Prob. 9.7.5
(solution by Michael Fisher)
1
We begin by expressing the initial state in a basis of the spherical harmonics, which will
allow us to apply the operators L̂2 and L̂z .
r
hθ, φ| φ(0)i =
3
1√
i
sin θ sin φ =
2(Y1,−1 − Y1,1 ) = √ (Y1,1 − Y1,−1 )
4π
2i
2
Now we apply the Hamiltonian to hθ, φ| φ(0)i.
hθ, φ| Ĥ |φ(0)i = (
L̂2
i ~2
~2
+ ωo L̂z ) hθ, φ| φ(0)i = √ (( + ~ωo )Y11 − ( − ~ωo )Y1,−1 )
2I
2I
2 2I
Now we can use these results to find hθ, φ| φ(t)i.
~
~
hθ, φ| φ(t)i = hθ, φ| e−iĤt/~ |φ(0)i = e−i( 2I +ωo )t Y11 − e−i( 2I −ωo )t Y1,−1
Finally, we compute hL̂x i.
hL̂x i =
1
1
hφ(t)| (L̂+ + L̂− ) |φ(t)i = hφ(t)| ((...)Y10 + (...)Y10 ) = 0
2
2
Maggie Regan
Physics 113
Boccio 9.7.7.
In this problem we are given the three operators on a 3-dimensional Hilbert space,


0 1 0
1
Lx = √  1 0 1  ,
2 0 1 0


0 −i 0
1
Ly = √  i 0 −i  ,
2 0 i
0


1 0 0
Lz =  0 0 0 
0 0 −1
.
In part (a) we want to find the possible values that we can obtain if Lz is measured.
Since Lz is already diagonalized, the possible values are the diagonal entries of 1, 0, and −1.
2
In part
p (b), a state of Lz = 1 is taken. Then, we are asked to find hLx i, hLx i, and
2
2
4Lx = hLx i − hLx i . So,
 
1

0 
|Lz = 1i =
0
.
Also,


1 0 1
1
L2x =  0 2 0  ,
2
1 0 1
q
1
2
Then, h1|Lx |1i = 0, h1|Lx |1i = 2 , and thus, 4Lx = 12 .
For part (c), we want to find the normalized eigenstates and eigenvectors of Lx in the
Lz basis. The eigenvalues of the Lx matrix are 1, 0, and -1. The corresponding normalized
eigenvectors are


1
2
q
1
2


q
− 12

v~2 = 
 q0


v~1 =  −
,
1
2

1
2
1



, and
1

q2

v~3 = 
1
2
.

1
2


These vectors and eigenvalues are already in the Lz basis, so no change of basis needs to
be done.
For part (d) we are given that the particle is in the state with Lz = −1. Then, Lx is
measured. We want to find the possible outcomes and their probabilities. To do this we find
the projection operators for the eigenvalues of the Lx matrix. So, |1ih1| = 41 , |0ih| = 21 , and
| − 1ih−1| = 14 .
In part (e) we have the state

1
|ψi = √ 
2
√1
2
√1
2


1
,
which is in the Lz basis. We have that L2z is measured and a result of 1 is obtained.
So, for L2z , 1 is a degenerate eigenvalue, so the state will be a linear combination of two
vectors. To find this linear combination we have that for Lz |ψi, prob(1) = 31 , prob(0) = 0,
and prob(-1) = 32 . We get these values by renormalizing the vector. This renormalization
factor is 34 because from the original state, we have that prob(1) = 14 + 24 = 34 .
Thus, the state is 13 ~v1 + 23 ~v3 , where
 
1

0 
~v1 =
0
and


0
~v3 =  0 
1
.
Thus,


1
1
|ψi = √  √0 
3
2
.
For part (f ), we have that a particle is in the state where prob(1) = prob(-1) = 14 and
prob(0) = 12 . This means that the particle is in the state with the basis of Lz . Then, we know
that the generalized state for this particle is |ψi = 21 |Lz = 1i + √12 |Lz = 0i + 21 |Lz = −1i. We
know this because the probabilities listed are simply the squares of the coefficients of the basis
states. However, we need to include phase factors for each term, because these phase factors
will cancel when Lz is operatingo n the state. So, it is easy to see that the most general,
iδ
−iδ
−iδ
normalized state with this property is |ψi = e 2 1 |Lz = 1i + e √22 |Lz = 0i + e 2 3 |Lz = −1i.
2
Then, we want to determine whether the phase factors are necessary. To see this, let’s
−iδ −e−iδ3
calculate P (Lx = 0). So, P (Lx = 0|ψ) = |hLx = 0|ψi|2 = | e 12√
|2 = 14 (1 − cos(δ1 − δ3 )).
2
Therefore, the relative phase differences are not irrelevant. The arbitrary phase factors must
be included when the state doesn’t commute with L̂z .
3
Maggie Regan
Physics 113
Boccio 9.7.7.
In this problem the z-component of the spin of an electron is measured and found to be
~
.
2
For part (a), we want to know that if a measurement is then made of the x-component
of the spin, what the possible values could be. If one component of the spin is known, then
the other components cannot be known. Also, there are only ever two values of spin that
.
are possible. Therefore, we have the possible values of ~2 and −~
2
Then, for part (b) we are asked to find the probabilities of finding these various results. So,
we have that there are equal possibilities for the two values, and therefore the probabilities
are both 21 . However, for a more mathematical method, we have that h+x| = √12 (1 1) and
h−x| = √12 (1 − 1). Thus, the probability of measuring the up spin of the x-component is
|| + zih+x||2 = 21 || + zi(1 1)|2 = 21 . Then, the probability of measuring the down spin of the
x-component is || + zih−x||2 = 12 || + zi(1 − 1)|2 = 12 .
In part (c), the axis defining the measured spin direction an angle θ with respect to the
z-axis. We then want to find the probabilities of the various possible results. So, we have
that
cos(θ/2)
| + n̂i =
eiφ sin(θ/2)
and
| − n̂i =
sin(θ/2)
−eiφ cos(θ/2)
.
Then, we have | + zi = cos(θ/2)|n̂i + sin(θ/2)|n̂i. Using this, we have that P ( ~2 |z) =
|h+n|zi|2 = cos2 (θ/2) and P ( −~
|z) = |h−n|zi|2 = sin2 (θ/2).
2
For part (d) we want to find the expectation value of the spin measurement of part
(c). We can find this by multiplying
eigenvalues by their respective probabilities. So,
the
2
~
~
2
hmeasurementi = 2 cos (θ/2) + − 2 sin (θ/2) = ~2 (cos2 (θ/2) − sin2 (θ/2)) = ~2 cos(θ).
1
Prob. 9.7.12
(solution by Michael Fisher)
1
a
We know that the probability of |0, 0i is
write this state as
3
4
and the probability of |1, −1i is 14 . So, we can
√
3
eiθ
|φi =
|0, 0i +
|1, −1i
2
2
where θ is an arbitary phase angle.
b
In order to determine the effect of a measurement of L̂x , we must switch to the basis of the
eigenvectors of Lx . From problem 9.7.7, we know that in an L = 1 state, the matrix Lx is
given by


0 1 0
1
Lx = √ 1 0 1
2
0 1 0
which has eigenvalues
λ1 = −1
λ2 = 1
λ3 = 0
and corresponding eigenvectors


1
√
1
~v1 = − 2
2
1
 
1
1 √ 
~v2 =
2
2
1
Prob. 9.7.12
(solution by Michael Fisher)
2

−1
1
~v3 = √  0 
2
1

in the Lz basis. Let
 
0

w
~ 1 = 0
1
This is the eigenvector associated with the eigenvalue of Lz of −1, which is the state we are
in. So, we can write the state |1, −1i in the Lx basis as
1
1
1
|1, −1i = |Lz = −1i = hw
~ 1 | ~v1 i ~v1 + hw
~ 1 | ~v2 i ~v2 + hw
~ 1 | ~v3 i ~v3 = ~v1 + ~v2 + √ ~v3
2
2
2
So,
1
4
1
prob(1) = prob(λ2 ) = prob(~v2 ) =
4
1
prob(0) = prob(λ3 ) = prob(~v3 ) =
2
prob(−1) = prob(λ1 ) = prob(~v1 ) =
c
We can write |1, −1i in the Lx basis as spatial wavefunctions by simply converting to the
spherical harmonics. So,
1
1
1
1
1
1
hθ, φ| ( ~v1 + ~v2 + √ ~v3 ) = Y1,−1 + Y1,1 + √ Y1,0
2
2
2
2
2
2
Prob. Z5-16
(solution by Alexandra Werth)
1
We are given
Ψ(θ, φ) =
q
3
Y (θ, φ)
8 11
+
q
1
Y (θ, φ)
8 10
+ AY1−1 (θ, φ)
We can quickly see that Y11 (θ, φ) is the Hilbert basis of Ψ(θ, φ) therefore we can rewrite Ψ
in terms of its ket vectors.
q
q
Ψ(θ, φ) = 38 |1, 1i + 18 |1, 0i + A |1, −1i
(a) To normalize Ψ the square of the coefficients infront of the kets must sum to one.
3
8
+ 18 + A2 = 1
q
A = 12
(b) Next, we want to find L̂+ Ψ(θ, ψ). First, lets define the operater L̂+ .
p
L̂+ Yl,m (θ, φ) = ~ l(l + 1) − m(m + 1)Yl,m+1 (θ, φ)
We can now apply this operator to Ψ(θ, φ) defined above.
q √
q √
L̂+ Ψ(θ, φ) = 18 ~ 2 |1, 1i + 12 ~ 2 |1, 0i
= 21 ~ |1, 1i + ~ |1, 0i
(c) Now, we want to solve for L̂x Ψ(θ, φ). The operator L̂x is defined as L̂x = 12 (L̂+ + L̂− .
The operator L̂− is,
p
L̂− Yl,m (θ, φ) = ~ l(l + 1) − m(m − 1)Yl,m−1 (θ, φ)
Therefore,
L̂− Ψ(θ, φ) =
q
√
3
~
8
√
=
2 |1, 0i +
3
~ |1, 0i
2
q
√
1
~
8
2 |1, −1i
+ 21 ~ |1, −1i
So,
L̂x = 12 (L̂+ + L̂− = 12 ( 21 ~ |1, 1i + ~ |1, 0i +
= 14 ~ |1, 1i +
√
3+2
~ |1, 0i
4
√
3
~ |1, 0i
2
+ 14 ~ |1, −1i
+ 21 ~ |1, −1i)
Prob. Z5-16
(solution by Alexandra Werth)
The probability of finding the angular momentum in the x direction while in state Ψ is
hΨ| L̂x |Ψi.
q
q
q
√
3+2
1
1
1
3
hΨ| L̂x |Ψi = 4
~ + 4 8 ~ + 4 12 ~
8
√
=
3+2
√ ~
4 2
We can also solve for hΨ| L̂2 |Ψi.
L̂2 Yl,m = ~2 l(l + 1)Yl,m (θ, φ)
q
q
q
L̂2 Ψ(θ, φ) = 2~2 38 Y11 + 2~2 18 Y10 + 2~2 12 Y1−1
hΨ| L̂2 |Ψi = 2~2 83 + 2~2 18 + 2~2 21
(d) We can solve for the probability of having angular momentum in the z direction.
Angular momentum in the z direction, Lz cooresponds to Y10 . Therefore,
| hY 10| Ψ(θ, φ)i |2 = 18 .
q
q
q
8
4
3
(e) Now, if we are given Φ(θ, φ) = 15
Y11 + 15
Y10 + 15
Y1−1 . We also know that the
operator L̂z Yl,m = m~Yl,m .
So,
q
hΦ| L̂z |Ψi = ~ 15
We can also solve,
q
hΦ| L̂− |Ψi = ~ 15
2
Maggie Regan
Physics 113
Zettili 5.31.
In this problem there is a spin 32 particle. We are given the Hamiltonian of Ĥ =
Ŝy2 − Ŝz2 ). From the text, we have that


0 0
0
1 0
0 

0 −1 0 
0 0 −3
3
~
0
Ŝz = 

2 0
0

√0
 3
Ŝ− = ~ 
 0
0

0
 0
Ŝ+ = ~ 
 0
0
0 0
0 0
2 √0
0
3
√
3
0
0
0

0
0 

0 
0

0 0
2 √0 

0
3 
0 0
Then, we can calculate Ŝx , Ŝy , Ŝx2 , Ŝy2 , and Ŝz2 .

√0
~
3
Ŝx = 

0
2
0
√
3 0
0
0
2 √0
2 √0
3
0
3 0




√
0
−
3 0
0
√
~ 3
0
−2
0
√
Ŝy = 

0
2
2
√0 − 3
0
0
3
0





√
3
0 2 3 √
0
2 
~  0
7
0 2 3
√
Ŝx2 =

2 3 √
0
7
0
4
0 2 3 0
3

1




0
(Ŝx2
~2
−
√
3
0
−2 3
0√
2 
~  0
7
0
−2 3
√
Ŝy2 =

−2 3
0√
7
0
4
0
−2 3
0
3


9

~ 0
Ŝz2 = 
2 0
0
0
1
0
0
0
0
1
0





0
0 

0 
9
Thus,
√
−9
0 4 3 √
0
0  0
1
0 4 3
√
Ĥ = 

4 3 √
0
1
0
4
0 4 3 0
−9





Using mathematica, we are then able to find the eigenvalues, which correspond to the
possible energy values of this Hamiltonian. These are −13
, −13
, 3 , and 43 0 . Therefore,
4 0
4 0 4 0
the eigenvectors are
 √ 
3

1
0 

v~1 = 
2  −1 
0
,


0
1  −1 

v~2 = 
2  √0 
3
,


−1
1 √
0 

v~3 = 
2 − 3 
0
, and


0
√
1
− 3 


v~4 = 
0 
2
−1
.
2
Prob. Z5-32
(solution by Alexandra Werth)
1
Given the hamiltonian
Ĥ =
0
(Ŝx2
~2
+ Ŝy2 ) + ~ Ŝz
A spin 25 particle has possible magnetic numbers of m = [− 52 , − 32 , − 12 , 21 , 32 , 52 ]. If we use the
identity that Ŝx2 + Ŝy2 = Ŝ 2 − Ŝz2 , we can write
Em = 25 , m ĥ 25 , m = ~02 (~2 ( 52 ( 25 + 1) − ~2 m2 ) + ~ ~m
= 0 ( 35
− m2 + m)
4
The possible energy eigenvaules can be solve for by plugging in for the various m values.
m
E
− 52
0
− 23
50
− 12
80
1
2
3
2
5
2
90
80
50