 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/34b/b/ii/solution

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

The degenerate

E_{0} = 9/2

subspace has ordered basis

∣3, 0 ⟩, ∣1, 1 ⟩ .

Using the ladder actions,

H^{'} ∣3, 0 ⟩ H^{'} ∣1, 1 ⟩ = A^{2} B^{†} ∣3, 0 ⟩ = 6 ∣1, 1 ⟩, = A^{† 2} B ∣1, 1 ⟩ = 6 ∣3, 0 ⟩ .

H^{'}_{E_{0}} = (0660) .

Its normalized eigenvectors and first-order energies are

∣ ψ_{+} ⟩ = \frac{∣3 , 0 ⟩ + ∣1 , 1 ⟩}{2}, E_{+} = \frac{9}{2} + 6 λ,

∣ ψ_{-} ⟩ = \frac{∣3 , 0 ⟩ - ∣1 , 1 ⟩}{2}, E_{-} = \frac{9}{2} - 6 λ .

There is no order-

λ

admixture from other unperturbed levels.

In fact,

H^{'}

preserves

n + 2 m

, so this two-dimensional eigenspace of

H_{0}

is invariant. Directly,

H^{'} ∣ ψ_{\pm} ⟩ = \pm 6 ∣ ψ_{\pm} ⟩, H_{0} ∣ ψ_{\pm} ⟩ = \frac{9}{2} ∣ ψ_{\pm} ⟩ .

Therefore

H ∣ ψ_{\pm} ⟩ = (\frac{9}{2} \pm 6 λ) ∣ ψ_{\pm} ⟩

for every real

λ

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