 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/34b/a/solution

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

c = ⟨ ψ ∣ ϕ ⟩

. The off-diagonal Hermiticity identity holds automatically, while

⟨ ψ ∣ H ∣ ψ ⟩ = g c

must be real. Thus

g ⟨ ψ ∣ ϕ ⟩ \in R

is necessary and sufficient. For

g \neq = 0

, put

r = g c /∣ g ∣ \in R

. Since the two normalized spanning vectors are linearly independent,

∣ r ∣ < 1

The normalized eigenvectors and eigenvalues are

∣ \pm ⟩ = \frac{( g ^{*} /∣ g ∣ ) ∣ ψ ⟩ \pm ∣ ϕ ⟩}{2 ( 1 \pm r )}, E_{\pm} = \pm ∣ g ∣.

Direct application of

H

verifies the eigenvalue equations. Their unnormalized inner product is

- \frac{g}{∣ g ∣} c + \frac{g ^{*}}{∣ g ∣} c^{*} = - r + r = 0,

so the distinct eigenvectors are orthogonal, as required for a Hermitian operator. If

g = 0

H = 0

and every vector is an eigenvector.

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