 Solution

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

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

Put

E_{1} = E_{2} = E

. Within the degenerate eigenspace spanned by

∣1 ⟩, ∣2 ⟩

, the perturbation is

V_{deg} = ε V_{0} (0110) .

(∣1 ⟩ \pm ∣2 ⟩) / 2

have first-order shifts

\pm ε V_{0}

. The coupling to

∣3 ⟩

has order

ε^{2}

, so it cannot affect an energy through order

ε

. Therefore

E_{+} = E + ε V_{0}, E_{-} = E - ε V_{0}, E_{3}^{'} = E_{3} through order ε .

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