Technical operator theory question on Albeverio's “Solvable Models in quantum mechanics”
$begingroup$
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
$$(vG_0 v)(x,y)=frac{v(x)v(y)}{4pi|x-y|}$$
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.
$$G_0(x,y)=frac{1}{4pi|x-y|},$$
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 $(f,vG_0vf)$, $finmathrm{L}^2(mathbb{R}^3)$, can be written as
$$(f,vG_0vf)=Vert G_0^{1/2}vfVert^2,$$
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 $Vert G_0^{1/2}vfVert^2=0$ 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 $(f,vG_0vf)=Vert G_0^{1/2}vfVert^2=0$ 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!
functional-analysis fourier-analysis operator-theory harmonic-analysis quantum-mechanics
asked Dec 17 '18 at 12:09
MrMatzetoniMrMatzetoni
1516
$endgroup$
add a comment |
$begingroup$
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
$$(vG_0 v)(x,y)=frac{v(x)v(y)}{4pi|x-y|}$$
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.
$$G_0(x,y)=frac{1}{4pi|x-y|},$$
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 $(f,vG_0vf)$, $finmathrm{L}^2(mathbb{R}^3)$, can be written as
$$(f,vG_0vf)=Vert G_0^{1/2}vfVert^2,$$
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 $Vert G_0^{1/2}vfVert^2=0$ 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 $(f,vG_0vf)=Vert G_0^{1/2}vfVert^2=0$ 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!
functional-analysis fourier-analysis operator-theory harmonic-analysis quantum-mechanics
asked Dec 17 '18 at 12:09
MrMatzetoniMrMatzetoni
1516
$endgroup$
$begingroup$
Kato pp. 281-334 discusses roots of unbounded operators.
$endgroup$
– Keith McClary
Dec 17 '18 at 17:49
add a comment |
$begingroup$
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
$$(vG_0 v)(x,y)=frac{v(x)v(y)}{4pi|x-y|}$$
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.
$$G_0(x,y)=frac{1}{4pi|x-y|},$$
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 $(f,vG_0vf)$, $finmathrm{L}^2(mathbb{R}^3)$, can be written as
$$(f,vG_0vf)=Vert G_0^{1/2}vfVert^2,$$
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 $Vert G_0^{1/2}vfVert^2=0$ 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 $(f,vG_0vf)=Vert G_0^{1/2}vfVert^2=0$ 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!
functional-analysis fourier-analysis operator-theory harmonic-analysis quantum-mechanics
asked Dec 17 '18 at 12:09
MrMatzetoniMrMatzetoni
1516
$endgroup$
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
$$(vG_0 v)(x,y)=frac{v(x)v(y)}{4pi|x-y|}$$
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.
$$G_0(x,y)=frac{1}{4pi|x-y|},$$
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 $(f,vG_0vf)$, $finmathrm{L}^2(mathbb{R}^3)$, can be written as
$$(f,vG_0vf)=Vert G_0^{1/2}vfVert^2,$$
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 $Vert G_0^{1/2}vfVert^2=0$ 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 $(f,vG_0vf)=Vert G_0^{1/2}vfVert^2=0$ 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!
functional-analysis fourier-analysis operator-theory harmonic-analysis quantum-mechanics
functional-analysis fourier-analysis operator-theory harmonic-analysis quantum-mechanics
asked Dec 17 '18 at 12:09
MrMatzetoniMrMatzetoni
1516
asked Dec 17 '18 at 12:09
MrMatzetoniMrMatzetoni
1516
asked Dec 17 '18 at 12:09
MrMatzetoniMrMatzetoni
1516
asked Dec 17 '18 at 12:09
MrMatzetoniMrMatzetoni
1516
asked Dec 17 '18 at 12:09
MrMatzetoniMrMatzetoni
1516
1516
$begingroup$
Kato pp. 281-334 discusses roots of unbounded operators.
$endgroup$
– Keith McClary
Dec 17 '18 at 17:49
add a comment |
$begingroup$
Kato pp. 281-334 discusses roots of unbounded operators.
$endgroup$
– Keith McClary
Dec 17 '18 at 17:49
$begingroup$
Kato pp. 281-334 discusses roots of unbounded operators.
$endgroup$
– Keith McClary
Dec 17 '18 at 17:49
$begingroup$
Kato pp. 281-334 discusses roots of unbounded operators.
$endgroup$
– Keith McClary
Dec 17 '18 at 17:49
add a comment |
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$begingroup$
Kato pp. 281-334 discusses roots of unbounded operators.
$endgroup$
– Keith McClary
Dec 17 '18 at 17:49