# Schrödinger operator with delta (zero range) interaction.

I am reading the book of Albeverio named Solvable models in quantum mechanics. In the first chapter it is explained how to realize the operator $"-\Delta+\delta_0"$ as a self adjoint operator on $L^2(\mathbb{R}^3)$. I explain the main idea. At the beginning one has to consider $H:=-\Delta_{|_{C^{\infty}_0}(\mathbb{R}^3\setminus \{0\})}$. One has to show that such an operator admits some self adjoint extensions on $L^2(\mathbb{R}^3)$, that will be, by definition, our operators $"-\Delta+\delta_0"$. There are two technical things that I didn't understand. The first one: the book says that one can show (it is not trivial) that the adjoint of $H$ is $H^*=-\Delta$ with domain $$\mathcal{D}(H^*)=\{g\in H^{2,2}_{loc}(\mathbb{R}^3\setminus \{0\})\cap L^2(\mathbb{R^3})\,\,s.t.\,\, \Delta g\in L^2(\mathbb{R^3})\}.$$ I really do not understand, maybe it is a very stupid thing, why this space is not equal to $H^2(\mathbb{R}^3)$.

I explain my doubt: I say that $g\in L^2(\mathbb{R}^3)$ and $\Delta g\in L^2(\mathbb{R}^3)$ allow me to consider their Fourier transform, getting $(1+|\xi|^2)\hat{g}\in L^2(\mathbb{R}^3)$, which says to me that $g\in H^2(\mathbb{R}^3)$. Why I'm wrong?

The second thing that i do not understand is the following. The autor says that a straightforward calculation shows that $$\psi(k,x)=\frac{e^{ik|x|}}{|x|},\,\, x\neq 0,\,\, \mathfrak{Im}(k)>0,$$ is the only one solution of $$H^* \psi(k)=k^2\psi(k),\,\, \psi(k)\in\mathcal{D}(H^*),\,\, k^2\in \mathbb{C}\setminus\mathbb{R},\,\, \mathfrak{Im}(k)>0.$$ I do not know how to procede in proving this fact.

Can someone help me to understand this two facts? I apologize for the technical, maybe stupid, question. Thanks.

The notation $\Delta g\in L^2(\mathbb R^3)$ is misleading: It should rather read $\Delta g\in L^2(\mathbb R^3 \setminus\{0\})$. The difference is just a null set, so negligible.
But it is fundamental for the definition of a weak derivative of $g$: By definition $$\int g \Delta \psi = \int \Delta g \psi$$ for all $\psi \in C_0^\infty(\mathbb R^3 \setminus\{0\})$. That is, the test function is zero in a neigborhood of the origin! Hence, this definition of weak derivatives cannot detect singularities at $x=0$. Thus $\Delta g$ is not the weak Laplacian of $g$ on all of $\mathbb R^3$.
Now look at those functions $\psi(\cdot, k)$. These functions are unbounded near the origin. Would $\psi(\cdot,k)\in H^2(\mathbb R^3)$ hold, then $\psi$ would be continuous and bounded (by Sobolev embedding) near the origin. Hence the domain of $H^*$ cannot be $H^2(\mathbb R^3)$.
• Note that the Sobolev embedding is into $C^{0,1/2}$ since $k-\frac{n}{p}=2-\frac{3}{2}=\frac{1}{2}$, where $k$ is the number of derivatives, $n$ is the dimension, and $p$ is the integral exponent for the Sobolev space. This is a subset of $C^0$, which is what is claimed here.