# Spin-orbit coupling perturbation

1. Nov 3, 2007

### noospace

1. The problem statement, all variables and given/known data

An electron in a hydrogen atom is in the n = 2, l = 1 state. It experiences a spin-orbit interaction $H' = \alpha \mathbf{L} \cdot \mathbf{S}$. Calculate the energy level shifts due to the spin-orbit interaction.

2. Relevant equations

Degenerate perturbation theory.

3. The attempt at a solution

This n,l state is triply degenerate due to the three possible values of m = -1,0,1.

The unperturbed Hamiltonian is just what goes in the Schrodinger equation right? In which case the eigenfunctions of the unperturbed hamiltonian are just the spherical harmonics $Y_{lm}$ multiplied by strictly radial functions. So I put

$\psi^{(0)} = \alpha Y_{10} + \beta Y_{1-1} + \gamma Y_{11}$

So I then write down the matrix $\langle Y_{1,i} |H'| Y_{1,j} \rangle$ and find the eigenvalues.

Am I getting warm?

2. Nov 3, 2007

### Meir Achuz

You have to evaluate $$L.S$$ for each angular state (J,L,S) by using
$$J^2=L^2+S^2+2L\cdot S$$.
You don't need the explicit wave functions.

3. Nov 3, 2007

### noospace

Hi Meir Achuz,

So for n = 2, l = 1 we have two possibilities for the total angular momentum corresponding to j = 1/2 and j=3/2 right?

But we also have some degeneracy coming from the possible values of m = -1,0,1.

Does this mean there will be a total of 6 energy shifts?

4. Nov 4, 2007

### Meir Achuz

L^2 and S^2 are known.
There are two energy levels.
One for J^2=(3/2)(5/2), and one for J^2=(1/2)(3/2).
There are 6 states, but because of rotational invariance, the 4 J=3/2 states are still, degenerate, as are the l2 J=1\/2 states.