= Solution
Define the <parity operator> by $(P\psi)(x)=\psi(-x)$. For the usual full-line <Hamiltonian operator>, or any parity-invariant operator domain and boundary conditions, an even potential gives $HP=PH$: the second derivative is unchanged by reflection and $V(-x)=V(x)$. If $H\psi=E\psi$, then $H(P\psi)=E(P\psi)$. A <nondegenerate energy eigenvalue> has a one-dimensional <eigenspace>, so $P\psi=\sigma\psi$ for some scalar $\sigma$. Since $P^2=I$, $\sigma^2=1$. Hence
$$
\boxed{P\psi=\psi\text{ or }P\psi=-\psi}.
$$
Thus the stationary state has even or odd <parity>. The symmetry must include the operator domain: an even potential alone would not imply this conclusion with asymmetric boundary conditions.
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