763313A QUANTUM MECHANICS II
Solutions 7
Spring 2017
1. Suppose we perturb the infinite cubical well by putting a delta function bump at the
point (a/4, a/2, 3a/4):
H 0 = a3 V0 δ(x − a/4)δ(y − a/2)δ(z − 3a/4) .
Find the first-order corrections to the energy of the ground state and the (triply degenerate) first excited states.
Solution :
The unperturbed states are (cf. QM1 lecture notes)
ψnx ,ny ,nz
3/2
n π n π n π 2
y
z
x
=
x sin
y sin
z .
sin
a
a
a
a
Since the ground state is nondegenerate the energy correction is
Z aZ aZ a
1
0
∗
E0 = hψ111 |H |ψ111 i =
ψ111
H 0 ψ111 dx dy dz
0
0
0
Z aZ aZ a
8
a a
3a
2 π
2 π
2 π
3
=
sin
x sin
y sin
z a V0 δ x −
δ y−
δ z−
dx dy dz
3
a
a
a
4
2
4
0
0
0 a
Z a
Z a
Z a
a
a
3a
2 π
2 π
2 π
sin
sin
x δ x−
dx
y δ y−
dy
z δ z−
dz
= 8V0
sin
a
4
a
2
a
4
0
0
0
π 3a
2 πa
2 πa
2
= 8V0 sin
sin
sin
a4
a2
a 4
= 2V0 .
There are three first excited states ψ112 , ψ121 and ψ211 , so we need to use the degenerate
perturbation theory (cf. lecture notes section 3.2). First we need to construct the W matrix. We will use the notation ψa0 = ψ112 , ψb0 = ψ121 and ψc0 = ψ211 . The matrix
elements of W are Wij = hψi0 |H 0 |ψj0 i. Now
Waa = hψa0 |H 0 |ψa0 i
Z aZ aZ
= 8V0
a
2
2
πy 2
2πz
a
sin
sin
a
a
a a
3a
δ x−
δ y−
δ z−
dx dy dz
4
2
4
3π
2 π
2 π
2
= 8V0 sin
sin
sin
= 4V0 .
4
2
2
0
0
sin
πx 0
1
×
The rest are calculated the same way. They are
Wbb = 0 ,
Wcc = 4V0 ,
Wac = −4V0 ,
Wab = 0 ,
Wbc = 0 .
Thus W is (W is hermitian)


1 0 −1
W = 4V0  0 0 0  .
−1 0 1
Now the first-order energy corrections are the eigenvalues of W .
4V0 − E 1
0
−4V0 = −E 1 (4V0 − E 1 )2 + 16V02 E 1 = 0
0
−E 1
0
det(W − E 1 I) = 1
−4V0
0
4V0 − E −E 1 [(4V0 − E 1 )2 − 16V02 ] = 0
E1 = 0
or
E1 = 0
or
E 1 = 8V0 .
So the corrections are 0, 0 and 8V0 .
2. Consider a quantum system with just three
Hamiltonian, in matrix form, is

(1 − )
H = V0  0
0
linearly independent states. Suppose the

0 0
1  ,
2
where V0 is a constant and 1.
a) Write down the eigenvectors and eigenvalues of the unperturbed Hamiltonian ( =
0).
b) Solve the exact eigenvalues of H. Expand each of them as a power series in , up
to second order.
c) Use the first- and second-order nondegenerate perturbation theory to find the approximate eigenvalue for the state that grows out of the nondegenerate eigenvector
of H 0 . Compare the exact result from a).
d) Use degenerate perturbation theory to find the first-order correction to the two
initially degenerate eigenvalues. Compare the exact results.
Solution:
a) Unperturbed Hamiltonian is


1 0 0
H 0 = V0 0 1 0 .
0 0 2
2
(1)
The eigenvalues and vectors are
 
1

|1i = 0 ,
0
 
0

|2i = 1 ,
0
 
0
|3i = 0 ,
1
E10 = V0
E20 = V0
E30 = 2V0 .
b) The eigenvalues are
V0 (1 − ) − λ
0
0
0
V0 − λ
V0 det(H − λI) = 0
V0 2V0 − λ
= [V0 (1 − ) − λ][(V0 − λ)(2V0 − λ) − V02 2 ] = 0
√
V0 2
3 ± 1 + 4 .
λ1 = V0 (1 − ) ,
λ2,3 =
2
Expanding these as a power series gives
λ1 = V0 (1 − )
λ2 = V0 (1 − 2 )
λ3 = V0 (2 + 2 ) .
c) The nondegenerate eigenstate is the state |3i = (0 0 1)T . The perturbed Hamiltonian is


−1 0 0
H 0 = V0  0 0 1 .
0 1 0
The first-order correction is

E31 = h3|H 0 |3i = V0
 
−1
0
0
0



0 0 1
0 0 1
0 = 0 .
0 1 0
1
Second-order correction is
E32
=
2
X
|hn|H 0 |3i|2
E30 − En0
n=1
.
Now
h1|H 0 |3i = 0 ,
h2|H 0 |3i = V0
3
and
E30 − E20 = V0 .
Therefore
E32 = 2 V0 .
The approximate eigenvalue of |3i is
E3 = 2V0 + 2 V0 = V0 (2 + 2 ) ,
which is the same as λ3 in b).
d) We will start by constructing the W -matrix in the {|1i, |2i} basis:
W11 = h1|H 0 |1i = −V0
W22 = h2|H 0 |2i = 0
W12 = h1|H 0 |2i = 0 .
We will use the result for the two-fold degeneracy derived in the lecture notes (p.
28)
V
p
1
0
1
2
2
W11 + W22 ± (W11 − W22 ) + 4|W12 | =
(−1 ± 1) .
E± =
2
2
So the first-order corrected eigenvalues are
E1 = E10 + E−1 = V0 (1 − )
E2 = E10 + E+1 = V0 ,
which are the same, to the first-order, as in b).
3. Suppose the Hamiltonian, for a particular system, is a function of some parameter λ;
let En (λ) and |ψn (λ)i be the eigenvalues and eigenstates of H(λ). Assuming that En is
non-degenerate (or that |ψn i are ”good” eigenstates), prove the Feynman-Hellmann
theorem:
∂H
∂En
= hψn |
|ψn i .
∂λ
∂λ
Solution:
This is a straight forward calculation
En |ψn i = H|ψn i
En hψn |ψn i = hψn |H|ψn i
∂En
∂
=
hψn |H|ψn i
∂λ
∂λ
∂En
∂hψn |
∂H
∂|ψn i
=
H|ψn i + hψn |
|ψn i + hψn |H
∂λ
∂λ
∂λ ∂λ
∂En
∂H
∂hψn |
∂|ψn i
= hψn |
|ψn i + En
|ψn i + hψn |
∂λ
∂λ
∂λ
∂λ
∂En
∂H
∂
= hψn |
|ψn i + En hψn |ψn i
∂λ
∂λ
∂λ
∂En
∂H
∂
∂H
= hψn |
|ψn i + En 1 = hψn |
|ψn i .
∂λ
∂λ
∂λ
∂λ
4
4. Show that
h[H, r · p]i = 0
in an eigenstate of H. Accordingly, show that for hydrogen
1
hT i = − hV i
2
and, thus,
1
1
h i= 2 .
r
na
Solution:
If we take the expectation value h[H, r · p]i with respect to an eigenstate of the Hamiltonian, say |φi, we obtain,
h[H, r · p]i = hφ|[H, r · p]|φi
= hφ|(Hr · p − r · pH)|φi
= hφ|(Er · p − r · pE)|φi
= Ehφ|(r · p − r · p)|φi = 0 .
The potential energy for hydrogen is
V (r) = −
e2 1
.
4π0 r
The Hamiltonian operator of the system is
H=
p2
+ V (r) .
2m
We consider the commutator
[H, r · p] =
1 2
[p , r · p] + [V (r), r · p] .
2m
(2)
We easily obtain
[p2 , r · p] = [p2 , xpx ] + [p2 , ypy ] + [p2 , zpz ]
= [p2x , xpx ] + [p2y , ypy ] + [p2z , zpz ] .
By employing the commutation relation [x, px ] = i~, we obtain
[p2x , xpx ] = p2x xpx − xpx p2x
= p2x xpx − (i~ + px x)p2x
= p2x xpx − i~p2x − px (i~ + px x)px
= −2i~p2x .
5
(3)
Similarly,
[p2y , ypy ] = −2i~p2y ,
[p2z , zpz ] = −2i~p2z .
By substituting these into Eq. (3), we obtain
[p2 , r · p] = −2i~p2 .
Let us next consider the last commutator at the right-hand side of Eq. (2). We easily
obtain
1
1
1
1
[ , r · p] = [ , xpx ] + [ , ypy ] + [ , zpz ] .
r
r
r
r
Recalling that px = −i~(∂/∂x), we obtain
x
1
1
[ , xpx ]ψ = px ψ − xpx ψ
r
r
r
2
x
x
x
= px ψ − i~ 3 ψ − px ψ
r
r
r
x2
= −i~ 3 ψ .
r
Thus
1
x2
[ , xpx ] = −i~ 3 .
r
r
Similarly,
y2
1
[ , ypy ] = −i~ 3 ,
r
r
1
z2
[ , zpz ] = −i~ 3 .
r
r
By substituting these into Eq. (4), we obtain
1
1
[ , r · p] = −i~ .
r
r
By multiplying both sides of this with −e2 /4π0 , we obtain
[V (r), r · p] = −i~V (r) .
Now we get Eq. (2) into a form
p2
− i~V (r)
2m
= −i~(2T + V ) ,
[H, r · p] = −2i~
6
(4)
where T = p2 /2m and V = V (r). Now
h[H, r · p]i = −i~(2hT i + hV i) = 0 .
That is,
1
hT i = − hV i .
2
If we take the expectation value of the Hamiltonian for hydrogen, we get
hHi = −
~2 1
,
2m (na)2
where a = 4π0 ~2 /me2 . On the other hand
1
e2 1
hHi = hT i + hV i = hV i = −
h i.
2
8π0 r
Thus we get
8π0 ~2 1
a
1
1
=
= 2 .
h i= 2
2
2
r
e 2m (na)
(na)
na
5. Let the unperturbed radial part of the Hamiltonian of hydrogen be
H0 = −
~2 d2
e2 1
~2 l(l + 1)
−
+
.
2m dr2
4π0 r 2m r2
Let us introduce a perturbation with
H = H0 +
λ
,
r2
where 0 ≤ λ ≤ 1 is the perturbation parameter. Thus, the new centrifugal term becomes
~2 l(l + 1)
λ
~2 l0 (l0 + 1)
+
≡
,
2mr2
r2
2mr2
defining l0 = l0 (λ). Correspondingly, the new eigenenergies are
2 2
m
e
1
0
E(λ) = E(l ) = − 2
2~ 4π0
(jmax + l0 + 1)2
(recall the definition of jmax from QMI, its value is not needed though). Use the FeynmanHellmann theorem (problem 3) at the limit λ = 0 to show that
h
1
1
i=
.
2
r
(l + 1/2)n3 a2
7
Solution:
Since
1
∂H
= 2
∂λ
r
we get
h
∂H
1
i.
i
=
h
r2
∂λ
By using Feynman-Hellmann theorem we can write
h
1
∂E(λ)
∂H
i=
i=h
2
r
∂λ
∂λ
∂E(l0 ) ∂l0
=
∂l0 ∂λ
2 2
m
e
1
∂l0
= 2
.
~ 4π0
(jmax + l0 + 1)3 ∂λ
Differentiating the new centrifugal term w.r.t λ gives
λ
∂ ~2 l(l + 1)
∂ ~2 l0 (l0 + 1)
+ 2 =
∂λ
2mr2
r
∂λ 2mr2
∂ ~2 l0 (l0 + 1) ∂l0
1
=
r2
∂l0 2mr2 ∂λ
~2 (2l0 + 1) ∂l0
1
=
r2
2mr2 ∂λ
0
∂l
2m
= 2 0
.
∂λ
~ (2l + 1)
Now
1
m
h 2i = 2
r
~
e2
4π0
2
1
2m
.
0
3
2
(jmax + l + 1) ~ (2l0 + 1)
In the case of λ = 0 which means l0 = l we get
1
h 2i =
r
me2
4π0 ~2
where we have used a =
2
4π0 ~2
me2
1
2
1
=
,
3
(jmax + l + 1) 2l + 1
(l + 1/2)n3 a2
and n = jmax + l + 1.
8