PX262 - G7a- pure spin-orbit coupling

setting the external magnetic field to zero:

{\hat{H}}_{L S} = f (r) ({\hat{J}}^{2} - {\hat{L}}^{2} - {\hat{S}}^{2})

there is no preferred axis and the energy eigenfunctions do not have to be eigenfunctions of ${\hat{L}}_{z}$ and ${\hat{S}}_{z}$
they will still be eigenfunctions of ${\hat{J}}^{2}$ , ${\hat{L}}^{2}$ and ${\hat{S}}^{2}$ operators
the energy levels of any total angular momentum operator have the form $j (j + 1) ℏ^{2}$ , where, $j = n / 2$ for $n \in Z$
the change in energy can therefore be approximated by its expectation value:

δ E = ⟨ f (r) ⟩ [j (j + 1) - l (l + 1) - s (s + 1)] ℏ^{2}

the three quantities are connected:

\begin{matrix} j = l + s & or, j = | l - s | & for s = \frac{1}{2} \\ j = l + \frac{1}{2} & or, j = l - \frac{1}{2} & unless l = 0 \end{matrix}

the energy levels will split into $2 (2 l + 1) = 6$ states

image: A. I. M. Rae, Quantum Mechanics 6th Edition

above image shows the effect of spin-orbit coupling on the energy levels of a one-electron atom
$l = 2$ states are split into a doublet, degeneracy of either state is $(2 j + 1) :$ $6$ for $j = 5 / 2$ , $4$ for $j = 3 / 2$
hence, the total number of states: $2 (2 l + 1) = 10$