Solutions to Problems in Merzbacher,
Quantum Mechanics, Third Edition
Homer Reid
May 13, 2001
Chapter 8
1
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
2
Problem 8.1
Apply the variational method to estimate the ground state energy of a particle
confined in a one-dimensional box for which V = 0 for −a < x < a, and Ψ(±a) = 0.
(a) First, use an unnormalized trapezoidal trial function which vanishes at ±a and
is symmetric with respect to the center of the well:
(
(a − |x|),
b≤x≤a
Ψt (x) =
(a − b),
|x| ≤ b.
(b) A more sophisticated trial function is parabolic, again vanishing at the end
points and even in x.
(c) Use a quartic trial function of the form
Ψt (x) = (a2 − x2 )(αx2 + β),
where the ratio of the adjustable parameters α and β is determined variationally.
(d) Compare the results of the different variational calculations with the exact
ground state energy,
R a and, using normalized wave functions, evaluate the mean
square deviation −a |Ψ(x) − Ψt (x)|2 dx for the various cases.
(e) Show that the variational procedure produces, in addition to the approximation
to the ground state, an optimal quartic trial function with nodes between the
endpoints. Interpret the corresponding stationary energy value.
First let’s observe that the exact expressions for the ground state wavefunction
and energy are
1
Ψn (x) = √ cos(kn x),
a
kn =
nπ
,
2a
En = n 2
~2
~2 π 2
≈
1.23
.
8ma2
ma2
(a) We need first to normalize the trial wavefunction. Taking
Ψt (x) =
(
γ(a − |x|),
γ(a − b),
b ≤ |x| ≤ a
x| ≤ b.
3
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
we have
Z
a
−a
Ψ2t (x)dx = 2
Z
= 2γ
a
Ψ2t (x)dx
0
(
Z
2
(a − b)
b
2
dx +
0
Z
a
2
b
(a − x) dx
)
1
3
= 2γ b(a − b) + (a − b)
3
1 3
3
2
2
2
2
2
= 2γ b(a + b − 2ab) + (a − b ) − a b + b a
3
2
= γ 2 a3 + 2b3 − 3ab2
3
2
2
so Ψt is normalized by taking
γ2 =
3
2
1
a3 + 2b3 − 3ab2
.
(1)
Now we can compute the energy expectation value of Ψt :
Z a
~2
d2
< Ψt |H|Ψt > = −
Ψt (x) 2 Ψt (x) dx
2m −a
dx
Integrating by parts,
=−
~2
2m
Z
a
Ψt (x)Ψ0t (x) −
−a
a
−a
Ψ02
t (x) dx
(the first integral vanishes since Ψt vanishes at the endpoints)
=+
~2
m
Z
a
Ψ02
t (x) dx
0
Z
~2 2 a
γ
dx
m
b
~2 2
γ (b − a).
=
m
=
Using (1), this is
< H >=
3~2
2m
(b − a)
a3 + 2b3 − 3ab2
.
To find the optimal value of b, we zero the derivative of this with respect to b:
0=
(a3
1
6b2 (b − a)
6ab(b − a)
− 3
+ 3
3
2
3
2
2
+ 2b − 3ab ) (a + 2b − 3ab )
(a + 2b3 − 3ab2 )2
= −4b3 + 9b2 a − 6a2 b + a3
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
(b) For a parabolic trial function we take
Ψt (x) = γ(a2 − x2 ).
The normalization integral is
Z a
Z a
(a2 − x2 )2 dx
Ψ2t (x) dx = 2γ 2
0
−a
Z a
2
(a4 + x4 − 2a2 x2 )dx
= 2γ
0
1
2
2
= 2γ a5 + a5 − a5
5
3
16 2 5
γ a
=
15
so Ψt (x) is normalized by taking
γ2 =
15
.
16a5
The expectation value of the energy is
Z a
d2
~2
Ψt (x) 2 Ψt (x) dx
<H >=−
2m −a
dx
Z a
2
~
= 2 γ2
(a2 − x2 )dx
m
0
4 ~2 2 3
γ a
=
3m
~2
5 ~2
≈
1.25
.
=
4 ma2
ma2
So this is in good agreement with the exact ground state energy.
(c) In this case we have
Ψt (x) = γ(a2 − x2 )(αx2 + β)
= γ[−αx4 + (αa2 − β)x2 + βa2 ]
The kinetic energy is
−
~2
~2 d 2
Ψt (x) = γ [6αx2 − (αa2 − β)].
2
2m dx
m
The expectation value of the energy is
4
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
5
Problem 8.2
Using scaled variables, as in Section 5.1, consider the anharmonic oscillator Hamiltonian,
1
1
H = p2ξ + ξ 2 + λξ 4
2
2
where λ is a real-valued parameter.
(a) Estimate the ground state energy by a variational calculation, using as a trial
function the ground state wave function for the harmonic oscillator
H0 (ω) =
1 2 1 2 2
p + ω ξ
2 ξ 2
where ω is an adjustable variational parameter. Derive an equation that
relates ω and λ.
(b) Compute the variational estimate of the ground state energy of H for various
positive values of the strength λ.
(c) Note that the method yields answers for a discrete energy eigenstate even if λ
is slightly negative. Draw the potential energy curve to judge if this result
makes physical sense. Explain.
(a) To find the ground state eigenfunction of the Hamiltonian Merzbacher proposes, it’s convenient to rewrite it:
1 2 1 2 2
p + ω ξ
2 ξ 2
1
1 ∂2
+ ω2 ξ 2
=−
2
2 ∂ξ
2
H0 (ω) =
Upon substituting u = ω 1/2 ξ we obtain
1 ∂2
1
=ω −
+ u2
2 ∂u2 2
and now this is just the ordinary harmonic oscillator Hamiltonian, scaled by a
constant factor ω, with ground-state eigenfunction
Ψ(ω) = Ce−u
2
/2
= Ce−ωξ
2
/2
Adding the normalization constant,
Ψ(ω) =
ω 1/4
π
e−ωξ
2
/2
.
.
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
6
Now we want to treat ω as a parameter and vary it until the energy expectation
value of Ψ(ω) is minimized. The energy expectation value is
hΨ| H |Ψi = hΨ| T |Ψi + hΨ| V |Ψi
where T = p2ξ /2 and V = ξ 2 /2 + λξ 4 . Let’s compute the two expectation values
separately. First of all, to compute the expectation value of T , we need to know
the result of operating on Ψ(ω) with p2ξ :
p2ξ Ψ(ω)
ω 1/4 ∂ ∂
−ωξ 2 /2
=−
e
π
∂ξ ∂ξ
i
ω 1/4 ∂ h
−ωξ 2 /2
−ωξe
=−
π
∂ξ
ω 1/4 2
−ω + ω 2 ξ 2 e−ωξ /2
=−
π
Then for the expectation value of T we have
Z
1 ∞
Ψ(ξ)p2ξ Ψ(ξ)dξ
hΨ| T |Ψi =
2 −∞
r Z
2
1 ω ∞
=−
−ω + ω 2 ξ 2 e−ωξ dξ
2 π −∞
r r
r 1 ω
π
π
1
−ω
=−
+ ω2
2 π
ω 2
ω3
ω
= .
4
(2)
On the other hand, for the expectation value of V we have
(3)
Z
Z ∞
ω 1 ∞ 2 −ωξ2
4 −ωξ 2
exptwoΨV Ψ =
ξ e
dξ + λ
ξ e
dξ
π 2 −∞
−∞
r r
r 3
ω 1 π
π
=
+ λ
3
π 4 ω
4
ω5
1
3λ
=
+
.
4ω 4ω 2
r
(4)
Adding (2) and (4),
exptwoΨHΨ =
1
3λ
1
ω+ + 2 .
2
ω ω
To minimize this with respect to ω we equate its ω derivative to 0:
0=1−
1
6λ
− 3
ω2
ω
(5)
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
7
or
ω 3 − ω − 6λ = 0.
(6)
We could then solve this equation for ω in terms of λ to obtain the energyminimizing value of ω for a given perturbing potential strength λ. But writing
down the full solution would be tedious. Instead let’s see what happens when
λ is small.
Evidently, when λ = 0 the Hamiltonian in this problem degenerates to the
normal harmonic oscillator Hamiltonian, for which the energy is minimized by
the (unscaled) ground state harmonic oscillator wavefunction, i.e. Ψ(ω) with
ω = 1. We can thus imagine that, for small λ, the energy-minimizing value of
ω will be close to 1, and we may write ω(λ) ≈ 1 + for some small . Inserting
this in (6),
(1 + 3 + 32 + 3 ) − (1 + ) = 6λ
Keeping only terms of zeroth or first order in the small quantity (which is
equivalent to keeping terms of lowest order in the perturbing potential strength
λ) we obtain from this
≈ 3λ,
so for λ 0 the minimizing value of ω is
ω ≈ 1 + 3λ.
Inserting this estimate into (5) and again keeping only terms of lowest order in
λ we find
(7)
exptwoΨHΨ =
≈
≈
≈
1
(1 + ) + (1 + )−1 + 3λ(1 + )−2
4
1
(1 + 3λ) + (1 + 3λ)−1 + 3λ(1 + 3λ)−2
4
1
[(1 + 3λ) + (1 − 3λ) + 3λ(1 − 6λ)]
4
1 3
+ λ.
2 4
(8)
Since the 1/2 term is the normal (unperturbed) energy of the state, the energy
shift caused by the perturbing potential is
∆E =
3
λ.
4
(9)
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
8
Problem 8.3
In first-order perturbation theory, calculate the change in the energy levels of a
linear harmonic oscillator that is perturbed by a potential gx4 . For small values of
the coefficient, compare the result with the variational calculation in Problem 2.
The energy shift to first order is
∆E = exptwoΨn (x)|gx4 |Ψn (x) = g hΨn | x4 |Ψn i .
I worked out this expectation value in Problem 5.3:
3g
∆E = gexptwoΨn x Ψn =
2
4
~
mω
2 1
+ n + n2
2
In particular, the energy shift of the ground state is
3g
∆E0 =
4
~
mω
2
which agrees with () (the difference in the factor (~/mω)2 just represents the
fact that in Problem 8.2 we used scaled variables, whereas in this problem we
inserted the units explicitly).
Problem 8.4
2
Using a Gaussian trial function, e−λx , with an adjustable parameter, make a variational estimate of the ground state energy for a particle in a Gaussian potential
well, represented by the Hamiltonian
H=
2
p2
− V0 e−αx
2m
(V0 > 0, α > 0).
For notational simplicity, I like to put β/2 = λ. Then
Ψ(x) = Ce−βx
2
/2
and the normalization constant is determined by
1=C
2
Z
∞
e
−∞
−βx2
dx
⇒
1/4
β
C=
.
π
9
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
The kinetic energy operator operating on this state yields
∂
p2
~2 ∂
TΨ =
Ψ(x) = −
Ψ(x)
2m
2m ∂x ∂x
1/4 2
i
2
~ ∂ h
β
−βxe−βx /2
=−
π
2m ∂x
1/4 2
2
~ β
=−
−β + β 2 x2 e−βx /2
π
2m
and its expectation value is
1/2 2 r
r β2
~
β
π
π
−
β
hT i =
π
2m
β
2
β3
~2 β
.
4m
The expectation value of the potential energy is
Z ∞
2
hV i = −V0
Ψ2 (x)e−αx dx
−∞
r Z ∞
2
2
β
e−βx e−αx dx
= −V0
π −∞
r r
β
π
= −V0
π (α + β)
s
β
.
= −V0
(α + β)
(10)
=
(11)
Combining (10) and (11),
~2 β
exptwoΨ|H|Ψ = hΨ| T |Ψi + hΨ| V |Ψi =
− V0
2m
s
β
.
(α + β)
(12)
To minimize with respect to β we equate the first β derivative of this to zero:
"
#
√
V0
1
β
~2
p
−
−p
0=
2m
2
β(α + β)
(α + β)3
1/2
~2
α2
V0
=
−
2m
2 β(α + β)3
2
mV0 α
3
= β(α + β) −
~2
2
mV0 α
= β 4 + 3β 3 α + 3β 2 α2 + βα3 −
~2
2
mV0
= x4 + 3x3 + 3x2 + x −
~2 α
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
10
where I put x = β/α. In theory we could write down an explicit expression for
the roots of this quartic in terms of mV0 /~2 α, and then insert said expression
into (12) to obtain the lowest energy attainable with this form of trial wave
function. In practice, however, this would be a mess, and I can’t see any way
to proceed other than numerically. Am I missing some kind of trick here?
Problem 8.5
Show that as inadequate a variational trial function as
( C 1 − |x|
|x| ≤ a
a
Ψ(x) =
0
|x| > a
yields, for the optimum value of a, an upper limit to the ground state energy of the
linear harmonic oscillator, which lies within less than 10 percent of the exact value.
The first task is to evaluate the normalization constant C.
Z a
2
1=C
Ψ(x)2 dx
−a
Z a
x 2
2
1−
= 2C
dx
a
0
Z a
x x2
2
1−2 + 2
= 2C
a
a
i
h0
a
= 2C 2 a − a +
3
so
C=
r
3
.
2a
The harmonic oscillator hamiltonian is
E =T +V =
p2
mω 2 x2
+
.
2m
2
(13)
2
a
~
2m
Z
~2
=−
2m
(
exptwoΨT Ψ = −
2
Ψ(x)
−a
∂
Ψ(x) dx
∂x2
Integrating by parts,
a
2 )
Z a
∂
∂
Ψ(x) dx
Ψ(x) Ψ(x) −
∂x
−a ∂x
−a
(14)
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
11
The first term vanishes...
~2
=
2m
3~2
=
2ma2
3
2a
Z
a
−a
1
dx
a2
(15)
(16)
exptwoΨV Ψ =
mω
2
2
2
Z
Z
a
x2 Ψ(x)2 dx
−a
a
x2 Ψ(x)2 dx
Z a
2x3
x4
3
2
2
x −
+ 2 dx
= mω
2a
a
a
0
a
3
4
5
x
x 3 x
−
+ 2 = mω 2
2a
3
2a 5a 0
= mω
=
0
mω 2 a2
20
3~2
mω 2 a2
+
.
2
2ma
20
To minimize with respect to a we set the a derivative of this to zero:
exptwoΨ|H|Ψ = hΨ| T |Ψi + hΨ| V |Ψi =
0=−
(17)
(18)
mω 2 a
3~2
+
ma3
10
or
30~2
m2 ω 2
√
~
.
a2 = 30
mω
a4 =
Inserting into (18),
3
exptwoΨHΨ = √ ~ω ≈ 0.547 · ~ω.
30
Of course the actual ground state energy is 0.5 · ~ω, so the fractional error is
0.047/0.5 < 10%.
Homer Reid’s Solutions to Merzbacher Problems: Chapter 8
12
Problem 8.6
A particle of mass m moves in a potential V (r). The n − th discrete energy eigenfunction of this system, Ψn (r), corresponds to the energy eigenvalue En . Apply the
variational principle by using as a trial function,
Ψt (r) = Ψn (λr),
where λ is a variational (scaling) parameter, and derive the virial theorem for stationary states.