Q.M3 - Tirgul 8

Q.M3 - Tirgul 8
Roee Steiner
Physics Department, Ben Gurion University of the Negev, BeerSheva 84105, Israel
16.12.2015
Contents
1 Particle on a sphere
2 Spin
1
1
1
1
2 +Spin 2 +Spin 2
2
3 Sakurai Q 3.27
3
4 Class Work
4
1
Particle on a sphere - problem 35
Particle with spin s = 1 on a sphere moves on the influence of the Hamiltonian:
H=
L2
LS
+
2I
I
(1)
Where S is the spin , L is the angular momentum, and I is the moment of
inertia. Find the energy correction which the LS
I gives. Find the energy levels
and the degeneracy of the system.
Solution We know that [L, S] = 0 and
J 2 = (L + S)2 = L2 + S 2 + 2LS
(2)
So
LS =
1 2
(J − L2 − S 2 )
2
We choose to work on the base:
|l, s; J, mj i
∗ e-mail:
[email protected]
1
(3)
∗
So
H|l, s; J, mj i =
l(l + 1) [J(J + 1) − l(l + 1) − s(s + 1)]
J(J + 1) − 2
+
|s=1 =
2I
2I
2I
(4)
the correction that the
LS
I
∆E =
gives to the energy are:
[J(J + 1) − l(l + 1) − s(s + 1)]
|s=1
2I
(5)
We can see that the Energy levels depends on J only , which has mj = 2J + 1,
Where J can have the following values J = |L + S|, |L + S − 1|, · · · , |L − S|
2
Spin 1+Spin
1
2
(1) Decompose the standard representation of Spin 1+Spin 12
(2) Find all the |J, mi representation states as sum of the standard representation states.
Solution
1⊗
1
3 1
= ⊕
2
2 2
(6)
The addition has built one J = 3/2 representation and one 1/2 representation.
(2)
b) The standard representation of Spin 1+Spin
1
2
gives us 6 states as follows:
|1 ↑i , |1 ↓i , |0 ↑i , |0 ↓i , |−1 ↑i , |−1 ↓i
Let us define a decending operator of the form:
J− = L− + S−
(7)
Where
S− |ml , ms , mn i =
p
l(l + 1) − ml (ml − 1) |ml − 1, ms , mn i
(8)
p
s(s + 1) − ms (ms − 1) |ml , ms − 1, mn i
(9)
L− |ml , ms , mn i =
The highest state is
|J = 3/2, m3/2 = 3/2i = |1 ↑i
(10)
To find the next step we operate the ladder operator:
J− |1 ↑i =
√
3|J =
3
1
, m 32 = i = L− |1 ↑i + S− |1 ↑i
2
2
2
(11)
i=
Using
the definitions of the L− , S− , we get the second level state(i.e L− |1 ↑
√
2|0 ↑i):
|J =
3
1
1 √
, m 32 = i = √ ( 2 |0 ↑i + |1 ↓i)
2
2
3
(12)
In order to get the third level state, we use the decending operator again
|J =
√
3
1
1
, m 32 = − i = √ (|−1 ↑i + 2 |0 ↓i)
2
2
3
(13)
The forth level state is simply
|J =
3
3
, m 32 = − i = |−1 ↓i
2
2
(14)
Now we going to find the J = 1/2 states. Because we like to find J = 1/2 state
it must be of the form:
C1 · |0 ↑i + C2 · |1 ↓i
(15)
and because orthogonality:
|J =
√
1
1
1
, m 12 = i = √ (− |0 ↑i + 2 · |1 ↓i)
2
2
3
(16)
√
1
1
1
, m 12 = − i = √ (|0 ↓i − 2 · |−1 ↑i)
2
2
3
(17)
the next state will be:
|J =
3
Sakurai Q 3.27
¬
Consider a spinless particle bound to a fixed center by a central force potential.
Relate, as much as possible, the matrix elements
1
hn0 , l0 , m0 | ∓ √ (x ± iy)|n, l, mi
2
and
hn0 , l0 , m0 |z|n, l, mi
using only the Wigner-Eckart theorem. Make sure to state under what conditions the matrix elements are non vanishing.
Solution
As we seen before
r
1
3 ±1
∓ √ (x ± iy) =
Y
(18)
4π 1
2
¬ In
the new addition 3.31
3
and
r
z=
3 0
Y
4π 1
(19)
So
r
3 ±1
1
0 0
0
Y |n, l, m1 i
hn , l
∓ √ (x ± iy)|n, l, m1 i = hn , l , m1 |
4π 1
2
q
3
Y1 ||n, li
hn0 , l0 || 4π
√
= hl, 1; m1 , ±1|l, 1; l0 , m01 i
2l + 1
0
0
, m01 |
and
r
0
hn , l
0
, m02 |z|n, l, m2 i
0
= hn , l
0
, m02 |
3 0
Y |n, l, m2 i
4π 1
q
3
Y1 ||n, li
hn0 , l0 || 4π
√
= hl, 1; m2 , 0|l, 1; l0 , m02 i
2l + 1
So
hn0 , l0 , m01 | ∓
√1 (x ± iy)|n, l, m1 i
2
hn0 , l0 , m02 |z|n, l, m2 i
=
hl, 1; m1 , ±1|l, 1; l0 , m01 i
hl, 1; m2 , 0|l, 1; l0 , m02 i
(20)
We didnt done all. We need to speak on selection roles.
m0 = m + q
(21)
l0 = |l ± 1|
(22)
m01 = m1 ± 1
(23)
in our case:
m02
= m2
(24)
where m1 corresponds to the ∓ √12 (x ± iy) vector and m2 corresponds to the z
vector.
4
Class Work
problem 21 and 36 from Gedalin lectern notes, and 3703 , 3402
4