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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/36b/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 36B / c / Solution by

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

In the basis

(∣ ψ_{1} ⟩, ∣ ψ_{2} ⟩)

,

H = (E_{1} h h E_{2}) .

Δ = E_{2} - E_{1} > 0

, its exact eigenvalues are

E_{\pm} = \frac{E _{1} + E _{2}}{2} \pm \frac{Δ ^{2}}{4} + h^{2} .

For the normalized trial state,

E (β) = E_{1} sin^{2} β + E_{2} cos^{2} β + 2 h sin β cos β = \frac{E _{1} + E _{2}}{2} + \frac{Δ}{2} cos 2 β + h sin 2 β .

The minimum of

A cos 2 β + h sin 2 β

is

- A^{2} + h^{2}

with

A = Δ/2

. Therefore

β min E (β) = \frac{E _{1} + E _{2}}{2} - \frac{Δ ^{2}}{4} + h^{2} = E_{-} .

The variational estimate is exact because the trial family ranges over all real normalized states in the two-dimensional Hilbert space, and the real symmetric Hamiltonian has a real ground eigenvector.

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