= Solution
For a bound <energy eigenstate> with a <nondegenerate energy eigenvalue> of an even one-dimensional potential, the <wavefunction> can be chosen with definite <parity>. A uniform <electric field> $\mathcal E$ gives the <electric-dipole interaction> $H'=-q\mathcal E x$, which is odd under reflection. Hence
$$
\Delta E_n^{(1)}=-q\mathcal E\langle n|x|n\rangle=0,
$$
since $x|\psi_n(x)|^2$ is odd. <Second-order nondegenerate perturbation theory> then gives the leading allowed <Stark effect> shift:
$$
\boxed{\Delta E_n=q^2\mathcal E^2\sum_{r\ne n}\frac{|\langle r|x|n\rangle|^2}{E_n^{(0)}-E_r^{(0)}}+O(\mathcal E^4).}
$$
Only opposite-<parity> intermediate states have nonzero <position> <matrix elements>. More generally, reflection maps $H(\mathcal E)$ to $H(-\mathcal E)$, so an analytic nondegenerate <energy level> is an even function of the field. This explains the absence of all odd orders and the usual <quadratic Stark effect>.
A level separation changes by the difference of the two quadratic coefficients. Symmetry alone does not guarantee a nonzero differential shift: in a <displaced quantum harmonic oscillator> every level moves by the same $-q^2\mathcal E^2/(2m\omega^2)$, leaving every spacing unchanged. Degenerate states would instead require <degenerate perturbation theory> and can split linearly; the <nondegeneracy of one-dimensional bound states> applies to ordinary isolated well levels.
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