# Spin orbit coupling

1. Mar 15, 2006

### sachi

I'm a little confused as to whether the following commutation relations still hold when spin orbit coupling occurs:
[Sx,Lx] = 0
[Sx,Ly] = 0
[S^2, Lx] = 0
[L^2, Sx] = 0
etc.

thanks very much for your help

2. Mar 16, 2006

### marlon

Err,

What have you done to solve this problem yourself ?

Helping you out does not imply that we will just "spoon feed" you the solution.

How do you think you can tackle this problem ?

marlon

3. Mar 19, 2006

### Meir Achuz

The CR's between the ops are not changed by the coupling.

4. Mar 23, 2006

### dextercioby

It doesn't matter how the hamiltonian looks, for a particle in quantum mechanics the Hilbert space has the structure

$$\mathcal{H} =L^{2}\left(\mathbb{R}^{3}\right) \otimes \mathbb{C}^{2n+1}$$

,where "n" is the spin of the particle.

Daniel.

5. Mar 23, 2006

### nrqed

They are still valid. But that's not the point. The point is whether the *perturbation* hamiltonian commutes with these operators. One finds that the spin orbit hamiltonian commutes with L^2, S^2, J^2 and J_z, but not with L_z and S_z. Therefore, m_s and m_l are not good quantum numbers but must be replaced by m_j and j. So the states of definite energy when the spin-orbit interaction are taken into account are the states labelled by the quantum numbers l,s,j, m_j (instead of the usual l,m_l,s,m_s that one uses to label the unperturbed hydrogenic wavefunctions).

Hope this helps. If it's not clear, write again.

Pat