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

Question

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 $V(x)\to 0$ 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 this should be true.

2. Can we say anything about the bottom of the continuous spectrum? E.g. if there are three discrete eigenvalues $\lambda_1 < \lambda_2 <\lambda_3 < 0$ corresponding to even $\phi_1$, odd $\phi_2$, and even $\phi_3$, will $0$ have an odd eigenfunction or an odd resonance?

I would also appreciate pointers to any references where I could read about these issues. Most of the resources I've seen treat the abstract theory of self-adjoint operators in a lot of detail, but don't go into much depth about specific examples like this.

Answer 1

If you were to consider two problems on $L^2[0,\infty)$ for $Lf = -f''+Vf$ with endpoint conditions $f(0)=0$ and $f'(0)=0$ respectively, then I would expect that the eigenvalues of one problem would interlace with the other (if there are any,) and the eigenfunctions of either would extend to the full interval (in the first case as odd functions and in the second case as even functions.) So that would suggest that what you're seeing is general. Typically the eigenfunction for the lowest eigenvalue of the full problem would be non-vanishing, which would make it even.

Unfortunately, I'm not sure I could fill in all the details for you.