Quantum Mechanics II angular Momentum II : Addition of Angular Momentum Clebsch-Gordan Coefficients
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- Example 1 - Adding two spin 1/2 angular momenta
- Solution
|
1 1 |
|
1 1 1 1 |
, 2 2 |
; 1, 1⟩ = (J1− + J2−) | |
, ; , 2 2 2 2 |
1 1 , |
|
|
2 2 |
; 1, 0⟩ =| 2 , 2 ; −2 , 2 ⟩+ | 2 , 2 ; 2 |
1 1 1
,
1
−2 ⟩
Thus the Clebsch-Gordon coefficients are
⟨1/2, 1/2; 1/2, 1/2 | 1/2, 1/2, 1, 1⟩ = 1
1
⟨1/2, 1/2; 1/2, −1/2 | 1/2, 1/2, 1, 0⟩ = √2
1
⟨1/2, 1/2; −1/2, 1/2 | 1/2, 1/2, 1, 0⟩ = √2
The coefficient for s = 1.m = −1 need not be calculated as it is obtained by symmetry to be also equal to 1. For s = 0, there is only state with m = 0. This state is orthogonal to the state | 1/2, 1/2, 1, 0⟩ and is given by
1
| 1/2, 1/2, 0, 0⟩ = √2 (| 1/2, 1/2, 1/2, −1/2⟩− | 1/2, 1/2, −1/2, 1/2⟩)
Thus
1
⟨1/2, 1/2; 1/2, −1/2 | 1/2, 1/2, 1, 0⟩ = √2
1
⟨1/2, 1/2; −1/2, 1/2 | 1/2, 1/2, 1, 0⟩ = −√2
It is conventional to indicate the state with m = 1/2 as | ↑⟩ or as α. The state with
m = −1/2 is then indicated as | ↓⟩ or as β. The triplet states are then s = 1 :
m = +1 :| ↑↑⟩
1
m = 0 : √2 [| ↑↓⟩+ | ↓↑⟩)
m = −1 :| ↓↓⟩
It may be noted that the triplets are symmetric with respect to the two spins. The singlet state s = 0 is given by
1
√2 [| ↑↓⟩− | ↓↑⟩)
which is antisymmetric under exchange of spins.
When the magnitudes of the two angular momenta are different, it is conventional to choose the one with larger magnitude as the first spin and the other one as the second. This convention has some repercussion in fixing the signs of Clebsch-Gordan coefficients, which we will talk about in the next example.
Example 2: A Hydrogen atom is known to be in the state n = 3, l = 2. Find all the
values of the total angular momentum j and for each j, the eigenvalue of J2 operator.
1 1 5 3
We have l = 2, s = 1/2. Thus allowed values of j are from 2 − 2 to 2 + 2 , i.e.
, . The
2 2
eigenvalue of J2 being j(j + 1)k2, we have, for j = 5/2 the eigenvalue is (35/4)k2 and
(15/4)k2 respectively.
Example 3: For two spins s = 1/2 each, show that 3 I + 1 Sˆ · Sˆ is a projection operator
4 k2 1 2
for S = 1 and 1 I − 1 Sˆ
· Sˆ
is a projection operator for S = 0
4
Solution:
k2 1 2
Note 3
I +
Sˆ1 · Sˆ2 = I +
1
[S2 − S2 − S2]
3
1
4 k2 4 2k2 1 2
For S = 1, S2 has eigenvalue 1(1 + 1)k2 = 2k2 and S2 and S2 have eigenvalue 3 k2 each.
1 2 4
Thus we get for S = 1 this expression to have value 1 and for S = 0 it has value 0. Thus the operator projects S = 1 state.
Example 4: Adding j1 = 1 with j2 = 1/2
In this case, in the j1, j2; m1, m2 representation there are 3 × 2 = 6 states. The resultant
angular momentum which satisfies | j1 − j2 |≤ j ≤ j1 + j2 can take values
3
and
2
1
, the
2
former has 4 states and the latter 2, making a total of 6 states as before. We proceed as
follows. Start with the state having maximum j value and maximum possible m value
corresponding to this j. In this case it is j =
3
and m =
2
3
. This state is obtained
2
uniquely from j1 = 1, m1 = 1 and j2 = 1/2, m2 = 1/2. Thus
| 1, 1/2; 3/2, 3/2⟩ =| 1, 1/2; 1, 1/2⟩
√
We apply J− on the ket on the left and J1− +J2− to the ket on the right, using the formula
J− | j, m⟩ = (j + m)(j − m + 1) | j, m − 1⟩. We then get
√3 | 1, 1/2; 3/2, 1/2⟩ = √2 | 1, 1/2; 0, 1/2⟩+ | 1, 1/2; 1, −1/2⟩ (18)
From this it follows that
3
⟨1, 1/2; 3/2, 1/2 | 1, 1/2; 0, 1/2⟩ = r2
3
⟨1, 1/2; 3/2, 1/2 | 1, 1/2; 1, −1/2⟩ = r1
The remaining coefficients belonging to the negative values of m are obtained by symme- try. We now consider the j = 1/2 state with = ±1/2. Once again, we will only compute m = 1/2 term. Note that the state | 1, 1/2; 1/2, 1/2⟩ is orthogonal to the state with the same m but j = 3/2. Thus this state is orthogonal to
3
3
| 1, 1/2; 3/2, 1/2⟩ = r2 | 1, 1/2; 0, 1/2⟩ + r1 | 1, 1/2; 1, −1/2⟩
The state which is orthogonal to the state on the right is obviously
3
3
| 1, 1/2; 1/2, 1/2⟩ = ± r1 | 1, 1/2; 0, 1/2⟩ − r2 | 1, 1/2; 1, −1/2⟩!
Thus, there is an ambiguity on the overall phase. In order to fix the sign, we use what is known as Condon-Shortley sign convention according which the state with the highest m of the larger component of the angular momentum that is being added is assigned a positive sign. In case where j1 = j2, as was the case in the previous example,
the first mentioned one, i.e. j1 will be taken to be the larger component for application of this convention. Thus in this case the state with j1 = 1, m1 = 1 will be taken to be positive. Thus we have
| 1, 1/2; 1/2, 1/2⟩ = r2 | 1, 1/2; 1, −1/2⟩ − r1 | 1, 1/2; 0, 1/2⟩
3 3
This gives two Clebsch-Gordan coefficients as follows
3
⟨1, 1/2; 1/2, 1/2 | 1, 1/2; 1, −1/2⟩ = r2
3
⟨1, 1/2; 1/2, 1/2 | 1, 1/2; 0, 1/2⟩ = −r1
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