====== Classes of static potentials ====== > Give a mathematician a situation which is the least bit ill-defined -- he will first of all make it well defined. Perhaps appropriately, but perhaps also inappropriately. The hydrogen atom illustrates this process nicely. The physicist asks: "What are the eigenfunctions of such-and-such a differential operator?" The mathematician replies: "The question as put is not well defined. First you must specify the linear space in which you wish to operate, then the precise domain of the operator as a subspace. Carrying all this out in the simplest way, we find the following result..." Whereupon the physicist may answer, much to the mathematician's chagrin: "Incidentally, I am not so much interested in the operator you have just analyzed as in the following operator, which has four or five additional small terms -- how different is the analysis of this modified problem?"[(Schwartz J, //The Pernicious Influence of Mathematics on Science//, in //Logic, Methodology and the Philosophy of Science//, p. 356-360 (Stanford University Press, 1962); also in //Discrete Thoughts: Essays in Mathematics, Science and Philosophy// (Springer, 1992))] To make the usual Schrödinger problem $i\frac{d}{dt}\psi=(-\Delta+V)\psi$ well-defined we have to answer: - for which potentials $V$ is the Hamiltonian self-adjoint and - what is the domain of such a Hamiltonian? In the free case $V=0$ we have $D(H)=H^2(\mathbb{R}^n)$ or for bounded space domains and zero boundary conditions $D(H)=H^2(\Omega)\cap H^1_0(\Omega)$. The full answer will be given by the Kato--Rellich theorem and potentials that are $\Delta$-bounded. **Definition.** Let $A,B$ be densely defined operators, then $B$ is called $A$-//bounded// if: - $D(A)\subseteq D(B)$ and - there are $a,b>0$ such that for all $\varphi \in D(A)$ it holds $\|B\varphi\| \leq a\|A\varphi\| + b\|\varphi\|$. The smallest such constant $a$ is called the //relative bound//. **Theorem (Kato--Rellich).** For $A$ self-adjoint, $B$ symmetric and $A$-bounded with relative bound $<1$ we have $A+B$ self-adjoint on $D(A)$. With tools of Fourier analysis one shows that for $A=-\Delta$ this indeed holds for $B=v \in L^2(\mathbb{R}^3)+L^\infty(\mathbb{R}^3)$, which includes the usual singular Coulomb potential, and it also holds for finite sums $B=V=\sum_i v(x_i) + \sum_{i> [[Time-dependent potentials]]