PRINCIPLES_OF_QUANTUM_MECHANICS_SOLUTIONS

Prévia do material em texto

SOLUTION MANUAL 
COMPILED BY YEMI BUKKY 
+234(0)8057474928; +234(0)8064974071 
(jjbukky@yahoo.com) 
Department of Physics, 
Federal University of Technology, 
Minna, NG 
Nigeria 
 
mailto:jjbukky@yahoo.com
PRINCIPLES OF QUANTUM 
MECHANICS 
BY R. SHANKAR 
SECOND EDITION 
SOLUTIONS 
COMPILED BY YEMI BUKKY (jjbukky@yahoo.com) 
Department of Physics, 
Federal University of Technology, Minna, NG 
Nigeria 
+2348057474928 
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mailto:jjbukky@yahoo.com
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Physics 710-712 due October 30, 2009
Problem Set 4
Problem 1: Do exercise 2.5.3 of the text.
Solution: The problem asks us to get the equations of motion using the Hamiltonian method
for the system shown in Figure 1.5 (p. 46) of the text. Using x1 and x2 shown there as
the generalized coordinates, the kinetic energy and potential energy are
T =
m
2
ẋ21 +
m
2
ẋ22, V =
k
2
x21 +
k
2
(x1 − x2)2 +
k
2
x22,
so the Lagrangian is L = T − V and thus the generalized momenta are
p1 =
∂L
∂ẋ1
=
∂T
∂ẋ1
= mẋ1, ⇒ ẋ1 =
p1
m
,
p2 =
∂L
∂ẋ2
=
∂T
∂ẋ2
= mẋ2, ⇒ ẋ2 =
p2
m
,
so the Hamiltonian is
H = T + V = p
2
1
2m
+
p22
2m
+
k
2
x21 +
k
2
(x1 − x2)2 +
k
2
x22,
and Hamilton’s equations are
ẋ1 =
∂H
∂p1
=
p1
m
, ṗ1 = −
∂H
∂x1
= −2kx1 + kx2,
ẋ2 =
∂H
∂p2
=
p2
m
, ṗ2 = −
∂H
∂x2
= −2kx2 + kx1.
Problem 2: Do exercise 2.7.2 of the text.
Solution: (i):
{qi, qj} :=
∑
k
(
∂qi
∂qk
· ∂qj
∂pk
− ∂qi
∂pk
·∂qj
∂qk
)
=
∑
k
(
∂qi
∂qk
·0− 0·∂qj
∂qk
)
= 0
{pi, pj} :=
∑
k
(
∂pi
∂qk
·∂pj
∂pk
− ∂pi
∂pk
·∂pj
∂qk
)
=
∑
k
(
0·∂pj
∂pk
− ∂pi
∂pk
·0
)
= 0
{qi, pj} :=
∑
k
(
∂qi
∂qk
·∂pj
∂pk
− ∂qi
∂pk
·∂pj
∂qk
)
=
∑
k
(δikδjk − 0·0) = δij ,
and
{qi,H} :=
∑
k
(
∂qi
∂qk
· ∂H
∂pk
− ∂qi
∂pk
· ∂H
∂qk
)
=
∑
k
(
δik·
∂H
∂pk
− 0· ∂H
∂qk
)
=
∂H
∂pi
= q̇i,
{pi,H} :=
∑
k
(
∂pi
∂qk
· ∂H
∂pk
− ∂pi
∂pk
· ∂H
∂qk
)
=
∑
k
(
0· ∂H
∂pk
− δik·
∂H
∂qk
)
= −∂H
∂qi
= ṗi,
where in the last steps I used Hamilton’s equations.
(ii): The Hamiltonian given is H = p2x + p2y + ax2 + by2. If a = b, H has a
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symmetry under simultaneous rotations in the x-y and px-py planes, under which `z (the
generator) is conserved. Therefore {`z,H} = 0.
We check this as follows:
{`z,H} =
∑
k
(
∂`z
∂qk
· ∂H
∂pk
− ∂`z
∂pk
· ∂H
∂qk
)
=
∂`z
∂x
· ∂H
∂px
+
∂`z
∂y
· ∂H
∂py
− ∂`z
∂px
·∂H
∂x
− ∂`z
∂py
·∂H
∂y
.
But
∂H
∂pk
= 2pk,
∂H
∂xk
=
(
∂H
∂x
,
∂H
∂y
)
= (2ax , 2by) ,
∂`z
∂pk
=
∂(xpy − ypx)
∂pk
=
(
∂`z
∂px
,
∂`z
∂py
)
= (−y , x) , ∂`z
∂qk
=
(
∂`z
∂x
,
∂`z
∂y
)
= (py , −px) ,
so
{`z , H} = py·2px + (−px)·2py − (−y)·2ax− x·2by = 2xy(a− b)
which vanishes if a = b.
Problem 3: Do exercise 2.8.1 of the text.
Solution: Since g = p1 + p2, it generates the infinitesimal transformations
δx1 = +�
∂g
∂p1
= +�, δp1 = −�
∂g
∂x1
= 0,
δx2 = +�
∂g
∂p2
= +�, δp2 = −�
∂g
∂x2
= 0.
So, to order �, these give the canonical transformations xi → x̄i(xj , pj) and pi → p̄i(xj , pj)
with
x̄1 = x1 + �, p̄1 = p1,
x̄2 = x2 + �, p̄2 = p2,
which is precisely a spatial translation of the whole system by an amount �.
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Physics 710-711-712 November 16, 2009
Problem Set 5
Problem 1: Do exercise 4.2.1 of the text.
(1)
Solution: The possible outcomes are Lz = {1, 0,−1}, which are the eigenvalues of Lz.
(2)
Solution: Lz|ψ〉 = 1 · |ψ〉 implies
|ψ〉 =
10
0
 .
(Note that I have normalized |ψ〉!) Then
〈Lx〉 = 〈ψ|Lx|ψ〉 =
(
1 0 0
) 1√
2
0 1 01 0 1
0 1 0
10
0
 = 1√
2
(
1 0 0
)01
0
 = 0.
〈L2x〉 = 〈ψ|L2x|ψ〉 =
(
1 0 0
) 1
2
0 1 01 0 1
0 1 0
0 1 01 0 1
0 1 0
10
0
 = 1
2
(
0 1 0
)01
0
 = 1
2
.
∆Lx =
√
〈L2x〉 − 〈Lx〉
2 =
√(
1
2
)
− 02 = 1√
2
.
(3)
Solution: The characteristic equation for Lx is
0 = det(Lx − λ) = det
−λ
1√
2
0
1√
2
−λ 1√
2
0 1√
2
−λ
 = λ− λ3, ⇒ λ ∈ {1, 0,−1}.
The corresponding eigenvectors, |λ〉, then satisfy
0 = (Lx − λ)|λ〉 =
−λ
1√
2
0
1√
2
−λ 1√
2
0 1√
2
−λ

ab
c
 =
 −λa+
b√
2
a√
2
− λb+ c√
2
b√
2
− λa

where we have parameterized the components of |λ〉 by (a b c). For λ = 1, we can solve
for b and c in terms of a, giving b =
√
2a and c = a. We then determine a by normalizing
|λ = 1〉:
|λ = 1〉 =
 a√2a
a
 , ⇒ 1 = 〈λ = 1|λ = 1〉 = (a∗ √2a∗ a∗)
 a√2a
a
 = 4|a|2, ⇒ a = 1
2
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(where I have chosen the arbitrary phase to be 1). Thus, and doing the same thing for
λ = 0 and λ = −1, gives
|λ = 1〉 = 1
2
 1√2
1
 , |λ = 0〉 = 1√
2
 10
−1
 , |λ = −1〉 = 1
2
 1−√2
1
 .
(4)
Solution: The possible outcomes are Lx = {1, 0,−1}, which are the eigenvalues of Lx.
|ψ〉 is the normalized eigenstate of Lz with eigenvalue Lz = −1, which is
|ψ〉 =
00
1
 .
So (here P stands for "probability of"):
P(Lx = 1) = |〈λ = 1|ψ〉|2 =
∣∣∣∣∣∣12 (1 √2 1)
00
1
∣∣∣∣∣∣
2
=
1
4
,
P(Lx = 0) = |〈λ = 0|ψ〉|2 =
∣∣∣∣∣∣12 (1 0 −1)
00
1
∣∣∣∣∣∣
2
=
1
2
,
P(Lx = −1) = |〈λ = −1|ψ〉|2 =
∣∣∣∣∣∣12 (1 −√2 1)
00
1
∣∣∣∣∣∣
2
=
1
4
.
(5)
Solution:
L2z =
1 0
1
 , ⇒ the possible outcomes are L2z = {0, 1}.
An eigenbasis of the L2z = 1 eigenspace is {|a〉, |b〉} with
|a〉 =
10
0
 , |b〉 =
00
1
 .
Therefore, upon measuring L2z = 1, the state collapses to
|ψ〉 −→ |ψ′〉 = (|a〉〈a|+ |b〉〈b|)|ψ〉
|(|a〉〈a|+ |b〉〈b|)|ψ〉|
.
But
[|a〉〈a|+ |b〉〈b|] |ψ〉 =
10
0
(1 0 0)+
00
1
(0 0 1)
 1
2
 11√
2
 =
10
0
 1
2
+
00
1
 1√
2
=
1
2
 10√
2
 ,
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has norm √√√√√1
2
(
1 0
√
2
) 1
2
 10√
2
 = √3
2
,
so
|ψ′〉 = 1√
3/2
1
2
 10√
2
 = 1√
3
 10√
2
 .
The probability of L2z = +1 is
P(L2z = 1) = 〈ψ| (|a〉〈a|+ |b〉〈b|) |ψ〉 = |〈a|ψ〉|2 + |〈b|ψ〉|2
=
∣∣∣∣∣∣12 (1 0 0)
 11√
2
∣∣∣∣∣∣
2
+
∣∣∣∣∣∣12 (0 0 1)
 11√
2
∣∣∣∣∣∣
2
=
1
4
+
1
2
=
3
4
.
If we measured Lz the posible outcomes are the eigenvalues Lz, {0,±1}, with probabilities
P(Lz = 1) =
∣∣(1 0 0) |ψ′〉∣∣2 =
∣∣∣∣∣∣ 1√3 (1 0 0)
 10√
2
∣∣∣∣∣∣
2
=
1
3
.
P(Lz = 0) =
∣∣(0 1 0) |ψ′〉∣∣2 =
∣∣∣∣∣∣ 1√3 (0 1 0)
 10√
2
∣∣∣∣∣∣
2
= 0.
P(Lz = −1) =
∣∣(0 0 1) |ψ′〉∣∣2 =
∣∣∣∣∣∣ 1√3 (0 0 1)
 10√
2
∣∣∣∣∣∣
2
=
2
3
.
(6)
Solution: In the Lz eigenbasis
|Lz = 1〉 =
10
0
 , |Lz = 0〉 =
01
0
 , |Lz = −1〉 =
00
1
 ,
write the unknown state as
|ψ〉 =
ab
c
 .
Then
P(Lz = 1) =
1
4
= |〈Lz = 1|ψ〉|2 =
∣∣∣∣∣∣(1 0 0)
ab
c
∣∣∣∣∣∣
2
= |a|2,
P(Lz = 1) =
1
2
= |〈Lz = 0|ψ〉|2 =
∣∣∣∣∣∣(0 1 0)
ab
c
∣∣∣∣∣∣
2
= |b|2,
P(Lz = 1) =
1
4
= |〈Lz = −1|ψ〉|2 =
∣∣∣∣∣∣(0 0 1)
ab
c
∣∣∣∣∣∣
2
= |c|2.
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The most general solution to these three equations is then
a =
1
2
eiδ1 , b =
1√
2
eiδ2 , c =
1
2
eiδ3 ,
for some arbitrary phases δi, which gives the desired answer.
The δi phase factors are not irrelevant. For example
P(Lx = 0) = |〈λ = 0|ψ〉|2 =
∣∣∣∣∣∣ 1√2 (1 0 −1) 12
 eiδ1√2eiδ2
eiδ3
∣∣∣∣∣∣
2
=
∣∣∣∣ eiδ12√2 − eiδ32√2
∣∣∣∣2
=
1
8
(
eiδ1 − eiδ3
) (
e−iδ1 − e−iδ3
)
=
1
8
(
1− ei(δ3−δ1) − e−i(δ3−δ1) + 1
)
=
1
4
(1− cos(δ3 − δ1)) ,
so something measurable (a probability) depends on the difference of the phases.
Problem 2: Do exercise 4.2.2 of the text.
Solution:
〈P 〉 = 〈ψ|P |ψ〉 =
∫ ∞
−∞
dx〈ψ|x〉〈x|P |ψ〉
=
∫ ∞
−∞
dxψ∗(x)
(
−i~ d
dx
)
ψ(x) = −i~
∫ ∞
−∞
dxψ(x)
dψ(x)
dx
= − i~
2
∫ ∞
−∞
dx
d
dx
(
ψ(x)2
)
= − i~
2
ψ2
∣∣∞
−∞ = 0
if ψ → 0 as |x| → ∞.
Alternatively, use the k-basis:
〈P 〉 = 〈ψ|P |ψ〉 =
∫ ∞
−∞
dk〈ψ|k〉〈k|P |ψ〉 =
∫ ∞
−∞
dk ~k 〈ψ|k〉〈k|ψ〉 =
∫ ∞
−∞
dk ~k ψ∗(k)ψ(k).
But
ψ(k) = 〈k|ψ〉 =
∫ ∞
−∞
dx 〈k|x〉〈x|ψ〉 = 1√
2π
∫ ∞
−∞
dx eikxψ(x),
therefore
ψ(−k) = 1√
2π
∫ ∞
−∞
dx e−ikxψ(x) = ψ∗(k)
since ψ(x) is real. So
〈P 〉 =
∫ ∞
−∞
dk ~kψ∗(k)ψ(k) =
∫ ∞
−∞
dk ~kψ(−k)ψ(k).
and under the change of variables k → −k, this becomes
〈P 〉 =
∫ ∞
−∞
dk ~(−k)ψ(k)ψ(−k) = −〈P 〉,
and so 〈P 〉 = 0.
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Problem 3: Do exercise 2.4.3 of the text.
Solution:〈
eip0x/~ψ
∣∣∣P ∣∣∣eip0x/~ψ〉 = ∫ ∞
−∞
dx 〈eip0x/~ψ|x〉〈x|P∣∣∣eip0x/~ψ〉 = ∫ ∞
−∞
dx
(
eip0x/~ψ(x)
)∗
(−i~) d
dx
(
eip0x/~ψ(x)
)
= −i~
∫ ∞
−∞
dxψ∗(x)e−ip0x/~
[
ip0
~
eip0x/~ψ(x) + eip0x/~
dψ
dx
]
=
∫ ∞
−∞
dxψ∗(x) p0 ψ(x)− i~
∫ ∞
−∞
dxψ∗(x)
dψ
dx
= p0
[∫ ∞
−∞
dx 〈ψ|x〉〈x|ψ〉
]
+
[∫ ∞
−∞
dx 〈ψ|x〉〈x|P |ψ〉
]
= p0〈ψ|ψ〉+ 〈ψ|P |ψ〉 = p0 + 〈P 〉.
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Problem 2
In[3]:= Psi@x_, t_D := HPi H∆^2 + hbar^2 t^2 êHm ^2 ∆^2LLL^H−1ê2L
Exp@H−Hx − Hp0ê mL tL^2LêH∆^2 + hbar^2 t^2 êHm ^2 ∆^2LLD;
PlotAEvaluate@Table@Psi@x, tD ê. 8p0 → 1, ∆ → 1, hbar → 1, m → 1<,
8t, 0, 14<DD, 8x, −20, 50<,
AxesLabel → 9"x", "»ψHxL»2"=, PlotRange → AllE
Out[4]=
-20 -10 10 20 30 40 50
x
0.1
0.2
0.3
0.4
0.5
»yHxL»2
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Physics 710-711-712 December 4, 2009
Problem Set 7
Problem 1: Exercise 5.3.1
The Hamiltonian is
H =
1
2m
P 2 + Vr(X)− iVi
where Vr is a real function and Vi a real constant. Therefore
H† =
1
2m
(P †)2 + Vr(X
†)− (i)∗Vi =
1
2m
P 2 + Vr(X) + iVi 6= H,
so H is not Hermitian.
Derivation of the continuity equation. Schrodinger’s equation and its complex con-
jugate in this case read
i~
∂ψ
∂t
= − ~
2
2m
∇2ψ + Vrψ − iViψ,
−i~∂ψ
∗
∂t
= − ~
2
2m
∇2ψ∗ + Vrψ∗ + iViψ∗.
Multiplying the first by ψ∗ and teh second by ψ and taking the difference, then dividing
by i~ gives
∂P
∂t
= −~∇·~j − 2
~
ViP,
where, as before, P = |ψ|2 and ~j = ~(ψ∗~∇ψ − ψ~∇)/(2mi) are the probability density
and current, respectively. Integrating this over all space, the ~∇·~j term vanishes (by
the divergence theorem, since we assume ~j → 0 at infinity), giving
dP
dt
= −2
~
ViP ,
where P =
∫
d3xP is the total probability. (I can pull Vi out of the integral since it is
assumed constant in the problem.) Integrating this differential equation gives
P(t) = P(0) e−2Vit/~.
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Problem 2: Exercise 5.3.4
With primes denoting derivative with respect to x,
j :=
~
2mi
[ψ∗ψ′ − ψ(ψ∗)′]
=
~
2mi
[
(A∗e−ixp/~ +B∗eixp/~)(Aeixp/~ +Be−ixp/~)′
− (Aeixp/~ +Be−ixp/~)(A∗e−ixp/~ +B∗eixp/~)′
]
=
1
2mi
[
(A∗e−ixp/~ +B∗eixp/~)(ipAeixp/~ − ipBe−ixp/~)
− (Aeixp/~ +Be−ixp/~)(−ipA∗e−ixp/~ + ipB∗eixp/~)
]
=
p
2m
[
|A|2 + AB∗e2ixp/~ − A∗Be−2ixp/~ − |B|2
+ |A|2 + A∗Be−2ixp/~ − AB∗e2ixp/~ − |B|2
]
=
p
m
[
|A|2 − |B|2
]
.
Problem 3: Exercise 5.4.2
(a) For x < 0, V = 0, so the general solution of the energy eigenstate equation is ψ< =
Aeikx + Be−ikx where ~k =
√
2mE. Similarly, for x > 0, ψ> = Ce
ikx + De−ikx.
Scattering boundary conditions means we set D = 0 (no incoming particles from
x = +∞).
Now we need to figure out the boundary conditions at x = 0. Look at the
time-independent Schrodinger equation,
− ~
2
2m
ψ′′ + V0aδ(x)ψ = Eψ. (1)
Since the potential has an infinite jump in it, ψ will be continuous, but ψ′ may
have a finite jump. To see how big the ψ′ jump is, integrate (1) from x = −� to
x = +� to get
~2
2m
[ψ′(−�)− ψ′(�)] + V0aψ(0) = E
∫ �
−�
dxψ.
In the limit as � → 0, the right hand side vanishes since ψ is continuous, from
which we learn that
ψ′>(0)− ψ′<(0) = (2maV0/~2)ψ(0).
Applying this boundary condition along with continuity of ψ to ψ< and ψ>
gives the two conditions
A+B = C
ikC − ikA+ ikB = (2maV0/~2)C.
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Dividing through by A, and solving for B/A and C/A gives B/A = maV0/(ik~2−
maV0) and C/A = ik~2/(ik~2 −maV0). Since R = |B/A|2 and T = |C/A|2, we
get
R =
m2a2V 20
k2~4 +m2a2V 20
, T =
k2~4
k2~4 +m2a2V 20
.
(b) Call x < −a region I, |x| < a region II, and x > a region III. Then solving for the
energy eigenstates, ~ψ′′ = 2m(E − V (x))ψ, of energy 0 < E ≤ V0 in each region
gives ψI = Ae
ikx + Be−ikx, ψII = Ce
−κx + Deκx, and ψIII = Ee
ikx + Fe−ikx,
with ~κ =
√
2m(V0 − E) and ~k =
√
2mE. Scattering boundary conditions
means we set F = 0 (no incoming particles from x = +∞). The incoming wave
has amplitude A, the reflected has amplitude B, the transmitted amplitude E.
Therefore R = |B/A|2 and T = |E/A|2, so we only need to solve for B/A and
E/A.
The boundary conditions at x = ±a are that ψ and ψ′ are continuous, implying
Ae−ika +Beika = Ceκa +De−κa,
ikAe−ika − ikBeika = −κCeκa + κDe−κa,
Eeika = Ce−κa +Deκa,
ikEeika = −κCe−κa + κDeκa.
Dividing by A and eliminating C/A and D/A gives
B
A
=
e−2iak(e4aκ − 1)(k2 + κ2)
(e4aκ − 1)(k2 − κ2) + 2i(e4aκ + 1)kκ
,
E
A
=
4ie2a(κ−ik)kκ
(e4aκ − 1)(k2 − κ2) + 2i(e4aκ + 1)kκ
,
so
R =
(e4aκ − 1)2(k2 + κ2)2
(e4aκ − 1)2(k2 − κ2)2 + 4(e4aκ + 1)2k2κ2
,
and T = 1−R.
85
Physics 711 January 15, 2010
Problem Set 8
Problem 1: Exercise 7.3.1
Plug the power series expansion ψ =
∑∞
n=0 cny
n into the equation ψ′′+(2ε−y2)ψ = 0
to get
∞∑
n=0
cn
[
n(n− 1)yn−2 + (2ε− y2)yn
]
= 0.
Shift n→ n+ 2 in the first term, and n→ n− 2 in the third term to get
∞∑
n=0
yn [(n+ 2)(n+ 1)cn+2 + 2εcn − cn−2] = 0
with the convention that
c−2 = c−1 = 0.
This implies
cn+2 =
cn−2
(n+ 1)(n+ 2)
− 2εcn
(n+ 1)(n+ 2)
for all n ≥ 0.
Problem 3: Exercise 7.3.7
In the momentum basis |ψ〉 → ψ(p), P → p, and X → i~(d/dp), so the energy
eigenvalue equation (
1
2m
P 2 +
mω2
2
X2
)
|E〉 = E|E〉
becomes
1
2m
p2ψ(p)− mω
2
2
~2ψ′′(p) = Eψ(p).
Compare this to the position-basis equation
mω2
2
x2ψ(x)− 1
2m
~2ψ′′(x) = Eψ(x).
These are the same equations with the substitutions x↔ p and m↔ 1/(mω2).
86
Problem 2: Exercise 7.3.5
〈n|X|n〉 =
∫ ∞
−∞
dxψ∗n(x)·x·ψn(x) =
∫ ∞
−∞
dx x·ψ2n(x) since ψn(x) is real
= 0 since x is odd and ψ2n(x) is even.
〈n|P |n〉 =
∫ ∞
−∞
dxψ∗n(x)(−i~)
d
dx
ψn(x) = (−i~)
∫ ∞
−∞
dxψnψ
′
n = −
i~
2
∫ ∞
−∞
dx (ψ2n)
′
= −i~
2
ψ2n
∣∣∞
−∞ = 0 since ψn → 0 as |x| → ∞.
〈1|X2|1〉 =
∫ ∞
−∞
dxψ∗1x
2ψ1 =
(mω
4π~
)1/2 ∫ ∞
−∞
dx x2
[
2x
(mω
~
)1/2]2
e−mωx
2/~
=
2√
π
(mω
~
)3/2 ∫ ∞
−∞
dx x4e−mωx
2/~ =
2√
π
(mω
~
)3/2 3√π
4
(mω
~
)−5/2
=
3~
2mω
.
〈1|P 2|1〉 =
∫ ∞
−∞
dxψ∗1(−i~)2ψ′′1 = −~2
2√
π
(mω
~
)3/2 ∫ ∞
−∞
dx xe−mωx
2/2~
(
xe−mωx
2/2~
)′′
= −~2 2√
π
(mω
~
)3/2 ∫ ∞
−∞
dx x2
(mω
~
) [(mω
~
)
x2 − 3
]
e−mωx
2/~
= −~2 2√
π
(mω
~
)5/2
·(−1)3
√
π
4
(mω
~
)−3/2
=
3mω~
2
.
∆X2 = 〈0|X2|0〉 =
√
mω
π~
∫ ∞
−∞
dx x2e−mωx
2/~ =
(mω
~
)1/2 √π
2
(mω
~
)−3/2
=
~
2mω
.
∆P 2 = 〈0|P 2|0〉 = −~2
√
mω
π~
∫ ∞
−∞
dx e−mωx
2/2~
(
e−mωx
2/2~
)′′
= − ~
2
√
π
(mω
~
)3/2 ∫ ∞
−∞
dx
(mω
~
x2 − 1
)
e−mωx
2/~
= − ~
2
√
π
(mω
~
)3/2
·(−1)
√
π
2
(mω
~
)−1/2
=
mω~
2
.
∴ ∆X ∆P =
√
~
2mω
·
√
mω~
2
=
~
2
.
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
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127
128
129
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131
132
133
134
135
136
137
138
139
140
141
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149
Physics 828: Homework Set No. 3
Due date: Friday, January 27, 2011, 1:00pm
in PRB M2043 (Biao Huang’s office)
Total point value of set: 60 points + 10 bonus points
Problem 1 (20 pts.): Exercise 10.3.5 (Shankar, p. 278)
Problem 2 (10 pts.): Exercise 10.3.6 (Shankar, p. 278)
Problem 3 (5 pts.): Exercise 11.2.2 (Shankar, p. 283)
Problem 4 (5 pts.): Exercise 11.4.1 (Shankar, p. 300)
Problem 5 (5 pts.): Exercise 11.4.2 (Shankar, p. 300). If you correctly derive in closed
form the explicit expression for [P̂ , Ĥ] you receive 10 bonus points.
Problem 6 (10 pts.): Exercise 11.4.3 (Shankar, p. 300)
Problem 7 (5 pts.): Exercise 11.4.4 (Shankar, p. 300)
1
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
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167
168
169
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171
172
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174
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184
Physics 828 Sketch of Solution to Set 2
Shankar 12.3.7
(1) The two-dimensional harmonic oscillator is obviously invariant under rotations
about the z-axis: the magnitude of the position and momenta are unaltered by rota-
tions around the z-axis. Therefore, the Hamiltonian commutes with the generatorof
rotations about the z-axis, Lz.
(2) So we write ψ(ρ, φ) = eimφ REm(ρ) where m is an integer, positive or negative.
In two dimensions since we know the Laplacian we have
− h̄
2
2µ
(
R′′Em +
1
ρ
R′Em
)
+
(
h̄2m2
2µρ2
+
1
2
µω2ρ2
)
REm(ρ) = EREm .
For small ρ assuming that REm(ρ) ∝ ρk (derivatives decrease the power by unity
increasing its importance for small ρ etc.) we can neglect the potential energy term
and the constant term on the right-hand side. Thus we have
R′′Em +
R′Em
ρ
∝ m
2
ρ2
REm ⇒ [k(k − 1) + k] = m2 ⇒ k2 = m2 .
For the wave function to be normalizable, for
∫∞
0 dρ ρR
2
Em(ρ) to be finite at the lower
limit k ≥ 0. Thus we have REm(ρ) ρ→0−→ ρ|m|.
(3) For large ρ, terms with inverse powers of ρ including the centrifugal term and
also the constant term on the right-hand side can be neglected. So we have
R′′Em =
µ2ω2ρ2
h̄2
REm ;
this is identical to the one-dimensional oscillator equation. See the careful analysis on
page 191. Up to powers of ρ the solution is1
REm(ρ)
ρ→∞−→ e−µω2h̄ ρ2 .
So we write as instructed
1
R′Em = −
µω
h̄
ρ e−
µω
2h̄ ρ
2
and R′′Em =
µω
h̄
(
−1 + µω
h̄
ρ2
)
e−
µω
2h̄ ρ
2
.
Thus we can neglect the constant term in R′′ and recover the solution. One can use this to check
that terms we have neglected are indeed small compared to the terms we have retained. When you
neglect terms it is a good idea to substitute the solution you have obtained and check that the terms
you have neglected are indeed smaller.
1
185
REm(ρ) = Um(ρ) ρ
|m| e−
µω
2h̄
ρ2 .
(4) Use the dimensionless variable ² = E
h̄ω
and y2 = µω
h̄
ρ2. Dividing the radial
equation by h̄ω we have dropping the annoying subscripts
[
−1
2
(
d2
dy2
+
1
y
d
dy
− m
2
y2
)
+
1
2
y2
]
R(y) = ²R(y) .
(5) We do the substitution and do elementary calculus and obtain the result given:
U ′′ +
[(
2|m|+ 1
y
)
− 2y
]
U ′ + (2²− 2|m| − 2) U = 0 .
(6) We substitute U(y) =
∑∞
r=0 Cry
r and collect the coefficient of yr. The second
derivative reduces the power by two and so we use the Cr+2 term etc.
(r + 2)(r + 1)Cr+2 + (2|m|+ 1)(r + 2)Cr+2 − 2rCr + (2²− 2|m| − 2)Cr = 0
yielding a two-term recursion relation.
(7) We write this as
Cr+2
Cr
= − (2(²− |m| − r − 1)
(r + 2)(2|m|+ r + 2) .
First if C0 is given C2 and the other even terms can be computed. As r →∞ we have
Cr+2
Cr
→ 2
r
.
This implies that U(y) grows as ey
2
which overwhelms the e−y
2/2 in R pushing it out
of the Hilbert space. So the series must terminate. Thus the boundary condition at
infinity leads as usual to energy quantization.
What about the odd terms? A series only with odd terms (set C0 = 0 so that all even
terms vanish) is inconsistent since then U(y) ∼ y for small y and thus R(ρ) ∼ ρ|m|+1
inconsistent with our earlier result in (2). This appears to be suggested as an argu-
ment. What if one starts with C0 and C1 non-zero? Substituting into the equation for
U we find that the (1/y)(dU/dy) leads to the term C1/y and there is no other source
of y−1 terms. Thus C1 = 0 and therefore, all odd terms vanish.
2
186
Therefore, we have r = 2k and the termination of the series condition yields
² = 1 + r + |m| = |m| + 2k + 1 ≡ (n + 1)
with k = 0, 1, , 2, · · ·.
(8) Since |m| = n − 2k for a given n (i.e., for a given energy) the maximum value
of m is n which occurs for k = 0, The azimuthal quantum number m decreases by steps
of 2 until m reaches the value −n. It is easy to se that there are n + 1 allowed values
of m yielding a degeneracy of n+1. In Cartesian coordinates the energy is nx +ny +1
in units of h̄ω. So the degeneracy corresponds to the number of ways in which we can
choose two non-negative integers to add up to n. We can choose nx to be any integer
from 0 to n and ny = n− nx. This yields the same degeneracy.
Shankar 12.6.1
(1) Since there is no angular dependence ` = 0.
(b) We have R(r) ∝ e−r/a0 and for large r (retaining only the dominant terms)
− h̄
2
2m
R′′ = ER(r) .
Substituting the given form we obtain E = − h̄2
2ma20
.
(c) Clearly the R′′ term and the energy term cancel for all r. If the equation is
valid for all r we must have
− h̄
2
2m
2
r
R′ + V (r) R(r) = 0 .
Substituting R(r) = e−r/a0 we obtain
h̄2
ma0r
+ V (r) = 0 ⇒ V (r) = − h̄
2
ma0r
.
Shankar 13.1.1 and 13.1.3 You should be able to fill in the steps. Here are some
steps dropping some subscripts for notational simplicty.
v =
∞∑
k=0
Ck ρ
k+`+1 .
3
187
Substituting into
v′′ − 2v′ +
(
e2λ
ρ
− `(` + 1)
ρ2
)
v = 0 (1)
we extract the coefficient of ρk+l carefully. Since two derivatives reduce the power of
ρ by two we should start from the term with ρk+`+2 with coefficient Ck+1 for the first
term and similarly for the last term. For the second and third terms which reduce the
power of ρ by unity we can start with the ρk+`+1 term with coefficient Ck. Thus we
have
(k + ` + 2)(k + ` + 1)Ck+1 − 2(k + ` + 1)Ck + e2λCk − `(` + 1)Ck+1 = 0 (2)
which yields
Ck+1
Ck
=
−q2λ + 2(k + ` + 1)
(k + ` + 2)(k + ` + 1) − `(` + 1) (3)
Since λ2 = 2m
h̄2W
from 13.1.9. For the numerator to vanish we have
e4λ2 = −2me
2
h̄2E
= 4(k + ` + 1)2 .
This yields 13.1.14.
Mathematical aside: Here is a different representation using classical mathemat-
ical physics. Note that the function L(ρ) ≡ ∑∞k=0 Ck ρk obeys the equation
ρL′′(ρ) + [ 2(` + 1) − 2ρ ]L′(ρ) + (q2λ − 2(` + 1))L(ρ) = 0 .
Substituting q2λ = 2n (where n is an integer eventually) we have
1
2
ρL′′ + [ (` + 1) − ρ ]L′ + (n − (` + 1))L = 0 .
Let z = 2ρ we have
z
d2L
dz2
+ [2(` + 1) − z]dL
dz
− (` + 1− n)L = 0 .
We know (with a good mathematical methods course) that the general solution to
zw′′ + (c− z) w′ − aw = 0
is given by the confluent hypergeometric function w = 1F1(a; c; z). Thus we find that
the solution L(ρ) is 1F1(`+1−n; 2`+2; 2ρ) . The particular terminating (for integer n)
4
188
confluent hypergeometric function can be related to the associated Laguerre polynomial.
Shankar 13.3
We are considering the case n = 2 and ` = 1 and thus k = 0 in the notation of the
text. So we have W = me
4
8h̄2
from 13.1.4 and form 13.1.6
ρ =
√
2mW
h̄2
r =
me2
2h̄2
r =
r
2a0
using 13.1.24. From 13.1.10 since k = 0 and ` = 1 we have v = C0ρ
2 and thus
R(r) =
U(r)
r
= C e
− r
2a0 r
where C is an overall constant. We know that Y10 =
√
3
4π
cos θ from 12.5.39 which is
normalized. So all we need is that Ce−r/(2a0)r is notmalized when integrated over the
radial coordinate. We have
C2
∫ ∞
0
dr r2 r2 e
− r
a0 = C2 24 a50 .
Thus C =
√
1
24a30
× 1/a0 and including the normalization from the spherical harmonic
yields the quoted answer.
Shankar 13.5
Since we are asked to compute 〈Ω〉 for stationary states |n`m〉 its time derivative
vanishes. Thus we have 〈[Ω, H]〉 = 0 by Ehrenfests’ theorem. So we compute the
commutator for Ω = ~R · ~P as ordered. We calculate (using the summation convention)
[
RjPj,
PiPi
2m
]
=
[
Rj,
PiPi
2m
]
Pj = 2ih̄
~P · ~P
2m
= 2ih̄ T .
We have used
[Rj, PiPi] = Pi[Rj, Pi] + [Rj, Pi]Pi = 2ih̄ Pj .
We consider the potential energy term next:
[RjPj, V (R)] = Rj[Pj, V (R)] = Rj
(
−ih̄ d
dRj
V (R)
)
.
5
189
Note that [Pi, V (R)] can be evaluated in the coordinate representation (in Cartesian
coordinates) by acting on a function f(r):
[Pi, V (r)]f(r) = −ih̄
(
∂
∂ri
V (r)f(r) − V (r) ∂
∂ri
V (r)f(r)
]
= −ih̄ ∂
∂ri
V (r) = −ih̄ ~R·~∇V.
Thus we can write formally [Pi, V (R)] = −ih̄∂V (R)/∂Ri. So we need to evaluate
~R · ~∇V (R). In the coordinate representation using spherical coordinates for central
potentials this is just rV ′(r). For the Coulomb potential we obtain−V (r) and including
the factor of −ih̄ we obtain ih̄ V (r). Substituting into the basic relation we have
〈2T + V 〉 = 0
as asserted. If V (R) ∝ Rn, rV ′(r) = nV and thus we obtain 〈T 〉 = n
2
〈V 〉.
6
190
Physics 710 March 12, 2010
Problem Set 15
Problem 12.6.1: ψE = Ae
−r/a0 .
(1) No (θ, φ)-dependence implies ψE ∝ Y 00 , so we must have ` = 0 and m = 0.
(2) Therefore ψE = RE,`=0 =
1
r
UE,0 which satisfieseqn. (12.6.5) with ` = 0:
(rψE)
′′ +
2µ
~2
[E − V (r)](rψE) = 0. (1)
As r → ∞, V (r) → 0, which implies in this limit (rψE)′ = A(1 − ra0 )e
−r/a0 ≈
−(A/a0)re−r/a0 , so (rψE)′′ ≈ (A/a20)re−r/a0 . Therefore, eqn. (1) reads in this
limit
A
a20
re−r/a0 = −2µE
~2
Are−r/a0 ,
from which it follows that
E = − ~
2
2µa20
.
(3) Now plug E into (1) and use (rψE)
′′ = A(− 2
a0
+ r
a20
)e−r/a0 to get
A
(
− 2
a0
+
r
a20
)
e−r/a0 +
2µ
~2
(
− ~
2
2µa20
− V (r)
)
Are−r/a0 = 0,
which gives
V (r) = − ~
2
µa0r
.
Problem 12.6.4:
(1) δ3(r − r′) is defined by the property that
∫
d3rδ3(r − r′)f(r) = f(r′). So simply
check: ∫
r2dr sin θdθdφ
1
r2 sin θ
δ(r − r′)δ(θ − θ′)δ(φ− φ′)f(r, θ, φ)
=
∫
drdθdφδ(r − r′)δ(θ − θ′)δ(φ− φ′)f(r, θ, φ) = f(r′, θ′, φ′).
(2) If r 6= 0 then∇2(1
r
) = 1
r2
∂
∂r
(r2 ∂
∂r
(1
r
))+(angular parts) = 1
r2
∂
∂r
(r2(−1
r2
)) = 1
r2
∂
∂r
(−1) =
0. When r = 0 the above calculation breaks down since terms are singu-
lar there. So consider an arbitrary continuous function f(r) and the integral∫
d3x∇2(1
r
)f(r) = lim�→0
∫ �
0
r2dr
∫
dΩ∇2(1
r
)f(r), since ∇2(1
r
) = 0 for r > 0.
Then, integrating by parts we get∫
d3x∇2
(
1
r
)
f(r) = lim
�→0
∫ �
0
r2dr
∫
dΩ
{
~∇·
(
f(r)~∇
(
1
r
))
−
(
~∇f(r)
)
·~∇
(
1
r
)}
.
191
Now recall that in spherical coordinates ~∇ = r̂ ∂
∂r
+ θ̂ 1
r
∂
∂θ
+ φ̂ 1
r sin θ
∂
∂φ
, implying
that ~∇(1/r) = r̂ ∂
∂r
(1/r) = −r̂/r2, so∫
d3x∇2
(
1
r
)
f = lim
�→0
∫ �
0
r2dr
∫
dΩ
{
~∇·
(
−r̂ f
r2
)
+
(
r̂·~∇f
) 1
r2
}
.
The second term on the right side vanishes, because as �→ 0, r̂·~∇f = (∂f/∂r)|r=0 =
const., so:
lim
�→0
∫ �
0
r2dr
∫
dΩ
(
r̂·~∇f
) 1
r2
= const.· lim
�→0
∫ �
0
r2dr
∫
dΩ
1
r2
= const.· lim
�→0
4π� = 0.
Therefore,∫
d3x∇2
(
1
r
)
f = lim
�→0
∫ �
0
r2dr
∫
dΩ~∇·
(
−r̂ f
r2
)
= lim
�→0
∫
dΩr2r̂·
(
−r̂ f
r2
)∣∣∣∣
r=�
= − lim
�→0
∫
dΩ f |r=� = −f(0) lim
�→0
(∫
dΩ
)
= −f(0) lim
�→0
4π
= −4πf(0).
In the second step I used the divergence theorem which states
∫
R
d3x~∇·~g =∫
∂R
d2a n̂·~g where R is any region, ∂R is its boundary, n̂ is the normal unit vector
to ∂R (pointing out of R) and d2a is the surface area element. In our case
R = {r < �}, d2a = dΩ, n̂ = r̂, and ~g = −r̂(f/r2). Thus we have shown that
∇2(1/r) = 0 for r 6= 0 and for any f(~r) that
∫
d3x∇2(1/r)f = −4πf(0). This is
the definition of the delta function, so
∇2
(
1
r
)
= −4πδ3(r).
Problem 12.6.9: Since ` = 0, ψ = R(r)Y 00 (θ, φ) = R(r). So the radial equation
becomes, with ψ(r) = (1/r)U(r),(
d2
dr2
+ k2
)
Uin = 0 r ≤ r0, k ≡
√
2µ(E + V0)
~2
,(
d2
dr2
− κ2
)
Uout = 0 r ≥ r0, κ ≡
√
−2µE
~2
,
where k and κ are defined to be the positive root. The solutions of these equations are
Uin = A sin kr + Ã cos kr,
Uout = Be
−κr + B̃e+κr.
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The boundary conditions are that Uin → 0 at r = 0, implying à = 0, and Uout → 0
at r =∞, implying that B̃ = 0. The matching conditions at r = r0 are the continuity
of ψ and its first derivative:
ψin = ψout ⇒ Uin = Uout|r=r0 , (2)
ψ′in = ψ
′
out ⇒ (Uin/r)′ = (Uout/r)′|r=r0 . (3)
Eqn. (2) implies A sin kr0 = Be
−κr0 , or,
B = A sin kr0 e
κr0 . (4)
Eqn. (3) implies d
dr
[(A/r) sin kr − (B/r)e−κr]r=r0 = 0, which gives
−A
r20
sin kr0 +
Ak
r0
cos kr0 +
B
r20
e−κr0 +
Bκ
r0
e−κr0 = 0.
Plugging (4) into this gives (Ak/r0) cos kr0 + (Aκ/r0) sin kr0 = 0, or
− tan kr0 =
k
κ
, (5)
which is what we wanted to show.
If V0 < π
2~2/(8µr20) and −V0 < E < 0 (for a bound state), then k2 =
2µ
~2 (E + V0) <
2µ
~2π
2~2/(8µr20) = π2/(4r20). Thus 0 < kr0 < (π/2), which implies that − tan kr0 < 0.
On the other hand, by definition, k/κ > 0. So there is no solution to eqn. (5).
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Physics 710-712 April 2, 2010
Problem Set 16
Problem 13.3.1: To say the pion has a range of λ ' 10−5 angstroms is to say that
a single pion can be localized on this scale: ∆X ∼ λ. From the uncertainty principle
∆X ∆P & ~, we then deduce ∆P & ~/λ. Since we assume λ is the smallest scale
on which we can localize the pion, it is plausible that the inequality is saturated, so
P ∼ ∆P ∼ ~/λ. The relation between the energy and momentum is (from special
relativity) E2 = m2c4 + c2P 2 ∼ m2c4 + c2~2/λ2. Now, for the notion of a “single
pion” to exist, we must have E . 2mc2. (See discussion in text on p. 363). So
4m2c4 & m2c4 + c2~2/λ2, or mc2 & c~/(
√
3λ). Again, since λ is the smallest scale, it is
plausible that the inequality is saturated, giving
mc2 ∼ c~√
3λ
∼ 2000 eV
◦
A
1.7× 10−5
◦
A
∼ 100 MeV.
(In reality mπc
2 = 140 MeV.)
Problem 13.3.2: Since the kinetic energy is T = 200 eV � 0.5 MeV ' mc2, the
electron is non-relativistic, so we can use T = p2/(2m) = (p2c2)/(2mc2) which implies
pc =
√
2mc2T . Then
λ =
2π~
p
=
2π~c
pc
=
2π~c√
2mc2T
'
√
2π(2000 eV
◦
A)√
(0.5× 106eV)(200 eV)
=
√
2π
10
◦
A' 1
◦
A .
Problem 13.3.3: Recalling En = −Ry/n2 ' −13eV/n2, we have
P (n = 2)
P (n = 1)
= 4e−(E2−E1)/kT = 4e−[(−1/4)−(−1)]13 eV/(kBT ) ' 4e−105/(T/K)
where I used (kB/eV) ∼ 9 × 10−5K−1. So it is clear that we need T & 105 K so that
the exponent is not very small. For example, if T = 6000 K, then
P (n = 2)
P (n = 1)
' 4e−105/(6000) ' 4e−16 ∼ 2× 10−7 � 1.
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209
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215
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Physics 710-712 May 14, 2010
Problem Set 21
Problem 17.2.1: With H0 = 1
2m
P 2 + 1
2
mω2X2 and H1 = λX4, and using that
X =
√
~/(2mω)(a+ a†), we have:
(1):
E1n = 〈n|H1|n〉 =
λ~2
4m2ω2
〈n|(a+ a†)4|n〉
=
λ~2
4m2ω2
〈n|(a2a†2 + aa†aa† + a†a2a† + aa†2a+ a†aa†a+ a†2a2)|n〉
=
λ~2
4m2ω2
〈n|(6a2a†2 − 12aa† + 3)|n〉
=
λ~2
4m2ω2
{6〈n+ 2|(n+ 1)(n+ 2)|n+ 2〉 − 12〈n+ 1|(n+ 1)|n+ 1〉+ 3〈n|n〉}
=
λ~2
4m2ω2
{6(n+ 1)(n+ 2)− 12(n+ 1) + 3} = 3~
2λ
4m2ω2
(2n2 + 2n+ 1),
where in the first line we dropped the zero superscripts from the unperturbed eigen-
states; in the second line we kept only terms with equal numbers of a’s and a†’s; in the
third line we used [a, a†] = 1; and in the fourth line we used a†|n〉 =
√
n+ 1|n+ 1〉.
(2): For any finite value of λ, as n gets large
E1n
∆E
∼ ~
2λn2/(m2ω2)
~ω
∼ n2 ~λ
m2ω3
� 1.
Physically, at large x, no matter how small λ is,
V (x) =
mω2
2
x2 + λx4 ∼ λx4
for λ 6= 0.
Problem 17.2.2: H = −~µ· ~B = −γ~S· ~B = H0 + H1 with H0 = −γB0Sz and
H1 = −γBSx. The H0 eigenvalues are E0± = ∓γB0~/2 with eigenstates |±〉
0, the Sz
eigenstates. Then
E1± = 〈±|
0H1|±〉0 = −γB〈±|0Sx|±〉0 = −
γB
2
〈±|0(S+ + S−)|±〉0 = 0,
and
E2± =
∑
m
′ |〈±|0H1|m〉0|2
E0± − E0m
=
|〈±|0H1|∓〉0|2
E0± − E0∓
=
γ2B2
4
|〈±|0(S+ + S−)|∓〉0|2
∓γB0~
= ∓ γB
2
4B0~
|〈±|0S±|∓〉0|2 = ∓
γB2
4B0~
|〈±|0~
√
(1
2
∓ (∓1
2
))(3
2
± (∓1
2
))|∓〉0|2 = ∓γ~B
2
4B0
,
261
and
|±〉′ =
∑
m
′ |m〉0〈m|0H1|±〉0
E0± − E0m
=
|∓〉0〈∓|0H1|±〉0
E0± − E0∓
= −γB
2
|∓〉0〈∓|0S∓|±〉0
∓γB0~
= ± B
2B0
|∓〉0,
where we used that 〈∓|0S∓|±〉0 = ~, as computed in the previous equation. Therefore
E± = E
0
± + E
1
± + E
2
± +O(B3) = ∓
γB0~
2
∓ γB
2~
4B0
= ∓γB0~
2
(
1 +
B2
2B20
)
,
and
|±〉 = |±〉0 + |±〉1 +O(B2) = |±〉0 ± B
2B0
|∓〉0.
Now compare to the exact answer. H = −γ(B0Sz + BSx) = −γ
√
B20 +B
2n̂·~S, where
n̂ = (B0k̂ + Bı̂)/
√
B20 +B
2. The eigenvalues of n̂·~S are ±~/2 with eigenvectors |±〉
with
|+〉 = cos θ
2
e−iφ/2|+〉0 + sin θ
2
eiφ/2|−〉0, (1)
|−〉 = − sin θ
2
e−iφ/2|+〉0 + cos θ
2
eiφ/2|−〉0,
where |±〉0 are the Sz eigenvectors. Eqn. (1) can be rewritten as
|±〉 = cos θ
2
e∓iφ/2|±〉0 ± sin θ
2
e±iφ/2|∓〉0,
and
cos θ
2
=
√
1 + cos θ
2
=
√
1 + nz
2
=
√√
B20 +B
2 +B0
2
√
B20 +B
2
= 1 +O(B2),
sin θ
2
=
√
1− cos θ
2
=
√
1− nz
2
=
√√
B20 +B
2 −B0
2
√
B20 +B
2
=
B
2B0
+O(B3),
and φ = 0, since n̂ is in the x-z plane. This implies
|±〉 = |±〉0 ± B
2B0
|∓〉0 +O(B2),in agreement with the perturbation theory result. Similarly, the exact eigenvalues are
E± = ∓
γ~
2
√
B20 +B
2 = ∓γ~B0
2
(
1 +
B2
B20
)1/2
= ∓γ~B0
2
(
1 +
B2
2B20
+O(B4)
)
,
again in agreement with the perturbation theory result.
262
Problem 17.3.2: H = AS2z + B(S
2
x − S2y) on the spin-1 Hilbert space. In the Sz
basis, |m〉 (m = ±1, 0), we have
Sz = ~
1 0 00 0 0
0 0 −1
 , Sx = ~√
2
0 1 01 0 1
0 1 0
 , Sy = i~√
2
0 −1 01 0 −1
0 1 0

(see p. 328 of the text). This implies
H = ~2
A 0 B0 0 0
B 0 A
 . (2)
Clearly
|m〉 = |0〉 =
01
0

is the eigenvector with E = 0. So we only need to look at the |m〉 = |±1〉 subspace
where H = H0 +H1 with
H0 = ~2A
(
1 0
0 1
)
, H1 = ~2B
(
0 1
1 0
)
on the |m〉 = {|±1〉} subspace.
H0 is degenerate: H0|m〉 = ~2A|m〉 for m = ±1. The basis stable under H1 is the one
which diagonalizes H1. Since(
0 1
1 0
)(
1
1
)
=
(
1
1
)
and
(
0 1
1 0
)(
1
−1
)
= −
(
1
−1
)
,
the eigenvectors of H1 are∣∣1〉 ≡ 1√
2
(|1〉+ |−1〉) and
∣∣−1〉 ≡ 1√
2
(|1〉 − |−1〉) .
To order O(B), the energy shifts of
∣∣1〉 and ∣∣−1〉 are
E11 =
〈
1
∣∣H1∣∣1〉 = ~2B 1√
2
(
1 1
)(0 1
1 0
)(
1
1
)
1√
2
= ~2B,
E1−1 =
〈
−1
∣∣H1∣∣−1〉 = ~2B 1√
2
(
1 −1
)(0 1
1 0
)(
1
−1
)
1√
2
= −~2B.
So, the eigenvalues, to O(B), are
E1 = ~2(A+B), E−1 = ~2(A−B), E0 = 0.
Compare this to the exact eigenvalues of (2) given by 0 = det(H−λ) = λ(~2(B−A) +
λ)(~2(B+A)−λ), which implies λ = {0, ~2(A−B), ~2(A+B)}. So, the O(B) results
are exact.
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Physics 710-712 May 26, 2010
Problem Set 22
Problem 18.2.2: To first order, the amplitude d2`m(∞) for the atom to be in the
|n = 2, `,m〉 state is
d2`m(∞) = −
i
~
∫ ∞
−∞
〈2`m|(eEZ)|100〉e−t2/τ2eiω21tdt
since the potential H1(t) = −~µ· ~E = eZEe−t2/τ2 . Here ω21 = (E2 − E1)/~, which from
now on we’ll just call ω. So we need to evaluate 〈2`m|Z|100〉. Z is a component
of a vector irreducible tensor operator, Z = T 01 , so by angular momentum selection
rules, only the 〈210|Z|100〉 matrix element is non-vanishing. This matrix element is
easily evaluated using the wave functions for the the |210〉 and |100〉 states, and that
Z = r cos θ:
〈210|Z|100〉 =
∫ ∞
0
r2dr
∫
dΩ
(
1
25πa30
)1/2(
r
a0
e−r/2a0 cos θ
)
(r cos θ)
(
1
πa30
)1/2
e−r/a0
= 2π
(∫ 1
−1
d(cos θ) cos2 θ
)
·
(∫ ∞
0
dr r4e−3r/2a0
)
· 1
25/2πa40
= 2π·2
3
·
(
2a0
3
)5(∫ ∞
0
dρ ρ4e−ρ
)
· 1
25/2πa40
= 215/2·3−5·a0.
Therefore
d2`m(∞) = δ`1δm0
(
−i
~
)
eE·215/2·3−5·a0
∫ ∞
−∞
e−t
2/τ2eiωtdt
= δ`1δm0
(
−ieE
~
)
215/2·3−5·a0
√
πτ 2e−ω
2τ2/2,
and so the probability for the transition is
P (n=2) =
∑
`,m
|d2`m|2 =
(
eE
~
)2(
215a20
310
)
πτ 2e−ω
2τ2/2.
This answer does not depend on the electron spin: since Sz is conserved by H
0 and
H1, there is still only a single final state it can go into.
Problem 18.2.4: The kinetic energy of the emitted electron is 16 keV = 1
2
mv2e =
1
2
mc2(ve/c)
2 = 1
2
(511keV)(ve/c)
2, which implies that ve/c ∼ 1/4. Therefore the time,
τ , for emission is τ ∼ a0/ve = (a0/c)/(ve/c) ∼ 4a0/c, since the typical size of the
1s electron orbit is r ∼ a0, the Bohr radius. In comparison, the characteristic time
scale of the 1s electron, T , is T ∼ a0/vs = (a0/c)/(vs/c) = (a0/c)/α ∼ 140a0/c, since
279
the typical velocity of an electron bound in the hydrogen atom is vs/c = α, the fine
structure constant. Therefore, T � τ , and the sudden approximation is appropriate.
In the sudden approximation, right after emission the 1s electron will be in the same
state, the |100〉(Z=1) state of hydrogen. Therefore, the amplitude for the electron to be
in the |100〉(Z=2) state of (He
3)+ is given by the overlap
(Z=2)〈100|100〉(Z=1) =
∫ ∞
0
r2dr
∫
dΩ
(
Z3
πa30
)1/2
e−rZ/a0·
(
1
πa30
)1/2
e−r/a0
= 4π·2
3/2
πa30
∫ ∞
0
r2dre−3r/a0 =
27/2
a30
(a0
3
)3 ∫ ∞
0
ρ2dρe−ρ = 29/2·3−3,
where in the first line Z = 2 and I used the fact that under changing the hydrogen
nucleus charge from 1 to Z, all that changes is a0 → a0/Z.
Finally, (Z=2)〈16, 3, 0|100〉(Z=1) = 0 since ` = 3 states are orthogonal to ` = 0 ones
(and the radial part, and therefore Z, does not affect this).
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Problem S18.4.3 (Shankar, page 494): 
 
(1) Show that a gauge transformation on potentials 0φ and 0A using 
 ( )0 , ( , )
t
r t c r t dtφ
−∞
′ ′Λ = − ∫ (1.1) 
gives ( )1 , 0r tφ = and 1 0 0( , )A r t A= −∇Λ 
Solution: This is problem 1 of HW8. 
 
 
(2) Show that if we transform once more to 2φ and 2A using 
 
 311
( , )1
4
A r t d r
r rπ
′ ′∇ ′Λ = −
′−∫
i (1.2) 
then 2( , ) 0A r t∇ =i . 
 
Solution: Using equation (18.4.12) from Shankar ( A A′ = −∇Λ ),we have 
 312 1
( , )1( , ) ( , )
4
A r tA r t A r t d r
r rπ
⎡ ⎤′ ′∇ ′= + ∇ ⎢ ⎥′−⎣ ⎦
∫
i (1.3) 
Note that the integrand is the gradient of a scalar function, which produces a vector. If 
we now take the divergence of this, we get 
 2 312 1
( , )1( , ) ( , )
4
A r tA r t A r t d r
r rπ
⎡ ⎤′ ′∇ ′∇ = ∇ + ∇ ⎢ ⎥′−⎣ ⎦
∫
ii i (1.4) 
Since the only r-dependence in the integrand comes from the denominator, we can use 
the identity 
 ( ) ( )2 31/ 4r r r rπδ′ ′∇ − = − − (1.5) 
Substituting this into (1.4) and doing the integrals over 3d r gives 
 [ ]2 1 1 1 1
1( , ) ( , ) 4 ( , ) ( , ) ( , ) 0
4 r r
A r t A r t A r t A r t A r tπ
π ′=
′ ′∇ = ∇ + − ∇ = ∇ −∇ =i i i i i (1.6) 
 
(3) Verify that 2φ is also zero using 0 0E∇ =i . 
 
Solution: 
 312 1 1
( , )1 1( , )
4
A r tr t d r
c t c t r r
φ φ φ
π
⎧ ⎫′ ′∇∂Λ ∂ ⎪ ⎪′= + = − ⎨ ⎬′∂ ∂ −⎪ ⎪⎩ ⎭
∫
i (1.7) 
Incorporating the results of part (1), this becomes 
285
 
( )
( )
0
0
3
2 1
0 3
( , ) ,
1 1( , ) 0
4
,1
4
A r t c r t
tr t d r
c t c t r r
cE r t
d r
c t r r
φ
φ φ
π
π
′⎧ ∂ ⎫⎡ ⎤′ ′ ′∇ + ∇⎪ ⎪⎢ ⎥∂Λ ∂ ⎪ ∂ ⎪⎣ ⎦ ′= + = − ⎨ ⎬′∂ ∂ −⎪ ⎪
⎪ ⎪⎩ ⎭
⎧ ⎫′ ′∇ −⎡ ⎤∂ ⎪ ⎪⎣ ⎦ ′= ⎨ ⎬′∂ −⎪ ⎪⎩ ⎭
∫
∫
i
i
 (1.8) 
If we are in free space, so 0ρ = (a condition that was implied but not explicitly stated), 
then Maxwell’s first equation tells us that the numerator of the integrand is zero; so 
2 ( , ) 0r tφ = . 
 
 
 
 
 
 
 
 
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Physics 215C Homework #1 Solutions
Richard Eager
Department of Physics
University of California; Santa Barbara, CA 93106
Shankar 20.1.1
Derive the continuity equation
∂P
∂t
+∇ · j = 0
where P = ψ†ψ and j = cψ†αψ
The Dirac equation takes the equivalent forms
i~
∂|ψ〉
∂t
=
(
cα · P + βmc2
)
|ψ〉
∂ψ
∂t
=
(
−cα · ∇ − i
~
βmc2
)
ψ
The conjugate equation is
∂ψ†
∂t
=
(
−cα · ∇+ i
~
βmc2
)
ψ†
∂P
∂t
= ψ†
∂ψ
∂t
+
∂ψ†
∂t
ψ
= −c
(
ψ†α · ∇ψ + (α · ∇ψ†)ψ
)
∇ · j = c
(
∇ψ†αψ + ψ†α · ∇ψ
)
Adding both equations together yields the desired result
∂P
∂t
+∇ · j = 0
Both terms in the Hamiltonian, cα · P and βmc2 are Hermitian, so where did
the relative minus sign come from? To show that P is a Hermitian operator you
need to integrate by parts. If you are working with the L2 inner product you
can drop boundary terms, but when working locally to derive the continuity
equation we don’t integrate by parts and get a relative minus sign.
1
317
2
Show that the probability current j of the previous exercise reduces in the non-
relativistic limit to Eq.(5.3.8) [which is the same as Sakurai Eq.(2.4.16)].
The probability current j = cψ†αψ and α =
(
0 σ
σ 0
)
We can write the wave
function ψ =
(
χ
Φ
)
in terms of its relativistic and non-relativistic components,
Φ and χ respectively.
j = c
(
χ† Φ†
)(0 σ
σ 0
)(
χ
Φ
)
(0.1)
= c
(
χ†σΦ + Φ†σχ
)
(0.2)
In the non-relativistic limit (20.2.13)
Φ ≈ σ · π
2mc
j = χ†
p̂
2m
χ+ (
p̂
2m
χ†)χ
which is the non-relativistic current (5.3.8)
j =
~
2mi
(ψ∗∇ψ − ψ∇ψ∗)
from the identification p̂ = −i~∇.
Shankar 20.2.1
Show that
π × π = iq~
c
B
where π = P − qAc .
π × π = P × P − qA
c
× P − P × qA
c
+qA
c
× qA
c
= −q
c
(A× P + P ×A)
=
iq~
c
(A×∇+∇×A)
The simplest way to manipulate operators is to act on a test function ψ
iq~
c
(A×∇+∇×A)ψ = iq~
c
(A×∇ψ +∇× (ψA))
=
iq~
c
(A×∇ψ + (∇ψ)A+ (∇×A)ψ)
=
iq~
c
(∇×A)ψ
=
iq~
c
Bψ
Therefore
π × π = iq~
c
B
2
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Shankar 20.1.1
Solve for the 4 spinors w that satisfy Shankar Eq. 20.3.3. You may assume that
the 3- momentum ~p is along the z-axis. Normalize them to unity, and show that
they are mutually orthogonal.
Equation (20.3.3) is
Ew = (α · p+ βm)w
In terms of the relativistic and non-relativistic components,(
E −m −σ · p
−σ · p E +m
)(
χ
Φ
)
=
(
0
0
)
The solutions are given in equations (20.3.7) and (20.3.8). Choosing the basis(
1
0
)
and
(
0
1
)
for Φ and letting p be in the z direction,
w1,3 =

p/(±E −m)
0
1
0
w2,4 =

0
p/(±E −m)
0
1

Orthogonality of the spinors is easy to see using E2 = p2 +m2.
5
The five terms in 20.2.28 are
P 2
2m
Hermitian
V Hermitian
− P
4
8m3c2
Hermitian
iσ · P × [P, V ]
4m2c2
Hermitian
P · [P, V ]
4m2c2
anti-Hermitian
Recall that P̂ is a Hermitian operator, the potential V is assumed to be
Hermitian (for conservation of probability) and the Pauli matrices σ are Her-
mitian. The commutator [X,Y ] of two Hermitian operators is anti-Hermitian
since [X,Y ]† = (XY )† − (Y X)† = Y †X† − X†Y † = Y X − XY = −[X,Y ].
The cross product of two Hermitian (vector) operators is again Hermitian since
(X × Y )† = X† × Y † = X × Y.
3
319
320
321
322
323
324
325
326
327
	a.pdf
	Shankar Solution COMPLETE - Copy - Copy.pdf
	PRINCIPLES OF QUANTUM MECHANICS
	Shankar Solution
	1and2
	1
	1.1.1 to 1.4.2
	1.6.1 to 1.8.10
	2
	hmwk3_solutions
	hmwk4_solutions
	shankar2
	1.1.2to1.6.1
	2.5.3to2.8.1
	2.7.1to2.8.7
	4.2.1to4.2.2
	4.2.1to10.1.1
	4.2.3to5.2.2
	5.1.4to5.2.5
	20091019125138059
	Problem 2
	5.3.1to5.4.2
	7.3.1to7.3.5
	7.3.1to7.3.6
	7.3.1to7.5.3
	7.4.1to7.5.4
	7.4.10to8.6.1
	9.4.1to9.4.4
	9.4.2to10.2.1
	10.2.1to10.3.6
	10.3.5to11.4.4
	hw3
	solution3
	11.2.1to12.4.1
	11.2.2to12.6.1
	12.3.3to12.5.13
	12.3.7to13.5
	12.6.1to12.6.9
	13.1.5to13.4.3
	13.3.1to13.3.3
	14.3.1to14.3.8
	14.3.3to15.1.2
	14.5.3to15.2.5
	15.2.1to15.2.2
	15a
	16.0
	17.2.1to17.3.2
	18.0
	18.2.1
	18.2.1to18.2.6
	18.2.2
	18.4.2
	18.4.3
	19.3.1 to 19.3.3
	19.5.4to19.6.2
	19.5.6to19.6.3
	20.1.1
	p632_wi11_hw6sol