= Stark effect of the hydrogen ground state
{c}
The even, spherically symmetric <hydrogen atom> ground state has no permanent <electric dipole>, so a uniform <electric field> gives no first-order energy shift. The <electric-dipole selection rules for hydrogen> allow odd-parity excited states to mix in at second order, yielding a negative shift quadratic in the field.
Here there is a <nondegenerate energy eigenvalue>, so <second-order nondegenerate perturbation theory> applies. With $H'=e\mathcal E z$, it gives
$$
\Delta E_{1s}^{(2)}
=e^2\mathcal E^2\sum_{k\ne1s}
\frac{|\langle k|z|1s\rangle|^2}{E_1-E_k}
=-\frac12\alpha_{1s}\mathcal E^2,
\qquad
\alpha_{1s}=\frac92\,4\pi\epsilon_0a_0^3.
$$
The notation includes integration over the <continuous spectrum>, as well as a sum over discrete <energy eigenstates>. All denominators are negative because $1s$ is the <ground state>. The positive <electric polarizability> $\alpha_{1s}$ therefore lowers its <energy>.<Quantum Mechanics by Richard Fitzpatrick/Quadratic Stark Effect>[Quantum Mechanics by Richard Fitzpatrick; Fitzpatrick, Quadratic Stark Effect, equations (7.57) and (7.65): quadratic convention and exact polarizability]{ref}
An independent way to see the coefficient avoids truncating that sum. Define $E_{\rm h}=e^2/(4\pi\epsilon_0a_0)$ and $f=ea_0\mathcal E/E_{\rm h}$. In coordinates $\boldsymbol\rho=\mathbf r/a_0$, the dimensionless unperturbed <Hamiltonian operator> and normalized <wavefunction> are $h_0=-\nabla_\rho^2/2-1/\rho$ and $\psi_0=\pi^{-1/2}e^{-\rho}$, with <energy eigenvalue> $-1/2$. Writing the first-order <wavefunction> correction as $f\chi$, substitution into the <time-independent Schrödinger equation> gives
$$
(h_0+1/2)\chi=-\rho\cos\theta\,\psi_0,
\qquad
\chi=-\tfrac12\rho(\rho+2)\cos\theta\,\psi_0.
$$
The second equality can be checked by differentiation. Its <parity> makes $\chi$ orthogonal to $\psi_0$. The resulting <integral> is
$$
\frac{\Delta E_{1s}^{(2)}}{E_{\rm h}}
=f^2\langle\psi_0|\rho\cos\theta|\chi\rangle
=-\frac{f^2}{6}\bigl(\langle\rho^3\rangle_0+2\langle\rho^2\rangle_0\bigr)
=-\frac94 f^2,
$$
since the normalized radial <probability density> is $4\rho^2e^{-2\rho}$, giving $\langle\rho^2\rangle_0=3$ and $\langle\rho^3\rangle_0=15/2$. Converting back to <SI> units gives the stated <electric polarizability>. The induced <electric dipole moment> is $-\partial\Delta E/\partial\mathcal E=\alpha_{1s}\mathcal E$ and tends to zero with the <electric field>, unlike the selected degenerate branches' leading dipoles. Both calculations describe weak-field level shifts; a spatially uniform static <electric field> also allows eventual <quantum tunnelling> and <ionization>, which these real <energy> corrections do not calculate.
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