= Solution
The total <angular momentum operator> is $\mathbf J=\mathbf L+\mathbf S$. Since orbital and spin operators act on different factors of the <tensor product>, they commute, and expanding $J^2$ gives
$$
\mathbf L\cdot\mathbf S=\frac12(J^2-L^2-S^2).
$$
A coupled <angular momentum eigenstate> $|l,s;j,m_j\rangle$ is a simultaneous <eigenstate> of these three squared operators, with <eigenvalues> $\hbar^2j(j+1)$, $\hbar^2l(l+1)$ and $\hbar^2s(s+1)$. Thus the angular factor in the <spin-orbit interaction> is the number
$$
\mathbf L\cdot\mathbf S\longmapsto
\frac{\hbar^2}{2}[j(j+1)-l(l+1)-s(s+1)].
$$
The other factor, $r^{-1}dV/dr$, is radial. It commutes with the angular operators, and a central-potential <energy eigenstate> separates into a radial <wavefunction> and an angular spinor. Taking its radial <expectation value> therefore gives
$$
\boxed{\langle H_1\rangle=\frac{\hbar^2}{4m^2c^2}
[j(j+1)-l(l+1)-s(s+1)]
\left\langle\frac1r\frac{dV}{dr}\right\rangle.}
$$
This <spin-orbit energy shift in a central potential> is independent of $m_j$, so rotational symmetry preserves the $2j+1$ magnetic degeneracy within a fixed $j$ multiplet. It is a first-order <fine structure> correction; the given coefficient already includes the relativistic factor needed for the electron's spin-orbit coupling.
Back to article page
