Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 33A (Principles of Quantum Mechanics)

$[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$, $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, $[L_i,S_j]=0$, and $J=L+S$ obeys the same angular-momentum algebra. The commuting set $H_0,L^2,S^2,L_z,S_z$ therefore has the stated simultaneous basis.

Spherical symmetry makes $H_0$ commute with every $L_i$ and it acts trivially on spin, so energy is independent of $m,\sigma$. For fixed $\ell$ and spin $1/2$, degeneracy is $2(2\ell +1)=4\ell +2$.

The alternative complete commuting set is $H_0,L^2,S^2,J^2,J_z$, since these operators mutually commute. Clebsch-Gordan coefficients give the unitary change of basis.

For $s=1/2$ and $\ell \ge 1$, $j=\ell \pm 1/2$. Their degeneracies are $2j+1=2\ell +2$ and $2\ell$, summing to $4\ell +2$ as before.

$L\cdot S=\frac{1}{2}(J^2-L^2-S^2)$. 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.