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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 Schrödinger 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 $-1$.

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 $H^{\prime }=-q\mathcal{E}x$, which is odd under reflection. Hencesince $x|{\psi }_n(x){|}^2$ is odd. Second-order nondegenerate perturbation theory then gives the leading allowed Stark effect shift: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.

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.

The even, spherically symmetric hydrogen atom ground state has no permanent electric dipole, so a uniform electric field gives no first-order energy shift. The electric-dipole selection rules for hydrogen allow odd-parity excited states to mix in at second order, yielding a negative shift quadratic in the field.

Here there is a nondegenerate energy eigenvalue, so second-order nondegenerate perturbation theory applies. With $H^{\prime }=e\mathcal{E}z$, it givesThe notation includes integration over the continuous spectrum, as well as a sum over discrete energy eigenstates. All denominators are negative because $1s$ is the ground state. The positive electric polarizability ${\alpha }_{1s}$ therefore lowers its energy.^[ref]

An independent way to see the coefficient avoids truncating that sum. Define $E_{\mathrm{h}}=e^2/(4\pi {ϵ}_0{a}_0)$ and $f=e{a}_0\mathcal{E}/E_{\mathrm{h}}$. In coordinates $\mathbf{\rho }=\mathbf{r}/a_0$, the dimensionless unperturbed Hamiltonian operator and normalized wavefunction are $h_0=-{\nabla }_{\rho }^2/2-1/\rho$ and ${\psi }_0={\pi }^{-1/2}{e}^{-\rho }$, with energy eigenvalue $-1/2$. Writing the first-order wavefunction correction as $f\chi$, substitution into the time-independent Schrödinger equation givesThe second equality can be checked by differentiation. Its parity makes $\chi$ orthogonal to ${\psi }_0$. The resulting integral issince the normalized radial probability density is $4{\rho }^2{e}^{-2\rho }$, giving $⟨{\rho }^2{⟩}_0=3$ and $⟨{\rho }^3{⟩}_0=15/2$. Converting back to SI units gives the stated electric polarizability. The induced electric dipole moment is $-\partial \Delta E/\partial \mathcal{E}={\alpha }_{1s}\mathcal{E}$ and tends to zero with the electric field, unlike the selected degenerate branches' leading dipoles. Both calculations describe weak-field level shifts; a spatially uniform static electric field also allows eventual quantum tunnelling and ionization, which these real energy corrections do not calculate.