Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 59 / a / Solution

Let $H^{(0)}|n⟩=E_n^{(0)}|n⟩$ with a nondegenerate reference eigenvalue, and choose intermediate normalization $⟨n|n^{(1)}⟩=0$. The first-order equation isProjection on $|j⟩$, $j\ne n$, gives $⟨j|n^{(1)}⟩=H_{jn}^{(1)}/(E_n^{(0)}-E_j^{(0)})$. The second-order projected equation gives $E_n^{(2)}=⟨n|H^{(1)}|n^{(1)}⟩$, henceThe sum runs over a complete orthonormal reference basis except the reference state; continuum states require integration if present. Degenerate states with zero denominators require degenerate perturbation theory first. These are expansion coefficients in $E_n=E_n^{(0)}+\lambda E_n^{(1)}+{\lambda }^2{E}_n^{(2)}+\cdots$.