= Solution
The unperturbed <infinite square well> is invariant under the <parity operator>. Its one-dimensional <energy eigenvalues> are nondegenerate, so each <energy eigenstate> has definite parity. Its position <expectation value> therefore vanishes: $|\phi_n(x)|^2$ is even, while $x$ is odd. The <first-order nondegenerate perturbation theory> correction is consequently
$$
\delta\epsilon_n^{(1)}=eE\langle\phi_n|x|\phi_n\rangle=0.
$$
The <electric field> does mix opposite-parity <quantum states>, and <second-order nondegenerate perturbation theory> contains two factors of $eEx$. Its leading contribution is thus proportional to $E^2$: this is a <quadratic Stark effect>. More generally, applying the <parity operator> changes $H(E)$ into $H(-E)$, so their corresponding <energy eigenvalues> agree. An analytic expansion about zero <electric field> contains only even powers of $E$. We assume $e|E|a$ is small compared with the unperturbed excitation scale.
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