This question has been bugging me for some time.


Take the hamiltonian for the hydrogen atom: $$\hat{H}=-\frac{1}{2}\nabla^2-\frac{1}{r},$$ acting on (a domain contained in) $L^2(\mathbb{R}^3)$. It is standard fact that this is an unbounded operator which has a countable infinity of eigenvalues, all of which are negative and which accumulate around 0, and has a continuous spectrum on the whole of $(0,\infty)$. Physically, the former are bound states which correspond to elliptic Keplerian orbits in the classical problem, and the latter are unbound states and correspond to hyperbolic orbits. I also know that the spectrum of all unbounded operators is a closed set, so that 0 is definitely in $\sigma\left(\hat{H}\right)$.


My question is then: to what part of the spectrum does 0 belong to (i.e. point, continuous, residual)? What are the corresponding eigenfunctions? What kind of degeneracy does it have? (I would expect it to admit a common eigenvector with any $l,m$ angular momentum numbers, but I'm far from sure.) How do the eigenfunctions correspond to the nearby bound and unbound states?



Is zero a hydrogen eigenvalue?
First of all, the Hamiltonian in question is defined on $L^2(\mathbb R^3)$, not on $L^2(\mathbb R)$. This is important because in the one-dimensional case the potential would have a non-integrable singularity which complicates things seriously. On $L^2(\mathbb R^3)$, the operator, defined as a closure from $C\_0^\infty$, is selfadjoint. This is proved, for example, in the book by


* T. Kato, *Perturbation theory for linear operators*, Springer, 1966.


Thus the residual spectrum is impossible. A rigorous calculation of eigenvalues and eigenfunctions can be found in the books


* L. D. Faddeev and O. A. Yakubovskii, *Lectures on quantum mechanics for mathematics students*. American Mathematical Society, 2009;
* L. A. Takhtajan, *Quantum mechanics for Mathematicians*, American Mathematical Society, 2008.


The point 0 is an accumulation point of negative eigenvalues and the limit point of continuous spectrum, thus it belongs to the essential spectrum.


