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# QM II Homework Assignment 8

## Time Dependent Perturbation Theory:

Chris Mueller
Dept. of Physics, University of Florida
26 March, 2009
Collaborators: Bobby Bond, Darsa Donelan, Lana Muniz, and Aaron Spector
1. Shankar 18.2.3.
Consider a particle in the ground state of a box of length L. Argue on semi-classical
grounds that the natural time period associated with it is T nL
2
/. If the box expands
symmetrically to double its size in time T what is the probability of catching the
particle in the ground state of the new box?
The relative semi-classical time scale is the time that it would take a particle to cross the well.
The quantum mechanical ground state energy of the innite square well is
E
1
=

2

2
2mL
2
.
The classical speed of a particle with this energy is
v =
_
2E
m
=

mL
.
The time to travel across the well is therefore
T =
v
L
=
mL
2

We can use this time scale to judge whether or not a perturbation to the potential is sudden or
adiabatic. For changes on much shorter timescales, the classical particle will not have crossed the
well and will therefore not know about the change. On the other hand, for changes much longer
than this timescale the particle will bounce o of the walls many times during the change and will
therefore not gain or lose energy during the perturbation.
For this problem, we are given that the time over which the perturbation happens is much
less than our timescale; T. This is therefore a sudden perturbation and we will expect the
wavefunction of state to be unchanged during the course of the perturbation. The initial (and
therefore nal) state of the particle is
(x) =
_
2
L
cos
_
x
L
_
.
The new ground state is

0
(x) =
_
1
L
cos
_
x
2L
_
.
1
Hence, the probability of nding the particle in the ground state is
P(
0
) =

_
L/2
L/2
_
2
L
2
cos
_
x
2L
_
cos
_
x
L
_
dx

2
=

2
L
_
4

2l
3
_

2
=
_
8
3
_
2
0.721.
2. Shankar 18.2.5
An oscillator is in the ground state of H = H
0
+ H
1
, where the time-independent per-
turbation H
1
is the linear potential V = fx. If at t = 0, H
1
is abruptly turned o,
show that the probability that the system is in the nthe eigenstate of H
0
is given by the
Poisson distribution
P(n) =
e

n
n!
where =
f
2
2m
3

We need to nd the eigenstates of the perturbed Hamiltonian before we can proceed with the
problem. We are fortunate in that this system is exactly solvable. Consider the Hamiltonian,
H =
p
2
2m
+
1
2
m
2
x
2
fx
=
p
2
2m
+
1
2
m
2
_
x
f
m
2
_
2

1
2
f
2
m
2
From this form we can see that the perturbed Hamiltonian is simply a harmonic oscillator with
a shifted constant energy and a translated center. The perturbed ground state wavefunction can
be found from the old unperturbed one by applying the translation operator.
|0 = T
_
f
m
2
_
|0
0

= exp
_
i
f
m
2
p
_
|0
0

= exp
_
i
f
m
2
i
_
m
2
(a

a)
_
|0
0

= exp
__
f
2
2m
3
(a

a)
_
|0
0
We want to expand the exponential into two terms, but we must be careful because non-
commuting operators do not satisfy the usual relation ship. Recall that for two operators A and
B,
e
A+B
= e
A
e
B
e

1
2
[A, B]
.
Before we return to the above equation, let us also compute a useful commutator. For operators
A and B and a scaler quantity a,
[aA, aB] = (aA)(aB) (aB)(aA) = a
2
(AB +BA) = a
2
[B, A].
2
Applying these two relations, the new state becomes
|0 = exp
__
f
2
2m
3
a

_
exp
_

_
f
2
2m
3
a
_
exp
_

f
2
4m
2
_
|0
0

= e

a
e

1
2

|0
0

We now want to expand the two exponential terms in terms of their Taylor expansions. Since
the operators will be acting on the ground state, all of the terms in the expansion of the second
operator will go to zero except for the identity. We can therefore drop it completely. Before we
expand the other term, it is helpful to recall that the eigenstates of the harmonic oscillator can be
expressed in terms of the ground state and the raising operator as
|n
0
=
(a

)
n

n!
|0
0

## Lets expand the exponential in the above expression.

|0 = e

1
2

)
n
n!
(a

)
n
|0
0

= e

1
2

)
n

n!
|n
0

Finally, since this perturbation is applied suddenly, the system will remain in this state even
after the perturbation is applied. The probability of being in one of the new eigenstates of the
unperturbed harmonic oscillator is simply its overlap with those states.
P(n) = |n
0
|0|
2
=
e

n
n!
3. For a spin
1
2
particle, consider the time dependent Hamiltonian,
H(t) = n(t) ,
where is the vector formed from the Pauli spin matrices. The unit vector n changes
with time. Suppose initially n = z, and the spin is in the ground state. Next, n is rotated
so that it points in the x direction. What is the wavefunction when n reaches x for the
case of (a) a rotation in time T
1
and (b) a rotation in time T
1
? (Note: you
do not need to have the correct phase for the wave function for case (b). The phase will
depend on the exact details of how the state is changed form z to x and we will discuss
this later.) For both cases, what is the probability that the spin will be in the new ground
state ( n = x)?
Lets begin this problem by determining the eigenvalues and eigenvectors of the initial and nal
Hamiltonians. At time t = 0, n points in the z direction and the Hamiltonian is
H(0) =
z
=
_
0
0
_
.
This is clearly already diagonal and so we can identify the eigenvectors and their energies as
| =
_
1
0
_
& | =
_
0
1
_
.
3
Now for the nal Hamiltonian. At time t = T, n points in the x direction and the Hamiltonian
has the form
H(T) =
x
=
_
0
0
_
.
We diagonalize this in the usual way, by calculating the characteristic determinant and setting it
equal to zero.

=
2

2
= ( )( + ) = 0 =
Plugging these eigenvalues back in to nd the eigenvectors reveals them to be
| =
1

2
_
1
1
_
& | =
1

2
_
1
1
_
.
Part a
Since the perturbation in this case is very sudden, we expect the wavefunction to remain
unchanged after the perturbation is applied. Hence, the perturbed wavefunction is
|
x
=
_
1
0
_
.
Hence, the probability of it being in the new ground state is
P() =

_
1 0
_
1

2
_
1
1
_

2
=
1
2
.
Part b
In this case, the approximation is adiabatic and we will expect the state of
z
to track to the
ground state of
x
. Hence, the state after the perturbation will be
|
x
=
1

2
_
1
1
_
.
Which means that the probability of nding the system in the ground state is one.
P()

2
_
1 1
_
1

2
_
1
1
_

2
= 1
4
4. A time dependent Hamiltonian has the form:
H(t) =
p
2
2m
+
1
2
m
2
(t)x
2
,
where the only time dependence comes from (t). This is equivalent to varying the spring
constant with time. In this problem, we will make use of equations 7.3.21 and 7.3.22 from
the book for the harmonic oscillator wavefunctions, together with the Gaussian integrals:
_

dxe
ax
2
=
_

a
_

dxx
2
e
ax
2
=
1
2

a
3
(a) From t = 0 to t = T the angular frequency changes from
0
to 2
0
. At t = 0 the
system is in the ground state of H(t = 0). If
0
T 1, what is the probability that
the system will still be in the ground state at t = T?
(b) What is the probability that it will be in the rst excited state at t = T? What is
the probability that it will be in the second excited state at t = T?
0
T 1. What is the probability that the system will still be
in the ground state at t = T?
(d) What is the probability that it will be in the rst excited state at t = T? What is
the probability that it will be in the second excited state at t = T?
Part a
Recall that the wavefunctions of the harmonic oscillator are Gaussians modulated by Hermite
polynomials. The ground state is
|
0
(
0
, x) =
_
m
0

_1
4
exp
_

m
0
x
2
2
_
.
Since the semi-classical time scale of the harmonic oscillator is T =
1
f

1

## and this perturbation

happens on a time scale much less than this, the wavefunction will remain unchanged through the
perturbation. Hence, the probability of the particle being in the new ground state is simply the
overlap with the wavefunction of the new ground state.
P (|
0
(2
0
, x)) = |
0
(
0
, x)|
0
(2
0
, x)|
2
=
_
_
_

dx

2m
0

exp
_

3m
0
x
2

_
_
_
2
=

2m
0

2
3m
0
=
2
3
2
3
0.9428
Part b
Since this is the same situation as in part a, the system will remain in the original ground
state after the perturbation is applied. The probability of being in the rst excited state of the
5
perturbed potential is
P (|
1
(2
0
, x)) =
_
m
0

2
5
4

dxx exp
_

m
0
x
2
2
_
_
2
= 0 .
So, the particle has a zero probability of ending up in the rst excited state of the perturbed
potential.
For the second excited state,
P (|
2
(2
0
, x)) =
__
m
0

2
_

dx
_
4m
0

x
2
1
_
exp
_

3m
0
x
2
2
__
2
=
m
0

2
_
_
2
3m
0
_

dx
_
8
3
u
2
1
_
e
u
2
du
_
2
=

2
3
_
8
3

_
2
=

2
27
0.0524
So we see that the particle is much more likely to end up in the ground state than in the rst
excited state.
As a check on our calculations we can make sure that these values do not sum up to more than
one. Upon checking we nd that
P (|
0
(2
0
, x)) +P (|
2
(2, x)) 0.9952
which conrms that our calculations are at least close. This also implies that the particle only has
a 0.48% chance of ending up in a state other than the ground state or second excited state.
Part c
Since this perturbation is applied on a much longer time scale than the natural time scale of the
system, the particle will end up in the ground state of the perturbed potential. The wavefunction
of this perturbed ground state is
|
0
(2
0
, x) =
_
2m
0

_1
4
exp
_

m
0
x
2

_
.
Hence, the overlap with the perturbed ground state is 1.
P (|
0
(2
0
, x)) = 1
Part d
As with part c, the slow perturbation will cause the particle to end up in the new ground state
with 100% probability. It will therefore have no overlap with the excited states.
P (|
1
(2
0
, x)) = 0 & P (|
2
(2
0
, x)) = 0
6
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