Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 33C / c / i / Solution

The states $|n,\ell ,m⟩$ diagonalize $L^2$:The perturbation therefore introduces no mixing. The orbital-angular-momentum-squared perturbation of hydrogen leavesat energywhileremain triply degenerate atThese statements are exact because the unperturbed Hamiltonian commutes with $L^2$.