 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/34b/a/solution

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

Projecting

H ∣Ψ ⟩ = E ∣Ψ ⟩

onto

⟨ ψ_{n} ∣

and using orthonormality gives

E_{0} c_{n} + \sum_{m = 0}^{2} ⟨ ψ_{n} ∣ V ∣ ψ_{m} ⟩ c_{m} = E c_{n} .

Thus, with

d = E_{0} + α

d - A - A - A d - A - A - A d c_{0} c_{1} c_{2} = E c_{0} c_{1} c_{2} .

The symmetric vector

(1, 1, 1)

E_{s} = d - 2 A = E_{0} + α - 2 A .

Let

ω = e^{2 πi /3}

, so

1 + ω + ω^{2} = 0

. For either vector

(1, ω, ω^{2}), (1, ω^{2}, ω),

the other two components in each row sum to minus the component in that row. Both therefore have eigenvalue

E_{d} = d + A = E_{0} + α + A .

The spectrum is consequently

E_{0} + α - 2 A, E_{0} + α + A, E_{0} + α + A,

with normalized eigenvectors obtained by multiplying the three displayed vectors by

1/ 3

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