps in your     calculation. The Pauli spin matrices  are:  0 1 0 −i 1 0 σx = , σy = , σz = 1 0 i 0 0 −1

Solution In the spirit of showing all the steps in our calculations, our first step would be to find the eigenvalues and eigenvectors of Sx , Sy and Sz . We know that S = ~/2σ, where σ represents the Pauli spin matrices. Starting with Sx first, 2 −λ ~/2 = λ2 − ~ = 0 −→ λ = ± ~ ~/2 −λ 4 2

And now we find the eigenvectors for each of the eigenvalues:        ~ ~ 1 −~/2 ~/2 α 0 1 (x) √ For λ = + ~/2 : = − α + β = 0 −→ χ+ = ~/2 −~/2 β 0 2 2 2 1        ~ 1 ~ ~/2 ~/2 α 0 1 (x) For λ = − ~/2 : = α + β = 0 −→ χ− = √ ~/2 ~/2 β 0 2 2 2 −1 Doing the same for Sy ,

Now the eigenvectors are For λ = + ~/2 : For λ = − ~/2 :

−λ i~/2



~ ~2 −i~/2 2 = λ − = 0 −→ λ = ± −λ 4 2

    α 0 = β 0      ~/2 −i~/2 α 0 = i~/2 ~/2 β 0 −~/2 −i~/2 i~/2 −~/2

i~ α− 2 i~ α+ 2

  ~ 1 1 (y) β = 0 −→ χ+ = √ 2 2 i   ~ 1 1 (y) β = 0 −→ χ− = √ 2 2 −i

And now for Sz

Whose eigenvectors are

~/2 − λ 0

For λ = + ~/2 : For λ = − ~/2 :

2 0 = − ~ + λ2 = 0 −→ λ = ± ~ −~/2 − λ 4 2 



    α 0 = β 0     0 α 0 = 0 β 0

0 0 0 −~

~ 0

  1 (z) − ~β = 0 −→ χ+ = 0   0 (z) ~α = 0 −→ χ− = 1

If, for the sake of exercise, we have a particle that is prepared in the z-direction with eigenvalue ~/2 and we wish to know the probability of obtaining ±~/2 in the x-direction,

W. Erbsen

QUANTUM MECHANICS

   †
α 1 (x) 1 = χ+ χ(z) = √ 1 β 2    †
α 1 (x) (z) = χ− χ = √ 1 −1 β 2

α + β 2 α+β √ −→ P = √ 2 2 α − β 2 α−β √ −→ P = √ 2 2

And to do the same for the y-direction, 



(y) χ+ (y)

χ−

†

†

χ

(z)

1 =√ 1 2

1 χ(z) = √ 1 2

  α − iβ 2
α α − iβ −i = √ −→ P = √ β 2 2   α + β 2
α α + iβ i = √ −→ P = √ β 2 2

However, if the system is initially prepared in the x-direction with an eigenvalue of ~/2, as is asked in the problem, the calculations are no different: 

(y) χ+



χ−

(y)

 
1 1 1 1 = √ 1 −i = √ −→ P = 0 2 2 2

†

(x) χ+

†

1 (x) χ+ = √ 1 2

 
1 1 1 i = √ −→ P = 0 2 2

Problem 26 a)

A two-dimensional harmonic oscillator has the potential V (x, y) 21 mω2 (x2 + 4y2 ). Calculate the energies of the first three lowest states, and identify the degrees of degeneracy for each energy.

b)

If there is an additional small coupling term W (x, y) = ax2 y present, where a is a small constant, calculate the first-order correction to each of the three levels.

Write down the matrix elements that you need to calculate. If the matrix element is zero, you have to say so and explain why. If it is not zero, say so and represent each integral by a constant and proceed further.

Solution a)

We find a solution by employing separation of variables. The Hamiltonian for our system is H =−

~2 ∂ 2 ~2 ∂ 2 mω2 2 − + x + 2mω2 y2 2m ∂x2 2m ∂y2 2

Substituting (5.1.107) into the time-independent Schr¨ odinger equation,   ~2 ∂ 2 ~2 ∂ 2 mω2 2 2 2 − − + x + 2mω y ψ(x, y) = Eψ(x, y) 2m ∂x2 2m ∂y2 2

(5.1.107)

405

CHAPTER 5: DEPARTMENTAL EXAMINATIONS



∂2 ∂2 m2 ω2 x2 4m2 ω2 y2 + 2 − − 2 2 ∂x ∂y ~ ~2



ψ(x, y) = −

2mE ψ(x, y) ~2

(5.1.108)

At this point we introduce the separable solutions ψ(x, y) = X(x)Y (y). Substituting this into (5.1.108),  2  ∂ ∂2 m2 ω2 x2 4m2 ω2 y2 2mE + − − X(x)Y (y) = − 2 X(x)Y (y) 2 2 2 2 ∂x ∂y ~ ~ ~   2   2 ∂ m2 ω2 x2 ∂ 4m2 ω2 y2 2mE − X(x)Y (y) + − X(x)Y (y) = − 2 X(x)Y (y) ∂x2 ~2 ∂y2 ~2 ~     1 ∂2 m2 ω2 x2 1 ∂2 4m2 ω2 y2 2mE X(x) − + Y (y) − =− 2 (5.1.109) 2 2 2 2 X(x) ∂x ~ Y (y) ∂y ~ ~ In our case the separation constant is E = E1 + E2 , and the separated equations become   mω 2 2 2mEx ∂2 X(x) − x − X(x) = 0 −→ Ex = ~ω(nx + 1/2 ) ∂x2 ~ ~2 " # 2 ∂2 2mω 2mEy 2 Y (y) − x − X(x) = 0 −→ Ey = 2~ω(ny + 1/2 ) ∂y2 ~ ~2 And we know that E = Ex + Ey , so: En = ~ω(nx + 1/2 ) + ~ω(2ny + 1) −→ En = ~ω nx + 2ny + 3/2 To find the degree of degeneracy for the first three energies, let’s make a table: nx 0 1 0 1 2 0 2 2

ny 0 0 1 1 0 2 1 2

E(~ω) 3 /2 5 /2 7 /2 9 /2 7 /2 11 /2 11 /2 15 /2




order 1 2 3 4 3 5 5 6

To display the degeneracies for the energies, let’s make another table, E(~ω) 3 /2 5 /2 7 /2 b)

d 1 1 2

Now the game has changed; we add a perturbation to our oscillator of the form W (x, y) = ax2 y. Before we can calculate the first order energy shift, we must recall some stuff about harmonic oscillators: r r ~ ~ † x= (ax + ax ), y= (a† + ay ) 2mω 4mω y √ √ a† ψn (x, y) = n + 1 ψn+1 (x, y), aψn (x, y) = n ψn−1 (x, y) With this, our perturbation becomes:

W. Erbsen

QUANTUM MECHANICS

!2 r ! ~ ~ † † W =a (a + ax ) (a + ay ) 2mω x 4mω y r ~ ~ =a (a† + ax )2 (a†y + ay ) 2mω 4mω x r

(5.1.110)

In general, the first order shift in energy is given by En(1) =hnx |hny |W (x, y)|ny i|nxi r a~ ~ = hnx |hny |(a†x + ax )2 (a†y + ay )|ny i|nxi 2mω 4mω r  a~ ~  = hnx |hny |(a†x + ax )2 a†y |ny i|nxi + hnx | hny |(a†x + ax )2 ay |ny i|nxi 2mω 4mω r  a~ ~ p √ = ny + 1 hnx |hny |(a†x + ax )2 |ny + 1i|nxi + ny hnx |hny |(a†x + ax )2 |ny − 1i|nxi 2mω 4mω (5.1.111)

Problem 28  well. That is A particle is in the ground state of an infinite square 0 0≤x≤L V (x) = ∞ otherwise

(5.1.112)

At t = 0 the wall at x = L is suddenly moved to x = 2L - this happens very fast, approximately instantaneously. a)

Calculate the probability that long after t = 0 the system is in the ground state of the new potential.

b)

How fast must the change take place for this “instantaneous” assumption to be good?

Solution The eigenstates for the ground states before and arbitrary eigenstates after the transition are r  πx  2 ψgs(x) = sin (5.1.113a) L L r  nπx  1 ψn (x) = sin (5.1.113b) L 2L

a)

We wish to express our initial ground state wave function ψgs (x) in terms of a linear combination of ψn (x) states: r r  πx  X  nπx  2 1 X ψgs = sin = cn ψn (x) = cn sin L L L n 2L n For the probability that we will be in the ground state, we calculate cn for n = 1:

407

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

cn =

Z

L

ψgs(x)ψn (x) dx

0

√ Z L  πx   πx  2 sin sin dx L 0 L 2L √ Z L h  πx πx   πx πx i 2 1 cos − − cos + dx = L 0 2 L 2L L 2L √   Z   πx  2 1 L 3πx = cos − cos dx L 2 0 2L 2L √    L  πx  2L 2 2L 3πx = sin − sin 2L π 2L 3π 2L 0 √     π  2L 3π 2 2L = sin − sin 2L π 2 3π 2 √   2 2L 2L = + 2L π 3π √ 4 2 = 3π =

So, the probability that we will remain in the ground state is 2

|c1 | = b)

32 9π 2

To find out how fast this process must take place, we use the time-energy uncertainty relation: ∆t∆E ∝ ~. We find the change in energy first ∆E =

n2 π 2 ~2 3n2 π 2 ~2 n2 π 2 ~2 − = 2 2 8mL 2mL 8mL2

So, the necessary time is (approximately) ∆t =

8mL2 3n2 π 2 ~

Problem 29 The scattering amplitude for a proton on a target atom is f(θ). That is, if the target is at rT = 0 then far from the target the wavefunction is eikr ψ(r) = eik·r + f(θ) (5.1.114) r a)

Suppose that there are N (N  1) target atoms located at the point r1 = A, r2 = 2A, r3 = 3A and so on. What is ψ(r) for positions r far from the line? That is, find ψ(r) for |r|  N |A|. It will have the form eikr ik·r ψ(r) = e + F (θ) (5.1.115) r Find F (θ) and dσ/dΩ = |F (θ)|2 . Is dσ/dΩ ∼ N ?

W. Erbsen

b)

QUANTUM MECHANICS

Suppose now that the N (N  1) target atoms are located at random positions, but all inside a small finite ball of radius R. Calculate F (θ) and dσ/dΩ. Is dσ/dΩ ∼ N ?

Solution a)

For some point P very far from the target, then we can make the approximation that r1 ≈ r2 ≈ r3 ≈ r. We start with the template provided, (5.1.115), but add to it an additional phase term: ψ(r) =eikr + f(θ) =eikr +

N X eikr−inkA r n=0

N X eikr f(θ) e−inkA r n=0

The sum in (5.1.117) is finite; to evaluate it, we let N X

P

e−inkr =

P

(5.1.116) an , where a = e−ikr . Then

an = 1 + a + a2 + a3 + ... + aN−1

n=0

1+

N X

an+1 = 1 + a + a2 + a3 + ... + aN−1 + aN

n=0

1+

N X

an+1 =

n=0

1 − aN =

n=0

an + aN

n=0

1 − aN =

N X

N X

an =

N X

n=0 N X

n=0

an −

N X

an+1

n=0

an (1 − a)

1 − aN 1−a

Using this, (5.1.117) becomes ψ(r) = eikr +

  eikr 1 − e−iNkr f(θ) r 1 − e−ikr

From (5.1.117), it is clear that F (θ) is simply the term in the brackets,

F (θ) = And now to find the differential cross section, dσ 2 = |F (θ)| dΩ

1 − e−iNkr f(θ) 1 − e−ikr

(5.1.117)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

From which it is clear that

2 1 − e−iNkr = f(θ) −ikr 1−e −iNkr/2  iNkr/2  2 e e − e−iNkr/2 2 ) |f(θ)| = −ikr/2 e eikr/2 − e−ikr/2 2 −iNkr e sin (N kr/2) 2 |f(θ)| = −ikr e sin (kr/2) sin2 (N kr/2) dσ 2 = |f(θ)| dΩ sin2 (kr/2)

No, the differential cross section is not proportional to N . b)

For the case of gathered target atoms we start in much the same way, ψ(r) =eikr + f(θ) =eikr +

N X eikr−iXn k r n=0

N X eikr−iXn k f(θ) e−iXn k r n=0

(5.1.118)

Where Xn represents the displacement of the nth target particle. From (5.1.118) it is clear that

F (θ) = f(θ)

N X

e−iXn k

(5.1.119)

n=0

To find the differential cross section, dσ 2 = |F (θ)| dΩ 2 N X = f(θ) e−iXn k n=0

= |f(θ)|

2

= |f(θ)|

2

N X

iXn k

e

n=0 N X

!

N X

−iXm k

e

m=0

!

eik(Xn −Xm )

n=0 m=0



2 = |f(θ)| 



2 = |f(θ)| 

N X

eik(Xn −Xm ) +

n=m=1 N X

n=m=1

N X

n6=m=1

e0 +

N X

n6=m=1



eik(Xn −Xm )  

eik(Xn −Xm ) 

W. Erbsen

ELECTRODYNAMICS



2 = |f(θ)| N +

N X

n6=m=1



eik(Xn −Xm ) 

(5.1.120)

The remaining sum is often referred to as random walk, where the trajectories are completely random and as an ensemble, average to zero. Thus, our differential cross section is dσ 2 = N |f(θ)| dΩ There is no escaping the fact that the differential cross section is proportional to the number of atoms in this particular configuration.

5.2

Electrodynamics

Problem 1 A conducting sphere of radius a is placed in a uniform electrostatic field E0 . If the sphere is well insulated, find the expressions for potential and electric fields V and E at the point P (r, θ, φ) lying outside the sphere. Also obtain the expression for the induced surface charge density σ and show that it forms a dipole distribution. Specifiy your boundary conditions clearly.

Solution The induced surface charge density is σ(θ) = ε0 E; the plan of action then is to calculate E, which can be readily done by intermediately solving for the potential, V :  ∞  X B` V (r, θ) = A` r ` + `+1 P` (cos θ) (5.2.1) r `=0

Where (5.2.1) stems from (3.63) in Griffiths. The problem, then, is to take (5.2.1) and adjust it so that it describes our problem. We do this by applying boundary conditions. It should be mentioned that as a conductor, the sphere has a constant potential, but we are only interested in the potential outside of the sphere, so we can choose it to be whatever we want. The most convenient choice would be to let Vsphere = 0. Additionally, we recognize that in the field (absent from the sphere) the potential is V (r, θ) = −E0 r cos θ. We can now form our boundary conditions:  0 at r = R V (r, θ) = −E0 r cos θ at r → ∞ Applying our first boundary condition to (5.2.1),  ∞  X B` B` V (R, θ) = A` R` + `+1 P` (cos θ) = 0 −→ A` R` + `+1 = 0 −→ B` = −A` R2`+1 R R `=0

(5.2.2)

411

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Substituting our new value for B` into (5.2.1), we have  ∞  X R2`+1 V (r, θ) = A` r ` − A` `+1 P` (cos θ) r

(5.2.3)

`=0

If we apply the second boundary condition to (5.2.3), we find that the second term in parenthesis approaches 0 as r → ∞, and we are left with V (r, θ) =

∞ X `=0


A` r ` P` (cos θ) = −E0 r cos θ

(5.2.4)

From (5.2.4) we can see that the second boundary condition is satisfied only if ` = 1 and A` = A1 = −E0 . With these modifications, (5.2.3) is now   R3 V (r, θ) = −E0 r + E0 2 cos θ (5.2.5) r From the potential, we can now find E, and in turn σ: ∂ V (r, θ) ∂r    ∂ R3 = − ε0 −E0 r − E0 2 cos θ ∂r r   R3 = − ε0 −E0 − 2E0 3 cos θ r r=R

σ(θ) = − ε0

From this it is easy to see that

σ(θ) = 3E0 ε0 cos θ

Problem 2 Two insulated square parallel conducting plates, of side L and separation d, are charged with surface densities +σ on the upper plate and −σ on the lower. Two dieletric slabs, each with thickness d/2 and area L × L, are inserted between the plates, one slab above the other. The dieletric constants are K1 and K2 . Assume that d  L. Determine: a)

D everywhere between the plates.

b)

E everywhere between the plates.

c)

The bound surface charge densities σb on the three dialetric surfaces.

d)

The capacitance.

Solution

W. Erbsen

ELECTRODYNAMICS

a)

We first recall that Gauss’ Law for for dielectrics can be written as I D · dA = Q Where D = ε0 KE is the electric displacement. In the current case, the area is just L · L = L2 , and the surface charge density defines the total charge as Q = σL2 , so that we have DL2 = σL2 −→ D = −σ ˆ z for

b)

− d/2 < z < d/2

We again use the fact that D = ε0 KE, such that Gauss’ Law becomes: σ ε0 K1,2 EL2 = −σL2 ˆ z −→ E = − ˆ z ε0 K1,2

(5.2.6)

(5.2.7)

And we now have

c)

 σ  z   −ε K ˆ 0 1 E= σ    (−ˆ z) ε0 K2

for 0 < z < d/2 for − d/2 < z < 0

The bound surface charge density is related to the polarization as σb = P · n ˆ, where we remember that P = ε0 χE. We also recall that the dielectric constant is defined as K = (1 + χ). The (general) polarization is then given by     σ 1 |P1,2 | = ε0 χ1,2 |E1,2 | = ε0 (K1,2 − 1) E = ε0 (K1,2 − 1) = 1− σ (5.2.8) ε0 K1,2 K1,2 The three surfaces the prompt is speaking about refers to the top surface, where slab 1 interfaces with the positive conducting plate (and analogously for slab 2 on the lower negative conducting plate), and the interface between the two dielectrics. The only tricky bit is in the interstitial region; but all we have to do is add the bound charge densities being contributed from both slabs. The result is

d)

   1   − 1 − σ   K1         1 1 σb = 1− σ− 1− σ  K1 K2         1   1− σ K2

for surface 1 for surface 2 for surface 3

In order to calculate the capacitance, we first recall the relation Q = C|∆V |. So, it seems like all we would have to do is calculate ∆V and we are home-free. We must integrate over both dielectrics, going from the negative terminal at the bottom of slab 2 to the positive terminal at the top of slab 1: Z 0 Z d/2 ∆V = − E2 dz − E1 dz − d/2



σ =− − ε0 K2

Z

0

0

− d/2

  Z d/2 σ dz − − dz ε0 K1 0

413

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

        σ d σ d 0− − + −0 ε0 K2 2 ε0 K1 2 dσ dσ + = 2ε0 K2 2ε0 K1   dσ 1 1 = + 2ε0 K1 K2 =



(5.2.9)

Using (5.2.9) we can now try to find C: C=

Q = ∆V

Which leads us to C= d



dσ 2ε0



σL2 1 K1

+

1 K2



2ε0 L2  1 1 + K1 K2

Problem 3 A hollow dieletric sphere is centered on the origin. It has dielectric constant K 6= 1. Its inner radius is a1 and its outer radius is a2 . It is uncharged. Now a point charge q > 0 is placed at the origin.

a)

Find the electric field strength E for r < a1 , a1 < r < a2 , and a2 < r.

b)

Find the surface charge density (charge/area) on the inside (at r = a1 ) and on the outside (at r = a2 ).

c)

Now the situation is changed. A thin conducting coating is applied to the outside of the sphere, and this surface is maintained at +V0 volts relative to infinity. Find E(r) for r > a2 .

Solution

W. Erbsen

ELECTRODYNAMICS

a)

b)

The electric field in all three regions can be found from Gauss’ Law as  1 q     4πε0 r 2    1 q E= 2  4πε K r 0 1     1 q   4πε0 r 2

for

0 < r < a1

for a1 ≤ r ≤ a2 for

r > a2

To find the bound surface charge density we must use the following relation σb = P · n ˆ, where P = ε0 χE. The general solution can be found by, 1 q 1 q σb = ε0 χ1,2 E1,2 · n ˆ = ε0 (K1,2 − 1) ˆ r·n ˆ = (K1,2 − 1) ˆ r·n ˆ (5.2.10) 2 4πε0 K1,2 r 4πK1,2 r 2 Our solution from (5.2.10) is applicable as long as we recognize that the induced charges on the inner surface are negative, while those on the outer surface are positive. Therefore, for the inner surface we have (−ˆ r) · n ˆ = −1, and for the outer ˆ r·n ˆ = 1. With this, our solutions become

c)

  1 q 1   − 1   K1 4π a21 σb =    1 1 q    1− K2 4π a22

at r = a1 at r = a2

Since we are only interested in the field outside the sphere, the particular configuration of all the interior charges is irrelevant; all that is important is the total charge. We know that there is one point charge at the center of the sphere, q, and that the conducting shell is held at V0 . What we don’t know is how much charge exists on the exterior of the conductor; we can find this out as follows: V (r = a2 ) = V0 =

1 (Q + q) −→ Q = 4πε0 a2 V0 − q 4πε0 a2

(5.2.11)

And now using (5.2.11) we can find the potential at a greater distance r > a2 : V (r > a2 ) =

1 (Q + q) 1 (4πε0 a2 V0 − q + q) V0 a2 = −→ V (r) = 4πε0 r 4πε0 r r

Recalling that E = −∇V , from (5.2.12) we have

1 ∂ V0 a2 1 E(r) = − ˆ r=− r ∂r r r

From which it is easy to see that the field is E(r) =

Problem 4

V0 a2 ˆ r r3

  V0 a2 − 2 ˆ r r

(5.2.12)

415

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

A linear molecule with a permanent electric dipole moment p0 and moment of inertia I (for example, HCL) is placed in a uniform electric field E. a)

Make a sketch showing the charges that make up the dipole, the dipole moment and the electric field when the dipole is in the equilibrium orientation.

b)

Derive an expression for the frequency ω0 of small amplitude oscillations about the equilibrium orientation.

c)

Describe the polarization and power of the radiation produced by this oscillating dipole, assuming it has been initially displaced by an angle θ  90 degrees from its equilibrium orientation. Make a sketch of the angular distribution of this power.

d)

Over what time scale will it continue radiating appreciably, and therefore what range of frequencies are emitted?

Solution a)

Picture picture picture

b)

To find an expression for ω0 , we begin with Newton’s Second Law applied to rotational kinematics: ¨ We note that the torque is also defined by τ = r × F, where r in our case is the τ = Iα = I θ. distance between the two charges in the dipole, let’s call this distance d. With this, we have q F τ = r × F = d × F = d × F = qd × =⇒ τ = P0 × E q q

(5.2.13)

Where in (5.2.13) I have used the fact that P0 = qd and also that F = qE. The magnitude of τ is of course |τ | = P0 E sin θ. Using this with Newton’s Second Law, we have I θ¨ = P0 E sin θ −→ θ¨ =

P0 Eθ P0 E sin θ ≈ =⇒ θ¨ = −ω2 θ I I

(5.2.14)

Where I have used the small angle approximation sin θ ≈ θ. We know that (5.2.14) is the differential equation for the harmonic oscillator, where we typically define ω as:

ω0 = c)

r

P0 E I

(5.2.15)

µ0 P02 ω4 12πc

(5.2.16)

Equation (11.22) from Griffiths is: hP i =

Which represents the total time averaged power radiated from an oscillating dipole. If we imagine our molecule rotating periodically (pendulum-like), it has a combined energy equal to E=

1 2 1 2 mv + Iω 2 2

(5.2.17)

And also sources of potential energy, which we neglect. When the molecule is perfectly aligned with the field, that is, θ = 0, then (5.2.17) reduces to E = 1/2 Iω2 . We also note that power is defined as the amount of work per unit time, and therefore from (5.2.16) we can say

W. Erbsen

ELECTRODYNAMICS

hP i =

dE µ0 P02 4 ENERGY = ω = Cω4 = 12πc TIME dt | {z } C

Where the time derivative of the rotational kinetic energy is simply Iωω. ˙ Recalling that C 0 = C/I, we have Iωω˙ = Cω4 −→

dω 1 = C 0 ω3 −→ 3 dω = C 0 dt dt ω

Whose solution is t=

1 2Cω2

Problem 5 A current I flows into a parallel plate capacitor with circular plates of radius R separated by a distance d. The current was 0 before t = 0 and I 6= 0 after. a)

What is the charge on the plates as a function of time?

b)

What is the electric field between the plates?

c)

What is the displacement current density between the plates?

d)

What is the magnetic field between the plates at r = R/2 from the center of the plates?

e)

What is the Poynting vector between the plates at R/2?

Solution a)

We note that the fundamental definition for current is I=

dq(t) dt

We can solve this differential equation, noting that I(0) = 0: Z I dt = q(t) −→ It + A = q(t) −→ q(t) = It b)

(5.2.18)

To find the electric field between the plates, we note that our plates have a (time dependent) surface charge density σ(t). Applying Gauss’ Law, I qenc E · dA = (−ˆ z) ε0 A σ(t)A EA = ε0

417

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

E=

1 q(t) (−ˆ z) ε0 πR2

(5.2.19)

From (5.2.19) and using (5.2.18), the electric field is given by E= c)

It (−ˆ z) ε0 πR2

(5.2.20)

We know that the displacement current density is denoted by Jd and is given by Jd = ε0

∂E ∂P + ∂t ∂t

(5.2.21)

The displacement current density is obtained simply by taking the time derivative of the electric displacement field D = ε0 E+P. In our case, we have no dielectric media so the last term in (5.2.21) is neglected. Using (5.2.20), Jd becomes   It ∂ Jd = (−ˆ z ) ∂t ε0 πR2 Which is simple enough to evaluate. The result is: Jd = d)

I (−ˆ z) ε0 πR2

(5.2.22)

The magnetic field at a position r = R/2 between the plates can be computed using Amp´ere’s Law: I B · d` =µ0 Ienc In we are allowed to switch Ienc with Id since the magnetic field within the plates is the same as between the plates. We also note that the displacement current density Jd is related to the displacement current Id by Jd = Id /πR2 (in the following case R is actually r). So, with (5.2.22), (d) becomes: B2πr =µ0 Jd πr 2   I B2πr =µ0 (−ˆ z) πr 2 πR2 µ0 Ir B =− ˆ z 2πR2

(5.2.23)

If we recall that r = R/2, it is easy to see that (5.2.23) can be written as: B=− e)

µ0 I ˆ φ 4πR

(5.2.24)

The fundamental formulation for the Poyinting Vector is S=

1 E×B µ0

(5.2.25)

Inserting our previously tabulated values for E from (5.2.20) and B from (5.2.24), (5.2.25) becomes:     1 It µ0 I ˆ S= − ˆ z × − φ µ0 ε0 πR2 4πR

W. Erbsen

ELECTRODYNAMICS

Which can be rewritten as S=

I2t ρ ˆ 4ε0 π 2 R3

Problem 6 A sphere of radius R carries charge Q distributed uniformly over its surface. The sphere is rotated at a constant angular velocity ω around the z-axis. Calculate the magnetic dipole moment. Find the magnetic field at the point (x, 0, 0) where x  R.

Solution The integral form of the magnetic dipole moment is Z Z µ = I dA = A dI

(5.2.26)

We also recall that the definition of current is I = dQ/dt, but if we want the differential element of the amount of current over some specified length of time (the period, T ), this becomes dI = dQ/T . The differential charge element in our case since we have a surface charge density is dQ = σ dA. So, theoretically we just insert our values into (5.2.26) and integrate. we begin by looking for the charge element dQ for a thin strip of the sphere, rotating around at some radius r = a sin θ with some angular velocity ω. This rotating charge, of course, produces current, and in turn a magnetic dipole moment. The thickness of the stripe is a dθ, and so we have:   Q 1 (5.2.27) dQ = σ dA = (a dθ) (2πa sin θ) = Q sin θ dθ 4πa2 2 Using (5.2.27) and remembering that the temporal period at some rotational frequency ω is T = 2π/ω, our differential current element becomes dI =

dQ ω 1 Qω = Q sin θ dθ = sin θ dθ 2π/ω 2π 2 4π

(5.2.28)

2

The loop area is A = πr 2 = π (a sin θ) , and the magnetic dipole moment becomes  Z Z π Z   Qω 4ωQa2 π 3 2 µ = AdI = π (a sin θ) ˆ z sin θ dθ sin θ = sin θ dθ ˆ z 4π 4 0 0 Straightforward integration yields Z
4ωQa2 π 1 − cos2 θ sin θ dθ ˆ z 4 0 Z π  Z π 4ωQa2 = sin θ dθ − cos2 θ sin θ dθ ˆ z 4 0 0

µ=

(5.2.29)

419

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

4ωQa2 4 4ωQa2 = 4 4ωQa2 = 4 =



 1  3 π cos θ 0 ˆ z 3   1 − (−1 − 1) + (−1 − 1) ˆ z 3   4 ˆ z 3 π

[− cos θ| 0 +

From which we conclude that µ=

ωQa2 ˆ z 3

(5.2.30)

To calculate the field B, we recount the following equation, the bastard love-child of the Biot-Sevart law: Bdip (r) =

µ0 1 [3 (µ · ˆ r) ˆ r − µ] 4π r 3

(5.2.31)

In our case we use ˆ r=x ˆ, and with (5.2.30), (5.2.31) becomes B(x) =

µ0 1 ωQa2 [3 (ˆ z·x ˆ) x ˆ−ˆ z] 4π x3 3

Where we note that ˆ z·x ˆ = 0, so we finally have B(x) = −

µ0 ωqa2 ˆ z 12πx3

Problem 8 Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a dielectric oil (dielectric constant ε, density ρ). The inner one is maintained at potential V and the outer one is grounded. To what height does the oil rise in the space between the tubes? (Hints: Find the electric field, then the electrical energy and then the force).

Solution To find the maximum height that the oil reaches, we must think about what forces are acting on the fluid. The force that is lifting the oil upwards is due to the capacitance between the two cylinders. The motion of the fluid stops when this force is exactly cancelled by the opposing force due to gravity. We can, then start from the end and work towards the beginning. Let’s say the fluid is currently at its maximum height, h. Then the mass density can be used as follows: ρ=


MASS m = −→ m = ρπ b2 − a2 h 2 2 VOLUME (πb − πa ) h

The force acting downward on the center of mass is then

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ELECTRODYNAMICS


ρgπ b2 − a2 h ˆ z Fg = 2

(5.2.32)

Where the factor of 2 in the denominator stems from the fact that we are interested in the average height. We also know that (5.2.32) must be equal to the force pulling up on the fluid. To find out what this opposing (lifting) force is, recall that 1 dC F = − V2 ˆ z 2 dz At this point we take a break to calculate what the capacitance of our system is. In general, I Q 1 Q Qenc −→ E2πrL = −→ E = ˆ r E · dA = ε0 ε0 2πε0 rL A And from (5.2.34) we now calculate the potential: Z a Z
1 Q a1 1 Q V =− E ·ˆ rdr = − ˆ r·ˆ rdr = log b/a 2πε0 L b r 2πε0 L b

(5.2.33)

(5.2.34)

(5.2.35)

Again using the handy relation Q = CV , (5.2.35) becomes C=

Q 2πε0 L = V log ( b/a )

(5.2.36)

What we have just found in (5.2.36) is the general relation for the capacitance of a cylindrical capacitor. In our case, part of the capacitor is filled with the dielectric, and the other with oil, so the total capacitance would just be the sum of the two: 2πε0 (L − z) log ( b/a) 2πε0 z (1 + χ) 2πε0 zK1 = = b log ( /a ) log ( b/a )

Cair = Coil So that the total capacitance is C = Cair + Coil =

2πε0 (L − z) 2πε0 z (1 + χ) 2πε0 + = (L + zχ) b b log ( /a ) log ( /a ) log ( b/a )

At this point we substitute (5.2.37) into (5.2.33):  1 2 d 2πε0 V 2 χπε0 F=− V − (L + zχ) ˆ z = ˆ z 2 dz log ( b/a ) log ( b/a ) z=h And now we set (5.2.32) equal to (5.2.38):


ρgπ b2 − a2 h V 2 χπε0 ˆ z= ˆ z 2 log ( b/a ) If we solve for the height h we finally arrive at h=

2V 2 (r − 1) ε0 ρg (b2 − a2 ) log ( b/a)

(5.2.37)

(5.2.38)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

The factor of 1/2 comes from the fact that h represents the average height of the oil, which is half-way up the vertical height.

Problem 9 A cylindrical capacitor consists of two long, concentric tubes of sheet metal of radii R1 and R2 , respectively. The space between the tubes is filled with a dielectric constant ε. a)

Find the capacitance of this capacitor.

b)

Suppose the potential difference between the shells is V0 , find the electrostatic energy.

Solution a)

The typical equation relating to capacitance is Q = CV → C = Q/V . We can firmly say that if the total height of the capacitor is L, and if the plates have a line charge density λ, then the total charge is Q = λL. We now calculate the potential, Z R1 V =− E · dr (5.2.39) R2

We now find E as follows E=

1 Q λL 1 = 4πε0 ε r 4πε0 ε r

(5.2.40)

Using (5.2.40), we can now calculate V from (5.2.39):     Z R1 1 λL λL R2 λL R1 λL dr = − (log [R2 ] − log [R1 ]) = − log = log V =− 4πε0 ε R2 r 4πε0 ε 4πε0 ε R1 4πε0 ε R2 (5.2.41) From (5.2.41) we can finally say that C= b)

λL 2π0 L h i = V 2 log R R1

(5.2.42)

We recall that the equation for the electrostatic energy is U = 1/2 CV 2 . From (5.2.41) and (5.2.42),     2 1  4πε0 ε  λL R2 h i U= log 2 log R2 4πε0 ε R1 R1

 2 2πε0 ε λ2 L2 R2 h i = log 2 2 2 16π ε0 ε R1 2 log R R1

From which it is easy to arrive at

W. Erbsen

ELECTRODYNAMICS

U=

  λ2 L2 R2 log 4πε0 ε R1

Problem 11 A long wire is bent into a hairpin-like shape shown in the figure. Find an exact expression for the magnetic field at the point P which lies at the center of the half circle.

Solution It is easy to show from Amp´ere’s Law that the magnetic fields produced by an infinite straight wire and an arc are, respectively: B=

µ0 I , 2πr

and

B=

µ0 Iφ 4πr

The two lengths of wire contribute each half of the magnetic field produced by an infinite straight wire, while the arc extends through an angle φ = π, so: µ0 I µ0 I µ0 I B= + −→ B = 2πR 4R 4R



2 +1 π



Problem 12 A copper wire with a diameter of 2 mm has a resistance of 1.6Ω for every 300 meters of length. It is carrying a current of 25 amps. Assuming that this DC current density is uniformly distributed over a cross section, a)

What is the numerical value of the resistivity ρ of the copper from the information given?

b)

Determine the electric field vector E, the magnetic induction vector B, and the Poynting vector S at the surface of the wire (be sure to state their directions and actual numerical values). µ0 /4π = 10−7 , 1/4πε0 = 9 × 109 in MKS units.

c)

How much energy (in Joules/meter) is stored in the electric field and in the magnetic field within the wire?

d)

Suppose the current is increasing at a rate of 1 Amp/s. What statement can be made about the radial dependence of the induced electric field that results within the wire?

Solution

423

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

a)

The fundamental relation relating resistance and resistivity is: R=

RA ρL −→ ρ = A L

(5.2.43)


2 In our case, A = πr 2 = π 10−3 m , so that the resistivity is given by

(1.6Ω)(π(10−3m)2 ) −→ ρ = 1.67 × 10−8 Ω · m 300m To find the electric field vector E, we must employ Gauss’ Law: I 1 λ 1 λ 1 λ −→ E · 2πr = −→ E = E · d` = ε0 L ε0 L 2πε0 r ρ=

b)

(5.2.44)

(5.2.45)

We notice that the equaled part in (5.2.45) looks suspiciously similar to the definition of the scalar potential V . And we can go ahead and find the scalar potential using the given information: V = IR = (25A)(1.6Ω) = 40v. Using this information, and manipulating (5.2.45), we have   1 λL 2 2V = (5.2.46) E= 4πε0 r L L Substituting the given values, (5.2.46) becomes: E=

2 v ˆ z 15 m

(5.2.47)

The magnetic field may similarly be found using Amp´ere’s Law: I µ0 I 2 × 10−7 · 25A B · d` =µ0 Ienc −→ B = = 2πr 10−3 m Evaluating this, we arrive at ˆ B = 5 × 10−3 T φ

(5.2.48)

Using (5.2.47) and (5.2.48), we can calculate the Poynting Vector S: S= c)

1 1 E×B = µ0 µ0



   4 v ˆ −→ S = −1061 W ρˆ ˆ z × 3 × 10−3 T φ 15 m m2

(5.2.49)

Finding the total energy stored in the both the electric and magnetic fields entails recalling the following relations: ηE =

energy 1 = ε0 |E|2 , volume 2

and

ηB =

energy 1 1 = |B|2 volume 2 µ0

Let’s say that we are interested in the amount of energy in a 1m length of wire. Then the volume is V = πr 2 = π(10−3 m)2 (1 m) = 3.14 × 10−6 m3 . Accordingly, the amount of energy stored in the electric and magnetic fields is given by 2 
A · s 2 v J 8.887 × 10 3.14 × 10−6 m3 −→ Energy in E = 2.473 × 10−19 V · m 15 m m   2
1 1 A·m ˆ 3.14 × 10−6 m3 −→ Energy in B = 1.562 × 10−5 J 5 × 10−3 T φ −7 2 4π10 V·s m 1 2



−12

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ELECTRODYNAMICS

d)

Qualitatively, when an time-varying current is present in a conductor (eg, AC current), the current has a greater tendency to exist near the surface of the conductor (or on the “skin”). This is called the skin effect.

Problem 13 A battery with emf V , a resistor with resistance R, and a capacitor with initial capacitance C have been connected in series as shown for a very long time. The dielectric, with permittivity ε, occupies 1/2 the gap in the parallel plate capacitor. Now suppose that the dielectric is removed very quickly, in a time short compared to any relevant time constants. The new capacitance is C 0 . a)

Give an explanation of what happens to each circuit element. After a very long time, a new equilibrium is reached. In terms of C, C 0, R and V , give expressions for: i)

The net change in energy in the capacitor

ii)

b)

The net change in energy in the battery

iii)

The total energy dissipated in the resistor

iv)

The amount of work that was done to remove the dielectric

In terms of ε, what is the ratio C/C 0?

Solution a)

We first recall that the capacitance for a parallel plate capacitor is given by C = εA/d, where A is the surface area of the plates and d is the distance between them, while ε corresponds to whatever material lies between. To find the capacitance of the capacitor before the dielectric is removed, we recognize that since the dielectric does not occupy the entire interstitial space between the plates we essentially have two capacitors in series, Cair and Cdia . Mathematically, these are (respectively) Cair =

ε0 A 2ε0 A = , d/ d 2

The total capacitance is then just

and

Cdia =

εA 2εA = d/ d 2

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

C=

1 1 CairCdia (2ε0 A/d)(2εA/d) 2εε0 A + = = = Cair Cdia Cair + Cdia 2ε0 A/d + 2εA/d d(ε + ε0 )

(5.2.50)

While after the dielectric is removed, the capacitance is simply C0 = i)

ε0 A d

The amount of energy in the capacitor decreases. Adding dielectrics between the plates of a capacitor act to decrease the magnitude of the electric field, but increase the capacitance (and therefore the energy stored) through the polarization of the static dielectric atoms. To calculate this change, we just have to remember that the amount of energy in a capacitor is just U = 1/2 CV 2 . So, ∆Uc = 1/2 C 0 V 2 − 1/2 CV 2 −→ ∆Uc = 1/2 V 2 (C 0 − C)

ii)

(5.2.52)

When the dielectric is removed, there is a drop in capacitance and charge flows from the plates of the capacitor towards the battery. The change in energy of the circuit occurs due to thermal losses in the resistor; the potential energy of a charge in a potential is U = QV , so the change in energy is then ∆Ub = Q0 V − QV = C 0 V 2 − CV 2 −→ ∆Ub = (C 0 − C) V 2

iii)

(5.2.51)

(5.2.53)

As mentioned previously, the total amount of energy dissipated in the circuit is done through the resistor. The basic equation to calculate this is P = I 2 R, however all we have to do is take the difference of (5.2.52) and (5.2.53): ∆Ur = ∆Uc − ∆Ub = 1/2 V 2 (C 0 − C) − (C 0 − C) V 2 −→ ∆Ur = 1/2 V 2 (−C 0 + C)

(5.2.54)

The ratio of the capacitances before and after the dielectric is removed is C C 2εε0 A d 2ε = · −→ = C0 d(ε + ε0 ) ε0 A C0 ε + ε0

Problem 14 A series LCR circuit with L= 2 H, C=2 µF, and R=20 Ω is driven by a generator with maximum emf of 100 V and variable frequency. Find the resonant frequency ω0 and the phase φ and maximum current Imax when the generator angular frequency is ω = 400 s−1 .

Solution √ We first recall that for a resonant RLC circuit, we have Z = R, ω = 1/ LC , XC = XL , and φ = 0. This is not the case, however; the first thing we want to do is find the resonant frequency ω0 of the circuit: 1 1 rad ω0 = √ = p −→ ω0 = 500 −6 sec LC (2 H)(2 × 10 F)

(5.2.55)

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ELECTRODYNAMICS

In order to find the phase angle, we must first recall that XC = 1/ωC, XL = ωL, and tan (φ) = (XL − XC ) /R. Using these, we have −1

φ = tan



ωL − R

1 ωC



−1

= tan

"

(400 s−1 )(2 H) −

1 (400 s−1 )(2×10−6 F)

20 Ω

The maximum current occurs at resonance, where R = Z = with this condition, and also the result from (5.2.55), Imax =

V V = q Z R2 + ωL −

−→ Imax = 0.222 A

1 ωC


2 = r

q

#

−→ φ = −87.46o 2

R2 + (XL − XC ) . Using Ohm’s Law 100 V



(20 Ω)2 + (400 s−1 ) (2 H) −

1 (400s−1 )(2×10−6

F)

2

Problem 15 A grounded conducting plane is connected to a conducting sphere with radius a through a battery of voltage V0 . The sphere is a distance d  a above the plane. a)

Find the charge on the sphere.

b)

Find the force between the plane and the sphere.

Solution a)

Since we are told that d  a, it is fair to go ahead and assume that the sphere can be thought of as isolated; and since it is connected to a battery of potential V0 , we could go ahead and assume that this is the potential of the sphere. The standard capacitance of a conducting sphere is C = 4πε0 r

(5.2.56)

427

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Where in our case r = a. Furthermore, we know that Q = CV , so the total charge on our sphere is just Q = 4πε0 aV0 b)

(5.2.57)

The force between the sphere and the grounded conducting plane can be calculated via the method of images, where we imagine an identical, albeit oppositely charged sphere a distance 2d away. Using (5.2.57) and Coulomb’s law, we have 2

F=

1 (4πε0 aV0 ) 1 Q2 = 2 4πε0 (2d) 4πε0 4d2

Which can be readily simplified to F =

πε0 a2 V02 d2

Problem 17 A conducting sphere of radius a and total charge Q is surrounded by a spherical shell of dielectric material (with permittivity ε) of inner radius a and outer radius b. Find the electrostatic energy of the system.

Solution The general form for the energy stored in any electrostatic system is Z ε W = E2 dτ 2

(5.2.58)

In our case, the electric field of the conducting sphere is given by E=

1 Q ˆ r 4πε r 2

Inserting this value for E into (5.2.58), we have 2 ZZZ  Z 2π Z π Z ∞ 1 Q  1 1 ε 2 W = ˆ r dV = Q · r 2 sin θ drdθdϕ 2 2 2 2 4πε r 2 16π ε r4 0 0 0

(5.2.59)

(5.2.60)

The radial integral in (5.2.60) must be divided according to each section: from 0 → a we are inside the conductor, so E = 0. However from a → b and b → ∞ we must evaluate the integral, and (5.2.60) becomes: " Z # Z 2π Z π Z ∞ Q2 ε20 b 1 1 W = dϕ sin θ dθ 2 dr + dr 32π 2 ε20 0 ε a r2 r2 0 b

W. Erbsen

ELECTRODYNAMICS

"  b  ∞ # Q2 ε20 1 1 = (2π)(2) 2 − + − 32π 2 ε20 ε r a r b  2   2 Q ε0 1 1 1 = − + + 2 2 8πε0 ε b a b From which is may be readily deduced that Q2 W = 8πε2



1 1 − a b



+

Q2 1 8πε20 b

Problem 18 The small loop of wire (radius a and resistance R) falls under gravity towards the larger loop (radius A), which has a constant current I. The small loop is contrained to move along the axis of the large loop and remains parallel to the large loop.

a)

Explain (in words) what happens to the small loop.

b)

For a particular height h and velocity v, what is the induced emf in the small loop? Assume a  A.

c)

What is the electromagnetic force acting on the small loop?

d)

Does the system radiate? If so, describe the radiation pattern qualitatively (e.g., frequency, direction, polarization).

Solution a)

Words words words

429

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

b)

The emf induced on the smaller loop will be due to the change of flux passing through it, changing as it falls: I dΦB d E=− =− B · dA (5.2.61) dt dt If we assume that the field is homogeneous throughout the smaller loop, we can expand on (5.2.61) as follows E=−

d B · πa2 dt

(5.2.62)

At this point all we have to do is find B. The best way to do this is with the Biot-Sevart Law, but before we do that we should define some needed quantities. d`, the length element of the source loop (the bigger one) is A dφ. Furthermore, the radial √ vector r is found by subtracting by boring trig stuff: r = z ˆ k − Aρ. ˆ The magnitude of r is just z 2 + A2 . Using all this in the Biot-Sevart Law: B= =

µ0 I d` × r 4π |r|3 ˆ × (z ˆ µ0 I A dφφ k − Aρ) ˆ 4π

(z 2 + A2 ) z ρˆ + Aˆ z

3/ 2

µ0 IA dφ 4π (z 2 + A2 ) 3/2  Z 2π  Z 2π µ0 IA 1 = z ρ ˆ dφ + A ˆ z dφ 4π (z 2 + A2 ) 3/2 0 0 µ0 I A = [0 + 2πA] ˆ z] 4π (z 2 + A2 ) 3/2 =

=

µ0 I A2 ˆ z 2 (z 2 + A2 ) 3/2

We can now use (5.2.63) in (5.2.62) to find E: " # µ0 I A2 2 d E = − πa dt 2 (z 2 + A2 ) 3/2 " # 1 πµ0 Ia2 A2 d dt dz =− 2 dt dz dt (z 2 + A2 ) 3/2 " # πµ0 Ia2 A2 d 1 =− z˙ 2 dz (z 2 + A2 ) 3/2 " 3z πµ0 Ia2 A2 d =− z˙ − 5 /2 2 2 2 dz (z + A ) z=h

(5.2.63)

(5.2.64)

By evaluating (5.2.64) at z = h, and recognizing that z˙ is the downward velocity, we have

E=

3hµ0 Iπa2 A2 ·v 5 2(h2 + A2 ) /2

(5.2.65)

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ELECTRODYNAMICS

c) d)

Not too sure on this one Yes , the system does radiate, in the ρ ˆ direction.

Problem 19 A cylindrically shaped conductor has length L, radius a, and conductivity σ. It carries a uniform time-independent current density J = Jˆ e0 , parallel to its long axis. a)

Determine the electric and magnetic fields within the conductor.

b)

Calculate the Poynting vector S within the conductor.

c)

Write out Poynting’s theorem for this situation. Verify that it is satisfied for any point within the conductor.

Solution a)

We first recall the fundamental relation for current density: J = σE. So, the electric field is given simply by

E=



J/σ ˆ z for 0≤ρ≤a 0 elsewhere

(5.2.66)

And to find the internal magnetic field B, we apply Amp´ere’s Law: I
B · d` =µ0 Iinc = µ0 πρ2 J 2πρB =µπρ2 J

Such that we have

B= b)



ˆ for µ0 Jρ/2 φ 0≤ρ≤a 0 elsewhere

(5.2.67)

We note that the Poynting vector takes the form S=

1 E×B µ0

And since we know that E = J/σ ρˆ from (5.2.66) and B = µ0 Jρ/2 ρ ˆ from (5.2.67), then the Poynting vector is simply

S=

J 2ρ (−ρ) ˆ 2σ

(5.2.68)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

c)

Poynting’s theorem states that ∂u +∇·S =−J·E ∂t Where u is the energy density. We can express the energy density as:   B2 1 2 ε0 E + u= 2 µ0 Substituting (5.2.66) and (5.2.67) into (5.2.70),   J2 µ20 J2 ρ2 1 ε0 2 + µ0 u= 2 σ 4   2 2 2 1 J µ0 J ρ = ε0 2 + 2 σ 4

(5.2.69)

(5.2.70)

(5.2.71)

And we note that u is time independent, so ∂u/∂t = 0. We also note that the divergence in cylindrical coordinates takes the form ∇·F =

1 ∂ 1 ∂Fϕ ∂Fz (ρFρ ) + + ρ ∂ρ ρ ∂ϕ ∂z

So finally we can calculate 1 ∂Sϕ ∂Sz 1 ∂ (ρSρ ) + + ρ ∂ρ ρ ∂ϕ ∂z 1 ∂ = (ρSρ ) ρ ∂ρ  2  J ρ 1 ∂ = ρ (−ρ) ˆ ρ ∂ρ 2σ 1 J2 ρ = (−ρ) ˆ ρ σ J2 =− ρ ˆ σ

∇·S=

(5.2.72)

So, using (5.2.72), Poynting’s theorem (5.2.69) is now −

J J2 J2 ρˆ = −J · E = −J ρˆ = − ρˆ X σ σ σ

Problem 21 An electron beam of uniform charge density ρ0 inside the beam radius r0 is traveling on the axis of a metal tube with an inner radius Ri and an outer radius of Ro which is kept at a potential φτ . a)

Calculate the potential at the center of the electron beam.

b)

Calculate the net charge on the tube per unit length `.

W. Erbsen

c)

ELECTRODYNAMICS

Calculate the energy that is required to double the tube potential to 2 × φτ assuming that the electron beam charge density remains constant.

Solution a)

Calculating the potential at the center of the beam poses no major obstacles. We imagine to be dragging a unit charge (“test” charge) from ∞ to the center of the beam . The total potential is how much work it took. So we start from the outside, from ∞, and boundary by boundary calculate the potentials and at the end add them together. Z 0 φ(r = 0) = − E · ds ∞ Ro

=−

Z

|∞

E4 · ds − {z } φτ

Z

Ri

Ro

E3 · ds −

Z

Ra

Ri

E2 · ds −

Z

0

E1 · ds

(5.2.73)

a

We are told that the potential of the shell is kept at φτ , so we needn’t even calculate E4 . The rest are ρ0 (πr 2 ) ρ0 ˆ r= rˆ r 2πε0 2ε0 ρ0 (πr02 ) ρ0 r02 E2 = ˆ r= ˆ r 2πε0 r 2ε0 r E3 =0

E1 =

(5.2.74a) (5.2.74b) (5.2.74c)

The reason why E3 = 0 is because this region corresponds to within a conductor, where the internal electric field is guaranteed to be zero. Using (5.2.74a), (5.2.74b) and (5.2.74c), we can evaluate our potential from (5.2.73): Z Ra Z 0 ρ0 r02 ρ0 φ(0) =φτ − ˆ r · dr − rˆ r · dr 2ε r 2ε 0 0 Ri a Z Z 0 ρ0 r02 Ro 1 ρ0 =φτ − dr − r dr 2ε0 Ri r 2ε0 a  0 ρ0 r02 ρ0 r 2 Ro =φτ − [log r|Ri − 2ε0 2ε0 2 a   2 ρ0 r0 Ri ρ0 a2 =φτ − log + (5.2.75) 2ε0 Ro 2ε0 2 From (5.2.75) it is easy to see that the potential is   ρ0 r02 Ri ρ0 a2 φ(0) = φτ + log + 2ε0 Ro 4ε0 b)

The net charge on the tube per unit length would just be the charge density times the area in question where we take the length to be equal to 1, λ = −ρ0 πa2

433

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Problem 22 A charge 2q is located at (0, 0, 0), a second charge, −q, is located at (0, −a, 0), and a third charge, −q, is located at (0, 0, −a). a)

Calculate the monopole and dipole moments.

b)

Calculate all quadrupole moments.

c)

What is the potential for r  a using multipole approximation?

Solution a)

The monopole moment is zero, since it is just the sum of the charges. The dipole moment for an array of charges is defined by p=

N X

ri q i

(5.2.76)

i=1

Applied to our geometric arrangement of charges, ˆ ˆ p = (−aˆj)(−q) + (0)(2q) + (−ak)(−q) −→ p = aq(ˆj + k) b)

(5.2.77)

The defining equation for the quadrupole moment is ∞


1X Q0 = qi 3xiα xiβ − ri2 δαβ 2 i=1

(5.2.78)

1 1 1 Q0 = − (a2 )(−q) + (0)(−2q) − (a2 )(−q) −→ Q0 = a2 q 2 2 2

(5.2.79)

Applying (5.2.78) to our case,

c)

The full equation for multipole expansion is Z ∞ 1 X 1 V (r) = (r 0 )n Pn (cos θ0 )ρ(r0 ) dτ 0 4πε0 n=0 r (n+1)  Z    Z Z 1 1 1 3 1 1 0 0 0 0 0 0 0 2 2 0 0 0 = ρ(r ) dτ + 2 r cos θ ρ(r ) dτ + 3 (r ) cos θ − ρ(r ) dτ + ... 4πε0 r r r 2 2 (5.2.80) The integrals represent the various -pole terms through quadrupole. We know that the monopole term is zero, so we simply substitute (5.2.77) and (5.2.79) into (5.2.80): 1 V (r) ≈ 4πε0

"

ˆ aq(ˆj + k) a2 q + 2 2 r r

#

W. Erbsen

ELECTRODYNAMICS

Problem 23 Consider a nonconducting sphere of radius R with a uniform charge density ρ, except for a cavity of radius a, a distance b > a from the center. a)

Find the electric field at the center of the cavity.

b)

Find the voltage potential (V (∞) = 0) at the center of the cavity.

Solution a)

We are given that the dielectric object has a charge density equal to ρ; and since the geometry of the problem is simple enough, from this we can readily calculate the total charge Q:
Q Q CHARGE = 4 = 4 −→ Q = 4/3 πρ R3 − a3 (5.2.81) ρ= VOLUME /3 πR3 − 4/3 πa3 /3 π (R3 − a3 ) And now we can apply Gauss’ Law to find the field E: E4πb2 =

Qenc 1 4 −→ E = ρ πb3 2 ε0 4πb 0 3

Which allows us to say E=

ρb ˆ r 3ε0

In order to find the voltage at the center of the cavity, we must imagine that we are dragging a unit charge to the center of the cavity from infinity. We do this in steps, addressing each boundary individually: Z a Z R Z a+b Z a V (r) = − E · ds = − E1 · ds − E2 · ds − E3 · ds (5.2.82) ∞

∞

R

a+b

The respective electric fields are E1 =

ρ(r 3 − a3 ) ρ(R3 − a3 ) ˆ r = ˆ r 3ε0 r 2 3ε0 r 2

(5.2.83a)

435

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

ρ(r 3 − a3 ) ˆ r 3ε0 r 2 ρb ˆ r E3 = 3ε0

E2 =

(5.2.83b) (5.2.83c)

We now substitute (5.2.83a)-(5.2.83c) into (5.2.82) and integrate. I will do each of the three integrals in turn and combine them later. The first is Z ρ(R3 − a3 ) R 1 dr V1 (r) = 2 3ε0 ∞ r R  ρ(R3 − a3 ) =− r 3ε0 ρ(R3 − a3 ) =− 3ε0 R

∞

(5.2.84)

And now for the second, Z

a+b

r 3 − a3 dr r2 R "Z # Z a+b a+b 1 ρ 3 = rdr − a dr 3ε0 R r2 R " a+b a+b #  ρ r 2 1 3 = −a − 3ε0 2 R r R   ρ (a + b)2 R2 1 1 − + − = 3ε0 2 2 a+b R

V2 (r) =

And that last little bitch is

ρ 3ε0

Z a ρb dr 3ε0 a+b ρb = [a − (a + b)] 3ε0 ρb2 =− 3ε0

(5.2.85)

V3 (r) =

We now take (5.2.84) - (5.2.86) and substitute it into (5.2.82):   ρ(R3 − a3 ) ρ (a + b)2 R2 1 1 ρb2 V (r) = + − + − + 3ε0 R 3ε0 2 2 a+b R 3ε0 Which finally leads us to   ρ (R3 − a3 ) (a + b)2 R2 1 1 2 V (r = 0) = + − + − +b 3ε0 R 2 2 a+b R

Problem 24

(5.2.86)

W. Erbsen

ELECTRODYNAMICS

A particle of charge +q and mass m has a velocity of V = (0, vy , vz ) when at a position of (x, 0, 0). There is a magnetic field of B = (0, 0, B). a)

Find an expression for the cyclotron period, i.e. the time to go once around a circular orbit.

b)

Find the position of the charge after one cyclotron period.

Solution a)

The cyclotron period can be found by: FC = Fm −→ mω2 r = qvB −→ ω =

qB m

So the period is τ= b)

2π 2πm −→ τ = ω qB

Most generally, we would approach the problem like this F = m¨ r = qv × B Which leads us to (¨ xx ˆ + y¨y ˆ + z¨ˆ z) =

qB [x(−ˆ ˙ y ) + yˆ ˙ x] = ω [yˆ ˙ x − xˆ ˙ y] m

This equation may be separated very easily; the resulting coupled differential equations are x ¨ =ωy˙ y¨ = − ωx˙ z¨ =0 We begin by integrating (5.2.87a) with respect to t:

(5.2.87a) (5.2.87b) (5.2.87c)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

x˙ = ωy + C(vx )

(5.2.88)

y¨ = −ω2 y − ωC(vx )

(5.2.89)

Substituting (5.2.88) into (5.2.87b),

Our initial conditions state that at t = 0 then vx = 0, so C(vx ) in (5.2.89) vanishes. We are left with a second order differential equation, with solutions of the form y(t) = A cos ωt + B sin ωt If we again apply our initial conditions, we note that at t = 0 then y = 0, so we must require that A = 0. Our solution is then y(t) = B sin ωt

(5.2.90)

We now substitute (5.2.90) into (5.2.88): x˙ = ωyB sin ωt −→ x(t) = −B cos ωt + C(x) −→ x(t) = −B cos ωt + x0

(5.2.91)

Where we note that initial conditions imply that C(x) is none other than the initial position of our particle. Additionally, we can unleash the true identity of the integration constant B everpresent in both (5.2.90) and (5.2.91). We do this by taking the derivative of (5.2.90) and setting it equal to vy : vy y(t) ˙ = Bω cos 0 = vy −→ B = ω And now we can rewrite (5.2.90) and (5.2.91): x(t) = − y(t) =

vy cos ωt + x0 ω

vy sin ωt ω

(5.2.92a) (5.2.92b)

And now, after one full period, t → τ , which implies that ωτ = 2π, it can be easily (very easily) seen that (5.2.92a) and (5.2.92b) reveal the location of the particle to be x(τ ) = x0 −

vy , ω

y(τ ) = 0

Problem 25 A long wire lies along the z-axis. A square coil of wire, each side of length a, lies in the xz-plane. The side nearest the wire is parallel to the wire and located at x = a. a)

Find the vector potential A if the wire is carrying a current I and there is no current in the loop.

b)

What is the flux through the loop, using the results from part a)?

c)

What is the mutual inductance of the loop and on the wire?

W. Erbsen

ELECTRODYNAMICS

d)

If the current I is varying at a rate dI/dt = K, what is the induced emf in the coil?

e)

If there is a current in the loop varying at a rate dI/dt = C, what is the size of the voltage across the ends of the long wire?

Solution a)

In order to find the magnetic vector potential A, we this by first setting up the determinant, ρˆ ˆ φ 1 ∂ ∂ ∂ρ ρ ∂φ Aρ Aφ

must “uncurl” it from B = ∇ × A. We do ˆ z ∂ ∂z Az

Due to the geometry of this particular problem, since I flows along the ˆ z direction and the face of the loop is adjacent aligned to the wire, we can make some shortcuts. We know that A only flows in the ˆ z direction in our case, so the other terms are zero. Additionally, we can see that A will depend only on ρ, so any derivatives of A that aren’t with respect to ρ will be zero. ρ ˆ z ˆ φ ˆ ∂ ∂ ˆ 0 0 = − Az φ ∂ρ ∂ρ 0 0 Az So, we now have

B=∇×A=−

∂ ˆ = µ0 I Az φ ∂ρ 2πρ

(5.2.93)

The differential equation in (5.2.93) can be solved straightforwardly,   ∂ µ0 I ˆ µ0 I a ˆ Az φ = φ −→ Az = log ∂ρ 2πρ 2π ρ b)

To find the flux, we use the typical definitions: Z Z Z Z Z a µ0 I 2a 1 µ0 I ΦB = B · dA = (∇ × A) · dA = dρdz = dρ dz 2πρ 2π a ρ 0

(5.2.94)

(5.2.95)

The integration in (5.2.95) is straightforward, and the result is ΦB = c)

µ0 Ia log [2] 2π

(5.2.96)

If we recall that the mutual inductance is qualitatively defined as M = ΦB /I, then the mutual inductance is simply: M=

µ0 a log [2] 2π

439

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

d)

Using (5.2.96), we can straightforwardly find the emf to be E=−

dΦB d µ0 Ia µ0 a dI =− log [2] = − log [2] dt dt 2π 2π dt

And using the handy substitution the prompt suggests, E=− e)

Kµ0 a log [2] 2π

With the given substitution, the voltage across the ends of the long wire is E=−

Cµ0 a log [2] 2π

Problem 26 A disc carrying a uniform surface charge density σ has radius a and rotates at an angular frequency ω. This disc lies in the xy-plane and rotates along the z-axis. A second identical disc rotating at the same frequency is located with its center on the z-axis at Z, but this one lies in the yz-plane about an axis parallel to the x-axis. Find the torque on the second disc. Indicate in a sketch the direction of the torque and the sense of rotation of the two discs.

Solution We begin by recanting a rather unfamiliar form of Biot-Sevart’s Law: dB =

µ0 KdS × r 4π |r|3

(5.2.97)

Where K is known as the surface charge density, and is described by K = σv. Similar to what we did in a previous problem, we define all the elements needed to compute the magnetic field from (5.2.97). The line element and its magnitude are, respectively p r = zˆ z − ρρ, ˆ and |r| = z 2 + ρ2 And we can also expand upon the surface charge density:

ˆ K = σv = σωρφ And we also recall that that the volume element for cylindrical coordinates is ds = ρ dρdφ. We can now calculate, Kds × r = (σωρρ) ˆ (ρ dρdφ) × (zˆ z − ρρ) ˆ 2 ˆ × (zˆ =σωρ dρdφφ z − ρρ) ˆ

W. Erbsen

ELECTRODYNAMICS

=σωρ2 dρdφ (z ρˆ + ρˆ z)

(5.2.98)

We can now substitute (5.2.98) into (5.2.97), µ0 σω ρ2 (z ρˆ + ρˆ z) dρdφ 3/ 4π (z 2 + ρ2 ) 2 ZZ 2 µ0 σω ρ (z ρˆ + ρˆ z) B= dρdφ 3/ 2 2 4π (z + ρ ) 2 " Z # Z a a µ0 σω ρ2 ρ3 z = ˆ dρ + z dρ 3/ ρ 3/ ˆ 2 2 2 0 (z 2 + ρ2 ) 0 (z 2 + ρ2 )

dB =

(5.2.99)

The first integral in (5.2.99) is neglected, due to symmetry. The second integral can be found in tables, and the result is   µ0 σω a2 + 2z 2 √ B= − 2z ˆ z (5.2.100) 2 z 2 + a2 Now, if we can reconcile with the fact that K = σv, then we can find the magnetic moment of the upper loop as follows Z a Z a Z a Z a Z a πσωa4 µ= I · dρ = K(πρ2 ) dρ = σv(πρ2 ) dρ = σωρ(πρ2 ) dρ = πσω ρ3 dρ = 4 0 0 0 0 0 (5.2.101) And if we recall that within this context τ = µ × B, then by using (5.2.100) and (5.2.101) we can find the torque:

τ=

  πµ0 σ 2 ω2 a4 a2 + 2z 2 √ − 2z (−ˆ y) 8 z 2 + a2

Problem 27 The KSU ultra-short pulse, ultra-high intensity laser in the JRM Lab has λ = 800 nm, pulse width of 20 femtoseconds, pulse energy of 6 mJ, and can be focused to a spot size of 100 µm. a)

What is the intensity of the light in the focus?

b)

What average electric field produces this intensity?

c)

To how many atomic units of the intensity and electric field does this correspond?

Solution

441

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

a)

Intensity is defined as the amount of energy per unit area per unit time and has units W/m2. Mathematically, I=

b)

ENERGY = (AREA) (TIME)

6 × 10−3 J 2 −→ I ≈ 3.8197 W/m 2 100×10−3 m π (20 × 10−13 s) 2 

(5.2.102)

The relationship between the field and intensity is

1 I = cε0 |E0 |2 −→ |E0| = 2

r

2I cε0

(5.2.103)

Where we can plug in our value from (5.2.102) and also for the various constants to arrive at a final value. It should be noted however that I ∝ |E|2 . c)

Donno

Problem 28 A parallel plate capacitor is being charged with a current I. The capacitor is circular and has an area of πa2 . a)

What is the displacement current between the plates?

b)

Referring to the drawing below, what is H in regions 1, 2, and 3 for ρ < a. (Neglect fringing fields)

c)

From the boundary condition on Ht , where H is evaluated at r = ρ(< a), what are the values on the surface currents K on the face of each of the capacitor plates?

d)

Make a sketch of all the current flows and label them.

Solution a)

Most formally, the displacement current is found by differentiating the electric displacement field

W. Erbsen

ELECTRODYNAMICS

D

=

ε0 E

↓ Jd

↓ ∂E ε0 ∂t

=

+

+

P ↓ ∂P ∂t

We have no dielectric so the polarization term is neglected in our case. One thing that should definitely be remembered is that the electric field between the plates of a parallel plate capacitor is E = σ/ε0 . We are now in a place to find the displacement current Jd :     ∂E ∂ σ ∂σ ∂ Q(t) = ε0 = = (5.2.104) Jd = ε0 ∂t ∂t ε0 ∂t ∂t πa2 ˙ If we recall that I(t) = Q(t), then (5.2.104) is Jd = b)

Recall that Amp´ere’s Law for H fields is

I

I ˆ z πa2

(5.2.105)

H · d` = Id

Within region 2, we can use (5.2.105) to find H: H2 (2πρ) =

I ˆ I ˆ φ −→ H2 = φ πa2 2πρ

(5.2.106)

And in regions 1 and 3, the value for H is simply H1,3 =

I ˆ φ 2πρ

(5.2.107)

Problem 29 A spherical conductor A contains two spherical cavities. The total charge on A itself is zero. However, there is a point charge qb at the center of one cavity, and qc at the center of the other. A considerable distance r away is another charge qd . What force acts on each of the four objects A, qb , qc, and qd ? Which answers, if any, are only approximate? If qb is displaced slightly away from the center of the cavity, will it experience a resistance force? (If so, what is it?)

Solution The forces on charges qb and qc themselves are, of course, zero. To calculate the force of qd acting on A, F=

1 qd (qb + qc ) 4πε0 r2

(5.2.108)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Which is the same as the net force of A acting on qd . We must recognize, however that (5.2.108) is an approximation. To get the exact answer you must employ the method of images with an image charge inside the sphere. The exact result turns out to be F=

1 qd (qb + qc) 1 q 2 Rr − 4πε0 r2 4πε0 (r 2 − R2 )2

Problem 30 A uniformily charged rod of length a resides along the z-axis as drawn. The linear charge density is λ. The Z a voltage potential at (x, 0, 0) is 1 λ √ V = dz (5.2.109) 2 4πε0 0 z + x2 a)

Expand the denominator of this expression for x  z to two nonzero terms.

b)

Substitute this expansion into the expression for V , work with two integrals.

c)

Identify the two results as monopole, dipole, quadrupole, octapole or whatever and give an expression for the two moments.

Solution

W. Erbsen

ELECTRODYNAMICS

a)

The series expansion around z is √

1 1 1 z2 3z 4 5z 6 = p = − 3+ 5− − ... x 2x 8x 16x7 z 2 + x2 x z 2 /x2 + 1

So, expanded to the first two nonzero terms:

1 1 z2 √ = − 3 x 2x z 2 + x2 b)

(5.2.110)

Substituting (5.2.110) into (5.2.109), we have  Z a  1 1 z2 V = λ − dz 4πε0 0 x 2x3  Z a  Z a 1 1 1 2 λ = dz − 3 z dz 4πε0 x 0 2x 0   1 a a3 = λ − 3 4πε0 x 6x So the result is V =

c)

λ a λ a3 − 4πε0 x 24πε0 x3

(5.2.111)

The first term in (5.2.111) is monopole, because of the 1/x factor, and the second is quadrupole, since it has a 1/x3 term.

Problem 31 a)

Find the voltage potential at point P a distance b from the right end of a uniform line charge with length ` and total charge Q as drawn.

b)

Using this voltage find the x-component of the electric field at point P .

Solution a)

We first note that our object has a charge given by dq = λ d`, and the total charge is Q, and the total length is `, so we know that λ = Q/`. Furthermore, we know that E = −∇V −→ dV = −Edx. We can take this further: Z 0 λ 1 V =− dx 4πε0 −` (b − x) 1 Q 0 =− [log (b − x)|−` 4πε0 `

445

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

1 4πε0 1 =− 4πε0 =−

Q [log (b) − log (b + `)] `   Q b log ` b+`

So that the potential is V = b)

1 Q log 4πε0 `



b+` b



(5.2.112)

E can be calculated rather straightforwardly as follows ∂ Vx ˆ ∂x    1 Q ∂ b+` log x ˆ =− 4πε0 ` ∂b b   1 Q 1 − (b+`)/b =− x ˆ 4πε0 ` b+`   1 Q1 b−b−` =− x ˆ 4πε0 ` b b+`

E=−

From which we arrive at 1 Q E= 4πε0 b



1 b+`



x ˆ

Problem 32 A dipole consistent of equal and opposite charges ±q separated by x resides a distance d/2 above a conducting, grounded, infinite plane as drawn. a)

On the drawing, accurately draw the image charge(s).

b)

Find the force between the dipole and the plane.

c)

Use the binomial expansion to expand this force to order (x/d)2 when d  x. With what power does the force depend on the distance d?

W. Erbsen

ELECTRODYNAMICS

Solution a)

Two charges are imagined to be symmetrically placed about the grounded conducting plane; the magnitudes of opposing charges are opposite.

b)

We use the following equation: F=

3(P · ˆ r)P0 + 3(P · ˆ r)P + 3(P · P0 )ˆ r − 15(P · ˆ r)(P0 · ˆ r)ˆ r r4

(5.2.113)

Which should probably be memorized. The vectors P and P0 represent the dipole moments of the charged, and uncharged dipoles (respectively). Their values are P = qxˆ x and P0 = qx(−ˆ x). The vector ˆ r represents the distance between the two moments r = 2d(−ˆ y). We can see that the majority of the terms in (5.2.113) vanish, F=

3(P · P0 ) ˆ r 3q 2 x2 3(qxˆ x · qx(−ˆ x)) ˆ r = −→ F = y ˆ 4 4 r r 16d4

(5.2.114)

Problem 33 a)

Two spheres initially uncharged are connected by a battery of voltage V . After the switch is closed, what is the charge on the larger sphere?

b)

What is the capacitance of the larger sphere?

Solution a)

When the switch is closed, the amount of charge that flows into the larger sphere (of radius r = b) can be found by V =

1 Q −→ Q = 4πεbV 4πε b

447

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

b)

To find the capacitance of the larger sphere we recall that Q = CV , so: C = 4πεb

Problem 34 A circular loop of wire with radius a and electrical resistance R lies in the xy-plane. A uniform mgnetic field  is turned on at time t = 0; for t > 0 the field is B   0 B(t) = √ ˆj + kˆ 1 − e−λt (5.2.115) 2 a)

Determine the current I(t) induced in the loop.

b)

Sketch a graph of I(t) versus t.

Solution a)

When the magnetic field is turned on, flux passes through the loop which causes current to flow. The amount of current is determined by how much flux passes through. We start from Ohm’s law,  = IR. Faraday’s Law says, E=−

dΦB dt

(5.2.116)

Where the magnetic flux, ΦB , is proportional to the amount of field lines that pass through a given area. In our case, this is I  πa2 B0 ˆ ˆ  ΦB = B · dA = √ j + k 1 − e−λt (5.2.117) 2 Substituting (5.2.117) into (5.2.116) we can find E:    d πa2 B0 ˆ ˆ  √ E=− j + k 1 − e−λt dt 2

W. Erbsen

ELECTRODYNAMICS

=−

λπa2 B0 ˆ ˆ −λt √ j+k e 2

(5.2.118)

Using (5.2.118) and also Ohm’s Law, we arrive at λπa2 B0 ˆ ˆ −λt I(t) = − √ j+k e 2R b)

(5.2.119)

Looks like an exponentially decaying signal. The negative sign denotes the direction, the coefficients the magnitude.

Problem 35 The solenoid pictured below is long, and has n turns per unit length and cross section A. a)

If the current in the solenoid changes from I1 to I2 , how much charge passes through the resistor?

b)

 If the current in the solenoid, as a function of timet t, is  t I(t) = I1 e− /τ + I2 1 − e− /τ

(5.2.120)

what is the current IR (t) in the resistor?

c)

Sketch separately the I1 and I2 terms and the total I vs. t/τ for I1 = 1A and I2 = 2A. On a separate graph sketch IR (t) vs. t/τ .

Solution a)

For discrete changes in quantities, there are no need for integrals or derivatives, eg dt → ∆t. Keeping this in mind, E=−

∆ΦB ∆ ∆B µ0 n(I2 − I1 )A = − BA = − A=− ∆t ∆t ∆t ∆t

(5.2.121)

449

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

We also know Ohm’s Law, which states that E = IR, where I = ∆Q/(∆t). With this knowledge, (5.2.121) becomes ∆Q µ0 n(I2 − I1 )A R=− ∆t ∆t From which it is clear that the ∆t on the denominator cancels, and we are left with ∆Q = − b)

µ0 n(I2 − I1 )A R

To calculate the current, E=−

∂ ∂ ∂I(t) BA = − µ0 nIA = −µ0 nA ∂t ∂t ∂t

(5.2.122)

Recalling again Ohm’s Law, IR (t) can be found by combining (5.2.122) and (5.2.120):    i t E µ0 nA ∂ h − t/τ µ0 nA I1 t I2 t IR (t) = =− I1 e =− − e− /τ + e− /τ + I2 1 − e− /τ R R ∂t R τ τ From which it is easily seen that IR (t) = −

5.3

µ0 nA − t/τ e [I2 − I1 ] Rτ

Modern Physics

Problem 1 13 Recently, 2000 rubidium atoms (87 atoms/cm3 , were ob37 Rb), which had been compressed to a density of 10 served to undergo a Bose-Einstein condensation as a function of temperature. This occurs when the deBroglie wavelengths overlap and the system forms a single macroscopic quantum state. Estimate the temperature at which Bose-Einstein condensation occurs for this system.

Solution We first recall that the deBroglie wavelength is given by λ=

h p

Where we are also told that N =2000 atoms ρ =1013 atoms · cm3

(5.3.1)

W. Erbsen

MODERN PHYSICS

So, we can say that the spacing is proportional to the deBroglie wavelength, and the spacing can be found by 2000 atoms = 1013 atoms · cm3 −→ L = L3



2000 atoms 101 3 atoms · cm3

 1/3

We now wish to set (5.3.2) equal to the deBroglie wavelength. But first, we recall that √  p p = 2mE → p ∼ 2mkB T E ∼ kB T

(5.3.2)

(5.3.3)

Where putting (5.3.3) into (5.3.1) of course yields h λ∼ √ 2mkB T

(5.3.4)

Setting (5.3.4) equal to (5.3.2) yields 

2000 atoms 13 10 atoms · cm3

 1/3

h = √ 2mkB T

Solving for T leaves us with 1

h /2 T ∼ 2mkB



2000 atoms 1013 atoms · cm3

 1/6

Problem 2 Rutile, TiO2 , is a tetragonal crystal a = 4.4923˚ A. The position of the Ti atoms in the unit cell are 0, 0, 0; 1 /2 , 1/2 , 1/2 and the position of the oxygen atoms are 0.31, 0.31, 0; 0.81, 0.19, 0.5; 0.69, 0.69, 0; 0.19, 0.81, 0.5. a)

What would be the d-spacing of the 310 planes of rutile?

b)

Using CuK x-radiation, λ = 1.54 ˚ A, what would be the diffraction angle from the (310) planes of Rutile?

c)

Assuming atomic scattering factors of fTi for the 2 titanium atoms and fO for the 4 oxygen atoms, what is the crystalline scattering factor from the (310) planes of TiO2 ?

Solution We first note that: a =4.4923 ˚ A Ti =

0, 0, 0 1 /2 , 1/2 , 1/2

451

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

0.31, 0.31, 0 0.81, 0.19, 0.5 O= 0.69, 0.69, 0 0.19, 0.81, 0.5 a)

In order to find the d-plane spacing, we recall that 1 h2 k2 `2 = 2 + 2 + 2 2 d a b c

(5.3.5)

Where h, k and ` correspond to the the indices of the plane (eg h = 3, k = 1, and ` = 0). Furthermore, the numbers a, b, and c correspond to the sides of the crystal. In our case a = b = c → a. Taking all this into account, (5.3.5) becomes   1 h2 k2 1 1 = + −→ d = a + (5.3.6) d 2 a2 a2 h k Plugging into (5.3.6) the appropriate values, we have   1 1 d = 4.4923 ˚ A + −→ d = 1.42 ˚ A 32 1 b)

(5.3.7)

We recall that the diffraction equation from a periodic structure is given by nλ = 2d sin θ

(5.3.8)

Where integer multiples of λ yield maxima. Since we are interested in the first maxima, then n = 1 in our case, and after applying this to (5.3.8) and solving for θ, we have   λ −1 θ = sin (5.3.9) 2d

c)

Putting in the correct numbers into (5.3.9) leaves us with " # 1.54 ˚ A −1
−→ θ = 32.8o θ = sin 2 1.42 ˚ A

In order to find the crystalline scattering factor for our lattice, we employ the following equation for reasons unknown: X Fhk` = fj exp [iφhk` (j)] (5.3.10) j

Where φhk` (j) = 2π (hxj + kyj + `zj ) Substituting (5.3.11) into (5.3.10) yields X Fhk` = fj exp [i2π (hxj + kyj + `zj )]

(5.3.11)

(5.3.12)

j

In our case, ` is always zero, so we can make our lives easier by ignoring the last term in (5.3.14): X Fhk` = fj exp [i2π (hxj + kyj )] j

W. Erbsen

MODERN PHYSICS

And we also know what h and k are, so let’s go ahead and put them in too: X F310 = fj exp [i2π (3xj + yj )]

(5.3.13)

j

Completing (5.3.10) and (5.3.11) for Titanium is the easiest, so let’s do that sum first: 
 Ti F310 =fTi exp [i2π (3 · 0 + 0)] + fTi exp i2π 3 · 1/2 + 1/2 =fTi [1 + exp [i4π]]

(5.3.14)

Now, let’s do the same for Oxygen: O F310 = fO exp [i2π (3xj + yj )] + fO exp [i2π (3xj + yj )] + fO exp [i2π (3xj + yj )] + fO exp [i2π (3xj + yj )] {z } | {z } | {z } | {z } | I

II

III

IV

So let’s do these one at a time..

I =fO exp [i2π (3 · 0.31 + 0.31)] ⇒ fO exp [2.48πi] II =fO exp [i2π (3 · 0.81 + 0.19)] ⇒ fO exp [5.24πi] III =fO exp [i2π (3 · 0.69 + 0.69)] ⇒ fO exp [5.52πi] IV =fO exp [i2π (3 · 0.19 + 0.81)] ⇒ fO exp [2.76πi]

(5.3.15a) (5.3.15b) (5.3.15c) (5.3.15d)

Combining (5.3.15a)-(5.3.15d) back into our original equation, O F310 =fO [exp [2.48πi] + exp [5.24πi] + exp [5.52πi] + exp [2.76πi]]

(5.3.16)

And now, combining (5.3.14) with (5.3.16), Ti O F310 = F310 + F310

−→ F310 = fTi [1 + exp [i4π]] + fO [exp [2.48πi] + exp [5.24πi] + exp [5.52πi] + exp [2.76πi]]

Problem 3 Copper is a monovalent metal of density 8 g/cm3 and atomic weight of 64 g/mole. Using the free electron approximation, and considering a cubic system of side L, a)

Find a formula for the number of states dN within an energy interval dε.

b)

Use your expression to evaluate the Fermi energy for copper in eV, at 0 K.

Solution The Free Electron Approximation assumes that the valence electrons are essentially detached from their parent ions. Subsequently, the free electrons are thought of as “conduction” electrons, and are able to move freely throughout the crystal. This approximation completely ignores all electron-electron and ion-electron interactions.

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

a)

In this part we are looking for the density of states for a single valence electron. Using the free electron approximation, we imagine the electron to be bound within a cube, each side of length L. The TISE equation says that   ~2 2 ~2 ∂ 2 ∂2 ∂2 − ∇ ψ = Eψ −→ − + + ψ = Eψ (5.3.17) 2m 2m ∂x2 ∂y2 ∂z 2 Whose corresponding eigenvalues are E=


~2 π 2 n2x + n2y + n2z 2 2mL

(5.3.18)

If we imagine the energy levels to be continuous rather than discrete, then the total number of   1/2 states can be approximated by the volume of a sphere of radius r = n2x + n2y + n2z . The total number of states, N , can then be found by integrating this radius within n-space: N =2·

1 4 3 · πr 8 3

(5.3.19)

Where in (5.3.19), the factor of 2 is due to spin degeneracy (two different spins with the same energy), while the factor of 1/8 is due to the fact that we are only interested in the quadrant where all n’s are positive (nx , ny , nz > 0), which happens only within one quadrant, which corresponds to 1 /8 of the total volume of the sphere. Substituting r in (5.3.19) for its equivalent form in terms of n gives  3/2 1 4  (5.3.20) N = 2 · · π n2x + n2y + n2z 8 3 Now, we solve (5.3.18) as follows

n2x + n2y + n2z =

2mL2 E ~2 π 2

(5.3.21)

Substituting (5.3.21) back into (5.3.20),   3/ 2mL2 E 2 1 4 N =2 · · π 8 3 ~2 π 2 3

/

π (2m) 2 L3 3/2 = · E 3 ~3 π 3 3

/

(2m) 2 L3 3/2 = E 3~3 π 2

(5.3.22)

We now recall that the density of states g(N ), is defined as the first derivative of N with respect to energy. Using (5.3.22), this becomes g(N ) =

dN dE "

# 3 / (2m) 2 L3 3/2 E 3~3 π 2 3   (2m) /2 L3 d 3/2 = E 3~3 π 2 dE d = dE

Carrying out the simple derivative in (5.3.23), we have

(5.3.23)

W. Erbsen

MODERN PHYSICS

3

/

g(N ) = b)

(2m) 2 L3 1/2 E 2~3 π 2

(5.3.24)

In order to find the Fermi Energy, EF , we first recall that the number of particles is given by Z EF N= g(E) · f(E) dE (5.3.25) 0

Where g(E) is the density of states, and f(E) is the Fermi-Dirac distribution function, which is given by f(E) =

1 exp [(E − µ) /kBT ] + 1

(5.3.26)

At T = 0 K, then µ(T = 0) = EF , where EF is the Fermi Energy. If E < µ, then (5.3.26) becomes f(E) =

1 −→ 1 exp [(E − µ) /kB T ] + 1

With this, (5.3.25) becomes N =

Z

EF

g(E) dE # 3 Z EF " (2m) /2 L3 1/2 E = dE 2~3 π 2 0 3 Z / (2m) 2 L3 EF 1/2 = E dE 2~3 π 2 0 0

3

/

=

(2m) 2 L3 2 3/2 · EF 2~3 π 2 3

=

3 (2m) 2 L3 / · EF2 3 2 3~ π

3

/

(5.3.27)

From the values given initially, we know that N N = 3 V L 3

=8

g 1 mol (100 cm) · ·· · NA 3 cm3 64 g (1 m)

g 1 mol (100 cm)3 atoms · ·· · 6.022 × 1022 3 3 cm 64 g mol (1 m) atoms =7.528 × 1028 m3 =8

Now, we wish to solve (5.3.27) for EF , which yields   2/ ~2 N 3 EF = 3π 2 3 2m L Substituting in (5.3.28) and the other appropriate values into (5.3.29) yields

(5.3.28)

(5.3.29)

455

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

"

# 2/3
3
3 3 6.582 × 10−16 eV · s π 2 8 28 atoms EF = · 3 × 10 m/s · 7.528 × 10 −→ EF = 6.4 eV (2 · 0.511 × 106 eV) m3

Problem 4 A white dwarf is a star which has used up its nuclear fuel and contracted under its own weight to a state of high density and arbitrarily low temperature. It is held up by the kinetic energy of its degenerate electrons. Assume that all electrons in the star behave like free electrons in a metal, that all the electron states are full up to the Fermi energy, that the star is spherical and made of helium, and that the electrons are non-relativistic. a)

Discuss briefly the physical principles needed to find the equilibrium radius of the star. Write down a set of equations which, when combined, will determine the radius. Don’t worry too much about the factors of 2.

b)

Show that this radius is given approximately by R=C

1 h2 M − /3 (Gme mp ) 5/3

(5.3.30)

Where C is a purely numerical constant, G is the gravitational constant, me and mp are the electron and proton masses, and M is the mass of the star.

Solution a)

The equilibrium radius is defined as the radius which minimizes the total energy of the star. In our case the energy comes from the gravitational energy, which acts to contract the star, and also the degenerate energy, which is attributed to the tendency of Fermions to repel one another. Mathematically, Etot = Egrav + Edeg

b)

(5.3.31)

Let’s find the gravitational energy in (5.3.31) first. In order to find the gravitational self-energy of some spherical mass with a density ρ, we recall that ρ = m/V → m = ρV . In other words,   4 4 m = πr 3 ρ recalling that V = πr 3 (5.3.32) 3 3 From which we can say that dm = 4πr 2 ρ dr

(5.3.33)

We also recall that the potential energy is given by Gm dm r R Substituting (5.3.33) into (5.3.34), and recalling that U = Egrav, we have Z R Gm Egrav = − dm r 0 dU = −

(5.3.34)

W. Erbsen

MODERN PHYSICS

=−G

Z

R

0

1 r



4πr 3 ρ 3




4πr 2 ρ dr

(5.3.35)

Where I have substituted m for the value from (5.3.32) and likewise for dm from (5.3.33). Continuing, Z 16π 2 Gρ2 R 4 Egrav = − r dr 3 0   16π 2 Gρ2 R5 =− 3 5 2 2   16π Gρ R5 =− 15   16π 2 Gρ2 3M =− 15 4πR3 2 3 GM (5.3.36) =− 5 R We now recall that the total energy of a degenerate electron gas is given by∗ 3 Edeg = N EF 5

(5.3.37)

Where the Fermi Energy, EF , is given by ~2 EF = 2me Substituting (5.3.38) into (5.3.37), Edeg



3π

2N

V

 2/3

(5.3.38)

"   2/3 # 3 ~2 2N = N 3π 5 2me V   2/3 3~2 2N N 3π = 10me V 2
5 3~2 1 / 3 3π 2 · 2/ · N /3 = 3 10me V

(5.3.39)

Recalling that V = 4/3πR3 , (5.3.39) becomes Edeg

  2/3
2 5 3~2 3 2 /3 = 3π · · N /3 3 10me 4πR   2/3 2 5 3~ 9π = · N /3 3 10me 4R   2/ 5 3~2 1 9π 3 = · N /3 2 10me R 4

(5.3.40)

Substituting (5.3.36) and (5.3.40) into (5.3.31), Etot ∗ Recall

that E =

R

E · d (ε) dε

3 GM 2 3~2 1 =− + 5 R 10me R2



9π 4

 2/3

N

5

/3

(5.3.41)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

And now minimizing the total energy with respect to R, we take the first derivative of (5.3.41) with respect to R and set it equal to zero: " #   2/ ∂Etot ∂ 3 GM 2 3~2 1 9π 3 5/3 = − + N ∂R ∂R 5 R 10me R2 4     2/3   5 3 1 3~2 9π 1 2 ∂ /3 ∂ = − GM + N 5 ∂R R 10me 4 ∂R R2     2/3   5 3 1 3~2 9π 2 = − GM 2 − 2 + N /3 − 3 5 R 10me 4 R 2       / 3 1 6~2 9π 3 5/3 1 = GM 2 − N (5.3.42) 2 5 R 10me 4 R3 Setting (5.3.42) equal to zero and solving for R, we find that   2/3 5 6~2 9π 3 2 GM R = N /3 5 10me 4   2/3 5 ~2 9π R= N /3 Gme M 2 4 Recalling that N = M/mp , (5.3.43) becomes   2/ ~2 9π 3 R= G 4   2/ ~2 9π 3 = G 4   2/ ~2 9π 3 = G 4   2/ ~2 9π 3 = G 4

1

5

2

me (mp · N )

N /3

1 1 me · m2p N 1/3  m  1/3 1 p

me · m2p

M

1

1 M 1/3

5/ 3

me · mp

And now recalling that ~ = h/2π, (5.3.44) becomes   2/3 h2 9π 1 1 R= 2 5/ 1 3 M /3 4π G 4 me · mp   2/3 2 1 9π h 1 1 = 2 4π 4 G me · mp5/3 M 1/3 And we are finally left with R=C

h

1

5/ 3

Gme mp

M − /3

Where C=

(5.3.43)

1 4π 2



9π 4

 2/3

(5.3.44)

(5.3.45)

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MODERN PHYSICS

Problem 5 A particular atom has two energy levels with a transition wavelength of 1064 nm. At 297 K there are 2.5 × 1021 atoms in the lower state. a)

How many atoms are in the upper state?

b)

Suppose that 1.8 × 1021 of the atoms in the lower state are pumped to the upper state. How much energy could this system release in a single laser pulse? How many photons are emitted in this pulse?

Solution a)

We first recall that the number of atoms in a particular state is   1 E N = exp − Z kB T

(5.3.46)

Taking the ratio of the two populations, and using (5.3.46), we have   NU (EU − EL) = exp − NL kB T

(5.3.47)

Where we note that ∆E =EU − EL hc = λ

4.136 × 10−5 eV · s 3 × 108 m · s−1 = (1064 × 10−9 m) =1.166 eV

(5.3.48)

Substituting (5.3.48) back into (5.3.47) and solving for NU yields " #
(1.166 eV)
NU = 2.5 × 1021 atoms · exp − 8.617 × 10−5 eV · K−1 (300 K) −→ NU = 40.87 atoms

b)

(5.3.49)

If 1.8 × 1021 atoms are pumped to the upper state, and we are interested in how much energy they could release in a single laser pulse, this is the same as asking what happens if all atoms simultaneously fall down to the lower state from the upper state. We know that the energy difference between the two levels is ∆E = 1.166 eV from (5.3.48), so the total energy that would be released is

eV Epulse = 1.8 × 1021 (1.166 eV) 1.602 × 10−19 J · eV−1 −→ Epulse = 2.099 × 1021 Pulse −→ Epulse = 336.292

Joules Pulse

459

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Problem 6 A helium-neon laser emits light with a wavelength of 6328 ˚ A in a transition between two states of neon atoms. What would be the ratio of population of the upper state relative to the lower state of the neon atoms if they were in thermal equalibrium at 300 K?

Solution At equilibrium (T = 300 K in this case), the population of the two states are equal. That is, N1 = N2 And the ratio would be   N2 −(E2 − E1 ) = exp N1 kB T

(5.3.50)

Where ∆E =E2 − E1 hc = λ

4.136 × 10−15 eV · s 3 × 108 m · s−1 = (632.8 × 10−9 m) =1.961 eV Applying this to (5.3.50) at T = 300 K yields " # N2 (1.961 eV) N2
= exp − −→ = 1.136 × 10−33 −2 5 N1 N1 8.617 × 10 eV · K (300 K)

Problem 7 The atomic number of Na is 11. a)

Write down the electronic configuration for the ground state of the Na atom showing in standard notation the assignment of all electrons to various one-electron states.

b)

Give the standard spectroscopic notation for the ground state of the Na atom.

c)

The lowest frequency line in the absorption spectrum of Na is a doublet. What are the spectroscopic designations of the pair of energy levels to which the atom is excited as a result of this absorption process?

d)

Discuss the mechanism responsible for the splitting between this pair of energy levels.

e)

Which of these levels lies the lowest? Discuss the basis on which you base your choice.

W. Erbsen

MODERN PHYSICS

Solution a)

The ground state electronic configuration of Sodium in standard notation is: 1s2 2s2 2p6 3s1

b)

Only the unpaired 3s electron contributes to the total angular momentum J: L = 0, S = 1/2 −→ J = 1/2

So, the spectroscopic notation is:

2

c)

S 1/2

The first excited state is the 3p state, which means that L = 1, S = 1/2 −→ J = 1/2 , 3/2

And therefore, the spectroscopic notation for the lowest excited states is 2

d)

2

P 3/2

The levels 32 P 1/2 and 32 P 3/2 are split when the spin-orbit interaction is taken into account. In the electron’s rest frame, it sees the nucleus orbiting it. This causes the electron to see a magnetic field caused by the nucleus “orbiting” the electron. This is seen as a current loop, and is proportional to the electron’s orbital angular momentum, Li . This field interacts with the electron’s magnetic dipole moment, µi , which is proportional to its spin angular momentum, si . This creates a perturbation term which is proportional to Li , Si for each electron. The perturbation term is given by H=

e)

P 1/2 ,

1 1 dV(ri ) Li · S i 2m2 c2 ri dri

The 32 P 1/2 state lies lower than the 32 P 3/2 when spin-orbit is taken into account. The reason is because of Hund’s rules, which states that for any two states with identical L and S values, the one with the lowest J is more tightly bound with the lowest energy shells that are less than half-full (as is the case with a single electron in a P -state).

Problem 8 A slab of lead shielding 1.0 cm thick reduces the intensity of 15 MeV γ rays by a factor of 2. a)

By what factor will 5.0 cm of lead reduce the beam?

b)

What is the assumption cross-section for 15 MeV γ’s in lead nuclei? Lead has a density of 11.4 g/cm3 , and an atomic weight of 207 g/mol.

Solution

461

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

The decrease in intensity of radiation passing through matter is mathematically given by I = I0 exp [−µd]

(5.3.51)

Where µ = nσ, and n is the number of atoms per cm3 , and σ is the absorption cross section. a)

First, we find µ for the initial case: 0.5I0 = I0 exp [−µ] −→ µ = log [2] So, for d = 5.0 cm, (5.3.51) becomes I = I0 exp [− log [2] · 5] −→ I =

b)

I0 32

Recalling that µ = nσ, and that for lead we found that µ = log [2], we have: log [2] = nσ −→ σ =

log [2] n

(5.3.52)

Where for lead, we find that g 1 mol NA atoms · · 3 cm 207 g 1mol 11.4 · NA atoms = 207 cm 3

n =11.4

(5.3.53)

Substituting (5.3.53) into (5.3.52), σ = log [2] ·

207 cm3 −→ σ = 2.090 × 10−23 cm3 11.4 · 6.022 × 1023

Problem 9 The potential energies of two diatomic molecules of the same reduced mass are shown. From the graph, determine which molecule has the larger a)

Inter-nuclear distance

b)

Rotational inertia (moment of inertia)

c)

Separation between rotational energy levels

d)

Binding energy

e)

Zero-point energy

f)

Separation between low-lying vibrational states

Solution

W. Erbsen

MODERN PHYSICS

a) b) c)

1 has a larger inter-nuclear distance 1 has a larger rotational intertia , because I = µR2 and since we have the same µ and 1 has a larger inter-nuclear distance, than 1 must also have a higher rotational inertia. 2 has a larger separation between rotational energy levels , because Erot ∝

` (` + 1) ~2 2I

And from part b), I1 > I2 so we must conclude that E1 < E2 . d) e)

2 has a larger binding energy , since the well is deeper. 2 has the largest zero-point energy . We recall that the zero-point energy is defined as the lowest energy possible e.g. the lowest vibrational state. We can approximate, using the simple harmonic oscillator, that r    1/2 2m Ki E1 = 1/2 ~ω1 T = (Vi − E) , ωi = E2 = 1/2 ~ω2 ~2 m And since V2 > V1 , then K2 > K1 and ω2 > ω1 , and therefore E2 > E1 .

f)

2 has the largest separation between the lowest-lying vibrational states , since ∆Evib ∝ ~ω −→ ω2 > ω1

Problem 10 A K0 meson (mass 497.7 MeV/c2 ) decays into a π + , π − meson pair with a mean life of 0.89−10 s. a)

Which one of the fundamental interactions is responsible for this decay?

463

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

b)

Suppose the K0 has a kinetic energy of 276 MeV when it decays, and that the two π Mesons emerge at equal angles to the original K0 direction. Find the kinetic energy of each π meson and the opening angle between them. The mass of a π meson is 139.6 MeV/c2 .

Solution a)

This decay is caused by the weak interaction . This is know because K o has strange quarks, while π-particles do not. Only in the weak interaction strangeness is not conserved.

b)

At this point, we must employ the wonders of conservation of momentum, which says pk = pπ + + pπ −

(5.3.54)

We now recall the handy formula p2 c2 = E 2 − mc2


2

(5.3.55)

First it would be nice to combine the two momenta from the Pions in (5.3.54): pk = 2pπ

(5.3.56)

However, we must be careful! The Pions do not both go in the same directions – they fly off at equal angles. Therefore, the momentum we concern ourselves with from with is pk = 2pπ cos θ

(5.3.57)

At this point we recognize that pπ in (5.3.57) can be rewritten using our good ol pal (5.3.55): q
2 2 (5.3.58) p2π c2 = Eπ2 − mπ c2 −→ pπ c = Eπ2 − (mπ c2 )

Now multiplying both sides of (5.3.58) by c,

pk c = 2pπ c cos θ

(5.3.59)

q 2 pk c = 2 Eπ2 − (mπ c2 ) cos θ

(5.3.60)

h
2 i p2k c2 = 4 Eπ2 − mπ c2 cos2 θ

(5.3.61)

And substituting (5.3.58) into (5.3.59),

And squaring both sides,

And substituting (5.3.55) into (5.3.61) for the LHS, h
2
2 i Ek2 − mk c2 = 4 cos2 θ Eπ2 − mπ c2

We know all the constants in this equation, except for Eπ . From conservation of energy, Ek = 2Eπ −→ Eπ = 1/2 Ek Substituting this back into (5.3.65), Ek2 − mk c2


2

=4 cos2 θ




2 Ek2 − mπ c2 4



(5.3.62)

W. Erbsen

MODERN PHYSICS

We can now solve (5.3.63) for θ, −1

θ = cos

h
2 i = cos2 θ Ek2 − 2mπ c2

(5.3.63)

"s

(5.3.64)

2

Ek2 − (mk c2 )

2

Ek2 − (2mπ c2 )

#

Before we go any further, we must recognize that Ek represents that total energy of the K o , so it includes both the mass energy term and the kinetic term. Using all the constants applied, (5.3.64) becomes "s # 2 2 (497.7 + 276 MeV) − (497.7 MeV) θ = cos−1 (5.3.65) 2 2 (497.7 + 276 MeV) − (2 · 139.6 MeV) Evaluating (5.3.65), we are left with θ = 34.82o HOWEVER, if we want the total angle between the two Pions, we must double this result, which yields θ = 69.34o The kinetic energy can just be found via Conservation of Energy, which says that Tk = 2Tπ −→ Tπ =

Tk 276 eV −→ Tπ = −→ Tπ = 138 MeV 2 2

Problem 11 ˚ spacing between the crystal planes. The A beam of 50 eV electrons is scattered from a cubic crystal with 1.2A electron beam is perpendicular to the crystal plane. At what angle will the first diffraction maximum be detected?

Solution We first recall that the deBroglie wavelength is given by h p

(5.3.66)

√ p2 −→ p = 2mT 2m

(5.3.67)

λ= And the momentum is T = Substituting (5.3.67) into (5.3.66) yields

465

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

λ= √

h 2mT

(5.3.68)

We also recall that the equation for diffraction from a 3D period structure is nλ = 2d sin θ

(5.3.69)

Where we recognize that the path length must be in integer multiples of λ for completely constructive interference. For the first maxima, we have n = 1, and with (5.3.68), we have   h h 1 −1 √ √ = 2d sin θ −→ θ = sin (5.3.70) 2d 2mT 2mT Substituting in the appropriate values into (5.3.70) yields   −34 6.626 × 10 J·s 1  q θ = sin−1  2 (1.2 × 10−10 m) 2 (9.109 × 10−31 Kg) (50 eV) 1.602 × 10−19 J · eV−1
−→ θ = 46.3o

Problem 12 One of the strongest emission lines observed from distant galaxies comes from Hydrogen and has a wavelength of 122 nm (in the ultraviolet region). a)

How fast must a galaxy be moving away from us in order for that line to be observed in the visible region of 366 nm?

b)

What would be the wavelength of the line if that galaxy were moving towards us at the same speed?

Solution a)

If the source and receiver are moving away from each other, the Doppler effect says that p 1 − u2 /c2 0 f =p fo 1 + u2 /c2 p 1 − u2 /c2 = fo 1 + u2 /c2 Recalling that f = c/λ, we can rewrite (5.3.71) in the following form s 1 1 − u/c 1 = λ0 1 + u/c λo

(5.3.71)

W. Erbsen

MODERN PHYSICS



λo λ0

2

1 − u/c 1 + u/c [1 − u/c] (c/c) = [1 + u/c] (c/c) c−u = c+u =

So, with (5.3.73) and the values given, we have  2  2  2 λ0 122 nm 1 1 c−u = = = = 0 c+u λ 366 nm 3 9

(5.3.72)

(5.3.73)

(5.3.74)

Rewriting, (5.3.74) becomes c−u 1 = c+u 9 9 (c − µ) =c + u

9c − 9u =c + u u + 9u =9c − c

From which it is easy to see that u= b)

8 c 10

If the galaxy is moving towards Earth at the same speed, we simply change the sign of u in (5.3.72): s 1 1 + u/c 1 = 0 λ 1 − u/c λ0 s 1 − u/c λ0 = λ0 1 + u/c r 1 − 0.8 = 122 nm 1 + 0.8 122 nm = 3 Which finally leads us to λ0 = 40.67 nm

Problem 13 Hall effect: Consider a free electron gas with an electric field E along the x-direction. The electrons are constantly undergoing collisions resulting in some average velocity E is applied. a)

The conductivity is defined using the current density j = Eσ. Find an expression for σ in terms of the relaxation time T which is half of the time between successive collisions. Let n be the density of electrons in

467

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

the material and m the electron mass. b)

Now assume the electron gas is confined to a wire lying along the x-direction (E is still applied). Under these circumstances the equation of motion of an electron is m

v dv = −e (E + v × B) − m dt T

(5.3.75)

Where the first term in the Lorentz force and the second term takes account of collisions. It is found that a voltage develops along the y-direction (across the wire). A measure of this voltage is given by the Hall constant R = Ey /(j · Bz ). Show that R = −1/(ne). c)

Will the Hall constant be larger in a metal or a semiconductor and why?

Solution a)

To find an expression for the conductivity, σ in terms of the relaxation time, T , we must employ several not-so-tedius substitutions. First, we write down the (given) Ohm’s Law: J = σE

(5.3.76)

We also recall that the drift velocity is defined by v=

J nq

(5.3.77)

Where v is the velocity vector. We also recall that v = a·T

(5.3.78)

Where a is the acceleration of course. Now, from Newton’s second law, F = ma

(5.3.79)

And we must also recall that the force on a charged particle is given by F = −qE

(5.3.80)

Setting (5.3.79) equal to (5.3.80), ma = −qE −→ a = −

qE m

We now take (5.3.81) and substitute it back into (5.3.78):   qE v= − ·T m

(5.3.81)

(5.3.82)

We now set (5.3.82) equal to (5.3.77), which yields J qE nq 2 ET =− · T −→ J = nq m m And finally, we set (5.3.83) and (5.3.83) equal to one another:

(5.3.83)

W. Erbsen

MODERN PHYSICS

σE = b)

nq 2 ET nq 2 T −→ σ = m m

(5.3.84)

The very first thing we need to do in this problem is to solve (5.3.75), which reads (for convienence) m

dv mv = −q (E + v × B) − dt T

(5.3.85)

Let’s evaluate the cross product first. We are told that the magnetic field is along the z-direction, and so ˆi v × B = vx 0

ˆj vy 0

ˆ z vz = vy Bz ˆi − vx Bz ˆj Bz

Substituting this in to (5.3.85) and breaking down the components,   m(v + v + v ) d(vx + vy + vz ) x y z m = −q Ex + Ey + Ez + vy Bz ˆi − vx Bz ˆj − dt T

(5.3.86)

(5.3.87)

A voltage only exists along the y-direction, so lets get rid of everything not in the y-direction, and (5.3.88) becomes   mv dvy y = −q Ey − vx Bz ˆj − m dt T And now,

  Ey ˆ 0 = −q Ey − vx Bz ˆj −→ vx = − j Bz

(5.3.88)

Recalling that J · R = vx , (5.3.88) becomes

R=

Ey ˆ j JBz

(5.3.89)

And now, recall that J · R = vx ,

J =q·v·n

Combining these, Rqvx n = −vx −→ R = − c)

1 nq

The hall constant will be larger .

Problem 14 The density of states g(E) is defined as the number of electronic states per unit energy range.

(5.3.90)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

a)

Write down an expression for the density of the electron states in metal.

b)

Find an expression for the Fermi energy from a).

c)

Find an expression for the fraction of the conduction electrons which are within kT of the Fermi energy.

d)

Evaluate this fraction for Copper at T = 300K. [Copper is a monovalent metal with ρ = 8.92g/cm3 and atomic weight M = 63.5g/mole.]

Solution a)

In order to find an expression for the density of states for a 3D electron gas, we start with the assumption that the metal is in the shape of a cube, each side having a length of L. Assuming for the time being hat we have one free electron and infinite potential at the cube boundaries, the Eigenenergies are given by E=

 ~2 π 2  2 n + n2y + n2z 2mL2 x

(5.3.91)

For macroscopic systems, we assume that L is large enough that we can assume that the energy spectrum is continuous. In this approximation, the total number of states (up to an energy E)   1/2 is proportional to the volume of a sphere of radius r = n2x + n2y + n2z . Since we are only interested in one quadrant, we concern ourselves with only 1/8th of the total volume of the sphere. Furthermore, we also have a 2-fold degeneracy, so we must also include a factor of 2. So, we have  2/2 1 4  N =2 · · π n2x + n2y + n2z 8 3  3/2 π 2 = nx + n2y + n2z (5.3.92) 3 From (5.3.91) we have

Substituting (5.3.93) into (5.3.92),

 2  2mL2 nx + n2y + n2z = 2 2 E ~ π   3/2 π 2mL2 N= E 3 ~2 π 2

(5.3.93)

(5.3.94)

Recalling that the density of states is defined by g(E) =

dN dE

(5.3.95)

We then substitute (5.3.94) into (5.3.95), yielding "   3/2 # d π 2mL2 g(E) = E dE 3 ~2 π 2 3

(2m) 2 L3 d h 3/2 i = E 3~3 π 2 dE 3   (2m) /2 L3 3 1/2 = E 3~3 π 2 2 /

(5.3.96)

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MODERN PHYSICS

Simplifying, (5.3.96) leads us to 3

(2m) /2 L3 1/2 E g(E) = 2~3 π 2 b)

We first recall that the number of particles in a system is given by Z EF N= g(E) · f(E) dE

(5.3.97)

(5.3.98)

0

Where g(E) is the density of states, and f(E) is the relevant distribution function, which in our case is the Fermi-Dirac function: 1 (5.3.99) f(E) = exp [(E − µ)/kB T ] + 1 For T = 0, this distribution gives the average number of particles in a state of energy E to be 1 for E < µ and 0 for E > µ. This maximum occupied energy (µ(T = 0)) is known as the Fermi Energy. So, letting f(E) → 1, and substituting (5.3.97) into (5.3.98), we have # 3 Z EF " / (2m) 2 L3 1/2 N = E dE 2~3 π 2 0 3 Z / (2m) 2 L3 EF 1/2 E dE = 2~3 π 2 0 3   / (2m) 2 L3 2 3/2 = E 2~3 π 2 3 F 3

/

(3m) 2 L3 3/2 = EF 2~3 π 2

(5.3.100)

Solving (5.3.100) for EF , EF =

"

3~3 π 2 (2m)

3 /2

L3

N

# 2/3

Which can be rewritten slightly nicer as   2/ ~2 3N π 2 3 EF = 2m L3 c)

(5.3.101)

In order to find the fraction of conduction electrons within kB T of the conduction band, we need to make a few assumptions. First, we assume that the density of states is constant over the entire kB T band, and secondly that the Fermi-Dirac distribution is still on the order of 1 for particles in this region. Then the number of particles within the kBT band is given by NkB T =g(EF ) · kB T Substituting (5.3.97) into (5.3.102), NkB T

"

# 3 / (2m) 2 L3 1/2 = E · kB T 2~3 π 2

(5.3.102)

(5.3.103)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

And also from (5.3.94) we know that N=

  3/2 π 2mL2 E 3 ~2 π 2

(5.3.104)

Taking the ratio of (5.3.103) and (5.3.104) leads us to NkB T 3 1 = kB T N 2 EF d)

(5.3.105)

We now wish to evaluate (5.3.105) for Copper. The first thing that we want to do is find the fraction N/V , which is:  3 N 8.92 mol 100 cm atoms = · (5.3.106) · 6.022 × 1023 3 V 63.5 cm 1m mol The Fermi Energy as found in (5.3.101) is " EF =

3~3 π 2

# 2/3

N 3 (2m) /2 L3 " # 2/3   2 / 3~3 π 2 N 3 = 3/ V (2m) 2

Where I have set L3 → V . Substituting (5.3.107) into (5.3.105), " # 2/3   2 −1 / 3 2 NkB T 3 3~ π N 3 = kB T  3 / N 2 V (2m) 2 "  2 3 # /3   2/3 /2 3 (2m) V  = kB T  2 3~3 π 2 N

(5.3.107)

(5.3.108)

Substituting in to (5.3.108) the appropriate values, 2 
2/3  /3   2/3 −31
2 · 9.109 × 10 Kg NkB T 3 63.5  · = 1.381 × 10−23 J · K−1 (300 K)  N 2 3 · 1.055 × 10−34 J · s 8.92 · 103 · 6.022 × 1023

Evaluating this, we are left with

NkB T = 5.5 × 10−3 N

Problem 15 A hypothetical atom has only two excited states, at 4.0 and 7.0eV, and has a ground-state ionization energy

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MODERN PHYSICS

of 9.0eV. If we used a vapor of such atoms from the Franck-Hertz experiment, for what voltages would we expect to see decreases in the current? List all voltages up to 20V.

Solution We first recall the principles behind the Franck-Hertz experiment. This seminal experiment demonstrated the quantization of atomic energy states by means of accelerating negatively charged electrons through a positively charged grid, surrounded by Mercury vapor. The resulting data takes the form of an increases oscillating signal, when plotting the net electron current vs the potential difference of the grid. The hypothetical atom described in the prompt has three energy levels: V3 =9 eV

[Continuum]

V2 =7 eV V1 =4 eV The electron current will drop whenever any multiple of V1 , V2 , or V3 is reached. Therefore, 1V1 =4 eV 2V1 3V1 4V1 5V1

=8 eV =12 eV =16 eV =20 eV

1V2 2V2 1V3 2V3

=7 eV =14 eV =9 eV =18 eV

Also, allowing for combinations of V1 , V2 and V3 , V1 + V2 =11 eV V1 + V3 =13 eV V2 + V3 =16 eV 2V1 + V2 =15 eV 2V1 + V3 =17 eV 3V1 + V2 =19 eV So, we expect the current to drop at the following values: V = 4, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, and 20 eV

Problem 16

473

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Consider two protons of mass mp and electric charge e, separated by a distance r. a)

Compute the ratio of the electromagnetic to gravitational forces acting on the protons.

b)

Given your every day experiences with bulk neutral bodies, what do you conclude from the numerical values that you computed in a)? In particular, what force is more important for small bodies? For very large bodies?

c)

Consider a large spherical body, of constant density, made up of N hydrogen atoms. Determine the critical mass of this body at which the binding energy of the body just balances the gravitational potential energy of teh body. Use the fact that the atom size is approximately the Bohr radius [= 1/(ame )] and the atomic binding energy is twice the Rydberg energy [=a2 me /2], where a is the fine structure constant and me is the electron mass. Also assume that the mass of the hydrogen atom is equal to mp .

d)

Determine the critical radius of the body, corresponding to the critical mass derived in c).

Solution a)

The electrostatic force between the two protons is Fe =

1 e2 e2 ⇒K 2 2 4πε0 r r

(5.3.109)

While the gravitational force acting on the bodies is FG = G

Mp2 r2

(5.3.110)

So that if we were to find the ratio of electromagnetic and gravitational forces acting on the protons, we would divide (5.3.109) by (5.3.110), which leads us to Fe Ke2 /r 2 Ke2 = = 2 2 Fg Gmp /r Gm2p

b)

Plugging into (5.3.111) the appropriate values,

2 8.988 × 109 N · m2 · C−2 1.6022 × 10−19 C Fe Fe 36 =
2 −→ F ≈ 10 −2 −11 2 −27 Fg g 6.674 × 10 N · m Kg (1.673 × 10 Kg)

(5.3.111)

(5.3.112)

Rewriting (5.3.111),

2 e mp  2 e 20 ≈10 mp

Fe K = Fg g



(5.3.113)

So, unless the mass of the particle is more than ten orders of magnitude larger than the charge, the electrostatic force will be more prevalent. However, in our everyday experiences, we have mostly bulk substances, being neutral in nature, so for really large objects (stars, planets, etc), the gravitational force wins, whereas on the atomic level (atoms, molecules), the electrostatic force still dominates.

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MODERN PHYSICS

c)

Here we must find the critical mass, in which the binding energy just balances the gravitational energy. In the following calculations, we will assume that the Hydrogen atom radius is on the order of the Bohr radius, and that the binding energy of Hydrogen is twice the Rydberg energy. We first recall that ρ=

m 4 −→ m = V · ρ = πr 3 ρ V 3

(5.3.114)

Taking the differential limit of (5.3.114), dm = 4πr 2 ρ dr

(5.3.115)

We also recall that the gravitational self energy, in differential form is Ug = −

GM dm r

Substituting (5.3.115) into (5.3.116) and carrying out the integration, Z R Gm Ug = − dm r 0  Z R 
G 4 3 =− πr ρ 4πr 2 ρ dr r 3 0 Z 2 16π G 2 R 4 =− ρ r dr 3 0   16π 2 G 2 R5 ρ =− 3 5 2 2 16π Gρ 5 R =− 15

(5.3.116)

(5.3.117)

And the density is (recall that M = mp N ): ρ=

M 3mp N = 4/3πR3 4πR3

And according to the prompt, the binding energy of one Hydrogen atom is  2  a m1 Ub = 2 2

(5.3.118)

(5.3.119)

For the entire mass, (5.3.119) becomes Ub = N A2 me Putting (5.3.118) into (5.3.117), 16π 2 GR5 9m2p N 2 · 15 16π 2 R6 2 5 2 G9mp R N =− 15R G9m2p N 2 =− 15R G3m2p N 2 =− 5R

UG = −

(5.3.120)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

3 m2p 2 =− G N 5 R

(5.3.121)

Recalling now that Ub = N a2 me

(5.3.122)

Setting (5.3.121) and (5.3.122) equal to one another, 3 m2p 2 G N 5 R 5 a2 me mp N = R 3 Gmp

N a2 me =

(5.3.123)

We now recall that M = mp · N , so that (5.3.123) becomes M= d)

5 a2 me R 3 Gmp

(5.3.124)

The critical radius is related to the critical mass as∗ R=N

1

/3

RBohr

Applying this to (5.3.124), we are left with† 

5 a2 me 2 R= R 3 Gm2p Bohr

 1/3

Problem 17 Use the Bohr theory to answer the following question: a)

Derive the energy levels of the hydrogen atom.

b)

The orbiting speed of the electron in the ground state.

c)

If an excitation laser pulse has a duration of one picosecond, what is the principal quantum number n for which the orbiting period of the electron is about one picosecond?

d)

If the electron is replaced by a negatively charged muon, what is the ground state energy of this muon? Use these facts to get your numerical answers: the ground state of atomic hydrogen has a binding energy of 13.6 eV, the mass of the proton is 1837 times that of en electron and teh mass of a muon is 206 times that of an electron. What is the photon energy for the n = 2 to n = 1 transition in this case?

e)

Use the uncertainty principle to estimate the ground state energy of atomic hydrogen and compare the results from the Bohr model.

∗ Hint: †R

volume of sphere = N · volume of hydrogen.

=N

1

/3

r0

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MODERN PHYSICS

Solution a)

The potential (energy) of a Hydrogen atom is given in its most general form by V =−

1 e2 4πε0 r

(5.3.125)

While the total energy of the system (kinetic plus potential) is E =T + V =

me v2 1 e2 − 2 4πε0 r

(5.3.126)

Assuming that we have a circular orbit, we may use the following formula for centripetal force: mv2 = r

F =

1 e2 4πε0 r 2 | {z }

(5.3.127)

Coulomb Force

Solving (5.3.127) for v2 , v2 =

1 e2 4πε0 me r

And now substituting (5.3.128) back into (5.3.126),   me 1 e2 1 − E= 2 4πε0 me r 4πε0    1 1 e2 2 1 = − 2 4πε0 r 2 4πε0 1 e2 =− 8πε0 r

(5.3.128)

e2 r  e2 r (5.3.129)

One of Bohr’s key assumptions is that circular orbit had quantized angular momentum. Mathematically, L = mvr = n~

(5.3.130)

Solving (5.3.130) for r, r=

n~ n2 ~2 −→ r 2 = 2 2 me v me v

(5.3.131)

Substituting our value for v2 from (5.3.128) back into (5.3.131), we have n2 ~2 me r 4πε0 2 m2e e 2 2 n ~ 4πε0 r= me e2

r2 =

(5.3.132)

Now substituting (5.3.132) into (5.3.129), E=−

1 2 me e2 e · 2 2 8πε0 n ~ 4πε0

(5.3.133)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

We can rearrange (5.3.133), which finally leads us to E =− b)

me e4 2

2 (4πε0 ) n2 ~2

(5.3.134)

For the ground state, n = 1, and using (5.3.128), and (5.3.132), we have   1/2 1 e2 me e2 v= 4πε0 me ~2 4πε0 " # 1/2 1 e4 = 2 (4πε0 ) ~2 =

1 e2 4πε0 ~

(5.3.135)

Plugging in the appropriate values into (5.3.135), 9

2

v = 98.987 × 10 N · m · C c)

−2





2 1.6023 × 10−19 C −→ v = 2.187 × 106 m · s−1 (1.055 × 10−34 J · s)

(5.3.136)

By default, the orbital period is given by Tn =

2πrn vn

(5.3.137)

Substituting our previous functional value of v into (5.3.137), as well as our value for r and also (5.3.137), we have  2 2   n ~ 4πε0 n~ Tn =2π 4πε · 0 me e2 e2 2

=

2π (4πε0 ) n3 ~3 me e4

(5.3.138)

Solving (5.3.138) for n, n=

Tn me e4

! 1/3

2π (4πε0 )3 ~3   1/ Tn me e4 K 2 3 = 2π~3

Substituting in the appropriate values into (5.3.139),


4
! 10−12 s 9.109 × 10−31 Kg 1.602 × 10−19 C 8.987 × 109 N · m2 · Kg−2 n= 3 2π (1.055 × 10−34 J · s)

(5.3.139)

(5.3.140)

Evaluating (5.3.144), we are left with, n = 18.73 −→ n = 19 d)

The energy of a Muon instead of an electron is found by replacing me in (5.3.136) by µ, which is the reduced mass of the atom, and is mathematically described by

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MODERN PHYSICS

µ=

mp · mµ mp + mµ

(5.3.141)

The relative mass of a proton to an electron is mp = 1838 me , so (5.3.141) becomes µ=

(1836 me ) (207 me ) = 186.0 me 1836 me + 207 me

(5.3.142)

With this new value of reduced mass, we must re-evaluate our results from (5.3.134), which leads us to E =− =− =

me e4 2 k 2n2 ~2

4
9.109 × 10−31 Kg 1.602 × 10−19 C 8.897 × 109 N · m2 · Kg−2 2n2 (1.055 × 10−34; J · s)

−13.6 eV n2

2

(5.3.143)

Applying our reduced mass from (5.3.145) to (5.3.146), −14.6 eVP n2 186.0 me me −13.6 eV · 186.0 = n2

E=

(5.3.144)

For the ground state, (5.3.144) becomes Egs(µ) = 2629 eV

(5.3.145)

And the photon energy for this case would just be E2 − E1 evaluated using (5.3.144), which yields   1 ∆E = E2 − E1 = −2529 eV − 1 −→ ∆E = 1896 eV 4 e)

From the uncertainty relation, we recall that ∆p∆r ∼ ~ −→ ∆p ∼

~ ∆r

Using these results, the total energy is E= =

p2 e2 −k 2m r ~2 2

2m (∆r)

−

ke2 r

(5.3.146)

We must minimize (5.3.146) in the usual fashion, ∂E(r) =0 ∂r ~2 ∂ 1 ∂ 1 − ke2 =0 2m ∂r r 2 ∂r r (5.3.147) Rearranging things a bit, (5.3.147) becomes

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

1 ~2 ·2· 2m Ke2 ~2 = kme2

r=

Substituting (5.3.148) back in to (5.3.146),     Kme2 ~2 K 2 m2 e4 2 − ke E= 2m ~4 ~2 me4 k 2 me4 k 2 = − 2 2~ ~2 4 2 me k = ~2

(5.3.148)

(5.3.149)

From (5.3.149), we can easily deduce that E=

me4 2 k 2~2

(5.3.150)

Which we can see is identical to (5.3.144).

Problem 18 The resistivity ρ of a metal due to scattering of the quasi-free electrons as per the Drude model is given by me vav (5.3.151) ρ= ne2 λ a)

Derive the expression above. What are the quantities n and λ, and explain qualitatively is ρ inversely proportional to them?

b)

The average electron velocity vav is very different in the classical and quantum-mechanical treatments. Explain the origin of the difference (why the quantum-mechanical velocity is much larger at room temperature, and is nearly independent of temperature).

c)

How is λ related to the cross-section σ for scattering of conduction electrons in the metal?

d)

Naively, one might take σ to be the geometrical area of the lattice ions. A more correct treatment gives a much smaller cross section which depends on temperature. Explain how this arises.

Solution a)

We first note that n is the number of electrons per unit volume, while λ is the mean free path. The electrons will move with a drift velocity, vd . In some time interval ∆t all charge that passes through cross sectional area A will be ∆Q = neAvd ∆t. We can find the current by taking the derivative, I=

dQ(t) d = [neAvd ∆t] dt dt

(5.3.152)

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MODERN PHYSICS

If we assume that the time interval is appropriately constant, (5.3.152) becomes I = neAvd

(5.3.153)

At this point, we must recall a number of seemingly arbitrary definitions. We of course recall Ohm’s law as V = IR, and the resistivity is given as R=

ρL A

(5.3.154)

We also recall that the electric field within the material is given by E=

V −→ V = EL L

(5.3.155)

Substituting (5.3.155) into Ohm’s Law, we have EL = IR We now substitute our expression for resistivity from (5.3.154) into (5.3.156),   ρL EA EL = I −→ I = A ρ

(5.3.156)

(5.3.157)

We now set (5.3.153) equal to (5.3.157), neAvd =

E EA −→ ρ = ρ nevd

(5.3.158)

In the same mantra as before, we recall that Newton’s second law says F = ma, F = eE, and v = aτ , so we can solve for v and arrive at eEτ m

(5.3.159)

m m E · −→ ρ = 2 ne eEτ ne τ

(5.3.160)

vd = Putting (5.3.159) back into (5.3.158), ρ=

But hold on there jack, τ is really the average time the particle goes without a collision, and we want our expression in terms of the mean free path. The formula which connects these two entities is simple: τ = λ/vavg . Substituting this back into (5.3.160), ρ= b)

m vavg mvavg · −→ ρ = ne2 λ ne2 λ

(5.3.161)

The average electron velocity, vavg is a very different animal when investigated in either the classical or quantum-mechanical regime. Classically, T =

mv2 3 = kB T 2 2

Playing with (5.3.162) a little bit, v=

r

3kB T m

(5.3.162)

(5.3.163)

Quantum-mechanically, only electrons that have energies within the kBT band contribute, so the average energy is the Fermi energy: Eavg = EF :

481

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

T = EF = Playing some more, vavg =

r

mv2 2

2EF m

(5.3.164)

(5.3.165)

Where we note that EF does not depend on T , so vavg does not depend on it either. Freaky! A qualitative summary, Ef = kB TF  kB Troom c)

−→

vavg (Quantum)  vavg (Classical)

To answer this we need only consider two elements. Firstly, we recognize that an electron entering on the left of the cylinder will be scattered before it reaches the right side of the same cylinder. We know that the volume of the cylinder is λσ, and that the density of scatterers is n = 1/λσ, so, our answer is then λ=

d)

1 nσ

The area swept out by the atomic vibrations perpendicular to the direction of propagating electrons is given by the cross section, σ. Increasing the temperature, T increases the vibrations. Therefore, as they both increase without limit, we have no choice but to say that σ, T → ∞ at the same rate.

Problem 19 The band theory of solids can be used to explain how a pure, intrinsic semiconductor (such as germanium) can become a p-type semiconductor. a)

How do conduction bands arise in a solid?

b)

Sketch an energy level diagram for electrons in a pure germanium crystal. Indicate both the conduction and valence bands and give order of magnitude estimates of the energies in your diagram.

Solution a)

This is a tricky question to answer, in the sense that the full explanation is a science in and of itself. Irregardless, we start with the idealized scenario of a repeating series of potentials (a “Dirac Comb”) that has a fixed height (V0 ) and each barrier is a fixed distance from one another a. The Barrier itself has a width, which we call b. These series of potentials go on forever, as to neglect any edging effects. At this point, we introduce our old friend the TISE in 1-D:   ~2 ∂ − + V (x) ψ(x) = Eψ(x) (5.3.166) 2m ∂x Whose realizable solutions end up begin

W. Erbsen

MODERN PHYSICS

ikx

ψ(x) = Ae

,

k=

p

2m (E − V (x)) ~

(5.3.167)

Our periodicity boundary condition also satisfies the TISE, but the solution will be shifted by some phase factor: ψ (x + N (a + b)) = eik(a+b) ψ(x)

(5.3.168)

Where (5.3.168) is lovingly referred to Bloch’s Theorem. Solving the TISE for repeating lattices using boundary conditions (eventually yields a transcidental equation, which can be used to visually determine the bands and band-gaps within a particular solid. The size, length, and location of the bands and bandgaps depends on the distance between the potentials, a, the width of the potentials, b, and the height of the potentials, V0 . From these three variables, we can plot the energy versus the wave number, and depending on the combination of the supplied variables, the plots may indicate a conductor, insulator, or a semiconductor.

b)

For a conductor, the Fermi Energy lies within a band and accordingly, the valence bands and conduction bands overlap. This implies that at T = 0, some of the electrons will be “free” to “conduct.” For an insulator, the Fermi energy lies either at the beginning, or the end of the gap (or somewhere inbetween). Subsequently, at T = 0 all orbitals below EF are filled, leaving no ”free“ electrons to “conduct.” For the case of a semi-conductor, the gaps are small enough for some electrons to jump across and become free. These particular characteristics can be controlled by way of introducing impurities to the pure semiconductor (known as “doping”). I can’t sketch here, but it should be fairly straightforward. Anyway, as far as an order of magnitude estimate for the gap spacing between the valence and conduction bands in a semi-conductor, the only way I can think to solve this problem is by experience, which tells me that the gap should be ∼ 1 eV

Problem 20 When sodium metal is illuminated with light of wavelength 4.20 × 102 nm, the stopping potential is found to be 0.64 V; when the wavelength is changed to 3.10 × 102 nm, the stopping potential becomes 1.69 V. Using only this data and the values of the speed of light and electronic charge, find the work function of sodium and the value of Planck’s constant.

Solution We first recall that the key equation for the photoelectric effect is Emax = hf − φ

(5.3.169)

The stopping potential is the potential that is required to stop an electron with some maximum kinetic energy Emax . Since the potential energy created by a potential difference V is U = qV , (5.3.169) becomes qV =hf − φ

483

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

=

hc −φ λ

(5.3.170)

Given that λ = 4.20 × 102 nm, and also that the stopping potential is V − 0.65 V, (5.3.170) becomes  hc V1 = 0.65 V qV1 = −φ (5.3.171) λ1 = 420 × 10−9 m λ1 And similarly, for λ = 3.10 × 102 nm and V = 1.69 V, we have  hc V2 = 1.69 V −φ qV2 = λ2 = 310 × 10−9 m λ2

(5.3.172)

Solving (5.3.171) and (5.3.172) for φ and setting them equal to one another leads us to hc hc − qV1 = − qV2 λ1 λ2   1 1 − hc =q (V1 − V2 ) λ1 λ2 q h = (V1 − V2 ) c



1 1 − λ1 λ2

−1

(5.3.173)

Substituting in the correct values into (5.3.173) leads us to  −1 1 1 1.602 × 10−19 C (0.65 V − 1.69 V) − h= 2.998 × 108 m · s−1 420 × 10−9 m 310 × 10−9 m −→ h = 6.578 × 10−34 J · s

(5.3.174)

Now, using either (5.3.171) or (5.3.172), we can calculate the work function φ. Using (5.3.172), φ=

hc − qV1 λ1

Substituting in the appropriate values,

6.578 × 10−34 J · s 2.998 × 108 m · s−1 φ= −→ (420 × 10−9 m)

φ = 3.661 × 10−19 J φ = 2.285 eV

Problem 21 Two identical photons are produced when a proton and an antiproton, both at rest, annihilate each other. What are the frequencies and corresponding wavelengths of the photons? Is it possible to produce a single photon in proton-antiproton annihilation? Why or why not?

Solution

W. Erbsen

MODERN PHYSICS

We first recall the following energy/momentum relations: Ep2 =m2 c4 + p2 c2

(5.3.175a)

Ep2

(5.3.175b)

2 4

=m c

We also recall that Ep = mc2 = 938.3 MeV. Now, recall from Einstein’s relation that Eγ h 938.3 × 106 eV = 4.136 × 10−15 eV · s =2.296 × 1023 Hz

Eγ = hf −→ f =

(5.3.176)

Now, we can say that λ=

c 3 × 108 m · s−1 = −→ λ = 1.322 × 10−15 m f 2.269 × 1023 Hz

It is not possible to produce a single photon in this particular interaction, since it would violate conservation of momentum (recall that the protons start off at rest )

Problem 22 A dust particle at rest relative to the sun is in equilibrium between the radiation pressure and gravitational attraction. What is the size of the dust particle? (For simplicity, you can assume that the dust particle in spherical)

Solution We first recall that the radiation pressure is defined by Prad =

hSi c

(5.3.177)

Where hSi is the time-averaged Poynting vector, aka the intensity. The power emitted by the sun is around P ≈ 3.8 × 1026 W, and so the intensity is I = hSi =

3.8 × 1026 W 4πr 2

(5.3.178)

Applying (5.3.178) to (5.3.177), Prad =

3.8 × 1026 W 4πr 2 c

Since pressure is defined as P = F/A, then (5.3.179) becomes

(5.3.179)

485

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Frad



3.8 × 1026 W 4πR2 = 4πr 2
3.8 × 1026 W 2 = R r2

Now, if we assume that the dust particles are perfectly reflective, then (5.3.180) becomes
3.8 × 1026 W 2 Frad =2 R r2

(5.3.180)

(5.3.181)

And we also recall that the gravitational force is given by Fgrav = G

mM r2

(5.3.182)

Equating (5.3.181) and (5.3.182),
3.8 × 1026 W 2 mM 2 R =G 2 r2 r Solving for R, this becomes R=

s

GmM 2 (3.8 × 1026 W)

(5.3.183)

Problem 23 What is the density of a neutron star? Evaluate the size (radius) of a neutron star as massive as the sun, −14 given that the radius 197 m, mass of the sun is 2 × 1030 kg. 79 Au nucleus is 7 × 10

Solution Density is of course given by ρ = m/v, and given the supplied constants, we have ρ=

197 × 1.67 × 10−27 Kg 3

4/3π (7 × 10−14 m)

−→ ρ = 2.194 × 1014 Kg · m−3

To find the radius of the neutron star, we recall that from the density, we can say that   1/ M 3M 3 ρ= −→ R = 4/3πR3 4πρ Plugging into (5.3.185) the appropriate values, R=

! 1/3
(3) 2 × 1030 Kg −→ R = 129589 m 4π (2.194 × 1014 Kg · m3 )

(5.3.184)

(5.3.185)

W. Erbsen

MODERN PHYSICS

Problem 24 A typical particle production process is

p + p −→ p + p + π 0

where mp = 938MeV/c2 , mass of π 0 = 135MeV/c2 . a)

In a standard accelerator one of the initial protons would be at rest. In this case what is the minimum laboratory energy for the moving particle so that the reaction listed above occurs?

b)

In colliding beam accelerators, such as the one proposed for the superconducting supercollider, the two initial particles are moving towards each other with equal speeds. Why do colliding beam accelerators have an energetic advantage over the type of accelerator from a)?

c)

What energy would be required of each proton in a colliding beams accelerator to produce a pion?

Solution We first recall the following: E 2 =p2 c2 + m0 c2 E =mc

2


2

(5.3.186a) (5.3.186b)

And that the invariant quantity is Center of mass frame invariant after collision


2 2

z }| { E 2 − p 2 c2

=

m0 c | {z }

Lab frame invariant before collision

Where in the center of mass frame (COM) the momentum is zero, so (5.3.187) becomes
2 m0 c2 = E 2

(5.3.187)

(5.3.188)

Let’s make a table:

Before • −→ • Σm a)

After •→

•→

2mp

•→

(5.3.189)

2mp + mπ

If we choose not to use the COM frame, then the momentum is conserved (let’s do it this way). We first apply conservation of energy to our system: Ei = Ef

(5.3.190)

Squaring (5.3.190) and substituting in (5.3.186a), p2i c2 + m2i c4 = p2f c2 + m2f c4 Conservation of momentum says that pi = pf , and so (5.3.191) becomes m2i c4 =m2f c4

(5.3.191)

487

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

2mp c2


2

 2 = (2mp + mπ ) c2

4m2p c4 =4m2p c4 + 2mp mπ c4 + 2mp mπ c4 + m2π c4 0 =4mp mπ c4 + m2π c4   mπ c2 + 2 0 =mπ c2 2mp c2   mπ 0 =mπ c2 +2 2mp

(5.3.192)

We can now say that this is the threshold energy, which is the minimum possible energy that will allow this reaction to happen. The RHS of (5.3.192) is this so-called threshold energy:   mπ c2 ET = mπ c2 + 2 2mp c2 Substituting in the appropriate values, we have   135 MeV ET = 135 MeV + 1 −→ ET = 279.7 MeV 2 · 938 MeV

(5.3.193)

And to find the total energy, we combine the results of (5.3.193) plus the rest mass energy of the proton: Etot = 279.7 MeV + 938 MeV −→ Etot = 1217.7 MeV b)

Since in the case of a), all the energy needed to create the π o is carried by the first proton, however if both were to move towards one another, then the energy would need to be much less in order to reach the threshold energy.

c)

If they are moving towards one another, we use the same logic as before - we want the threshold energy, that is the energy that is just barely enough to complete the reaction. In this case, we care only about the creation of three particles: two Protons, and one Pion. Therefore, each incident proton must have an initial energy equal to Ep =

1 (2mp + mπ ) 2

Plugging in the appropriate number, Ep =

1 [2 (938 MeV) + 135 MeV] −→ Ep = 1005.5 MeV 2

Problem 25 In a muonic hydrogen atom, the electron is replaced by a muon, which is 207 times heavier. a)

What is the wavelength of the n = 2 to n = 1 transition in Muonic hydrogen? In what region of the electromagnetic spectrum does this wavelength belong?

b)

How would the fine structure and Zeeman effects differ from those in ordinary atomic hydrogen?

W. Erbsen

MODERN PHYSICS

Solution a)

We first recall the Rydberg formula, which reads 1 1 − 2 n2f ni

1 =R λ

!

(5.3.194)

While the Rydberg constant is given by me R= 4πc~3 R ∼C · me



e2 4πε0

2

= 1.097 × 107 m−1

(5.3.195)

Where me is the electron mass. For Muonic Hydrogen, we replace me with µ/me , where µ is the reduced mass, where mµ = 207 me ,

mp = 1836 me

So, the reduced mass is µ=

(207 me ) (1836 me ) = 186.0 me (207 me ) + (1836 me )

(5.3.196)

Substituting (5.3.196) into (5.3.194), 1 µ =R· λ me

1 1 − 2 n2f ni

!

Substituting in the appropriate values, (5.3.197) becomes   
1 me 1 = 1.097 × 107 m−1 186.0 −1 λ me 4 =1.530 × 109 m−1

(5.3.197)

(5.3.198)

From which it is easy to see that λ = 6.535 × 10−10 m Which are x-rays. b)

The only appreciable difference between Hydrogen and Muonic Hydrogen is the difference in the masses of the electron and the Muon, respectively. Therefore, we must find the splitting as a function of mass for both fine structure and Zeeman effects.

The fine structure is of order: Efs ∼ α4 me2 −→ Efs ∼ m While the Zeeman effect is of order: Ez ∼ me ·

e~ 1 · B −→ Ez ∼ 2m m

489

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Problem 26 For an intrinsic semiconductor, the number of electrons per unit volume in the conduction band and the number of holes in the valence band are given by   (EC − EF ) n =NC exp − (5.3.199a) kB T   (EF − EV ) (5.3.199b) p =NV exp − kB T Where 3

2(2πmkB T ) /2 NC = NV = h3

(5.3.200)

a)

Find the position of the Fermi energy in an intrinsic semiconductor.

b)

Show that the product of the number of electrons in the conduction band and the number of holes in the valence band depends only on temperature and the gap energy, Eg .

c)

Discuss how one can use the temperature (T ) dependence of the conductivity σ to approximately measure the gap energy provided that T is not too high.

Solution We first recall that the Fermi Energy is defined as the highest occupied state at T = 0 K. We also recall that the Intrinsic Fermi Energy (Ei ) is defined as the Fermi Energy in the absence of doping. a)

In an intrinsic semiconductor, the number of electrons is equal to the number of holes (n = p), since the thermal excitation of an electron leaves behind a hole in the valence band. The very first thing that we wish to do is multiply (5.3.199a) and (5.3.199b):   (Ec − Ev ) (5.3.201) np = Nc Nv exp − kB T And furthermore, from (5.3.200), we may deduce that   3/ 2πme kB T 2 NC = 2 h2  3/  2πmh kB T 2 NV = 2 h2

(5.3.202a) (5.3.202b)

Where mh is the electron effective mass near the top of the valence band. Using (5.3.202a)(5.3.202b), we can now rewrite (5.3.199a) and (5.3.199b), respectively:   3/2   2πme kB T (Ec − EF ) n =2 · exp − (5.3.203a) h2 kB T   3/   2πmh kB T 2 (EF − Ev ) p =2 · exp − (5.3.203b) h2 kB T Multiplying (5.3.203a) and (5.3.203b) leads us to

W. Erbsen

MODERN PHYSICS

np = 4



2πkBT h2

3

(me mh )

3

/2

  (Ec − Ev ) exp − kB T

(5.3.204)

The intrinsic carrier concentration depends exponentially on (Ec − EV ) /2kBT . At the Fermi Energy, n = p because the number of electrons and the number of holes are created in pairs for intrinsic semiconductors, as is what we have here. So,     (Ec − Ef ) (EF − Ev ) Nc exp − =Nv exp − kB T kB T Since Nc and Nv are equal in this case, this becomes (EF − Ev ) (Ec − Ef ) =− kB T kB T Ec − Ef = − EF − Ev From which it is easy to gather that EF =

Ev + Ec 2

In other words, the Fermi Energy is located exactly inbetween the valence band and the conduction band. b)

From the results of the previous problem, we found that np = 4

c)



2πkBT h2

3

(mh me )

3

/2



Eg exp − kB T



(5.3.205)

Where Eg = Ec − Ev is the gap energy.

We recall from Ohm’s law that

σ=

j , E

Vd =

j nq

Combining these, we find that σ=

nqv E

(5.3.206)

nq 2 v F

(5.3.207)

Also recalling that F = qE, (5.3.206) becomes σ=

Now, we recall that v = aτ , where τ is the time between collisions. Combining this with Newton’s second law, F = ma, we have F = vm/τ , and (5.3.207) becomes σ= Recalling from before that σ =σc + σv

nq 2 τ m

(5.3.208)

491

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

q2τ = m

(   3/     3/  ) 2πme kB T 2 (Ec − EF ) 2πmh kB T 2 (EF − Ev ) 2 · exp − +2 · exp − h2 kB T h2 kB T

From this we can easily say that 

(Ev + Ec ) σ ∼ C exp − kB T



So, as long as the temperature is low, the gap will dominate this equation, and we can solve for the gap as needed.

Problem 27 The potential energy between two atoms in a molecule can often be described rather well by the Lennard-Jones potential, which can be written    a 6  a 12 U(r) = U0 −2 (5.3.209) r r

Where U0 and a are constants. a)

Find the interatomic separation r0 in terms of a for which the potential energy is minimum.

b)

Find the corresponding value of Umin .

c)

Use the figure below to obtain numerical values for r0 and U0 for the H2 molecule. Express your answer in nanometers and electron volts.

d)

Make a plot of the potential energy U (r) versus the internuclear separation r for the H2 molecules. Plot each term separately, together with the total U (r).

Solution a)

To find the value of the interatomic separation r0 when the potential energy is at a minimum, we must minimize (5.3.209). The general prescription for minimizing any function is to take the derivative with respect to the minimization variable and set it equal to zero, solving for the appropriate minimized value. In our case, ∂U(r) =0 (5.3.210) ∂r r=r0 So, applying (5.3.210) to (5.3.209), we have  ∂  a 12

 a 6  −2 =0 ∂r r r     ∂ 1 ∂ 1 6 a12 − 2a =0 ∂r r 12 ∂r r 6

W. Erbsen

MODERN PHYSICS

    12 6 a12 − 13 − 2a6 − 7 = 0 r r 12 a6 a12 13 = 12 6 r r a6 =r 6

(5.3.211)

Applying r = r0 at (5.3.211), we have r0 = a b)

The corresponding value of Umin can be found by applying (5.3.212) to (5.3.209): "   6 # 12 a a U(r0 ) =U0 −2 r0 r0    a 6  a 12 =U0 −2 a a

(5.3.212)

(5.3.213)

Which leads us to Umin = −U0 c)

(5.3.214)

The values for r0 and U0 can be obtained quite simply from the figure. The value of r0 is the horizontal position of the deepest point in the potential, while the minimal energy is simply the depth from the bottom:

493

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

r0 = 0.074 nm Umin = −|27.2 eV − 4.5eV| −→ Umin = −31.7 eV

Problem 28 The measured conductivity of copper at room temperature is 5.88 × 107 Ω−1 · m−1 , its Fermi energy is 7.03eV, density is 8.96g m/cm3 and molar mass is 63.5gm/mole. a)

Calculate the density of free electrons in copper.

b)

Calculate the average time between collisions of the conduction electrons.

c)

Calculate the mean free path of electrons in copper.

d)

Calculate the smallest possible deBroglie wavelength of electrons in copper at T = 0K. How does this value compare with the atomic separation in copper?

Solution a)

Copper has only one free (valence) electron. According to the Free Electron Approximation, all the conductivity is due to this one electron per copper atom. So, if we wanted to calculate the density of free electrons in copper, we recall that ρ = M/V , and since we have only one free electron per atom, we find the density as ρ  Cu n= · NA m g  g −1 atoms =8.96 · 63.5 · · 6.022 × 1023 3 cm  mol  mol g 1 mol 23 atoms =8.96 · · · 6.022 × 10 (5.3.215) cm3 63.5 g mol From (5.3.215), it is easy to see that we are left simply with n = 8.497 × 1022

b)

atoms cm3

(5.3.216)

In order to calculate the average time between collisions of the conduction electrons, we first must consider the kinetic energy of these electrons, and from that find the average velocity: r mv2 2E E= −→ v = (5.3.217) 2 m Where we take E in (5.3.217) to be the Fermi Energy, EF , which is given in the prompt. Accordingly, we can evaluate v in (5.3.217) as s 2 (7.03 eV) m 2 v= · (3 × 108 m/s) −→ v = 1.574 × 106 (5.3.218) (0.511 × 106 eV) s

W. Erbsen

MODERN PHYSICS

We now recall that the conductivity is defined as σ=

ne2 d mv

(5.3.219)

Where we recognize that the relationship between velocity, the time between collisions and the distance traveled between collisions is v = d/τ → d/v = τ . Substituting this into (5.3.219),   ne2 d σ= · m v ne2 ·τ (5.3.220) = m Solving (5.3.220) for τ yeilds mσ τ= 2 ne

9.109 × 10−31 Kg 5.880 × 107 Ω−1 · m−1 = 3 2 (8.497 × 1022 atoms · cm−3 ) (100 cm · m−1 ) (1.602 × 10−19 C)

(5.3.221)

Evaluating all the goodness in (5.3.221) leaves us with τ = 2.462 × 10−14 s c)

The mean free path is found from the velocity found in part b) as d =v · τ

= 1.574 × 106 m · s−1

d)

(5.3.222)





2.462 × 10−14 s −→ d = 38.75 nm

(5.3.223)

The deBroglie wavelength can be found rather straightforwardly using (5.3.218) as λ=

h 6.626 × 10−34 J · s = p (9.109 × 10−31 Kg) (1.574 × 106 m · s−1 )

(5.3.224)

From (5.3.224) we see that λ = 4.621 × 10−10 m

(5.3.225)

The atomic separation of copper can be found from the density, ρ, which is given in the problem:   1/3 m m m ρ= = 3 −→ L = (5.3.226) v L ρ Substituting in the appropriate values into (5.3.226) yields "

# 63.5 g · mol−1 6.022 × 1022 atoms · mol−1 L= −→ L = 2.275 × 10−10 m 3 (8.96 g · cm3 ) (100 cm · m−1 )

Problem 29 The proper mean lifetime of π Mesons (Pions) is 2.6 × 10−8 s. If a beam of such particles has a speed of 0.9 c:

495

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

a)

What would their mean life be as measured in the laboratory?

b)

How far would they travel (on the average) before they decay?

c)

What would your answer be to part b) if you neglect time dilation?

d)

What is the interval in space-time between creation of a typical pion and its decay?

Solution We first recall that the equation for time-dilation is T = γT0

(5.3.227)

Where the Lorentz factor, γ, is equal to

a)

1 γ= p 1 − v2 /c2

In the lab frame, using (5.3.227)-(5.3.228) we have

2.6 × 10−8 s T = p −→ T = 5.965 × 10−8 s 1 − (0.9)2

c)

In the lab frame, we know that ∆x = v · T , and using (5.3.229), we have

∆x = (0.9) 3 × 108 m · s−1 5.965 × 10−8 s −→ ∆x = 16.105 m

d)

The Space-time interval is defined as

b)

(5.3.228)

(5.3.229)

(5.3.230)

If time-dilation is neglected, we can find ∆x as

∆x = (0.9) 3 × 108 m · s−1 2.6 × 10−8 s −→ ∆x = 7.020 m ∆s2 = −∆x2 + c2 ∆T 2

(5.3.231)

Substituting in the appropriate values into (5.3.231),

2 2 ∆s2 = 3 × 108 m · s−1 5.965 × 10−8 s − (16.105 m) −→ ∆s2 = 60.86 m2

So, the space-time interval is

∆s =

√ 60.86 m2 −→ ∆s = 7.801 m

Problem 30 Four identical, non-interacting particles are placed in a one dimensional box of length L. Find the lowest total energy of the system and list the quantum numbers of all occupied states (including ms ) if: a)

The particles are electrons

W. Erbsen

MODERN PHYSICS

b)

The particles are neutral Pions, π 0 .

c)

Two of the particles are protons and the other two are neutrons.

Solution We first recall that the 1-D, time independent SE reads −

~2 ∂ 2 ψ(x) = Eψ(x) 2m ∂x2

(5.3.232)

Solving (5.3.232) with the appropriate boundary conditions, we find that the corresponding Eigenfunction and Eigenvalues are, respectively r  nπx  2 ψ(x) = sin (5.3.233a) L L n2 ~2 π 2 (5.3.233b) En = 2mL2 a)

Since electrons are Fermions, no two electrons can be described by the same set of quantum numbers, as necessitated by the Pauli Exclusion Principle. Therefore, the lowest occupied states are: |n, ms i = |1, 1/2 i, |1, −1/2 i, |2, 1/2 i, |2, −1/2 i And from (5.3.233b), the lowest energy corresponding to this state is Egs =

b)

5~2 π 2 ~2 π 2 [1 + 1 + 4 + 4] −→ Egs = 2 2mL me L2

The neutral pion has a spin of zero, making it a Boson. Since the Pauli Exclusion principle does not apply to Bosons, all the particles may lie in the lowest state: |n, ms i = |1, 0i, |1, 0i, |1, 0i, |1, 0i While the energy is, following (5.3.233b), Egs =

c)

~2 π 2 2~2 π 2 [1 + 1 + 1 + 1] −→ Egs = 2 2mL mπ L2

Both Protons and Neutrons are Fermions, but they are distinguishable from one another. Therefore, the lowest state is then |n, ms i = |1, 1/2 i, |1, −1/2 i, |1, 1/2 i, |1, −1/2 i And the Eigenenergies are Egs

~2 π 2 ~2 π 2 ~2 π 2 E = = [1 + 1] + [1 + 1] −→ gs 2mp L2 2mn L2 L2



1 1 + mp mn



497

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

5.4

Statistical Mechanics

Problem 1 a)

Define the chemical potential µ. Show that two systems are in diffusive equilibrium if µ1 = µ2 . You may start with F = F1 + F2 (free energy) and use the fact that µ1 = µ2 should minimize F .

b)

Now consider a 1 − D gas of length L and number of distinguishable and non-interacting particles N . Assume that temperature T is high enough that we are in the classical regime. Find the chemical potential µ in terms of T and other parameters of the problem.

c)

Now find the free energy F and the specific heat at constant volume CV .

Solution a)

The chemical potential, µ, can be thought of as a number that allows us to quantify whether or not two (or more) systems are in diffusive equilibrium. If they are equal, then there will be no (net) flux of particles moving from one system to another. If they are not equal, then particles will tend to flow from a region of high chemical potential to a region of low chemical potential. We also recall that the Helmholtz Free Energy (F ) measures the amount of “useful” work that is extractable from a system at constant temperature and volume. The Free Energy has several important characteristics: i) ii)

At equilibrium, the Free Energy is at a minimum An important form of the second law of thermodynamics is U = F + TS Where U is the internal energy, and S is the final entropy. At diffusive equilibrium, we know that dF = 0, and the number of particles in each system remains constant. From this, we may also deduce that the total number of particles remains constant. Quantitatively, N = N1 + N2 In fact, whether we are in diffusive equilibrium or not, the total number of particles must remain constant! That is, 0 = dN1 + dN2 −→ dN1 = −dN2

(5.4.1)

We can also think of the free energy of each system, as we did with the particle number previously. The total free energy would be described by F = F1 + F2

(5.4.2)

And when we are in diffusive equilibrium, according to our first stipulation, F must be minimized with respect to the particle number:

W. Erbsen

STATISTICAL MECHANICS

dF =0 dN

(5.4.3)

Combining (5.4.1) and (5.4.2), dF =

∂F1 ∂F2 dN1 + dN2 ∂N1 ∂N2

(5.4.4)

Setting (5.4.4) equal to zero, as demonstrated in (5.4.3), we have ∂F2 ∂F1 dN1 = − dN2 ∂N1 ∂N2

(5.4.5)

Applying (5.4.1) once again to (5.4.5), we are left with ∂F2 ∂F1 ∂F2 ∂F1 dN1 = − (−dN1 ) −→ = ∂N1 ∂N2 ∂N1 ∂N2 We now recall that the chemical potential is defined by   ∂F µ= ∂N V,T

(5.4.6)

(5.4.7)

Applying the definition of chemical potential from (5.4.7) to our result (5.4.6), it is easy to see that at equilibrium, µ1 = µ2 b)

We are tasked with finding the chemical potential for distinguishable, non-interacting particles in the classical regime for a 1-D gas. As we do in all of statistical mechanics, we begin by finding the (single particle) partition function:   X Ei Z1 = exp − (5.4.8) kB T i

For a classical gas, the energy will come in the form of the kinetic energy, which is E = p2 /(2m). Furthermore, within the classical regime, we may take our energy distribution to be continuous, and as such may be integrated in the following way:   Z Z 1 p2 Z1 = 3 exp − d3 p d3 x (5.4.9) h 2mkB T For a 1-D gas, as is the case here, (5.4.9) becomes   Z Z L 1 ∞ p2 Z1 = exp − dp dx h −∞ 2mkB T 0 i 1 hp = · 2πmkB T · [L] h Lp = 2πmkB T h

Within the classical approximation, the full partition function can be approximated by ZN ∼ = Z1N Applying this to the single particle partition function of (5.4.10), we have

(5.4.10)

499

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

ZN =

 p N L 2πmkB T h

(5.4.11)

We now recall that the Free energy takes the form

F = −kB T log [ZN ]

(5.4.12)

Substituting (5.4.11) into (5.4.12), ( N ) Lp F = − kB T log 2πmkB T h  p  L = − N kBT log 2πmkB T h

(5.4.13)

Now, we recall that the chemical potential, given by (5.4.7), can be calculated from (5.4.13) as   p  ∂ L −N kB T log µ= 2πmkB T (5.4.14) ∂N h The answer is then

 p  L µ = −kB T log 2πmkB T h c)

From (5.4.13), F was found to be F = −N kB T log

 p  L 2πmkB T h

To find the specific heat capacity at constant volume, we have to get back to work. Recall that this quantity is defined as   ∂U CV = (5.4.15) ∂T V We also recall that the internal energy is given by U =−

∂ log (ZN ) ∂β

Now substituting (5.4.11) into (5.4.16), ( N ) Lp 2πmkB T h  p  L log 2πmkB T h s " # 2πmL2 log βh2 (  1/ ) 2πmL2 2 log βh2

∂ U =− log ∂β =−N

∂ ∂β

=−N

∂ ∂β

=−N

∂ ∂β

(5.4.16)

W. Erbsen

STATISTICAL MECHANICS

N 2 N =− 2 =−

  ∂ 2πmL2 log ∂β βh2   ∂ 2πmL2 1 log 2 ∂β {z } β | h

(5.4.17)

Garbage

The derivative of (5.4.17) comes out being very simple, N 1 · 2 β N kB T = 2

U=

(5.4.18)

Putting our cleaned up expression back into (5.4.15), we find CV :   ∂ N kB T N kB CV = −→ CV = ∂T 2 2

Problem 2 Consider a particle in a 1-dimensional box of side L with available energy levels (in the presence of a magnetic field B along ˆ z) ~2  π 2 εn = γ 2 n2 + m · B; γ2 = (5.4.19) 2m L ˆ Where n is an integer greater than zero and m = ±mz k. a)

What is the partition function for this particle?

b)

What is the partition function for N classical, identical, noninteracting particles in the same box in the presence of a magnetic field? Hence find an expression for the internal energy U , and the Helmholtz free energy F .

c)

What is the equilibrium pressure and magnetization of this gas?

Solution From the information supplied, En = γ 2 n2 + m · B ~  π 2 γ2 = 2m L 2

a)

    

→ En =

n2 ~2 π 2 +m·B 2mL2

(5.4.20)

The first step in this problem, as is in most problems in this field, is to find the partition function. Let’s ride this pony:   X X En Z1 = exp − = exp [−En β] (5.4.21) kB T n n

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Substituting (5.4.20) into (5.4.21),   2 2 2   n ~ π Z1 = exp − +m·B β 2mL2 n   2 2 2 X n ~ π β − m · Bβ = exp − 2mL2 n  2 2 2  X n ~ π = exp − β − ±m B β z 0 2mL2 n X

(5.4.22)

I changed m · B → ±mz B0 . Because of the ± in front of the spin quantum number, we need to split (5.4.22) into two distinct sums, each addressing its respective parity. With this, (5.4.22) is now  2 2 2   2 2 2  X n ~ π n ~ π β − m B β + exp − β + m B β (5.4.23) Z1 = exp − z 0 z 0 2mL2 2mL2 n At this point, we turn to our calculus loving friends and with a tote on their pipe I decide to go the way of the integral. And it was so, the sum was transferred into an integral, and (5.4.23) becomes  2 2 2   2 2 2  Z ∞ n ~ π n ~ π Z1 = exp − β − mz B0 β + exp − β + mz B0 β dn 2mL2 2mL2 0  2 2 2   2 2 2   Z ∞ n ~ π n ~ π β exp [−m B β] + exp − β exp [m B β] dn = exp − z 0 z 0 2mL2 2mL2 0  2 2 2   2 2 2  Z ∞ Z ∞ n ~ π n ~ π β exp − β dn = exp [−mz B0 β] exp − dn + exp [m B β] z 0 2 2mL 2mL2 0 0     Z ∞ Z ∞ β~2 π 2 2 β~2 π 2 2 = exp [−mz B0 β] exp − n dn + exp [m B β] exp − n dn z 0 2mL2 2mL2 0 0   Z ∞ β~2 π 2 2 = {exp [−mz B0 β] + exp [mz B0 β]} exp − n dn 2mL2 0   Z ∞ β~2 π 2 2 n dn =2 cosh [mz B0 β] exp − 2mL2 0 " s # 1 2πmL2 =2 cosh [mz B0 β] (5.4.24) 2 β~2 From (5.4.24), it is easy to see that the partition function becomes s 2πmL2 Z1 = cosh [mz B0 β] β~2 b)

(5.4.25)

But what if the partition function is to be applied to a classical, indistinguishable, non-interacting system? From this we must find the internal energy U , as well as the free energy F . We first recall that since we have indistinguishable particles, the full partition function can be found from the single-particle partition function as ZN ZN ∼ = 1 N! Rearranging the single particle partition function (5.4.25) and applying (5.4.26), we have

(5.4.26)

W. Erbsen

STATISTICAL MECHANICS

ZN

s " #N 2πmL2 1 cosh [mz B0 β] = N! β~2

(5.4.27)

Now, the free energy is F = −kB T log [ZN ]

(5.4.28)

And plugging (5.4.27) into (5.4.28),  "  s #N    2 2πmL  − log [N !] F = −kB T log  cosh [mz B0 β]   β~2

(5.4.29)

At this point, we recall Stirling’s Approximation:

log [N !] ∼ = N log [N ] − N Applying this to (5.4.29),  

 #N   2 2πmL  − (N log [N ] − N ) F = − kB T log  cosh [mz B0 β]   β~2 s ( " # ) 2πmL2 = − kB T N log cosh [mz B0 β] − N log [N ] + N β~2 s ( " # ) 2πmL2 = − N kB T log cosh [mz B0 β] − log [N ] + 1 β~2 "

s

Simplifying (5.4.30) further leads us to ( F = −N kB T

"

cosh [mz B0 β] log N

s

2πmL2 β~2

#

+1

)

(5.4.30)

(5.4.31)

To find the internal energy, U , we first recall that U =−

∂ log [ZN ] ∂β

Substituting the partition function from (5.4.27) into (5.4.32),  s " #N  2 ∂ 1 2πmL  U =− log  cosh [mz B0 β] ∂β N! β~2 s " " ## 1 ∂ 2πmL2 =−N log cosh [mz B0 β] 1 ∂β β~2 N ! /N r   ∂ 1 2πmL2 cosh [mz B0 β] √ =−N log 1 2 /N ∂β β |N ! {z ~ } Garbage

Proceeding,

(5.4.32)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

  ∂ cosh [mz B0 β] √ U =−N log ∂β β n hp io ∂ =−N log [cosh [mz B0 β]] − log β ∂β   ∂ 1 ∂ =−N log [cosh [mz B0 β]] − log [β] ∂β 2 ∂β Playing around with (5.4.33) a bit we can see that we have our answer:   1 U = −N mz B0 tanh [mz B0 β] − 2β c)

(5.4.33)

(5.4.34)

In order to find the equilibrium pressure, we first recall that P =−

∂F ∂F ⇒− ∂V ∂L

(5.4.35)

Where I took V → L since we are only in 1-D. Anyway, we can substitute our value for the Free Energy from (5.4.31) into (5.4.35): s " ( " # )# ∂ cosh [mz B0 β] 2πmL2 P =− −N kB T log +1 ∂L N β~2 s # # " " ∂ cosh [mz B0 β] 2πmL2 =N kB T +1 log ∂L N β~2 r     ∂ cosh [mz B0 β] 2πm =N kB T (5.4.36) log L + 1 ∂L N β~2 | {z } Garbage

Eww smelly garbage yuck yuck. Anyway, ignoring the icky garbage, (5.4.36) becomes P =N kB T

N kB T ∂ [L] −→ P = ∂L L

(5.4.37)

The Magnetization, however, is a different story entirely. The relationship we are interested in is M =−

∂F ∂B

Substituting our previous value of F into (5.4.38) yields s " ( " #)# ∂ cosh [mz B0 β] 2πmL2 M =− −N kB T log +1 ∂B N β~2 s " # ∂ cosh [mz B0 β] 2πmL2 =N kB T log +1 ∂B N β~2 s   ∂ 1 2πmL2 =N kB T log cosh [mz B0 β] +1 ∂B N β~2   ∂ M =N kB T log cosh [mz B0 β] ∂B

(5.4.38)

W. Erbsen

STATISTICAL MECHANICS

=N kB T [mβ tanh [mz B0 β]] Which finally leaves us with M = N mz tanh [mz B0 β]

Problem 4 a)

From the condition that two coexisting phases of a given chemical substance must have the same chemical ptential µ1 = µ2 , derive the Claussius-Clayperon equation S2 − S1 dP = dT V2 − V1

b)

(5.4.39)

Drawn below is the phase diagram for He4 . Carefully explain why the slope of the liquid-solid phase boundary is zero at T = 0.

Solution a)

We first recall that a common form of the first law of thermodynamics is dU = T dS − P dV

(5.4.40)

While the Helmholtz and Gibbs free energy are given, respectively, by F =U − T S G =F + P V

(5.4.41a) (5.4.41b)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

If we make the substitution (5.4.41a) into (5.4.41b), we have G = U − TS + PV

(5.4.42)

Taking (5.4.42) to the differential limit, we have dG =dU − T dS − S dT + P dV + V dP =dU − (T dS − P dV ) − s dT + V dP =dU − dU − S dT + V dP = − S dT + V dP

(5.4.43)

Where I substituted in (5.4.40) in the second to last step arriving at (5.4.43). We now recall that the chemical potential is given by   ∂G ∂F µ= ⇒ (5.4.44) ∂N V,T ∂N We can now play around with (5.4.44) as follows µ=

∂G G −→ dµ = −→ dG = N dµ ∂N N

(5.4.45)

Applying (5.4.45) to (5.4.43), n dµ =V dP − S dT V S dµ = dP − dT N N

(5.4.46)

We now temporarily recall that since we are at diffusive equilibrium, then µ1 = µ2 and furthermore, dµ1 = dµ2 . Ergo, (5.4.46) is transformed into V1 S1 V2 S2 dP − dT = dP − dT N N N N V1 V2 S1 S2 dP − dP = dT − dT N N N N dP (V1 − V2 ) =dT (S1 − S2 )

(5.4.47)

From (5.4.47), it is most easily seen that dP S1 − S2 = dT V1 − V2 b)

(5.4.48)

We are given a phase diagram for He4 , which we immediately recognize as a Boson. This is especially interesting because Bosons do not obey the Pauli Exclusion Principle, and can therefore all occupy the same state at an arbitrarily low temperature. The multiplicity, that is, the number of states available to a system without changing anything, is only 1. e.g., at low temperatures, there is only 1 way that the system can be configured (all in the ground state), and subsequently the multiplicity is Ω = 1. This is particularly interesting, since He4 has a multiplicity of 1, then the entropy is 0: S = kBT log [Ω] −→ S = kB T log [1] −→ S = 0 The Claussius-Clayperon equation that we derived earlier in (5.4.48) describes the slope of the phase boundaries. In the case that we have low temperatures (approaching zero), where the multiplicity,

W. Erbsen

STATISTICAL MECHANICS

Ω, is 1, while the entropy, S, is 0. Plugging this in to (5.4.48), we get dP 0−0 dP = −→ =0 dT V1 − V2 dT The Claussius-Clayperon equation can be thought of as the “slope equation” for phase diagrams. When the result is zero, that implies that there is no slope, and so you get a straight line.

Problem 5 Consider an ideal gas of diatomic molecules. The rotational motion is quantized according to εj = j(j + 1)ε0 , j = 0, 1, 2, ... The multiplicity of each rotational level is gj = 2j + 1. a)

Find the rotational part of the partition function Z for one molecule.

b)

Convert the sum (for Z) to an integral and evaluate the integral for kT  ε0 . From this find an expression for the specific heat capacity at constant volume at low temperatures.

Solution a)

For a system consisting of degeneracy, the form of the partition function that we will employ is   X Ej Z1 = gj exp − (5.4.49) kB T j

In our case, we have gj = 2j + 1 ,

εj = j(j + 1) ε0

Applying both of these to our single-particle partition function in (5.4.49), we have   X j(j + 1) ε0 Z1 = (2j + 1) exp − kB T

(5.4.50)

j

b)

Converting the sum in (5.4.50) into an integral:   Z ∞ j(j + 1) ε0 Z1 = (2j + 1) exp − dj kB T 0 u = j(j + 1) = j 2 + j du = 2j + 1 dj And using this du substitution, we can transform (5.4.51) into   Z ∞ uε0 Z1 = exp − du kB T 0    ∞ kB T uε0 =− exp − ε0 k B T 0

(5.4.51)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

From which it is easy to deduce that Z1 =

kB T ε0

(5.4.52)

We need to move away from the single-particle partition function, and since we are dealing indistinguishable entities: ZN ZN ∼ = 1 N!

(5.4.53)

Substituting the single particle partition function from (5.4.52) into (5.4.53),  N 1 kB T ZN = N! ε0 Now hold on to yer britches, and recall that   ∂U , CV = ∂T V

U =−

(5.4.54)

∂ log [ZN ] ∂β

First, let’s evaluate the internal energy U , by substituting into (5.4.54) this expression: "  N # ∂ 1 1 U =− log ∂β N ! βε0    ∂ 1 1 =−N log ∂β N ! 1/N βε0     1 1 1 ∂ =−N log 1/ N ∂β ε0 β N! {z } |

(5.4.55)

And (5.4.55) becomes



(5.4.56)

N −→ u = N kB T β

(5.4.57)

U =−N

∂ log ∂β



1 β

Carrying out the simple derivative in (5.4.56) leads us to U =

To find the specific heat, we substitute the internal energy (5.4.57) into the appropriate formula CV =

∂ (nkB T ) −→ CV = N kB ∂T

For an ideal gas with three degrees of freedom, we find that U=

3 N kB T 2

And so the specific heat at constant volume becomes CV =

5 N kB 2

W. Erbsen

STATISTICAL MECHANICS

Problem 6 a)

Write down the Grand Partition Function (Gibbs sum), Ω. Define all quantities used and explain what the summations are over.

b)

Show that the average number of particles in a Grand Canonical Ensemble is given by hN i = τ

∂ log Ω ∂µ

(5.4.58)

where τ = kT and µ is the chemical potential. c)

Now consider a dilute gas of hydrogen atoms. Assume each atom may have the following states: State Ground Positive ion Negative ion

No. of Electrons 1 0 2

Energy 0 −∆ +δ

Find the condition that the average number of electrons per atom is 1.

Solution a)

We recall that within the framework of the Grand Canonical Ensemble, the system in question is in both thermal and diffusive equilibrium. The sum itself takes the form   ∞ X X N µ − Es,N Ω (µ, T ) = exp (5.4.59) kB T N=0 s(N)

Ω = The Gibbs Sum µ = The Chemical Potential N = The Number of Particles Es,N = The energy of N-particles in S-states b)

We recall that, in general, the average number of particles is X hN i = N · P (N, Es,N )

(5.4.60)

s,N

Where the probability function is given by

  1 N µ − Es,N P (N, Es,N ) = · exp Ω (µ, T ) kB T

(5.4.61)

And substituting the probability function (5.4.61) back into (5.4.60) to find the average number of particles,   X 1 N U − Es,N hN i = N· · exp (5.4.62) Ω (µ, T ) kB T s,N

For the time being, let’s take the following (seemingly) arbitrary derivative:

509

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

    ∂Ω (µ, T ) ∂  X N µ − Es,N  = exp ∂µ ∂µ kB T N,s(N)   X N N µ − Es,N exp = kB T kB T N,s(N)   X 1 N µ − Es,N = N exp kB T kB T

(5.4.63)

N,s(N)

Solving (5.4.63) for the portion that resembles a part of (5.4.62),   X N µ − Es,N ∂ Ω (µ, T ) = N exp kB T · ∂µ kB T

(5.4.64)

We can substitute the RHS of (5.4.64) into (5.4.62): X kB T ∂ hN i = · Ω (µ, T ) Ω (µ, T ) ∂µ

(5.4.65)

N,s(N)

s,N

Let’s rewrite (5.4.65): hN i =

X

kB T

s,N

1 ∂ · Ω (µ, T ) Ω (µ, T ) ∂µ {z } |

(5.4.66)

Let’s look a little closer at the Gibbs function from (5.4.59) (ignore sums momentarily):   N µ − Es,N Ω (µ, T ) = exp kB T Let’s take the first derivative with respect to µ:

  ∂ ∂ N µ − Es,N Ω (µ, T ) = exp ∂µ ∂µ kB T   N N µ − Es,N = exp kB T kB T

(5.4.67)

If we take the first derivative of the Gibbs function with respect to µ, we get some coefficient times the original function. Now, there’s two ways of getting rid of the original function: assuming that you just want to play around with the coefficients for some reason. You could a) simply divide your answer by the original function, or b), you could take the same derivative you took before, but had the natural log of the function: 1 ∂ ∂ · Ω (µ, T ) = log [Ω (µ, T )] Ω (µ, T ) ∂µ ∂µ

(5.4.68)

We note that the LHS of (5.4.68) is identical to the underbraced term in (5.4.66). Substituting the RHS of (5.4.68) into (5.4.66), replacing the underbraced term, we have: X ∂ hN i = kB T log [Ω (µ, T )] (5.4.69) ∂µ s,N

Rearranging, (5.4.69) becomes

W. Erbsen

STATISTICAL MECHANICS

hN i = kB T

∂ log [Ω (µ, T )] ∂µ

(5.4.70)

Where we note that the summation is implicit in the Gibbs function. c)

The first thing that we have to do is find the Gibbs sum. Using the probability function (5.4.61) as our outline, the sum itself takes the form   ∞ X X N µ − Es,N (5.4.71) Ω (µ, T ) = exp kB T N=0 s(N)

There are only three unique states, so the second sum in (5.4.71) only goes to 3. We also know that we have three states of electrons, of occupancy 0, 1, or 2 as given in the provided table. Accordingly, (5.4.71) becomes       (1)µ − 0 (0)µ − (−∆) (2)µ − (δ) Ω (µ, T ) = exp + exp + exp kB T kB T kB T       µ ∆ 2µ − δ = exp + exp + exp kB T kB T kB T = exp [µβ] + exp [∆β] + exp [(2µ − δ) β] (5.4.72) Substituting (5.4.72) into the equation for average number of particles from (5.4.70), 1 ∂ log [exp [µβ] + exp [∆β] + exp [(2µ − δ) β]] β ∂µ   1 β exp [µβ] + 2β exp [(2µ − δ)β] = β exp [µβ] + exp [∆β] + exp [(2µ − δ)β] exp [µβ] + 2 exp [(2µ − δ)β] = exp [µβ] + exp [∆β] + exp [(2µ − δ)β]

hN i =

(5.4.73)

We are actually looking for a condition where hN i = 1, so setting this in (5.4.73) yields exp [µβ] + exp [∆β] + exp [(2µ − δ)β] = exp [µβ] + 2 exp [(2µ − δ)β] exp [∆β] = exp [(2µ − δ)β] ∆β = (2µ − δ)β

(5.4.74)

From (5.4.74), one of the conditions that might be met is in terms of the chemical potential, which reads µ=

∆+δ 2

Problem 7 A system of N distinguishable noninteracting particles has one-particle energy levels of En = nε, where the degeneracy of the nth level is equal to n, and n = 1, 2, 3, ...

511

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

a)

Show that the partition function of the system is" Z=

−1

where β = (kB T )

eβε 2

(eβε − 1)

#N

(5.4.75)

.

b)

Calculate the entropy of the system of particles.

c)

Determine the behavior of the entropy at high temperature.

d)

Determine the behavior of the specific heat at high temperatures.

Solution We first recognize that since we have distinguishable particles, then the partition function as a function of the single-particle partition function would be ZN ∼ = Z1N . a)

The standard form of the partition function (with degeneracy) is   X En Z1 = n · exp − kB T n We now substitute En = nε into (5.4.76), Z1 =

(5.4.76)

X



nε n · exp − kB T



(5.4.77)

X



n

(5.4.78)

n

We can rewrite (5.4.77) slightly, Z1 =

n

ε n · exp − kB T

And now, we see that if we were to take some arbitrary derivative of some function then the answer could be (5.4.78). One such form is 1 ∂ X n Z1 = − n · exp [−εβ] (5.4.79) ε ∂β n Where I have changed form kB T to β −1 . We now recall the following summation identity: X r rn = 1−r

Applying this to (5.4.79),   1 ∂ exp [−εβ] Z1 = − ε ∂β 1 − exp [−εβ]    1 ∂ exp [−εβ] =− ε ∂β 1 − exp [−εβ]      1 ∂ 1 1 ∂ =− exp [−εβ] + exp [−εβ] ε ∂β 1 − exp [−εβ] 1 − exp [−εβ] ∂β

W. Erbsen

STATISTICAL MECHANICS

1 =− ε = = = =

(

exp [−εβ]

exp [−2εβ]

"

(1 − exp [−εβ])2

2

+

2

+

(1 − exp [−εβ]) exp [−2εβ]

#

−ε exp [−εβ]

1 + [−ε exp [−εβ]] 1 − exp [−εβ]

)

exp [−εβ] 1 − exp [−εβ]

exp [−εβ] 1 − exp [−εβ] · 1 − exp [−εβ] 1 − exp [−εβ]

(1 − exp [−εβ]) exp [−2εβ] + exp [−εβ] − exp [−2εβ] 2

(1 − exp [−εβ]) exp [−εβ] 2

(1 − exp [−εβ]) exp [−εβ] exp [2εβ]] = · 1 − 2 exp [−εβ] + exp [−2εβ] exp [2εβ]] exp [εβ] = exp [2εβ] − 2 exp [εβ] + 1

(5.4.80)

We an finally say that (5.4.80) can be reduced to exp [εβ]

Z1 =

(exp[εβ] − 1)2

(5.4.81)

Converting (5.4.81) into the full partition function, ZN = b)

"

exp [εβ] 2

(exp[εβ] − 1)

#N

(5.4.82)

To calculate the entropy for a system of particles, we first recall that S=−

∂F ∂T

(5.4.83)

So, we first need to find F : F = −kB T log [ZN ] Substituting the partition function (5.4.82) into (5.4.84), " #N    exp [εβ] F = − kBT log  (exp[εβ] − 1)2  " # exp [εβ] = − N kBT log 2 (exp[εβ] − 1) n h io 2 = − N kBT log [exp [εβ]] − log (exp [εβ] − 1) = − N kBT {log [exp [εβ]] − 2 log [(exp [εβ] − 1)]}

Substituting (5.4.85) back into (5.4.83), we can find S, S =−

∂ [−N kB T {log [exp [εβ]] − 2 log [(exp [εβ] − 1)]}] ∂T

(5.4.84)

(5.4.85)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

        ∂ ε ε T log exp + 2 log exp −1 ∂T kB T kB T      ∂ ε ε +2 log exp −1 =N kB |{z} ∂T kB kB T |{z} Don’t matter =N kB

(5.4.86)

Garbage

Rewriting (5.4.86):

    ∂ ε log exp −1 S =2N kB ∂T kB T h i   exp kBεT ε  h i =2N kB − kB T exp ε − 1

(5.4.87)

kB T

Rearranging (5.4.87),

h i ε exp 2N kB T h i S =− ε T exp kB T − 1

(5.4.88)

Problem 8 In a mono-atomic crystalline solid each atom can occupy either a regular lattice site or an interstitial site. The energy of an atom at an interstitial site exceeds the energy of an atom at a lattice site by ε. Assume that the number of atoms, lattice sites, and interstitial sites are all equal in number. a)

Calculate the entropy of the crystal in the state where exactly n of the N atoms are at interstitial sites.

b)

What is the temperature of the crystal in this state, if the crystal is at thermal equilibrium?

c)

If ε = 0.5 eV and teh temperature of the crystal is 273 K, what is the fraction of atoms at interstitial sites?

Hint: You have two choices to make in this problem - which atoms to put at interstitial sites, and which interstitial sites to put them in.

Solution a)

To calculate the entropy, we recall that S=−

∂F ∂T

(5.4.89)

And also that the Free energy is F = −kB T log [ZN ] While the single-particle partition function in the case of degeneracy is

(5.4.90)

W. Erbsen

STATISTICAL MECHANICS

Z1 =

X i



Ei gi · exp − kB T



(5.4.91)

Now, there are two different sources of degeneracy in this problem. There is one source coming from the n interstitial lattices, and then there is N for the regular lattices. So, combining these into (5.4.91), yields   X   X En EN Z1 = n exp − + N exp − (5.4.92) kB T kB T n N

Depending on where we are on the lattice we have a different energy, such that we should rewrite (5.4.92) as   X   X (E0 + ε) E0 Z1 = n exp − + N exp − kB T kB T n N   (X   X ) E0 ε = exp − n exp − + N (5.4.93) kB T k BT n N

At this point, we say that the degeneracy for this particular two-level case can be expressed in the following more compact form: g(N, n) =

N! (N − n)!n!

Using this equation, we can completely transform (5.4.93), which now reads      N! E0 ε ZN = · exp − exp − +1 (N − n)!n! kB T kB T Using (5.4.90), we can now find the Free Energy by way of (5.4.95) as follows       N! E0 ε F = − kB T log · exp − exp − +1 (N − n)!n! kB T kB T           N! E0 ε = − kB T log + log exp − + log exp − +1 (N − n)!n! kB T kB T        N! E0 ε = − kB T log − + log exp − +1 (N − n)!n! kB T kB T And to calculate the entropy, we now substitute (5.4.96) back in to (5.4.89):         ∂ N! E0 ε S =− −kB T log − + log exp − +1 ∂T (N − n)!n! kB T kB T        ∂ N! E0 ε =kB T log − + T log exp − +1 ∂T (N − n)!n! kB kB T            N! ε ∂ ε =kB log + log exp − +1 +T log exp − +1 (N − n)!n! kB T ∂T kB T          N! ε ε exp [−ε/(kB T )] =kB log + log exp − +1 +T (N − n)!n! kB T T 2 exp [−ε/(kB T )] + 1 We can simplify (5.4.97) again:

(5.4.94)

(5.4.95)

(5.4.96)

(5.4.97)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS



     N! ε ε exp [−ε/(kB T )] S = kB log + kB log exp − + 1 + kB (N − n)!n! kB T T exp [−ε/(kB T )] + 1 b)

(5.4.98)

We recall that from the first law of thermodynamics, dU = T dS − P dV

(5.4.99)

At constant volume, (5.4.99) becomes dU = T dS −→

dS 1 = dU T

(5.4.100)

Applying the chain rule to (5.4.100), dS dS dU dn dS dn 1 = · · = · = dU dU dn dU dn dU T And now, U =n (E + ε) + (N − n) E =nE + nε + N E − nE =nε + N E Taking the differential limit, dU = ε dn −→

dn 1 = dε ε

Substituting this back in to our chain rule, 1 dS ε dS 1 · = −→ = dn ε T dn T

(5.4.101)

At this point, we apply (5.4.101) to (5.4.98):         d N! ε ε exp [−ε/(kB T )] dS = kB log + kB log exp − + 1 + kB dn dn (N − n)!n! kB T T exp [−ε/(kB T )] + 1   d N! =kB log (5.4.102) dn (N − n)!n! We can further rewrite (5.4.102), dS d =kB {log [N !] − log [(N − n)!] + log [n!]} dn dn

(5.4.103)

log [N !] = N log [N ] − N

(5.4.104)

Recall the identity

Using (5.4.104), we can rewrite (5.4.103) as: dS d =kB {log [(N − n)!] + n log [n] − n} dn dn We can apply the same logic to the mixed term in (5.4.105): log [(N − n)!] = (N − n) log [N − n] − (N − n)

(5.4.105)

W. Erbsen

STATISTICAL MECHANICS

=N log [N − n] − n log [N − n] − N + n

(5.4.106)

Substituting (5.4.106) back into (5.4.105), d dS =kB {N log [N − n] − n log [N − n] − N + n + n log [n] − n} dn dn d =kB {N log [N − n] − n log [N − n] + n log [n]} dn   N n =kB − log [N − n] − + log [n] + 1 N −n N −n     N −n N n + − +1 =kB log n N −n N −n     N −n =kB log +2 n We now set (5.4.107) to ε/T :

    ε N −n kB log +2 = n T

(5.4.107)

(5.4.108)

Solving (5.4.108) for T , T = c)

ε kB

   −1 N −n log +2 n

(5.4.109)

To find the fraction of atoms in the interstitial sites, we simply solve (5.4.109) for N/n:   ε N −n = log +2 kB T n   ε N −n exp −1 = (5.4.110) kB T n From (5.4.110) we can see that   N ε = exp −1 +1 n kB T

(5.4.111)

Problem 9 Calculate the magnetic susceptibility χ=



∂M ∂H H=0

(5.4.112)

Where M is the magnetic moment of the sample and H is the applied field. Calculate χ as a function of T of a gas of N permanent dipoles, each of moment µ, a)

If any direction is allowed (classical spines).

b)

If the dipole is only allowed to assume 2 directions, parallel and opposite to the applied field (Ising spins).

517

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Solution a)

The first thing that we recall is the energy of a dipole in an external field, which is given by U = −µ · H

(5.4.113)

And the magnetization is given by M =−

∂F ∂H

(5.4.114)

Now, if we have a whole bunch of dipoles, then they are not all perfectly aligned (or antialigned), but rather at some random angle θi . Then the total energy from the entirety of the molecules can be found as X U =−µ·H cos θi i

= − µH

X

cos θi

(5.4.115)

i

Since the particles in this case are distinguishable, the full partition function can be found from the single particle partition function as ZN ∼ = Z1N . The single-particle partition function is   X µH cos θi (5.4.116) Z1 = exp kB T i Turning the sum in (5.4.116) into an integral, we have   Z 2π Z π µH cos θ Z1 = exp sin θ dθdφ kB T 0 0 Letting u = cos θ and du = − sin θ dθ, (5.4.117) becomes   Z −1 µHu Z1 = − 2π exp du kB T 1    −1 µHu kB T exp = − 2π · µH k B T 1      kB T −µH µH = − 2π · exp − exp µH kB T kB T   kB T −µH =4π · sinh µH kB T The full partition function can be found from (5.4.118) easily as   N kB T −µH ZN = 4π · sinh µH kB T We now use (5.4.119) to find the Free Energy, F : F = − kB T log [ZN ] (  N ) kB T µH = − kB T log −4π · sinh µH kB T

(5.4.117)

(5.4.118)

(5.4.119)

W. Erbsen

STATISTICAL MECHANICS

   kB T µH = − N kB T log −4π · sinh µH kB T

(5.4.120)

We now substitute F from (5.4.120) into (5.4.114), the equation for the Magnetization, M :     ∂ kB T µH M =− −N kB T log −4π · sinh ∂H µH kB T    ∂ kB T µH =N kB T log −4π · sinh ∂H µH kB T       4πkB T µH ∂ log − log [H] + log − sinh =N kB T ∂H µ kB T     1 ∂ µH =N kB T − + log − sinh H ∂H kB T    1 µ µH =N kB T − + cosh H kB T kB T    kB T µH + µ cosh =N − (5.4.121) H kB T Now, according to the prompt, to find the susceptibility χ, all we must do is differentiate M with respect to H. Doing this, we take the derivative of (5.4.121) with respect to H: ∂M ∂H     ∂ kB T µH + µ cosh = N − ∂H H kB T       ∂ 1 ∂ µH +µ cosh =N −kB T ∂H H ∂H kB T       1 µ 2 µH =N −kB T − 2 − µ − csch H kB T kB T    2 kB T µ µH =N + csch2 H2 kB T kB T

χ=

(5.4.122)

While (5.4.122) is the exact answer, and I believe that we were supposed to use an approximation. We make this approximation in (5.4.121), where we expand the cosh term in a series expansion:   ∞ X kB T µH µ3 H 3 µH cosh = + − 3 T 3 ) + ... kB T µH 3kBT 45 (kB H=0

If we assume that the field is weak, then we can approximate this function by its first two expansion terms. Substituting them back into the Magnetization from (5.4.121),    kB T kB T µH N µ2 H M =N − +µ + −→ M = H µH 3kB T 3kBT It is trivial then, to find the susceptibility in the weak-field regime:   ∂ N µ2 H N µ2 χ= −→ χ = ∂H 3kB T 3kBT b)

We now need to see how things change if the dipoles are only allowed to assume 2 directions, parallel and anti-parallel. The only difference is that here our energies are discrete, so our problem

519

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

is actually easier. I will fly through it rather quickly. Firstly, the single particle partition function is     µH µH + exp Z1 = exp − kB T kB T Whose full counterpart is then ZN



   N µH µH = exp − + exp kB T kB T

And (5.4.123), we can find the Fermi energy (    N ) µH µH F = − kB T log exp − + exp kB T kB T      µH µH + exp = − N kB T log exp − kB T kB T    µH = − N kB T log 2 cosh kB T

(5.4.123)

(5.4.124)

Finding the magnetization of (5.4.124),    µH ∂ log 2 cosh M =N kB T ∂H kB T    µ µH =N kB T tanh kB T kB T 

(5.4.125)

Approximating the first two terms in tanh in very much the same way we did before,   ∞ X µH µH µ3 H 3 2µ5 H 5 tanh = − + 3 5 T 5 + ... 3 kB T kB T 3 (kB T ) 15kB H=0

Substituting in the first two terms back into the Magnetization equation (5.4.125),    µH µ3 H 3 µ M =N kB T − 3 T 3) kB T kB T 3 (kB   µH µ3 H 3 =N µ − 3 T 3) kB T 3 (kB And finally, to find the susceptibility we differentiate (5.4.126) with respect to H:    ∂ µH µ3 H 3 χ= Nµ − 3 3 ∂H kB T 3 (kB T ) Playing around with this a little, we are finally left with χ = Nµ



µ µ3 H 2 − 3 3 kB T (kB T )



(5.4.126)

W. Erbsen

STATISTICAL MECHANICS

Problem 10 Similar to the van der Waals equation of state is the Dieterici equation of state, p(V − b) = RT e−a/RT V

(5.4.127)

Find the critical constants pc, V c and T c in this model of a weakly interacting gas. This equation of state was proposed to account for the interaction of gas atoms with walls.

Solution We first recall what a typical phase diagram looks like, which is of P vs. T . The critical point is where the phase boundaries cease to exist, and all phases coexist simultaneously. This point has the coordinates PC and TC . We can also plot a phase diagram of P vs. V , plotting a variety of different isotherms (lines of constant temperature). We find that in plotting one particular isotherm, that has a point which coincides with Pc and VC from the P T diagram plot. At this point, the slope is zero. The isotherms that lie inside the shaded area are in stable equilibrium of gas and liquid. At the critical point, on the isotherm at the critical temperature, the first (and therefore, the second) derivative with respect to V must be zero:    2  ∂P ∂ P = =0 ∂V Tc ∂V 2 Tc Solving the equation of state given from (5.4.127) for P yields h RT a i P = · exp − V −b RT V Taking the first derivative of (5.4.128) yields     h ∂P ∂ RT a i = · exp − ∂V Tc ∂V V − b RT V   h h RT ∂ a i a i ∂ RT = exp − + exp − V − b ∂V RT V RT V ∂V V − b   h h RT h a a ii a i RT = exp − + exp − − V − b RT V 2 RT V RT V (V − b)2 h i h i a a RT a = 2 exp − − exp − V (V − b) RT V (V − b)2 RT V    exp − RTa V a RT = − V −b V2 V −b And setting (5.4.129) equal to zero,   ∂P a RT =0 −→ 2 = −→ V 2 RT − aV + ab = 0 ∂V Tc V V −b

(5.4.128)

(5.4.129)

(5.4.130)

Taking the second derivative of P is the same as taking the first derivative of (5.4.129), so let’s do that:

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS



∂2P ∂V 2



Tc

  # exp − RTa V a RT − V −b V2 V −b    2  exp − RTa V a (b − V )2 − 2aRT V (b − 2V )(b − V ) + 2R2 T 2 V 4 = 4 3 RT V (V − b) ∂ = ∂V

"

Setting (5.4.131) equal to zero,  2  ∂ P = 0 −→ a2 (b − V )2 = 2aRT V (b − 2V )(b − V ) + 2R2 T 2 V 4 ∂V 2 Tc

(5.4.131)

(5.4.132)

We now have two equations, and two unknowns. Let’s now solve (5.4.130) for T T =

a(V − b) RV 2

(5.4.133)

And substituting this into our second equation (5.4.132),   2 a(V − b) 2 a(V − b) a (b − V ) =2aRV (b − 2V )(b − V ) + 2R V4 RV 2 RV 2 (b − 2V ) + 2a2 (V − b)2 −a2 (V − b)2 =2a2 (V − b)2 V (b − 2V ) −1 =2 +2 V 2(2V − b) 3= V 2

2



From (5.4.134), we can see that Vc = 2b . Plugging in this value into (5.4.133) yields T =

a a(2b − b) −→ Tc = 2 R(2b) 4bR

Substituting both Vc and Tc into P from (5.4.128),   R h a i a 4bR 1 a P = · exp − · · −→ Pc = 2 · exp [−2] 2b − b 4bR R a 2b 4b

Problem 11 For a gas of molecules with diameter d, number density n and at temperature T , find a)

The mean free path

b)

Their average speed

c)

The pressure of the gas using kinetic arguments

(5.4.134)

W. Erbsen

STATISTICAL MECHANICS

Solution a)

The mean free path is defined as the average distance that a gas particle transverses between collisions. If we imagine two gas particles, each of diameter d, passing close to one another, the question is: what is the effective collisional diameter? In other words, if two individual particles happen to collide at one point in space, then it is easy to see that if any particle approaches any other particle, then the net new particle diameter would have to be the sum of the two collided molecules, or 2d. This illustration works particularly well if we make a number of critical assumptions: i)

Only one particle is moving, while all other molecules (target molecules) are stationary. If the moving particle comes within a distance of 2d of a stationary particle, then a collision occurs. → The effective collision area is then A = πd2 .

If the particle travels through space some distance, and in its wake it sweeps out a volume proportional to its cross sectional area and the distance that it traveled. The volume of such a sphere is V = `A = `πd2 , where ` is the length of the swept out cylinder. We must also recognize that we must harness the time that passes while the particle snakes its path through space. Recalling that velocity is defined as v = ∆x/∆t, and in our case ∆x = ` and ∆t is just plain old t. Then we can say that v is the average velocity, and t is the time between successive collisions.

When the volume of this imaginary cylinder equals the average volume per particle, then we are likely to get a collision. → The mean free path (`) is the length of the cylinder when this condition is met Quantitatively, this condition is Volume of Cylinder =Avg Volume per Particle V πd2 ` = N

(5.4.135)

We can solve (5.4.135) for the mean free path, `: `=

1 V · πd2 N

(5.4.136)

It is important that we recognize that (5.4.136) is only an approximation, since according to stipulation i), we have assumed that the target molecules are stationary, which is most certainly never true. To remedy this, we can rewrite ` from (5.4.136) in terms of the average velocity, v: `=v·t=

1 V · 2 πd N

(5.4.137)

In truth, what we are interested in is the relative velocity between the various particles, not the average velocity. Luckily, this question has been asked before, and a quantitative link has been defined relating the two quantities. The savior comes in the form of the Maxwell-Boltzmann Speed Distribution Function, which I shall derive later. The punch-line is √ 1 v rel = 2 v −→ v = √ vrel (5.4.138) 2

523

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Substituting (5.4.138) into (5.4.137), 1 1 V ` = √ v rel · t −→ ` = √ · 2 2 2 πd N b)

(5.4.139)

We recall that any probability can be obtained by integrating the probability distribution function over a particular range: Z v2 P (v) = f(v) dv v1

Our goal is to find some distribution function that allows us to integrate and find a value for speed. We also note that there are many different speeds that can correspond to any particular velocity. In general,     Probability of having Number of velocity vectors f(v) ∝ · (5.4.140) a velocity v corresponding to the same speed The first factor in (5.4.140) is simply the Boltzmann factor, so we can begin building our distribution function:   mv2 f(v) ∼ exp − (5.4.141) 2mkB T And for the second term in (5.4.140), we need to get a little more abstract.. we imagine a 3-D coordinate system in which the axes correspond to different velocity components vx , vy , and vz . Some speed then, will act to mark an invisible sphere around the origin of our coordinate system, such that if one end of the vector is planted at the origin, it can trace out the sphere in any direction it pleases. This “velocity space” analogy convinces us that the second factor in (5.4.140) is only a proportionality factor, equal to the area that is swept out in v-space. This area is A = 4πv2 . Therefore, from (5.4.141), our distribution function is now   mv2 f(v) ∼ exp − · 4πv2 · C (5.4.142) 2mkB T Where C normalizes our distribution function. Now, Z ∞ f(v) dv = 1 0   Z ∞ mv2 2 4πC v exp − dv = 1 2mkB T 0  "   3 # 1/2  1 2kB T 4πC π =1 4  m 

 3/ 2πkB T 2 C· =1 m   3/2 m C= 2πkBT

Substituting (5.4.143) back into our function (5.4.142)

(5.4.143)

W. Erbsen

STATISTICAL MECHANICS



m f(v) = 4πv · 2πkB T 2

 3/2



mv2 · exp − 2mkB T



(5.4.144)

This, (5.4.144), is the famous Maxwell-Boltzmann Distribution Function. We will now use it. But first, going back to the fundamentals: an average quantity can be found by Z x hxi = x · f(x) dx 0

Following this example, we now wish to find the average speed, v, using our distribution function from (5.4.144): Z ∞ v= v · f(v) dv 0



m =4π 2πkB T

 3/2 Z

0

∞



mv2 v · exp − 2mkB T 3



dv

(5.4.145)

The integral in (5.4.145) is tedious , and I would recommend doing it in Mathematica or possibly a table. Anyway, the result is r 8kB T v= (5.4.146) πm c)

What is pressure? Well, kinetically speaking, pressure is the average force exerted on the walls of some container. Mathematically, Favg = N ·

mv 2x L

(5.4.147)

Now, if the speed is the same in all directions, then (5.4.147) becomes Favg = 3N ·

mv2 L

(5.4.148)

And we recall that the pressure is defined as P = F/A, in our case, the force is the average force applied to the sides of the container. Using (5.4.148), the pressure becomes P =

Favg 3mv2 −→ P = N · Area L·A

Problem 12 Consider a large reservoir energy U0 − ε in thermal contact with a system with energy ε. a)

Give an argument based on the multiplicity (number of distinguishable ways a state with a given energy may  ε  be obtained) which leads to the result P (ε) ∼ exp − (5.4.149) kT

Where P (ε) is the probability of finding the system in a particular state with energy ε. Hint: Recall that the entropy S = k log (multiplicity).

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

b)

A system of N magnetic moments (¯ µ) may P orient themselves parallel or antiparallel to an applied field. Deduce from a) the total magnetization M ≡ 1 µ ¯i as a function of H and T .

Solution a)

Multiplicity refers to the number of configurations that a system has available to it while staying essentially the same. The entropy can be linked to the multiplicity as S = kB log [Ω]

(5.4.150)

And the probability of being in one particular state is P =

1 Ω

(5.4.151)

And we recall that the first law says that dU = T dS − P dV

(5.4.152)

Since the volume in our case is constant, (5.4.152) reduces to dU = T dS −→

dS 1 = dU T

(5.4.153)

Applying the chain rule to (5.4.153), dS dU dΩ dS dΩ 1 · · = · = dU dΩ dU dΩ dU T

(5.4.154)

Sweet. Now, substitute our relation for S from (5.4.150) into (5.4.154) and let’s see what happens:   dS dΩ d dΩ · = kB log [Ω] · dΩ dU dΩ dU kB dΩ · (5.4.155) = Ω dU And setting (5.4.155) equal to 1/T , as originally intended, kB dΩ 1 1 1 · = −→ dΩ = dU Ω dU T Ω kB T

(5.4.156)

Integrating both sides of (5.4.156), we find that log [Ω] =

E kB T



E Ω = exp kB T



(5.4.157)

And applying (5.4.151) to (5.4.157), we have 

E P = exp − kB T b)



We first recall that for a two-state paramagnet, we can either be in the direction of the field, or anti-parallel to it:

W. Erbsen

STATISTICAL MECHANICS

E = µ·B

(5.4.158)

While the total number of domains is given by N = N↑ + N↓ Each with respective energies E↑ =µB

E↓ = − µB We also recall that the magnetization is defined as the density of magnetic moments. In our case, M = µ (N↑ + N↑ )

(5.4.159)

The single-particle partition function is then given by     µB µB Z1 = exp + exp − kB T kB T   µB =2 cosh kB T And assuming that our particles are distinguishable, then the full partition function can be found easily:   N µB ZN = 2 cosh (5.4.160) kB T Now finding the Helmholtz Free Energy, we use the partition function from (5.4.160), and so: (  N ) µB F = − kBT log 2 cosh kB T    µB = − N kBT log 2 cosh (5.4.161) kB T The magnetization, M , is then found using (5.4.161) as     ∂ µB M =− −N kB T log 2 cosh ∂B kB T     ∂ µB =N kB T log 2 cosh ∂B kB T    µB µ =N kB T tanh kB T kB T Simplifying things a bit, the magnetization becomes M = N µ tanh



µB kB T



527

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Problem 13 Consider a two-state system with energies 0 and ε. a)

Calculate the partition function

b)

Find the average energy at temperature T

c)

Find the heat capacity and graph it versus temperature

Solution a)

We first recall that the plain old partition function is given in its most basic form as   X Es Z= exp − kB T s

(5.4.162)

We only have a two-level system, so it is easy to expand (5.4.162) to suit our needs:   ε Z = exp [0] + exp − kB T Which can be rewritten as:   ε Z = 1 + exp − kB T b)

In order to find the average energy, we recall that 1 X U= Es exp [−Es /(kB T )] Z s

(5.4.163)

(5.4.164)

Evaluating the internal energy from (5.4.164) out to the two allowed terms, much as we did before: U=

0 · exp [0] + ε exp [−ε/(kB T )] 1 + exp [−ε/(kB T )]

(5.4.165)

Getting rid of the null terms in (5.4.165) leads us to U=

ε exp [−ε/(kB T )] 1 + exp [−ε/(kB T )]

(5.4.166)

We can actually nicen (5.4.166) up a bit (for reasons that will become clear momentarily) by letting β = 1/(kBT ): U= c)

ε exp [−εβ] 1 + exp [−εβ]

To find the heat capacity at constant volume, we first recall:   ∂U CV = ∂β V Substituting the internal energy from (5.4.167) into (5.4.168),

(5.4.167)

(5.4.168)

W. Erbsen

STATISTICAL MECHANICS

  ∂ ε exp [−εβ] CV = ∂β 1 + exp [−εβ]     ∂ 1 1 ∂ + [ε exp [−εβ]] =ε exp [−εβ] ∂β 1 + exp [−εβ] 1 + exp [−εβ] ∂β " #    2  εβ exp [−εβ] 1 =ε exp [−εβ] + −ε β exp [−εβ] 2 1 + exp [−εβ] (1 + exp [−εβ]) #  "  εβ exp [−εβ] −ε2 β exp [−εβ] =ε exp [−εβ] 2 + 1 + exp [−εβ] (1 + exp [−εβ]) (" # ) exp [−2εβ] (− exp [−εβ]) (1 + exp [−εβ]) 2 =ε β 2 + (1 + exp [−εβ]) (1 + exp [−εβ]) (1 + exp [−εβ]) ) ( exp [−2εβ] − exp [−εβ] − exp [−2εβ] =ε2 β (1 + exp [−εβ])2 We can now deduce that 2

CV = −ε β

"

exp [−εβ] (1 + exp [−εβ])

2

#

Problem 14 An ideal diatomic gas has rotational energy levels given by Ej = a)

h2 j(j + 1), 8π 2 I

with degeneracies

(5.4.169)

For Oxygen, what fraction of the molecules is in the lowest rotational energy state at T = 50K? Recall that θrot =

b)

gj = 2j + 1

h2 = 2K 8π 2 IkB

(5.4.170)

Repeat this for Hydrogen, with θrot = 85K.

Solution a)

The single particle partition function (with degeneracy) is by default given by   X Ej Z1 = gj · exp − kB T

(5.4.171)

Substituting in the appropriate values into (5.4.171),   X 1 h2 Z1 = (2j + 1) · exp − · 2 j(j + 1) kB T 8π I j

(5.4.172)

j

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

At this point, we can rewrite Ej and Θrot as follows kB h2 · j(j + 1) 8π 2 I kB =Θrot · kB · j(j + 1)

Ej =

Substituting (5.4.173) back into our partition function from (5.4.172),   X 1 · Θrot · kB · j(j + 1) Z1 = (2j + 1) · exp − kB T j   X Θrot = (2j + 1) · exp − j(j + 1) T j We now wish to convert the sum in (5.4.174) into an integral:   Z Θrot Z1 = (2j + 1) · exp − j(j + 1) dj T Using u = j(j + 1) = j 2 + 1 and du = (2j + 1) du, (5.4.175) becomes   Z ∞ Θrot Z1 = exp − u du T 0    ∞ Θrot T u exp − =− Θrot T 0 T = Θrot

(5.4.173)

(5.4.174)

(5.4.175)

(5.4.176)

From the provided information, we have T = 50 K, and (5.4.176) becomes Z1 = b)

T 1 Θrot 1 1 −→ = −→ = Θrot Z1 T Z1 25

(5.4.177)

We recall the single particle partition function from (5.4.175) and using the fact that Θrot = 85 K, and T = 50 K, (5.4.175) becomes   X 85 Z1 = (2j + 1) · exp − j(j + 1) (5.4.178) 50 j The sum in (5.4.178) seems to converge very quickly. Let’s look at the first few terms: Z1 (0) =1



 85 Z1 (1) =3 · exp − · 2 = 0.11 50   85 Z1 (2) =5 · exp − · 6 = 0.00 50 So, taking only the first two terms, we have Z1 ≈ 1.100, and so the inverse of this yields 1 ≈ 0.909 Z1

W. Erbsen

STATISTICAL MECHANICS

Problem 15 On a glass surface, water molecules can hydrogen bond to either 1 or 2 OH groups as shown below with energies ε or 2ε, respectively. The molecule may also be weakly bound to the surface with energy ε1 ( ε) (not shown).

a)

Find an expression for the partition function for a water molecule on the surface.

b)

Find the limiting behavior of the internal energy U and the specific heat C at i) high temperatures and ii) low temperatures.

c)

Hence, sketch the behavior of U and C as a function of temperature.

Solution a)

The single-particle partition function is easily found to be Z1 = exp [εβ] + exp [−2εβ] + exp [−ε1 β]

b)

(5.4.179)

The relevant expression for the internal energy is U =kB T 2

∂ ∂ log [Zn ] = − log [ZN ] ∂T ∂β

(5.4.180)

Since our particles in this case are distinguishable, the partition function in (5.4.179) becomes ZN = Z1N = [exp [εβ] + exp [−2εβ] + exp [−ε1 β]]N Substituting (5.4.181) into (5.4.180), n o ∂ N U =− log [exp [εβ] + exp [−2εβ] + exp [−ε1 β]] ∂β ∂ =−N log [exp [εβ] + exp [−2εβ] + exp [−ε1 β]] ∂β   ε exp [εβ] − 2ε exp [−2εβ] − ε1 exp [−ε1 β] =−N exp [εβ] + exp [−2εβ] + exp [−ε1 β] We now recall that the specific heat at constant volume is given by

(5.4.181)

(5.4.182)

531

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

CV =



∂U ∂T



(5.4.183) V

Rewriting (5.4.182) and substituting into (5.4.183), h i h i h i  ε1 2ε ε   − 2ε exp − − ε exp − ε exp 1 kB T kB T kB T ∂ h i h i h i CV = − N  ∂T  exp ε + exp − 2ε + exp − ε1 kB T kB T kB T 2  h ih h h i h ii h h i i h h iii  3ε ε+ε1 3ε 3ε   2ε+ε1 2 2   ε 4 + exp kB T + 9 exp kB T + 2εε1 exp kB T − 1 + ε1 exp kB T exp kB T N =  h i h i h i2  kB T 2  ε1 3ε+ε   exp k2ε + exp + exp T k T k T B B B (5.4.184)

Here, (5.4.184) represents the full expression for the specific heat. See attached plots of U vs. T as well as CV vs. T below. Cv

U 0.35

0.5 0.4

0.30

0.3 0.25 0.2 0.1 T 20

40

60

80

100 T 2

4

6

8

10

Problem 16 The Diesel engine cycle consists of several processes Suppose the cycle is applied to n moles of an ideal diatomic gas. In terms of the given pressures and volumes and the gas constant R, give expressions for a)

The heat QH given to the gas during combustion.

b)

The heat QC removed from the gas during the cooling process.

c)

The entropy change of the gas in the process BC.

d)

The entropy change of the gas in process CD.

AB:

Adiabatic compression from volume V0 to volume V1 , where pressure = p1 . The compression ratio rc = V0 /V1 is high enough that it causes ignition of the air-fuel mixture without needing a spark.

BC:

Expansion at constant pressure p1 from volume V1 to V2 < V0 , during the burning of the fuel.

CD:

Adiabatic expansion from V2 to V0 , with expansion ration = V0 /V2 .

W. Erbsen

DA:

STATISTICAL MECHANICS

Constant volume (V0 ) cooling back to the original temperature at point A.

Treat the gas as an ideal gas, and suppose that the constant pressure (CP ) and constant volume (CV ) heat capacities are known. a)

Calculate the efficiency of the cycle, defined as the ratio of the work output to the heat absorbed per cycle, ε = W/QH .

b)

Express ε as a function of γ = CP /CV , re , and rc.

c)

Calculate the entropy changes in each process of the cycle, and sketch the cycle in a S-T diagram.

Solution a)

To find the heat given off during ignition, QH, we are looking at the B→C isobar. We recall that the heat is given by Q = nC∆T

(5.4.185)

QBC = nCP [TC − TB ]

(5.4.186)

Applying this to the B→C isobar,

We now recall that the ideal gas law says that P V = nRT −→ T =

PV nR

Substituting this equation for T into (5.4.186),     P1 V1 P1 V2 CP P1 QBC =nCP − −→ QBC = [V1 − V2 ] nR nR R

(5.4.187)

(5.4.188)

533

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

b)

We now wish to find the heat QC removed during the cooling process, which along the D→A isochor. The process is very much the same as in the last part, adopting (5.4.186) for our purposes, QDA = nCV [TA − TD ]

(5.4.189)

And substituting (5.4.187) into (5.4.189),     PD V0 PA V0 CV V0 QDA = nCV − −→ QDA = [PD − PA ] nR nR R c)

(5.4.190)

Now we want to calculate the change in entropy along B→C. We first recall that the entropy is defined by Q T

(5.4.191)

CP P1 [V1 − V2 ] RT

(5.4.192)

∆S = Substituting (5.4.188) into (5.4.191), ∆S =

And also that the specific heat at constant pressure is CP = T

∂S ∂T

(5.4.193)

Treating (5.4.193) like a differential equation,

  1 1 S V2 dS = dT −→ = log CP T CP V1

(5.4.194)

Solving (5.4.194) for S, S = CP log



V2 V1



(5.4.195)

d)

To find the change in entropy along the isochore D→A, we can go ahead and use (5.4.195) and say that   V0 S = CV log −→ S = 0 (5.4.196) V0

e)

We are now asked to find the efficiency of the entire cycle, which is given as ε = W/QH . We also recall that W = Qin − Qout , so that in our case, the efficiency is given by ε= =

QBC − QDA QBC
CP P1 R [V1 − V2 ] − CP P1 R

CV V 0 R

[V1 − V2 ]

[PD − PA]


Which leads us to ε=

CP P1 [V1 − V2 ] − CV V0 [PD − PA ] CP P1 [V1 − V2 ]




W. Erbsen

STATISTICAL MECHANICS

f)

We now wish to rewrite ε in terms of some new variables, which are γ=

CP , CV

re =

V0 , V2

rc =

V0 V1

The result is ε=1− g)

1 P1 γ



PD − PA 1/ − 1/ rc re



−→ ε = 1 −

  1 PD − PA rc re P1 γ rc − re

The entropy for both cycles are ∆SBC = ∆SDA



V2 = CP log V1



,

∆SAB = ∆SCD = 0

And the net change in entropy is of course ∆Snet = 0

Problem 17 A classical monatomic ideal gas of N particles (each of mass m) is confined to a cylinder of radius r and infinite height. A gravitational field points along the axis of the cylinder downwards. a)

Determine the Helmholtz energy A.

b)

Find the internal energy U and the specific heat CV .

c)

Why is CV 6= 3/2 N k?

Solution a)

There are two components of energy in this system: kinetic (p2 /2m) and potential (mgh). The Hamiltonian for this system is then p2 + mgz 2m p2x + p2y + p2z = + mgz 2m

H=

While the single-particle partition function takes the form   Z Z 1 H Z1 = 3 exp − d3 pd3 r h kB T

(5.4.197)

(5.4.198)

Substituting our Hamiltonian from (5.4.197) into the partition function from (5.4.198) yields

535

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

# p2x + p2y + p2z mgz exp − − d3 pd3 r 2mkB T kB T " #   Z ∞ Z ∞ p2x + p2y + p2z mgz 1 exp − dpx dpy dpz exp − dz = 3 h −∞ 2mkB T kB T 0 " #       Z ∞ Z ∞ Z ∞ Z ∞ p2y 1 p2x p2z mgz = 3 exp − dpx exp − dpy exp − dpz exp − dz h −∞ 2mkB T 2mkB T 2mkB T kB T −∞ −∞ 0   i hp i hp iZ ∞ 1 hp mgz = 3· 2πmkB T · 2πmkB T · 2πmkB T exp − dz h kB T 0 3   / Z [2πmkB T ] 2 ∞ mgz exp − dz (5.4.199) = h3 kB T {z } |0

1 Z1 = 3 h

Z Z

"

We actually want to solve this last integral in polar coordinates:     Z ∞ Z ∞ Z 2π Z R mgz mgz exp − dz = exp − rdrdθdz kB T kB T 0 0 0 0   Z ∞ Z 2π Z R mgz = exp − dz dθ r dr kB T 0 0 0    ∞  2 kB T mgz R =− exp − · [2π] · mg kB T 0 2 R2 kB T · 2π · mg 2 R2 πkB T = mg =

(5.4.200)

Substituting (5.4.200) back into the original integral of our function (5.4.199), 3

[2πmkB T ] /2 R2 πkB T · Z1 = h3 mg

(5.4.201)

And for classical particles, we recall that ZN ∼ = Z1N , so applying this to the single particle partition function (5.4.201) yields " #N 3 / [2πmkB T ] 2 R2 πkB T ZN = · h3 mg " #N 3 / 2πmkB T 2 R2 πkB T = · h2 mg " # 3 N   2πmk T   R2 πk T  2/3 /2  B B = · (5.4.202)   h2 mg

We now recall that the internal energy is defined by U =−

∂ log [Zn ] ∂β

(5.4.203)

W. Erbsen

STATISTICAL MECHANICS

Substituting the partition function (5.4.202) into the internal energy (5.4.203),   3 N "    2  2/3 # /2      ∂ 2πmkB T R πkB T U =− log ·    ∂β h2 mg     2  2/3 # R π · βmg  ! 2/3  3 1 /2 /2 5/2 3N ∂ 2 m π  log  =− 2 ∂β β 5/2 gh3

3N ∂ log =− 2 ∂β

=−N

"

2πm βh2

 3/2 1/2 5/2  ∂ 2 m π 1 log ∂β gh3 β 5/2 | {z }

(5.4.204)

Garbage

Carrying out the derivative in (5.4.204), U =N

h 5 i 5N ∂ 5N ∂ log β /2 = log [β] −→ U = kB T ∂β 2 ∂β 2

(5.4.205)

Recalling that the specific heat at constant volume is none other than the first derivative of internal energy with respect to temperature, we have from (5.4.205)   ∂ 5N 5N CV = kB T −→ CV = kB ∂T 2 2 b)

We recognize that CV 6= 3/2 N kB because we have more degrees of freedom in this particular system, which stems from our original Hamiltonian.

Problem 18 A thin-walled vessel of volume V is kept at a constant temperature T . A gas leaks slowly out of the vessel through a small hole of area A into surrounding vacuum. Find the time required for the pressure in the vessel to drop to half of its original value.

Solution We first note that Pi = P0 Pf = 1/2 P0

Ni = N Nf = 1/2 N

The number of particles leaving the container over a time interval dt is dN = −nAv dt

(5.4.206)

537

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Where we recall that n = N/V . Furthermore, the average velocity is given by r 8kBT v= πm Substituting v from (5.4.207) into (5.4.206), and solving for dt, yields r 8kB T dN = − nA dt πm r V 1 πm dt = − · dN N A 8kB T

(5.4.207)

(5.4.208)

Integrating (5.4.208), r Z N/2 1 πm V ; dN A 8kB T N N r  V πm   N  =− log /2 − log [N ] A 8kB T

t=−

(5.4.209)

Simplifying (5.4.209) gives us a neat expression for t: r V πm t= log [2] A 8kB T

Problem 19 Consider a paramagnetic substance with the equation of state M = AH/(T − T0 ). Here M is the magnetization, H is the applied magnetic field, A and T0 are constants, and T is the temperature. The equation of state is valid only for T > T0 . Show that CM , the heat capacity at constant magnetization, is independent of M .

Solution We first recall that specific heat (in general) is defined as ∂Q ∂T ∂S C =T ∂T C=

(5.4.210a) (5.4.210b)

From the 2nd Law, if the heat capacity at constant M is indeed independent of M itself, then taking the derivative with respect to it should be equal to zero: ∂CM =0 ∂M

(5.4.211)

This is the condition that we are trying to prove. One possible way to do this is to work backwards, hoping to arrive at the same result. Plugging (5.4.210b) into (5.4.211), we have

W. Erbsen

STATISTICAL MECHANICS

  ∂CM ∂ ∂S = T ∂M ∂M ∂T 2 ∂ S =T ∂M ∂T

(5.4.212)

Now, we switch the order of the derivatives in (5.4.212), yielding ∂CM ∂ ∂S =T ∂M ∂M ∂T ∂ ∂S =T · ∂T ∂M

(5.4.213)

And now, we recall a few seemingly unrelated identities: dU =T dS + HM

(5.4.214a)

dU =T dS + dF dF = − SdT + HdM

(5.4.214b) (5.4.214c)

Now, don’t forget the function that was given in the prompt: M=

AH M −→ H = (T − T0 ) · T − T0 A

(5.4.215)

We now recall one of the least-studied areas within the realm of statistical mechanics. One of them is particularly useful for this problem, and that is     ∂S ∂H =− (5.4.216) ∂M T ∂T M We now substitute H from (5.4.215) into the RHS of (5.4.216)     ∂S ∂ M =− (T − T0 ) · ∂M T ∂T A M = A

(5.4.217)

We now wish to substitute the LHS of (5.4.216) into the RHS of (5.4.213): ∂CM ∂ ∂S =T · ∂M ∂T ∂M ∂ M =T · ∂T A

(5.4.218)

And since neither M nor A has any time dependence, (5.4.218) is clearly null, and the result could not be more clear: ∂CM =0 ∂M

539

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Problem 20 A sample of ideal gas is taken through the cyclic process abca shown in the figure. At point a, T = 300K.

a)

What are the temperatures of the gas at points b and c?

b)

Complete the table by inserting a plus sign, a minus sign, or a zero in each indicated cell. Note that Q is positive when heat is absorbed by the gas and W is positive when work is done by the gas. ∆E is the change in internal energy of the gas. Q

W

∆E

a −→ b b −→ c c −→ a

Solution a)

We recall from the ideal gas law that P V = N kB T

(5.4.219)

From the figure, we are able to deduce that both P and V strictly by visual inspection, while T is given only at point a. We can now solve for N kB in (5.4.219): N kB =

PV T

(5.4.220)

Throughout the cycle, the fraction in (5.4.220) is constant, so to find the temperature at some other point when the other variables are given, PaVa Pb Vb Pb Vb Ta = −→ Tb = Ta Tb Pa Va

(5.4.221)

W. Erbsen

STATISTICAL MECHANICS

Plugging in the appropriate values into (5.4.221),

7.5 × 103 N · m2 3.0 m3 (300 K) Tb = −→ Tb = 2700 K (2.5 × 103 N · m2 ) (1.0 m3 ) We can repeat the same process to find Tc by making a change of variables from b → c: Tc =

Pc Vc Ta Pa Va

(5.4.222)

And substituting in the correct values into (5.4.222), we have

2.5 × 103 N · m2 3.0 m3 (300 K) −→ Tc = 900 K Tc = (2.5 × 103 N · m2 ) (1.0 m3 ) b)

We note that Q is positive when heat is absorbed, while W is positive when work is done by the gas. Accordingly, From a→b b→c c→a

∆t ↑ ↓ ↓

Q ↑ ↓ ↑

W ↑ = ↓

Problem 21 A sample of helium gas inside a cylinder terminated with a piston doubles its volume from Vi = 1m3 to Vf = 2m3 . 6 During this process the pressure and volume are related by P V /5 = A = constant. Assume that the product P V 2 always equals /3 U , where U is the internal energy. a)

What is the change in energy of the gas?

b)

What is the change in entropy of the gas?

c)

How much heat was added to or removed from the gas?

Solution a)

Given that the original formula P V

6

/5

= A, we can solve for P before and after this expansion:

Pi =

A Vi

6/ 5

,

Pf =

A 6/ 5

Vf

And the change in energy can be found using the second given expression, PV =

2 3 U −→ U = P V 3 2

And so the change in energy, ∆U , can be found by 3 ∆U = (Pf Vf − Pi Vi ) 2

(5.4.223)

541

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

3 = 2

A

Vf −

6 /5

Vf

1

3 = A 2

1 /5

Vf

−

A Vi

6/ 5

1 Vi

1/ 5

Vi

!

! (5.4.224)

Substituting in the appropriate values into (5.4.224), 3 ∆U = A 2 b)

1 (2 m3 )

1/ 5

−

1 (1 m3 )

1/ 5

!

(5.4.225)

To find the change in entropy of the gas, we first recall that a common form of the first law of thermodynamics is dU = T dS − P dV

(5.4.226)

1 P dU + dV T T

(5.4.227)

Solving (5.4.226) for dS, dS =

We now recall from the ideal gas law, and also from the information given in the problem, that P V = N kB T =

2 U 3

3 N kB T 2 2 U −→ T = 3 N kB 2 U N kB T = −→ P = V 3 V −→ U =

Substituting (5.4.228a)-(5.4.228c) into (5.4.227),     3 N kB 1 N kB T dS = dU + dV 2 U T V     3 1 1 = N kB dU + nkB dV 2 U V

(5.4.228a) (5.4.228b) (5.4.228c)

(5.4.229)

Integrating, (5.4.229) becomes 3 U Vf S = N kB [log U |Ufi + nkB [log V |Vi 2     3 Uf Vf = N kB log + N kB log 2 Ui Vi

(5.4.230)

And we now innocently recall that 3 1 3 1 A = A 1/ 1 / 5 2 (2 m3 ) 2 Vf 5 3 1 3 1 = A 1/ Ui = A 1 / 5 3 2 (1 m ) 2 V 5

Uf =

i

And substituting (5.4.231a) and (5.4.231b) into S from (5.4.230),

(5.4.231a) (5.4.231b)

W. Erbsen

STATISTICAL MECHANICS

    − 1/ Vf 5 3 Vf S = N kB log  − 1/  + N kB log 5 2 Vi Vi     3 Vi Vf + N kB log = N kB log 10 Vf Vi   7 Vf = N kB log 10 Vi

(5.4.232)

Substituting the appropriate values into (5.4.232), we are left with S= c)

7 N kB log [2] 10

We first recall that dQ = dW + dU

(5.4.233)

To find out what W is in (5.4.233), we recall that work is defined as W = 6 6 that P V /5 = A → P = AV − /5 , so we have Z Vf 6 W =A V − /5 dV Vi

V 1 f =−A·5 1 V /5 

=−A·5

Vi

1

1/ 5

Vf

−

1 Vi

1/ 5

R

P dV . Also remembering

!

(5.4.234)

Now, substituting ∆U from (5.4.225) and W from (5.4.234) into (5.4.233), we have ! ! 1 1 3 1 1 − 1/ + · A − 1/ Q=−A·5 1 1 / / 2 Vf 5 Vi 5 Vf 5 Vi 5 ! 7 1 1 =− A − 1/ 1 /5 2 V V 5 f

i

And with the given values, this becomes 7 Q= A 2

1 (1 m3 )

1/ 5

−

1 (2 m3 )

1/ 5

!

Problem 24 An arbitrary network of resistors, capacitors, and inductors is in thermal equilibrium at temperature T . There are no sources in the network.

543

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

a)

By considering a capacitor C as a subsystem, find the probability density P (V ) that a voltage V exists across it. Find the mean and the root-mean-square voltages on a 10 pF capacitor when T = 300 K and when T = 77 K.

b)

Similarly, find the probability density P (I) that a current I flows through an inductor L at temperature T . What is the rms current in a 1 mH inductor when T = 300 K and when T = 77 K?

Solution a)

We recall that the energy stored in a capacitor is given by E(V ) =

CV 2 2

(5.4.235)

While the probability function is most generally given by   εs 1 P (εs ) = exp − Z kB T

(5.4.236)

Where the partition function in this case is   εs exp − kB T

(5.4.237)

And substituting (5.4.235) into (5.4.237) yields   X CV 2 Z= exp − 2kB T

(5.4.238)

Z=

X s

V

Passing this to the integral limit,(5.4.238) becomes r   Z ∞ CV 2 2πkB T dV −→ Z = Z= exp − 2k T C B −∞

(5.4.239)

Substituting (5.4.235) and (5.4.239) into (5.4.236), r   C CV 2 P (V ) = exp − 2πkBT 2kB T To find the mean voltage, hV i, we recall that a common way to define this quantity is   Z 1 E(x) hxi = x · exp − dx Z kB T In our case, (5.4.240) becomes r hV i =

C 2πkB T

Z

∞



CV 2 V · exp − 2kBT −∞



(5.4.240)

dV −→ hV i = 0

This is because we were trying to integrate an even function multiplied by an odd function over a symmetric domain. The RMS, however, does not fit this criteria, and as such we would expect a non-zero result:

W. Erbsen

STATISTICAL MECHANICS

2

r

  CV 2 V 2 · exp − dV 2kBT −∞  s  r  3 C 2k T 1 B   = π· 2πkBT 2 C s C 8k 3 T 3 π · · B3 = 4 2πkBT C r 2T2 kB = C2

hV i =

Z

C 2πkBT

∞

(5.4.241)

From (5.4.241) it is easy to deduce that hV 2 i = b)

kB T C

Finding the probability density for current flowing through an inductor is identical to the process just underwent, therefore I will only summarize the results: r   I IL2 P (I) = · exp − 2πkB T 2kB T hIi = 0 hI 2 i =

kB T L

Problem 25

a)

Compare the 4 level systems shown below. More than one particle may occupy a level. Which system has the Highest temperature Lowest specific heat

Lowest temperature Highest entropy

545

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

b)

Consider the system shown below. Explain why it can be thought of as having a negative temperature. (Hint: Consider the Boltzmann factor e−E/kT ).

c)

If this system (from part b)) is brought into contact with a large reservoir at temperature T R (positive) draw a graph indicating how its temperature will change as a function of time as it comes to equilibrium with the reservoir. (Take the initial temperature of the system to be T S a negative number).

d)

Explain why no problems with the third law are encountered in part c).

Solution a)

Highest temp: (ii) Lowest temp: (iii) Lowest specific heat: (ii) Highest entropy: (ii)

b)

If the temperature is increased slightly, then it is possible that the total energy will decrease, which is the opposite of the typical behavior.

c)

The third law states that as the entropy in a perfect crystal approaches zero, then the temperature approaches zero as well.

This example does not violate the third law of thermodynamics because system at a negative temperature is further away from absolute zero than a system at positive temperature (from the point of view of energy, anyway).

W. Erbsen

STATISTICAL MECHANICS

Problem 26 Equipartition. A classical harmonic oscillator H=

p2 Kq 2 + 2m 2

(5.4.242)

is in thermal contact with a heat bath at temperature T . Calculate the partition function for the oscillator in the canonical ensemble and show explicitly that hEi = kB T,

and

2

2

h(E − hEi) i = (kB T )

(5.4.243)

Solution We first recall that the single-particle partition function is given by Z Z 1 Z1 = 3 exp [−βH] d3 pd3 q h

(5.4.244)

However, in our case we are assuming that the Hamiltonian is only in 1-D, so that substituting our Hamiltonian from (5.4.242) into (5.4.244), we have   2  Z Z 1 p kq 2 Z1 = + exp −β dpdq h 2m 2     Z Z ∞ 1 ∞ p2 Kq 2 = exp − dp exp dq h −∞ 2mkB T 2 −∞ !  r 2πk T 1 p B = 2mπkB T h K r 2πkBT m = (5.4.245) h K We now recall that the internal energy, hEi, is defined by hEi = −

∂ log Z1 ∂β

Rearranging (5.4.245) and substituting into (5.4.246), we have  r  2π m ∂ log hEi = − −→ hEi = kB T ∂β βh K

(5.4.246)

(5.4.247)

Now, we recall that 2

h(E − hEi) i = hE 2 i − hEi2 And we can also express hEi in yet another way Z Z 1 hEi = E · exp [−βH] dpdq −→ hEi = kB T Z1

(5.4.248)

(5.4.249)

547

CHAPTER 5: DEPARTMENTAL EXAMINATIONS

As before, We can extend this to find hE 2 i, however, and (5.4.249) becomes Z Z 1 hE 2 i = E 2 exp [−βH] dpdq Z1

(5.4.250)

And now, seemingly arbitrarily, we take the first derivative of hEi with respect to β in (5.4.247): ∂ ∂ 1 1 hEi = =− 2 ∂β ∂β β β

(5.4.251)

And now, stay with me here, we wish to do the same thing for the primary expression in (5.4.249):   Z Z ∂ ∂ 1 hEi = E · exp [−βH] dpdq (5.4.252) ∂β ∂β Z1 Recognizing that Z1 has a dependence on β, we continue with (5.4.252), yielding   Z Z   Z Z ∂ 1 ∂ ∂ 1 hEi = E · exp [−βH] dpdq + E · exp [−βH] dpdq ∂β Z1 ∂β ∂β Z1   Z Z Z Z 1 ∂ 1 2 E · exp [−βH] dpdq + E · exp [−βH] dpdq = Z1 ∂β Z | {z 1}

(5.4.253)

Let’s take a closer look at the underbraced term in (5.4.253), using the value of Z1 found in (5.4.245): " r # r ∂ 1 ∂ hβ K h K = ⇒ (5.4.254) ∂β Z1 ∂β 2π m 2π m Substituting (5.4.254) back in to (5.4.253), r Z Z Z Z ∂ 1 h K 2 hEi = E · exp [−βH] dpdq + E · exp [−βH] dpdq ∂β Z1 2π m Inserting in to (5.4.255) our expression for Z1 found in (5.4.245), r r Z Z Z Z ∂ h K h K 2 hEi = E · exp [−βH] dpdq + E · exp [−βH] dpdq ∂β 2πkB T m 2π m | {z }

Let us know rewrite the underbraced term in (5.4.256): r r Z Z h K h K E · exp [−βH] dpdq = · Z1 hEi 2π m 2π m r r   h K 2πkB T m = · hEi 2π m h K 1 = hEi β

(5.4.255)

(5.4.256)

(5.4.257)

Plugging (5.4.257) back into (5.4.256), ∂ 1 hEi = ∂β Z1 |

Z Z

1 E 2 · exp [−βH] dpdq + hEi β {z }

(5.4.258)

W. Erbsen

STATISTICAL MECHANICS

We immediately recognize the underbraced term to be none other than our good friend hE 2 i. Using this knowledge, we can now rewrite (5.4.258), ∂ 1 hEi =hE 2 i + hEi ∂β β Rearranging this a bit, ∂ 1 hEi hE 2 i = hEi − β ∂β     1 1 ∂ 1 = − β β ∂β β 1 1 = 2+ 2 β β 2 = 2 β

(5.4.259)

Substituting (5.4.247) and (5.4.259) into (5.4.248), hE 2 i − hEi2 =

2 1 − 2 −→ hE 2 i − hEi2 = (kB T )2 β2 β

Problem 27 Consider two single-particle states and two particles. Calculate the entropy of this system when the particles have the following statistics: a)

Maxwell-Boltzmann (i.e. classical)

b)

Bose-Einstein

c)

Fermi-Dirac

d)

If two particles of mass m are confined to a box of volume V , approximately what temperature do classical statistics lose their validity?

Solution a)

We first recall how the entropy depends on the multiplicity S = kB log [Ω]

(5.4.260)

We first need to know what the multiplicity, Ω is before we can use this tool though. The multiplicity refers to the number of available states in the system. It’s as simple as that. For Maxwell-Boltzmann statistics, we recognize that this is a classical system, and both particles may occupy any of the states available with no restrictions what-so-ever:

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

• •

•

•

•

•

•

  X 

•

So, the entropy in this case is

(Ω) = 4

S = kB log [4] b)

Within the framework of hte Bose-Einsein universe, we recognize that the particles may still occupy the same orbital, however being indistinguishable particles, there is no difference between states with similar occupancies. Eg,  • • •  X (Ω) = 3  • • • And the entropy ends up being

S = kB log [3] c)

For the Fermi-Direac distribution, we recognize that we are constrained clenched fist of Pauli’s Exclusion Principle. According to Mr Pauli, there is only one possible configuration:  •  X (Ω) = 1  • And, as expected, the entropy, which is given as

S = kB log [1] −→ S = 0 d)

Within the classical regime, the single-particle partition function is given by "
# Z Z p2x + p2y + p2z 1 Z1 = exp − d3 p d3 r h 2mkB T "
# Z p2x + p2y + p2z V = exp − d3 p h 2mkB T =

3 V (2πmkB T ) /2 h

(5.4.261)

And the classical limit is defined as Z1 1 N

(5.4.262)

Substituting (5.4.261) into (5.4.262), V h

(2πmkB T ) N

(2πmkB T ) 2πmkB T 

3

/2



3

/2

1 

Nh V

Nh V  2/3

(5.4.263)

W. Erbsen

STATISTICAL MECHANICS

From (5.4.263), it is easy to see that the condition is 1 T  2πmkB



Nh V

 2/3

Problem 28 A zipper has N links; each link has a state in which it is closed with energy 0 and a state in which it is open with energy ε. We require, however, that the zipper can only unzip from the left end, and that the link number, s, can only open if all links to the left (1, 2, ..., s − 1) are already open. a)

Show that the partition function can be summed in the form: QN =

b)

1 − exp [−(N + 1)βε] 1 − exp[−βε]

(5.4.264)

In the limit ε  kT , find the average number of open links.

The above model is a very simplified model of the unwinding of two-stranded DNA molecules.

Solution Ooooh the zipper problem. I zip you so much. a)

The standard form of the single-particle partition function is   ∞ X εs Z1 = exp − kB T s=0

(5.4.265)

Because the “particles” are distinguishable, then we can gain the full partition from (5.4.265) quite easily  s ∞ X εs Zs = exp − (5.4.266) kB T s=0 We now take a field trip. Let’s take a close look at the following series: N X

xs = 1 + x + x2 + x3 + ... + xN

(5.4.267)

s=0

We now multiply both sides of (5.4.267) by x. This looks like x

N X

xs = x + x2 + x3 + x4 + ... + xN+1

s=0

We now take (5.4.267) and subtract from it (5.4.268). This yields

(5.4.268)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

(1 − x)

N X s=0

xs = 1 − xN+1

(5.4.269)

If we were to rearrange (5.4.269), then it would look like N X

xs =

s=0

1 − xN+1 1−x

(5.4.270)

This exact sum is what we need in order to evaluate (5.4.266)! Zs = b)

1 − exp [− (N + 1) εβ] 1 − exp [−εβ]

(5.4.271)

In general, the way to find the average number of open links, we use the form hsi =

N 1 X s · exp[−sεβ] Zs s=0

(5.4.272)

Now, notice that we can arrive at (5.4.272) if we are very tricky and take a derivative of the following form: hsi =

N ∂ X 1 exp [−sεβ] Zs ∂ (εβ) s=0 | {z }

(5.4.273)

Zs

Rewriting (5.4.273) once more,

hsi =

1 ∂ Zs Zs ∂x

(5.4.274)

Where I have made the temporary substitution x = εβ for convenience. Carrying out the first derivative in (5.4.274),   ∂ ∂ 1 − exp [− (N + 1) x] Zs = ∂x ∂x 1 − exp [−x]     1 ∂ ∂ 1 = [1 − exp [− (N + 1) x]] + [1 − exp [− (N + 1) x]] 1 − exp [−x] ∂x ∂x 1 − exp [−x] " # 1 exp [−x] = [(N + 1) exp [−(N + 1)x]] − [1 − exp [−(N + 1)x]] 2 1 − exp [−x] (1 − exp [−x]) " # exp [−(N + 1)x] 1 − exp [−(N + 1)x] =(N + 1) − exp [−x] 1 − exp [−x] (1 − exp [−x])2 exp [−(N + 1)x] exp [−x] exp [−x] exp [−(N + 1)x] − 2 + 2 1 − exp [−x] (1 − exp [−x]) (1 − exp [−x])   exp [−(N + 1)x] 1 exp [−x] exp [−(N + 1)x] exp [−x] =(N + 1) − − + 1 − exp [−x] 1 − exp [−x] 1 − exp [−x] 1 − exp [−x]   1 exp [−x] exp [−(N + 1)x] exp [−x] = (N + 1) exp [−(N + 1)x] − + 1 − exp [−x] 1 − exp [−x] 1 − exp [−x] =(N + 1)

W. Erbsen

STATISTICAL MECHANICS

=

   1 1 − exp [−(N + 1)x] (N + 1) exp [−(N + 1)x] − exp [−x] 1 − exp [−x] 1 − exp [−x]

(5.4.275)

Where we note that the expression in brackets in (5.4.275) is none other than Zs . Rewriting, 1 ∂ Zs = [(N + 1) exp [−(N + 1)x] − exp [−x] Zs ] ∂x 1 − exp [−x]

(5.4.276)

At this point we are ready to substitute (5.4.276) and (5.4.271) into (5.4.274): 1 1 · [(N + 1) exp [−(N + 1)x] − exp [−x] Zs ] Zs 1 − exp [−x] (N + 1) exp [−(N + 1)x] − exp [−x] = 1 − exp [−x] (N + 1) exp [−(N + 1)x] exp [−x] = − 1 − exp [−x] 1 − exp [−x]

hsi =

(5.4.277)

Putting the numbers back in to (5.4.277) leads us to hsi =

exp [−εβ] (N + 1) exp [−(N + 1)εβ] − 1 − exp [−εβ] 1 − exp [−εβ]

(5.4.278)

We are asked to evaluate the average number of open links from (5.4.278) given the following stipulation: ε  kBT −→ εβ  1 If εβ  1, then all the exponential terms approach zero, and so the average number of open links is hsi = 0

Problem 29 a)

Obtain the Van der Waals equation of state   N 2a p+ (V − N b) = N kT V

(5.4.279)

by making the following two corrections to the ideal gas: i) ii) b)

Use an “effective volume,” instead of the usual volume Include interactions in a mean field form. State clearly where the mean field assumptions come in.

What is the work done by the gas if it expands isothermally from V1 to V2 at some temperature T ?

Solution

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

a)

The ideal gas law is, of course, given by P V = N kB T

(5.4.280)

Let’s apply ii) first to (5.4.280), where we are including interactions from the “mass field form.” Mean field theory essentially consists of replacing all interactions attributed to any one body, with an average, or “effective” interaction. → This reduces a complicated multi-body problem to an effectively single-body problem, whose solutions are easier to find.

But what is the probability that two particles will be in the same place at the same time? → The density (ρ = m/v) is sort of like a probability, in the sense that if the density is large, the probability that two particles will try to be in the same space at the same time increases (this probability actually goes like P ∝ ρ2 ). → Think of two coins: if the probability that one is heads up is ρ, then the probability that they will both be heads up is ρ2 . This adds an “effective pressure” term to (5.4.280). The addition of a new term indicates that due to the chance of spatially co-existing molecules, the pressure is actually more, proportional to several constants. Anyway, with this correction, (5.4.280) becomes
P + aρ2 V = N kB T (5.4.281)

Where the constant a is a proportionality constant, which depends on how much the particles interact with one another. We now recall how we define the particle density: ρ=

N V

Substituting this in to (5.4.281) yields   N2 P + a 2 V = N kB T V

(5.4.282)

Next, we wish to address i), which is attributed to the “effective volume,” which basically means that the two particles cannot coexist at the same place at the same time. → Therefore, the amount of space available for these molecules flying around is actually less then the total volume of the container, by an amount proportional to the number of molecules present: Veff = V − N b

(5.4.283)

Where b is the volume of the individual gas particle, and N is the number of particles in the volume. Therefore, substituting (5.4.283) into (5.4.282), we have   N2 P + a 2 (V − N b) = N kB T (5.4.284) V

W. Erbsen

STATISTICAL MECHANICS

b)

To find the work done by a gas which expands isothermally (constant temperature), from some initial volume Vi to some final volume Vf , we must first recall that Z Vf W = P dV (5.4.285) Vi

At this point, we solve (5.4.284) for P , which will subsequently be substituted into (5.4.285). But first things first: P =

N kB T aN 2 − 2 V − Nb V

And now substituting (5.4.286) back into (5.4.285),  Z Vf  N kB T aN 2 W = − 2 dV V − Nb V Vi Z Vf Z Vf 1 1 dV − aN 2 dV =N kB T 2 V − N b V Vi Vi V  1 f Vf 2 =N kB T [log [V − N b]|Vi − aN − V Vi

=N kB T [log [Vf − N b] − log [Vi − N b]] + aN 2

(5.4.286)



1 1 − Vf Vi



(5.4.287)

We can simplify (5.4.287) slightly, leaving us with     Vf − N b 1 1 W = N kB T log + aN 2 − Vi − N b Vf Vi

Problem 30
A quantum harmonic oscillator has energy levels En = n + 1/2 ~ω0 , where n = 0, 1, 2, ... Treat this single oscillator to be a small system coupled to a heat bath at temperature T . What is the probability of finding the oscillator in its nth quantum state?

Solution The Quantum Harmonic Oscillator (or simply QHO, as we lovingly call it) has the following eigenenergies:
En = n + 1/2 ~ω

(5.4.288)

And if we dutifully recall, the partition function is given by Z=

  En exp − kB T n=0 ∞ X

(5.4.289)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Now, we are interested in finding the QHO in some other state, let’s call it n0 . How do we find this out? Well, we first recall that P (En) =

exp [−En /(kB T )] Z

(5.4.290)

exp [−En0 /(kB T )] Z

(5.4.291)

Now let’s tailor it to our system in need: P (En0 ) = Now, we must find Z: "

#
n + 1/2 ~ω0 Z= exp − kB T n=0 ∞ X

(5.4.292)

Now, the nth excited state is n0 , and (5.4.291) becomes 
 exp − n0 + 1/2 ~ω0 /(kB t) 0 P P (En) = ∞ 1 n=0 exp [− (n + /2 ) ~ω0 /(kB T )] 
 0 1 exp − n + /2 ~ω0 β = P∞ 1 n=0 exp [− (n + /2 ) ~ω0 β] ∞ 
 X 
 0 1 = exp − n + /2 ~ω0 β · exp n + 1/2 ~ω0 β n=0

∞    X = exp −n0 ~ω0 β − 1/2 ~ω0 β · exp 1/2 ~ω0 β exp [n~ω0 β] n=0

∞   X = exp −n0 ~ω0 β − 1/2 ~ω0 β + 1/2 ~ω0 β · exp [n~ω0 β] n=0

= exp [−n0 ~ω0 β] ·

∞ X

exp [n~ω0 β]

(5.4.293)

n=0

At this point we recall the following arcane summation identity ∞ X

ean =

n=0

1 1 − ea

Applying this to (5.4.293), P (En0 ) =

exp [−n0 ~ω0 β] 1 − exp [~ω0 β]

Problem 31 The surface temperature of the sun is T0 (= 5800o K); its radius is R(= 7 × 108 m) while the radius of the

W. Erbsen

STATISTICAL MECHANICS

Earth is r(= 6.37 × 106 m). The mean distance between the sun and Earth is L(= 1.5 × 1013 m). In first approximation, one can assume that both the sun and the Earth absorb all electromagnetic radiation incident upon them. Assume that Earth has reached a steady state at some temperature T . a)

Find an approximate expression for the temperature T of Earth in terms of the astronomical parameters mentioned above.

b)

Calculate the temperature T numerically.

Solution a)

The Stefan-Boltzmann Law gives the total power that the Sun is emitting, and is given by PS = 4πR2 σT04

(5.4.294)

Since the sun emits radiation in all directions, the Earth only absorbs a small fraction of the total output, and depends on the cross sectional area of the Earth and the distance it is from the emitter. Quantitatively, A 4πL2 πr 2 =PS 4πL2 r2 =PS 4L2

PE =PS

(5.4.295)

As suggested in the prompt, if we assume in the first approximation that the Earth is a perfect absorber, then (5.4.295) holds, and we can use (5.4.294) to come up with another expression which describes the power absorbed by the Earth. We start by slightly rewriting (5.4.294) PE = 4πr 2 σT 4

(5.4.296)

At this point, we substitute both (5.4.294) and (5.4.296) into (5.4.295): 4πr 2 σT 4 = 4πR2 σT04 ·

r2 1 −→ T 4 = T04 · R2 · 4L2 4L2

(5.4.297)

From (5.4.297) it is easy to see that T = T0 b)

r

R 2L

Substituting in the appropriate values into (5.4.298), we solve s 7 × 108 m T = 5800 K −→ T = 28.017 K 2 (1.5 × 1015 ) m

(5.4.298)

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CHAPTER 5: DEPARTMENTAL EXAMINATIONS

Problem 32 A 1.00 gram drop of water is supercooled to −5.00o C (remains in liquid state). Then suddenly and irreversibly it freezes and becomes solid ice at the temperature of the surrounding air, 5.00o C. The specific heat of ice is 2.220 J · g−1 · K, and of water is 4.136 J · g−1 · K, and the heat of fusion of water is 333 J · g−1 . a)

How much heat leaves the drop as it freezes?

b)

What is the change in entropy of the drop as it freezes?

Solution m = 1.00 g T1 = −5.00o T2 = −5.00o a)

b)

Cw = 2.220 J · g−1 · K Ci = 4.186 J · g−1 · K



Lf = 3.333 J · g−1

Two equations that have been since long forgotten are Q =mC∆T

(5.4.299a)

Q =mLf

(5.4.299b)

Where we recall that the heat, Q is the key quantity in all this. If a cup of ice is sitting in room temperature, then heat is entering the vessel, causing it to melt. If, however you have the same cup of ice cooled to exactly 32o F (remember, ice doesn’t normally freeze lower than this), and it starts to turn to water of the same temperature. The heat required to make this phase transition may not contribute towards a change in ∆T , however energy is being applied because it takes energy to undergo a phase change. In our case, the sample does not undergo any change in temperature, and the amount of heat that leaves the drop is due only to the change of phase. Therefore, we only need to use (5.4.299b), and applying the appropriate constants, yields
Q = (1.00 g) 333 J · g−1 −→ Q = 333 J (5.4.300) A common form of entropy is

dS =

∆Q T

And substituting in the appropriate values into (5.4.301), we have S=

333 J −→ S = 1.243 J · K−1 273 − 5 K

(5.4.301)