Let $H_0:D(H_0)\subset\mathcal H\to\mathcal H$ be a self-adjoint operator on a [Hilbert space](/page/Hilbert%20Space) $\mathcal H$, and let $E_0$ be an isolated simple eigenvalue of $H_0$ with normalised eigenvector $\psi_0\in D(H_0)$. Write $\mathcal E_0=\ker(H_0-E_0I)=\operatorname{span}\{\psi_0\}$. Let $V:D(V)\subset\mathcal H\to\mathcal H$ be a symmetric operator with $D(H_0)\subset D(V)$, and suppose $V$ is $H_0$-bounded with relative bound strictly smaller than $1$, so that \begin{align*} H(\lambda):D(H_0)\subset\mathcal H\to\mathcal H,\qquad H(\lambda)=H_0+\lambda V,\qquad D(H(\lambda))=D(H_0), \end{align*} is self-adjoint for all sufficiently small real $\lambda$. Suppose there is a normalised eigenbranch with $\psi(\lambda)\in D(H_0)$ and expansions \begin{align*} E(\lambda)=E_0+\lambda E_1+\lambda^2E_2+O(\lambda^3), \end{align*} and \begin{align*} \psi(\lambda)=\psi_0+\lambda\psi_1+O(\lambda^2), \end{align*} where the remainders are in the graph norm of $H_0$ and $(\psi_0,\psi(\lambda))_{\mathcal H}=1$. Then $V\psi_0$ is defined and \begin{align*} E_1=(V\psi_0,\psi_0)_{\mathcal H}. \end{align*} If, in addition, the reduced resolvent contribution comes only from a discrete orthogonal eigenspace with orthonormal eigenbasis $\{\psi_n:n\ge 1\}\subset D(H_0)$, eigenvalues $E_n\ne E_0$, no continuous spectral component of $(V-E_1I)\psi_0$ in $\mathcal E_0^\perp$, and \begin{align*} \sum_{n\ge 1}\frac{|(V\psi_0,\psi_n)_{\mathcal H}|^2}{|E_0-E_n|}<\infty, \end{align*} then the second-order coefficient determined by this discrete expansion is \begin{align*} E_2=\sum_{n\ge 1}\frac{|(V\psi_0,\psi_n)_{\mathcal H}|^2}{E_0-E_n}. \end{align*}