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PHY662, Spring 2004
Examples for Adding Angular Momentum
6th February 2004
The ideas you need for the homework have been presented in lecture and/or in Shankar’s text, but here are a couple of clarifying comments. All of this is standard material, which can also be found in the texts on QM on reserve in the library.
Looked at mathematically, Clebsch-Gordon coefficients represent the transformation matrix from the product representation to a representation where the states are eigen- vectors of J^2 and Jz_._
Important relations
Let |jm〉 be an eigenstate of J^2 and Jz , with J = j¯h (i.e., J^2 = j(j + 1)¯h^2 ) and Jz = m¯h. Now note that J+J− + J−J+ = (Jx + iJy )(Jx − iJy ) + (Jx − iJy )(Jx + iJy ) = 2 J x^2 + 2J y^2 , as the cross terms all cancel. But in the commuttor, the squares cancel and the cross terms add up, so [J+, J−] = 2(iJy Jx − iJxJy ) = − 2 i¯h[Jx, Jy ] = 2¯hJz , so J+J− = J−J+ + 2¯hJz. Then
Jz J−|jm〉 = Jz (Jx − iJy )|jm〉 = (Jz Jx − iJz Jy )|jm〉 = (JxJz − (−i¯hJy ) − iJy Jz − i(−i¯hJx))|jm〉 = J−(Jz − ¯h)|jm〉 = J−(m − 1)¯h|jm〉 = [(m − 1)¯h]J−|jm〉 ,
and
J^2 J−|jm〉 = (J z^2 + J x^2 + J y^2 )(Jx − iJy )|jm〉
= (J z^2 +
(J+J− + J−J+))J−|jm〉
= {Jz J−(Jz − ¯h) +
(J−J+J− + 2¯hJz J− + J−J−J+ + 2¯hJ−Jz )}|jm〉
= {J−(Jz − ¯h)^2 +
J−(J+J− + 2¯hJ−(Jz − ¯h) + J−J+ + 2¯hJz )}|jm〉
= J−{J z^2 − 2¯hJz + ¯h^2 +
(J+J− + J−J+) − ¯h^2 + 2¯hJz }|jm〉
= J−J^2 |jm〉 = j(j + 1)¯h^2 J−|jm〉
which implies that J−|jm〉 = α|j, m − 1 〉, as it has the proper eigenvalues for J^2 and Jz. What is the normalization constant α? This can be determined only up to a phase (which is set to 1 here), by computing the norm of J−|jm〉 (note that the adjoint state to J−|jm〉 is 〈jm|J+), with J−|jm〉 =
j(j + 1) − m(m − 1)|j, m − 1 〉, by the notes from Feb. 03 (class 07).
Computing Clebsch-Gordon coefficients
We add together two angular momenta (for example, spins of two particles, or the spin of a particle and its angular momentum). What we want to do is find the states with definite Jtot^ and definite J ztot , where “tot” refers to the total angular momentum operators, with (Jtot)^2 = ( J~ 1 I 2 + I 1 J~ 2 )^2 and J ztot = (J 1 ,z I 2 + I 1 J 2 ,z ), where the subscript refers to which angular momentum is being operated on. We can also define J ±tot = (J 1 ,±I 2 + I 1 J 2 ,±), where Ji,± = Ji,x ± iJi,y.
There is a well defined procedure for constructing the |jm〉 states found by adding two angular momentum, using the product states |j 1 m 1 〉|j 2 m 2 〉 ≡ |j 1 m 1 ; j 2 m 2 〉 as a basis.
In brief, start with a state that has definite quantum numbers for total angular momen- tum and the z-component of angular momentum - express this state as a product state. Then apply total momentum lowering operators to this state to find the linear combina- tions of product states that give the other total J, Jz eigenstates.
- START. Note that there is only one state with Jz = j 1 + j 2. That is the state |j 1 j 1 ; j 2 j 2 〉. What is J^2? Well, one can write
J^2 = J 12 + J 22 + 2J 1 ,z J 2 ,z + J 1 ,+J 2 ,− + J 1 ,−J 2 ,+ ;
applying this operator to this state gives (note J 1 ,+|j 1 j 1 〉 = 0, etc.)
J^2 |j 1 j 1 ; j 2 j 2 〉 = [j 1 (j 1 + 1) + j 2 (j 2 + 1) + 2j 1 j 2 + 0 + 0]|j 1 j 1 ; j 2 j 2 〉 = (j 1 + j 2 )(j 1 + j 2 + 1)|j 1 j 1 ; j 2 j 2 〉 ,
so this state has J = j 1 + j 2. One can then conclude that
|jm〉tot = |j 1 j 1 ; j 2 j 2 〉.
- APPLY J− TO LOWER Jz. Given a state |kk〉tot defined in terms of the product states, construct |k, k − 1 〉, |k, k − 2 〉,.. ., |k, −k〉, by repeatedly applying J −tot , using J−|k, r〉 =
k(k + 1) − r(r − 1)|k, r − 1 〉
j=∑j 1 +j 2
j=
j − 2
j=|j (^1) ∑−j 2 |− 1
j=
j + j 1 + j 2 − j 1 + j 2 + 1
= (j 1 + j 2 + 1)(j 1 + j 2 ) − (j 1 − j 2 )(j 1 − j 2 − 1) + 2j 2 + 1 = j 12 + j^22 + 2j 1 j 2 + j 1 + j 2 − j^21 − j 22 + 2j 1 j 2 + j 1 − j 2 + 2j 2 + 1 = 4 j 1 j 2 + 2j 1 + 2j 2 + 1 = (2j 1 + 1)(2j 2 + 1).
This shows that only the states with j = |j 1 − j 2 |,... , j 1 + j 2 show up in the total angular momentum representation. This is an important result.
Part of a sample calculation
Here j 1 = 2 and j 2 = 1.
First, | 33 〉 = | 22 〉| 11 〉 = |22; 11〉.
Then apply lowering operators:
J−| 33 〉 =
12 − 6 | 32 〉 = (J 1 ,− + J 2 ,−)|22; 11〉 =
so
| 32 〉 =
Applying J− again to | 32 〉 gives
√ 12 − 2 | 31 〉 =
or
| 31 〉 =
What does this mean? It means, for example, that if I know the total J = 3 and the total Jz = 1, then the probability that the first particle has m 1 = 1 is 158. One could continue, but you don’t need to if you are interested in the higher Jz states.
For example, | 22 〉 must be orthogonal to | 32 〉 yet have Jz = 2. The only way to get Jz = 2 is by combining |21; 11〉and |22; 10〉. By inspection, it follows that
(up to a constant phase factor which is naturally chosen here). The other J = 2 states can then be found by applying J− to | 22 〉.
In summary, this calculation was done in the following steps, where for each ket in the total angular momentum representation, we compute its representation as a sum of
states in the product of the two component representations:
| 33 〉 ⇐ (Step1, Start) ⇓ (2, using J−) | 32 〉 ⇒ (4, orthogonality) | 22 〉 ⇓ (3, using J−) | 31 〉
