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

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.