Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 33A / b

$L\cdot S=\frac{1}{2}(J^2-L^2-S^2)$. Hence the coupled $|n\ell sj{j}_z⟩$ basis diagonalizes the spin-orbit term.

For $j=\ell +1/2$, $L\cdot S={\hslash }^2\ell /2$, giving $E_n+\lambda {\hslash }^2\ell /2$ with degeneracy $2\ell +2$. For $j=\ell -1/2$, it is $-{\hslash }^2(\ell +1)/2$, giving $E_n-\lambda {\hslash }^2(\ell +1)/2$ with degeneracy $2\ell$. Thus the original multiplet splits into two.

The weighted sum is $(2\ell +2)(\ell /2)+(2\ell )(-(\ell +1)/2)=0$, times $\lambda {\hslash }^2$, so the centre of gravity is unchanged.