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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?

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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 $(0,\infty)$.

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  • $\begingroup$ thank you for the first part of your answer. However, I cannot follow your second paragraph. What is now the precise argument why there is no point spectrum for the Coulomb potential embedded in the continuous one? $\endgroup$
    – Xing Wang
    Commented Aug 31, 2018 at 16:32
  • $\begingroup$ @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 $(0,\infty)$. $\endgroup$
    – Christian Remling
    Commented Aug 31, 2018 at 20:34
  • $\begingroup$ @thank you for the clarification. is there a reference of this fact that you are aware of? I must admit that I still only have a rather vague idea of what is happening here. $\endgroup$
    – Xing Wang
    Commented Aug 31, 2018 at 21:54
  • $\begingroup$ The result from my last comment is discussed in Weidmann's Lecture Notes: springer.com/us/book/9783540179023 $\endgroup$
    – Christian Remling
    Commented Aug 31, 2018 at 22:18
  • $\begingroup$ 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 $(0,\infty)$. $\endgroup$
    – Christian Remling
    Commented Aug 31, 2018 at 22:21

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