Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 1 / A1 / a / ii / 2

For two eigenfunctions, apply the same Hermitian operator identity across the two states:Hence $(q_n-q_m)⟨{\psi }_m,{\psi }_n⟩=0$. Distinct eigenvalues therefore imply orthogonal eigenfunctions. Within a degenerate eigenspace, orthogonality is a possible basis choice, rather than a consequence for every pair of vectors.