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

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.

Multiply the eigenfunction expansion by ${\psi }_m^{\ast }$ and integrate over the common domain. Orthogonality and wavefunction normalization giveThus $\boxed{c_m=⟨{\psi }_m,f⟩}$. For a normalized $f$, the Born rule identifies $|c_m{|}^2$ as the probability of this basis outcome.

The decisive observation is that $x$ is itself an eigenfunction:On the regular Legendre polynomial domain, integration by parts has endpoint factor $x^2-1$, so the boundary form vanishes for regular functions at $x=\pm 1$. The interval alone is not an endpoint condition; regularity is the usual condition selecting the Legendre polynomials. Their eigenvalues are $n(n+1)$, and the $n=1$ normalized function isThereforeEquivalently, the Hermitian operator identity gives $(q_n-2)c_n=0$, and nondegeneracy leaves only one possible nonzero coefficient. The unnormalized function $x$ has squared norm $2/3$, consistent with this coefficient.

The quantum variational principle states that the energy expectation of an admissible trial wavefunction is at least the ground state energy:For the discrete, complete orthonormal basis assumed here, expand in Hamiltonian operator eigenfunctions, $f={\sum }_n{c}_n{\varphi }_n$, with $\hat{H}{\varphi }_n=E_n{\varphi }_n$ and $E_n\ge E_0$. ThenEvery summand is nonnegative. Equality holds precisely when all occupied components lie in the ground eigenspace. Thus minimizing trial energies supplies upper bounds on $E_0$, with exact equality if the trial family contains a ground state.