Graduate Quantum Mechanics
Solution
HW #6 ~HW #8
(Sakurai Ch.2 #21~#37)
Solved by Sunghyon Kyeong
(starinphysics@yonsei.ac.kr)
Modfied by Chul Kim
(creativefe@phya.yonsei.ac.kr)
May 25, 2008
Send me a mail. If you find any mistakes in this document
2. J.J. Sakurai 2.21
A particle in one dimension is trapped between two rigid walls:
0, for 0 < x < L
V =
∞, for x < 0, x > L
At t=0 it is known to be exactly at x=L/2 with certainty. What are the relative probabilities for
the particle to be found in various energy eigenstates? Write down the wave function for t ≥ 0. (You
need not worry about absolute normalization, convergence, and other mathematical subtleties.)
(Solution)
1D-box problem is very well known. So, I will start the problem assuming we have already known
the eigen ftn and eigen energy without calculating.
2 nπ sin
x
ψn =
L
L
π 2 h̄2 n2
,
where n ∈ Z.
En =
2mL2
At, t=0, particle is located at x =
L
2
which can be represented as follow:
L =
δ x−
cn ψn
2
n
L
2
L nπ
sin x
cn =
dxδ x −
L 0
2
L
2
nπ
sin
=
L
2
Therefore, δ-function can be expanded by complete set of eigen functions as follow:
∞
2 nπ
nπ
L
=
sin sin x
δ x−
2
L n=1
2
L
i. relative probability for the particle to be found in various energy eigenstates.
P (n) = |cn |2 =
2 2 nπ L
L
= (1 − cosnπ) = .
sin
L
2
4
2
for odd n
ii. time evolution for δ x − 12 .
∞
E2n+1 (2n + 1)π
H L 2 (2n + 1)π
=
exp − i
exp − i t δ x −
t sin
sin
x
h̄
2
L
h̄
2
L
n=0
3. J.J. Sakurai 2.22
Consider a particle in one dimension bound to a fixed center by a δ-function potential of the form
V (x) = −v0 δ(x), (v0 real and positive).
Page 2
Find the wave function and the binding energy of the ground state. Are there excited bound states?
(Solution)
We will start from the schrödinger equation
−
h̄2 d2
ψ(x) − v0 δ(x)ψ(x) = Eψ(x)
2m dx2
It yields bound state(E < 0)
We’ll look first at region I (x < 0, V (x) = 0)
−
h̄2 d2
d2 ψ(x)
2mE
ψ(x) = Eψ(x) =⇒
= − 2 ψ(x) = κ2 ψ(x)
2
2m dx
dx2
h̄
after solving above differential equation, we can get following:
ψI = Ae−κx + Beκx , for x < 0
ψ=
ψII = Ge−κx + F eκx , for x > 0
,κ > 0
where κ2 = − 2mE
h̄2
We can get an appropriate solutions ψ (at both regions) by applying boundary conditions ψI (x →
−∞) = 0 and ψII (x → ∞) = 0 with ψI (x = 0) = ψII (x = 0)
for x < 0
ψI = Beκx ,
ψ=
ψII = Be−κx , for x > 0
To find the remaining coefficients κ and B, we have to integrate the schrödinger equation from −
to + and then take the limit as || → 0.
+
+
+
h̄2
d2 ψ(x)
−
dx
−
v
dxδ(x)ψ(x)
=
dxEψ(x)
0
2m −
dx2
−
−
(RHS) term is vanished by integrating process.
−
h̄2 dψ(x) dψ(x) − v0 ψ(0) = 0
−
2m
dx +
dx −
h̄2 − κB − (−κB − v0 B = 0
2m
Finally, we can get the bound state energy
=⇒ −
κ2 =
m2 v02
2mE
mv 2
= − 2 =⇒ E = − 20
4
h̄
h̄
2h̄
There exists only one bound state.
+∞
0
dx|ψ|2 =
dx|B|2 e2κx +
−∞
−∞
Therefore, the last coefficient is B =
ψ=
√
+∞
dx|B|2 e−2κx =
0
|B|2
=1
κ
κ and the final solution is given by.
ψI =
ψII =
mv
mv0 h̄20 x
e
,
h̄2
mv
mv0 − h̄20 x
e
,
h̄2
for
for
x<0
x>0
These two solution can be combined into one compact form as follow:
mv
mv0
0
ψ(x) =
2 exp −
2 |x|
h̄
h̄
Page 3
4. J.J. Sakurai 2.23
A particle of mass m in one dimension is bound to a fixed center by an attractive δ-function potential:
V (x) = −λδ(x), (λ > 0).
At t=0, the potential is suddenly switched off(that is, V = 0 for t> 0). Find the wave function for
t> 0).(Be quantitative! But you need not attempt to evaluate an integral that may appear.)
(Solution)
I will start from schrödinger equation
−
h̄2 d2 φ(x)
− λδ(x)φ(x) = Eφ(x)
2m dx2
(1)
The fourier transformation bewteen real space and momentum space is given as follow:
+∞
1
dk φ̃(k)eikx
φ(x) =
2π −∞
(2)
We can get fourier transform of φ(x) as plugging equation(2) into equation(1) and by using the
δ-function property.
+
h̄2 k 2
φ̃(k) − λφ(x = 0) = E φ̃(k)
2m
From the above equation, we can get φ̃(k) and φ(x)as follows:
3
φ̃(k)
=
φ(x)
=
2mλφ(0)/h̄2
2κ 2
λφ(x = 0)
=
= 2
2 2
2 + κ2
k
k
+ κ2
h̄ k /2m − E
+∞
3
1
κ2
dk 2
eikx
2π −∞
k + κ2
At t > 0, there exist only kinetic energy in hamiltonian
U (t) = exp −i
H
t
h̄
= exp −i
1 p2 t
h̄ 2m
= exp
ih̄ ∂ 2
t
2m ∂x2
We can get the time-dependent function by applying the above time-evolution operator.
φ(x, t)
= U φ(x, 0)
= exp
=
1
2π
ih̄ d2
t
2m dx2
1
2π
+∞
3
2
dk
−∞
3
+∞
dk
−∞
2κ 2
expikx
2
k + κ2
2κ
k 2 h̄
exp
−i
t + ikx
k 2 + κ2
2m
You need not attempt to evaluate an integral that may appear.
5. J.J. Sakurai 2.24
A particle in one dimension(−∞ < x < ∞) is subjected to a constant force derivable from
V = λx, (λ > 0).
(a) Is the energy spectrum continuous or discrete? Write down an approximate expression for the
energy eigenfunction specified by E. Also sketch it crudely.
Page 4
(Solution)
In the case under construction there is only a continuous spectrum and the eigenfunctions are non
degenerates. From the discussion on WKB approximation.
region I : E > V (x)
A
i x2
exp
+
dx 2m(E − V (x))
ψI (x) =
1/4
h̄ x1
[E − V (x)]
√
π
2mλ x2 =E/λ
A
exp
dx (E/λ − x) −
=
1/4
h̄
4
x1
[E − V (x)]
3/2
A
2mλ π
2 E
=
exp
−
−
x
−
1/4
3 λ
h̄
4
[E − V (x)]
Similarly, we can easily find a wave function at the opposite region
region II : E < V (x)
ψII (x)
=
=
=
Where, q = α
E
λ
−x
A2
1
exp −
1/4
h̄
[V (x) − E]
x2
x1 =E/λ
1
exp −
1/4
2mλh̄
[V (x) − E]
c2
2
3/2
exp − (−q)
3
[−q]1/4
1/3
and α = 2mλ
h̄2
A2
dx 2m(V (x) − E)
x2
x1 =E/λ
d(2mλx) 2m(V (x) − E)
(b) Discuss briefly what changes are needed if V is replaced by
V = λ|x|.
(Solution)
We have bound states when we take the potential V = λ|x|. Therefore, the energy spectrum is
discrete. So, in this case the energy eigenstates have to satisfy the consistency relation.
x2
1
πh̄
dx 2m(E − λ|x|) = n +
2
x1
Where, n is non-negative integer.
E
The turning points are x1 = − E
λ , x2 = λ
So,
+E/λ
+E/λ
1
πh̄ =
dx 2m(E − λ|x|) = 2
dx 2m(E − λ|x|)
n+
2
−E/λ
0
√
2
= 2 2mλ
3
E
λ
3/2
We can finally get the discrete eigen energy for nth state as follow:
2/3
3 n + 12 πh̄λ
√
En =
4 2m
Page 5
6. J.J. Sakurai 2.25
Consider an electron confined to the interior of a hollow cylinderical shell whose axis coincides
with the z-axis. The wave function is requared to vanish on the inner and outer wall, ρ = ρa and ρb ,
and also at the top and bottom, z = 0 and L.
(a) Find the energy eigenfunctions. (Do not bother with normalization.)
Show that the energy eigenvalues are given by
2
h̄2
lπ
2
kmn +
(l = 1, 2, 3, ..., m = 0, 1, 2, 3, ...),
Elmn =
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.
(Solution)
Consider a wave function at the inner and outer wall.
−
h̄2
2 ϕ =
2m2
Eϕ
2 ϕ + k 2 ϕ =
0
k2 =
where,
2me E
h̄2
The schrödinger equation at the both inner and outer wall is
∂2
1 ∂
∂2
1 ∂2
+
+ 2 2 + 2 + k2 ϕ = 0
2
∂ρ
ρ ∂ρ ρ ∂φ
∂z
(3)
with the separation of coordinates
ϕ(ρ, φ, z) = R(ρ)Φ(φ)Z(z)
Then, equation (3) becomes
1
R
d2 R 1 dR
+
dρ2
ρ dρ
+
1 1 d2 Φ 1 d2 Z
+
+k 2
ρ2 Φ dφ2 Z dz 2
1
R
d R 1 dR
+
dρ2
ρ dρ
ρ2 R (ρ) + ρR (ρ) + ρ2
0
−kz2
−m2
2
=
−m
− kz2 + k 2 = 0
ρ2
m2
2
kmn
− 2 R(ρ) = 0
ρ
2
+
2
where, kmn
= k 2 − kz2
Where a prime denotes differentiation with respect to ρ. We are now in a position to apply boundary
conditions.
(i) z-direction : general solution is Z(z) = Az sin(kz z) + Bz cos(kz z)
Z(z = 0) = Z(z = L) = 0
=⇒
Bz
=
kz
=
(ii) φ-direction : general solution is Φ(φ) = Aφ eimφ
and boundary condition in φ-direction is given as follow:
Φ(φ) = Φ(φ + 2π)
Therefore, m should be integer.(i.e. m ∈ Z)
Page 6
0
lπ
L
l = 1, 2, 3, ...
(iii) ρ-direction : general solution at the finite region as following. ρz < ρ < ρb
R(ρ) = Aρ Jm (kmn ρ) + Bρ Nm (kmn ρ)
R(ρ = ρa ) = R(ρ = ρb ) = 0
Jm (kmn ρa ) Nm (kmn ρa )
Jm (kmn ρb ) Nm (kmn ρb )
Aρ
Bρ
=0
To get a non-trivial solution below determinant is automatically satisfied for arbitrary Aρ and Bρ (I
mean, determine Aρ and Bρ with the above matrix equation.)
Jm (kmn ρa ) Nm (kmn ρa ) Jm (kmn ρb ) Nm (kmn ρb ) = 0
→ Jm (kmn ρb )Nm (kmn ρa ) − Nm (kmn ρb )Jm (kmn ρa ) = 0
Therefore, we can get the wave function.
ϕ(ρ, φ, z) = Cρ (Nm (kmn ρb )Jm (kmn ρ) − Jm (kmn ρb )Nm (kmn ρ)) sin
lπ
z eimφ
L
Also, we can get the energy eigenvalue.
Enml
h̄2 2
h̄2
k2 +
=
k =
2me
2me mn
lπ
L
2
where, l = 1, 2, 3, ...
(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 tough the electron never
“touches” the magnetic field.
(Solution)
The magnetic field is related to vector potential A through the relation B = ∇ × A.
We can find the a vector potential by using stokes’ theorem.
2
ΦB =
B · dS = πρa B =
A · dl = 2πρAφ
S
A=
Bρ2a
2ρ
C
φ̂ =
ΦB
2πρ
φ̂
The Schrödinger equation in the presence of the magnetic field B can be written as follows
e e 1 −ih̄∇ − A · −ih̄∇ − A ϕ = Eϕ
2me
c
c
2
2 ∂
∂
h̄
1 ∂
ie ΦB
+ ẑ
ρ̂
−
ϕ = Eϕ
+ φ̂
−
2m2
∂ρ
ρ ∂φ h̄c 2π
∂z
We can see the change only in φ-component compare with B = 0 case.
d
ie ΦB
−
dφ h̄c 2π
2
Φ(φ) = −m2 Φ(φ)
Trial solution of Φ(φ) can be written as follows:
Φ(φ) = Cφ eiM φ
Page 7
(4)
By putting a trial solution into equation (4), we can get
d
ie ΦB
−
dφ h̄c 2π
ie ΦB
iM −
h̄c 2π
ie ΦB
h̄c 2π
ie ΦB
iM −
h̄c 2π
iM −
iM −
Cφ eiM φ
=
−m2 Cφ eiM φ
Cφ eiM φ
=
−m2 Cφ eiM φ
=
−m2
ie ΦB
h̄c 2π
2
So, we can get the relation between M and m.
M = ±m +
ie ΦB
h̄c 2π
Applying the boundary condition,
Φ(φ) = Φ(φ + 2π)
=⇒
M = ±m +
ie ΦB
∈Z
h̄c 2π
This means that the energy eigenfunctions will
ϕ(ρ, φ, z) = Cρ (Nm (kmn ρb )Jm (kmn ρ) − Jm (kmn ρb )Nm (kmn ρ)) sin
lπ
z eiM φ
L
But now m is not an integer. As a result the energy of the ground state will be
2
h̄2 2
h̄2
lπ
k2 +
where, l = 1, 2, 3, ...
k =
EnM l =
2me
2me mn
L
(c) Compare, in particular, the ground state of the B = 0 problem with that of the B = 0 problem.
Show that if we require the ground-state energy to be unchanged in the presence of B, we obtain
“flux quantization”
2πN h̄c
πρ2a B =
,
(N = 0, ±1, ±2, ...).
e
(Solution)
ie ΦB
ie ΦB
Now, m = M − h̄c
2π is not zero in general but it corresponds to M ∈ Z such that 0 ≤ M − h̄c 2π < 1.
Notice also that if we require the ground state to be unchanged in the presence of B, we obtain “flux
quantization”
M−
ie ΦB
=0
h̄c 2π
=⇒
ΦB = πρ2a B =
Page 8
2πM h̄c
e
where, M ∈ Z.
7. J.J. Sakurai 2.26
Consider a particle moving in one dimension under the influence of a potential V (x). Suppose its
wave function can be written as exp[iS(x, t)/h̄]. Prove that S(x, t) satisfies the classical HamiltonJacobi equation to the extent that h̄ can be regarded as small in some sense.
Show how one may obtain the correct wave function for a plane wave by starting with the solution of
the classical Hamilton-Jacobi equation with V (x) set equal to zero. Why do we get the exact wave
function in this particular case?
(Solution)
Hamilton-Jacobi equation in the classical limit is the following.
∂S(x, t)
1
| ∇S(x, t) |2 +V (x) +
=0
2m
∂t
Use time dependent Schrodinger equation.
[
···
−h̄2 2
∇ + V (x)]ψ
2m
=
···ψ
=
iS(x, t)
−h̄2 2
∇ exp[
] =
2m
h̄
iS(x, t)
∂
] =
· · · exp[
∂t
h̄
∂
ψ
∂t
iS(x, t)
]
exp[
h̄
iS(x, t)
1
| ∇S(x, t) |2 exp[
]
2m
h̄
i ∂S(x, t)
iS(x, t)
exp[
]
h̄ ∂t
h̄
ih̄
Then the time dependent Schrodinger equation becomes,
iS(x, t)
iS(x, t)
∂S(x, t)
iS(x, t)
1
| ∇S(x, t) |2 exp[
] + V (x)exp[
]=−
exp[
]
2m
h̄
h̄
∂t
h̄
Divide both side with the exponential term then one can see that satisfies the Hamilton-Jacobi
equation.
⇒
∂S(x, t)
1
| ∇S(x, t) |2 +V (x) +
=0
2m
∂t
If V(x) = 0 ,
∂S(x, t)
1
| ∇S(x, t) |2 +
= 0 · · · S(x, t) ≡ S1 (x) + S2 (t)
2m
∂t
1 ∂ 2 S1
∂S2
⇒
=−
=C
2
2m ∂x
∂t
Note that C is a constant. We can find the solution of S as following.
S1 =
√
2mcx + const.S2 = −ct + const. ⇒ S =
√
2mcx − ct
Insert this S into the exponential term of ψ,
√
⇒ ψ = exp[i( 2mcx − ct)/h̄]
Now the wave function is a plane wave. Physically, if V(x) = 0 , then plane wave should come out
because it is a free particle case. Mathematically, the second derivative of S is 0.
Page 9
8. J.J. Sakurai 2.27
Using spherical coordinates, obtain an expression for j for the ground and excited states of the
hydrogen atom. Show, in particular, that for ml = 0 states, there is a circulating flux in the sense
that j is in the direction of increasing or decreasing φ, depending on whether ml is positive or
negative.
(Solution)
General solution of 3D schrödinger equation is given by
|m|
ϕnlm = Rnl (r)Yml (θ.φ) = Rnl (r)Nlm Pl
(cosθ)eimφ
current density j can be expressed as follow.
j=
ih̄
(ϕnlm ∇ϕ∗nlm − ϕ∗nlm ∇ϕnlm )
2me
and differential operator in spherical coordinate is given by
∇=
∂
1 ∂
1 ∂
r̂ +
θ̂ +
φ̂
∂r
r ∂θ
rsinθ ∂φ
We can split into current density j by parts.
jr
=
jθ
=
ih̄
(ϕnlm ∇r ϕ∗nlm − ϕ∗nlm ∇r ϕnlm ) = 0
2me
ih̄
(ϕnlm ∇θ ϕ∗nlm − ϕ∗nlm ∇θ ϕnlm ) = 0
2me
These two term are automatically vanished because there are no imaginary components in the (r, θ)
parts.
jφ
ih̄
(ϕnlm ∇φ ϕ∗nlm − ϕ∗nlm ∇φ ϕnlm )
2me
ih̄ 2
1 |m|
∂
∂
2
=
R (r)Nlm
p (cosθ) eimφ e−imφ − e−imφ eimφ
2me nl
sinθ l
∂φ
∂φ
h̄ 1
= 2m
|ϕnlm |2
2me sinθ
mh̄
=
|ϕnlm |2
me rsinθ
=
Finally, we can get the φ̂-direction current density as follow:
j = jφ φ̂ =
mh̄
|ϕnlm |2 φ̂
me rsinθ
There survives only φ̂-component, which means current flow in φ̂-direction depending on the sign of
magnetic moment m
Page 10
9. J.J. Sakurai 2.28
Derive (2.5.16) and obtain the three-dimensional generalization of (2.5.16).
(Solution)
K(x , t; x , t0 )
=
x |exp −i
+∞
=
−∞
=
=
=
=
=
H(t − t0 )
h̄
|x dp x |p p |exp −i
p2 h̄
(t − t0 ) |x 2m
p2 h̄
p
dp exp −i
(t − t0 ) exp i (x − x )
2m
h̄
−∞
+∞
2
i(t − t0 )
x − x
x − x
1
p −
dp exp −
m −
m
2πh̄ −∞
2mh̄
t − t0
t − t0
+∞
1
i(t − t0 ) 2
i m(x − x )2
dp exp −
exp
p
2πh̄
h̄ 2(t − t0 )
2mh̄
−∞
1
2mh̄π
i m(x − x )
exp
2πh̄
h̄ 2(t − t0 )
i(t − t0 )
m
im(x − x )
exp
2πih̄(t − t0 )
2h̄(t − t0 )
1
2πh̄
+∞
Generalization in three dimension
K(x , t; x , t0 )
m
2πih̄(t − t0 )
=
3/2
im|x − x |
exp
2h̄(t − t0 )
10. J.J. Sakurai 2.29
Define the partition function as
d3 x K(x , t; x , 0)|β=it/h̄ ,
Z=
as in (2.5.20)-(2.5.22). Show that the ground-state energy is obtained by taking
−
1 ∂Z
,
Z ∂β
(β → ∞).
(Solution)
Ground state energy ≡ EG .
1.
Z
=
=
=
d3 x K(x , t; x , 0)
d3 x
| x |n|2 exp(−βEn )
n
exp(−βEn )
n
−
1 ∂Z
Z ∂β
=
=
1 ∂ 1 [
exp(−βEn )] = − [ (−En )exp(−βEn )]
Z ∂β n
Z n
n (En )exp(−βEn )
n exp(−βEn )
−
Page 11
2
In the limit of β → ∞,
n (En )exp(−βEn )
⇒ EG
n exp(−βEn )
(5)
2. In 1 − D box, Density of state is L/2π
Z=
π 2 h̄2
2mL2
exp(−β
n
n2 π 2 h̄2
)
2mL2
2
2
2
n π h̄
2
1 ∂Z
n n exp(−β 2mL2 )
=
n2 π 2 h̄2
Z ∂β
n exp(−β 2mL2 )
2
π 2 h̄2
n2 π 2 h̄2
π 2 h̄2
n n exp(−β 2mL2 )
2mL2
=
= EG
lim
2
2
2
n π h̄
2mL2
n exp(−β 2mL2 )
⇒−
11. J.J. Sakurai 2.30
The propagator in momentum space analogous to (2.5.26) is given by p , t|p , t0 . Derive an explicit
expression for p , t|p , t0 for the free particle case.
(Solution)
Propagator in momentum space.
p , t|p , t0 =
=
=
=
H
(t − t0 ) |p h̄
2
p
p |exp −i
(t − t0 ) |p 2mh̄
p
exp −i
(t − t0 ) δ 3 (p − p )
2mh̄
p |exp −i
K(p , t; p , t0 )
Page 12
12. J.J. Sakurai 2.31
(a) Write down an expression for the classical action for a simple harmonic oscillator for a finite
time interval.
(Solution)
General expression of x(t) and ẋ(t) for a simple harmonic oscillator is given by:
x(t) =
ẋ(t) =
The action can be written as follow:
t2
t2
dt£(x, ẋ; t) =
dt
S =
t1
=
mw2
2
t2
t1
t1
Acosw(t − t1 ) + Bsinw(t − t1 )
−wAsinw(t − t1 ) + wBcosw(t − t1 )
1
1
mẋ2 − mw2 x2
2
2
dt [−Asinw(t − t1 ) + Bcosw(t − t1 )]2 − [Acosw(t − t1 ) + Bsinw(t − t1 )]2
To calculate the above action, we first find coefficients A and B by applying boundary conditions:
x(t = t1 − t1 = 0)
= A
=⇒ A = x1
x(t = t2 − t1 = T ) = AcoswT + BsinwT = x1
x2 − x1 coswT
=⇒ B =
sinwT
We are now in a position to calculate the action as following procedure.
mw2 t2 dt [−Asinw(t − t1 ) + Bcosw(t − t1 )]2 − [Acosw(t − t1 ) + Bsinw(t − t1 )]2
S =
2
t1
2 T
mw
dt [−Asinwt + Bcoswt]2 − [Acoswt + Bsinwt]2
=
2
0
mw2 T 2
dt (A − B 2 )sin2 wt − (A2 − B 2 )cos2 wt − 4ABsinwtcoswt
=
2
0
mw2 T 2
dt (B − A2 )cos2wt − 2ABsin2wt
=
2
0
T
2
1
mw
1
(B 2 − A2 ) sin2wt + 2AB
=
cos2wt
2
2w
2w
0
mw 2
2
(B − A )sin2wT + 2AB(cos2wT − 1)
=
4
mw 2
(B − A2 )2sinwT coswT − 4ABsin2 wT
=
4
mw
=
sinwT (B 2 − A2 )coswt − 2ABsinwT
2
where coefficients are given by
B 2 − A2
=
=
AB
=
x2 − x1 coswT
− x21
sinwT
2
1
x2 + x21 (cos2 wT − sin2 wT ) − 2x1 x2 coswT
2
sin wT
x2 − x1 coswT
x1
sinwT
Page 13
We can finally get the action in terms of x1 and x2 by plugging in the coefficients as shown in above:
S
=
=
=
coswT mw
x22 + x21 (cos2 wT − sin2 wT ) − 2x1 x2 coswT
sinwT
2
2
sin wT
x2 − x1 coswT
−2x1
sinwT
sinwT
mw 2
2 2
(x2 + x21 − 2x
1 sin wT − 2x1 x2 coswT )coswT
2sinwT
(
2((((
2
(
2x(
−2x1 x2 sin2 wT + (
sin
wT
coswT
1
mw 2
(x2 + x21 )coswT − 2x1 x2
2sinwT
(b) Construct xn , tn |xn−1 , tn−1 for a simple harmonic oscillator using Feynman’s prescription for
tn − tn−1 = Δt small. Keeping only terms up to order (Δt)2 , show that it is in complete
agreement with the t − t0 → 0 limit of the propagator given by (2.5.26).
(Solution)
S(n, n − 1)
tn
dt
=
tn−1
m
= Δt
2
=
m
2Δt
xn , tn |xn−1 , tn−1 =
=
mẋ2
− V (x)
2
xn − xn−1
Δt
2
−V
x2n − 2xn xn−1 + x2n−1 −
m
exp
2πih̄Δt
i
h̄
xn + xn−1
2
w2
(xn + xn−1 )2 Δt
2
dt£
m
iΔt
exp
2πih̄Δt
h̄
m
2
xn − xn−1
Δt
2
m
− w2
2
xn + xn−1
2
2
Therefore, it agrees with the t-t0 ⇒ 0 limit of the propagator given by (2.5.26) and (2.5.18).
13. J.J. Sakurai 2.32
State the Schwinger action principle (see Finkelstein 1973, 155). Obtain the solution for x2 t2 |x1 t1 by integrating the Schwinger principle and compare it with the corresponding Feynman expression
for x2 t2 |x1 t1 .
Describe the classical limits of these two expressions.
(Solution)
i. Schiwinger action principle
In quantum mechanics we are interested in computing the transition amplitude x2 , t2 |x1 , t1 where |x1 , t1 represents the quantum state at time t1 , and |x2 , t2 represents the state at time
t2 ≥ t1 . These states are chosen to be eigenstates of the position operator x̂i :
x̂i |xα , tα = xi (tα )|xα , tα (α = 1, 2)
The Schwinger action principle states that
δ x2 , t2 |x1 , t1 =
Page 14
i
x2 , t2 |δS|x1 , t1 h̄
(6)
where S represents the action obtained by the replacement of xi in the action for the classical theory with x̂i , along with an operator ordering which leads to S being self-adjoint. δ in
equation (6) represents any possible variation, including variations with respect to the times
t1 , t2 , the dynamical variables q i , or the structure of the Lagrangian. The variations of the
dynamical variables δq i will be chosen to be c-numbers, appropriate to bosonic theories.
ii. Obtain the solution for x2 , t2 |x1 , t1 We have to integrate the equation from the statement of schwinger’s action principle referred
from the Finkelstein’s Nonrelativistic Mechanics
δ x2 , t2 |x1 , t1 =
i
x2 , t2 |δS|x1 , t1 h̄
Let’s consider a action operator which acts from (x1 , t1 ) to (x2 , t2 ) sequently. We call it δS21 .
Then, we get
δ x2 , t2 |x1 , t1 i
i
x2 , t2 |δS|x1 , t1 = x2 , t2 |δS21 |x1 , t1 h̄
h̄
i
δS21
x2 , t2 |x1 , t1 h̄
i δ x2 , t2 |x1 , t1 = δS21
x2 , t2 |x1 , t1 h̄
=
=
=⇒
By integrating above equation, we get
i x2 , t2 |x1 , t1 = exp
S
h̄ 21
(7)
Let’s suppose that h̄ can, in some sense, be regarded as a small quantity. Then S21
which appears
in the equation (7) should ssatisfy the Hamilton-Jacobi equation in the classical mechanics.
∂ S21
1
(x, t)|2 + V (x) +
|∇S21
=0
2m
∂t
iii. Compare the transition amplitude with the corresponding Feynman expression
Feynman expression for x2 , t2 |x1 , t1 is
i
x2 , t2 |x1 , t1 =
exp
S21
h̄
all path
By taking the classical limit h̄ → 0, then the dominant path is the only classical path. Therefore,
transition amplitude x2 , t2 |x1 , t1 is exactly the same result for both of approaching.
14. J.J. Sakurai 2.33
Show that the wave-mechanics approach to the gravity-induced problem discussed in Section 2.6
also leads to phase-difference expression (2.6.17).
(Solution)
Hamiltonian of this system is given by
H=
p2
p2
+ mn gz =
+ mn gl2 sinδ
2mn
2mn
Potential has different value at different position.
VAC
VBD
=
=
0
mn gl2 sinδ
Page 15
if z = 0
if z = l2 sinδ
Then, we can calculate the phase difference as follow:
1 lf dl
equation(2.6.9)
Δφ = −
[VBD − VAC ]
h̄ li v
1
= − mn gl2 sinδl1
h̄v
mn gl1 l2 sinδ 1
mn gl1 l2 sinδ mn λ̄
m2 gl1 l2 λ̄sinδ
= −
=−
=− n
h̄
v
h̄
h̄
h̄2
Where, λ̄ is the deBroglie’s (matter) wave length which has to satisfy the following relation.
p =
mn v =
v
h̄
mn λ̄
=
h̄
λ̄
15. J.J. Sakurai 2.34
(a) Verity (2.6.25) and (2.6.27).
(Solution)
We can prove the equation (2.6.25) simply by using commutation relation.
e
Πi = pi − Ai
c
e
e [Πi , Πj ] = pi − Ai , pj − Aj
c
c
e e
= pi , − Aj + − Ai , pj
c
c
∂ e ∂ e − Aj + ih̄
− Ai
= −ih̄
∂xi
c
∂xj
c
∂
ih̄e
∂
=
Aj −
Ai
c
∂xi
∂xj
ih̄e
=
εijk Bk
c
We can prove the equation (2.6.27) by using heisenberg equation of motion and commutation relation.
m
dΠi
d2 x i
=
2
dt
dt
=
=
=
=
1
[Πi , H]
ih̄ 1
Π2
Πi ,
+ eφ(x)
ih̄
2m
⎡
⎤
1 ⎣
1
Πi ,
Π2j ⎦
[Πi , eφ(x)] +
ih̄
2mih̄
j
1
1 ([Πi , Πj ] Πj + Πj [Πi , Πj ])
[pi , eφ(x)] +
ih̄
2mih̄ j
∂φ(x)
+
= −e
∂xi
1
= e Ei +
2c
1
= e E+
2c
ih̄e
1
(εijk Bk Πj + εijk Πj Bk )
2mih̄
c
Πj
Πj
εijk Bk − εikj Bk
m
m
dx
dxi
dx
(· · · Πj =
×B−B×
)
dt
dt i
dt
Page 16
(b) Verity continuity equation (2.6.30) with j given by (2.6.31).
(Solution)
The time-dependent schrödinger equations are
∂ψ
∂t
∂ψ ∗
−ih̄
∂t
ih̄
= Hψ
= H ∗ ψ∗
Where, hamiltonian H is given by
e2
e
1
e
−h̄2 ∇2 + ih̄(∇ · A) + 2ih̄ A · ∇ + 2 A2 + eφ
H=
2m
c
c
c
The derivation of ρ with respect to time t is
ih̄
∂ρ
∂t
∂ψ
∂ψ ∗ ψ
∂ψ ∗
= ih̄ψ ∗
+ ih̄ψ
∂t
∂t
∂t
= (ψ ∗ Hψ − ψHψ ∗ )
=
ih̄
Multiplying ψ ∗ and ψ at the left and right side of the hamiltonian as shown in the equation (8).
1
e
e
e2
−h̄2 ψ ∗ ∇2 ψ + ih̄(∇ · A)|ψ|2 + 2ih̄ A · ψ ∗ ∇ψ + 2 A2 |ψ|2 + eφ|ψ|2
ψ ∗ Hψ =
2m
c
c
c
The complex conjugate of the above equation is
1
e
e
e2 2 2
2
∗
2 ∗
2
∗
−h̄ ψ∇ ψ − ih̄(∇ · A)|ψ| − 2ih̄ A · ψ∇ψ + 2 A |ψ| + eφ|ψ|2
ψHψ =
2m
c
c
c
Then substracting the last two equations we get
e h̄2 ∗ 2
ψ ∇ ψ − ψ∇2 ψ ∗ +
ih̄(∇ · A)|ψ|2
2m
mc
e ∂
∂
ih̄A · (ψ ∗ ∇ψ − ψ∇ψ ∗ ) = ih̄ ψ ∗ ψ + ψ ψ ∗
+
mc
∂t
∂t
e h̄2
∂
∇ · ih̄A|ψ|2 = ih̄ |ψ|2
−
∇ · [ψ ∗ ∇ψ − ψ∇ψ ∗ ] +
2m
mc
∂t
−
=⇒
Dividing ih̄ for both side of the last equation and rearranging process, we can get
e ∂
h̄
∇ · A|ψ|2 = 0
|ψ|2 + ∇ · Im(ψ ∗ ∇ψ) −
∂t
m
mc
e ∂
h̄
· A|ψ|2
= 0
|ψ|2 + ∇ ·
Im(ψ ∗ ∇ψ) −
∂t
m
mc
Finally, we get the continuity equation
with j =
h̄ m
Im(ψ ∗ ∇ψ) −
e
2
mc A|ψ|
∂ρ
+∇·j=0
∂t
and ρ = |ψ|2
Page 17
(8)
16. J.J. Sakurai 2.35
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 interaction of the orbital magnetic moment (e/2mc)L with the magnetic field
B. Show that there is also an extra term proportional to B 2 (x2 + y 2 ), and comment briefly on its
physical significance.
(Solution)
We have to choose an appropriate gauge to describe the system easily.
y x 1
A = B − , ,0 = B × ρ
2 2
2
By using this gauge, we can construct the hamiltonian as follow:
H
=
=
=
1 e 2
p̂ − A
2m
c
e2
1
e
p̂2 −
p̂ · A + A · p̂ + 2 A2
2m
c
c
2
e2 B 2 2
p̂
e
p̂ · (B × ρ) + (B × ρ) · p̂ +
−
(x + y 2 )
2
2m 4mc ⎛
⎞ 2mc 4
=
p̂2
e2
e ⎜
⎟
B 2 (x2 + y 2 )
−
⎝B · (ρ × p̂) + (ρ × p̂) ·B ⎠ +
2m 4mc
8mc2
=
e2
p̂2
e
B·L+
B 2 (x2 + y 2 )
−
2m 2mc
8mc2
L
L
(comment)
The hamiltonian can have maximum energy(unstable state) when the two vectors(B and L) are
anti-parallel and minimum energy(stable state) when the two vectors are parallel.
17. J.J. Sakurai 2.36
An electron moves in the presence of a uniform magnetic field in the z-direction (B = Bẑ).
(a) Evaluate
[Πx , Πy ] ,
where
Πx ≡ px −
eAx
,
c
Page 18
Πy ≡ py −
eAy
.
c
(Solution)
[Πx , Πy ] =
=
=
=
e
e px − Ax , py − Ay
c
c
e e
px , − Ay + − Ax , py
c
c
∂
e ∂
Ay −
Ax
ih̄
c ∂x
∂y
e
e
ih̄ ∇ × A = ih̄ B
c
c
ẑ
Note that B only has z component.
(b) By comparing the Hamiltonian and the commutation relation obtained in (a) with those of the
one-dimensional oscillator problem, show how we can immediately write the energy eigenvalues
as
1
h̄2 k 2
|eB|h̄
n+
,
Ek,n =
+
2m
mc
2
where h̄k is the continuous eigenvalue of the pz operator and n is a nonnegative integer including
zero.
(Solution)
Hamiltonian for this system is given by
e 2
1 p̂ − A
H =
2m
c
1 2
1 2
1 2
=
Π +
Π +
Π
2m x 2m y 2m z
Π2y
p2z
Π2x
= Hfree + HB−field
=
+
+
2m
2m 2m
Where, HB−field =
Π2x
2m
+
Π2y
2m
and Hfree =
p2z
2m
Let’s check a commutator relation between Hfree and HB−field .
1 e 2
1 e 2 p2
px − Ax +
py − Ay , z = 0
[Hfree , HB−field ] =
2m
c
2m
c
2m
(9)
Where, we use basic commutator relation as follow:
[pi , pj ] = 0
∂
pi , A (x) = −ih̄
A(x)
∂xi
Since, Hfree and HB−field are commute as shown in equation (9). There exits a set of simultaneous eigenkets |k, n of the operators (Hfree and HB−field ).
If h̄k is the continuous eigenvalue of the operator pz and its eigenkets |k, n, we will have
Hfree |k, n =
h̄2 k 2
|k, n
2m
On the other hand HB−field is similar to the hamiltonian of the one-dimensional harmonic
oscillator problem which is given by
H=
p2
1
+ mw2 x2
2m 2
Page 19
with [x, p] = ih̄. Inorder to get the similar form with the eigen energy of the 1 − D harmonic
oscillator En = h̄w n + 12 we should have the same commutator relation between the squared
operators in the hamiltonian. From (a) we have
e
cΠx
[Πx , Πy ] = ih̄ B =⇒
, Πy = ih̄
c
eB
Considering the commutator relation between p and x, [p,x] and comparing it with the above
commutator relation, one can write HB−field as the following form.
HB−field =
Where, the frequency is given w =
1 2 1
Π + mw2
2m y 2
|eB|
mc
eΠx
|eB|
2
By combining two separate hamiltonian, we get
HB−field |k, n + Hfree |k, n
H|k, n =
1
h̄2 k 2
|eB|
n+
|k, n +
|k, n
mc
2
2m
1
h̄2 k 2
|eB|
n+
+
|k, n
mc
2
2m
=
=
18. J.J. Sakurai 2.37
Consider the neutron interferometer.
Prove that the difference in the magnetic fields that produce two successive maxima in the counting
rates is given by
4πh̄c
ΔB =
|e|gn λ̄l
where gn (=-1.91) is the neutron magnetic moment in units of −eh̄/2mn c. [If you had solved this
problem in 1967, you could have published it in Physical Review Letters!]
(Solution)
First, we have to know the fact that
“A nearly monoenergetic beam of thermal neutrons is split into two parts - path A and path B. Path A
always goes through a magnetic-field-free region; in contrast, path B enters a small region where a static
magnetic field is present.”
Now, Let’s consider a hamiltonian for neutron with spin angular momentum in z-direction
H = −μ · B = wSz
Where w is the spin-precession frequency
w = gn
|eB|
mn c
for the neutron with a magnetic moment of gn eh̄/2mn c
Page 20
We now consider a simultaneous eigenket of energy and spin angular momentum satisfying below
properties.
h̄w
|Sz ; ±
2
h̄
Sz |Sz ; ± = ± |Sz ; ±
2
H|Sz ; ± = ±
We are now in a position to find the time evolution operator.
U (T )
wT
H
wT
T = exp −i
Sz = exp −i
σz
h̄
h̄
2
−iwT
(−iwT /2)2
(−iwT /2)3
= 1+
σz +
+
σz + ...
2
2!
3!
wT
wT
wT
0
cos( wT
2 ) − isin( 2 )
= cos(
) − iσz sin(
)=
0
cos( wT
)
+
isin( wT
2
2
2
2 )
= U (T ) = exp −i
=
0
exp(−i wT
2 )
0
exp(+i wT
2 )
=
n λ̄
0
exp(−i wlm
2h̄ )
n λ̄
0
exp(+i wlm
2h̄ )
Where T is the time spent in the B = 0 region and it is given by
T =
lmn λ̄
l
l
=
=
v
p/mn
h̄
We now have to think about the state ket at the interference region. It is combined with two kets
which went through different path. Only two of one ket is affected by the magnetic field.
So, we get
|β = C (|α + U (T )α) = C[I + U (T )]|α
To find maxima in the counting rates, we have to calculate norm of |β and check its behaviors.
Norm of |β
⎡
β|β =
⎤
C 2 ⎣I 2 + U + U † + U
U † ⎦
I
=
=
0
2cos( wT
2 )
C 2 (2 + U + U † ) = C 2 2 +
0
2cos( wT
2 )
wT
wT
= 4C 2 cos2
C 2 2 + 2cos
2
4
Counting rate is related to the frequency change as follow:
Δw = gn
There are two solutions in the range of 0 ≤
ΔwT
4
|e|ΔB
mn c
< 2π
ΔwT
= 0, π
4
However, 0 gives a trivial solution. Therefore, the counting rates is calculated as follow:
π
=
ΔB
=
|e|ΔB
1
ΔwT
gn
=
4
4
mn c
4πmn c
4πh̄c
=
|e|gn T
|e|gn λ̄l
Page 21
T