Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 36B / c / Solution

In the basis $(|{\psi }_1⟩,|{\psi }_2⟩)$,Writing $\Delta =E_2-E_1>0$, its exact eigenvalues are

For the normalized trial state,The minimum of $A\cos 2\beta +h\sin 2\beta$ is $-\sqrt{A^2+h^2}$ with $A=\Delta /2$. ThereforeThe variational estimate is exact because the trial family ranges over all real normalized states in the two-dimensional Hilbert space, and the real symmetric Hamiltonian has a real ground eigenvector.