# Why do the eigenfunctions of a 1D Schroedinger operator with even potential alternate in parity?

Let $\mathcal L$ be a Schroedinger operator on the real line of the form

$\mathcal L = -\frac {d^2} {dx^2} + V(x),$

where $V$ is an even, smooth function. I am interested in the case where $V(x)\to 0$ sufficiently fast as $x\to \pm \infty$, so that the continuous spectrum of $V$ is equal to $[0,\infty)$, but this may not be important to the issue at hand. Since $\mathcal L$ commutes with the parity operator $P: f(x)\mapsto f(-x)$, they are simultaneously diagonalizable, and a basis for each eigenspace can be chosen so that each basis element has definite parity.

My questions about the spectrum of $\mathcal L$ relative to $L^2(\mathbb R)$:

1. If all the discrete eigenvalues of $V$ are simple--so that, if I understand correctly, the eigenfunctions must have definite parity--does the parity of the eigenfunctions necessarily alternate? This seems to be the case in every example I've seen, but I don't know why it should be true.

2. Can we say anything about the bottom of the continuous spectrum? E.g. if there are two discrete eigenvalues $\lambda_1 < \lambda_2 < 0$ corresponding to even $\phi_1$ and odd $\phi_2$, will $0$ have an even eigenfunction or an even resonance?

3. Would any of this change if $\mathcal L$ were replaced by a general Sturm-Liouville operator on $\mathbb R$?

• The ev's are simple automatically, but there may not be any (for example, if $V\equiv 0$). Also, $0$ will never be an eigenvalue for rapidly decaying $V$, so it's unclear what you mean by Q2. – Christian Remling Nov 15 '16 at 7:53

The eigenvalues with even/odd eigenfunctions can equivalently be described as the eigenvalues of the half line problem $$-y''(x)+V(x)y(x)=Ey(x), \quad 0\le x<\infty ,$$ with boundary conditions $y'(0)=0$ and $y(0)=0$, respectively.