I've been looking at one of the simplest systems I can think of: a one-dimensional infinite square well on $[0,1]$ with Hamiltonian given by the following: $$\hat{H}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \alpha \, \delta\left(x-a\right)$$ with $\alpha>0, \; 0<a<1$. The typical way to solve this would be through eigenfunction expansion. For this question, it is possible to find the eigenfunctions explicitly; they solve the time-independent Schrödinger equation in a distributional sense. The conditions I've applied to look for the eigenfunctions are: $\psi(x)$ continuous on $[0,1]$, $\psi(x)=0$ at $x=0$ and $x=1$, and $$\psi'(a^+)-\psi'(a^-)=\frac{2m\alpha}{\hbar^2}\psi(a)$$ (found by integrating the time-independent Schrödinger equation from $a-\varepsilon$ to $a+\varepsilon$ and sending $\varepsilon\rightarrow 0$, assuming $\psi\in L^2[0,1]$).

I have plenty of questions about this system, but my most pressing is the following: for most systems (such as the infinite square well without delta potential), Sturm-Liouville theory can be applied to show the closure in $L^2[0,1]$ of the span of the eigenfunctions of $\hat{H}$ is all of $L^2[0,1]$, and so I can write any initial condition as an eigenfunction expansion that time-evolves trivially and have solved the full time-dependent PDE. Is the closure of the span of my eigenfunctions all of $L^2[0,1]$? I can write down the explicit form of my eigenfunctions if necessary, but I don't think that will help solve the question.

As an extension to the above, what if I consider $H_0^1[0,1]$ with $H^1$ norm?

  • $\begingroup$ Probabistically, this is the generator of the Brownian motion killed upon leaving the interval, and additionally killed at an exponential rate according to the local time at $a$. The corresponding heat semigroup consists of compact operators, so yes, they should form a complete orthogonal set in $L^2$. $\endgroup$ – Mateusz Kwaśnicki Mar 5 '19 at 9:16
  • $\begingroup$ You can still associate a self-adjoint operator with your expression, so completeness of the eigenfunctions (assuming you have pure point spectrum, as you will on a bounded interval) is guaranteed. $\endgroup$ – Christian Remling Mar 5 '19 at 15:41
  • $\begingroup$ There is a huge literature on SO with measure valued potentials, so maybe you should start out by doing some searching. $\endgroup$ – Christian Remling Mar 5 '19 at 15:42
  • $\begingroup$ Thanks for your responses, they're reassuring. I do have a point spectrum and my operator is self-adjoint. I have a book that attacks semigroup theory from a functional analysis perspective, so I'll read that. I'd be happy if you could direct me to some literature on the topic of the Schrödinger equation with measure-valued potentials; I only see one paper that is far from what I am looking for. $\endgroup$ – S. Thornton Mar 5 '19 at 17:08

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