GRADUATE QUANTUM MECHANICS: 502 Spring 2002
Solutions to Assignment 2.
1. To find the Clebsch Gordan coefficients, we start with the state formed by combining parallel spins, then we
apply the lowering operator successively. In the following solutions, the Clebsh-Gordan coefficients can be read
off from the corresponding expansion of the angular momentum states.
When we make the combination 12 × 32 we generate a j = 2 and a j = 1 state. The total number of states is
2 × 4 = 8 = 5 + 3, so the j = 1 and j = 2 state exhaust the total number of states.
1 3
⊗ =2⊕1
2 2
The state with maximum mz corresponding formed from j =
(1)
1
2
and j =
3
2
is
3 3 1 1
|2, 2i = | , i| , i
2 2 2 2
(2)
If we apply the lowering operator to this state we obtain
J− |2, 2i =
3 3 1 1
J− | , i| , i =
2 2 2 2
=
2|2, 1i
3 3
1 1
3 3 1 1
| , i(J− | , i + (J− | , i| , i
2 2
2 2
2 2 2 2
√ 3 1 1 1
3 3 1 1
| , i| , − i + 3| , i| , i,
2 2 2 2
2 2 2 2
(3)
(4)
(5)
so that
√
3 3 1 1 1
1 3 3 1 1
| , i| , i
|2, 1i = | , i| , − i +
2 2 2 2 2
2 2 2 2 2
(6)
Repeating this process, we obtain
|2, 2i =
|2, 1i =
|2, 0i =
|2, −1i =
|2, −2i =
3 3 1 1
| , i| , i
2 2 2 2
√
1 3 3 1 1
3 3 1 1 1
| , i| , − i +
| , i| , i
2 2 2 2 2
2 2 2 2 2 3 1 1 1
1
3 1 1 1
√ | , i| , − i + | , − i| , i
2
2
2
2
2
2 2 2
2
√
3 3 1 1
3 1 1 1
1
3| , − i| , − i + | , − i| , − i
2
2 2 2 2
2 2 2 2
3 3 1 1
| , − i| , − i
2 2 2 2
(7)
(8)
(9)
(10)
(11)
Now, to obtain the spin-1 states, we write
3 3 1 1
3 1 1 1
|1, 1i = α| , i| , i + β| , i| , − i
2 2 2 2
2 2 2 2
In order that this state be orthogonal to |2, 1i, we choose
3 1 1 1
1 √ 3 3 1 1
3| , i| , − i − | , i| , i
|1, 1i =
2
2 2 2 2
2 2 2 2
1
(12)
(13)
Now, successive application of the lowering operator yields
|1, 0i =
|1, −1i =
3 1 1 1
1 3 1 1 1
√ | , i| , − i − | , − i| , i
2
2
2
2
2
2 2 2
2
√
1 3 1 1 1
− 3 3 3 1 1
| , − i| , i + | , − i| , − i
2 2 2 2 2
2 2 2 2 2
(14)
(15)
2. Sakurai, problem 3.18. The probability that the rotated state D(R)|l = 2, m = 0i has Jz = m0 is given by
(2)
(2)
But since Dm0 0 =
q
p(m0 ) = |hl = 2, m0 |D(R)|l = 2, m = 0i|2 = |Dm0 0 (R)|2
4π ∗
5 Y2m0 (R),
(16)
we obtain
p(m0 ) =
4π ∗
|Y 0 (θ, φ)|2
5 2m
(17)
Writing this out explicitly, we obtain
3
sin4 θ,
8
3
sin2 θ cos2 θ,
2
1
(3 cos2 θ − 1).
4
p(±2) =
p(±1) =
p(0) =
(18)
(19)
(20)
3. Sakurai, problem 3.24. The singlet wavefunction of the two spins can be written
1
|ψi = √ (|n̂, +i1 |n̂, −i2 − |n̂, −i1 |n̂, +i2 )
2
(21)
where n̂ is an arbitrary observation axis.
(a) When observer B makes no measurement, the state remains in a linear superposition of the two states
|z, ±i1 |z, ∓i2 with an amplitude of magnitude √12 to be either state. The probability for A to obtain
s1z = h̄/2 is thus precisely 12 . This holds for any quantization axis, so the probability for measuring
s1x = h̄/2 is also precisely 12 .
(b) If B measures the spin of the particle to be s2z = h̄/2, this projects the system into the state |z, −i1 |z, +i2 ,
so that A must measure s1z = −h̄/2. If A measures s1x , then since
1
|z, −i1 |z, +i2 = √ (|x, +i1 |z, +i2 + |x, −i1 |z, +i2 )
2
(22)
there is an equal chance that A will measure s1x = ±h̄/2.
4. (a) Now using the Wigner Eckart theorem, we can write
1
λ
1
Qa b = ehjm|xa xb − r2 |jmi = 2 hjm|Ja Jb − J 2 |jmi
3
3
h̄
(23)
Now if we evaluate hjj|Q33 |jji, we obtain
hjj|Q33 |jji =
e
hjj|z 2 − r2 |jji = 3Q
3
2
(24)
on the other hand, from the r.h.s. of the first expression, we obtain
1
j(2j − 1)
λhjj|Jz2 − J 2 |jji = λ
3
3
(25)
Comparing these two results, we obtain
λ=
Q
2(2j − 1)
(26)
(b) From the previous part of the question, we have
ehjm|x2 − y 2 |jm0 i
=
=
Q
hjm|Jx2 − Jy2 |jm0 i
h̄ j(2j − 1)
Q
2
2
hjm|J+
− J−
|jm0 i
2
h̄ 2j(2j − 1)
2
(27)
We can write this in the form
ehjm|x2 − y 2 |jm0 i =
Q
2
2
δm,m0 −2 hjm|J+
|jm + 2i + δm,m0 +2 hjm|J−
|jm − 2i
h̄ 2j(2j − 1)
2
(28)
For the specific case of j = 1, we have
h11|J+ 2|1, −1i = h1 − 1|J+ 2|1, +1i = 2h̄2
so that

0
eh1m|x2 − y 2 |1m0 i = Q  0
2
3

0 2
0 0
0 0
(29)
(30)