# Schrödinger operator with Coulomb potential

The free Laplacian $-\Delta$ has absolutely continuous spectrum $[0,\infty).$ The Coulomb Hamiltonian $H=-\Delta-\frac{1}{\vert x\vert}$ on $L^2(\mathbb R^3)$ has absolutely continuous spectrum $[0,\infty)$ and discrete spectrum below zero.

It is known that the essential spectrum is preserved under relative compact perturbations and the Coulomb potential is an example of this. So this implies that since the essential spectrum of the free Laplacian is $[0,\infty)$ it will also be the essential spectrum of the Coulomb Hamiltonian.

However, is there also a theorem that tells us that there are no eigenvalues embedded in the a.c. spectrum for the Coulomb Hamiltonian or is this just somehow known to be true?

This has to be shown separately. There are potentials with this decay $V(x)=O(|x|^{-1})$ that have embedded (in the ac spectrum) eigenvalues. The most famous of these is the von Neumann-Wigner potential (search for it for more information).
This potential will be oscillating. The fact that for the Coulomb potential $V(x)=V(|x|)$ is a radial function and decreasing does imply that there is no singular spectrum on $(0,\infty)$.
• @XingWang: It's not meant to be a complete argument, I just mentioned (without proof) a general fact, to put things into context: if (in one dimension) $V$ is of bounded variation and $V\to 0$, then the spectrum is purely ac on $(0,\infty)$. – Christian Remling Aug 31 '18 at 20:34
• I have to say, though, that the proof given there is a bit old-fashioned and quite a bit longer than it should be. The standard argument would go like this roughly: (1) Show by ODE techniques that the solutions of $-y''+Vy=k^2y$ look asymptotically like the free solutions $e^{\pm ikx}$; (2) use a suitable criterion (subordinacy theory, for example) that shows that then the spectrum is purely ac on $(0,\infty)$. – Christian Remling Aug 31 '18 at 22:21