# Technical operator theory question on Albeverio's “Solvable Models in quantum mechanics”

I'm currently studying S. Albeverio's book "Solvable models in quantum mechanics" where some technical things are used that I don't fully understand. It is a general technical operator theory question, but I will introduce the setting.

General setting:

Looking at the Hamiltonian $$-\Delta + V$$ with the underlying Hilbert space $$\mathrm{L}^2(\mathbb{R}^3)$$, where $$V$$ is a real potential, the aim in the first chapter of the book is to approximate a $$\delta$$-Potential in 3D by scaling $$V$$. We denote $$v:=|V|^{1/2}$$, where the potential $$V$$ is an element of the Rollnik-class, i. e. real functions for which $$\int_{\mathbb{R}^6}\frac{|V(x)||V(y)|}{|x-y|^2}dxdy < \infty$$ The operator $$vG_0 v: \mathrm{L}^2(\mathbb{R}^3)\rightarrow \mathrm{L}^2(\mathbb{R}^3)$$ defined by the kernel $$(vG_0 v)(x,y)=\frac{v(x)v(y)}{4\pi|x-y|}$$ plays an important role in this setting (again $$v:=|V|^{1/2}$$). Note that the kernel is pointwise positive and the Rollnik-condition ensures that $$vG_0v$$ is Hilbert-Schmidt. Furthermore $$G_0$$ denotes the Operator given by convolution with the fundamental solution of the Laplace operator, i.e. $$G_0(x,y)=\frac{1}{4\pi|x-y|},$$ and $$G_0(-\Delta \varphi)=\varphi$$ for all $$\varphi \in C_c^\infty$$.

The (technical) problem:

On p.21 and p.22 he uses the fact, that $$(f,vG_0vf)$$, $$f\in\mathrm{L}^2(\mathbb{R}^3)$$, can be written as $$(f,vG_0vf)=\Vert G_0^{1/2}vf\Vert^2,$$ so in a sense he uses that $$vG_0v$$ is a positive Operator and can be written as $$vG_0v=vG_0^{1/2}G_0^{1/2}v=(G_0^{1/2}v)^*(G_0^{1/2}v)$$. Also he uses that $$\Vert G_0^{1/2}vf\Vert^2=0$$ implies $$vf=0$$.

I know that if an bounded linear operator $$A$$ is positive, it can be decomposed as $$A=B^*B$$ with $$B$$ also bounded. But I'm having great trouble in understanding how $$G_0^{1/2}v$$ is defined, as the operators $$v$$ (multiplication) and $$G_0$$, which $$vG_0v$$ is composed of, are unbounded. Why does $$G_0^{1/2}v$$ even exist and why is it bounded? How can I conclude $$(f,vG_0vf)=\Vert G_0^{1/2}vf\Vert^2=0$$ implies $$vf=0$$ and in what sense? Why is $$vG_0v$$ (obviously?) positive?

I thought about this a long time but couldn't come up with a satisfying solution. I would appreciate your help and comments!

• Kato pp. 281-334 discusses roots of unbounded operators. – Keith McClary Dec 17 '18 at 17:49

## 1 Answer

Theorem 9.8 in

Elliott H. Lieb und Michael Loss. Analysis. Second Ed. Bd. 14. Graduate Studies in Mathematics. American Mathematical Society, 2001.

gives a important result regarding this issue. It resolved all my doubts.