 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2/35b/i/solution

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

Let

Δ = E_{+} - E_{-} > 0

. Hermiticity gives

⟨ - ∣Δ H ∣ + ⟩ = - iλ,

so in the ordered basis

(∣ - ⟩, ∣ + ⟩)

H_{0} + Δ H = (E_{-} iλ - iλ E_{+}) .

E_{n}^{(1)} = ⟨ n ∣Δ H ∣ n ⟩, ∣ n^{(1)} ⟩ = \sum_{m \neq = n} ∣ m ⟩ \frac{⟨ m ∣Δ H ∣ n ⟩}{E _{n} - E _{m}} .

The diagonal matrix elements vanish, so both linear energy corrections are zero. The state corrections are

∣ -^{(1)} ⟩ = ∣ + ⟩ \frac{iλ}{E _{-} - E _{+}} = - \frac{iλ}{Δ} ∣ + ⟩,

∣ +^{(1)} ⟩ = ∣ - ⟩ \frac{- iλ}{E _{+} - E _{-}} = - \frac{iλ}{Δ} ∣ - ⟩ .

Thus, through linear order,

E_{-}^{pert} E_{+}^{pert} = E_{-} + O (λ^{2}), = E_{+} + O (λ^{2}), ∣ - ⟩_{pert} ∣ + ⟩_{pert} = ∣ - ⟩ - \frac{iλ}{Δ} ∣ + ⟩ + O (λ^{2}), = ∣ + ⟩ - \frac{iλ}{Δ} ∣ - ⟩ + O (λ^{2}) .

Normalization changes only at quadratic order.

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