= Solution
The <Stark effect> changes atomic <energy levels> in a static <electric field>. For $\mathbf E=\mathcal E\hat{\mathbf z}$, the <electric-dipole interaction> is $H'=-d_z\mathcal E=e\mathcal E z$ for an electron, with $e>0$. A nondegenerate state of definite <parity> has $\langle z\rangle=0$, so its leading <quadratic Stark effect> is
$$
\Delta E_n^{(2)}=e^2\mathcal E^2\sum_{r\ne n}\frac{|\langle r|z|n\rangle|^2}{E_n-E_r}.
$$
The <ground state> shift is negative, conventionally $-\alpha\mathcal E^2/2$ with positive <electric polarizability>.<Quantum Mechanics by Richard Fitzpatrick/Quadratic Stark Effect>[Equations(7.40),(7.57),(7.65): parity selection and polarizability]{ref}
Degenerate opposite-parity states require <degenerate perturbation theory>. In ideal Coulomb <hydrogen>, the $2s,2p_0$ subspace has an off-diagonal matrix element of magnitude $3ea_{\rm B}\mathcal E$, where $a_{\rm B}$ is the <Bohr radius>. Its mixed states shift by
$$
\boxed{\Delta E=\pm3ea_{\rm B}\mathcal E,\qquad\Delta E_{2p_{\pm1}}=0\ \text{to first order}.}
$$
This <Linear Stark effect in the first excited hydrogen eigenspace> assumes fine-structure and Lamb splittings are negligible compared with the mixing. In an isolated parity eigenstate with such degeneracies resolved, sufficiently weak fields restore quadratic behaviour. Strong fields require further mixing and can ionize the atom.<Quantum Mechanics by Richard Fitzpatrick/Linear Stark Effect>[Equations(7.82)–(7.94): n=2 mixing and shifts]{ref}
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