Graduate Quantum Mechanics
Instructor: Professor Pengfei Zhuang
Homework 4
TA: Tiecheng Guo∗
November 5, 2017
”Live each day as if it is your last,” said Mahatma Gandi. ”Learn as if you’ll live forever.” This is
what I’m passionate about. It is precisely this. It is this inextinguishable, undaunted appetite for
learning and experience, no matter how risible, no matter how esoteric, no matter how seditious it
might seem. This defines the imagined futures of our fellow Hungarians
—Bernie Dunlap
2.28 Consider an electron confined to the interior of a hollow cylindrical shell whose axis coincides with the
z-axis. The wave function is required to vanish on the inner and outer walls, ρ = ρa and ρb , and also at the
top and bottom, z = 0 and L.
(a) Find the eigenfunctions. (Do not bother with normalization.) Show that the energy eigenvalues are given
by
2 #
2 "
lπ
~
2
(l = 1, 2, 3, · · · , m = 0, 1, 2, · · · ),
kmn +
Enlm =
2me
L
where kmn is the nth root of the transcendental equation
Jm (kmn ρb )Nm (kmn ρa ) − Nm (kmn ρb )Jm (kmn ρa ) = 0.
(b) Repeat the same problem when there is a uniform magnetic field B = B ẑ for 0 < ρ < ρa . Note that
the energy eigenvalues are influenced by the magnetic field even though the electron never ”touches” the
magnetic field.
(c) Compare, in particular, the ground state of the B = 0 problem with that of the B 6= 0 problem. Show that
if we require the ground-state energy to be unchanged in the presence of B, we obtain ”flux quantization”
πρ2a B =
2πN ~c
,
e
(N = 0, ±1, ±2, · · · ).
(a) Recall the experssion for Laplace operator in cylindrical coordinates:
1 ∂
∂f
1 ∂2f
∂2f
∆f =
ρ
+ 2 2 + 2.
ρ ∂ρ
∂ρ
ρ ∂φ
∂z
The Schrödinger equation in the cylindrical shell region is
−
~2
∆Ψ(ρ, φ, z) = EΨ(ρ, φ, z).
2me
Using the method of separation of variables, denote E = E12 +E3 and Ψ(ρ, φ, z) = ψ1 (ρ)ψ2 (φ)ψ3 (z).
We have
d2 ψ3
2mE3
+
ψ3 = 0
2
dz
~2 dψ1
d2 ψ2 1
d2 ψ1
2
+
(kρ)
+
(kρ)
+
ψ1 = 0,
(kρ)2
d(kρ)2
d(kρ)
dφ2 ψ2
∗ Time
is kind,
Modern Quantum Mechanics
1
Second Edition
(1)
(2)
where we have denoted k ≡
q
2mE12
~2 .
Combine with the boundary condition ψ3 (0) = ψ3 (L) =
2
~2
lπ 2
0, we have ψ3 = A3 sin( lπ
(l = 1, 2, 3, · · · ). Denote ddφψ22 ψ12 = −ν 2 ,
L z) and E3 = 2m L
we have the solution of Eq.(2),
ψ1 = A1 Jν (kρ) + B1 Nν (kρ),
where Jν (x) and Nν (x) are the first and the second Bessel functions respectively.
Considering the 2π periodicity, we can know that ν is a non-negative integer and we denote it
as m. ψ2 = A2 eimφ +B2 e−imφ . Now we consider the boundary condition ψ1 (ρa ) = ψ1 (ρb ) = 0,
we can get
A1 Jm (kρa ) + B1 Nm (kρa ) = 0,
A1 Jm (kρb ) + B1 Nm (kρb ) = 0.
Eliminate the constant coefficiants, we have
Jm (kρb )Nm (kρa ) − Nm (kρb )Jm (kρa ) = 0.
If we denote k ≡ kmn , the nth root of this transcendental equation. We finally have the energy
eigenvalues
lπ
~2
2
Elmn = E12 + E3 =
kmn
+ ( )2
(l = 1, 2, 3, · · · , m = 0, 1, 2, · · · ).
2me
L
The wave function is
Ψ(ρ, φ, z) = (A1 Jm (kmn ρ) + B1 Nm (kmn ρ))(A2 eimφ + B2 e−imφ ) sin(
lπ
z).
L
ie
)A,where
(b) In the presence of magnetic field Bẑ, we need only to repace the gradient ∇ by ∇−( ~c
e is the charge of the particle. Here the vector potential is
A=
Bρ2a
φ̂.
2ρ
Follow the same procedure, we find that Eq.(2) becomes
dψ1
d2 ψ2 1
d2 ψ1
2
+(kρ)
+
(kρ)
+
(kρ)
−
d(kρ)2
d(kρ)
dφ2 ψ2
2
ie
~c
1
+
dφ ψ2
dψ2
Bρ2a
ie
~c
2 Bρ2a
2
2 !
ψ1 = 0.
(3)
Then we get a new differential equation for ψ2 ,
d2 ψ2
−
dφ2
ie
~c
dψ2
Bρ2a
+
dφ
The solution is
ψ2 = ei
eBρ2
a
2~c
φ
ν2 −
eBρ2a
2~c
2 !
ψ2 = 0.
A2 eiνφ + B2 e−iνφ .
Here ν isn’t necessarily an integer.
Hence we get the similar eigenvalues
"
2 #
~2
lπ
E=
k2 +
2me νn
L
The wave function is
ΨB (ρ, φ, z) = ei
eBρ2
a
2~c
(l = 1, 2, 3, · · · ).
φ
Ψ(ρ, φ, z).
(c) If we require the ground state energy to be unchanged in the presence of B, then we have
m = ν = 0.
So we have
ψ2 = ei
Modern Quantum Mechanics
2
eBρ2
a
2~c
φ
.
Second Edition
Due to the 2π periodicity of ψ2 , we can obtain that
eBρ2a
= N,
2~c
where N is an integer. It’s clear that we have the ”flux quantization”
2πN ~c
,
e
πρ2a B =
(N = 0, ±1, ±2, · · · ).
2.32 Define the partition function as
Z
d3 x0 K(x0 , t; x0 , 0)|β=it/~ .
Z=
Show that the ground-state energy is obtained by taking
−
1 ∂Z
,
Z ∂β
(β → ∞).
Illustrate this for a particle in a one-dimensional box.
X
exp[−βEa0 ]
Z=
a0
The ground state energy
E0 =
X
Ea0
a0
1 ∂Z
exp[−βEa0 ]
=−
.
Z
Z ∂β
For a particle in a one dimensional box, the periodic boudary condition gives that dn =
we get
r
Z ∞
X
L
L
m
p2
β]
dp =
.
Z=
exp[−βEn ] =
exp[−
2m 2π~
~ 2πβ
−∞
n
L
2π~ dp.
So
The ground state energy is
−
1 ∂Z
1
=
.
Z ∂β
2β
2.34
(a) Write down an expression for the classical action for a simple harmonic oscillator for a finite time interval.
(b) Construct hxn , tn |xn−1 , tn−1 i for a simple harmonic oscillator using Feynman’s prescription for tn − tn−1 =
∆t small. Keeping only terms up to order (∆t)2 , show that it is in complete agreement with the t − t0 → 0
limit of the propagator given by (2.6.26).
(a) For a simple harmonic oscillator,
Z
tn
m 2 m 2 2
ẋ − ω x ]
2
2
tn−1
"
#
2
m
xn − xn−1
2 2
= ∆t
− ω xn
2
∆t
S(n, n − 1) =
dt[
(b) Use (2.6.46), we have
r
hxn , tn |xn−1 , tn−1 i =
m
iS(n, n − 1)
exp
.
2πi~∆t
~
Use (2.6.18), we have
r
K(xn , tn ; xn−1 , tn−1 ) =
mω
imω
2
2
exp
(xn + xn−1 ) cos(ω∆t) − 2xn xn−1 .
2πi~ sin(ω∆t)
2~ sin(ω∆t)
Due to the fact that ∆t is small, we have
sin(ω∆t) = ω∆t
ω∆t
cos(ω∆t) = 1 −
.
2
x2 +x2
And use n 2 n−1 ∆t ≈ x2n ∆t, it’s easy to see that the two ways to deal with hxn , tn |xn−1 , tn−1 i
is equivalent.
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Second Edition
2.38 Consider the Hamiltonian of a spinless particle of charge e. In the presence of a static magnetic field, the
interaction terms can be generated by
poperator → poperator −
eA
,
c
where A is the appropriate vector potential. Suppose, for simplicity, that the magnetic field B is uniform
in the positive z-direction. Prove that the above prescription indeed leads to the correct expression for the
e
L with the magnetic field B. Show that there is also an
interaction of the orbital magnetic moment 2mc
2 2
2
extra term proportional to B (x + y ), and comment briefly on its physical significance.
p2
The Hamiltonian is H0 = 2m
+ φ(r). In the presence of a static magnetic field, we have
H0 → H =
(p − eA/c)2
+ φ(r).
2m
The magnetic vector potential for a uniform magnetic field in Coulomb gauge ∇ · A = 0 is
A=
1
B × r.
2
Note that we can write
A·p=
1
B
1
(B × r)p = (r × p) = B · L.
2
2
2
When we choose B to be an uniform magnetic field along z direction, we have
A2 =
1
1
(B × r)2 = B 2 (x2 + y 2 ).
4
4
Thus we have
e2 B 2 2
eBLz
+
(x + y 2 ).
2mc
8mc2
e
It’s easy to see that the orbital magnetic moment is 2mc
L. There is also the quadratic Zeeman
2 2
2
effect contribution proportional to B (x +y ) which contributes to the ”diamagnetic suseptibility”
χ appearing as an energy shift.
Anyway, I don’t know the physical meaning! Do you know?
2.39 An electron moves in the presence of a uniform magnetic field in the z-direction (B = B ẑ).
H = H0 −
(a) Evaluate
[Πx , Πy ],
where
Πx ≡ px −
eAx
,
c
Πy ≡ py −
eAy
.
c
(b) By comparing the Hamiltonian and the commutation relation obtained in (a) with those of the onedimensional oscillator problem, show how we can immediately write the energy eigenvalues as
~2 k 2
|eB|~
1
Ek,n =
+
n+
,
2m
mc
2
where ~k is the continuous eigenvalue of the pz operator and n is a nonnegative integer including zero.
(a)
e
i~eB
[Πx , Πy ] = − ([Ax , py ] + [px , Ay ]) =
.
c
c
c
(b) From (a), we can know [− eB
Πy , Πx ] = i~. The Hamiltonian can be witten as
H=
Π2y
Π2x
p2
Π2
m e2 B 2
p2z
c
2
+
+ z = x +
Π
)
+
.
(−
y
2m 2m 2m
2m
2 m2 c2 eB
2m
Compare with the one-dimensional oscillator hamiltonian, we can write the energy eigen value
easily
~2 k 2
|eB|~
1
Ek,n =
+
n+
.
2m
mc
2
Here
|eB|
mc
is seen as ω in oscillator problem.
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Second Edition
2.40 Consider the neutron interferometer.
Prove that the difference in the magnetic fields that produce two successive maxima in the counting rates
is given by
4π~c
,
∆B =
|e|gn λ̄l
where gn (= −1.9) is the neutron magnetic moment in units of −e~/2me c. (If you had solved this problem
in 1967, you could have published your solution in Physical Review Letters! )
T . Two successive maxima in the
The phase difference of two different path is ∆φ = gn µ∆B
~
counting rates imply that the phase difference is 2π, where T is the time taking for the neutron to
pass the magnetic region. So we have
∆φ = 2π =
e~
∆B lmλ̄
gn 2mc
,
~
~
where e is the element charge, λ̄ is the reduced wave length.
Hence we get
4π~c
∆B =
,
egn λ̄l
which is consistent with the conclusion.
Modern Quantum Mechanics
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Second Edition