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For a normalized eigenstate $|n⟩$ belonging to a nondegenerate energy eigenvalue,

Two bound-state solutions of the same one-dimensional time-independent Schrodinger equation have constant Wronskian. Their decay at infinity makes this constant zero, so the two solutions are linearly dependent. Every bound-state energy is therefore nondegenerate.

The parity operator acts on a spatial wavefunction by $(P\psi )(x)=\psi (-x)$ and satisfies $P^2=I$. For a Hamiltonian operator with an even potential and parity-invariant domain, $HP=PH$. A nondegenerate energy eigenvalue has a one-dimensional eigenspace, so its eigenfunction is an eigenvector of $P$ with eigenvalue $+1$ or $-1$.

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.

For a normalized eigenstate $|n⟩$ with a nondegenerate energy eigenvalue, first-order nondegenerate perturbation theory and second-order nondegenerate perturbation theory give

The parity of a wavefunction is its eigenvalue under spatial reflection: $\psi (-x)=+\psi (x)$ for even parity and $\psi (-x)=-\psi (x)$ for odd parity. If $V(-x)=V(x)$, the Hamiltonian commutes with reflection, so its energy eigenspaces admit a parity eigenbasis. In particular, every nondegenerate bound-state energy eigenfunction has definite parity; within a degenerate scattering eigenspace one must choose the even and odd linear combinations.

Let $\Delta =E_{+}-E_{-}>0$. Hermiticity givesso in the ordered basis $(|-⟩,|+⟩)$,For a nondegenerate unperturbed level, the first-order nondegenerate perturbation theory formulas areThe diagonal matrix elements vanish, so both linear energy corrections are zero. The state corrections areThus, through linear order,Normalization changes only at quadratic order.

Separation with Dirichlet walls givesFor $a<b<c$, the ground state $(1,1,1)$ has a nondegenerate energy. The cheapest excitation changes the quantum number in the longest direction, so the first excited state $(1,1,2)$ also has a nondegenerate energy.

If $a<b=c$, the ground-state energy remains nondegenerate, but $(1,2,1)$ and $(1,1,2)$ have equal first-excited energy, giving degeneracy two.