 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/33b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Now put

E_{2} = E_{3} = E

and

E_{1} \neq = E

. All diagonal matrix elements of

V

vanish. The leading coupling is

⟨ 1∣ V ∣2 ⟩ = ε V_{0}

, whereas

⟨ 1∣ V ∣3 ⟩ = ε^{2} V_{0}

. Applying the second-order formula gives

E_{1}^{'} E_{2}^{'} E_{3}^{'} = E_{1} + \frac{ε ^{2} V _{0}^{2}}{E _{1} - E} + O (ε^{4}), = E + \frac{ε ^{2} V _{0}^{2}}{E - E _{1}} + O (ε^{4}), = E + O (ε^{4}) .

The degeneracy causes no singular denominator through this order because

⟨ 2∣ V ∣3 ⟩ = 0

. Equivalently, the effective perturbation on the degenerate subspace begins as

\frac{V _{0}^{2}}{E - E _{1}} (ε^{2} ε^{3} ε^{3} ε^{4}) .

Its off-diagonal mixing is only order

ε^{3}

, so degenerate perturbation theory is unnecessary for energies through order $ε^{2}$ .

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