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 numberkx and a Gaussian amplitude is given by

Ψ(x, 0) = C exp

[

− x2

4(∆x)2+ ikxx

]

.

Calculate the corresponding kx distribution and Ψ(x, t), assumingfree particle motion. Plot |Ψ(x, t)|2 as a function of x for severalvalues of t, choosing ∆x small enough to show that the wave packetspreads in time, while it advances according to the classical laws.Apply the results to calculate the effect of spreading in some typicalmicroscopic and macroscopic experiments.

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

Φ(k) = (2π)−1/2

∫

∞

−∞

Ψ(x, 0)e−ikxdx

= (2π)−1/2C

∫

∞

−∞

exp

[

− x2

4(∆x)2+ i(k0 − k)x

]

dx (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

4(∆x)2+ i(k0 − k)x = −

[

x2

4(∆x)2− i(k0 − k)x − (k0 − k)2(∆x)2 + (k0 − k)2(∆x)2

]

= −[

x

2(∆x)− i(k0 − k)∆x

]2

− (k0 − k)2(∆x)2

= − 1

4(∆x)2[

x − 2i(k0 − k)(∆x)2]2 − (k0 − k)2(∆x)2 (2)

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

Φ(k) = (2π)−1/2C exp[−(k0−k)2(∆x)2]

∫

∞

−∞

exp

{

− 1

4(∆x)2[x − 2i(k0 − k)(∆x)2]2

}

dx

In the integral we can make the shift x → x − 2i(k0 − k)(∆x)2 and use thestandard 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 inconfiguration space, we can write

Φ(k) =√

2C∆x exp

[

− (k0 − k)2

4(∆k)2

]

where ∆k = 1/(2∆x). This is the minimum possible k width attainable fora wave packet with x width ∆x, which is why the Gaussian wave packet issometimes referred to as a minimum uncertainty wave packet.

The next step is to compute Ψ(x, t) for t > 0. Since we are talking abouta free particle, we know that the momentum eigenfunctions are also energyeigenfunctions, which makes their time evolution particularly simple to writedown. 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 termsexp(ikx), with the kth term weighted in the sum by the factor Φ(k). The wavepacket at a later time t > 0 will be given by the same linear combination, butnow with the kth term multiplied by a phase factor exp[−iω(k)t] describing itstime evolution. In symbols we have

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) =hk2

2m.

2

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

Ψ(x, t) =√

2C∆x

∫

∞

−∞

exp

[

−(k0 − k)2(∆x)2 + ikx − ihk2

2mt

]

dk. (3)

Again we complete the square in the exponent:

−(k0 − k)2(∆x)2 + ikx − ihk2

2mt = −

{

[(∆x)2 +ih

2mt]k2 − [2k0(∆x)2 + ix]k + k2

0(∆x)2}

= −{

α2k2 − βk + γ}

= −α2

{

k2 − β

α2k +

β2

4α4− β2

4α4

}

− γ

= −α2[k − β

2α2]2 +

β2

4α2− γ

where we have defined some shorthand:

α2 = (∆x)2 +ih

2mt β = 2k0(∆x)2 + ix γ = k2

0(∆x)2.

Using this in (??) we find:

Ψ(x, t) =√

2C∆x exp

[

β2

4α2− γ

]∫

∞

−∞

exp

[

−α2(k − β

2α2)2

]

dk.

The integral evaluates to π1/2/α. We have

Ψ(x, t) =√

2πC∆x

(

1

α

)

exp

[

β2

4α2− γ

]

=√

2πC

[

(∆x)2

(∆x)2 + ih2m t

]1/2

e−k2

0(∆x)2 exp

[

(ix + 2k0(∆x)2)2

4[(∆x)2 + ih2m t]

]

This is pretty ugly, but it does display the relevant features. The importantpoint is that term iht/2m adds to the initial uncertainty (∆x)2, so that thewave packet spreads out with time.

In the figure, I’ve plotted this function for a few values of t, with the follow-ing parameters: m=940 Mev (corresponding to a proton or neutron), ∆x=3A,k0=0.8 A−1. This value of k0 corresponds, for a neutron, to a velocity of about5 · 104 m/s; and note that, sure enough, the center of the wave packet travelsabout 5 nm in 100 fs. The time scale of the spread of this wave packet is ≈ 100fs.

3

0

50

100

150

200

250

300

350

-1e-08 -5e-09 0 5e-09 1e-08

Wav

efun

ctio

n (a

rbitr

ary

units

)

Distance (meters)

Gaussian wave packet example

t=0 st=30 fst=70 fs

t=100 fs

4

Problem 2.2

Express the spreading Gaussian wave function Ψ(x, t) obtainedin Problem 1 in the form Ψ(x, t) = exp[iS(x, t)/h]. Identify thefunction S(x, t) and show that it satisfies the quantum mechanicalHamilton-Jacobi equation.

In the last problem we found

Ψ(x, t) =√

2πC

(

∆x

α

)

exp

[

β2

4α2− γ

]

= exp

[

ln

(√2πC

∆x

α

)

+β2

4α2− γ

]

(1)

where we’ve again used the shorthand we defined earlier:

α2 = (∆x)2 +ih

2mt β = 2k0(∆x)2 + ix γ = k2

0(∆x)2.

From (??) we can identify (neglecting an unimportant additive constant):

S(x, t) =h

i

[

− lnα +β2

4α2

]

.

Things are actually easier if we define δ = α2. Then

δ = (∆x)2 +ih

2mt S(x, t) =

h

i

[

−1

2ln δ +

β2

4δ

]

Now computing partial derivatives:

∂S

∂t=

h

i

[

− 1

2δ− β2

4δ2

]

∂δ

∂t

= − h2

2m

[

1

2δ+

β2

4δ2

]

(2)

∂S

∂x=

h

i

β

2δ

∂β

∂x

=hβ

2δ(3)

∂2S

∂x2=

ih

2δ(4)

The quantum-mechanical Hamilton-Jacobi equation for a free particle is

∂S

∂t+

1

2m

[

∂S

∂x

]2

− ih

2m

∂2S

∂x2= 0

5

Inserting (??), (??), and (??) into this equation, we find

− h2

4mδ− h2β2

8mδ2+

h2β2

8mδ2+

h2

4mδ= 0

so, sure enough, the equation is satisfied.

Problem 2.3

Consider a wave function that initially is the superposition of twowell-separated narrow wave packets:

Ψ1(x, 0) + Ψ2(x, 0)

chosen so that the absolute value of the overlap integral

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? Justifyyour answer.

It seems to me that the answer to this problem depends entirely on thespecifics 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 plottedin the figure from problem 2.1 If one of those wave packets were centered inChile and another in China, the overlap integral would be tiny, since the wavepackets only have appreciable value within a few angstroms of their centers.Furthermore, even if the neutrons are initially moving toward each other, theirwave packets spread out on a time scale of ≈ 100 fs, long before their centersever 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 collapsingentirely.

Problem 2.4

A high resolution neutron interferometer narrows the energy spreadof thermal neutrons of 20 meV kinetic energy to a wavelength dis-persion level of ∆λ/λ= 10−9. Estimate the length of the wavepackets in the direction of motion. Over what length of time willthe 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 usabout the momentum dispersion level?

p =2πh

λ

dp =−2πh

λ2dλ

so

∣

∣

∣

∣

dp

p

∣

∣

∣

∣

=

∣

∣

∣

∣

dλ

λ

∣

∣

∣

∣

= 10−9

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 · s6.1 · 10−6 ev / c

≈ c · 10−10 s

≈ 30 cm.

This is HUGE! So the point is, if we know with this precision how quicklyour thermal neutrons are moving, we have only the most rough indication ofwhere in the room they might be.

To estimate the time scale of spreading of the wave packets, we can imaginethat they are Gaussian packets. In this case we start to get appreciable spreadingwhen

h

2mt ≈ (∆x)2

or

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



Homer Reid

March 8, 1999

Chapter 3

Problem 3.1

If the state Ψ(r) is a superposition,

Ψ(r) = c1Ψ1(r) + c2Ψ2(r)

where Ψ1(r) and Ψ2(r) are related to one another by time reversal,show that the probability current density can be expressed withoutan interference term involving Ψ1 and Ψ2.

I found this to be a pretty cool problem! First of all, we have the probabilityconservation equation:

d

dtρ = −~∇ · ~J.

To show that ~J contains no cross terms, it suffices to show that its divergencehas no cross terms, and to show this it suffices (by probability conservation) toshow that dρ/dt has no cross terms. We have

ρ = Ψ∗Ψ

= [c∗1Ψ∗

1 + c∗2Ψ∗

2] · [c1Ψ1 + c2Ψ2]

= |c1||Ψ1| + |c2||Ψ2| + c1c∗

2Ψ1Ψ∗

2 + c∗1c2Ψ∗

1Ψ2 (1)

1

Problem 3.2

For a free particle in one dimension, calculate the variance at timet, (∆x)2t ≡

⟨

(x − 〈x〉t)2⟩

t=

⟨

x2⟩

t−〈x〉2t without explicit use of the

wave function by applying (3.44) repeatedly. Show that

(∆x)2t = (∆x)20 +2

m

[

1

2〈xpx + pxx〉0 − 〈x〉0 〈px〉

]

t +(∆px)2

m2t2

and(∆px)2t = (∆px)20 = (∆px).2

I find it easiest to use a slightly different notation: w(t) ≡ (∆x)2t . (The wreminds me of “width.”) Then

w(t) = w(0) + tdw

dt

∣

∣

∣

∣

t=0

+1

2t2

d2w

dt2

∣

∣

∣

∣

t=0

+ · · · (2)

We have

dw

dt=

d

dt

[

⟨

x2⟩

− 〈x〉2]

=d

dt

⟨

x2⟩

− 2 〈x〉 d

dt〈x〉 (3)

d2w

dt2=

d2

dt2⟨

x2⟩

− 2

[

d

dt〈x〉

]2

− 2 〈x〉 d2

dt2〈x〉 (4)

We need to compute the time derivatives of 〈x〉 and⟨

x2⟩

. The relevantequation is

d

dt〈F 〉 =

1

ih〈FH − HF 〉 +

⟨

∂F

∂t

⟩

for any operator F . For a free particle, the Hamiltonian is H = p2/2m, andthe all-important commutation relation is px = xp − ih. We can use this tocalculate the time derivatives:

d

dt〈x〉 =

1

ih〈[x, H ]〉

=1

2imh

⟨

xp2 − p2x⟩

=1

2imh

⟨

xp2 − p(xp − ih)⟩

=1

2imh

⟨

xp2 − pxp + ihp⟩

2

=1

2imh

⟨

xp2 − (xp − ih)p + ihp⟩

=1

2imh〈2ihp〉

=〈p〉m

(5)

d2

dt2〈x〉 =

1

m

d

dt〈p〉 = 0 (6)

d

dt

⟨

x2⟩

=1

ih

⟨

[x2, H ]⟩

=1

2imh

⟨

x2p2 − p2x2⟩

=1

2imh

⟨

x2p2 − p(xp − ih)x⟩

=1

2imh

⟨

x2p2 − pxpx + ihpx⟩

=1

2imh

⟨

x2p2 − (xp − ih)2 + ih(xp − ih)⟩

=1

2imh

⟨

x2p2 − xpxp + 2ihxp + h2 + h2 + ihxp⟩

=1

2imh

⟨

x2p2 − x(xp − ih)p + 3ihxp + 2h2⟩

=1

2imh

⟨

2h2 + 4ihxp⟩

= − ih

m+

2

m〈xp〉 (7)

d2

dt2⟨

x2⟩

=2

m

d

dt〈xp〉

=2

ihm〈[xp, H ]〉

= − 1

ihm2

⟨

xp3 − p2xp⟩

= − 1

ihm2

⟨

xp3 − p(xp − ih)p⟩

= − 1

ihm2

⟨

xp3 − pxp2 + ihp2⟩

= − 1

ihm2

⟨

xp3 − (xp − ih)p2 + ihp2⟩

=1

ihm2

⟨

2ihp2⟩

=2

m2

⟨

p2⟩

(8)

d3

dt3⟨

x2⟩

=2

ihm2

⟨

[p2, H ]⟩

= 0 (9)

Now that we’ve computed all time derivatives of 〈x〉 and⟨

x2⟩

, it’s time to

3

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

dw

dt=

d

dt

⟨

x2⟩

− 2 〈x〉 d

dt〈x〉

= − ih

m+

2

m〈xp〉 − 2

m〈x〉〈p〉

=2

m

⟨

− ih

2+ xp

⟩

− 2

m〈x〉〈p〉

=2

m

⟨

px − xp

2+ xp

⟩

− 2

m〈x〉〈p〉

=2

m

⟨

px + xp

2

⟩

− 2

m〈x〉〈p〉 (10)

d2w

dt2=

d2

dt2⟨

x2⟩

− 2

[

d

dt〈x〉

]2

− 2 〈x〉 d2

dt2〈x〉

=2

m2

⟨

p2⟩

− 2

m2〈p〉2 =

2

m2(∆p)2 (11)

Finally, we plug these into the original equation (2) to find

w(t) = w(0) +2

m

[

1

2〈px + xp〉 − 〈x〉〈p〉

]

t +(∆p)2

m2t.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 =p2

2m+

1

2mω2x2.

(a) Derive the equation of motion for the expectation value 〈x〉t,and show that it oscillates, similarly to the classical oscillator,as

〈x〉t = 〈x〉0 cosωt +〈p〉0mω

sinωt.

(b) Derive a second-order differential equation of motion for theexpectation value 〈T − V 〉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

(∆x)2t ≡⟨

x2⟩

t− 〈x〉2t = (∆x)20 cos2 ωt +

(∆p)20m2ω2

sin2 ωt

+

[

1

2〈xp + px〉0 − 〈x〉0 〈p〉0

]

sin 2ωt

mω

Verify that this reduces to the result of Problem 2 in the limitω → 0.

(d) Work out the corresponding formula for the variance (∆p)2t .

(a) Again I like to use slightly different notation: e(t) = 〈x〉t. Then

d

dte(t) =

1

ih〈xH − Hx〉

=1

2ihm

⟨

xp2 − p2x⟩

=1

2ihm

⟨

xp2 − p(xp − ih)⟩

=1

2ihm

⟨

xp2 − (xp − ih)p + ihp⟩

=1

2ihm〈2ihp〉

=〈p〉m

.

d2

dt2e(t) =

d

dt

〈p〉m

5

=1

ihm〈pH − Hp〉

=ω2

2ih

⟨

px2 − x2p⟩

=ω2

2ih

⟨

(xp − ih)x − x2p⟩

=ω2

2ih

⟨

x(xp − ih) − ihx − x2p⟩

=ω2

2ih〈−2ihx〉

= −ω2 〈x〉 .

So we haved2

dt2e(t) = −ω2e(t)

with general solution e(t) = A cos ωt+B sin ωt. The coefficients are determinedby the boundary conditions:

e(0) = 〈x〉0 → A = 〈x〉0e′(0) =

〈p〉0m

→ B =〈p〉0mω

.

(b) Let’s define v(t) = 〈T − V 〉t. Then

d

dtv(t) =

1

ih〈(T − V )H − H(T − V )〉

=1

ih〈(T − V )(T + V ) − (T + V )(T − V )〉

=2

ih〈TV − V T 〉

=ω2

2ih

⟨

p2x2 − x2p2⟩

.

We already worked out this commutator in Problem 2:⟨

p2x2 − x2p2⟩

= −⟨

4ihxp + 2h2⟩

so

d

dtv(t) = −2ω2 〈xp〉 + ihω2.

= −2ω2 〈xp〉 + ω2 〈xp − px〉= −ω2 〈xp + px〉 (12)

Next,

d2

dt2v(t) = −2ω2

ih〈xpH − Hxp〉

= −2ω2

ih

[

1

2m

⟨

xp3 − p2xp⟩

+mω2

2

⟨

xpx2 − x3p⟩

]

(13)

6

The bracketed expressions are

⟨

xp3 − p2xp⟩

=⟨

xp3 − p(xp − ih)p⟩

=⟨

xp3 − (xp − ih)p2 + ihp2⟩

=⟨

2ihp2⟩

⟨

xpx2 − x3p⟩

=⟨

x(xp − ih)x − x3p⟩

=⟨

x2(xp − ih) − ihx2 − x3p⟩

=⟨

−2ihx2⟩

and plugging these back into (13) gives

d2

dt2v(t) = = −4ω2

[

⟨

p2⟩

m− mω2

2

⟨

x2⟩

]

= −4ω2v(t)

with solutionv(t) = A cos 2ωt + B sin 2ωt. (14)

Evaluating at t = 0 givesA = 〈T 〉0 − 〈V 〉0 .

Also, we can use (12) evaluated at t = 0 to determine B:

−ω2 〈xp + px〉0 + ihω2 = 2ωB

so

B = −ω 〈xp + px〉02

.

The next task is to compute⟨

x2⟩

t:

⟨

x2⟩

t=

2

mω2〈V 〉t

=1

mω2〈H − (T − V )〉t

=1

mω2[〈H〉t − v(t)] .

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

⟨

x2⟩

t=

1

mω2

(

〈T 〉0 + 〈V 〉0 − [〈T 〉0 − 〈V 〉0] cos 2ωt +ω 〈xp + px〉0

2sin 2ωt

)

=1

mω2

(

2 〈T 〉0 sin2 ωt + 2 〈V 〉0 cos2 ωt +ω 〈xp + px〉0

2sin 2ωt

)

=

⟨

p2⟩

0

m2ω2sin2 ωt +

⟨

x2⟩

0cos2 ωt +

1

2〈xp + px〉0

sin 2ωt

mω. (15)

7

(c) Earlier we found that

〈x〉t = 〈x〉0 cosωt +〈p〉0mω

sin ωt

〈x〉2t = 〈x〉20 cos2 ωt +〈p〉20

m2ω2sin2 ωt + 〈x〉0 〈p〉0 sin 2ωt.

Subtracting from (15) gives

(∆x)2t =⟨

x2⟩

− 〈x〉2 =[

⟨

x2⟩

0− 〈x〉20

]

cos2 ωt

+1

m2ω2

[

⟨

p2⟩

0− 〈p〉20

]

+

[

1

2〈xp + px〉0 − 〈x〉0 〈p〉0

]

sin 2ωt

= (∆x)20 cos2 ωt +(∆p)20m2ω2

sin2 ωt +

[

1

2〈xp + px〉0 − 〈x〉0 〈p〉0

]

sin 2ωt

mω.

As ω → 0, cos2 ωt → 1, (sin2 ωt/ω2) → 1, and (sin 2ωt/ω) → 2, as neededto ensure matchup with the result of Problem 2.

Problem 3.4

Prove that the probability density and the probability current den-sity at position r0 can be expressed in terms of the operators r andp as expectation values of the operators

ρ(r0) → δ(r − r0) j(r0) →1

2m[pδ(r − r0) + δ(r − r0)p] .

Derive expressions for these densities in the momentum represen-tation.

The first one is trivial:

〈δ(r − r0)〉 =

∫

Ψ∗(r)δ(r − r0)Ψ(r)dr = Ψ∗(r0)Ψ(r0) = ρ(r0).

For the second one,

1

2m〈pδ(r − r0) + δ(r − r0)p〉 = − ih

2m

∫

[Ψ∗∇δ(r − r0)Ψ + Ψ∗δ(r − r0)∇Ψ] dr

The gradient operator in the first term operates on everything to its right:

= − ih

2m

∫

[Ψ∗Ψ∇δ(r− r0) + 2δ(r− r0)Ψ∗∇Ψ] dr.

8

Here we can use the identity∫

f(x)δ′(x − a)dx = −f ′(a) :

= − ih

2m|−∇(Ψ∗Ψ) + 2Ψ∗∇Ψ|

r=r0

=ih

2m|Ψ∇Ψ∗ − Ψ∗∇Ψ|

r=r0

= j(r0).

Problem 3.5

For a system described by the wave function Ψ(r′), the Wigner

distribution function is defined as

W (r′,p′) =1

(2πh)3exp(−ip′·x′′/h)Ψ∗

(

r′ − r

′′

2

)

Ψ

(

r′ +

r′′

2

)

dr′′.

(a) Show that W (r′,p′) is a real-valued function, defined over thesix-dimensional “phase space” (r′,p′).

(b) Prove that∫

W (r′,p′)dp′ = |Ψ(r′)|2

and that the expectation value of a function of the operatorr in a normalized state is

〈f(r)〉 =

∫ ∫

f(r′)W (r′,p′)dr′dp′.

(c) Show that the Wigner distribution function is normalized as

∫

W (r′,p′)dr′dp′ = 1.

(d) Show that the probability density ρ(r0) at position r0 is ob-tained from the Wigner distribution function with

ρ(r0) → f(r) = δ(r − r0).

(a)

9



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 ofthe matrix elements of px?

For the harmonic oscillator, we have

H =1

2mp2

x +1

2mω2x2

sop2

x = 2mH − m2ω2x2

and

< Ψn|p2x|Ψk >= 2mhω(n +

1

2)δnk − m2ω2 < Ψn|x2|Ψk > . (1)

The nth eigenfunction is

Ψ(x) =

(

1

2nn!

)1/2(mω

hπ

)1/4

exp(−mω

2hx2)Hn(

√

mω

hx).

The matrix element of x2 is then

< Ψn|x2|Ψk >=

(

1

2n+kn!k!

)1/2(mω

hπ

)1/2∫

∞

−∞

x2 exp(mω

hx2)Hn(

√

mω

hx)Hk(

√

mω

hx) dx.

1

Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 2

The obvious substitution is u = (mω/h)x, with which we obtain

< Ψn|x2|Ψk >=

(

1

2n+kn!k!π

)1/2 (

h

mω

)∫

∞

−∞

u2e−u2

Hn(u)Hk(u) du. (2)

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

∑

n,k,p

Inkpsntk(2λ)p

n! k! p!=

√π eλ2+2λ(s+t)+2st.

=√

π

(

1 + λ2 +1

2λ4 + · · ·

) (

1 + 2λ(s + t) +1

2(2λ)2(s + t)2 + · · ·

) (

1 + (2st) +1

2(2st)2 + · · ·

)

(3)There are two ways to get a λ2 term out of this. One way is to take the λ2 termfrom the first series and the 1 from the second series, together with any termfrom the last series. The second way is to take the 1 from the first series andthe λ2 term from the second series, along with any term from the last series.Writing down only terms obtainable in this way, we have

= · · · +√

πλ2(

1 + 2(s + t)2)

(

1 + (2st) +1

2(2st)2 + · · ·

)

+ · · ·

= · · · +√

πλ2(

1 + 2s2 + 2t2 + 4st)

∞∑

j=0

1

j!(2st)j + · · ·

= · · · +√

πλ2∞∑

j=0

{

2j

j!sjtj +

2j+1

j!sj+2tj +

2j+1

j!sjtj+2 +

2j+2

j!sj+1tj+1

}

+ · · ·

Comparing termwise with (3), we can read off

Ink2 =

(n + 2)!2n√

π , n = k − 2n!2n−1√π(1 + 2n) , n = k0 , otherwise.

Plugging this into (2), we have

< Ψn+2|x2|Ψn > =1

2[(n + 2)(n + 1)]

1/2

(

h

mω

)

< Ψn|x2|Ψn > =1

2(2n + 1)

(

h

mω

)

.

Finally, from (1),

< Ψn+2|p2|Ψn > = −1

2[(n + 2)(n + 1)]

1/2(mωh)

< Ψn|x2|Ψn > =1

2(2n + 1)(mωh).


I find it kind of confusing that the matrix element for p2 comes out negativein the first case. It would be absurd for the expectation value (i.e., diagonalmatrix 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 off-diagonal matrix elements of an operator, but it’s still weird.

However, in another sense it seems inescapable that p2 should have a negativeoff-diagonal matrix element here, because the off-diagonal matrix elements of Hmust vanish in the energy eigenfunction basis, but x2 has a nonvanishing matrixelement, and H is just a sum of x2 and p2 terms, so p2 must have a negativematrix 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 station-ary state of the harmonic oscillator. Compare with the results of the virial theorem.

The potential energy operator is U = mω2x2/2. We found the expectationvalues of x2 in the last problem, so

〈U〉 =1

2mω2

⟨

x2⟩

=hω

2(n +

1

2)

which is just half the energy expectation value. The kinetic energy expectationvalue must of course make up the difference, so we have 〈T 〉 = 〈U〉.

On the other hand, the virial theorem is supposed to be saying

2 〈T 〉 =

⟨

xd

dxV (x)

⟩

.

In this case,d

dxV (x) = mω2x,

so the virial theorem says that

〈T 〉 =

⟨

1

2mω2x2

⟩

= 〈U〉

in accord with what we concluded earlier.

Problem 5.3

Calculate the expectation value of x4 for the nth energy eigenstate of the harmonicoscillator.

⟨

Ψn

∣

∣x4∣

∣Ψn

⟩

=1

n! 2n

(mω

hπ

)1/2∫

∞

−∞

x4 exp(−mω

hx2)H2

n(

√

mω

hx) dx


=1

n! 2n√

π

(

h

mω

)2 ∫

∞

−∞

u4e−u2

H2n(u) du (4)

For the integral we want to use (3) again, but this time we’ll need to writeout the expansion a little further than before.

∑

n,k,p

Inkpsntk(2λ)p

n! k! p!=

√π eλ2+2λ(s+t)+2st. (5)

=√

π

(

1 + λ2 +1

2λ4 + · · ·

) (

1 + · · · + 1

2(2λ)2(s + t)2 + · · · + 1

4!(2λ)4(s + t)4 + · · ·

)

∞∑

j=0

(2st)j

j!

=√

π

(

1 + λ2 +1

2λ4 + · · ·

)

(

1 + · · · + 4λ2st + · · · + 4λ4(st)2 + · · ·)

∞∑

j=0

(2st)j

j!

= · · · +√

π

∞∑

j=0

(λ4)(st)j

{

2j−1

j!+ 4

2j−1

(j − 1)!+ 4

2j−2

(j − 2)!

}

+ · · ·

= · · · +√

π

∞∑

j=0

(λ4)(st)j 2j

j!

{

1

2+ 2j + j(j − 1)

}

= · · · +√

π

∞∑

j=0

(λ4)(st)j 2j

j!

{

1

2+ j + j2

}

In the first line, I only wrote out terms that can be combined to give a factor ofλ4. In the second line, I further limited it to terms that also contain the samenumber of powers of s as t. Equating powers in (5),

Inn4 =3

22n n!

√π(

1

2+ n + n2),

so (4) is

⟨

Ψn

∣

∣x4∣

∣ Ψn

⟩

=3

2

(

h

mω

)2

(1

2+ n + n2).

Problem 5.4

For the energy eigenstates with n=0, 1, and 2, compute the probability that thecoordinate of a linear harmonic oscillator in its ground state has a value greaterthan the amplitude of the classical oscillator of the same energy.

The classical amplitude is A =√

(2E)/(mω2). The probability of findingthe particle with coordinate greater than this is

P (|x| > A) =

∫

−A

−∞

Ψ2n(x) dx +

∫

∞

A

Ψ2n(x) dx


= 2

∫

∞

A

Ψ2n(x) dx

=2

n! 2n

(mω

hπ

)1/2∫

∞

A

exp(−mω

hx2)H2

n(

√

mω

hx) dx

=2

n! 2n√

π

∫

∞

√2E/hω

e−u2

H2n(u) du

=1

n! 2n−1√

π

∫

∞

√

2n+1

e−u2

H2n(u) du

In going from the first line to the second we invoked the fact that Ψn has eithereven or odd parity, so Ψ2

n has even parity. In going from the second to last lineto the last line, we noted that the energy of the nth eigenstate is hω(n + 1/2).

In particular,

Pn=0(|x| > A) =1√π

∫

∞

1

e−u2

du

=1√2π

{

∫

∞

0

e−p2/2dp −∫

√

2

0

e−p2/2dp

}

=1√2π

{√

π

2−

√2π · erf(

√2)

}

=1

2− erf(

√2) ≈ 0.31

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 findinga particle with displacement x is a Gaussian distribution. Plot the width of thedistribution as a function of temperature. Check the results in the classical andlow-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 oscillatorsin the n′th state to the number of oscillators in the nth state is

Nn′

Nn= e−(n′

−n)hω/kT .

In particular, for any n,Nn = N0e

−nhω/kT .


The probability of finding a particle between x and dx is

P (x)dx =

∞∑

n=0

Cn|Ψn(x)|2dx

= C0

(mω

hπ

)1/2

exp(−mω

hx2)

∞∑

n=0

e−nhω/kT

n! 2nH2

n(

√

mω

hx)dx

This can be summed using the Mehler formula with t = exp(−hω/kT ) :

P (x) = C0

(mω

hπ

)1/2

exp(−mω

hx2)

(

1√1 − t2

)

exp

[(

2t

1 + t

)

mω

hx2

]

= C0

(mω

hπ

)1/2(

1√1 − t2

)

exp

[

−(

1 − t

1 + t

)

mω

hx2

]

This is a Gaussian distribution with variance

σ2 =h

2mω

(

1 + t

1 − t

)

=h

2mω

(

1 + e−hω/kT

1 − e−hω/kT

)

=h

2mωcoth

(

hω

2kT

)



Homer Reid

June 24, 2000

Chapter 6

Problem 6.1

Obtain the transmission coefficient for a rectangular potential barrier of width 2aif the energy exceeds the height V0 of the barrier. Plot the transmission coefficientas a function E/V0 (up to E/V0 = 3), choosing (2ma2V0)

1/2 = (3π/2)~.

In the text, Merzbacher treats this problem for the case where the particle’senergy is less than the potential barrier. He obtains the result

M11 =

(

cosh 2κa +iε

2sinh 2κa

)

e2ika (1)

where

κ =

√

2m(V0 − E)

~2

and

ε =κ

k− k

κ. (2)

We can re-use the result (1) for the case where the energy is greater thanthe potential barrier. To do this we note that κ becomes imaginary in this case,and we write

κ = iβ = i

√

2m(E − V0)

~2

so that (2) becomes

ε =iβ

k− k

iβ= i

(

β

k+

k

β

)

≡ iλ

1


Plugging into (1) and noting that cosh ix = cosx, sinh ix = i sinx we have

M11 =

(

cos 2βa − iλ

2sin 2βa

)

e2ika.

and

|M11| =

(

cos2 2βa +λ2

4sin2 βa

)1/2

=

[

1 +

(

λ2

4− 1

)

sin2 βa

]1/2

=

[

1 +

(

β4 + k4

4β2k2− 1

2

)

sin2 βa

]1/2

=

[

1 +

(

(β2 − k2)2

4β2k2

)

sin2 βa

]1/2

=

[

1 +

(

V 20

4E(E − V0)

)

sin2 βa

]1/2

=

[

1 +

(

1

4γ(γ − 1)

)

sin2 βa

]1/2

=

[

4γ(γ − 1) + sin2 βa

4γ(γ − 1)

]1/2

where γ = E/V0.We have

βa =

[

2m

h2(E − V0)

]1/2

a

=

(

2mV0a2

h2

)1/2

(γ − 1)1/2

=3π

2(γ − 1)1/2

so the transmission coefficient is

T =1

|M11|2=

[

4γ(γ − 1)

4γ(γ − 1) + sin2[

3π2 (γ − 1)1/2

]

]

This is plotted in Figure 1.


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 1 1.5 2 2.5 3 3.5 4

Tra

nsm

issi

on C

oeffi

cien

t

E/V0

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


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 Schrodingerequation 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 andbroad.

We have

Ψ(x) =

0, x ≤ 0

Aeik1x + Be−ik1x, 0 ≤x ≤ a

Ceik2x + De−ik2x, a ≤ x

with

k1 =

√

2m(E + V0)

~2k2 =

√

2mE

~2.

Applying the requirement that Ψ be continuous at x = 0, we see we must takeA = −B, so Ψ(x) = γ sin k1x for 0 ≤ x ≤ a. The other standard requirement,that the derivative of Ψ also be continuous, does not hold at x = 0 because thepotential is infinite there. Hence γ is undetermined as yet. Eventually, we couldapply 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 k1a = Ceik2a + De−ik2a

k1γ cos k1a = ik2[Ceik2a − De−ik2a]

Combining these yields

C =1

2iγe−ik2a

[

k1

k2cos k1a + i sin k1a

]

(3)

D = − 1

2iγe+ik2a

[

k1

ik2cos k1a − i sin k1a

]

(4)

It’s now convenient to write

k1

k2cos k1a + i sin k1a = αeiδ (5)

where

α2 =

(

k1

k2

)2

cos2 k1a + sin2 k1a (6)


and

δ = tan−1

[

k2

k1tan k1a

]

so that α contains magnitude information, while δ represents phase information.Then we can rewrite (3) and (4) as

C =γα

2ie−ik2aeiδ

D = −γα

2ie+ik2ae−iδ

Then the expression for the wavefunction to the left of x = a becomes

Ψ(x) = Ceik2x + De−ik2x (x > a)

=γα

2i

[

eik2(x−a)eiδ − e−ik2(x−a)e−iδ]

=γα

2ie−iδ

[

eik2(x−a)e2iδ − e−ik2(x−a)]

.

Using (5) and (6), the scattering coefficient is

|1 − e2iδ |2 =

∣

∣

∣

∣

∣

∣

∣

1 −

(

k1

k2

)2

cos2 k1a + 2ik1

k2

cos k1a sin k1a − sin2 k1a(

k1

k2

)2

cos2 k1a + sin2 k1a

∣

∣

∣

∣

∣

∣

∣

2

=

∣

∣

∣

∣

∣

∣

∣

2 sin k1a[

sin k1a − ik1

k2

cos k1a]

(

k1

k2

)2

cos2 k1a + sin2 k1a

∣

∣

∣

∣

∣

∣

∣

2

=4 sin2 k1a

(

k1

k2

)2

cos2 k1a + sin2 k1a(7)

We have(

k1

k2

)2

=E + V0

E=

(

1 +1

λ

)

k1a =

√

2ma2(E + V0)

~2= β(λ + 1)1/2

where λ = E/V0 and β = (2ma2V0/~2)1/2 in Merzbacher’s notation. Then the

scattering coefficient (7) is

scattering coefficient =4 sin2[β(λ + 1)1/2]

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

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


0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5 3

Sca

tterin

g co

effic

ient

E/V0

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


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 ofthe system. Compute and plot the energy levels in units of ~

2/ma2 as a function ofthe dimensionless parameter mag/~

2. Explain the features of the plot. In the limitof 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 (oddparity) 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 Schrodinger equation,but with energy E < 0 since we’re looking for bound states. Putting k =√

2mE/~2, we have

Ψ(x) =

Aekx + Be−kx, x ≤ −aCekx + De−kx, −a ≤ x ≤ aEekx + Fe−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, thewavefunction will have definite parity. Considering first the even parity solution,

Ψ(x) =

Aekx, x ≤ −aB cosh(kx), −a ≤ x ≤ aAe−kx, a ≤ x.

(8)

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-continuitycondition doesn’t hold there. Instead, we can write down the Schrodinger equa-tion,

d2

dx2Ψ(x) =

2m

~2V (x)Ψ(x) − 2m

~2EΨ(x),

then integrate from −a− ε to −a + ε and take the limit as ε → 0. This gives

dΨ

dx

∣

∣

∣

∣

−a+

−a−

= −2mg

~2Ψ(−a). (10)

Applying this condition to the wavefunction (8) yields

kB sinh(−ka) − kAe−ka = −2mg

~2B cosh(−ka).


Substituting from (9),

kB sinh(−ka) − kB cosh(−ka) = −2mg

~2B cosh(−ka)

or

tanh(ka) =2mg

~2k− 1 =

β

ka− 1

with β = 2mag/~2. This equation determines the energy eigenvalue of the even-

parity state, which will be the ground state. On the other hand, the odd paritystate looks like

Ψ(x) =

Aekx, x ≤ −aB sinh(kx), −a ≤ x ≤ a−Ae−kx, a ≤ x.

(11)

Matching values at x = −a gives

Ae−ka = B sinh(−ka)

and applying condition (10) gives

kB cosh(−ka) − kAe−ka = −2mg

~2B sinh(−ka)

kB cosh(ka) + kB sinh(ka) =2mg

~2B sinh(ka)

coth(ka) =β

ka− 1

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 valueof ka than the tanh curve; that means that the energy eigenvalue for the oddparity 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 moleculewith interatomic distance 2a = |X |, if the potential energy of the “molecule” istaken 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 totalpotential energy of the system as a function of |X |.


0

0.5

1

1.5

2

1 1.2 1.4 1.6 1.8 2P

Sfra

grep

lacem

ents

tanh(ka)

coth(ka)

β/(ka) − 1

ka

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


Problem 6.5

If the potential in Problem 3 has g < 0 (double barrier), calculate the transmissioncoefficient and show that it exhibits resonances. (Note the analogy between thesystem and the Fabry-Perot etalon in optics.)

Now we’re assuming that the energy E is positive, so

Ψ(x) =

Aeikx + Be−ikx x ≤ −a

Ceikx + De−ikx − a ≤ x ≤ a

Eeikx + Fe−ikx a ≤ x

(12)

with k =√

2mE/~2. Matching values at x = −a, we have

Ae−ika + Beika = C−ika + Dika (13)

Also, as before, we have the derivative condition

dΨ

dx

∣

∣

∣

∣

x=−a+

x=a−

= −2mg

~2Ψ(−a)

where g is now negative. Applying this to the wavefunction in (12), we have

ik[Ce−ika − Deika − Ae−ika + Beika] = −2mg

~2[Ae−ika + Beika]. (14)

Combining (13) and (14) yields

C =

(

1 +β

ika

)

A +β

ikae2kaB (15)

D = − β

ikae−2kaA +

(

1 − β

ika

)

B (16)

with β = mag/~2 as before.

Now, applying the matching conditions to the wavefunction at x = +a willgive 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

E =

(

1 − β

ika

)

C − β

ikae2kaD (17)

F = +β

ikae2kaC +

(

1 +β

ika

)

D (18)


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 5 10 15 20 25 30

PSfra

grep

lacem

ents

ka

T(k

)

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

Combining equations (15) through (18), we have

E =

[

1 +

(

β

ika

)2

(e−4ika − 1)

]

A + +

[

2β

ka

(

1 − β

ika

)

sin 2ka

]

B

F =

[

2β

ka

(

1 +β

ika

)

sin 2ka

]

A +

[

1 +

(

β

ika

)2

(e4ika − 1)

]

B

This is the M matrix, and the transmission coefficient is given by T = 1/|M11|2,or

T =1

1 + 2(

βka

)2[

(

βka

)2

+ 1

]

(1 − cos(4ka))

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


Problem 6.6

A particle moves in one dimension with energy E in the field of a potential definedas the sum of a Heaviside step function and a delta function:

V (x) = V0η(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 andreflected plane waves on the left of the origin (x < 0) to the amplitudes onthe right (x > 0).

(b) Derive the elements of the matrix S, which relates incoming and outgoingamplitudes.

(c) Show that the S matrix is unitary and that the elements of the S matrix satisfythe properties expected from the applicable symmetry considerations.

(d) Calculate the transmission coefficient for particles incident from the right andfor particles incident from the left, which have the same energy (buf differentvelocities).

We have

Ψ(x) =

{

Aeik1x + Be−ik1x, x ≤ 0

Ceik2x + De−ik2x, x ≥ 0

with

k1 =

√

2m

~2E k2 =

√

2m

~2(E − V0).

Matching values at x = 0 gives

C + D = A + B (19)

Also, the delta function at the origin gives rise to a discontinuity in the derivativeof the wavefunction as before:

dΨ

dx

∣

∣

∣

∣

0+

0−

=2mg

~2Ψ(0)

so

ik2(C − D) − ik1(A − B) =2mg

~2(A + B)

or

C − D =k1

k2(A − B) +

2mg

ik2~2(A + B). (20)


Adding and subtracting (19) and (20), we can read off

C =1

2

[

1 +k1

k2+

2mg

ik2~2

]

A +1

2

[

1 − k1

k2+

2mg

ik2~2

]

B

D =1

2

[

1 − k1

k2− 2mg

ik2~2

]

A1

2

[

1 +k1

k2− 2mg

ik2~2

]

B.

We could also write this as(

CD

)

=

(

M11 M12

M∗

12 M∗

11

) (

AB

)

Or instead of the M matrix we could use the S matrix, which is defined by

(

BC

)

=

(

S11 S12

S21 S22

) (

AD

)

Since we already know the M coefficients, we can calculate the elements of theS matrix from the formula

S11 = −M∗

12

M∗

11

S12 = 1M∗

11

S21 = 1M∗

11

S22 = +M12

M∗

11

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



Homer Reid

April 5, 2001

Chapter 7

Before starting on these problems I found it useful to review how the WKBapproximation works in the first place. The Schrodinger equation is

− ~2

2m

d2

dx2Ψ(x) + V (x)Ψ(x) = EΨ(x)

ord2

dx2Ψ(x) + k2(x)Ψ(x) = 0, k(x) ≡

√

2m

~2[E − V (x)].

We postulate for Ψ the functional form

Ψ(x) = AeiS(x)/~

in which case the Schrodinger equation becomes

i~S′′(x) = [S′(x)]2 − ~2k2(x). (1)

This equation can’t be solved directly, but we obtain guidance from the obser-vation that, for a constant potential, S(x) = ±kx, so that S ′′ vanishes. For anonconstant but slowly varying potential we might imagine S ′′(x) will be small,and we may take S′′ = 0 as the seed of a series of successive approximationsto the exact solution. To be specific, we will construct a series of functionsS0(x), S1(x), · · · , where S0 is the solution of (1) with 0 on the left hand side;S1 is a solution with S′′

0 on the left hand side; and so on. In other words, atthe nth step in the approximation sequence (by which point we have computedSn(x)), we compute S′′

n(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) anduse this as the source term for calculating Sn+2, and so on ad infinitum. In

1


symbols,

0 = [S′

0(x)]2 − ~2k2(x) (2)

i~S′′

0 = [S′

1(x)]2 − ~2k2(x) (3)

i~S′′

1 = [S′

2(x)]2 − ~2k2(x) (4)

· · · ·

Equation (2) is clearly solved by taking

S′

0(x) = ±~k(x) ⇒ S0(x) = S00 ± ~

∫ x

−∞

k(x′)dx′ (5)

for any constant S00. Then S′′

0 (x) = ±~k′(x), so (3) is

S′

1(x) = ±~

√

k2(x) ± ik′(x).

With the two ± signs here, we appear to have four possible choices for S ′

1. Butlet’s think a little about the ± signs in this equation. The ± sign under theradical comes from the two choices of sign in (5). But if we chose, say, the plussign in that equation, so that S ′

0 > 0, we would also expect that S ′

1 > 0. Indeed,if we choose the plus sign in (5) but the minus sign in (3), then S ′

0 and S′

1 haveopposite sign, so S′

1 differs from S′

0 by an amount at least as large as S ′

0, inwhich 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 twochoices are

S′

1 = +~

√

k2(x) + ik′(x) or S′

1 = −~

√

k2(x) − ik′(x). (6)

If V (x) is constant, k(x) is constant, and, as we observed before, the sequenceof approximations terminates at 0th order with S0 being an exact solution. Byextension, if V (x) is not constant but changes little over one particle wavelength,we have k′(x)/k2(x) � 1, so we may expand the radicals in (6):

S′

1 ≈ ~k(x)

{

1 +ik′(x)

2k2(x)

}

or S′

1 ≈ −~k(x)

{

1 − ik′(x)

2k2(x)

}

or

S′

1 ≈ ±~k(x) +i~k′(x)

2k(x). (7)

Integrating,

S1(x) = S1(a) ± ~

∫ x

a

k(u)du +i~

2

∫ x

a

k′(u)

k(u)dx′

= S1(a) ± ~

∫ x

a

k(u)du +i~

2ln

k(x)

k(a)

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


The wavefunction at this order of approximation is

Ψ(x) = exp(iS1(x)/~) =(

eiS1(a)/~

)(

e±iR

xa

k(u)du)(

eln k(x)/k(a))

−1/2

= Ψ(a)

√

k(a)

k(x)e±i

R

xa

k(u)du

= Ψ(a)G±(x; a)

where

(8)

G±(x; a) ≡√

k(a)

k(x)e±i

R xa

k(u)du. (9)

We have written it this way to illustrate that the function G(x, a) is kind of likea Green’s function or propagator for the wavefunction, in the sense that, if youknow what Ψ is at some point a, you can just multiply it by G±(x; a) to findout what Ψ is at x. But this doesn’t seem quite right: Schrodinger’s equationis a second-order differential equation, but (8) seems to be saying that we needonly one initial condition—the value of Ψ at x = a—to find the value of Ψ atother points. To clarify this subtle point, let’s investigate the equations leadingup to (8). If the approximation (7) makes sense, then there are two solutionsof Schrodinger’s equation at x = a, one whose phase increases with increasingx (positive derivative), and one whose phase decreases. Equation (8) seems tobe saying that we can use either G+ or G− to get to Ψ(x) from Ψ(a); but therequirement the dΨ/dx be continuous at x = a means that only one or theother will do. Indeed, in using (8) to continue Ψ from a to x we must choosethe appropriate propagator—either G+ or G−, according to the derivative ofΨ at x = a; otherwise the overall wave function will have a discontinuity in itsfirst derivative at x = a. So to use (8) to obtain values for Ψ at a point x, weneed to know both Ψ and Ψ′ at a nearby point x = a, as should be the case fora second-order differential equation.

If we want to de-emphasize this nature of the solution with the propagatorwe may write

Ψ(x) = C

√

1

k(x)e±i

R xa

k(u)du (10)

where C = Ψ(a)√

k(a). In regions where V (x) > E, k(x) is imaginary, so it’suseful to define

κ(x) = −ik(x) =

√

2m

~2[V (x) − E] (11)

and

Ψ(x) = C

√

1

κ(x)e±

R xa

κ(u)du. (12)


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 isequivalent 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 WKBmethod, however, gives us no way of determining the value of Ψ at that onestarting point. Furthermore, even if we know Ψ at one point within a region ofvalidity, we can’t use (8) to determine Ψ in other, nonadjacent regions, becausewe can’t carry the propagator across regions of invalidity. So basically what weneed is a way of finding one starting value for Ψ(x) in every region of validityof the WKB approximation.

How do we find such points? Well, one sure-fire way to get starting pointsin regions of validity is to identify regions of invalidity, of which there will beat least one adjacent to each region of validity, and then get values of Ψ at theboundaries of the regions of invalidity—which will also count as values in theregions of validity. So we need to identify the regions of invalidity of the WKBapproximation and do a more accurate solution of the Schrodinger equationthere.

The WKB approximation breaks down when k′/k2 � 1 ceases to hold, whichis true when k ≈ 0 but k′ 6= 0, which happens near a classical turning point ofthe motion—i.e., a point x0 at which V (x0) = E. But near such a point we mayexpand V (x) − E in a Taylor series around the point x0; if we keep only thefirst (linear in x) term in the series, we arrive at a Schrodinger equation whichwe can solve exactly in the vicinity of x0. To do this, suppose the point x0 is aclassical turning point of the motion, so that V (x0) = E. In the neighborhoodof x0 we may expand V (x):

V (x) = E + (x − x0)V′(x0) + · · · (13)

Then the Schrodinger equation becomes

d2

dx2Ψ(x) − 2m

~2V ′(x0)(x − x0)Ψ(x) = 0. (14)

The useful substitution here is

u(x) = γ(x − x0) γ ≡[

2m

~2V ′(x0)

]1/3

sox(u) =

u

γ+ x0.

If we defineΦ(u) = Ψ(x(u))

then

dΦ

du=

dΨ

dx

dx

du=

1

γΨ′(x(u))

d2Φ

du2=

1

γ2Ψ′′(x(u))


so (14) becomes

γ2 d2

du2Φ(u) − γ3(x − x0)Φ(u) = 0

ord2

du2Φ(u) − uΦ(u) = 0.

The solution to this differential equation is

Φ(u) = β1Ai(u) + β2Bi(u) (15)

so the solution to the Schrodinger equation (14) is

Ψ(x) = β1Ai(

γ(x − x0))

+ β2Bi(

γ(x − x0))

.

For γ(x − x0) � 1 we have the asymptotic expression

Ψ(x) ≈ π−1/2 [γ(x − x0)]−1/4

(β1

2e−

2

3|γ(x−x0)|

3/2

+ β2e+ 2

3|γ(x−x0)|

3/2)

(16)

and for γ(x − x0) � −1 we have

Ψ(x) ≈ π−1/2 |γ(x − x0)|−1/4[

β1 cos(2

3|γ(x − x0)|3/2 − π

4

)

− β2 sin(2

3|γ(x − x0)|3/2 − π

4

)]

. (17)

To simplify these, we need to consider two possible kinds of turning point.

Case 1: V ′(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 regionto the left of the turning point is the classically accessible region, while the rightof the turning point is classically forbidden. Since V ′(x0) > 0, γ > 0, so forx < x0 (??) holds. For points close to the turning point on the left side,

k(x) =

√

2m

~2[E − V (x)] ≈

[

2m

~2V ′

]1/2

(x0 − x)1/2 = γ3/2(x0 − x)1/2

so

|γ(x − x0)|−1/4=

√

γ

k(x)(18)

and∫ x0

x

k(u)du = γ3/2

∫ x0

x

(x0 − x)1/2du =2

3γ3/2(x0 − x)3/2

=2

3|γ(x − x0)|3/2 . (19)

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


κ(x) =

√

2m

~2

[

V (x) − E]

≈[

2m

~2V ′(x0)(x − x0)

]1/2

= γ3/2(x − x0)1/2 (20)

so, for x near x0,

∫ x

x0

κ(u)du = γ3/2

∫ x

x0

(u − x0)1/2 du =

2

3γ3/2(x − x0)

3/2 (21)

and also

|γ(x − x0)|−1/4=

√

γ

κ(x). (22)

Using (18) and (19) in (17), and (21) and (22) in (16), the solutions to theSchrodinger equation on either side of a classical turning point x0 at whichV ′(x0) > 0 are

Ψ(x) =

√

1

k(x)

[

2β1 cos(

∫ x0

x

k(u)du − π

4

)

− β2 sin(

∫ x0

x

k(u)du − π

4

)]

, x < x0 (23)

Ψ(x) =

√

1

κ(x)

[

β1eR

xx0

κ(u)du+ β2e

−

R

xx0

κ(u)du]

, x < x0 (24)

(we redefined the β constants slightly in going to this equation).

Case 2: V ′(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 forbiddenregion to the left. Since V ′(x0) < 0, γ < 0. That means that the regions ofapplicability of (16) and (17) are on opposite sides of the turning points as theywere in the previous case. The solutions to the Schrodinger equation on eitherside of the turning point are

Ψ(x) =

√

1

κ(x)

[

β1eR

x0

xκ(u)du + β2e

−

R

xx0

κ(u)du]

, x < x0 (25)

Ψ(x) =

√

1

k(x)

[

2β1 cos(

∫ x

x0

k(u)du − π

4

)

− β2 sin(

∫ x

x0

k(u)du − π

4

)]

, x > x0 (26)

(27)

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


up into regions bounded by turning points, within which regions the WKBapproximation (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 pointswithin 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 forpoints x0 ± ε at which to make the “handoff” from approximations (16) and(17) to the WKB approximation These points must satisfy several conditions.First, the approximate Schrodinger equation (14) is only valid as long as we canneglect the quadratic and higher-order terms in the expansion (13), so we musthave

ε |V ′′(x0)| � |V ′(x0)| ⇒ ε �∣

∣

∣

∣

V ′(x0)

V ′′(x0)

∣

∣

∣

∣

. (28)

But at the same time, γε must be sufficiently greater than 1 to justify theapproximation (16) (or sufficiently less than -1 to justify (17)); the conditionhere is

∣

∣

∣

∣

2m

~2V ′(x0)

∣

∣

∣

∣

1/3

ε � 1 ⇒ ε �∣

∣

∣

∣

2m

~2V ′(x0)

∣

∣

∣

∣

−1/3

. (29)

Finally, the points x±ε must be sufficiently far away from the turning points thatthe approximation (7) is valid for the derivative of the phase of the wavefunction;the condition for this to be the case was

∣

∣

∣

∣

k′(x)

k2(x)

∣

∣

∣

∣

� 1 ⇒ 1

2

(

~2

2m

)

∣

∣

∣

∣

∣

V ′(x ± ε)

[E − V (x ± ε)]3/2

∣

∣

∣

∣

∣

� 1. (30)

If there are no points x0 ± ε satisfying all three conditions, the WKB approxi-mation 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 pointsat x = a and x = b. Although there are no discontinuities in the potentialhere, the problem may be analyzed in a manner similar to that used in theconsideration of one-dimensional piecewise constant potentials, as in Chapter6: we divide space into a number of distinct regions, obtain solutions of theSchrodinger equation in each region, and then match values and derivatives atthe region boundaries.

To divide space into distinct regions in this case, we begin by identifyingnarrow regions around the turning points a and b in which the linear approx-imation (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 regionaround a be a− ε1 < x < a+ ε1, and that around b be b− ε2 < x < b+ ε2. Thenspace divides naturally into five regions: (a) x < a − ε

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


PSfra

grep

lacem

ents

a b

E

V (x)

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


must throw out the term that grows exponentially as x → −∞, so we are leftwith

Ψ(x) = A

√

1

κ(x)e−

R

(a−ε)x

κ(u)du, x < a − ε. (31)

(b) a − ε < x < a + εIn this region we are close enough to the turning point that (13) is valid, so

(25) and (26) may be used.

Ψ(x) =

√

1

κ(x)

(

β1e−

R

ax

κ(u)du + β2e+

R

ax

κ(u)du)

, x < (a − ε). (32)

From (31) and (32) we see that continuity of both the value and first derivative ofΨ(x) at x = a−ε 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 ofall higher derivatives, as must be the case since there is no discontinuity in thepotential.

But now that we know the value of Ψ at x = a − ε, we also know it atx = a + ε, because of course the solution of the Schrodinger equation in thenarrow 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− ε is validat x = a + ε. With β1 = A and β2 = 0, (26) becomes

Ψ(x) = 2A

√

1

k(x)cos(

∫ x

a

k(u)du − π

4

)

= A

√

1

k(x)

[

e+i(R x

ak(u)du−π/4) + e−i(

R xa

k(u)du−π/4)]

, x = (a + ε)−

(33)

(c) a + ε < x < b − εIn 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) forthe wavefunction at x = a+ ε, integrating from a+ ε to x in the propagator (9),and using G+ and G−, respectively, to propagate the first and second terms in


(33), we obtain for the wavefunction at a point x in this region

Ψ(x) = A

√

1

k(x)

[

e+i

“

R

(a+ε)a

k(u)du+R x(a+ε)

k(u)du−π/4”

+ e−i

“

R

(a+ε)a

k(u)du+R x(a+ε)

−π/4”]

= A

√

1

k(x)

[

e+i(R x

ak(u)du−π/4) + e−i(

R xa

k(u)du−π/4)]

= 2A

√

1

k(x)cos

(∫ x

a

k(u)du − π

4

)

= 2A

√

1

k(x)cos

(

∫ b

a

k(u)du −∫ b

x

k(u)du − π

4

)

, a + ε < x < b − ε

(34)

Okay, I have now carried this analysis far enough to see for myself exactlywhere the Bohr-Sommerfeld quantization condition

∫ b

a

k(u)du =

(

n +1

2

)

π, n = 1, 2, · · · (35)

comes from, which was my original goal, so I am now going to stop this exerciseand proceed directly to the problems.

Problem 7.1

Apply the WKB method to a particle that falls with acceleration g in a uniformgravitational field directed along the z axis and that is reflected from a perfectlyelastic plane surface at z = 0. Compare with the rigorous solutions of this problem.

We’ll start with the exact solution to the problem. The requirement ofperfect elastic reflection at z = 0 may be imposed by taking V (x) to jumpsuddenly to infinity at z = 0, i.e.

V (x) =

{

mgz, z > 0

∞, z ≤ 0.

For z > 0, the Schrodinger equation is

0 =d2

dx2Ψ(x) +

2m

~2[E − mgz]Ψ(x)

=d2

dx2Ψ(x) +

2m2g

~2

[

E

mg− z

]

Ψ(x)

=d2

dx2Ψ(x) − 2m2g

~2[z − z0] Ψ(x) (36)


where z0 = E/mg. With the substitution

u = γ(z − z0) γ =

(

2m2g

~2

)1/3

and taking Φ(u) = Ψ(x(u)), we find that (36) is just the Airy equation for Φ(u),

d2

du2Φ(u) − uΦ(u) = 0

with solutionsΦ(u) = β1Ai(u) + β2Bi(u).

Since we require a solution that remains finite as z → ∞, we must take β2 = 0.The solution to (36) is then

Ψ(x) = β1Ai(

γ(z − z0))

, (z > 0). (37)

For z < 0, I wasn’t quite sure how to account for the infinite potentialjump 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 Schrodinger equationfor z < 0 is

d2

dz2Ψ(z) − 2m

~2[V0 − E]Ψ(z) = 0

with solution

Ψ(z) = Ae−kz, k =

√

2m

~2[V0 − E]. (38)

Matching values and derivatives of (37) and (38) at z = 0, we have

β1Ai(−γz0) = A

γβ1Ai′(−γz0) = −kA

Dividing, we obtain1

γ

Ai(−γz0)

Ai′(−γz0)= −1

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 theenergy eigenvalues En are given by

(

2m2g

~2

)1/3En

mg= xnm ⇒ En =

(

mg2~

2

2

)1/3

xn (39)

where xn is the nth root of the equation

Ai(−xn) = 0.


So that’s the exact solution. In the WKB approximation, the spectrum ofenergy eigenvalues is determined by the condition (35). In this case the classicalturning points are at z = 0 and z = z0, so we have

(

n +1

2

)

π =

∫ z0

0

k(z) dz

=

√

2m

~2

∫ z0

0

[E − mgz]1/2 dz

=

√

2m2g

~2

∫ z0

0

[z0 − z]1/2 du

=

√

2m2g

~2

∣

∣

∣

∣

−2

3(z0 − z)3/2

∣

∣

∣

∣

z0

0

=2

3

√

2m2g

~2

(

E

mg

)3/2

so the nth eigenvalue is given by

En =

(

mg2~

2

2

)1/3 [3

2

(

n +1

2

)

π

]2/3

.



Homer Reid

May 13, 2001

Chapter 8

1


Problem 8.1

Apply the variational method to estimate the ground state energy of a particleconfined 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 andis symmetric with respect to the center of the well:

Ψt(x) =

{

(a − |x|), b ≤ x ≤ a

(a − b), |x| ≤ b.

(b) A more sophisticated trial function is parabolic, again vanishing at the endpoints 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 variation-ally.

(d) Compare the results of the different variational calculations with the exactground state energy, and, using normalized wave functions, evaluate the meansquare deviation

∫ a

−a|Ψ(x) − Ψt(x)|2dx for the various cases.

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

First let’s observe that the exact expressions for the ground state wavefunctionand energy are

Ψn(x) =1√a

cos(knx), kn =nπ

2a, En = n2 ~

2π2

8ma2≈ 1.23

~2

ma2.

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

Ψt(x) =

{

γ(a − |x|), b ≤ |x| ≤ a

γ(a − b), x| ≤ b.


we have∫ a

−a

Ψ2t (x)dx = 2

∫ a

0

Ψ2t (x)dx

= 2γ2

{

(a − b)2∫ b

0

dx +

∫ a

b

(a − x)2 dx

}

= 2γ2

{

b(a − b)2 +1

3(a − b)3

}

= 2γ2

{

b(a2 + b2 − 2ab) +1

3(a3 − b3) − a2b + b2a

}

= γ2 2

3

(

a3 + 2b3 − 3ab2)

so Ψt is normalized by taking

γ2 =3

2

(

1

a3 + 2b3 − 3ab2

)

. (1)

Now we can compute the energy expectation value of Ψt:

< Ψt|H |Ψt > = − ~2

2m

∫ a

−a

Ψt(x)d2

dx2Ψt(x) dx

Integrating by parts,

= − ~2

2m

{

Ψt(x)Ψ′

t(x)∣

∣

∣

a

−a−

∫ a

−a

Ψ′2t (x) dx

}

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

= +~

2

m

∫ a

0

Ψ′2t (x) dx

=~

2

mγ2

∫ a

b

dx

=~

2

mγ2(b − a).

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 =1

(a3 + 2b3 − 3ab2)− 6b2(b − a)

(a3 + 2b3 − 3ab2)2+

6ab(b− a)

(a3 + 2b3 − 3ab2)2

= −4b3 + 9b2a − 6a2b + a3


(b) For a parabolic trial function we take

Ψt(x) = γ(a2 − x2).

The normalization integral is

∫ a

−a

Ψ2t (x) dx = 2γ2

∫ a

0

(a2 − x2)2dx

= 2γ2

∫ a

0

(a4 + x4 − 2a2x2)dx

= 2γ2

(

a5 +1

5a5 − 2

3a5

)

=16

15γ2a5

so Ψt(x) is normalized by taking

γ2 =15

16a5.

The expectation value of the energy is

< H > = − ~2

2m

∫ a

−a

Ψt(x)d2

dx2Ψt(x) dx

= 2~

2

mγ2

∫ a

0

(a2 − x2)dx

=4

3

~2

mγ2a3

=5

4

~2

ma2≈ 1.25

~2

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

2m

d2

dx2Ψt(x) = γ

~2

m[6αx2 − (αa2 − β)].

The expectation value of the energy is


Problem 8.2

Using scaled variables, as in Section 5.1, consider the anharmonic oscillator Hamil-tonian,

H =1

2p2

ξ +1

2ξ2 + λξ4

where λ is a real-valued parameter.

(a) Estimate the ground state energy by a variational calculation, using as a trialfunction the ground state wave function for the harmonic oscillator

H0(ω) =1

2p2

ξ +1

2ω2ξ2

where ω is an adjustable variational parameter. Derive an equation thatrelates ω and λ.

(b) Compute the variational estimate of the ground state energy of H for variouspositive values of the strength λ.

(c) Note that the method yields answers for a discrete energy eigenstate even if λis slightly negative. Draw the potential energy curve to judge if this resultmakes physical sense. Explain.

(a) To find the ground state eigenfunction of the Hamiltonian Merzbacher pro-poses, it’s convenient to rewrite it:

H0(ω) =1

2p2

ξ +1

2ω2ξ2

= −1

2

∂2

∂ξ2+

1

2ω2ξ2

Upon substituting u = ω1/2ξ we obtain

= ω

{

−1

2

∂2

∂u2+

1

2u2

}

and now this is just the ordinary harmonic oscillator Hamiltonian, scaled by aconstant factor ω, with ground-state eigenfunction

Ψ(ω) = Ce−u2/2 = Ce−ωξ2/2.

Adding the normalization constant,

Ψ(ω) =(ω

π

)1/4

e−ωξ2/2.


Now we want to treat ω as a parameter and vary it until the energy expectationvalue of Ψ(ω) is minimized. The energy expectation value is

〈Ψ|H |Ψ〉 = 〈Ψ|T |Ψ〉 + 〈Ψ|V |Ψ〉

where T = p2ξ/2 and V = ξ2/2 + λξ4. Let’s compute the two expectation values

separately. First of all, to compute the expectation value of T , we need to knowthe result of operating on Ψ(ω) with p2

ξ :

p2ξΨ(ω) = −

(ω

π

)1/4{

∂

∂ξ

[

∂

∂ξe−ωξ2/2

]}

= −(ω

π

)1/4{

∂

∂ξ

[

−ωξe−ωξ2/2]

}

= −(ω

π

)1/4[

−ω + ω2ξ2]

e−ωξ2/2

Then for the expectation value of T we have

〈Ψ|T |Ψ〉 =1

2

∫

∞

−∞

Ψ(ξ)p2ξΨ(ξ)dξ

= −1

2

√

ω

π

∫

∞

−∞

(

−ω + ω2ξ2)

e−ωξ2

dξ

= −1

2

√

ω

π

[

−ω

√

π

ω+

1

2ω2

√

π

ω3

]

=ω

4. (2)

On the other hand, for the expectation value of V we have

(3)

exptwoΨV Ψ =

√

ω

π

{

1

2

∫

∞

−∞

ξ2e−ωξ2

dξ + λ

∫

∞

−∞

ξ4e−ωξ2

dξ

}

=

√

ω

π

{

1

4

√

π

ω3+

3

4λ

√

π

ω5

}

=1

4ω+

3λ

4ω2. (4)

Adding (2) and (4),

exptwoΨHΨ =1

2

[

ω +1

ω+

3λ

ω2

]

. (5)

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

0 = 1 − 1

ω2− 6λ

ω3


or

ω3 − ω − 6λ = 0. (6)

We could then solve this equation for ω in terms of λ to obtain the energy-minimizing value of ω for a given perturbing potential strength λ. But writingdown the full solution would be tedious. Instead let’s see what happens whenλ is small.

Evidently, when λ = 0 the Hamiltonian in this problem degenerates to thenormal harmonic oscillator Hamiltonian, for which the energy is minimized bythe (unscaled) ground state harmonic oscillator wavefunction, i.e. Ψ(ω) withω = 1. We can thus imagine that, for small λ, the energy-minimizing value ofω will be close to 1, and we may write ω(λ) ≈ 1 + ε for some small ε. Insertingthis in (6),

(1 + 3ε + 3ε2 + ε3) − (1 + ε) = 6λ

Keeping only terms of zeroth or first order in the small quantity ε (which isequivalent to keeping terms of lowest order in the perturbing potential strengthλ) we obtain from this

ε ≈ 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 inλ we find

(7)

exptwoΨHΨ =1

4

[

(1 + ε) + (1 + ε)−1 + 3λ(1 + ε)−2]

≈ 1

4

[

(1 + 3λ) + (1 + 3λ)−1 + 3λ(1 + 3λ)−2]

≈ 1

4[(1 + 3λ) + (1 − 3λ) + 3λ(1 − 6λ)]

≈ 1

2+

3

4λ. (8)

Since the 1/2 term is the normal (unperturbed) energy of the state, the energyshift caused by the perturbing potential is

∆E =3

4λ. (9)


Problem 8.3

In first-order perturbation theory, calculate the change in the energy levels of alinear harmonic oscillator that is perturbed by a potential gx4. For small values ofthe 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 〈Ψn|x4 |Ψn〉 .

I worked out this expectation value in Problem 5.3:

∆E = gexptwoΨnx4Ψn =3g

2

(

~

mω

)2(1

2+ n + n2

)

In particular, the energy shift of the ground state is

∆E0 =3g

4

(

~

mω

)2

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

Problem 8.4

Using a Gaussian trial function, e−λx2

, with an adjustable parameter, make a vari-ational estimate of the ground state energy for a particle in a Gaussian potentialwell, represented by the Hamiltonian

H =p2

2m− V0e

−αx2

(V0 > 0, α > 0).

For notational simplicity, I like to put β/2 = λ. Then

Ψ(x) = Ce−βx2/2

and the normalization constant is determined by

1 = C2

∫

∞

−∞

e−βx2

dx ⇒ C =

(

β

π

)1/4

.


The kinetic energy operator operating on this state yields

TΨ =p2

2mΨ(x) = − ~

2

2m

∂

∂x

[

∂

∂xΨ(x)

]

= −(

β

π

)1/4~

2

2m

∂

∂x

[

−βxe−βx2/2]

= −(

β

π

)1/4~

2

2m

[

−β + β2x2]

e−βx2/2

and its expectation value is

〈T 〉 =

(

β

π

)1/2~

2

2m

[

β

√

π

β− β2

2

√

π

β3

]

=~

2β

4m. (10)

The expectation value of the potential energy is

〈V 〉 = −V0

∫

∞

−∞

Ψ2(x)e−αx2

dx

= −V0

√

β

π

∫

∞

−∞

e−βx2

e−αx2

dx

= −V0

√

β

π

√

π

(α + β)

= −V0

√

β

(α + β). (11)

Combining (10) and (11),

exptwoΨ|H |Ψ = 〈Ψ|T |Ψ〉 + 〈Ψ|V |Ψ〉 =~

2β

2m− V0

√

β

(α + β). (12)

To minimize with respect to β we equate the first β derivative of this to zero:

0 =~

2

2m− V0

2

[

1√

β(α + β)−

√β

√

(α + β)3

]

=~

2

2m− V0

2

[

α2

β(α + β)3

]1/2

= β(α + β)3 −(

mV0α

~2

)2

= β4 + 3β3α + 3β2α2 + βα3 −(

mV0α

~2

)2

= x4 + 3x3 + 3x2 + x −(

mV0

~2α

)2


where I put x = β/α. In theory we could write down an explicit expression forthe roots of this quartic in terms of mV0/~

2α, and then insert said expressioninto (12) to obtain the lowest energy attainable with this form of trial wavefunction. In practice, however, this would be a mess, and I can’t see any wayto proceed other than numerically. Am I missing some kind of trick here?

Problem 8.5

Show that as inadequate a variational trial function as

Ψ(x) =

{

C(

1 − |x|a

)

|x| ≤ a

0 |x| > a

yields, for the optimum value of a, an upper limit to the ground state energy of thelinear harmonic oscillator, which lies within less than 10 percent of the exact value.

The first task is to evaluate the normalization constant C.

1 = C2

∫ a

−a

Ψ(x)2dx

= 2C2

∫ a

0

(

1 − x

a

)2

dx

= 2C2

∫ a

0

(

1 − 2x

a+

x2

a2

)

= 2C2[

a − a +a

3

]

so

C =

√

3

2a.

The harmonic oscillator hamiltonian is

E = T + V =p2

2m+

mω2x2

2.

(13)

exptwoΨTΨ = − ~2

2m

∫ a

−a

Ψ(x)∂2

∂x2Ψ(x) dx (14)

Integrating by parts,

= − ~2

2m

{

Ψ(x)∂

∂xΨ(x)

∣

∣

∣

∣

a

−a

−∫ a

−a

[

∂

∂xΨ(x)

]2

dx

}


The first term vanishes...

=~

2

2m

(

3

2a

)∫ a

−a

1

a2dx

=3~

2

2ma2(15)

(16)

exptwoΨV Ψ =mω2

2

∫ a

−a

x2Ψ(x)2 dx

= mω2

∫ a

0

x2Ψ(x)2 dx

= mω2

(

3

2a

)∫ a

0

(

x2 − 2x3

a+

x4

a2

)

dx

= mω2

(

3

2a

)∣

∣

∣

∣

x3

3− x4

2a+

x5

5a2

∣

∣

∣

∣

a

0

=mω2a2

20(17)

exptwoΨ|H |Ψ = 〈Ψ|T |Ψ〉 + 〈Ψ|V |Ψ〉 =3~

2

2ma2+

mω2a2

20. (18)

To minimize with respect to a we set the a derivative of this to zero:

0 = − 3~2

ma3+

mω2a

10

or

a4 =30~

2

m2ω2

a2 =√

30~

mω.

Inserting into (18),

exptwoΨHΨ =3√30

~ω ≈ 0.547 · ~ω.

Of course the actual ground state energy is 0.5 · ~ω, so the fractional error is0.047/0.5 < 10%.


Problem 8.6

A particle of mass m moves in a potential V (r). The n − th discrete energy eigen-function of this system, Ψn(r), corresponds to the energy eigenvalue En. Apply thevariational principle by using as a trial function,

Ψt(r) = Ψn(λr),

where λ is a variational (scaling) parameter, and derive the virial theorem for sta-tionary states.

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

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