 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/33c/c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The states

∣ n, ℓ, m ⟩

diagonalize

L^{2}

L^{2} ∣ n, ℓ, m ⟩ = ℏ^{2} ℓ (ℓ + 1) ∣ n, ℓ, m ⟩ .

The perturbation therefore introduces no mixing. The orbital-angular-momentum-squared perturbation of hydrogen leaves

∣2, 0, 0 ⟩

E_{2},

while

∣2, 1, - 1 ⟩, ∣2, 1, 0 ⟩, ∣2, 1, 1 ⟩

remain triply degenerate at

E_{2} + 2 g ℏ^{2} .

These statements are exact because the unperturbed Hamiltonian commutes with

L^{2}

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