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?
Best Answer
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
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.