 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 $P^2=I$, ${\sigma }^2=1$. HenceThus 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.