A very specific question, but posted on the offchance that someone may be able to help. If we have a Schrödinger equation with arbitrary "superpotential" \begin{equation} \frac{d^2 \psi}{dx^2} +(W^2 \pm W') \psi = E \psi \end{equation} satisfying periodic boundary conditions $\psi(x) = \psi(x+a)$, how can we put a lower bound on $E_0$ where $E_0$ is the closest eigenvalue to 0 (assuming $W(x) = W(x+a)$ is such that $E_n \neq 0$ for all $n$)?
1 Answer
This is not a very explicit answer, just an attempt to put things into context.
First of all, by periodic boundary conditions, one usually means $\psi(0)=\psi(a)$, $\psi'(0)=\psi'(a)$ (without the condition on the derivatives, you are not getting a selfadjoint operator). Next, let's recall that a Schrödinger operator $H_{DD}=d^2/dx^2+V(x)$ with Dirichlet boundary conditions $u(0)=u(a)=0$ has positive spectrum precisely if we can write $V=W^2+W'$. Moreover, we can take $W=u'/u$, where $u$ is a positive solution of $u''+Vu=0$ (which exists because we are below the Dirichlet spectrum).
Now your assumption that $W(a)=W(0)$ becomes $(u'/u)(0)=(u'/u)(0)$. In other words, $0\in\sigma (H_{\alpha\alpha})$; here, I denote by $H_{\beta\gamma}$ the operator with boundary conditions $(u'/u)(0)=\tan\beta$, $(u'/u)(a)=\tan\gamma$. The $\alpha$ from above is defined as the common value of $\arctan (u'/u)$ at the boundary.
We are in the spectrum of the periodic problem if $E\in\sigma (H_{\beta\beta})$ for some $\beta$ and, moreover, the solution to $y''+Vy=Ey$ that satisfies the boundary conditions has the additional property that $y(a)=y(b)$. In particular, your assumption that $0$ is not in the spectrum of the periodic problem becomes $u(a)\not= u(0)$ (or, equivalently, $\int_0^a W(t)\, dt\not= 0$).
So, to sum this up, at $E=0$ we have $0\in\sigma (H_{\alpha\alpha})$, and to find the nearest periodic eigenvalue, we will now change $\alpha$ and monitor the condition that $y(0)=y(a)$. Notice that $E(\beta)$, defined as this moving eigenvalue of $H_{\beta\beta}$ (and starting with $E(\alpha)=0$) is monotone in $\alpha$; in fact, this property holds if I vary just one boundary condition. In particular, the sought periodic eigenvalue lies somewhere between the Neumann eigenvalue $E(\pi/2)<0$ and the Dirichlet eigenvalue $E(0)>0$. We can obtain lower bounds on the distance to zero from $u(a)u(0)$; we have to control how fast this can change in energy, and for this, we can use Gronwall style estimates from the DE, but this will get more technical and I'll stop here.

$\begingroup$ But Christian, doesn't $\psi (x+a)=\psi(x)$, as written by the OP, imply both $\psi (a)=\psi(0)$ and $\psi' (a)=\psi'(0)$ ? ;) $\endgroup$ Sep 22, 2019 at 1:44

$\begingroup$ @MichaelEngelhardt: Yes, it does, if the condition was intended to hold for all $x$. I can't remember my original thoughts now, but probably I interpreted it to be imposed for one fixed $x$. $\endgroup$ Sep 23, 2019 at 14:05

$\begingroup$ yes, probably ... physicists sometimes talk a bit funny. I just ran across this old post by chance yesterday, and I couldn't resist a tongueincheek comment since you sounded so curmudgeonly. $\endgroup$ Sep 23, 2019 at 17:54