Download Homework Set III

Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
Homework Set III
1. Modified Exercise 8.6.2 of Shankar
(a) The classical Lagrangian of the harmonic oscillator is
1
1
L = mẋ2 − ω 2 x2 .
2
2
(1)
The corresponding propagator can be written in the form of
K(x2 , T ; x1 , 0) = A(T ) exp(iScl /~)
(2)
where Scl is the action along the classical path. Find Scl .
Answer:
We choose the classical path to be
x(t) = x0 sin(ωt − δ)
(3)
So
1
L = mω 2 x20 cos2 (ωt + δ) − sin2 (ωt + δ)
2
1
= mω 2 x20 cos(2(ωt + δ))
2
(4)
(5)
So we find the classical action as
Z T
1
2 2
Scl = mω x0
cos [2(ωt + δ)] dt
2
0
1
= mωx20 (sin(2(ωT + δ)) − sin(2δ))
4
(6)
(7)
We solve our path for x1 and x2
x1 = x(0) = x0 sin δ
x2 = x(T ) = x0 sin(ωT + δ)
= x0 sin ωT cos δ + x1 cos ωT
Solving for x0 cos δ
x0 cos δ =
x2 − x1 cos ωT
sin ωT
(8)
(9)
(10)
(11)
Putting this into the action
1
Scl = mωx20 (sin(2(ωT + δ) − sin(2δ))
4
1
= mωx20 (sin(ωT + δ) cos(ωT + δ) − sin δ cos δ)
2
1
x2 − x1 cos ωT
= mω x2 x0 cos(ωT + δ) − x1
2
sin ωT
September 27, 2013
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(12)
(13)
(14)
HW 3
Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
Now we expand the cos.
x0 cos(ωT + δ) = x0 cos ωT cos δ − x0 sin ωT sin δ
x2 − x1 cos ωT
− x1 sin ωT
= cos ωT
sin ωT
1
x2 − x1 cos ωT
x2 − x1 cos ωT
Scl = mω x2 cos ωT
− x1 x2 sin ωT − x1
2
sin ωT
sin ωT
mω
Scl =
x22 cos ωT − x1 x2 cos2 ωT − x1 x2 sin2 ωT − x1 x2 + x21 cos ωT
2 sin ωT
mω
Scl =
(x22 + x21 ) cos ωT − 2x1 x2
2 sin ωT
(15)
(16)
(17)
(18)
(19)
(b) Show that
Z
∞
X
∞
K(x, T ; x, 0)dx =
−∞
exp (−iEn T /~) ,
(20)
n=0
where En is the energy eigenvalue of the quantum mechanical oscillator.
Answer:
We know that
X
ψn (x)ψn∗ (x0 ) exp (−iEn T /~)
K(x, T ; x0 , 0) =
(21)
n
So
Z
∞
K(x, T ; x, 0)dx =
−∞
X
Z
∞
exp (−iEn T /~)
−∞
n
ψn (x)ψn∗ (x) =
X
exp (−iEn T /~) (22)
n
(c) Use Part 1b to find A(T ).
Answer:
From Part 1a we know that
K(x2 , T ; x1 , 0) = A(T ) exp(iScl /~)
imω(cos ωT − 1)
K(x, T ; x, 0) = A(T ) exp
x2
~ sin ωT
Z ∞
Z ∞
imω(cos ωT − 1)
K(x, T ; x, 0)dx = A(T )
exp
x2 dx
~
sin
ωT
−∞
s−∞
X
π~ sin ωT
exp (−iEn T /~) = A(T )
imω(cos ωT − 1)
n
September 27, 2013
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(23)
(24)
(25)
(26)
HW 3
Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
Looking at the infinite sum we can simplify that to
X
X
exp (−iEn T /~) =
exp (−iωT (n + 1/2))
n
(27)
n
= exp (−iωT /2)
X
[exp (−iωT )]n
(28)
n
1
1 − exp (−iωT )
1
=
exp (iωT /2) − exp (−iωT /2)
1
=
2i sin ωT /2
= exp (−iωT /2)
(29)
(30)
(31)
Putting this into Equation 26 we get
s
1
π~ sin ωT
= A(T )
2i sin ωT /2
imω(cos ωT − 1)
(32)
Solving for A(T )
r
imω(cos ωT − 1)
1
π~ sin ωT
2i sin ωT /2
r
mω
A(T ) =
2πi~ sin ωT
A(T ) =
(33)
(34)
2. Modified Exercise 8.6.3 of Shankar.
(a) Use the result from Part 1 to give K(x, T ; x, 0) for the harmonic oscillator.
Answer:
K(x2 , T ; x1 , 0) = A(T ) exp(iScl /~)
r
mω
imω
2
2
K(x2 , T ; x1 , 0) =
exp
(x2 + x1 ) cos ωT − 2x1 x2
2πi~ sin ωT
2~ sin ωT
r
mω
imω(cos ωT − 1) 2
K(x, T ; x, 0) =
exp
x
2πi~ sin ωT
~ sin ωT
(35)
(36)
(37)
(b) Express K(x, T ; x, 0) in terms of the wave functions φn (x) for the energy eigenstates of
the harmonic oscillator.
Answer:
K(x, T ; x, 0) =
X
|ψn (x)|2 exp (−iEn T /~)
(38)
X
(39)
n
K(x, T ; x, 0) = exp(−iωT /2)
|ψn (x)|2 [exp (−iωT )]n
n
September 27, 2013
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HW 3
Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
(c) By expanding K(x, T ; x, 0) in powers of α = exp(−iωT ), derive φ0 (x) and φ1 (x).
Answer:
The first two terms are
X
K(x, T ; x, 0) =
|ψn (x)|2 αn+1/2
(40)
n=0,1
= |ψ0 (x)|2 α1/2 + |ψ1 (x)|2 α3/2
Taking Equation 37 we write it in terms of α.
r
mω
imω(cos ωT − 1) 2
exp
x
K(x, T ; x, 0) =
2πi~ sin ωT
~ sin ωT
r
mω
imω(α + α−1 − 1) 2
=
exp
x
2πi~(α − α−1 )
~(α − α−1 )
(41)
(42)
(43)
(44)
The square root becomes
r
mω
≈
2πi~(α − α−1 )
r
imω 1/2
α
2π~
and the exponent becomes
imω(α + α−1 − 1) 2
−imωx2
imωx2
x ≈ exp
α
exp
1+
~(α − α−1 )
~
~
(45)
(46)
So we can write
r
K(x, T ; x, 0) =
imω
exp
2π~
−imωx2
~
imωx2 3/2
1/2
α
α +
~
Setting this equal to Equation 41 gives the two equations below
r
imω
−imωx2
exp
= |ψ0 (x)|2
2π~
~
r
imωx2
imω
−imωx2
exp
= |ψ1 (x)|2
~
2π~
~
(47)
(48)
(49)
3. Modified Exercise 8.6.4 of Shankar
~ = Bxŷ corresponds to the uniform magnetic field
(a) Show that the vector potential A
~ = B ẑ.
B
Answer:
~ =∇
~ ×A
~ = ∂Ay ẑ = B ẑ
B
∂x
September 27, 2013
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(50)
HW 3
Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
(b) The classical Lagrangian for a particle of charge q in the above field is
q
1
~
L(~x, ~x˙ ) = m~x˙ 2 + ~x˙ · A,
2
c
(51)
where ~x = xx̂ + y ŷ + z ẑ. Consider the wave function obtained from the path integral:
Z
iε
ψ(~x, t + ε) = C exp
L(~x + ~η /2, −~η /ε) ψ(~x + ~η , t)d3 ~η ,
(52)
~
where C is a constant, ε → 0+ , and −~η /ε → ~x˙ . Derive the Shroedinger equation for
ψ(~x, t).
Answer:
mη 2 q ~η
+
· Bxŷ,
2ε2
cε
mη 2 qBxηy
=
+
2ε2
cε
L(~x + ~η /2, −~η /ε) =
(53)
(54)
(55)
Z
ψ(~x, t + ε) = C
iε
exp
~
mη 2 qBxηy
+
2ε2
cε
ψ(~x + ~η , t)d3 ~η ,
(56)
(c) Show that the result agrees with the quantum mechanical Hamiltionian
1 ~ q ~ 2
P− A ,
H=
2m
c
(57)
~
where P~ = −i~∇.
Answer:
September 27, 2013
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HW 3