Quantum Mechanics - Homework Assignment 4

Alejandro Gomez Espinosa

2
_
_
0 1 0
1 0 1
0 1 0
_
_
, L
y
=
1

2
_
_
0 i 0
i 0 i
0 i 0
_
_
, L
z
=
_
_
1 0 0
0 0 0
0 0 1
_
_
a) Obtain eigenvalues and normalized eigenvectors for all the three operators L
x
, L
y
and L
z
.
To calculate the eigenvalues we use det(L
i
I) = 0:
0 = det(L
x
I) = det

2
0
1

2

1

2
0
1

=
3
+

2
+

2
=
3
= (
2
1)
Then, the eigenvalues are: = 1, 0, 1. For the eigenvectors:
_
_
_
0
1

2
0
1

2
0
1

2
0
1

2
0
_
_
_
_
_
x
y
z
_
_
=
_
_
x
y
z
_
_
Solving this problem for each eigenvalue, we found:
= 1
y

2
= x,
x

2
+
z

2
= y,
y

2
= z
= 0
y

2
= 0,
x

2
+
z

2
= 0,
y

2
= 0
= 1
y

2
= x,
x

2
+
z

2
= y,
y

2
= z
Taking the basis 1| = (100), 2| = (010) and 3| = (001)., the normalized
eigenvectors that satisfy this conditions are:
|L
x
, 1 =
1

2
_

2
|1 +|2
1

2
|3
_
|L
x
, 0 =
1

2
(|1 |3)
|L
x
, 1 =
1

2
_
1

2
|1 +|2 +
1

2
|3
_

gomez@physics.rutgers.edu
1
Now for L
y
:
0 = det(L
y
I) = det

2
0
i

2

i

2
0
i

= (
2
1)
Then, the eigenvalues are: = 1, 0, 1. For the eigenvectors:
_
_
_
0
i

2
0
i

2
0
i

2
0
i

2
0
_
_
_
_
_
x
y
z
_
_
=
_
_
x
y
z
_
_
Solving this problem for each eigenvalue, we found:
= 1
iy

2
= x,
ix

iz

2
= y,
iy

2
= z
= 0
iy

2
= 0,
ix

iz

2
= 0,
iy

2
= 0
= 1
iy

2
= x,
ix

2
+
iz

2
= y,
iy

2
= z
The normalized eigenvectors that satisfy this conditions are:
|L
y
, 1 =
1

2
_
i

2
|1 +|2
i

2
|3
_
|L
y
, 0 =
1

2
(|1 +|3)
|L
y
, 1 =
1

2
_

2
|1 +|2 +
i

2
|3
_
Now for L
z
:
0 = det(L
z
I) = det

1 0 0
0 0
0 1

= ( 1)( 1)
Then, the eigenvalues are: = 1, 0, 1. For the eigenvectors:
_
_
_
0
i

2
0
i

2
0
i

2
0
i

2
0
_
_
_
_
_
x
y
z
_
_
=
_
_
x
y
z
_
_
Solving this problem for each eigenvalue, we found:
= 1 x = x, z = z
= 0 x = 0, z = 0
= 1 x = x, z = z
The normalized eigenvectors that satisfy this conditions are:
|L
z
, 1 = |3
|L
z
, 0 = |2
|L
z
, 1 = |1
2
b) Take the state in which L
z
= 1. In this state what are L
x
, L
2
x
, and L
x
.
If the system is in the state where L
z
= 1, then the system is in the state:
|L
z
, 1 = |1.
L
x
= L
z
, 1|L
x
|L
z
, 1 =
_
1 0 0
_
_
_
_
0
i

2
0
i

2
0
i

2
0
i

2
0
_
_
_
_
_
1
0
0
_
_
= 0
L
2
x
=
_
1 0 0
_
_
_
_
0
i

2
0
i

2
0
i

2
0
i

2
0
_
_
_
_
_
_
0
i

2
0
i

2
0
i

2
0
i

2
0
_
_
_
_
_
1
0
0
_
_
=
1
2
L
x
=
_
L
2
x
L
x

2
=
_
1
2
0 =
1

2
c) Suppose the system is prepared in the state with L
x
= 0, and then L
z
is measured.
What are the possible outcomes and their probabilities?
If the system is prepared in the state with L
x
= 0, then
|

= |L
x
, 0 =
1

2
(|1 |3)
Then, if L
z
is measured, we know from (a) that the possible outcomes are:
L
z
= 1, 0, 1. Now, lets calculate the probabilities:
P(L
z
= 1) =

|P(Lz = 1)|

=
1
2
(1| 3|)(|33|)(|1 |3) =
1
2
P(L
z
= 0) =
1
2
(1| 3|)(|22|)(|1 |3) = 0
P(L
z
= 1) =
1
2
(1| 3|)(|11|)(|1 |3) =
1
2
d) Suppose the system is prepared in the state with L
x
= 0, and then L
z
and L
y
are measured, in that order. What is the probability that the results of both
measurements will be +1?
From the previous problem (c), we found that after measured L
z
the system is
in the state: |1 or in |3). If we measure L
y
= 1 then the probability is:
P(L
y
= 1, L
z
= 1) = L
x
, 0|P(L
z
= 1)P(L
y
= 1)P(L
z
= 1)|L
x
, 0
=
1
2
(1|3|)|11|
_
1
4
(i|1 +

2|2 i|3)(i1| +

22| + i3|)
_
|11|(|1|3)
=
1
8
e) Consider the state
| =
_
_
1/2
1/2
_
1/2
_
_
3
in the L
z
basis. If L
2
z
is measured in this state and a result +1 is obtained,
what is the state after the measurement? How probable was this result? If L
z
is
measured, what are the outcomes and respective probabilities?
The eigenvalues of the operator L
z
is:
0 = det(L
2
z
I) = det

1 0 0
0 0
0 1

= (1 )
2
where the eigenvalues are: = 0, 1 and the eigenvectors are: |L
2
z
, 0 = |2 and
|L
2
z
, 1 =
1

2
(|1 +|3). The state with L
2
z
= 1 is:
|

= |L
2
z
, 1L
2
z
, 1| = (|11| +|33|)
_
1
2
(|1 +|2 +

2|3)
_
=
1
2
|1+

2
2
|3
and the normalized new state is:
|

=
2

3
_
1
2
|1 +

2
2
|3
_
=
1

3
_
|1 +

2|3
_
Then, the probability that the system is in the state L
2
z
= 1 is:
P(L
2
z
= 1) = |P(L
2
z
= 1)| =
=
1
4
_
1| +2| +

23|
_
(|11| +|33|)
_
|1 +|2 +

2|3
_
=
3
4
Finally, if we measure L
z
, we possible values are L
z
= 1, 0, 1 and the proba-
bilities are:
P(Lz = 1) = |L
z
, 1|

|
2
=
1
3
_
3|
_
|1 +

2|3
__
2
=
2
3
P(Lz = 0) = |L
z
, 0|

|
2
=
1
3
_
2|
_
|1 +

2|3
__
2
= 0
P(Lz = 1) = |L
z
, 1|

|
2
=
1
3
_
1|
_
|1 +

2|3
__
2
=
1
3
f ) A particle is in a state for which the probabilities are P(L
z
= 1) = 1/4, P(L
z
=
0) = 1/2, and P(L
z
= 1) = 1/4. Convince yourself that the most general,
normalized state with this property is
| =
e
i
1
2
|L
z
= 1 +
e
i
2

2
|L
z
= 0 +
e
i
3
2
|L
z
= 1
It was stated earlier on that if | is a normalized state then the state e
i
|
is a physically equivalent normalized state. Does this mean that the factors e
i
i
multiplying the L
z
eigenstates are irrelevant? (Calculate for example P(L
x
=
0).)
4
Lets take a general state | = a|1 +b|2 +c|3 and calculate the probabilities:
P(Lz = 1) = |L
z
, 1||
2
= (3| (a|1 + b|2 + c|3))
2
= |c|
2
=
1
4
P(Lz = 0) = |L
z
, 0||
2
= (2| (a|1 + b|2 + c|3))
2
= |b|
2
=
1
2
P(Lz = 1) = |L
z
, 1||
2
= (3| (a|1 + b|2 + c|3))
2
= |a|
2
=
1
2
where the most general solution for the coecients are:
c =
_
1
4
=
e
i
3
2
, b =
_
1
2
=
e
i
2

2
, , a =
_
1
4
=
e
i
1
2
Therefore, the most general state is:
| =
e
i
1
2
|L
z
= 1 +
e
i
2

2
|L
z
= 0 +
e
i
3
2
|L
z
= 1
Now, lets calculate the probability of L
x
= 0:
P(Lx = 0) = |L
x
, 0||
2
=
1
8
_
1| 3|
_
e
i
1
|1 + e
i
2

2|2 + e
i
3
|3
__
2
=
1
8
_
e
i
1
e
i
3
__
e
i
1
e
i
3
_
=
1
8
_
1 e
i(
3

1
)
e
i(
3

1
)
+ 1
_
=
1
4
(1 cos(
3

1
))
then, we know that this factors are not irrelevant.
2) Read the discussion on pp. 133-134 of Shankar about the statistical operator (or den-
sity matrix). Show that the time-dependent statistical operator (t) obeys the evolu-
tion equation
d(t)
dt
=
i

[H, (t)]
You may start from (t) =

i
p
i
|
i
(t)
i
(t)|, where the statistical weights p
i
are
independent of time, and where each |
i
(t) obeys the time-dependent Schrodinger
equation.
Lets calculate the LHS:
d(t)
dt
=
d
dt
_

i
p
i
|
i

i
|
_
=

i
p
i
_
d (|
i
)
dt

i
| +|
i

d (
i
|)
dt
_
Then we now that the time-dependent Schrorinder equation is:
H| = i
d
dt
|
i

H| =
d
dt
|
and the adjoint Schrorinder equation:
H| = i
d
dt
|
i

|H =
d
dt
|
5
Replacing this in the previous relation:
d
dt
=

i
p
i
_

H|
i

i
| +|
i

i
|H
_
=
i

_
H

i
p
i
|
i

i
|

i
p
i
|
i

i
|H
_
=
i

[H, (t)]
Shankar Ex. 4.2.2 Show that for a real wave function (x), the expectation value mo-
mentum P = 0. (Hint: Show that the probabilities for the momenta p are equal.)
Generalize this result to the case = c
r
, where
r
is real and c an arbitrary (real
or complex) constant. (Recall that | and | are physically equivalent.)
Lets compute the probabilities for p:
P(p) = |p||
2
= |pp|
=
_

dx (|xx|p)p|xx|
=
1
2
_

dx

(x)e
ipx
e
ipx
(x)
=
1
2
_

dx
2
(x) since

(x) = (x)
P(p) = |p||
2
= | pp|
=
_

dx (|xx| p)p|xx|
=
1
2
_

dx

(x)e
ipx
e
ipx
(x)
=
1
2
_

dx
2
(x) since

(x) = (x)
As the probabily for p and p are equal, then P = 0. Also, for the previous
relations, is easy to see that the result is maintained if we replace by c.
Shankar Ex. 4.2.3 Show that if (x) has mean momentum P, e
ipox/
(x) has momen-
tum P + p
0
.
First compute P
P = |P|
=
_

dx |xx|P|xx|
= i
_

dx

6
Then, lets calculate this mean value when = e
ipox/
(x)
P = i
_

dx
_
exp
_

ip
o
x

_

x
_
exp
_
ip
o
x

_
= i
_

dx
_
exp
_

ip
o
x

__

x
exp
_
ip
o
x

_

ip
0

exp
_
ip
o
x

_
= i
_

dx

x
i
ip
0

dx

(x)(x)
= P + p
0
4) A particle in 1D has a wavefunction (x) = x| = 1/(x
2
+ a
2
) where a is a positive
real constant. Obtain a correctly normalized version of this wavefunction and com-
pute X, X
2
, P, and P
2
. Also compute the correctly normalized (p) = p|.
(Note: In this kind of problem, you are allowed to quote results fro denite or inde-
nite integrals that you obtain from books, or from programs like Maple or Mathemat-
ica, but please briey cite your source.)
To correctly normalized this wavefunction, rst calculate |:
| =
_

dx |xx| =
_

dx

(x)(x)
=
_

dx
1
(x
2
+ a
2
)
2
=

2a
3
where here, and in the following, all the solutions of the integrals are taken from
Mathematica
1
. Thus, the system is in the normalized state:
x| =
_
2a
3

1
x
2
+ a
2
Now, lets calculate the X, X
2
, P, and P
2
:
X = |X|
=
_

dx |xx|X|xx|
=
_

dx

(x)x(x)
=
2a
3

dx
x
(x
2
+ a
2
)
2
= 0
1
Wolfram Alpha LLC. 2012. WolframAlpha. http://www.wolframalpha.com (access July 10, 2012).
7
X
2
= |X
2
|
=
_

dx |xx|X
2
|xx|
=
_

dx

(x)x
2
(x)
=
2a
3

dx
x
2
(x
2
+ a
2
)
2
=

2a
_
2a
3

_
= a
2
P = |P|
=
_

dx |xx|P|xx|
=
_

dx

(x)
_
i
d
dx
_
(x)
=
2ia
3

dx
_
1
x
2
+ a
2
_
d
dt
_
1
x
2
+ a
2
_
=
2ia
3

dx
_
1
x
2
+ a
2
__
2x
(x
2
+ a
2
)
2
_
=
2ia
3

dx
2x
(x
2
+ a
2
)
3
= 0
P
2
= |P
2
|
=
_

dx |xx|P
2
|xx|
=
_

dx

(x)
_

2
d
2
dx
2
_
(x)
=
2a
3

dx
_
1
x
2
+ a
2
_
d
dt
_
2x
(x
2
+ a
2
)
2
_
=
2a
3

dx
_
1
x
2
+ a
2
__
2
(x
2
+ a
2
)
2

2x
2
(x
2
+ a
2
)
3
_
=
2a
3

_
3
4a
3


8a
5
_
Finally, the wavefunction in the momentum representation is:
(p) = p| =
_

dx p|xx| =
_

dx
p
(x)

(x) =
1

2
_
2a
3

dx
exp
_

ipx

_
x
2
+ a
2
8