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PHY 389K Quantum Mechanics, Homework
Set 5 Solutions
Matthias Ihl
03/03/2008
Note: I will post updated versions of the homework solutions on my home-
page: http://zippy.ph.utexas.edu/~msihl/teaching.html
1 Problem 1
Consider a system whose density operator is ρ(t), evolving under the influence
of a hamiltonian H(t). Show that the trace of ρ2 does not vary with time.
Can the system evolve so as to be successively in a pure state and in a
statistical mixture of states?
Ans. Let U(t) ≡ U(t, 0) be a time-evolution operator. Then,
ρ(t) ≡
∑
i
wi|αi, t〉〈αi, t| =
∑
i
wiU(t)|αi, 0〉〈αi, 0|U †(t)
= U(t)
[
∑
i
wi|αi, 0〉〈αi, 0|
]
U †(t) = U(t)ρ0U †(t)
Because U is unitary,
ρ(t)2 = U(t)ρ0U †(t)U(t)ρ0U †(t)
= U(t)ρ2
0
U †(t)
and
Tr(ρ(t)2) =
∑
j
〈bj , t|ρ(t)2|bj , t〉 =
∑
j
〈bj, 0|U †(t)U(t)ρ20U †(t)U(t)|bj , 0〉
=
∑
j
〈bj , 0|ρ20|bj, 0〉 = Tr(ρ20)
1
Pure states satisfy Tr(ρ2) = 1 and mixed states satisfy Tr(ρ2) < 1. There-
fore, the system cannot evolve successively in a pure state and in a mixture
state.
2 Problem 2
Compute the propagator K(xf , tf ; xi, ti) for the simple harmonic oscillator.
Ans. For the simple harmonic oscillator,
S =
m
2
∫
(ẋ2 − ω2x2)dt,
and x(t) satisfies the classical eqution
ẍ+ ω2x = 0
Assume that u(t) is a classical solution of this equation, and define a new
variable
X ≡ u−1x.
Then,
S =
m
2
∫ tf
ti
[
u2Ẋ2 +
d
dt
(uu̇X2) +
(
u̇2 − d
dt
(uu̇) − ω2u2
)
X2
]
dt
=
m
2
( u̇f
uf
x2f −
u̇i
ui
x2i
)
+
m
2
∫ f
i
u2Ẋ2dt (ẍ+ ω2x = 0)
where ui = u(ti), uf = u(tf). Define one more new variable, a new time
variable
η ≡
∫
dt
u2
Then
S =
m
2
( u̇f
uf
x2f −
u̇i
ui
x2i
)
+
1
2
∫ ηf
ηi
X ′2(η)η (1)
where the prime denotes the derivative with respect to η. From the relation
∆tk =
∫ ηk+1
ηk
u2dη = u2(ηk)∆ηk +
1
2
du2
dη
∣
∣
ηk
∆η2k +O(∆η
3
k)
= u2k∆ηkexp
[ 1
2u2
du2
dη
∣
∣
∣
k
∆ηk
]
+O(∆η3k),
we get the following relation
D[x] = lim
N→∞
exp(−1
2
d lnu
dτ
|0∆η0)
ui
√
2πi~∆η0
ΠNk=1
exp(−1
2
d ln u
dτ
|k∆ηk)u−1k dxk√
2πi~∆ηk
=
1
ui
exp
[
− 1
2
∫ ηf
ηi
d ln u
dη
dη
]
DX = 1√
uiuf
D[X]
(2)
So, using (1) and (2),
K(xf , tf ; xi, ti) =
1
√
uiuf
exp
[mi
2~
( u̇f
uf
x2f −
u̇i
ui
x2i
)]
∫
D[X]exp
(mi
2~
∫
X ′2dη
)
(3)
Using the fact of the free particle problem to get the integral over X(Sakurai
2.5.16),
∫
D[X]exp
(mi
2~
∫
X ′2dη
)
=
√
m
2πi~(ηf − ηi)
exp
[mi(Xf −Xi)2
2~(ηf − ηi)
]
Let’s find an explicit expression for ηf − ηi in terms of the original time
t. Let u(t), v(t) be two independent solutions of the classical equation of a
harmonic oscillator. Then the Wronskian
W ≡ uv̇ − u̇v
is contant in time and
ηf − ηi =
1
W
∫ tf
ti
(uv̇ − u̇v
u2
)
dt =
1
W
∫ tf
ti
d
dt
(
v
u
)dt =
uivf − ufvi
Wuiuf
If we normalize u, v in such a way that
uivf − ufvi = 1, (4)
(3) becomes
K(xf , tf ; xi, ti) =
√
mW
2πi~
exp
[mi
2~
(Wui + u̇f
uf
x2f +
Wuf − u̇i
ui
x2i − 2Wxixf
)]
Moreover, from the fact that
W = uiv̇i − u̇ivi = uf v̇f − u̇fvf ,
we get
K(xf , tf ; xi, ti) =
√
mW
2πi~
exp
[mi
2~
((uiv̇f−viu̇f)x2f +(uf v̇i−vf u̇ix2i −2Wxixf )
]
(5)
For a harmonic oscillator, the two independent solutions staisfying (4) are
u(t) =
cosωt√
sinωT
, v(t) =
sinωt√
sinωT
where T = tf − ti. Finally, substituting these solutions in (5), we have
K(xf , tf ; xi, ti) =
√
mω
2πi~ sinωT
exp
[ miω
2~ sinωT
((x2f + x
2
i ) cosωT − 2xixf )
]
3 Problem 3
Consider a harmonic oscillator of mass m and angular frequency ω. At time
t = 0, the state of this oscillator is given by:
|ψ(0)〉 =
∑
n
cn|n〉
where |n〉 are the stationary states with energy ~ω(n + 1
2
). What is the
probability P that a measurement of the oscillator’s energy performed at an
arbitrary time t > 0 will yield a result greater than 2~ω? When P = 0, what
are the non-zero coefficients cn?
Ans.
|ψ(t)〉 = U(t)|ψ(0)〉 = exp
(−iHt
~
)
|ψ(0)〉 =
∑
n
cnexp
(−iEnt
~
)
|n〉
∑
n
cnexp
(
− iω(n+ 1
2
)
)
|n〉
Therefore, the probability to get ~ω(n+ 1
2
) equals to |〈n|ψ(t)〉|2 = |cn|2. For
n > 2, the energy is greater than 2~ω. Hence,
P =
∞
∑
n=2
|〈n|ψ(t)〉|2 =
∞
∑
n=2
|cn|2 = 1 − (|c0|2 + |cn|2).
• If P = 0, then |c0|2 + |c1|2 = 1. So if we take c0 be real, then
c0 = cos θ and c1 = sin θe
iφ, φ ∈ R
Special thanks to Yoonsang Lee (ylee@math.utexas.edu) for providing these
solutions.