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 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.

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 3

we have∫ a

−a

Ψ2

t (x)dx = 2

∫ a

0

Ψ2

t (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

}

= γ22

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

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 4

(b) For a parabolic trial function we take

Ψt(x) = γ(a2 − x2).

The normalization integral is

∫ a

−a

Ψ2

t (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

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 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.

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 expectationvalue of Ψ(ω) is minimized. The energy expectation value is

〈Ψ|H |Ψ〉 = 〈Ψ|T |Ψ〉 + 〈Ψ|V |Ψ〉

where T = p2

ξ/2 and V = ξ2/2 + λξ4. Let’s compute the two expectation valuesseparately. First of all, to compute the expectation value of T , we need to knowthe result of operating on Ψ(ω) with p2

ξ :

p2

ξΨ(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)p2

ξΨ(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

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 7

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)

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 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

.

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 9

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

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 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

}

Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 11

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%.

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 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.