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

Take $\hat{V}\psi =V(x)\psi$ with real $V$, and $\hat{T}=-({\hslash }^2/2m)d^2/d{x}^2$. The operator domain matters: use sufficiently smooth wavefunctions for which the following boundary terms vanish at both infinities. A real multiplication operator givesTwo applications of integration by parts giveThus the potential energy and kinetic energy operators obey the stated Hermitian operator identity on this common domain. For $D=d/dx$, one application of integration by parts instead givesThe derivative is anti-Hermitian, rather than Hermitian; the minus sign cannot generally be removed. Multiplication by $-i\hslash$ makes the corresponding momentum operator Hermitian. These integrations establish the requested symmetry identities; full self-adjointness of an unbounded operator additionally requires equality of its domain with its adjoint's domain.

Let $\hat{Q}{\psi }_n=q_n{\psi }_n$ with ${\psi }_n\ne 0$. The Hermitian operator identity and conjugate linearity of the first slot of the inner product implyThe squared norm is positive, so $q_n=q_n^{\ast }$: every such eigenvalue is real.

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.