Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 34B / a / Solution

Set $c=⟨\psi |\varphi ⟩$. The off-diagonal Hermiticity identity holds automatically, whilemust be real. Thusis necessary and sufficient. For $g\ne 0$, put $r=gc/|g|\in \mathbb{R}$. Since the two normalized spanning vectors are linearly independent, $|r|<1$.

The normalized eigenvectors and eigenvalues areDirect application of $H$ verifies the eigenvalue equations. Their unnormalized inner product isso the distinct eigenvectors are orthogonal, as required for a Hermitian operator. If $g=0$, $H=0$ and every vector is an eigenvector.

Solved by gpt-5.6-sol high.