Solutions to Problems in Merzbacher Quantum Mechanics 3rd Ed-reid-p59

Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid November 20, 1999

Chapter 2

Problem 2.1 A one-dimensional initial wave packet with a mean wave number kx and a Gaussian amplitude is given by   x2 + ik x . Ψ(x, 0) = C exp − x 4(∆x)2 Calculate the corresponding kx distribution and Ψ(x, t), assuming free particle motion. Plot |Ψ(x, t)|2 as a function of x for several values of t, choosing ∆x small enough to show that the wave packet spreads in time, while it advances according to the classical laws. Apply the results to calculate the effect of spreading in some typical microscopic and macroscopic experiments.

The first step is to compute the Fourier transform of Ψ(x, 0) to find the distribution of the wave packet in momentum space:

Φ(k) = (2π)−1/2

Z

∞

Ψ(x, 0)e−ikx dx   x2 + i(k − k)x dx = (2π)−1/2 C exp − 0 4(∆x)2 −∞ −∞ Z ∞

1

(1)

(I have dropped the x subscripts, and I write k0 instead of k). To proceed we need to complete the square in the exponent:

−

x2 + i(k0 − k)x 4(∆x)2

 x2 2 2 2 2 − i(k − k)x − (k − k) (∆x) + (k − k) (∆x) 0 0 0 4(∆x)2 2  x − i(k0 − k)∆x − (k0 − k)2 (∆x)2 = − 2(∆x)  2 1 x − 2i(k0 − k)(∆x)2 − (k0 − k)2 (∆x)2 (2) = − 2 4(∆x)

= −



Now we plug (??) into (??) to find:

Φ(k) = (2π)−1/2 C exp[−(k0 −k)2 (∆x)2 ]

Z

∞ −∞

 exp −

 1 2 2 ] [x − 2i(k − k)(∆x) dx 0 4(∆x)2

In the integral weR can make the shift x → x − 2i(k0 − k)(∆x)2 and use the ∞ standard formula −∞ exp(ax2 )dx = (π/a)1/2 . The result is Φ(k) =

√

  2C∆x exp −(k0 − k)2 (∆x)2

To put this into direct correspondence with the form of the wave packet in configuration space, we can write   √ (k0 − k)2 Φ(k) = 2C∆x exp − 4(∆k)2 where ∆k = 1/(2∆x). This is the minimum possible k width attainable for a wave packet with x width ∆x, which is why the Gaussian wave packet is sometimes referred to as a minimum uncertainty wave packet. The next step is to compute Ψ(x, t) for t > 0. Since we are talking about a free particle, we know that the momentum eigenfunctions are also energy eigenfunctions, which makes their time evolution particularly simple to write down. In the above work we have expressed the initial wave packet Ψ(x, 0) as a linear combination of momentum eigenfunctions, i.e. as a sum of terms exp(ikx), with the kth term weighted in the sum by the factor Φ(k). The wave packet at a later time t > 0 will be given by the same linear combination, but now with the kth term multiplied by a phase factor exp[−iω(k)t] describing its time evolution. In symbols we have Z ∞ Ψ(x, t) = Φ(k)ei[kx−ω(k)t] dk. −∞

For a free particle the frequency and wave number are connected through ω(k) = 2

h ¯ k2 . 2m

Using our earlier expression for Φ(k), we find

Ψ(x, t) =

Z √ 2C∆x

  h ¯ k2 t dk. exp −(k0 − k)2 (∆x)2 + ikx − i 2m −∞ ∞

(3)

Again we complete the square in the exponent:   h ¯ k2 i¯ h −(k0 − k)2 (∆x)2 + ikx − i t = − [(∆x)2 + t]k 2 − [2k0 (∆x)2 + ix]k + k02 (∆x)2 2m 2m  2 2 = − α k − βk + γ   β β2 β2 = −α2 k 2 − 2 k + 4 − 4 − γ α 4α 4α 2 β β = −α2 [k − 2 ]2 + 2 − γ 2α 4α where we have defined some shorthand: α2 = (∆x)2 +

i¯ h t 2m

β = 2k0 (∆x)2 + ix

γ = k02 (∆x)2 .

Using this in (??) we find: √ Ψ(x, t) = 2C∆x exp



β2 −γ 4α2

Z

 β 2 exp −α (k − 2 ) dk. 2α −∞ ∞



2

The integral evaluates to π 1/2 /α. We have

Ψ(x, t) = =

√

   2  1 β 2πC∆x exp −γ α 4α2 " #1/2 # " √ (∆x)2 (ix + 2k0 (∆x)2 )2 −k02 (∆x)2 2πC e exp i¯ h i¯ h (∆x)2 + 2m t 4[(∆x)2 + 2m t]

This is pretty ugly, but it does display the relevant features. The important point is that term i¯ ht/2m adds to the initial uncertainty (∆x)2 , so that the wave packet spreads out with time. In the figure, I’ve plotted this function for a few values of t, with the following parameters: m=940 Mev (corresponding to a proton or neutron), ∆x=3 ˚ A, k0 =0.8 ˚ A−1 . This value of k0 corresponds, for a neutron, to a velocity of about 5 · 104 m/s; and note that, sure enough, the center of the wave packet travels about 5 nm in 100 fs. The time scale of the spread of this wave packet is ≈ 100 fs.

3

Gaussian wave packet example 350 t=0 s t=30 fs t=70 fs t=100 fs

4

Wavefunction (arbitrary units)

300

250

200

150

100

50

0 -1e-08

-5e-09

0 Distance (meters)

5e-09

1e-08

Problem 2.2 Express the spreading Gaussian wave function Ψ(x, t) obtained in Problem 1 in the form Ψ(x, t) = exp[iS(x, t)/¯ h]. Identify the function S(x, t) and show that it satisfies the quantum mechanical Hamilton-Jacobi equation. In the last problem we found

Ψ(x, t)

√

  2  ∆x β = 2πC exp −γ α 4α2     √ β2 ∆x + 2 −γ 2πC = exp ln α 4α 

(1)

where we’ve again used the shorthand we defined earlier: α2 = (∆x)2 +

i¯ h t 2m

β = 2k0 (∆x)2 + ix

γ = k02 (∆x)2 .

From (??) we can identify (neglecting an unimportant additive constant):   β2 h ¯ − ln α + 2 . S(x, t) = i 4α Things are actually easier if we define δ = α2 . Then   i¯ h 1 h ¯ β2 2 δ = (∆x) + − ln δ + t S(x, t) = 2m i 2 4δ Now computing partial derivatives: ∂S ∂t

= =

∂S ∂x

= =

∂2S ∂x2

=

  h ¯ 1 β 2 ∂δ − − 2 i 2δ 4δ ∂t   β2 h ¯2 1 + 2 − 2m 2δ 4δ h ¯ β ∂β i 2δ ∂x h ¯β 2δ i¯ h 2δ

The quantum-mechanical Hamilton-Jacobi equation for a free particle is  2 1 ∂S i¯ h ∂2S ∂S + − =0 ∂t 2m ∂x 2m ∂x2 5

(2)

(3) (4)

Inserting (??), (??), and (??) into this equation, we find h ¯2 h ¯ 2β2 h ¯ 2β2 h ¯2 − + + =0 4mδ 8mδ 2 8mδ 2 4mδ so, sure enough, the equation is satisfied. −

Problem 2.3 Consider a wave function that initially is the superposition of two well-separated narrow wave packets: Ψ1 (x, 0) + Ψ2 (x, 0) chosen so that the absolute value of the overlap integral Z +∞ γ(0) = Ψ∗1 (x)Ψ2 (x)dx −∞

is very small. As time evolves, the wave packets move and spread. Will |γ(t)| increase in time, as the wave packets overlap? Justify your answer. It seems to me that the answer to this problem depends entirely on the specifics of the particular problem. One could well imagine a situation in which the overlap integral would not increase with time. Consider, for example, the neutron wave packets plotted in the figure from problem 2.1 If one of those wave packets were centered in Chile and another in China, the overlap integral would be tiny, since the wave packets only have appreciable value within a few angstroms of their centers. Furthermore, even if the neutrons are initially moving toward each other, their wave packets spread out on a time scale of ≈ 100 fs, long before their centers ever come close to each other. On the other hand, if the two neutron wave packets were each centered, say, 20 angstroms apart, then they would certainly overlap a little before collapsing entirely.

Problem 2.4 A high resolution neutron interferometer narrows the energy spread of thermal neutrons of 20 meV kinetic energy to a wavelength dispersion level of ∆λ/λ= 10−9 . Estimate the length of the wave packets in the direction of motion. Over what length of time will the wave packets spread appreciably? 6

First let’s compute the average momentum of the neutrons. p0

= [ 2mE ]1/2 ≈ [ 2 · (940Mev · c−2 ) · (20mev) ]1/2 = 6.1 kev / c

We’re given the fractional wavelength dispersion level; what does this tell us about the momentum dispersion level? p = dp = so dp = p

so the momentum uncertainty is

2π¯ h λ −2π¯ h dλ λ2 dλ = 10−9 λ

∆p = p0 · 10−9 = 6.1 · 10−6 ev. This implies a position uncertainty of ∆x

≈ =

h ¯ ∆p 6.6 · 10−16 ev · s 6.1 · 10−6 ev / c

≈ c · 10−10 s ≈ 30 cm.

This is HUGE! So the point is, if we know with this precision how quickly our thermal neutrons are moving, we have only the most rough indication of where in the room they might be. To estimate the time scale of spreading of the wave packets, we can imagine that they are Gaussian packets. In this case we start to get appreciable spreading when

or

h ¯ t ≈ (∆x)2 2m t ≈ 2m(∆x)2 /¯ h 2 · 9.4 · 108 ev · (0.3 m)2 ≈ c2 · 6.6 · 10−16 ev · s ≈ 32 ks ≈ 10 hours. 7

Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid March 8, 1999

Chapter 3 Problem 3.1 If the state Ψ(r) is a superposition, Ψ(r) = c1 Ψ1 (r) + c2 Ψ2 (r) where Ψ1 (r) and Ψ2 (r) are related to one another by time reversal, show that the probability current density can be expressed without an interference term involving Ψ1 and Ψ2 . I found this to be a pretty cool problem! First of all, we have the probability conservation equation: d ~ · J. ~ ρ = −∇ dt To show that J~ contains no cross terms, it suffices to show that its divergence has no cross terms, and to show this it suffices (by probability conservation) to show that dρ/dt has no cross terms. We have ρ = Ψ∗ Ψ = [c∗1 Ψ∗1 + c∗2 Ψ∗2 ] · [c1 Ψ1 + c2 Ψ2 ]

= |c1 ||Ψ1 | + |c2 ||Ψ2 | + c1 c∗2 Ψ1 Ψ∗2 + c∗1 c2 Ψ∗1 Ψ2

1

(1)

Problem 3.2 For a free particle in one calculate the variance at time



 dimension, t, (∆x)2t ≡ (x − hxit )2 t = x2 t − hxi2t without explicit use of the wave function by applying (3.44) repeatedly. Show that   2 1 (∆px )2 2 2 2 (∆x)t = (∆x)0 + hxpx + px xi0 − hxi0 hpx i t + t m 2 m2 and (∆px )2t = (∆px )20 = (∆px ).2

I find it easiest to use a slightly different notation: w(t) ≡ (∆x)2t . (The w reminds me of “width.”) Then dw 1 2 d2 w +··· (2) t w(t) = w(0) + t + dt t=0 2 dt2 t=0

We have

i d h 2  2 x − hxi dt d d 2 x − 2 hxi hxi = (3) dt dt  2 d2 d2 2  d d2 w − 2 hxi = x − 2 hxi hxi (4) dt2 dt2 dt dt2

 We need to compute the time derivatives of hxi and x2 . The relevant equation is   d 1 ∂F hF i = hF H − HF i + dt i¯ h ∂t dw dt

=

for any operator F . For a free particle, the Hamiltonian is H = p2 /2m, and the all-important commutation relation is px = xp − i¯ h. We can use this to calculate the time derivatives: d hxi dt

= = = =

1 h[x, H]i i¯ h  1 2 xp − p2 x 2im¯ h  1 2 xp − p(xp − i¯ h) 2im¯ h  1 2 xp − pxp + i¯ hp 2im¯ h 2

 1 2 xp − (xp − i¯ h)p + i¯ hp 2im¯ h 1 = h2i¯ hpi 2im¯ h hpi (5) = m d2 1 d hxi = hpi = 0 (6) dt2 m dt  d 2 1 2 x = [x , H] dt i¯ h  1 2 2 x p − p 2 x2 = 2im¯ h  1 2 2 = x p − p(xp − i¯ h)x 2im¯ h  1 2 2 = x p − pxpx + i¯ hpx 2im¯ h  1 2 2 = x p − (xp − i¯ h)2 + i¯ h(xp − i¯ h) 2im¯ h  1 2 2 = x p − xpxp + 2i¯ hxp + h ¯2 + h ¯ 2 + i¯ hxp 2im¯ h  1 2 2 = x p − x(xp − i¯ h)p + 3i¯ hxp + 2¯ h2 2im¯ h  1 2 = 2¯ h + 4i¯ hxp 2im¯ h i¯ h 2 = − + hxpi (7) m m 2 d d2 2  x = hxpi 2 dt m dt 2 = h[xp, H]i i¯ hm  1 3 = − xp − p2 xp i¯ hm2  1 3 = − xp − p(xp − i¯ h)p i¯ hm2  1 3 = − xp − pxp2 + i¯ hp2 i¯ hm2  1 3 = − xp − (xp − i¯ h)p2 + i¯ hp2 i¯ hm2  1

= 2i¯ hp2 i¯ hm2 2 2 p (8) = m2  d3 2  2 2 x = [p , H] = 0 (9) 3 2 dt i¯ hm

 Now that we’ve computed all time derivatives of hxi and x2 , it’s time to =

3

plug them into (3) and (4) to compute the time derivatives of w. dw dt

= = = = =

d2 w dt2

= =

d 2 d x − 2 hxi hxi dt dt i¯ h 2 2 − + hxpi − hxi hpi m  m  m i¯ h 2 2 − + xp − hxi hpi m 2 m   2 px − xp 2 + xp − hxi hpi m 2 m   2 2 px + xp − hxi hpi m 2 m 2  d2 2  d2 d hxi hxi x − 2 − 2 hxi dt2 dt dt2 2 2 2 2 p − 2 hpi2 = 2 (∆p)2 m2 m m

(10)

(11)

Finally, we plug these into the original equation (2) to find   2 1 (∆p)2 2 w(t) = w(0) + hpx + xpi − hxi hpi t + t. m 2 m2 The other portion of this problem, the constancy of (∆p)2 , is trivial, since (∆p)2 contains expectation values of p and p2 , which both commute with H.

4

Problem 3.3 Consider a linear harmonic oscillator with Hamiltonian H =T +V =

p2 1 + mω 2 x2 . 2m 2

(a) Derive the equation of motion for the expectation value hxi t , and show that it oscillates, similarly to the classical oscillator, as hpi0 sin ωt. hxit = hxi0 cos ωt + mω (b) Derive a second-order differential equation of motion for the expectation value hT − V it by repeated application of (3.44) and use of the virial theorem. Integrate this equation and,

 remembering conservation of energy, calculate x2 t . (c) Show that

 2 (∆x)2t ≡ x2 t − hxit

(∆p)2 = (∆x)20 cos2 ωt + 2 20 sin2 ωt m ω   sin 2ωt 1 + hxp + pxi0 − hxi0 hpi0 2 mω

Verify that this reduces to the result of Problem 2 in the limit ω → 0. (d) Work out the corresponding formula for the variance (∆p)2t .

(a) Again I like to use slightly different notation: e(t) = hxi t . Then d e(t) = dt = = = = = d2 e(t) = dt2

1 hxH − Hxi i¯ h  1 2 xp − p2 x 2i¯ hm  1 2 xp − p(xp − i¯ h) 2i¯ hm  1 2 xp − (xp − i¯ h)p + i¯ hp 2i¯ hm 1 h2i¯ hpi 2i¯ hm hpi . m d hpi dt m 5

= = = = = =

1 hpH − Hpi i¯ hm 2

 ω px2 − x2 p 2i¯ h  ω2

(xp − i¯ h)x − x2 p 2i¯ h  ω2

x(xp − i¯ h) − i¯ hx − x2 p 2i¯ h ω2 h−2i¯ hxi 2i¯ h −ω 2 hxi .

So we have

d2 e(t) = −ω 2 e(t) dt2 with general solution e(t) = A cos ωt + B sin ωt. The coefficients are determined by the boundary conditions: e(0) = hxi0 hpi0 e0 (0) = m

→ →

A = hxi0 hpi0 B= . mω

(b) Let’s define v(t) = hT − V it . Then 1 h(T − V )H − H(T − V )i i¯ h 1 = h(T − V )(T + V ) − (T + V )(T − V )i i¯ h 2 = hT V − V T i i¯ h 2

 ω p 2 x2 − x 2 p 2 . = 2i¯ h We already worked out this commutator in Problem 2:

2 2 

 p x − x2 p2 = − 4i¯ hxp + 2¯ h2 d v(t) dt

=

so

d v(t) = −2ω 2 hxpi + i¯ hω 2 . dt = −2ω 2 hxpi + ω 2 hxp − pxi = −ω 2 hxp + pxi

(12)

Next, d2 2ω 2 v(t) = − hxpH − Hxpi dt2 i¯ h    mω 2

 2ω 2 1 3 = − xp − p2 xp + xpx2 − x3 p i¯ h 2m 2 6

(13)

The bracketed expressions are

3  xp − p2 xp =

= =



2

3

xpx − x p



= = =



xp3 − p(xp − i¯ h)p



 xp3 − (xp − i¯ h)p2 + i¯ hp2

 2i¯ hp2

 x(xp − i¯ h)x − x3 p

2  x (xp − i¯ h) − i¯ hx2 − x3 p

 −2i¯ hx2

and plugging these back into (13) gives

d2 v(t) = = −4ω 2 dt2

# "  p2 mω 2 2  − x m 2

= −4ω 2 v(t) with solution

v(t) = A cos 2ωt + B sin 2ωt.

(14)

Evaluating at t = 0 gives A = hT i0 − hV i0 . Also, we can use (12) evaluated at t = 0 to determine B: −ω 2 hxp + pxi0 + i¯ hω 2 = 2ωB so

ω hxp + pxi0 . B=− 2

2 The next task is to compute x t :

x2



t

=

= =

2 hV it mω 2 1 hH − (T − V )it mω 2 1 [hHit − v(t)] . mω 2

Since H does not depend explicitly on time, hHi is constant in time. For v(t) we can use (14):  

2 ω hxp + pxi0 1 sin 2ωt x t = hT i + hV i − [hT i − hV i ] cos 2ωt + 0 0 0 0 mω 2 2   ω hxp + pxi0 1 2 2 = 2 hT i sin ωt + 2 hV i cos ωt + sin 2ωt 0 0 mω 2 2

2

 p 0 1 sin 2ωt sin2 ωt + x2 0 cos2 ωt + hxp + pxi0 . (15) = m2 ω 2 2 mω 7

(c) Earlier we found that hxit hxi2t

hpi0 sin ωt mω 2 hpi = hxi20 cos2 ωt + 2 02 sin2 ωt + hxi0 hpi0 sin 2ωt. m ω = hxi0 cos ωt +

Subtracting from (15) gives h  i

 x2 0 − hxi20 cos2 ωt (∆x)2t = x2 − hxi2 = i 1 h  + 2 2 p2 0 − hpi20 m   ω 1 hxp + pxi0 − hxi0 hpi0 sin 2ωt + 2   1 sin 2ωt (∆p)2 = (∆x)20 cos2 ωt + 2 20 sin2 ωt + hxp + pxi0 − hxi0 hpi0 . m ω 2 mω As ω → 0, cos2 ωt → 1, (sin2 ωt/ω 2 ) → 1, and (sin 2ωt/ω) → 2, as needed to ensure matchup with the result of Problem 2.

Problem 3.4 Prove that the probability density and the probability current density at position r0 can be expressed in terms of the operators r and p as expectation values of the operators ρ(r0 ) → δ(r − r0 )

j(r0 ) →

1 [pδ(r − r0 ) + δ(r − r0 )p] . 2m

Derive expressions for these densities in the momentum representation. The first one is trivial: Z hδ(r − r0 )i = Ψ∗ (r)δ(r − r0 )Ψ(r)dr = Ψ∗ (r0 )Ψ(r0 ) = ρ(r0 ).

For the second one,

1 i¯ h hpδ(r − r0 ) + δ(r − r0 )pi = − 2m 2m

Z

[Ψ∗ ∇δ(r − r0 )Ψ + Ψ∗ δ(r − r0 )∇Ψ] dr

The gradient operator in the first term operates on everything to its right: Z i¯ h =− [Ψ∗ Ψ∇δ(r − r0 ) + 2δ(r − r0 )Ψ∗ ∇Ψ] dr. 2m 8

Here we can use the identity

R

f (x)δ 0 (x − a)dx = −f 0 (a) :

i¯ h |−∇(Ψ∗ Ψ) + 2Ψ∗ ∇Ψ|r=r0 2m i¯ h = |Ψ∇Ψ∗ − Ψ∗ ∇Ψ|r=r0 2m = j(r0 ). = −

Problem 3.5 For a system described by the wave function Ψ(r0 ), the Wigner distribution function is defined as     1 r00 r00 0 0 0 0 0 00 ∗ W (r , p ) = Ψ r + dr00 . exp(−ip ·x /¯ h)Ψ r − (2π¯ h)3 2 2 (a) Show that W (r0 , p0 ) is a real-valued function, defined over the six-dimensional “phase space” (r0 , p0 ). (b) Prove that

Z

W (r0 , p0 )dp0 = |Ψ(r0 )|2

and that the expectation value of a function of the operator r in a normalized state is Z Z hf (r)i = f (r0 )W (r0 , p0 )dr0 dp0 . (c) Show that the Wigner distribution function is normalized as Z W (r0 , p0 )dr0 dp0 = 1. (d) Show that the probability density ρ(r0 ) at position r0 is obtained from the Wigner distribution function with ρ(r0 ) → f (r) = δ(r − r0 ). (a)

9

Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid June 24, 2000

Chapter 5

Problem 5.1 Calculate the matrix elements of p2x with respect to the energy eigenfunctions of the harmonic oscillator and write down the first few rows and columns of the matrix. Can the same result be obtained directly by matrix algebra from a knowledge of the matrix elements of px ? For the harmonic oscillator, we have H=

1 2 1 p + mω 2 x2 2m x 2

so p2x = 2mH − m2 ω 2 x2 and

1 < Ψn |p2x |Ψk >= 2m¯ hω(n + )δnk − m2 ω 2 < Ψn |x2 |Ψk > . 2 The nth eigenfunction is Ψ(x) =



1 n 2 n!

1/2 

mω 2 mω 1/4 exp(− x )Hn ( h ¯π 2¯ h

r

(1)

mω x). h ¯

The matrix element of x2 is then r r 1/2   Z mω 1/2 ∞ 2 mω 2 mω mω 1 2 x exp( < Ψn |x |Ψk >= x )Hn ( x)Hk ( x) dx. n+k 2 n!k! h ¯π h ¯ h ¯ h ¯ −∞ 1

2

Homer Reid’s Solutions to Merzbacher Problems: Chapter 5

The obvious substitution is u = (mω/¯ h)x, with which we obtain < Ψn |x2 |Ψk >=



1 2n+k n!k!π

1/2 

h ¯ mω

Z

∞ 2

u2 e−u Hn (u)Hk (u) du.

(2)

−∞

The integral is what Merzbacher calls Inkp with p = 2. The useful formula is X

n,k,p

Inkp

√ 2 sn tk (2λ)p = π eλ +2λ(s+t)+2st . n! k! p!

√

    1 1 1 π 1 + λ 2 + λ4 + · · · 1 + 2λ(s + t) + (2λ)2 (s + t)2 + · · · 1 + (2st) + (2st)2 + · · · 2 2 2 (3) There are two ways to get a λ2 term out of this. One way is to take the λ2 term from the first series and the 1 from the second series, together with any term from the last series. The second way is to take the 1 from the first series and the λ2 term from the second series, along with any term from the last series. Writing down only terms obtainable in this way, we have  
√ 1 = · · · + πλ2 1 + 2(s + t)2 1 + (2st) + (2st)2 + · · · + · · · 2 ∞
X 1 √ (2st)j + · · · = · · · + πλ2 1 + 2s2 + 2t2 + 4st j! j=0  ∞  j j+1 X √ 2 2 j j 2 2j+1 j j+2 2j+2 j+1 j+1 j+2 j = · · · + πλ s t + s t + s t + s t +··· j! j! j! j! j=0 =

Comparing termwise with (3), we can read off  √ n π , n=k−2  (n + 2)!2 √ n!2n−1 π(1 + 2n) , n = k Ink2 =  0 , otherwise.

Plugging this into (2), we have

< Ψn+2 |x2 |Ψn > = < Ψn |x2 |Ψn > =

1 1/2 [(n + 2)(n + 1)] 2   h ¯ 1 (2n + 1) . 2 mω



h ¯ mω



Finally, from (1), 1 1/2 h) < Ψn+2 |p2 |Ψn > = − [(n + 2)(n + 1)] (mω¯ 2 1 < Ψn |x2 |Ψn > = (2n + 1)(mω¯ h). 2

Homer Reid’s Solutions to Merzbacher Problems: Chapter 5

3

I find it kind of confusing that the matrix element for p2 comes out negative in the first case. It would be absurd for the expectation value (i.e., diagonal matrix element) of the square of an observable operator to come out negative. In this case it is less absurd since there’s no classical interpretation of the offdiagonal matrix elements of an operator, but it’s still weird. However, in another sense it seems inescapable that p2 should have a negative off-diagonal matrix element here, because the off-diagonal matrix elements of H must vanish in the energy eigenfunction basis, but x2 has a nonvanishing matrix element, and H is just a sum of x2 and p2 terms, so p2 must have a negative matrix element to cancel out the positive matrix element of x2 .

Problem 5.2 Calculate the expectation values of the potential and kinetic energies in any stationary state of the harmonic oscillator. Compare with the results of the virial theorem.

The potential energy operator is U = mω 2 x2 /2. We found the expectation values of x2 in the last problem, so

 h 1 ¯ω 1 hU i = mω 2 x2 = (n + ) 2 2 2 which is just half the energy expectation value. The kinetic energy expectation value must of course make up the difference, so we have hT i = hU i. On the other hand, the virial theorem is supposed to be saying   d 2 hT i = x V (x) . dx In this case,

d V (x) = mω 2 x, dx

so the virial theorem says that hT i =



1 mω 2 x2 2



= hU i

in accord with what we concluded earlier.

Problem 5.3 Calculate the expectation value of x4 for the nth energy eigenstate of the harmonic oscillator.



 Ψ n x4 Ψ n

=

r Z 1  mω 1/2 ∞ 4 mω mω 2 2 x exp(− x )Hn ( x) dx n n! 2 h ¯π h ¯ h ¯ −∞

Homer Reid’s Solutions to Merzbacher Problems: Chapter 5

=

1 √ n! 2n π



h ¯ mω

2 Z

4

∞ 2

u4 e−u Hn2 (u) du

(4)

−∞

For the integral we want to use (3) again, but this time we’ll need to write out the expansion a little further than before. X √ 2 sn tk (2λ)p = π eλ +2λ(s+t)+2st . (5) Inkp n! k! p! n,k,p

= = = = =

 X ∞ 1 4 1 1 (2st)j 2 2 4 4 π 1+λ + λ + ··· 1 + · · · + (2λ) (s + t) + · · · + (2λ) (s + t) + · · · 2 2 4! j! j=0   ∞
X (2st)j √ 1 π 1 + λ2 + λ4 + · · · 1 + · · · + 4λ2 st + · · · + 4λ4 (st)2 + · · · 2 j! j=0   j−1 ∞ √ X 2 2j−1 2j−2 (λ4 )(st)j ··· + π +··· +4 +4 j! (j − 1)! (j − 2)! j=0   ∞ j √ X 1 4 j2 (λ )(st) ··· + π + 2j + j(j − 1) j! 2 j=0   ∞ √ X 2j 1 (λ4 )(st)j + j + j2 ··· + π j! 2 j=0 √



2

In the first line, I only wrote out terms that can be combined to give a factor of λ4 . In the second line, I further limited it to terms that also contain the same number of powers of s as t. Equating powers in (5), √ 1 3 Inn4 = 2n n! π( + n + n2 ), 2 2 so (4) is

 3 Ψ n x4 Ψ n = 2



h ¯ mω

2

1 ( + n + n2 ). 2

Problem 5.4

For the energy eigenstates with n=0, 1, and 2, compute the probability that the coordinate of a linear harmonic oscillator in its ground state has a value greater than the amplitude of the classical oscillator of the same energy. p The classical amplitude is A = (2E)/(mω 2 ). The probability of finding the particle with coordinate greater than this is Z −A Z ∞ P (|x| > A) = Ψ2n (x) dx + Ψ2n (x) dx −∞

A

5

Homer Reid’s Solutions to Merzbacher Problems: Chapter 5

= 2

Z

∞

A

= = =

Ψ2n (x) dx

r Z 2  mω 1/2 ∞ mω 2 2 mω exp(− x )Hn ( x) dx n n! 2 h ¯π h ¯ h ¯ A Z ∞ 2 2 √ e−u Hn2 (u) du n! 2n π √2E/¯hω Z ∞ 2 1 √ e−u Hn2 (u) du √ n−1 n! 2 π 2n+1

In going from the first line to the second we invoked the fact that Ψn has either even or odd parity, so Ψ2n has even parity. In going from the second to last line to the last line, we noted that the energy of the nth eigenstate is h ¯ ω(n + 1/2). In particular,

Pn=0 (|x| > A) = = = =

1 √ π

Z

1 √ 2π

∞

1 ( Z

2

e−u du ∞

e 0

−p2 /2

dp −

Z

√ 2

e 0

−p2 /2

dp

)

 r √ π √ 1 √ − 2π · erf( 2) 2 2π √ 1 − erf( 2) ≈ 0.31 2

Problem 5.5 Show that if an ensemble of linear harmonic oscillators is in thermal equilibrium, governed by the Boltzmann distribution, the probability per unit length of finding a particle with displacement x is a Gaussian distribution. Plot the width of the distribution as a function of temperature. Check the results in the classical and low-temperature limits. [Hint: Equation (5.43) may be used.] Suppose we denote the number of oscillators in the nth energy state by Nn . If the ensemble is in thermal equilibrium, the ratio of the number of oscillators in the n0 th state to the number of oscillators in the nth state is 0 Nn 0 = e−(n −n)¯hω/kT . Nn

In particular, for any n, Nn = N0 e−n¯hω/kT .

6

Homer Reid’s Solutions to Merzbacher Problems: Chapter 5

The probability of finding a particle between x and dx is P (x)dx

=

∞ X

n=0

= C0

Cn |Ψn (x)|2 dx

 mω 1/2 h ¯π

exp(−

∞ mω 2 X e−n¯hω/kT 2 x ) Hn ( h ¯ n! 2n n=0

r

mω x)dx h ¯

This can be summed using the Mehler formula with t = exp(−¯ hω/kT ) :       mω 1/2 mω 2 mω 2 1 2t P (x) = C0 x ) √ x exp exp(− h ¯π h ¯ 1+t h ¯ 1 − t2       mω 1/2  1 1 − t mω 2 √ exp − x = C0 h ¯π 1+t h ¯ 1 − t2 This is a Gaussian distribution with variance   h ¯ 1+t σ2 = 2mω 1 − t   1 + e−¯hω/kT h ¯ = 2mω 1 − e−¯hω/kT   h ¯ω h ¯ coth = 2mω 2kT

Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid June 24, 2000

Chapter 6

Problem 6.1 Obtain the transmission coefficient for a rectangular potential barrier of width 2a if the energy exceeds the height V0 of the barrier. Plot the transmission coefficient as a function E/V0 (up to E/V0 = 3), choosing (2ma2 V0 )1/2 = (3π/2)~. In the text, Merzbacher treats this problem for the case where the particle’s energy is less than the potential barrier. He obtains the result   i M11 = cosh 2κa + sinh 2κa e2ika (1) 2 where κ= and

r

2m(V0 − E) ~2

κ k − . (2) k κ We can re-use the result (1) for the case where the energy is greater than the potential barrier. To do this we note that κ becomes imaginary in this case, and we write r 2m(E − V0 ) κ = iβ = i ~2 so that (2) becomes   iβ k k β = ≡ iλ − =i + k iβ k β =

1

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

Plugging into (1) and noting that cosh ix = cos x, sinh ix = i sin x we have   iλ sin 2βa e2ika . M11 = cos 2βa − 2 and 1/2 λ2 2 |M11 | = cos 2βa + sin βa 4   2  1/2 λ = 1+ − 1 sin2 βa 4  1/2   4 1 β + k4 2 sin βa − = 1+ 4β 2 k 2 2  1/2   2 2 2 (β − k ) 2 sin βa = 1+ 4β 2 k 2  1/2   V02 2 = 1+ sin βa 4E(E − V0 )    1/2 1 = 1+ sin2 βa 4γ(γ − 1)  1/2 4γ(γ − 1) + sin2 βa = 4γ(γ − 1) 

2

where γ = E/V0 . We have 1/2 2m a (E − V ) 0 h2  1/2 2mV0 a2 = (γ − 1)1/2 h2 3π (γ − 1)1/2 = 2

βa =



so the transmission coefficient is " # 1 4γ(γ − 1)   T = = 1/2 |M11 |2 4γ(γ − 1) + sin2 3π 2 (γ − 1) This is plotted in Figure 1.

2

3

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

1 0.9 0.8

Transmission Coefficient

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.5

1

1.5

2 E/V0

2.5

3

3.5

Figure 1: Transmission coefficient versus E/V0 for Problem 6.1.

4

4

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

Problem 6.2 Consider a potential V = 0 for x > a, V = −V0 for a ≥ x ≥ 0, and V = +∞ for x < 0. Show that for x > a the positive energy solutions of the Schr¨ odinger equation have the form ei(kx+2δ) − e−ikx

Calculate the scattering coefficient |1 − e2iδ |2 and show that it exhibits maxima (resonances) at certain discrete energies if the potential is sufficiently deep and broad. We have

with

 0,   Ψ(x) = Aeik1 x + Be−ik1 x ,   ik2 x Ce + De−ik2 x ,

x ≤0

0 ≤x ≤ a≤

a x

r

r 2m(E + V0 ) 2mE k2 = . k1 = ~2 ~2 Applying the requirement that Ψ be continuous at x = 0, we see we must take A = −B, so Ψ(x) = γ sin k1 x for 0 ≤ x ≤ a. The other standard requirement, that the derivative of Ψ also be continuous, does not hold at x = 0 because the potential is infinite there. Hence γ is undetermined as yet. Eventually, we could apply the normalization condition on Ψ to find γ if we wanted to. Next applying continutity of Ψ and its derivative at x = a, we obtain γ sin k1 a = Ceik2 a + De−ik

2

a 2

k1 γ cos k1 a = ik2 [Ceik2 a − De−ik a ] Combining these yields   1 −ik2 a k1 γe cos k1 a + i sin k1 a 2i k2   1 k1 D = − γe+ik2 a cos k1 a − i sin k1 a 2i ik2 C=

(3) (4)

It’s now convenient to write k1 cos k1 a + i sin k1 a = αeiδ k2 where α2 =



k1 k2

2

cos2 k1 a + sin2 k1 a

(5)

(6)

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

and δ = tan

−1



k2 tan k1 a k1

5



so that α contains magnitude information, while δ represents phase information. Then we can rewrite (3) and (4) as γα −ik2 a iδ e e 2i γα D = − e+ik2 a e−iδ 2i C=

Then the expression for the wavefunction to the left of x = a becomes Ψ(x) = Ceik2 x + De−ik2 x (x > a) i γα h ik2 (x−a) iδ −ik2 (x−a) −iδ e e −e e = 2i i γα −iδ h ik2 (x−a) 2iδ e = e e − e−ik2 (x−a) . 2i

Using (5) and (6), the scattering coefficient is 2  2 2 k1 k1 2 cos k a sin k a − sin k a cos k a + 2i 1 1 1 1 k2 k2 2iδ 2 |1 − e | = 1 −  2 2 k1 cos2 k1 a + sin k1 a k2 2 h i 2 sin k1 a sin k1 a − i kk1 cos k1 a 2 =  2 k1 cos2 k1 a + sin2 k1 a k2 We have

= 

4 sin2 k1 a

k1 k2

2

2

(7)

cos2 k1 a + sin2 k1 a

  E + V0 1 = 1+ E λ r 2ma2 (E + V0 ) k1 a = = β(λ + 1)1/2 ~2

k1 k2

=

where λ = E/V0 and β = (2ma2 V0 /~2 )1/2 in Merzbacher’s notation. Then the scattering coefficient (7) is scattering coefficient =

(1 +

1 2 λ)

4 sin2 [β(λ + 1)1/2 ] cos2 [β(λ + 1)1/2 ] + sin2 [β(λ + 1)1/2 ]

In Figure 6.2 I have plotted this for β = 25.

6

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

4

3.5

Scattering coefficient

3

2.5

2

1.5

1

0.5

0 0

0.5

1

1.5 E/V0

2

2.5

Figure 2: Scattering coefficient versus E/V0 for Problem 6.2.

3

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

7

Problem 6.3 A particle of mass m moves in the one-dimensional double well potential V (x) = −gδ(x − a) − gδ(x + a). If g > 0, obtain transcendental equations for the bound-state energy eigenvalues of the system. Compute and plot the energy levels in units of ~2 /ma2 as a function of the dimensionless parameter mag/~2 . Explain the features of the plot. In the limit of large separation, 2a, between the wells, obtain a simple formula for the splitting ∆E between the ground state (even parity) energy level, E+ , and the excited (odd parity) energy level, E− . In this problem, we can divide the x axis into three regions. In each region, the wavefunction is just the solution to the free-particle Schr¨ odinger equation, but with energy E < 0 since we’re looking for bound states. Putting k = p 2mE/~2 , we have  x ≤ −a  Aekx + Be−kx , Cekx + De−kx , −a ≤ x ≤ a Ψ(x) =  Eekx + F e−kx , a ≤ x.

Now, first of all, the wavefunction can’t blow up at infinity, so B = E = 0. Also, since the potential in this problem has mirror-reversal symmetry, the wavefunction will have definite parity. Considering first the even parity solution,  x ≤ −a  Aekx , B cosh(kx), −a ≤ x ≤ a Ψ(x) = (8)  Ae−kx , a ≤ x. Matching the value of the wavefunction at x = −a gives Ae−ka = B cosh(−ka).

(9)

Since the potential becomes infinite at x = −a, the normal derivative-continuity condition doesn’t hold there. Instead, we can write down the Schr¨ odinger equation, d2 2m 2m Ψ(x) = 2 V (x)Ψ(x) − 2 EΨ(x), 2 dx ~ ~ then integrate from −a −  to −a +  and take the limit as  → 0. This gives −a+ dΨ 2mg = − 2 Ψ(−a). (10) dx −a− ~ Applying this condition to the wavefunction (8) yields kB sinh(−ka) − kAe−ka = −

2mg B cosh(−ka). ~2

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

8

Substituting from (9), kB sinh(−ka) − kB cosh(−ka) = −

2mg B cosh(−ka) ~2

or tanh(ka) =

β 2mg −1= −1 ~2 k ka

with β = 2mag/~2 . This equation determines the energy eigenvalue of the evenparity state, which will be the ground state. On the other hand, the odd parity state looks like  x ≤ −a  Aekx , B sinh(kx), −a ≤ x ≤ a (11) Ψ(x) =  −Ae−kx , a ≤ x. Matching values at x = −a gives

Ae−ka = B sinh(−ka) and applying condition (10) gives 2mg B sinh(−ka) ~2 2mg kB cosh(ka) + kB sinh(ka) = 2 B sinh(ka) ~ β −1 coth(ka) = ka kB cosh(−ka) − kAe−ka = −

so this is the condition that determines the energy of the odd parity state. In Figure (3) I have plotted tanh(ka), coth(ka), and β/(ka) − 1 for the case β = 3. As expected, the coth curve crosses the β/(ka) − 1 curve at a lower value of ka than the tanh curve; that means that the energy eigenvalue for the odd parity state is smaller in magnitude (less negative) than the even parity state.

Problem 6.4 Problem 3 provides a primitive model for a one-electron linear diatomic molecule with interatomic distance 2a = |X|, if the potential energy of the “molecule” is taken as E± (|X|), supplemented by a repulsive interaction λg/|X| between the wells (“atoms”). Show that, for a sufficiently small value of λ, the system (“molecule”) is stable if the particle (“electron”) is in the even parity state. Sketch the total potential energy of the system as a function of |X|.

9

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

PSfrag replacements

2 tanh(ka) coth(ka) β/(ka) − 1

1.5

1

0.5

0 1

1.2

1.4

1.6

1.8

ka

Figure 3: Graphical determination of energy levels for Problem 6.3 with β = 3.

2

10

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

Problem 6.5 If the potential in Problem 3 has g < 0 (double barrier), calculate the transmission coefficient and show that it exhibits resonances. (Note the analogy between the system and the Fabry-Perot ´etalon in optics.) Now we’re assuming that the energy E is positive, so  ikx −ikx x ≤ −a   Ae + Be ikx −ikx Ψ(x) = Ce + De −a≤x≤a   ikx −ikx Ee + F e a≤x

with k =

p 2mE/~2 . Matching values at x = −a, we have Ae−ika + Beika = C −ika + Dika

(12)

(13)

Also, as before, we have the derivative condition x=−a+ dΨ 2mg = − 2 Ψ(−a) dx x=a− ~

where g is now negative. Applying this to the wavefunction in (12), we have ik[Ce−ika − Deika − Ae−ika + Beika ] = −

2mg [Ae−ika + Beika ]. ~2

Combining (13) and (14) yields   β 2ka β e B A+ C = 1+ ika ika   β −2ka β D=− B e A+ 1− ika ika

(14)

(15) (16)

with β = mag/~2 as before. Now, applying the matching conditions to the wavefunction at x = +a will give two equations exactly like (13) and (14), but with the substitutions A → C, B → D, C → E, D → F , and a → −a. Making these substitutions in (15) and (16) we obtain   β β 2ka e D (17) E = 1− C− ika ika   β 2ka β F =+ D (18) e C + 1+ ika ika

PSfrag replacements

11

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

1 0.9 0.8 0.7

T (k)

0.6 0.5 0.4 0.3 0.2 0.1 0 0

5

10

15

20

25

ka

Figure 4: Transmission coefficient in Problem 6.4 with β = 15.

Combining equations (15) through (18), we have # "   2    2β β β −4ika (e − 1) A + + E = 1+ 1− sin 2ka B ika ka ika # "      2 2β β β 4ika (e − 1) B F = 1+ sin 2ka A + 1 + ka ika ika This is the M matrix, and the transmission coefficient is given by T = 1/|M11 |2 , or 1  T =  2  2 β β + 1 (1 − cos(4ka)) 1 + 2 ka ka

In Figure 4 I have plotted this for β = 15.

30

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

12

Problem 6.6 A particle moves in one dimension with energy E in the field of a potential defined as the sum of a Heaviside step function and a delta function: V (x) = V0 η(x) + gδ(x)

(with V0 and g > 0)

The particle is assumed to have energy E > V0 . (a) Work out the matrix M , which relates the amplitudes of the incident and reflected plane waves on the left of the origin (x < 0) to the amplitudes on the right (x > 0). (b) Derive the elements of the matrix S, which relates incoming and outgoing amplitudes. (c) Show that the S matrix is unitary and that the elements of the S matrix satisfy the properties expected from the applicable symmetry considerations. (d) Calculate the transmission coefficient for particles incident from the right and for particles incident from the left, which have the same energy (buf different velocities). We have Ψ(x) = with

(

r

Aeik1 x + Be−ik1 x , Ce

ik2 x

2m E k1 = ~2 Matching values at x = 0 gives

+ De

k2 =

−ik2 x

r

,

x≤0

x≥0

2m (E − V0 ). ~2

C +D =A+B

(19)

Also, the delta function at the origin gives rise to a discontinuity in the derivative of the wavefunction as before: 0+ dΨ 2mg = 2 Ψ(0) dx 0− ~

so

ik2 (C − D) − ik1 (A − B) =

or C −D =

2mg (A + B) ~2

k1 2mg (A − B) + (A + B). k2 ik2 ~2

(20)

Homer Reid’s Solutions to Merzbacher Problems: Chapter 6

13

Adding and subtracting (19) and (20), we can read off     1 2mg 2mg 1 k1 k1 A+ B + + C= 1+ 1− 2 k2 ik2 ~2 2 k2 ik2 ~2     2mg 2mg 1 1 k1 k1 − − D= A 1+ B. 1− 2 k2 ik2 ~2 2 k2 ik2 ~2 We could also write this as    C M11 = ∗ D M12

M12 ∗ M11



A B



Or instead of the M matrix we could use the S matrix, which is defined by      A S11 S12 B = S21 S22 D C Since we already know the M coefficients, we can calculate the elements of the S matrix from the formula S11 S21

M∗

= − M12 ∗ 11 1 = M∗ 11

S12 = M1∗ 11 M12 S22 = + M ∗ 11

However, this is tedious and long and boring and I don’t want to do it.

Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid April 5, 2001

Chapter 7 Before starting on these problems I found it useful to review how the WKB approximation works in the first place. The Schr¨ odinger equation is −

~2 d 2 Ψ(x) + V (x)Ψ(x) = EΨ(x) 2m dx2

or

d2 Ψ(x) + k 2 (x)Ψ(x) = 0, dx2 We postulate for Ψ the functional form

k(x) ≡

r

2m [E − V (x)]. ~2

Ψ(x) = AeiS(x)/~ in which case the Schr¨ odinger equation becomes i~S 00 (x) = [S 0 (x)]2 − ~2 k 2 (x).

(1)

This equation can’t be solved directly, but we obtain guidance from the observation that, for a constant potential, S(x) = ±kx, so that S 00 vanishes. For a nonconstant but slowly varying potential we might imagine S 00 (x) will be small, and we may take S 00 = 0 as the seed of a series of successive approximations to the exact solution. To be specific, we will construct a series of functions S0 (x), S1 (x), · · · , where S0 is the solution of (1) with 0 on the left hand side; S1 is a solution with S000 on the left hand side; and so on. In other words, at the nth step in the approximation sequence (by which point we have computed Sn (x)), we compute Sn00 (x) and use that as the source term on the LHS of (1) to calculate Sn+1 (x). Then we compute the second derivative of Sn+1 (x) and use this as the source term for calculating Sn+2 , and so on ad infinitum. In

1

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

2

symbols, 0 = [S00 (x)]2 − ~2 k 2 (x)

i~S000 i~S100

=

[S10 (x)]2 [S20 (x)]2

= ····

(2)

2 2

(3)

2 2

(4)

− ~ k (x)

− ~ k (x)

Equation (2) is clearly solved by taking S00 (x)

= ±~k(x)

⇒

S0 (x) = S00 ± ~

Z

x

k(x0 )dx0

(5)

−∞

for any constant S00 . Then S000 (x) = ±~k 0 (x), so (3) is p S10 (x) = ±~ k 2 (x) ± ik 0 (x).

With the two ± signs here, we appear to have four possible choices for S10 . But let’s think a little about the ± signs in this equation. The ± sign under the radical comes from the two choices of sign in (5). But if we chose, say, the plus sign in that equation, so that S00 > 0, we would also expect that S10 > 0. Indeed, if we choose the plus sign in (5) but the minus sign in (3), then S00 and S10 have opposite sign, so S10 differs from S00 by an amount at least as large as S00 , in which case our approximation sequence S0 , S1 , · · · has little hope of converging. So we choose either both plus signs or both minus signs in (3), whence our two choices are p p S10 = +~ k 2 (x) + ik 0 (x) or S10 = −~ k 2 (x) − ik 0 (x). (6)

If V (x) is constant, k(x) is constant, and, as we observed before, the sequence of approximations terminates at 0th order with S0 being an exact solution. By extension, if V (x) is not constant but changes little over one particle wavelength, we have k 0 (x)/k 2 (x)  1, so we may expand the radicals in (6):     ik 0 (x) ik 0 (x) S10 ≈ ~k(x) 1 + 2 or S10 ≈ −~k(x) 1 − 2 2k (x) 2k (x) or i~k 0 (x) S10 ≈ ±~k(x) + . (7) 2k(x) Integrating, Z i~ x k 0 (u) 0 dx 2 a k(u) Za x i~ k(x) = S1 (a) ± ~ k(u)du + ln 2 k(a) a

S1 (x) = S1 (a) ± ~

Z

x

k(u)du +

where a is some point chosen such that the approximation (7) is valid in the full range a < x0 < x. We could go on to compute S2 , S3 , etc., but in practice it seems the approximation is always terminated at S1 .

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

3

The wavefunction at this order of approximation is −1/2    Rx Ψ(x) = exp(iS1 (x)/~) = eiS1 (a)/~ e±i a k(u)du eln k(x)/k(a) s k(a) ±i R x k(u)du = Ψ(a) e a k(x) = Ψ(a)G± (x; a) where (8) G± (x; a) ≡

s

k(a) ±i R x k(u)du e a . k(x)

(9)

We have written it this way to illustrate that the function G(x, a) is kind of like a Green’s function or propagator for the wavefunction, in the sense that, if you know what Ψ is at some point a, you can just multiply it by G± (x; a) to find out what Ψ is at x. But this doesn’t seem quite right: Schr¨ odinger’s equation is a second-order differential equation, but (8) seems to be saying that we need only one initial condition—the value of Ψ at x = a—to find the value of Ψ at other points. To clarify this subtle point, let’s investigate the equations leading up to (8). If the approximation (7) makes sense, then there are two solutions of Schr¨ odinger’s equation at x = a, one whose phase increases with increasing x (positive derivative), and one whose phase decreases. Equation (8) seems to be saying that we can use either G+ or G− to get to Ψ(x) from Ψ(a); but the requirement the dΨ/dx be continuous at x = a means that only one or the other will do. Indeed, in using (8) to continue Ψ from a to x we must choose the appropriate propagator—either G+ or G− , according to the derivative of Ψ at x = a; otherwise the overall wave function will have a discontinuity in its first derivative at x = a. So to use (8) to obtain values for Ψ at a point x, we need to know both Ψ and Ψ0 at a nearby point x = a, as should be the case for a second-order differential equation. If we want to de-emphasize this nature of the solution with the propagator we may write s 1 ±i R x k(u)du Ψ(x) = C e a (10) k(x) p where C = Ψ(a) k(a). In regions where V (x) > E, k(x) is imaginary, so it’s useful to define r 2m κ(x) = −ik(x) = [V (x) − E] (11) ~2 and s 1 ± R x κ(u)du e a . (12) Ψ(x) = C κ(x)

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

4

If we have a region of space in which the WKB approximation is valid, knowing the value of Ψ (and its derivative) at one point within the region is equivalent to knowing it everywhere, because we can use the propagator (9) to get from that one point to every other point within the region. The WKB method, however, gives us no way of determining the value of Ψ at that one starting point. Furthermore, even if we know Ψ at one point within a region of validity, we can’t use (8) to determine Ψ in other, nonadjacent regions, because we can’t carry the propagator across regions of invalidity. So basically what we need is a way of finding one starting value for Ψ(x) in every region of validity of the WKB approximation. How do we find such points? Well, one sure-fire way to get starting points in regions of validity is to identify regions of invalidity, of which there will be at least one adjacent to each region of validity, and then get values of Ψ at the boundaries of the regions of invalidity—which will also count as values in the regions of validity. So we need to identify the regions of invalidity of the WKB approximation and do a more accurate solution of the Schr¨ odinger equation there. The WKB approximation breaks down when k 0 /k 2  1 ceases to hold, which is true when k ≈ 0 but k 0 6= 0, which happens near a classical turning point of the motion—i.e., a point x0 at which V (x0 ) = E. But near such a point we may expand V (x) − E in a Taylor series around the point x0 ; if we keep only the first (linear in x) term in the series, we arrive at a Schr¨ odinger equation which we can solve exactly in the vicinity of x0 . To do this, suppose the point x0 is a classical turning point of the motion, so that V (x0 ) = E. In the neighborhood of x0 we may expand V (x): V (x) = E + (x − x0 )V 0 (x0 ) + · · ·

(13)

Then the Schr¨ odinger equation becomes d2 2m Ψ(x) − 2 V 0 (x0 )(x − x0 )Ψ(x) = 0. 2 dx ~ The useful substitution here is  1/3 2m 0 u(x) = γ(x − x0 ) γ≡ V (x0 ) ~2

so

x(u) =

u + x0 . γ

If we define Φ(u) = Ψ(x(u)) then dΦ dΨ dx 1 = = Ψ0 (x(u)) du dx du γ 1 00 d2 Φ = 2 Ψ (x(u)) du2 γ

(14)

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

5

so (14) becomes γ2

d2 Φ(u) − γ 3 (x − x0 )Φ(u) = 0 du2

or

d2 Φ(u) − uΦ(u) = 0. du2 The solution to this differential equation is Φ(u) = β1 Ai(u) + β2 Bi(u)

(15)

so the solution to the Schr¨ odinger equation (14) is     Ψ(x) = β1 Ai γ(x − x0 ) + β2 Bi γ(x − x0 ) .

For γ(x − x0 )  1 we have the asymptotic expression   3/2 2 −1/4 β1 − 23 |γ(x−x0 )|3/2 Ψ(x) ≈ π −1/2 [γ(x − x0 )] e + β2 e+ 3 |γ(x−x0 )| 2

(16)

and for γ(x − x0 )  −1 we have Ψ(x) ≈ π −1/2 |γ(x − x0 )|

−1/4

h

2

π 3/2 |γ(x − x0 )| − 3 4 2 π i 3/2 − β2 sin . |γ(x − x0 )| − 3 4

β1 cos

(17)

To simplify these, we need to consider two possible kinds of turning point. Case 1: V 0 (x0 ) > 0. In this case the potential is increasing through the turning point at x0 , which means that V (x) < E for x < x0 , and V (x) > E for x > x0 . Hence the region to the left of the turning point is the classically accessible region, while the right of the turning point is classically forbidden. Since V 0 (x0 ) > 0, γ > 0, so for x < x0 (??) holds. For points close to the turning point on the left side, r  1/2 2m 0 2m [E − V (x)] ≈ V k(x) = (x0 − x)1/2 = γ 3/2 (x0 − x)1/2 ~2 ~2 so |γ(x − x0 )|

−1/4

and Z

x0

k(u)du = γ x

3/2

Z

=

r

γ k(x)

x0 x

(x0 − x)1/2 du =

2 = |γ(x − x0 )|3/2 . 3

(18)

2 3/2 γ (x0 − x)3/2 3 (19)

On the other hand, for points close to the turning point on the left side we have x > x0 , so γ(x − x0 ) > 0. In this region,

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

κ(x) =

r

1/2   2m  2m 0 = γ 3/2 (x − x0 )1/2 V (x0 )(x − x0 ) V (x) − E ≈ ~2 ~2

so, for x near x0 , Z x Z x 2 κ(u)du = γ 3/2 (u − x0 )1/2 du = γ 3/2 (x − x0 )3/2 3 x0 x0 and also |γ(x − x0 )|

−1/4

=

r

γ . κ(x)

6

(20)

(21)

(22)

Using (18) and (19) in (17), and (21) and (22) in (16), the solutions to the Schr¨ odinger equation on either side of a classical turning point x0 at which V 0 (x0 ) > 0 are s  Z x0 1 h π k(u)du − 2β1 cos Ψ(x) = k(x) 4 x  Z x0 π i k(u)du − − β2 sin , x < x0 (23) 4 x s i R 1 h Rxx κ(u)du − x κ(u)du Ψ(x) = β1 e 0 + β 2 e x0 , x < x0 (24) κ(x) (we redefined the β constants slightly in going to this equation). Case 2: V 0 (x0 ) < 0. In this case the potential is decreasing through the turning point, so the classically accessible region is to the right of the turning point, and the forbidden region to the left. Since V 0 (x0 ) < 0, γ < 0. That means that the regions of applicability of (16) and (17) are on opposite sides of the turning points as they were in the previous case. The solutions to the Schr¨ odinger equation on either side of the turning point are s i R 1 h R x0 κ(u)du − x κ(u)du β1 e x + β 2 e x0 , x < x0 (25) Ψ(x) = κ(x) s Z x π 1 h k(u)du − Ψ(x) = 2β1 cos k(x) 4 x0 Z x π i − β2 sin , x > x0 (26) k(u)du − 4 x0 (27)

So, to apply the WKB approximation to a given potential V (x), the first step is to identify the classical turning points of the motion, and to divide space

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

7

up into regions bounded by turning points, within which regions the WKB approximation (7) is valid. Then, for each turning point, we write down (23) and (24) (or (25) and (26)) at nearby points on either side of the turning point, and then use (10) to evolve the wavefunction from those points to other points within the separate regions. We should probably quantify the meaning of “nearby” in that last sentence. Suppose x0 is a classical turning point of the motion, and we are looking for points x0 ±  at which to make the “handoff” from approximations (16) and (17) to the WKB approximation These points must satisfy several conditions. First, the approximate Schr¨ odinger equation (14) is only valid as long as we can neglect the quadratic and higher-order terms in the expansion (13), so we must have 0 V (x0 ) . (28)  |V 00 (x0 )|  |V 0 (x0 )| ⇒   00 V (x0 ) But at the same time, γ must be sufficiently greater than 1 to justify the approximation (16) (or sufficiently less than -1 to justify (17)); the condition here is 1/3 −1/3 2m 0 2m 0   1 ⇒   . (29) V (x ) V (x ) 0 0 ~2 ~2 Finally, the points x± must be sufficiently far away from the turning points that the approximation (7) is valid for the derivative of the phase of the wavefunction; the condition for this to be the case was 0  2  0 k (x) ~ V (x ± ) 1 1⇒ (30)  1. k 2 (x) 2 2m [E − V (x ± )]3/2

If there are no points x0 ±  satisfying all three conditions, the WKB approximation cannot be used.

To apply all of this to the problem of bound states in a potential well, consider a potential like that shown in Figure 1, with two classical turning points at x = a and x = b. Although there are no discontinuities in the potential here, the problem may be analyzed in a manner similar to that used in the consideration of one-dimensional piecewise constant potentials, as in Chapter 6: we divide space into a number of distinct regions, obtain solutions of the Schr¨ odinger equation in each region, and then match values and derivatives at the region boundaries. To divide space into distinct regions in this case, we begin by identifying narrow regions around the turning points a and b in which the linear approximation (13) is valid. In the narrow region around x = a, we may use (25) and (26); around x = b we may use (23) and (24). Let the narrow such region around a be a − 1 < x < a + 1, and that around b be b − 2 < x < b + 2. Then space divides naturally into five regions: (a) x < a − 

In this region we are far enough to the left of the turning point that the WKB approximation is valid, and the wavefunction takes the form (10). However, we

8

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

PSfrag replacements V (x) E

a

b

Figure 1: A potential V (x) with two classical turning points for an energy E.

9

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

must throw out the term that grows exponentially as x → −∞, so we are left with s 1 − R (a−) κ(u)du e x , x < a − . (31) Ψ(x) = A κ(x) (b) a −  < x < a +  In this region we are close enough to the turning point that (13) is valid, so (25) and (26) may be used. s  Ra 1  − R a κ(u)du Ψ(x) = β1 e x + β2 e+ x κ(u)du , x < (a − ). (32) κ(x) From (31) and (32) we see that continuity of both the value and first derivative of Ψ(x) at x = a −  requires taking β1 = A, β2 = 0. With this choice of constants, we achive continuity not only of the value and first derivative of Ψ but also of all higher derivatives, as must be the case since there is no discontinuity in the potential. But now that we know the value of Ψ at x = a − , we also know it at x = a + , because of course the solution of the Schr¨ odinger equation in the narrow strip around a (to which (32) is an asymptotic approximation for x < a) is valid throughout the strip; the same solution that’s valid at x = a −  is valid at x = a + . With β1 = A and β2 = 0, (26) becomes

Ψ(x) = 2A =A

s

s

 1 cos k(x)

Z

x a

k(u)du −

π 4

i Rx 1 h +i(R x k(u)du−π/4) a + e−i( a k(u)du−π/4) , e k(x)

x = (a + )− (33)

(c) a +  < x < b −  In this region the WKB approximation (7) is valid, so we may use (8) to find the wavefunction at any point within the region. Using the expression (33) for the wavefunction at x = a + , integrating from a +  to x in the propagator (9), and using G+ and G− , respectively, to propagate the first and second terms in

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

10

(33), we obtain for the wavefunction at a point x in this region s  “R “R ” ” Rx Rx 1 −i a(a+) k(u)du+ (a+) −π/4 +i a(a+) k(u)du+ (a+) k(u)du−π/4 +e Ψ(x) = A e k(x) s i Rx 1 h +i(R x k(u)du−π/4) a =A e + e−i( a k(u)du−π/4) k(x) s  Z x 1 π = 2A cos k(u)du − k(x) 4 a s ! Z b Z b 1 π , a+ 0, the Schr¨ odinger equation is d2 Ψ(x) + dx2 d2 = 2 Ψ(x) + dx d2 = 2 Ψ(x) − dx

0=

2m [E − mgz] Ψ(x) ~2   2m2 g E − z Ψ(x) ~2 mg 2m2 g [z − z0 ] Ψ(x) ~2

(36)

11

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

where z0 = E/mg. With the substitution u = γ(z − z0 )

γ=



2m2 g ~2

1/3

and taking Φ(u) = Ψ(x(u)), we find that (36) is just the Airy equation for Φ(u), d2 Φ(u) − uΦ(u) = 0 du2 with solutions Φ(u) = β1 Ai(u) + β2 Bi(u). Since we require a solution that remains finite as z → ∞, we must take β2 = 0. The solution to (36) is then
Ψ(x) = β1 Ai γ(z − z0 ) , (z > 0). (37)

For z < 0, I wasn’t quite sure how to account for the infinite potential jump at z = 0, so instead I supposed the potential for z < 0 to be a constant, V (z) = V0 , where eventually I’ll take V0 → ∞. Then the Schr¨ odinger equation for z < 0 is d2 2m Ψ(z) − 2 [V0 − E] Ψ(z) = 0 2 dz ~ with solution r 2m −kz [V0 − E]. (38) Ψ(z) = Ae , k= ~2 Matching values and derivatives of (37) and (38) at z = 0, we have β1 Ai(−γz0 ) = A γβ1 Ai0 (−γz0 ) = −kA Dividing, we obtain 1 Ai(−γz0 ) 1 =− γ Ai0 (−γz0 ) k Now taking V0 → ∞, we also have k → ∞, so the RHS of this goes to zero; thus the condition is that −γz0 be a zero of the Airy function, which means the energy eigenvalues En are given by 

2m2 g ~2

1/3

En = xnm ⇒ En = mg

where xn is the nth root of the equation Ai(−xn ) = 0.



mg 2 ~2 2

1/3

xn

(39)

Homer Reid’s Solutions to Merzbacher Problems: Chapter 7

12

So that’s the exact solution. In the WKB approximation, the spectrum of energy eigenvalues is determined by the condition (35). In this case the classical turning points are at z = 0 and z = z0 , so we have   Z z0 1 n+ π= k(z) dz 2 0 r Z 2m z0 = [E − mgz]1/2 dz ~2 0 r Z 2m2 g z0 [z0 − z]1/2 du = ~2 0 r z0 2m2 g 2 3/2 − (z0 − z) = 2 ~ 3 0 r  3/2 2 2 2m g E = 3 ~2 mg so the nth eigenvalue is given by

En =



mg 2 ~2 2

1/3    2/3 1 3 n+ π . 2 2

Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid May 13, 2001

Chapter 8

1

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

2

Problem 8.1 Apply the variational method to estimate the ground state energy of a particle confined in a one-dimensional box for which V = 0 for −a < x < a, and Ψ(±a) = 0. (a) First, use an unnormalized trapezoidal trial function which vanishes at ±a and is symmetric with respect to the center of the well: ( (a − |x|), b≤x≤a Ψt (x) = (a − b), |x| ≤ b. (b) A more sophisticated trial function is parabolic, again vanishing at the end points and even in x. (c) Use a quartic trial function of the form Ψt (x) = (a2 − x2 )(αx2 + β), where the ratio of the adjustable parameters α and β is determined variationally. (d) Compare the results of the different variational calculations with the exact ground state energy, R a and, using normalized wave functions, evaluate the mean square deviation −a |Ψ(x) − Ψt (x)|2 dx for the various cases.

(e) Show that the variational procedure produces, in addition to the approximation to the ground state, an optimal quartic trial function with nodes between the endpoints. Interpret the corresponding stationary energy value.

First let’s observe that the exact expressions for the ground state wavefunction and energy are 1 Ψn (x) = √ cos(kn x), a

kn =

nπ , 2a

En = n 2

~2 ~2 π 2 ≈ 1.23 . 8ma2 ma2

(a) We need first to normalize the trial wavefunction. Taking

Ψt (x) =

(

γ(a − |x|), γ(a − b),

b ≤ |x| ≤ a x| ≤ b.

3

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

we have Z

a

Ψ2t (x)dx = 2

−a

Z

= 2γ

a

Ψ2t (x)dx 0 ( Z

2

(a − b)

b

2

dx + 0

Z

a b

2

(a − x) dx

)

 1 3 = 2γ b(a − b) + (a − b) 3   1 3 3 2 2 2 2 2 = 2γ b(a + b − 2ab) + (a − b ) − a b + b a 3
2 = γ 2 a3 + 2b3 − 3ab2 3 2



2

so Ψt is normalized by taking

γ2 =



3 2

1 a3 + 2b3 − 3ab2



.

(1)

Now we can compute the energy expectation value of Ψt : Z a ~2 d2 < Ψt |H|Ψt > = − Ψt (x) 2 Ψt (x) dx 2m −a dx Integrating by parts, =−

~2 2m



Z a Ψt (x)Ψ0t (x) − −a

a

Ψ02 t (x) dx −a



(the first integral vanishes since Ψt vanishes at the endpoints) =+

~2 m

Z

a 0

Ψ02 t (x) dx

Z ~2 2 a γ dx m b ~2 2 γ (b − a). = m =

Using (1), this is < H >=

3~2 2m



(b − a) a3 + 2b3 − 3ab2



.

To find the optimal value of b, we zero the derivative of this with respect to b: 0=

(a3

1 6b2 (b − a) 6ab(b − a) − 3 + 3 3 2 3 2 2 + 2b − 3ab ) (a + 2b − 3ab ) (a + 2b3 − 3ab2 )2

= −4b3 + 9b2 a − 6a2 b + a3

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

(b) For a parabolic trial function we take Ψt (x) = γ(a2 − x2 ). The normalization integral is Z a Z a (a2 − x2 )2 dx Ψ2t (x) dx = 2γ 2 0 −a Z a 2 (a4 + x4 − 2a2 x2 )dx = 2γ 0  1 2 2 = 2γ a5 + a5 − a5 5 3 16 2 5 γ a = 15 so Ψt (x) is normalized by taking γ2 =

15 . 16a5

The expectation value of the energy is Z a d2 ~2 Ψt (x) 2 Ψt (x) dx =− 2m −a dx Z a 2 ~ = 2 γ2 (a2 − x2 )dx m 0 4 ~2 2 3 γ a = 3m ~2 5 ~2 ≈ 1.25 . = 4 ma2 ma2 So this is in good agreement with the exact ground state energy. (c) In this case we have Ψt (x) = γ(a2 − x2 )(αx2 + β)

= γ[−αx4 + (αa2 − β)x2 + βa2 ]

The kinetic energy is −

~2 ~2 d 2 Ψt (x) = γ [6αx2 − (αa2 − β)]. 2 2m dx m

The expectation value of the energy is

4

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

5

Problem 8.2 Using scaled variables, as in Section 5.1, consider the anharmonic oscillator Hamiltonian, 1 1 H = p2ξ + ξ 2 + λξ 4 2 2 where λ is a real-valued parameter. (a) Estimate the ground state energy by a variational calculation, using as a trial function the ground state wave function for the harmonic oscillator H0 (ω) =

1 2 1 2 2 p + ω ξ 2 ξ 2

where ω is an adjustable variational parameter. Derive an equation that relates ω and λ. (b) Compute the variational estimate of the ground state energy of H for various positive values of the strength λ. (c) Note that the method yields answers for a discrete energy eigenstate even if λ is slightly negative. Draw the potential energy curve to judge if this result makes physical sense. Explain.

(a) To find the ground state eigenfunction of the Hamiltonian Merzbacher proposes, it’s convenient to rewrite it: 1 2 1 2 2 p + ω ξ 2 ξ 2 1 1 ∂2 + ω2 ξ 2 =− 2 2 ∂ξ 2

H0 (ω) =

Upon substituting u = ω 1/2 ξ we obtain 1 ∂2 1 =ω − + u2 2 ∂u2 2 



and now this is just the ordinary harmonic oscillator Hamiltonian, scaled by a constant factor ω, with ground-state eigenfunction Ψ(ω) = Ce−u

2

/2

= Ce−ωξ

2

/2

Adding the normalization constant, Ψ(ω) =

 ω 1/4 π

e−ωξ

2

/2

.

.

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

6

Now we want to treat ω as a parameter and vary it until the energy expectation value of Ψ(ω) is minimized. The energy expectation value is hΨ| H |Ψi = hΨ| T |Ψi + hΨ| V |Ψi where T = p2ξ /2 and V = ξ 2 /2 + λξ 4 . Let’s compute the two expectation values separately. First of all, to compute the expectation value of T , we need to know the result of operating on Ψ(ω) with p2ξ : p2ξ Ψ(ω)

  ω 1/4  ∂  ∂ −ωξ 2 /2 =− e π ∂ξ ∂ξ  i  ω 1/4 ∂ h −ωξ 2 /2 −ωξe =− π ∂ξ  ω 1/4   2 −ω + ω 2 ξ 2 e−ωξ /2 =− π

Then for the expectation value of T we have Z 1 ∞ Ψ(ξ)p2ξ Ψ(ξ)dξ hΨ| T |Ψi = 2 −∞ r Z
2 1 ω ∞ =− −ω + ω 2 ξ 2 e−ωξ dξ 2 π −∞ r  r r  1 ω π π 1 −ω =− + ω2 2 π ω 2 ω3 ω = . 4

(2)

On the other hand, for the expectation value of V we have (3) r

 Z  Z ∞ ω 1 ∞ 2 −ωξ2 4 −ωξ 2 exptwoΨV Ψ = ξ e dξ + λ ξ e dξ π 2 −∞ −∞ r  r r  3 ω 1 π π = + λ 3 π 4 ω 4 ω5 1 3λ = + . 4ω 4ω 2

(4)

Adding (2) and (4), exptwoΨHΨ =

  1 3λ 1 ω+ + 2 . 2 ω ω

To minimize this with respect to ω we equate its ω derivative to 0: 0=1−

1 6λ − 3 ω2 ω

(5)

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

7

or ω 3 − ω − 6λ = 0.

(6)

We could then solve this equation for ω in terms of λ to obtain the energyminimizing value of ω for a given perturbing potential strength λ. But writing down the full solution would be tedious. Instead let’s see what happens when λ is small. Evidently, when λ = 0 the Hamiltonian in this problem degenerates to the normal harmonic oscillator Hamiltonian, for which the energy is minimized by the (unscaled) ground state harmonic oscillator wavefunction, i.e. Ψ(ω) with ω = 1. We can thus imagine that, for small λ, the energy-minimizing value of ω will be close to 1, and we may write ω(λ) ≈ 1 +  for some small . Inserting this in (6), (1 + 3 + 32 + 3 ) − (1 + ) = 6λ Keeping only terms of zeroth or first order in the small quantity  (which is equivalent to keeping terms of lowest order in the perturbing potential strength λ) we obtain from this  ≈ 3λ, so for λ  0 the minimizing value of ω is ω ≈ 1 + 3λ. Inserting this estimate into (5) and again keeping only terms of lowest order in λ we find (7) exptwoΨHΨ = ≈ ≈ ≈

 1 (1 + ) + (1 + )−1 + 3λ(1 + )−2 4  1 (1 + 3λ) + (1 + 3λ)−1 + 3λ(1 + 3λ)−2 4 1 [(1 + 3λ) + (1 − 3λ) + 3λ(1 − 6λ)] 4 1 3 + λ. 2 4

(8)

Since the 1/2 term is the normal (unperturbed) energy of the state, the energy shift caused by the perturbing potential is ∆E =

3 λ. 4

(9)

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

8

Problem 8.3 In first-order perturbation theory, calculate the change in the energy levels of a linear harmonic oscillator that is perturbed by a potential gx4 . For small values of the coefficient, compare the result with the variational calculation in Problem 2. The energy shift to first order is ∆E = exptwoΨn (x)|gx4 |Ψn (x) = g hΨn | x4 |Ψn i . I worked out this expectation value in Problem 5.3: 3g ∆E = gexptwoΨn x Ψn = 2 4



~ mω

2 

 1 + n + n2 2

In particular, the energy shift of the ground state is 3g ∆E0 = 4



~ mω

2

which agrees with () (the difference in the factor (~/mω)2 just represents the fact that in Problem 8.2 we used scaled variables, whereas in this problem we inserted the units explicitly).

Problem 8.4 2

Using a Gaussian trial function, e−λx , with an adjustable parameter, make a variational estimate of the ground state energy for a particle in a Gaussian potential well, represented by the Hamiltonian H=

2 p2 − V0 e−αx 2m

(V0 > 0, α > 0).

For notational simplicity, I like to put β/2 = λ. Then Ψ(x) = Ce−βx

2

/2

and the normalization constant is determined by 1=C

2

Z

∞

e −∞

−βx2

dx

⇒

 1/4 β C= . π

9

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

The kinetic energy operator operating on this state yields   ∂ p2 ~2 ∂ TΨ = Ψ(x) = − Ψ(x) 2m 2m ∂x ∂x  1/4 2 i 2 ~ ∂ h β −βxe−βx /2 =− π 2m ∂x  1/4 2  2 ~  β =− −β + β 2 x2 e−βx /2 π 2m and its expectation value is  1/2 2  r r  β2 ~ β π π − β hT i = π 2m β 2 β3 ~2 β . 4m The expectation value of the potential energy is Z ∞ 2 hV i = −V0 Ψ2 (x)e−αx dx −∞ r Z ∞ 2 2 β e−βx e−αx dx = −V0 π −∞ r r β π = −V0 π (α + β) s β . = −V0 (α + β)

(10)

=

(11)

Combining (10) and (11), ~2 β exptwoΨ|H|Ψ = hΨ| T |Ψi + hΨ| V |Ψi = − V0 2m

s

β . (α + β)

(12)

To minimize with respect to β we equate the first β derivative of this to zero: " # √ V0 1 β ~2 p − −p 0= 2m 2 β(α + β) (α + β)3  1/2 ~2 α2 V0 = − 2m 2 β(α + β)3  2 mV0 α 3 = β(α + β) − ~2 2  mV0 α = β 4 + 3β 3 α + 3β 2 α2 + βα3 − ~2 2  mV0 = x4 + 3x3 + 3x2 + x − ~2 α

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

10

where I put x = β/α. In theory we could write down an explicit expression for the roots of this quartic in terms of mV0 /~2 α, and then insert said expression into (12) to obtain the lowest energy attainable with this form of trial wave function. In practice, however, this would be a mess, and I can’t see any way to proceed other than numerically. Am I missing some kind of trick here?

Problem 8.5 Show that as inadequate a variational trial function as  (  C 1 − |x| |x| ≤ a a Ψ(x) = 0 |x| > a yields, for the optimum value of a, an upper limit to the ground state energy of the linear harmonic oscillator, which lies within less than 10 percent of the exact value.

The first task is to evaluate the normalization constant C. Z a 2 1=C Ψ(x)2 dx −a Z a x 2 2 1− = 2C dx a 0  Z a x x2 2 1−2 + 2 = 2C a a i h0 a = 2C 2 a − a + 3 so

C=

r

3 . 2a

The harmonic oscillator hamiltonian is E =T +V =

p2 mω 2 x2 + . 2m 2 (13)

2

a

~ 2m

Z

~2 =− 2m

(

exptwoΨT Ψ = −

2

Ψ(x) −a

∂ Ψ(x) dx ∂x2

Integrating by parts, a 2 ) Z a ∂ ∂ Ψ(x) dx Ψ(x) Ψ(x) − ∂x −a ∂x −a

(14)

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

11

The first term vanishes... ~2 = 2m



3~2 = 2ma2

3 2a

Z

a −a

1 dx a2 (15)

(16) exptwoΨV Ψ =

mω 2

2

2

Z

Z

a

x2 Ψ(x)2 dx

−a a

x2 Ψ(x)2 dx   Z a 2x3 x4 3 2 2 x − + 2 dx = mω 2a a a 0 a   3 4 5 x x 3 x − + 2 = mω 2 2a 3 2a 5a 0 = mω

=

0

mω 2 a2 20

3~2 mω 2 a2 + . 2 2ma 20 To minimize with respect to a we set the a derivative of this to zero: exptwoΨ|H|Ψ = hΨ| T |Ψi + hΨ| V |Ψi =

0=−

(17)

(18)

mω 2 a 3~2 + ma3 10

or 30~2 m2 ω 2 √ ~ . a2 = 30 mω

a4 =

Inserting into (18), 3 exptwoΨHΨ = √ ~ω ≈ 0.547 · ~ω. 30 Of course the actual ground state energy is 0.5 · ~ω, so the fractional error is 0.047/0.5 < 10%.

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8

12

Problem 8.6 A particle of mass m moves in a potential V (r). The n − th discrete energy eigenfunction of this system, Ψn (r), corresponds to the energy eigenvalue En . Apply the variational principle by using as a trial function, Ψt (r) = Ψn (λr), where λ is a variational (scaling) parameter, and derive the virial theorem for stationary states.