The Spin-Orbit Correction

We calculate the classical Hamiltonian for the spin-orbit interaction which we will later apply as a perturbation. The B field from the proton in the electron's rest frame is

$$\bgroup\color{black} \vec B = -{\vec v \over c} \times \vec E. \egroup$$

Therefore the correction is

$$\bgroup\color{black} H_2 = {ge\over {2mc}}\vec S \cdot\vec B = - {ge\over {2mc^2}}\vec S \cdot\vec v \times\vec E \egroup$$

$$\bgroup\color{black} = {ge\over {2m^2c^2}}\vec S \cdot\vec p \times\vec\nabla\phi .\egroup$$

$\phi$ only depends on $r\Rightarrow\nabla\phi =\hat r {d\phi\over{dr}}= {\vec r\over r}{d\phi\over {dr}}$

$$\bgroup\color{black} H_2 = {ge\over {2m^2c^2}}\vec S \cdot\ve... ...{2m^2c^2}}\vec S \cdot\vec L {1\over r}{d\phi\over{dr}} \egroup$$

$$\bgroup\color{black} \phi = {e\over r} \qquad\Rightarrow\qquad {d\phi\over {dr}}= -{e\over {r^2}} \egroup$$

$$\bgroup\color{black} H_2 = {1\over 2}{ge^2\over {2m^2c^2r^3}}\vec L \cdot\vec S \egroup$$

Note that this was just a classical calculation which we will apply to quantum states later. It is correct for the EM forces, but, the electron is actually in a rotating system which gives an additional $\vec L\cdot\vec S$ term (not from the B field!). This term is 1/2 the size and of opposite sign. We have already included this factor of 2 in the answer given above.

Recall that

$$\bgroup\color{black} H_2 \propto \vec L \cdot\vec S = {1\over 2}\left[J^2 - L^2 - S^2\right] \egroup$$

and we will therefore want to work with states of definite $j$, $\ell$, and $s$.