= Induced dipole moment and quadratic energy shift
{title2=$\delta E=-\tfrac12\alpha E^2,\quad d=\alpha E$}
For a nondegenerate parity-symmetric ground state in an external <electric field>, write the perturbation as $V=-E\hat d$. The first-order state correction gives
$$
\langle\hat d\rangle=2E\sum_{k\ne0}\frac{|\langle k|\hat d|0\rangle|^2}{E_k^{(0)}-E_0^{(0)}}+O(E^3)=\alpha E+O(E^3).
$$
The positive coefficient $\alpha$ is the <electric polarizability> along the applied direction. Meanwhile <second-order nondegenerate perturbation theory> gives $\delta E_0^{(2)}=-\alpha E^2/2$. Thus the induced <electric dipole moment> satisfies $d=-2\delta E_0^{(2)}/E$, not $-\delta E_0^{(2)}/E$. The <Hellmann–Feynman theorem> independently gives $d=-\partial E_0/\partial E$.
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