I know that the Coulomb-Hamiltonian $H=-\Delta - |\cdot|^{-1}$ is self-adjoint with $\operatorname{dom}(H)=H^2(\mathbb R^3)$. This follows by the Kato-Rellich-Theorem.

Has the corresponding quadratic form a form domain of $Q(H) = H^1(\mathbb R^3)$?

Is it also true for dimensions which are bigger that $3$? How to see it?

Best wishes :)

  • $\begingroup$ Has noone any idea? Is the question not clear? I would also be glad if one could answer one the questions. :) $\endgroup$
    – Peter
    Jan 30, 2014 at 23:11

2 Answers 2


The general theorem is XIII.96 in Reed/Simon "Methods of Modern Mathematical Physics, Vol. IV: Analysis of Operators" (Academic Press, 1978). A multiplication operator $V$ is $-\Delta$-bounded with relative bound zero if it is uniformly locally $L^p$ where $p = 2$ if dimension $n \leq 3$ and $p>n/2$ if $n \geq 4$. Then you can apply Kato-Rellich (Th. X.12 in Reed/Simon Vol. II) to get a self-adjoint operator $H = -\Delta + V$ with domain $H^2(\mathbb{R}^n)$ as you state correctly.

For your singular Coulomb potential the $L^p$-condition means the following integral (over a small bounded region around the origin, so we choose a ball $B$ with radius 1) has to converge. $$ \int_B |V(x)|^p d^n x $$

For $V(x)=-|x|^{-1}$ this is equivalent to $$ \int_0^1 r^{-p+n-1} dr < \infty. $$

We get the condition $-p+n-1 > -1$ thus $n > p$. So your statement is true in dimensions $n \geq 3$. For potentials bounded from below, you have even more options (i.e. locally $L^2$ for all dimensions), see http://www.math.caltech.edu/papers/bsimon/xliv.pdf, p.3 for an introduction and references.

The idea using quadratic forms could be used to include even more singular potentials. But I would be careful about the domain (you need something like $\sqrt{|V|} \psi \in L^2$ for $\psi \in H^1$) of the quadratic form, maybe you can find some hints in another paper by B. Simon: http://www.math.caltech.edu/SimonPapers/13.pdf

Enjoy! ;-)


This is related to a problem posed by Tosio Kato in 1953 and not resolved until 2002, a few years after Kato's death. Take a look at the references on this page to Kato's Conjecture:


I believe that you are essentially correct because the form domain is related to the square root of the elliptic operator, and that is related to Kato's conjecture.


I don't think that the Coulombic singularity is a problem for what you are wanting to say.

  • $\begingroup$ Thanks a lot. :) I already figured out that in $d\geq 4$ dimensions the Hydrogen-Hamiltonian is self-adjoint on $H^2(\mathbb R^d)$. This follows also by the Kato-Rellich-Theorem. Does anyone know if this Hamiltonian is still self-adjoint in 1 or 2 dimensions? $\endgroup$
    – Peter
    Feb 11, 2014 at 14:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .