Induced dipole moment and quadratic energy shift ($\delta E=-\frac{1}{2}\alpha E^2,d=\alpha E$)

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 givesThe 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 $d=-\partial E_0/\partial E$.

A neutral isotropic atom with static electric polarizability $\alpha >0$ in the field of a distant point charge $Q$ has induced dipole $\mathbf{p}=\alpha \mathbf{E}$ and energy $U=-\alpha E^2/2$. Since $E=|Q|/(4\pi {ϵ}_0{r}^2)$,The negative radial force is attractive for either sign of $Q$. Reversing $Q$ reverses the induced dipole but not the energy or leading force. The factor $1/2$ accounts for the work of inducing the dipole; using the permanent-dipole energy $-\mathbf{p}\cdot \mathbf{E}$ without that work would double the answer. The approximation assumes weak fields, a slowly varying field across the atom and negligible electronic overlap.