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 $L^2(\mathbb R)$:

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.

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?

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