Solutions to Levi Applied Quantum Mechanics 2nd Ed

Applied Quantum Mechanics

Chapter 1 Problems and Solutions

LAST NAME

FI RS T NAME

Useful constants

MKS (SI) 8

Speed of light in free space

c = 2.99792458 × 10 m s

Planck’s constant

h

= 6.58 6.5821 2118 1889 89 ( 26 ) × 10

–1

–16

eV s

Electron mass

× 10 – 3 J s – 19 e = 1.602 1.602176 17646 462 2 (63 ) × 10 C – 31 m 0 = 9.10 9.1093 9381 8188 88 ( 72 ) × 10 kg

Neutron mass

1.6749 4927 2716 16 ( 13 ) m n = 1.67

× 10 – 27 kg

Proton mass

m p = 1.67 1.6726 2621 2158 58 ( 13 )

× 10 – 27 kg

Boltzmann constant

k  B = 1.38 1.3806 0650 503 3 ( 24 )

h

Electron charge

= 1.054 1.05457 5715 1596 96 (82 )

8.6173 7342 42( 15) k  B = 8.61

× 10 –23

× 10 –5 e V

Permittivity of free space

ε0

= 8.8541878

× 10 – 12 F

Permeability of free space

µ0

= 4π

Speed of light in free space

c = 1  ⁄ 

Avagadro’s number

6.0221 2141 4199 99 ( 79 )  N  A = 6.02

Bohr radius

a B = 0.52 0.5291 9177 7721 21 ( 19 ) ×10

× 10 – 7 H

JK

m

m

K

–1 –1

–1

–1

ε 0µ0 × 10 23 mo l – 1 –10

m

4 πε 0 h a B = ---------------m 0e 2

2

Inverse fine-structure constant

α– 1

α –1

= 137.03 137.03599 59976 76 ( 50)

4 πε 0 h c = ----------------e2

Applied quantum mechanics

1

PROBLEM 1

A metal ball is buried in an ice cube that is in a bucket of water. (a) If the ice cube with the metal ball is initially under water, what happens to the water level when the ice melts? (b) If the ice cube with the metal ball is initially floating in the water, what happens to the water level when the ice melts? (c) Explain how the Earth’s average sea level could have increased by at least 100 m compared to about 20,000 years ago. (d) Estimate the thickness and weight per unit area of the ice that melted in (c). You may wish to use the fact that the density of ice is 920 kg m -3, today the land surface area of the Earth is about 148,300,000 km2 and water area is about 361,800,000 km 2. PROBLEM 2

Sketch and find the volume of the largest and smallest convex plug manufactured from a sphere of radius r  = 1 cm to fit exactly into a circular hole of radius r  = 1 cm, an isosceles triangle with base 2 cm and a height h = 1 cm, and a half circle radius r  = 1 cm and base 2 cm. PROBLEM 3

An initially stationary particle mass m1 is on a frictionless table surface and another particle mass m2 is positioned vertically vertically below the edge of the table. The distance from from the particle mass particles are connected by a taught, light, inextensible m 1 to the edge of the table is l. The two particles string of length L > l. (a) How much time elapses before the particle mass m1 is launched off the edge of the table? (b) What is the subsequent motion of the particles? (c) How is your answer for (a) and (b) modified if the string has spring constant

κ ?

PROBLEM 4

The velocity of water waves in shallow water may be approximated as v =

g h where g is the

acceleration due to gravity and h is the depth of of the water. Sketch the lowest lowest frequency standing standing water wave in a 5 m long garden pond that is 0.9 m deep and estimate its frequency. PROBLEM 5

(a) What is the dispersion relation of a wave whose group velocity is half the phase velocity? (b) What is the dispersion relation of a wave whose group velocity is twice the phase velocity? (c) What is the dispersion relation when the group velocity is four times the phase velocity? PROBLEM 6

A stationary ground-based radar uses a continuous electromagnetic wave at 10 GHz frequency to measure the speed of a passing airplane moving at a constant altitude and in a straight line at 1000 km hr-1. What is the maximum beat beat frequency between the out going going and reflected radar beams? Sketch how the beat frequency varies as a function of time. What happens to the beat frequency if the airplane moves in an arc? 2

PROBLEM 1

A metal ball is buried in an ice cube that is in a bucket of water. (a) If the ice cube with the metal ball is initially under water, what happens to the water level when the ice melts? (b) If the ice cube with the metal ball is initially floating in the water, what happens to the water level when the ice melts? (c) Explain how the Earth’s average sea level could have increased by at least 100 m compared to about 20,000 years ago. (d) Estimate the thickness and weight per unit area of the ice that melted in (c). You may wish to use the fact that the density of ice is 920 kg m -3, today the land surface area of the Earth is about 148,300,000 km2 and water area is about 361,800,000 km 2. PROBLEM 2

Sketch and find the volume of the largest and smallest convex plug manufactured from a sphere of radius r  = 1 cm to fit exactly into a circular hole of radius r  = 1 cm, an isosceles triangle with base 2 cm and a height h = 1 cm, and a half circle radius r  = 1 cm and base 2 cm. PROBLEM 3

An initially stationary particle mass m1 is on a frictionless table surface and another particle mass m2 is positioned vertically vertically below the edge of the table. The distance from from the particle mass particles are connected by a taught, light, inextensible m 1 to the edge of the table is l. The two particles string of length L > l. (a) How much time elapses before the particle mass m1 is launched off the edge of the table? (b) What is the subsequent motion of the particles? (c) How is your answer for (a) and (b) modified if the string has spring constant

κ ?

PROBLEM 4

The velocity of water waves in shallow water may be approximated as v =

g h where g is the

acceleration due to gravity and h is the depth of of the water. Sketch the lowest lowest frequency standing standing water wave in a 5 m long garden pond that is 0.9 m deep and estimate its frequency. PROBLEM 5

(a) What is the dispersion relation of a wave whose group velocity is half the phase velocity? (b) What is the dispersion relation of a wave whose group velocity is twice the phase velocity? (c) What is the dispersion relation when the group velocity is four times the phase velocity? PROBLEM 6

A stationary ground-based radar uses a continuous electromagnetic wave at 10 GHz frequency to measure the speed of a passing airplane moving at a constant altitude and in a straight line at 1000 km hr-1. What is the maximum beat beat frequency between the out going going and reflected radar beams? Sketch how the beat frequency varies as a function of time. What happens to the beat frequency if the airplane moves in an arc? 2

PROBLEM 7

How would Maxwell’s equations be modified if magnetic charge g (magnetic monopoles) were discovered? Derive an expression for conservation conservation of magnetic current and write down a generalized Lorentz force law that includes magnetic charge. Write Maxwell’s equations equations with magnetic

ε E + i µH .

charge in terms of a field G = PROBLEM 8

The capacitance of a small metal sphere in air is C 0 = 1.1 with relative permittivity 2. 2 × 10

–18

ε r 

1

× 10 –18

F . A thin dielectric film

= 10 uniformly coats the sphere and the capacitance increases to

F . What is the thickness of the dielectric film film and what is the single electron charging charging

energy of the dielectric coated metal sphere? PROBLEM 9

(a) A diatomic molecule has atoms with mass m1 and m2. An isotopic form of the molecule has atoms with mass m'1 and m'2. Find the ratio of vibration oscillation oscillation frequency ω  /  ω' of the two molecules. 12

16

(b) (b) What What is is the the rati ratio o of vibr vibrat atio iona nall freq freque uenc ncie iess for for carb carbon on mon monox oxid idee isot isotop opee 12 ( C O ) and and 13

16

carbon monoxide isotope 13 ( C O )? PROBLEM 10

(a) Find the frequency frequency of oscillation of the particle particle of mass m illustrated in the Fig. The particle is only free to move along a line and is attached to a light spring whose other end is fixed at point  A located a distance l perpendicular to the line. A force F 0 is required to extend the spring to length l . (b) Part (a) describes a new type type of child’s swing. If the child weighs 20 kg, the length length l = 2.5 m, and the force F 0 = 450 N, what is the period of oscillation? Fixed point A

Spring

Length, l

Mass, m

Displacement, - x

Applied quantum mechanics

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SOLUTIONS Solution 1

(a) The water level decreases. If ice has volume V  then net change in volume of water in the bucket is

∆V

= V ( ρic e  ⁄ ρ water  – 1 ) .

(b) Again, the mass of the volume of liquid displaced equals the mass of the floating object. Assuming the metal has a density greater than that of water, the water level decreases when the ice melts. (c) If there is just ice floating in the bucket, the water level does not change when the ice melts. This fact combined with the results from part (a) and (b) allows us to conclude that only the ice melting over land contributes to increasing the sea level. (d) Today 71% water, 29% land, ratio is 2.45. The average thickness of ice on land is simply 100 m × 2.45 = 245 m. If one assumes half the land area under ice, then average thickness of ice is 490 m. If this is distributed uniformly from thin to thick ice, then one might expect maximum ice thicknesses near 1000 m (i.e. ~ 3300 ft high mountains of ice). Ice weighs one metric tone per cubic meter so the weight per unit area is the thickness in meters multiplied by tones. Solution 2

We are asked to sketch and find the volume of the largest and smallest convex plug manufactured from a sphere of radius r  = 1 cm to fit exactly into a circular hole of radius r  = 1 cm, an isosceles triangle with base 2 cm and a height h = 1 cm, and a half circle radius r  = 1 cm and base 2 cm. The minimum volume is 1.333 cm 3 corresponding to a geometry consisting of triangles placed on a circular base as shown in the Fig.

2r  = 2 cm

h = 1 cm

r  = 1 cm

2 cm

To calculate this minimum volume of the plug consider the triangle at position  x from the origin has height k , base length 2 l, and area kl.

4

h r 

r  r 

 x k 

l

l

( r - x)

l

2

2

2

= k  = ( r + x ) ( r – x ) = r  –  x Volume of the plug is r 

∫ 

r 

2

Vo l = 2 k  d   x = 2 0

2 2 ∫ ( r  –  x ) d x 0

2

3 r 

x 2 = 2 r  x – ---3

0

4 3 = - r  3

Since r = 1 cm the total volume is exactly 4/3 = 1.333 cm 3. To find the maximum volume of the plug manufactured from a sphere of radius r  = 1 cm we note that the geometry is a half sphere with two slices cut off as shown in the following Fig. The geometry is found by passing the sphere along the three orthogonal directions  x, y , and z and cutting using a circle, a triangle, and a half circle, and a triangle respectively. As will be shown, the volume is 1.608 cm3. This is the maximum convex volume of the plug manufactured from a sphere of  radius r = 1 cm.

2 r = 2 cm

h = 1 cm

r = 1 cm

2 cm

To calculate the volume we first calculate the volume sliced off the half sphere as illustrated in following Fig.

Applied quantum mechanics

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l l

( r - x) r 

(r -  x)  x

l r 

r   x r 

r   /21/2

2

2

2

= ( r + x ) ( r – x ) = r  –  x so that area of disk radius l at position  x is l

π ( r 2 – x 2 )

 Area =

The volume of the disk is the area multiplied by the disk thickness dx . The total volume is the integral from  x = r  ⁄  2 to x = r . Remembering to multiply by 2 because there are two slices gives total volume sliced off  r 

Vo l off  = 2 π

∫  ( r  – x 2

2

) d x

= 2π

r  ⁄  2

3 r 

x r  x – ---3 2

0

1 1 1 3 = 2 π r    1 – ---  –  ------- – ----------     3   2 6 2  

5 3 2 Vo l off  = 2 π r   - – ----------    3 6 2   The volume of the plug is just the volume of a half sphere minus Vo l of f . 3

2 π r  3 2 5 3 1 2 5 3 5 1 Vo l = ----------- – 2 π r   -- – ----------   = 2 π r   -- – -- + ----------   = 2 π r   ---------- – ---           3 3 6 2 3 3 6 2 6 2 3 Since r = 1 cm the total volume is 1.608 cm 3. If we lift the restriction that the plug is manufactured from a sphere of radius 1 cm, then calculating the maximum convex volume turns out to be a bit complicated as may be seen in the following Fig. The geometry is found by cutting along the three orthogonal directions  x , y , and z using a circle, a triangle, and a half circle, and a triangle respectively. The volume is 1.659 cm 3.

2 r = 2 cm

h = 1 cm  z

 y r = 1 cm  x

2 cm

6

Solution 3

(a) and (b). Force due to gravity acts on particle mass m2. This force is F  = gm2 where g is the acceleration due to gravity. The light string transmits the force to the second particle mass m1 that is free to slide horizontally on the table surface. If  x is the vertical position of the particle with mass 2

g m2 = ------------------m 1 + m2 d t  dx

m 2 then

2

and, starting from rest, speed

gm 2 t  dx = ------------------- with displacement d t  m 1 + m2

2

gm 2 t   x = --------------------------- . Eventually, particle m 1 is launched with velocity v 1 from the end of the table, at 2 ( m 1 + m2 )

which point gravity causes it to accelerate in the x -direction under its own weight. If the distance of  2 l ( m 1 + m 2) ----------------------------- for m1 to reach m 2g

particle mass m1 to the edge of the table is l, then it takes time t  =

gm2 2 l (m 1 + m 2 ) The launch velocity is v 1 = ------------------- ----------------------------. The initial angular momentum of  m1 + m2 m 2g the system when particle mass m1 is launched from the edge of the table is Lv 1m1m2 /(m1+m2).

the edge.

(c) If the string has a spring constant then part of the energy of the system can be stored in the string during its trajectory. For example, this will happen during the initial acceleration phase. Solution 4

We are given wave velocity, v =

gh . This is a constant because h and g are fixed. Hence,

group velocity and phase velocity are the same and lowest frequency standing wave =

10

× 0.9  ⁄ 10

λ

ω

= 2π f =

gh× k = 

g h × 2 π ⁄ λ . For the

= 2 × 5 m and using g ~ 10, then the frequency of the wave is

= 0.3 H z .

Solution 5

ω

(a)

0.5

= Ak  .

(b) Because the group velocity v g = and

so

d ω dk  ------- = 2 ------ . ω k 

∂ω is twice the phase velocity ∂ k 

Integrating both

sides of

this

ln ω = 2 ln k + ln A = ln Ak  so that the dispersion relation is 2

(c)

ω

v p =

equation

ω

ω

ω  ⁄ k  we have d ω

= 2 ---d k  k 

allows

one to

write

2

= Ak  where A is a constant.

= Ak  .

Solution 6

For a source moving at velocity v, the Doppler shifted frequency  f ’ is  f ′ = f 0 ⁄ ( 1 ± v c ) where  f 0 is the original frequency. The ± sign refers to the source moving towards or away from the detec-

tor.

∆ f

Hence, = f 0 ( 1 – 1

and  f 0 = 10 v

×

10

the

(1 ± v

c) ) =

maximum

change

±f 0 ( v  ⁄ ( c + v) ) .

in

frequency

In our case, v = 3.6

× 10

relative 8

-1

to

ms ,c=3

 f 0 8

× 10

is -1

ms ,

Hz giving ∆ f = 9.26 kHz. The component of velocity in the direction of the detector is

cos(θ ) where

θ

is the angle between the ground and the plane flying at height h . A simple

expression for the time dependence of the angle

θ = θ( t ) is found knowing that the aircraft is mov-

ing at a constant velocity (and assuming the airplane flies right over the ground station). If the dis-

Applied quantum mechanics

7

2

2

2 2

tance to the plane (range) is l then l = h + v t  (assuming we set t  = 0 when the plane is 2

2 2

overhead). Then cos ( θ ( t ) ) = v t  ⁄ l = v t ⁄  h + v t  = 1 ⁄  h

2

 ⁄ v 2 t 2 + 1 .

The Doppler shift as a

function of time is ∆ f  × cos( θ (t )). If the plane does not fly overhead it is necessary to introduce another angle

φ to describe its position.

If the plane is flying in an arc the Doppler shift can decrease. For example, it will be zero if the plane is flying in a circle centered on the ground station. Assuming the plane flys at a constant speed, whatever trajectory the plane describes, it cannot have a Doppler shift that is greater than the maximum value we calculated ∆ f  for that velocity. The reason why a typical radar antenna at an airport is highly directional and rotates is to find the angular position of the aircraft with respect to the radar detector. Pulsed radar can be used to find the range (distance) to the aircraft. The trajectory of the aircraft can be obtained by comparing sequential position updates. Doppler radar is not required. Solution 7

If magnetic charge g existed Maxwell’s equations would have to include magnetic charge density

ρm

and magnetic current density J m as well as the usual electron charge density

ρ e and elec-

tron current density J e . The new equations would be

∇ ⋅ D = ρe ∇ ⋅ B = ρm ∇× E = – ∂----B--- – Jm ∂ t  ∇× H = J e + ∂----D --∂ t  where D = ε E and

B =

µH .

Conservation of magnetic current J m =

ρ mv

is expressed as

∂ρ m + ∇ ⋅ Jm = 0 ∂ t  For a particle carrying both electric charge e and magnetic charge g , the Lorentz force is F = e( E + v

× B ) + g( B – v × E)

Reminding ourselves of the SI units used for magnetic intensity H[A m -1], magnetic flux density

µH, electric field strength E[V m -1],the displacement vector field D = ε E, permia2 bility µ [N A-2] or [H m -1], permittivity ε [A2 s2 N-1 m-2] or [F m -1], and ε 0 = 1 ⁄ µ 0 c where c is B [T] where B =

*

the speed of light in free-space. If G G corresponds to energy density U  = then it makes sense to write G =

ε

µ

-- E + i --- H since 2 2

ε E2

and

µH2

( E ⋅ D + B ⋅ H )  ⁄ 2

have dimensions of energy

density U [J m -3]. Maxwell’s equations in terms of E and H with magnetic charge can be written

ε

ρe

-- ∇ ⋅ E = ---------2 2ε

µ--- ∇ ⋅ H = ---ρ----m---2 2µ H ε- ∇× E = – ε--- µ ∂------+ J    2 2 ∂ t  m  8

µ--- ∇× H = 2

µ---  ε ∂-----E-- + J   e   2  ∂ t 

It follows that

ρm

ρ

∇⋅G

= -------e--- + i ----------2ε 2µ

and

∇ ×G

=

ε-- ∇× E + i µ--- ∇× H

2 2 which can be rewritten as 1 ---------- ∇

εµ

×G

= i

ε ∂H

= – --- µ ------- – 2 ∂ t 

ε- J 2

µ ∂E

m

µ

+ i --- ε ------- + i --- J e 2 ∂ t  2

J iJ d G – -------m---- + --------e-d t  2µ 2ε

since i

µ ∂H ε∂E d G = – --- ------- + i -- ------d t  t  ∂ 2 2 ∂ t  The complex field G can be used to simplify the usual four Maxwell equations to just two equa-

tions. Solution 8

We are given the information that a small metal sphere has capacitance C 0 = 1.1

× 10– 18 F  in air.

First, we calculate the radius r 0 of the sphere using the formula C 0 = 4 πε 0 r 0

It follows that –18 C 0 1.1 × 10 r 0 = ----------= --------------------------------------–-- = 10 n m 12 4 πε 0 4 π × 8.85 × 10 Now consider the case when the metal sphere radius r 0 with charge Q is coated with a dielectric of 

relative permittivity

εr 

to radius r 1 and

ε r 

from r 1 radius to r 2. The voltage drop from r 0 through the dielectrics to a much larger metal sphere of radius r 2 is found by integrating 1

2

r 0

r 1

r 0

r 1

∫ 

∫ 

∫ 

1

2

1

d r  ∫ ---------------------4 πε ε r 

Q V = – E r d r – E r d r  = – ----------------------2 d r  – 4 πε 0 ε r 1 r  r  r  r 

Q

2

r 2

0 r 2

Q 1 1 1 1 V  = -----------  ---------- – ---------- + ---------- – ----------   4 πε 0  ε r 1 r 0 ε r 1 r 1 ε r 2 r 1 ε r 2 r 2 

For this particular problem we now take the limit r 2 →

∞.

Capacitance

4 πε 0 4 πε 0 Q = ------------------------------------------------C  = ---- = ---------------------------------------------------------V  1 1  -----1----- – ---------  ε r 2 – ε r 1    – 0    ε r  r  ε r  r  + ---------   -----1----- – --1--  ----------------   ε r  r  1 0 1 1 2 1  ε r 1 r 0 r 1  ε r 1 ε r     2

C  ε r 2 – ε r 1   C  ---------- – ----  ----------------  = 4 πε 0 ε r 1 r 0 r 1  ε r 1 ε r    2

Applied quantum mechanics

9

C  C  ε r 2 – ε r 1  ---------- – 4 πε 0 = ----  ----------------  r 1  ε r 1 ε r    ε r 1 r 0 2

 ε r 2 – ε r 1   ε r 1 r 0 r 1 = C  ---------------- -------------------------------- =  ε r 1 ε r   C – 4 πε 0 ε r 1 r 0 2

r  – ε r  r 0  ε---------------- ------------------------- ε r    r  C 0    1 – ε---------- C   

Putting in the numbers. We have

2

1

2

ε r 

1

1

= 10 ,

εr 

2

= 1 , C 0 = 1.1 × 10

–18

F , C  = 2.2 × 10

–18

F ,

and r 0 = 10 nm , so that r 1 =

 1------–-----10   r 0 =  1 ----  ------------------- 1 – 10    ---2---  

9 -- r 0 = 22.5 nm 4

and we may conclude that the thickness of the dielectric film is r 1 – r 0 = 12.5 n m . Charging energy ∆ E = e

2

 ⁄ 2 C  =

1.6

× 10 – 19 ⁄ 4.4 × 10– 18

= 36 m eV . This value is similar to

ambient thermal energy k  B T  = 25 m eV . Solution 9

(a) The molecule consists of two particles mass m 1 and mass m 2 with position r 1 and r 2 respectively. The center of mass coordinate is R and relative position vector is r . We assume that the potential depends only on the difference vector r = r 2 – r 1 and if the origin is the center of  mass then m1 r 1 + m 2 r 2 = 0 , so that m2 ( m1 + m 2 )

r 1 = ----------------------- r

and m1 ( m1 + m 2 )

r 2 = ----------------------- r

Since, for example, – m2 m1

–m 2 m1

r 1 = --------- r 2 = --------- ( r 1 – r )

m2

2 r 1  1 + ------   = ------ r



m m 1  

m1

r 1 ( m 1 + m2 ) = m 2 r

m2

r 1 = ----------------------- r

( m1 + m 2 )

Now, combining center of mass motion and relative motion, the Hamiltonian is the sum of kinetic and potential energy terms 2 m2 m1 1 · · 1 · · 2 r  + -- m 2 R + ----------------------- r   + V ( r H = T+ V =   - m 1  R – ----------------------2  ( m 1 + m 2)   2  ( m 1 + m2 )  

)

where the total kinetic energy is 2

2

m1 m m m 1 ·2 ·2 ·2 1 · 2 1 m 1 m2 · 2 r T  = -- ( m 1 + m 2 ) R + ---------------------2----- 2 r + -------------1--------2----- 2 r = -- ( m 1 + m 2 ) R + -- ----------------------2 2 2 ( m1 + m 2 ) ( m1 + m 2 ) ( m 1 + m2 ) or

1 · 2 1 ·2 T  = -- M R + -- m r 2 2

10

where  M = m 1 + m 2

and the reduced mass is m 1 m2 m = ----------------------( m1 + m 2 )

m1 r

m2

R

Because the potential energy of the two interacting particles depends only on the separation between them, V = V ( r 2 – r 1

)

= V ( r ) . The frequency of oscillation of the molecule is deter-

mined by the potential and given by

ω

=

κ  ⁄ m

κ  is

where

the spring constant and m is the

reduced mass 1  ⁄ m = 1 ⁄ m 1 + 1  ⁄ m2 . Since the isotope form of the molecule interacts in the same

κ

way,

ω′ ----ω

=

κ′ , and the ratio of oscillation frequency is given by

m 1 m 2 ( m ′1 + m ′ 2 ) ----------------------------------------m ′1 m ′ 2 ( m 1 + m 2 ) We could get the same result quite simply if we assume motion is restricted to one dimension.

=

In that case the force on mass m1 and mass m2 is given by their relative displacement multiplied by the spring constant κ .

m1 u

κ 

m2 v

2

m1

du 2

d t 

=

κ ( v – u )

=

κ ( u – v )

2

m2

d v 2

d t 

which has solution of the form e

( κ  – m 1 ω 2) u – κ v = 0 2 – κ u + ( κ – m2 ω ) v =

–i ωt 

giving

0

giving a characteristic equation

Applied quantum mechanics

11

κ – m1 ω 2 – κ  κ – m 2ω 2 – κ  m 1 m 2   2 - ω κ  =  -----------------m 1 + m 2 

(κ  – m 1 ω 2 ) (κ  – m 2 ω 2 ) – κ 2

=

= mω

= 0

2

and

ω

m 1 + m 2  κ  ------------------ = m 1 m 2  

=

κ  ⁄ m κ

and, as before, since the isotope form of the molecule interacts in the same way,

=

κ′ , and

the ratio of oscillation frequency is given by

ω′ ----ω

m 1 m 2 ( m ′1 + m ′ 2 ) ----------------------------------------m ′1 m ′ 2 ( m 1 + m 2 )

=

(b) To find the ratio of vibrational frequencies we use the result in part (a) and put in the atomic

ω ω

12

C 

masses to give --------- = 13

C 

13 × 16 × ( 12 + 16 ) ----------------------------------------------- = 1.02 . As expected, the lighter 12 × 16 × ( 13 + 16 )

vibrates at a higher frequency (by about 2%) than the

13

12

16

C O molecule

16

C O molecule.

Solution 10

(a) For small displacement  x we assume F 0 doesn’t change much so for extension

∆l

the work 

done is F 0 ∆ l . Hence, the potential energy of the spring is equal to the force F 0 multiplied by the

∆ l . For x « l we have ∆ l = l 2 + x 2 – l = x 2 ⁄ 2 l , so that the potential energy is 2 2 F 0 x  ⁄ 2 l = κ  x  ⁄  2 . This is the potential energy of a harmonic oscillator with spring constant F 0  ⁄  l . This oscillator has frequency ω = κ  ⁄  m = F 0 ⁄  m l .

extension V =

κ  =

(b)

The

frequency

ω = F 0  ⁄ ml = τ = 1 ⁄ f  = 2 π ⁄ ω

12

450  ⁄ 20

of

× 2.5

oscillation =

in

radians –1

9 = 3 rad s ,

= 2.09 s , i.e., about 2 seconds.

so

per the

second period

of

is

given

oscillation

by is

Applied Quantum Mechanics

Chapter 2 Problems and Solutions

LAST NAME

FIRST NAME

Useful constants

MKS (SI)

Electron mass

× 1 08 m s –1 – 16 h = 6.58211889 ( 26 ) × 10 eV s –3 h = 1.054571596 ( 82 ) × 10 J s – 19 e = 1.602176462 (63 ) × 10 C –31 m 0 = 9.10938188 ( 72 ) × 10 kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10

–27

kg

Boltzmann constant

k  B = 1.3806503 ( 24 )

Speed of light in free space Planck’s constant

Electron charge

c = 2.99792458

k  B = 8.617342 (15 )

× 10 –23

× 10 – 5

Permittivity of free space

ε0

= 8.8541878

× 10 – 12 F

Permeability of free space

µ0

= 4 π × 10

H m

Speed of light in free space

c = 1  ⁄ 

Avagadro’s number

–7

eV K m

–1

–1

–1

–1

ε 0µ0

 N  A = 6.02214199 ( 79 )

× 10 23 mo l –1

a B = 0.52917721 ( 19 ) ×10

Bohr radius

JK

–10

m

4 πε 0 h a B = ---------------m 0e2

2

Inverse fine-structure constant

α– 1

α –1

= 137.0359976 ( 50)

4 πε 0 h c = ----------------e2

Applied quantum mechanics

1

PROBLEM 1

(a) The Sun has a surface temperature of 5800 K and an average radius 6.96 the mean Sun-Mars distance is 2.28

×

11

10

× 108 m.

Assuming

m, what is the total radiative power per unit area inci-

dent on the upper Mars atmosphere facing the Sun? (b) If the surface temperature of the Sun was 6800 K, by how much would the total radiative power per unit area incident on Mars increase? PROBLEM 2

If a photon of energy 2 eV is reflected from a metal mirror, how much momentum is exchanged? Why can the reflection not be modeled as a collision of the photon with a single electron in the metal? PROBLEM 3

Consider a lithium atom (Li) with two electrons missing. (a) Draw an energy level diagram for the Li ++ ion. (b) Derive the expression for the energy (in eV) and wavelength (in nm) of emitted light from transitions between energy levels. (c) Calculate the three longest wavelengths (in nm) for transitions terminating at n = 2 . (d) If the lithium ion were embedded in a dielectric with relative permittivity

εr

= 10, what

would be the expression for the energy (in eV) and wavelength (in nm) of emitted light from transitions between energy levels. PROBLEM 4

(a) There is a full symmetry between the position operator and the momentum operator. They form a conjugate pair . In real-space momentum is a differential operator. Show that in k -space position is a differential operator, i h

∂ , by evaluating expectation value ∂ p x

∞

〈 xˆ 〉

=

*

∫  dx ψ  ( x ) x ψ ( x )

–∞

in terms of φ (k  x) which is the Fourier transform of  ψ ( x ). (b) The wave function for a particle in real-space is

ψ ( x , t ) .

Usually it is assumed that position

 x and time t  are continuous and smoothly varying. Given that particle energy is quantized such that  E  =

h

ω , show that the energy operator for the wave function ψ ( x, t )

is i h

∂. ∂ t 

PROBLEM 5

A simple model of a heterostructure diode predicts that current increases exponentially with increasing forward voltage bias. Under what conditions will this predicted behavior fail? PROBLEM 6

Write down the Hamiltonian operator for (a) a one-dimensional simple harmonic oscillator, (b) a Helium atom, (c) a Hydrogen molecule, (d) a molecule with nn nuclei and ne electrons.

2

PROBLEM 7

Calculate the classical velocity of the electron in the n-th orbit of a Li ++ ion. If this electron is described as a wave packet and its position is known to an accuracy of  ∆ x = 1 pm, calculate the characteristic time ∆τ ∆ x for the width of the wave packet to double. Compare ∆τ∆ x with the time to complete one classical orbit. PROBLEM 8

What is the Bohr radius for an electron with effective electron mass m * = 0.021 × m 0 in a medium with low-frequency relative permittivity

ε r 0

= 14.55 corresponding to the conduction

band properties of single crystal InAs? PROBLEM 9

Because electromagnetic radiation possesses momentum it can exert a force. If completley absorbed by matter, the absorbed electromagnetic radiation energy per unit time per unit area is a pressure called radiation pressure. (a) If the maximum radiative power per unit area incident on the upper Earth atmosphere facing the Sun is 5.5 kW m -2, what is the corresponding radiation pressure? (b) Estimate the photon flux needed to create the pressure in (a). (c) Compare the result in (a) with the pressure due to one atmosphere.

Applied quantum mechanics

3

SOLUTIONS

Solution 1

(a) 2.4 kW m -2. (b) 4.5 kW m -2. Solution 2

θ normal to a flat h ω  ⁄  c where frequency ω is

Assume the photon of energy  E  = 2 eV is incident from free-space at angle metal mirror. The magnitude of photon momentum is  p ph = given by the quanta of photon energy  E  =

∆ p

( 2 h ω c ) cos θ which, –19 8 2 E ⁄ c = 2 × 2 × 1.6 × 10  ⁄ 3 × 10 =

h

ω.

= 2.1 × 10

=

The momentum change upon reflection is

θ

for –27

h k 

= 0,

kg m s

–1

gives

a

value

. The reason why this reflection can

not be modeled as a collision of the photon with a single electron in the metal is that in such a situation it is not possible to conserve both energy and momentum and have the photon maintain its wavelength. Collision with an electron would take away energy from the photon and cause a change in photon wavelength. Since, by definition, reflected light has the same wavelength as the incident light, photon energy cannot change. Another way to express this is that the difference in dispersion relation for free electrons and photons is such that light cannot couple directly to a free electron. Solution 3 2

 Z   E n =  Ry -----2 n

122.4 For Li++ Z = 3 and so  E n = ---------2---- e V . n Emission wavelength is

λ 5, 2

λ

2πh c – 10.1 = -------------------- = ---------------- nm , giving 1 1  E n 2 – E n 1 ----- – ----n2 n1

λ 3, 2

= 72.7 nm ,

λ 4, 2

= 53.8 n m ,

= 48.1 n m . 2 4

– m0 Z  e  1 1 = 10 we have  E n 2 –  E n 1 = -------------------------------- – -----  , so divide energy differences by 2 2 ( 4 π ε0 ε r ) h n 2 n 1  

For

ε r 

ε 2r 

= 100

Solution 4

(a) The expectation value of the position operator is ∞

〈 xˆ 〉 =

*

∫  dx ψ  ( x ) x ψ ( x )

–∞

〈 xˆ 〉 = ---1--2π

∞

1 = -----2π

∞ *

∫ dx ∫ d k ′φ  ( k ′ ) e

–∞

–∞

∞

∞ *

∫  dx ∫ dk ′φ  ( k ′ ) e

–∞

– ik ′ x

–i k ′ x

–∞

∞

x

∫ d k φ  (k ) e

ikx

–∞

∞

ikx e   ∂  ------x ∫  dk φ  ( k ) ∂ k  ix   –∞

4

〈 xˆ 〉 〈 xˆ 〉

∞

1 = -----2π

∞

ikx

d k ′φ 

∫  ∫  dx

–∞

–∞

∞

∞

–1 = -------2πi

*

( k ′ ) e – ik ′ x x  φ ( k ) e-------

∫ d x ∫ dk ′φ  ( k ′ )e

–∞

– i k ′ x

–∞

–∞

–∞

∂ i kx ∫ dk  ∂ k φ ( k )  e

i = -----2π

–∞

∂ ∫  ∫ d k dk ′φ ( k ′ ) ∂ k φ( k )δ ( k – k ′ ) ∞ ∞ *

= i

ik x ∂ φ (k )  e------  ∂ k    ix  

dk 

∞

∞ ∞

〈 xˆ 〉

∫ 

–

ix

*

∞

∞

∞

∫  dx e

∞ ∞

i( k – k ′ ) x

–∞

*

∫ d k φ ( k ′ ) ∂ k φ( k )

Hence, in k -space position is a differential operator, i h

–

∞

∂

*

= ih

–∞

– –

∂ ∫  ∫ dk d k ′φ (k ′ ) ∂ k φ (k ) ∞ ∞

–

∂

*

= i

∞

∫ d k φ ( k ′ ) ∂ p φ( k )

–∞

∂. ∂ p

(b) The expectation value of the energy operator is ∞

ˆ〉 〈 E 

=

∫ 

–∞

ˆ〉 〈 E 

1 * d ωψ  ( ω ) h ω ψ ( ω ) = -----2π

1 = -----2π

∞

∫  ∞

∞

∫ 

–∞

–h = -------2 πi

∫ 

*

dt ′φ

*

∫ 

( t ′ )e

∞

d ω

–∞

= ih

∫  ∫ 

∫ 

*

–∞

(t ′ ) e –i ωt ′h ω ∫ d t φ ( t ) e i ωt  –∞

iω t ′

i ω t 

e h ω  φ ( t ) ------ iω

∞

∞

– –∞

∫ 

–∞

iω t  ∂ φ ( t )  e------   ∂ t   i ω  

d t 

∞

∫ 

dt ′φ

–∞

*

*

dt d t ′ φ ( t ′ )

–∞ –∞

∞

∞ ∞

ih * ∂ i ωt  i ω( t – t ′ ) ( t ′ ) e – iω t ′ ∫ dt  ∂ φ ( t )   ∫  ∫ d t d t ′φ (t ′ ) ∂ t φ( t )  e = -2----π- ∫ d ω e t  ∂ –∞ –∞ –∞ – ∞

∞ ∞

ˆ〉 〈 E 

–iω t ′

–∞

∞

∫ 

–∞

∞

d t ′φ

(t ′ ) e –i ωt ′h ω ∫ d t φ ( t ) ∂  e---------   ∂ t  i ω   –∞

d t ′φ

∫ 

–∞

d ω

∞

d ω

∞

–∞

ˆ 〉 = ---1--〈 E  2π ˆ〉 〈 E 

∞

d ω

∞

∂ φ( t )δ (t – t ′ ) ∂ t 

∞

= ih

*

d t φ ( t ′ )

∫ 

–∞

Hence, in t -space energy is a differential operator, i h

∂ φ ( t ) ∂ t 

∞

= ih

*

∂

∫ d t φ ( t ′ ) ∂ t φ ( t )

–∞

∂ . ∂ t 

Solution 5

Series resistance. Space charging at current density j > nev . For bipolar devices with direct a band gap can have stimulated emission. Solution 6

We are asked to write down the Hamiltonian operator for (a) a one-dimensional simple harmonic oscillator, (b) a Helium atom, (c) a Hydrogen molecule, (d) a molecule with nn nuclei and ne electrons. (a) The Hamiltonian for a particle mass m restricted to harmonic oscillatory motion of frequency ω in the x -direction is 2

2

2

mω 2 –h d H = T + V =  -------+ ---------- x 2 2 m d  x 2

(b) and (c) may be found from the general solution (d) for a molecule with n n nuclei and ne electrons in which the Hamiltonian is  j = nn

2

–h 2 --------- ∇ +  H  = 2 M  j  j = 1

∑

i = ne

2

–h 2 --------- ∇ + V ( r ) 2 m0 i=1

∑

Applied quantum mechanics

5

Here, the first term is the contribution to the kinetic energy from the n n nuclei of mass M  j , the second term is the kinetic energy contribution from the n e electrons of mass m0, and V ( r ) is the potential energy given by 2

V ( r ) =

2

2  Z  j Z  j ′ e  Z  j e e ------------------ + ------------------ – ---------------i > i′ 4 πε 0 r ii ′  j > j ′ 4 πε 0 r  i, j 4 πε 0 r ij  jj ′

∑

∑

∑

The first term in the potential is the electron-electron coulomb repulsion between the electrons i and i'. The second term is the nucleus-nucleus coulomb repulsion between the nuclei  j and  j ' with

charges  Z   j and  Z   j'  respectively. The third term is the coulomb attraction between electron i and nucleus j. Solution 7

For n = 1, orbit time ~ 10 -17 s, dispersion time ~ 10 -20 s. Solution 8

4 πε 0 ε r h ε r  * = a B -------= 0.529 a B = --------------------2 m 0 m eff e me ff  2

× 1 0–10 × 693

= 36.7 nm

Solution 9

(a) The maximum radiative power per unit area incident on the upper Earth atmosphere facing the Sun is Stotal = 5.5 kW m -2. The radiation pressure us just S total ⁄ c = 1.8 × 10 (b) From Exercise 1 Chapter 2, unit area is S total ⁄ e h ω average = 1.8

h

ω average

× 10

22

s

–1

–5

–2

N m .

= 1.92 eV  so the number of photons per second per –2

m .

(c) The radiation pressure in (a) is a small value compared to one atmosphere which is about 5

–2

10 N m . Never-the-less, electromagnetic radiation from the Sun can be absorbed, exert a force, and change the direction of small particles in space. For example, it is responsible for continuously sweeping dust particles out of the solar system. As another example, uncharged dust particles from a comet can form a dust tail whose direction and shape is determined by electromagnetic radiation pressure from the Sun. Interestingly, comets can also shed charged dust particles to form what is called a gas tail. The gas tail is swept along by a stream of charged particles and magnetic field lines emitted by the Sun that is called the solar wind. Often a comet will have two separate tails because the solar wind often does not point radially outward from the Sun.

6

Applied Quantum Mechanics

Chapter 3 Problems and Solutions

LAST NAME

FIRST NAME

Useful constants

MKS (SI)

Electron mass

× 1 08 m s –1 – 16 eV s h = 6.58211889 ( 26 ) × 10 –3 h = 1.054571596 ( 82 ) × 10 J s – 19 e = 1.602176462 (63 ) × 10 C –31 kg m 0 = 9.10938188 ( 72 ) × 10

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10

–27

kg

Boltzmann constant

k  B = 1.3806503 ( 24 )

Speed of light in free space Planck’s constant

Electron charge

c = 2.99792458

k  B = 8.617342 (15 )

× 10 –23

× 10 – 5

Permittivity of free space

ε0

= 8.8541878

× 10 – 12 F

Permeability of free space

µ0

= 4 π × 10

H m

Speed of light in free space

c = 1  ⁄ 

Avagadro’s number

–7

eV K m

–1

–1

–1

–1

ε 0µ0

 N  A = 6.02214199 ( 79 )

× 10 23 mo l –1

a B = 0.52917721 ( 19 ) ×10

Bohr radius

JK

–10

m

2

a B

4 πε 0 h = ---------------m 0e2

Inverse fine-structure constant

α– 1

α –1

= 137.0359976 ( 50)

4 πε 0 h c = ----------------e2

Applied quantum mechanics

1

PROBLEM 1

Prove that particle flux (current) is zero if the one-dimensional exponential decaying wave func– κ  x – i ωt  tion in tunnel barrier of energy V 0 and finite thickness  L is ψ ( x, t ) = Be , where κ  is a real positive number and particle energy  E  = h ω < V 0 . PROBLEM 2

(a) Use a Taylor expansion to show that the second derivative of a wavefunction

ψ ( x ) sampled

at positions  x j = jh 0 , where  j is an integer and h 0 is a small fixed increment in distance  x, may be approximated as 2

ψ ( x j – 1 ) – 2 ψ ( x j ) + ψ ( x j + 1 ) = -----------------------------------------------------------------2 d  x h0 (b) By keeping additional terms in the expansion, show that a more accurate approximation of  d 

2

ψ (x j )

the second derivative is 2

d  d  x

2

ψ (x j )

– ψ ( x j – 2 ) + 16 ψ ( x j – 1 ) – 30 ψ ( x j ) + 16 ψ ( x j + 1 ) – ψ ( x j + 2 ) = --------------------------------------------------------------------------------------------------------------------------------------2 12 h 0

PROBLEM 3

Using the method outlined in Exercise 7 of Chapter 3 as a starting point, calculate numerically the first four energy eigenvalues and eigenfunctions for an electron with effective mass * m e = 0.07 × m0 confined to a potential well V ( x ) = V 0 of width  L = 10 nm with  periodic boundary conditions. Periodic boundary conditions require that the wave function at position  x = 0 is connected (wrapped around) to position  x =  L. The wave function and its first derivative are continuous and smooth at this connection. Your solution should include plots of the eigenfunctions and a listing of the computer program you used to calculate the eigenfunctions and eigenvalues. PROBLEM 4

Using the method outlined in Exercise 7 of Chapter 3 as a starting point, calculate numerically the first six energy eigenvalues and eigenfunctions for an electron with effective mass * m e = 0.07 × m0 confined to a triangular potential well of width  L = 20 nm bounded by barriers of  infinite energy at  x < 0 and  x > L . The triangular potential well as a function of distance  x is given by V ( x ) = V 0 × x  /  L where V 0 = 1 eV. Explain the change in shape of the wave function with increasing eigenenergy. Your solution should include plots of the eigenfunctions and a listing of the computer program you used to calculate the eigenfunctions and eigenvalues. PROBLEM 5

Calculate the transmission and reflection coefficient for an electron of energy  E , moving from left to right, impinging normal to the plane of a semiconductor heterojunction potential barrier of  energy V 0 , where the effective electron mass on the left-hand side is m 1 and the effective electron mass on the right-hand side is m 2 . If the potential barrier energy is V 0 = 1.5 eV and the ratio of effective electron mass on either side of the heterointerface is m1 /  m2 = 3, at what particle energy is the transmission coefficient unity? What is the transmission coefficient in the limit E → ∞ ?

2

SOLUTIONS Solution 1

To prove that particle flux (current) is zero if the one-dimensional exponential decaying wave – κ  x – iω t  function in tunnel barrier of energy V 0 and finite thickness  L is ψ ( x , t ) = Be , where κ  is a real positive number and particle energy  E  = h ω < V 0 , we substitute the wave function into the current operator J = ----------- ψ  ( x , t )

– i eh 2m 

– i eh 2m

*

d  ψ ( x, t ) – d  x

ψ ( x, t ) d  ψ * ( x , t )     d   x

* – κ  x + i ωt 

J = ----------- ( B e

κ i eh

J = ----------- B

2m

2

( – κ  B e – κ  x – i ωt ) – Be – κ  x – i ωt ( – κ  B * e – κ  x + i ω t ))

( e –2 κ  x – e – 2κ  x)

= 0

Solution 2

(a) Consider the Taylor expansion for the function  f ( x ).

( x j + 1 )

 f ′′ ( x j ) 2 f ′′′ ( x j ) 3 = f ( x j ) + f ′ ( x j ) h 0 + -------------- h 0 + ---------------- h 0 + 2! 3!

…

=

∑ -n-1---! ⋅ f ( )( x ) ⋅ h n

 j

n 0

n

where j is an integer and h0 is a small fixed increment in distance  x . Keeping terms to order h 0 , we see that the first derivative is

′ ( x j )

f ( x j + 1 ) – f ( x j ) = --------------------------------h0

Hence, the first derivative of a wavefunction

ψ (x )

sampled at positions x j = j h 0 may be approxi-

mated as

ψ ( x j + 1 ) – ψ ( x j ) d  ψ ( x j ) = -------------------------------------d  x h0 2

To find the second derivative we keep terms to order h 0 so that

( x j + 1 )

 f ′′ ( x j ) 2 = f ( x j ) + f ′ ( x j ) h 0 + -------------- h 0 2

and  f ′′ ( x j ) 2 = f ( x j ) – f ′ ( x j ) h 0 + -------------- h 0 2 Adding these two equations gives

( x j – 1 )

( x j – 1 ) + f ( x j + 1 )

2

= 2 f ( x j ) + f ′′ ( x j ) h 0

or

′′ ( x j )

f ( x j – 1 ) – 2 f ( x j ) + f ( x j + 1 ) = ----------------------------------------------------------2 h0

so that the second derivative of a wavefunction

ψ (x )

sampled at positions x j = j h 0 may be

approximated as 2

d  d  x

ψ ( x j ) 2

ψ ( x j – 1 ) – 2 ψ ( x j ) + ψ ( x j + 1 ) = -----------------------------------------------------------------2 h0

This is the three-point approximation to the second derivative accurate to second order, 0 ( h 0 ) . 2

(b) By keeping five terms in the expansion instead of three one may obtain a more accurate approximation of the second derivative accurate to fourth order, 0 ( h 0 ) . To see how, we use the following Taylor expansions with derivatives up to fourth-order.

Applied quantum mechanics

3

( x j + 2 )

4 2 2 3 4 = f ( x j ) + 2 f ′ ( x j ) h 0 + 2 f ′ ′ ( x j ) h 0 + -- f ′′′ ( x j ) h 0 + - f ′′′′ ( x j ) h 0 3 3

( x j + 1 )

1 1 1 2 3 4 = f ( x j ) + f ′ ( x j ) h0 + - f ′′ ( x j ) h 0 + -- f ′′′ ( x j ) h 0 + ------ f ′′′′ ( x j ) h 0 2 6 24

( x j – 1 )

1 1 1 2 3 4 = f ( x j ) – f ′ ( x j ) h 0 + -- f ′′ ( x j ) h0 – -- f ′′′ ( x j ) h 0 + ------ f ′ ′′′ ( x j ) h0 2 6 24

4 2 2 3 4 = f ( x j ) – 2 f ′ ( x j ) h 0 + 2 f ′′ ( x j ) h 0 – -- f ′ ′′ ( x j ) h 0 + -- f ′ ′′′ ( x j ) h 0 3 3 To eliminate the terms in the first, third, and fourth derivative we multiply each equation by coefficients a, b , c, d , respectively, to obtain

( x j – 2 )

a f ( x j + 2 ) + bf ( x j + 1 ) + c f ( x j – 1 ) + d f ( x j – 2 ) =

(a + b + c + d ) f  ( x j ) + ( 2 a + b – c – 2 d ) f ′ ( x j ) h 0 +

c 3 2a b 2 d  4  2 a + b- + -c- + 2 d   f ′′ ( x j ) h 20 +  4-----a- + b- – --c- – 4----d  --  f ′′′ ( x j ) h 0 +  ------ + ------ + ------ + ------   f ′ ′′′ ( x j ) h 0          2 2 3 6 6 3 3 24 24 3

The coefficients that eliminate terms in the first, third, and fourth derivative must satisfy 0 =

( 2 a + b – c – 2 d )

0 =

 4-----a- + b- – --c- – 4----d     3 6 6 3-- 

0 =

 2-----a- + ---b--- + ---c--- + 2----d     3 24 24 3-- 

The solution to this set of linear equations is a = d  b = – 16 d  c = – 16 d 

Setting a = 1 we obtain

( x j + 2 ) – 16 f ( x j + 1 ) – 16 f ( x j – 1 ) + f( x j – 2 )

2

= –30 f ( x j) – 12 f ′′ ( x j ) h 0

so that

′′ ( x j )

– f ( x j + 2 ) + 16 f ( x j + 1 ) – 30 f ( x j ) + 16 f ( x j – 1 ) – f ( x j – 2 ) = --------------------------------------------------------------------------------------------------------------------------2 12 h 0

so that the second derivative of a wavefunction

ψ (x )

sampled at positions x j = jh 0 may be

approximated as 2

d  d  x

2

ψ (x j )

– ψ ( x j – 2 ) + 16 ψ ( x j – 1 ) – 30 ψ ( x j ) + 16 ψ ( x j + 1 ) – ψ ( x j + 2 ) = --------------------------------------------------------------------------------------------------------------------------------------2 12 h 0

Solution 3

To find numerically the first four energy eigenvalues and eigenfunctions for an electron with * effective mass m e = 0.07 × m 0 confined to a potential well V ( x) = V 0 of width  L = 10 nm with  periodic boundary conditions we descretize the wavefunction and potential appearing in the timeindependent Schrödinger equation 2

 H ψ n ( x ) =

h 2 d   – ------ 2 m d  x 2

+ V ( x )  ψ n ( x ) = E n ψ n ( x )

 

using a discrete set of  N + 1 equally-spaced points such that position x j = j × h 0 where the index = 0, 1 , 2 … N , and h 0 is the interval between adjacent sampling points so one may define  x j + 1 ≡ x j + h 0 . The region in which we wish to solve the Schrödinger equation is of length L=

Nh0 . At each sampling point the wave function has value

ψ  j

=

ψ ( x j )

and the potential is 4

V  j = V ( x j ) . The second derivative of the discretized wave function we use the three-point finite-

difference approximation which gives 2

d  d  x

2

ψ ( x j )

ψ ( x j – 1 ) – 2 ψ ( x j ) + ψ ( x j + 1 ) = -----------------------------------------------------------------2 h0

Substitution into the Schrödinger equation gives a matrix equation  H ψ ( x j ) = – u j ψ ( x j – 1 ) + d ψ ( x j ) – u j + 1 ψ ( x j + 1 ) = E ψ ( x j )

where the Hamiltonian is a symmetric tri-diagonal matrix. The diagonal matrix elements are h

2

d  j = ---------2 + V  j mh 0 and the adjacent off-diagonal matrix elements are h

2

u j = -------------2 2 mh 0 As usual, the wavefunction must be continuous and smooth. In addition, the periodic boundary

conditions require

ψ ( x )

=

ψ ( x + L )

so that

ψ 0 ( x 0 )

=

ψ  N ( x N ) .

In this situation the Hamiltonian

becomes a  N × N  matrix H and the Schrödinger equation is

( H – E I )ψ  =

( d 1 – E )

–u2

0

0

. .

.

–u1

–u2

( d 2 – E )

– u3

0

. .

.

.

0

–u3

( d 3 – E )

–u 4

. .

.

.

0

0

– u4

( d 4 – E )

. .

.

.

.

.

.

.

. .

.

.

.

.

.

.

.

. .

– u N – 2

0

.

.

.

.

. . ( d  N – 1 – E )

–u1

.

.

.

. .

ψ  N  – 2 ψ  N  – 1 ψ  N 

– u N 

– u N – 1

( d  N  – E )

.

ψ 1 ψ 2 ψ 3 ψ 4

= 0

where I the identity matrix. The matrix elements in the upper right and lower left corners are due to the periodic boundary conditions. For the case we are interested in, the first four eigenenergies are:  E 1 = 0 eV,  E 2 = 0.215 eV,  E 3 = 0.215 eV,  E 4 = 0.860 eV

Note the trivial solution with a constant wavefunction has a zero energy eigenvalue. Eigenenergy  E 2 and  E 3 are degenerate with orthogonal (sine and cosine) wavefunctions.

Applied quantum mechanics

5

Solution 4

The first six eigenenergies are:  E 1 = 0.259 eV,  E 2 = 0.453 eV,  E 3 = 0.612 eV,  E 4 = 0.752 eV,  E 5 = 0.884 eV, E 6 = 1.02 eV Solution 5

For a potential step V 0, impedance matching occurs when the ratio of effective electron masses is m2 E – V  ------ = ---------------0 m1  E  or V 0  E  = -----------------------------1 – ( m2 ⁄ m 1 ) If the potential barrier energy is V 0 = 1.5 eV and the ratio of effective electron mass on either

side of the heterointerface is m1 / m2 = 3, then 1.5  E  = ----------------------- = 2.25 eV  1 – ( 1 ⁄ 3 ) and the transmission coefficient in the limit  E → ∞ is found by noting that k 2 m ---= ------1 k 1  E → ∞ m2

so that Trans  E → ∞ =

m1 4 ------ ---------------------------2 m2 1 + m ------1   m 2 

For m 1 /  m 2 = 3 this gives 0.928 transmission and 0.072 reflection. One does not expect a particle with infinite energy to be reflected by a finite potential step!

6

Applied Quantum Mechanics

Chapter 4 Problems and Solutions

LAST NAME

FIRST NAME

Useful constants

MKS (SI) 8

–1

Speed of light in free space

c = 2.99792458 × 1 0 m s

Planck’s constant

h

= 6.58211889 ( 26 ) × 10

h

= 1.054571596 ( 82 ) × 10

– 16

eV s

–3

J s

– 19

Electron charge

e = 1.602176462 (63 ) × 10

Electron mass

m 0 = 9.10938188 ( 72 ) × 10

–31

kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10

–27

kg

Boltzmann constant

k  B = 1.3806503 ( 24 ) × 10 k  B = 8.617342 (15 ) × 10

Permittivity of free space

ε 0 = 8.8541878 × 10

Permeability of free space

µ 0 = 4 π × 10

Speed of light in free space

c = 1  ⁄  ε 0 µ 0

Avagadro’s number

–7

H m

– 12

–5

JK

eV K

F m

–1

–1

–1

–1

 N  A = 6.02214199 ( 79 ) × 10 a B = 0.52917721 ( 19 ) ×10

Bohr radius

–23

C

23

–10

mo l

–1

m

2

4 πε 0 h a B = ---------------m 0e 2

Inverse fine-structure constant α –1 = 137.0359976 ( 50) 4 πε 0 h c α – 1 = ----------------2 e

Applied quantum mechanics

1

PROBLEM 1

Write a computer program in Matlab that uses the propagation matrix method to find the transmission resonances of a particle of mass m = m 0 (where m 0 is the bare electron mass). (a) Use your computer program to find transmission as a function of energy for a particle mass m0 through 12 identical one-dimensional potential barriers each of energy 1 eV, width 0.1 nm, sequentially placed every 0.5 nm (so that the potential well between each barrier has width 0.4 nm). What are the allowed (band) and disallowed (band gap) ranges of energy for particle transmission through the structure? How do you expect the velocity of the transmitted particle to vary as a function of energy? (b) How do these bands compare with the situation in which there are only three barriers, each with 1 eV barrier energy, 0.1 nm barrier width, and 0.4 nm well width? Your solution should include plots of transmission as a function of energy and a listing of the computer program you used. PROBLEM 2

Vertical-cavity surface-emitting lasers (VCSELs) operating at wavelength λ = 1300 nm are needed for local area fiber optic applications. (a) Use the propagation matrix to design design a high-reflectivity Bragg mirror for for electromagnetic radiation with center wavelength λ0 = 1300 nm incident normal to the surface of an AlAs /  GaAs periodic dielectric layer stack consisting consisting of 25 identical layer-pairs. Each individual dielectric layer layer has a thickness λ  / 4n , with n being the refractive index of the dielectric. Use n AlAs = 3.0 for the refractive index of AlAs and n GaAs = 3.5 for GaAs. Calculate and plot optical reflectivity in the wavelength range 1200 nm > λ > 1400 nm . (b) Extend the design of your Bragg reflector reflector to a two-mirror structure similar similar to that used in the design of a VCSEL. This may be achieved by increasing the number of pairs to 50 and making the thickness of the central GaAs GaAs layer one wavelength long. Recalculate and plot the reflectivity reflectivity over the same wavelength range as in (a). Using high wavelength resolution to find the the bandwidth of this optical pass band filter near λ = 1300 nm . VCSEL structure center wavelength λ0 nAlAs nGaAs

Bragg mirror center wavelength λ 0

25 identical dielectric layer pairs

Center region thickness

λ0 / nGaAs 25 identical dielectric layer pairs

25 identical dielectric layer pairs n GaAs n AlAs

 z

Incident el electromagnetic fi field, wavelength λ

nGaAs nAlAs

Incident el electromagnetic field, wavelength λ

Your results should include a printout of the computer program you used and a computer-generated plot of particle transmission as a function of incident wavelength.

2

PROBLEM 3

1.0

3 n m 2.7 nm 2. 25 n m

0.8    )    V   e 0.6    (    )     x    (      V 0.4  ,   y   g   r 0.2   e   n   e    l   a    i    t 0.0   n   e    t   o -0.2    P -0.4

0.5 eV 1.13 eV 0.6 eV

-0.6 0

5

10

Distance, x (nm)

Write a computer program in Matlab that uses the propagation matrix method to find the transmission resonances of a particle of mass m = 0.07 × m 0 (where m 0 is the bare electron mass) for the following one-dimensional one-dimensional potentials. (a) A uniform electric field falls across the double barrier and single well structure as shown in the Fig. The right-hand edge of the 2.25-nm-thick 2.25-nm-thick barrier of energy energy 1.13 eV is at a potential -0.6 eV below the left-hand edge of the 3-nm-thick 3-nm-thick barrier of energy 0.5 eV. The well width is 2.7 nm. Comment on the changes in transmission you observe. (b) Rewrite your program to calculate transmission of a particle as a function of potential drop caused by the application of an electric field across the structure. Calculate the specific case of initial particle energy E = 0.025 eV with the particle incident on the structure from the left-hand side. PROBLEM 4

Write a computer program to solve the Schrödinger wave equation for the first 17 eigenvalues * of an electron with effective mass m e = 0.07 × m 0 confined to the periodic potential sketched in with  periodic boundary conditions. Each of the eight quantum wells is of width the following Fig. with periodic 6.25 nm. Each quantum well is separated separated by a potential potential barrier of thickness thickness 3.75 nm. nm. The barrier potential energy is 0.9 eV. How many energy bandgaps bandgaps are present in the first 17 eigenvales eigenvales and what are their values? Plot the highest energy eigenfunction eigenfunction of the first band and the lowest energy eigenfunction of the second band.

3.75 nm

   )    V   e 1.0    (    )     x    (      V 0.8     e  ,   y   g   r 0.6   e   n   e    l   a 0.4    i    t   n   e    t   o    P 0.2

6.25 nm

0.0 0

20

40

60

80

100

Distance, x (nm)

Applied quantum mechanics

3

PROBLEM 5

Use the results of Problem 4 with periodic boundary conditions to approximate a periodic onedimensional delta function potential with period 10 nm by considering 8 potential barriers with energy 20 eV and width 0.25 nm. Plot the lowest and highest energy eigenfunction eigenfunction of the first band. Explain the difference difference in the wave functions functions you obtain.

4

SOLUTIONS

Solution 1

Essentially the same as exercise 3 in chapter 4 (Chap4Exercise3 Matlab file) of the book. Solution 2

Essentially the same as exercise 4 in chapter 4 (Chap4Exercise4 Matlab file) of the book. Solution 3

Essentially the same as exercise 7 in chapter 4 (Chap4Exercise7 Matlab file) of the book. Solution 4

This requires putting in periodic boundary conditions, but is otherwise the same as exercise 9 in chapter 4 of the book. There are eight quantum wells and so there are eight eigenfunctions associated with each band. The first seventeen eigenfunctions include two complete bands each containing eight states. The eigenvalues and band gaps are:  E 1 = 0.086418 eV  E 2 = 0.086616 eV  E 3 = 0.086616 eV  E 4 = 0.087097 eV  E 5 = 0.087097 eV  E 6 = 0.087584 eV  E 7 = 0.087584 eV  E 8 = 0.087787 eV

Band gap = E 9 -  E 8 = 0.247613 eV  E 9 = 0.3354 eV  E 10 = 0.33669 eV  E 11 = 0.33669 eV  E 12 = 0.33987 eV  E 13 = 0.33987 eV  E 14 = 0.34314 eV  E 15 = 0.34314 eV  E 16 = 0.34453 eV

Band gap = E 17 - E 16 = 0.35326 eV  E 17 = 0.69779 eV

Applied quantum mechanics

5

   )    V   e    (    )     x    (      V  ,   y   g   r   e   n   e    l   a    i    t   n   e    t   o    P

0.08

0.9 0.8

ψ 8

ψ 9

0.06     ψ

0.7

 , 0.04   n   o    i 0.02    t   c   n 0.00   u    f   e -0.02   v   a    W-0.04 -0.06

0.6 0.5 0.4 0.3 0.2 0.1

-0.08

0.0 0

10

20

30

40

50

60

70

80

0

10

20

Distance, x (nm)

30

40

50

60

70

80

Distance, x (nm)

Solution 5

Use code from Problem 4. The difference in form of the wavefunction for lowest eigenenergy value in the band ( E 1 = 0.049316 eV) and highest eigenenergy value in the band ( E 8 = 0.053691 eV) is due to value of Bloch wave vector k in the wavefunction ψ k ( x ) = uk ( x ) exp( ik.x ) that describes a particle in a potential of period  L. For E 1 the value of k  = 0 and for E 8 the value of  k  =  L . π / 

0.08 20

   )    V   e    ( 16    )     x    (      V  ,   y 12   g   r   e   n   e 8    l   a    i    t   n   e 4    t   o    P 0

ψ 1

0.06 0.04  ,   n 0.02   o    i    t   c   n 0.00   u    f   e -0.02   v   a     ψ

   W-0.04

ψ 8

-0.06 -0.08 0

10

20

30

40

50

Distance, x (nm)

60

70

80

0

10

20

30

40

50

60

70

80

Distance,  x (nm)

6

Applied Quantum Mechanics

Chapter 5 Problems and Solutions

LAST NAME

FIRST NAME

Useful constants

MKS (SI)

Electron mass

× 1 08 m s –1 – 16 eV s h = 6.58211889 ( 26 ) × 10 –3 h = 1.054571596 ( 82 ) × 10 J s – 19 e = 1.602176462 ( 63 ) × 10 C –31 m 0 = 9.10938188 ( 72 ) × 10 kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10

–27

kg

Boltzmann constant

k  B = 1.3806503 ( 24 )

Speed of light in free space Planck’s constant

Electron charge

c = 2.99792458

k  B = 8.617342 (15 )

Permittivity of free space

ε0

= 8.8541878

Permeability of free space

µ0

= 4 π × 10

Speed of light in free space

c = 1  ⁄ 

Avagadro’s number

–7

× 10 –23

× 10 – 5

× 10 – 12 F H m

eV K m

–1

–1

–1

–1

ε 0µ0

 N  A = 6.02214199 ( 79 )

× 10 23 mo l –1

a B = 0.52917721 ( 19 ) ×10

Bohr radius

JK

–10

m

4 πε 0 h a B = ---------------m 0e 2

2

Inverse fine-structure constant

α– 1

α –1

= 137.0359976 ( 50)

4 πε 0 h c = ----------------e2

Applied quantum mechanics

1

PROBLEM 1

An electron in an infinite, one-dimensional, rectangular potential well of width  L is in the simple superposition state consisting of the ground and third excited state so that 1 = ------- ( ψ 1 ( x, t ) + 2 Find expressions for:

ψ ( x, t )

ψ 4 ( x, t ) )

ψ ( x , t ) 2 . (b) The average particle position, 〈 x ( t ) 〉 . 2 (c) The momentum probability density, ψ ( p x, t ) (d) The average momentum, 〈 p x ( t ) 〉 . (e) The current flux, J ( x, t ) . (a) The probability density,

PROBLEM 2

(a) Show that the density of states for a free-particle of mass m in two-dimensions is m  D 2 ( E ) = ------------2 2πh (b) At low temperature, electrons in two electrodes occupy states up to the Fermi energy,  E F . The two electrodes are connected by a two dimensional conductance

region. Derive an expression for the conductance of electrons flowing between the two electrodes as a function of applied voltage V , assuming the transmission coefficient through the two-dimensional region is unity. Consider the two limiting cases eV  >> E F  and eV