Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
Homework Set II
1. Modified Exercise 7.5.4 of Shankar.
(a) The classical partition function for one-dimensional systems is
Z Z
dxdp −βH(x,p)
Zcl =
e
,
2π~
(1)
where β = (kB T )−1 ,kB is the Boltzman constant, and ~ is introduced for comparison
with quantum mechanical results in the appropriate limit. Find Zcl for a classical
oscillator with
p2
1
H(x, p) =
+ mω 2 x2 ,
(2)
2m 2
and show that the average energy
Ēcl = −
∂
ln Zcl
∂β
(3)
is independent of m and ω.
Answer:
Z Z
dxdp −βH(x,p)
e
2π~
Zcl =
Z Z
2
p
+ 12 mω 2 x2
dxdp −β 2m
e
=
2π~
r
r
1
2π
2πm
=
2π~ mω 2 β
β
1
=
~ωβ
∂
ln Zcl
∂β
∂
=
ln (~ωβ)
∂β
1
=
β
Ēcl = −
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(b) Now consider the quantum mechanical partition function
Zqm = Tr e−βH
(11)
Find Zqm for a quantum mechanical oscillator and the corresponding average energy
Ēqm . When does Ēqm approach Ēcl ?
Answer:
Zqm = Tr e−βH
∞
X
=
e−β~ωn
(12)
(13)
n=0
September 20, 2013
1
HW 2
Yong-Zhong Qian
Advanced Quantum Mechanics
Ēqm
Peter Hansen
∞
X
∂
=−
ln
e−β~ωn
∂β
n=0
P∞
−β~ωn
~ωne
n=0
= P
∞
−β~ωn
n=0 e
~ωeβ~ω eβ~ω − 1
=
(eβ~ω − 1)2 eβ~ω
~ω
= β~ω
e
−1
!
(14)
(15)
(16)
(17)
If we take the limit of ~ → 0 then we would expect the classical limit.
lim Ēqm =
~→0
1
β
(18)
(c) Use the result in 1b to find the specific heat per oscillator
C(T ) =
∂ Ē
.
∂T
(19)
What limiting forms does C(T ) have for T → 0 and ∞, respectively?
Answer:
∂ Ē
∂ Ē
= −kB β 2
∂T
∂β
~ω
2 ∂
= kB β
∂β eβ~ω − 1
eβ~ω
C(β) = kB β 2 ~2 ω 2
(eβ~ω − 1)2
C(T ) =
If we let β → 0 (T → ∞) letting β~ω = x,
x2 e x
C(β) = kB
(ex − 1)2
x2
≈ kB 1 −
12
~2 ω 2 β 2
= kB 1 −
12
(20)
(21)
(22)
(23)
(24)
(25)
(26)
If we let β → ∞ (T → 0) letting β~ω = x,
C(β) = kB
≈ kB
x2 e x
(ex − 1)2
x2 e−x
= kB ~2 ω 2 β 2 e−~ωβ
September 20, 2013
2
(27)
(28)
(29)
HW 2
Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
2. Modified Exercise 5.4.3 from Shankar
(a) Consider a quantum mechanical particle subject to a constant force f in one dimension
with the Hamiltonian
P2
− fX
(30)
H=
2m
Find the energy eigenstates hp|Ei in momentum space.
Answer:
In momentum space our Hamiltonian H is
d
p2
− i~f
H=
2m
dp
(31)
The eigenvalue problem is then
2
p
d
− i~f
ψ̃(p) = Ep ψ̃(p)
2m
dp
d
p2
ψ̃(p) − i~f ψ̃(p) = Ep ψ̃(p)
2m
dp
i
d
p2
ψ̃(p) =
ψ̃(p)
Ep −
dp
~f
2m
i
p3
ψ̃(p) = A exp
Ep p −
~f
6m
From Shankar equation 1.10.26 we know that
Z ∞
1
0
dpeip(E−E )/~f = ~f δ(E 0 − E)
2π −∞
(32)
(33)
(34)
(35)
(36)
So to normalize, we require
δ(E − E 0 ) = hE|E 0 i
Z
= hE|pi hp|E 0 i dp
= |A|2
2
Z
e
Z
= |A|
i
~f
3
p
E 0 p− 6m
(37)
(38)
e
−i
~f
p3
Ep− 6m
0
eip(E −E)/~f dp
dp
(39)
(40)
= |A|2 2π~f δ(E 0 − E)
(41)
1
= |A|2
2π~f
(42)
(b) Use the result from Problem 2a and the completeness relation
the propagator in momentum space
K̃(p, t; p0 , 0) ≡ hp| exp(−iHt/~)|p0 i
R
dE |Ei hE| = I to find
(43)
Note that E is continuous and hE|E 0 i = δ(E − E 0 ).
September 20, 2013
3
HW 2
Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
Answer:
K̃(p, t; p0 , 0) ≡ hp| exp(−iHt/~)|p0 i
Z Z
=
hp|Ei hE| exp (−iHt/~) |E 0 i hE 0 |p0 i dEdE 0
Z Z
=
ψE? (p) hE|E 0 i exp (−iE 0 t/~) ψE 0 (p0 )dEdE 0
Z
= ψE? (p) exp (−iEt/~) ψE (p0 )dE
Z
p3 − p03
i
0
2
E(p − p ) −
exp [−iEt/~] dE
= |A|
exp
~f
6m
3
Z
−i p − p03
iE
2
0
= |A| exp
(p − p − tf ) dE
exp
~f
6m
~f
3
−i p − p03
2
= |A| exp
2π~f δ(p − p0 − tf )
~f
6m
3
−i p − p03
= exp
δ(p − p0 − tf )
~f
6m
(c) Use the result from Problem 2b and the completeness relation
the propagator in position space
R
hx|pi = √
(45)
(46)
(47)
(48)
(49)
(50)
(51)
dp |pi hp| = I to find
K(x, t; x0 , 0) ≡ hx| exp(−iHt/~)|x0 i
Answer:
We know that
(44)
1
eipx/~
2π~
K(x, t; x0 , 0) ≡ hx| exp(−iHt/~)|x0 i
Z Z
=
hx|pi hp| exp(−iHt/~)|p0 i hp0 |x0 i dpdp0
3
Z Z
−i p − p03
=
hx|pi exp
δ(p − p0 − tf ) hp0 |x0 i dpdp0
~f
6m
3
Z Z
i
1
−i p − p03
0 0
=
exp (px − p x ) exp
2π~
~
~f
6m
δ(p − p0 − tf )dpdp0
3
Z
1
i
−i p − (p − tf )3
0
=
exp (px − (p − tf )x ) exp
dp
2π~
~
~f
6m
(52)
(53)
(54)
(55)
(56)
(57)
(58)
We need to complete the square in p of the exponential to find the solution. Just looking
September 20, 2013
4
HW 2
Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
at the polynomial in p (factoring out i/~) we have
1
p3 − (p3 − 3p2 f t + 3pf 2 t2 − f 3 t3 )
(59)
6f m
ft
= p(x − x0 ) + tf x0 −
3p2 − 3pf t + f 2 t2
(60)
6f m
t
2
2 2
0 2m
0
(61)
p − pf t + f t /3 + p(x − x )
+ 2mf x
=−
2m
t
t
2m
2
0
0
2 2
=−
p + p(
(x − x ) − f t) + 2mf x + f t /3
(62)
2m
t
"
#
2
2
t
m(x − x0 ) f t
1 2m
=−
p+
−
−
(x − x0 ) − f t) + 2mf x0 + f 2 t2 /3
2m
t
2
4
t
(63)
"
#
2
t
f 2 t2
m(x − x0 ) f t
m2
=−
−
p+
− 2 (x − x0 )2 + mf (x + x0 ) +
(64)
2m
t
2
t
12
p(x − x0 ) + tf x0 −
Having completed the square we can finish the integration
it
f 2 t2
1
m2
0 2
0
0
exp −
K(x, t; x , 0) =
− 2 (x − x ) + mf (x + x ) +
2π~
2m~
t
12
#
"
Z
2
m(x − x0 ) f t
it
p+
−
dp
exp −
2m~
t
2
|
}
√ {z
2πm~/it
r
m
it
m2
f 2 t2
0
0 2
0
K(x, t; x , 0) =
exp −
− 2 (x − x ) + mf (x + x ) +
2iπ~t
2m~
t
12
r
m
i m
f t(x + x0 ) f 2 t3
=
exp
(x − x0 )2 −
−
2iπ~t
~ 2t
2
24m
(65)
(66)
(67)
3. Modified Excercise 8.6.1 from Shankar
Show that the result of Problem 2 is equivalent to
K(x, t; x0 , 0) =
m 1/2
eiScl /~ ,
2πi~t
(68)
where Scl is the action calculated for the classical path of the particle.
Answer:
To calculate the classical action we know that the path is due to a constant force f .
ẍ =
September 20, 2013
f
m
ẋ = v 0 +
f
f t2
t x = x0 + v 0 t +
m
2m
5
(69)
HW 2
Yong-Zhong Qian
Advanced Quantum Mechanics
Peter Hansen
Substituting these into the classical action give
Z t
L(x, ẋ, t)dt
Scl =
0
Z t
m 2
=
ẋ + f x dt
2
0
2
#
Z t" f t2
m
f
0
0
0
=
v + t +f x +vt+
dt
2
m
2m
0
Z t 02
mv
f 2 t2
f 2 t2
0
0
0
+ v ft +
+ fx + fv t +
dt
=
2
2m
2m
0
Z t 02
mv
f 2 t2
0
0
=
+ 2v f t +
+ f x dt
2
m
0
mv 02
f 2 t3
0 2
=
t + v ft +
+ f x0 t
2
3m
(70)
(71)
(72)
(73)
(74)
(75)
Now we substitute in for v 0 = (x − x0 − f t2 /2m)/t.
mv 02
f 2 t3
0 2
t + v ft +
+ f x0 t
Scl =
2
3m
2 mt x − x0
ft
x − x0
ft
f 2 t3
=
−
+
−
f t2 +
+ f x0 t
2
t
2m
t
2m
3m
f 2 t3 f 2 t3
f 2 t3 f t(x − x0 ) m(x − x0 )2
−
+
+ (x − x0 )f t −
+
+ f x0 t
=
8m
2
2t
2m
3m
f 2 t3 f t(x + x0 ) m(x − x0 )2
=−
+
+
24m
2
2t
(76)
(77)
(78)
(79)
which is the same as the above exponent in Equation 67 up to the sign of the second term.
I haven’t found a mistake to give the difference yet.
September 20, 2013
6
HW 2