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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.

For $g\ge 0$, the Hamiltonian operator is nonnegative: its quadratic form is ${\hslash }^2\int |{\psi }^{\prime }{|}^{2}dx/(2m)+g(|\psi (a){|}^2+|\psi (-a){|}^2)\ge 0$. Thus it has no negative-energy bound states. A nonnegative-energy exterior wavefunction cannot be square integrable unless it vanishes identically, so there are no other bound states either.

For $g<0$, set $\nu =-mg/{\hslash }^2>0$. The even equation is the zero of $f(\kappa )=\nu (1+e^{-2a\kappa })-\kappa$. Since $f(0)=2\nu >0$, $f(\kappa )\to -\infty$ and $f^{\prime }(\kappa )=-2a\nu e^{-2a\kappa }-1<0$, the intermediate value theorem and strict monotonicity give exactly one positive equation root.

The odd equation is the zero of $h(\kappa )=\nu (1-e^{-2a\kappa })-\kappa$. Here $h(0)=0$, $h^{\prime }(0)=2a\nu -1$, $h^{\prime \prime }(\kappa )=-4a^2\nu e^{-2a\kappa }<0$, and $h(\kappa )\to -\infty$. If $2a\nu \le 1$, strict concavity makes $h(\kappa )<0$ for every $\kappa >0$. If $2a\nu >1$, $h$ initially increases, has exactly one maximum, and then decreases through zero exactly once. This proves the bound-state count of an attractive symmetric double-delta potential:At the threshold $g=-{\hslash }^2/(2ma)$, the odd zero-energy solution is proportional to $x$ inside $(-a,a)$ and to $a\mathrm{sgn}x$ outside. It is bounded but not a square-integrable function, so it is a zero-energy scattering resonance rather than a second bound state. Each negative-energy bound state is either even or odd by nondegeneracy of one-dimensional bound states; the two branches above exhaust the count.

For $V(-x)=V(x)$, the parity operator $P\psi (x)=\psi (-x)$ commutes with the Hamiltonian operator. Also $P^2=I$, so its eigenvalues are $\pm 1$ and the energy eigenspaces admit a basis of definite parity. More strongly, nondegeneracy of one-dimensional bound states says that each bound state energy here is nondegenerate: the Wronskian of two solutions at the same energy is constant by the Schrödinger equation, and decay at infinity makes it zero, forcing linear dependence. Thus each bound state itself is even or odd.

An even function obeys the Neumann boundary condition ${\psi }^{\prime }(0)=0$; an odd function obeys the Dirichlet boundary condition $\psi (0)=0$. WriteSmooth reflection symmetry gives ${\chi }^{\prime }(0)=0$, while $K^{\prime }(0)=-\chi (0)/\hslash$. Hence, for the central WKB approximation from part (c),The two boundary conditions consequently give the WKB quantization condition for a symmetric double well:The full wavefunction is ${\psi }_{+}(x)={\psi }_R(|x|)$ in the even case and ${\psi }_{-}(x)=\mathrm{sgn}(x){\psi }_R(|x|)$ in the odd case, with the value at zero fixed continuously. The relevant boundary condition makes these extensions differentiable at zero. Both action integrals $\theta ,\varphi$ depend on $E$ through the integrands and the classical turning points.

Suppose ${\chi }_1$ and ${\chi }_2$ are two bound-state solutions with the same energy. Their Wronskiansatisfies $W^{\prime }=0$ because the Schrödinger equation contains no first-derivative term. Hence $W$ is constant. Both wavefunctions and their derivatives decay at infinity, so $W\to 0$ there and therefore $W\equiv 0$. The two solutions are linearly dependent. Thus every one-dimensional bound-state energy is nondegenerate.

The Sturm oscillation theorem gives three nodes of a wavefunction for the fourth bound state. reflection symmetry and nondegeneracy of one-dimensional bound states make it odd: one node of a wavefunction is at the centre, and the other two lie at opposite positions. In each flat allowed region the wavefunction is a sinusoid, with $k=\sqrt{2m(E-V)}/\hslash$; its wavelength is shortest in the central, deepest region. Outside the well it decays exponentially, with opposite signs on the two sides. Both $\Psi$ and ${\Psi }^{\prime }$ remain continuous at every finite step, as required by wavefunction matching at a finite potential step; only the curvature jumps. In particular, the wavefunction need not vanish at a step.

This original wavefunction matching at a finite potential step illustration uses freely chosen well dimensions and energies. It shows the required parity, node count, wavelength ordering and decaying tails; the source drawing supplies no numerical scale from which to fix exact node positions.

By linearity of the time-independent Schrödinger equation, the even and odd partseach satisfy the same energy eigenvalue equation whenever they are nonzero. They obey $R{\psi }_{+}={\psi }_{+}$ and $R{\psi }_{-}=-{\psi }_{-}$, so they have definite parity. At least one is nonzero, and $\psi ={\psi }_{+}+{\psi }_{-}$. Each nonzero part can be normalized separately.

Therefore a basis of stationary solutions can be chosen to have definite parity, and solving the even and odd sectors loses no energy eigenvalues. This does not assert that every possible quantum superposition has definite parity; a superposition of the two sectors need not. For one-dimensional bound states in the usual regular setting, nondegeneracy of one-dimensional bound states further implies that each individual energy eigenstate has definite parity.