= Solution
The electron's <electric dipole moment> is $\mathbf d=-e\langle x\rangle\hat{\mathbf x}$. It vanishes in the unperturbed ground <energy eigenstate>. The first correction to the <wavefunction> is linear in the <electric field> and has the opposite parity; its interference with the unperturbed state therefore gives a nonzero position <expectation value> of first order.
In detail, the normalized ground <quantum state>, to the order needed, is
$$
|g\rangle=|\phi_0\rangle+
eE\sum_{k\geq1}\frac{x_{k0}}{\epsilon_0^{(0)}-\epsilon_k^{(0)}}|\phi_k\rangle+O(E^2).
$$
The normalization correction starts at second order and does not affect the linear term below. The two cross terms give
$$
\langle x\rangle=2eE\sum_{k\geq1}
\frac{|x_{k0}|^2}{\epsilon_0^{(0)}-\epsilon_k^{(0)}}+O(E^3).
$$
The absence of an $E^2$ term follows from the <parity operator>: $\langle x\rangle$ changes sign under field reversal. Comparing with the <second-order nondegenerate perturbation theory> energy sum yields
$$
\boxed{\mathbf d=-\frac{2\delta\epsilon_0^{(2)}}{E^2}\mathbf E+O(E^3)
=\alpha_{\mathrm{pol}}\mathbf E+O(E^3),\qquad
\alpha_{\mathrm{pol}}=-\frac{2\delta\epsilon_0^{(2)}}{E^2}>0.}
$$
Thus the induced <electric dipole moment> points along the <electric field>, with positive <electric polarizability>. The <induced dipole moment and quadratic energy shift> obey $\delta\epsilon_0^{(2)}=-\alpha_{\mathrm{pol}}E^2/2$. This factor of one half accounts for the energy needed to induce the dipole. As a check, the <Hellmann–Feynman theorem> gives $d_x=-\partial\epsilon_0/\partial E$, with the same result.
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