# Spin orbit coupling

1. Feb 25, 2013

### aaaa202

I have a question about how my book derives a formula. It starts with:
We have an electron free too move on a cylinder and from the cylinder there is an electric field pointing radially outwards.
Now for an electron moving in an electric field it sees local magnetic field given by:

B = 1/c^2 v x E (1)
And from spin pertubation theory we now get the following pertubation of the Hamiltonian:
H' = -μ $\bullet$ B = C L$\bullet$S

I don't understand the last equality. I know μ(the magnetic moment) is proportional to the spin but how do we get L from (1)? - and how does it even make sense to use "v" the velocity in classical electrodynamics in a quantum mechanical system?

2. Feb 25, 2013

### DrDu

I would argue like this: use $v=i\hbar [H,\vec{r}]$ (holds also quantum mechanically), express $\vec{r}$ in polar coordinates and use $H=L^2/2mr^2$.