Solutions to problems in merzbacher quantum mechanics 3rd ed-reid-p59 PDF

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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 Ψ(x, 0) = C exp − + ik x . 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

−∞ Z ∞

(2π)−1/2 C

−∞

 exp − 1

 x2 + i( k − k ) x dx 0 4(∆x)2

(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 − i(k0 − k )x − (k0 − k )2 (∆x)2 + (k0 − k )2 (∆x)2 2 4(∆x) 2  x − i(k0 − k )∆x − (k0 − k)2 (∆x)2 − 2(∆x)  2 1 (2) x − 2i(k0 − k)(∆x)2 − (k0 − k )2 (∆x)2 − 4(∆x)2 −



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

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

Z

∞

−∞

 exp −

 1 [x − 2i(k0 − k )(∆x)2 ]2 dx 2 4(∆x)

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 − 2 4(∆k )

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

¯hk 2 . 2m

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

Ψ(x, t) =

Z √ 2C ∆x

∞

  ¯hk2 t dk. exp −(k0 − k )2 (∆x)2 + ikx − i 2m −∞

(3)

Again we complete the square in the exponent:   ¯hk 2 i¯ h t]k 2 − [2k0 (∆x)2 + ix]k + k02(∆x)2 t = − [(∆x)2 + −(k0 − k)2 (∆x)2 + ikx − i 2m 2m   = − α2 k 2 − βk + γ   β2 β2 β −γ = −α2 k 2 − 2 k + 4 − 4 4α α 4α β2 β = −α2 [k − 2 ]2 + 2 − γ 4α 2α 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

∞

  β 2 exp −α2 (k − ) dk. 2α2 −∞

Z

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

Ψ(x, t) = =

√ 2πC∆x

    2 1 β −γ exp 2 4α α " #1/2 " # 2 √ 2 (ix + 2k0 (∆x)2 )2 (∆x) −k20 (∆x) exp 2πC e i¯ h i¯ h 4[(∆x)2 + 2m (∆x)2 + 2m t 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 followA, ing parameters: m=940 Mev (corresponding to a proton or neutron), ∆x=3 ˚ A−1 . This value of k0 corresponds, for a neutron, to a velocity of about k0 =0.8 ˚ 4 5 · 10 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

4

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

Ψ(x, t) =

√

(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):   h ¯ β2 S(x, t) = − ln α + . 2 i 4α Things are actually easier if we define δ = α2 . Then   β2 i¯ h 1 ¯h δ = (∆x)2 + − ln δ + t S(x, t) = 2 i 4δ 2m Now computing partial derivatives: ∂S ∂t

= =

∂S ∂x

= =

2

∂ S ∂x2

=

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

(2)

(3) (4)

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

Inserting (??), (??), and (??) into this equation, we find h ¯2β2 ¯h2 β 2 ¯h2 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) = Ψ1∗(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

2π¯ h λ −2π¯ h dλ λ2

        −9  dp   dλ  p  =  λ  = 10

so the momentum uncertainty is

∆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 h ¯ t ≈ (∆x)2 2m

or t

2m(∆x)2 /¯ h 2 · 9.4 · 108 ev · (0.3 m)2 ≈ c 2 · 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) = c 1 Ψ1 (r) + c 2 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 = [c1∗Ψ∗1 + c2∗Ψ2∗] · [c 1 Ψ1 + c 2 Ψ2 ] = |c 1 ||Ψ1 | + |c 2 ||Ψ2 | + c 1 c2∗Ψ1 Ψ2∗ + c1∗c 2 Ψ1∗Ψ2

1

(1)

Problem 3.2 For a free particle in one calculate the variance at time     dimension, t, (∆x)t2 ≡ (x − hxit )2 t = x2 t −hxit2 without explicit use of the wave function by applying (3.44) repeatedly. Show that   2 1 (∆px )2 2 hxpx + px xi0 − hxi0 hpx i t + t (∆x)2t = (∆x)20 + m2 m 2 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   1 2 d 2 w  dw  + w(t) = w(0) + t t + ··· (2) 2 dt2 t=0 dt t=0 We have

i d h 2  2 x − hxi dt d  2 d = (3) x − 2 hxi hxi dt dt 2  2   2 2 d d w d d = hxi − 2 hxi 2 hxi x2 − 2 (4) dt2 dt dt2 dt   We need to compute the time derivatives of hxi and x2 . The relevant equation is   1 ∂F d hF i = hF H − H F 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 − ih) ¯ 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 h pi = (5) m 2 1 d d hxi = hpi = 0 (6) dt2 m dt  1 2 d  2 [x , H ] x = i¯ h dt  1  2 2 = x p − p2 x 2 2im¯ h  1  2 2 = x p − p(xp − ih)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 − ih)p ¯ + 3i¯hxp + 2¯ h2 2im¯ h  1  2 = 2¯ h + 4i¯hxp 2im¯ h 2 i¯ h = − + hxpi (7) m m d2  2  2 d hxpi x = m dt dt2 2 = h[xp, H ]i i¯ hm  1  3 = − xp − p2 xp i¯ h m2  1  3 = − xp − p(xp − i¯ h)p i¯ h m2  1  3 = − xp − pxp2 + i¯hp2 i¯ h m2  1  3 = − xp − (xp − i¯ h)p2 + i¯hp2 i¯ h m2  1  2i¯hp2 = i¯ h m2 2  2 = p (8) m2 3     2 d x2 = [p2 , H ] = 0 (9) dt3 i¯ h m2  2 Now that we’ve computed all time derivatives of hxi and x , it’s time to =

3

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

d2 w dt2

d d  2 x − 2 hxi hxi dt dt i¯ h 2 2 = − + hxpi − hxi hpi m  m  m 2 i¯ h 2 = − + xp − hxi hpi m 2 m   2 px − xp 2 = + xp − h x i h pi 2 m m   2 px + xp 2 = − hxi hpi 2 m m   2 d2 d2  2  d = x −2 hxi − 2 hxi 2 hxi dt dt2 dt 2 2 2  2 2 p − 2 hpi = 2 (∆p)2 = m m2 m

=

(10)

(11)

Finally, we plug these into the original equation (2) to find   (∆p)2 2 2 1 t. w(t) = w(0) + hpx + xpi − hxi hpi t + m2 m 2 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 =

1 p2 + 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 i t 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)t2 ≡ x2 t − hxi t

(∆p)2 = (∆x)02 cos2 ωt + 2 20 sin2 ωt m ω   sin 2ωt 1 hxp + pxi0 − hxi0 hpi0 + mω 2

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 1 d e(t) = hxH − H xi dt 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 h pi . = m d2 d hpi e(t) = dt2 dt m 5

= = = = = =

1 hpH − H pi i¯ hm 2   ω px2 − x2 p 2i¯ h  ω2  (xp − ih ¯ )x − x2 p 2i¯ h  ω2  x(xp − ih) ¯ − i¯hx − x2 p 2i¯ h ω2 h−2i¯hxi 2i¯ h 2 −ω 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 e (0) = m

→

′

→

A = hxi0 hpi0 . mω

B=

(b) Let’s define v(t) = hT − V i t . 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 p2 x 2 − x 2 p2 . = 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 2 = −2ω hxpi + ω 2 hxp − pxi =

−ω 2 hxp + pxi

(12)

Next, 2ω 2 d2 v(t) = − hxpH − H xpi i¯ h  dt2    mω 2  2ω 2 1  3 = − xpx2 − x3 p xp − p2 xp + i¯ h 2m 2 6

(13)

The bracketed expressions are  3  xp − p2 xp = xpx2 − x3 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¯ hx 2 − x 3 p   = −2i¯ hx 2 =









and plugging these back into (13) gives

d2 v(t) = = −4ω 2 dt2 =

"  # mω 2  2  p2 − 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 = 2 ωB 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 [hH it − v(t)] . mω 2

Since H does not depend explicitly on time, hH i is constant in time. For v(t) we can use (14):    2 ω hxp + pxi0 1 hT i0 + hV i0 − [hT i0 − hV i0 ] cos 2ωt + sin 2ωt x t = 2 2 mω   1 ω hxp + pxi0 2 2 = 2 hT i0 sin ωt + 2 hV i0 cos ωt + sin 2ωt 2 2 mω  2 p 0 2  2 sin 2ωt 1 2 = sin ωt + x 0 cos ωt + hxp + pxi0 . (15) m2 ω 2 mω 2 7

(c) Earlier we found that hxit

=

2

=

hxi t

h pi 0 sin ωt mω 2 h p i 2 hxi0 cos2 ωt + 2 02 sin2 ωt + hxi0 hpi0 sin 2ωt. m ω

hxi0 cos ωt +

Subtracting from (15) gives i h    2 (∆x)t2 = x2 − hxi2 = x2 0 − hxi0 cos2 ωt i 1 h  2 + 2 2 p2 0 − hpi0 m ω   1 + hxp + pxi0 − hxi0 hpi0 sin 2ωt 2   (∆p)2 1 sin 2ωt hxp + pxi0 − hxi0 hpi0 . = (∆x)20 cos2 ωt + 2 20 sin2 ωt + mω 2 m ω 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 )

1 [pδ (r − r0 ) + δ (r − r0 )p] . 2m

j(r0 ) →

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] d r. 2m 8

Here we can use the identity

R

f (x)δ ′ (x − a)dx = −f ′ (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(r′ ), the Wigner distribution function is defined as     r′′ r′′ 1 Ψ r′ + exp(−ip′ ·x′′ /h)Ψ ¯ ∗ r′ − dr′′. W (r′ , p′ ) = 3 (2πh) ¯ 2 2 (a) Show that W (r′ , p′ ) is a real-valued function, defined over the six-dimensional “phase space” (r′ , p′ ). (b) Prove that Z

W (r′ , p′ )dp′ = |Ψ(r′ )|2

and that the expectation value of a function of the operator r in a normalized state is Z Z hf (r)i = f (r′ )W (r′ , p′ )d r′ dp′ . (c) Show that the Wigner distribution function is normalized as Z W (r′ , p′ )d r′ d p′ = 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 hω(n + )δnk − m2 ω 2 < Ψn |x2 |Ψk > . < Ψn |p2x |Ψk >= 2m¯ 2 The nth eigenfunction is Ψ(x) =



1 2n n!

1/2 

mω h ¯π

1/4

exp(−

(1)

r mω 2 mω x). x )Hn ( h ¯ 2¯ h

The matrix element of x2 is then r r 1/2   Z mω 1/2 ∞ 2 mω mω 2 mω 1 x) dx. x exp( x)Hk ( x )Hn ( < Ψn |x2 |Ψk >= n+k h ¯ 2 n!k! h ¯ h ¯ h ¯π −∞ 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 1 + 2λ(s + t) + (2λ)2 (s + t)2 + · · · 1 + (2st) + (2st)2 + · · · = π 1 + λ2 + λ4 + · · · 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 th...