Consider the Schroedinger operator

$$H = -\Delta + \frac{c}{\vert x \vert^2}$$ in two dimensions with $c >0$

This operator has a self-adjoint realization, since it is a positive symmetric operator on $C_c^{\infty}( \mathbb R^2 \setminus \{0\}).$ I would like to understand three things:

0.) What are possible domains $D(H)$ on which $H$ is self-adjoint?

1.) Does there exist a domain $D(H)$ of $H$ on which $-\Delta$ and $H$ are both self-adjoint.

2.) On these domains $D(H)$, can we expect $$\Vert \frac{f}{\vert x \vert^2}\Vert_{L^2} \le \alpha_1 \Vert f \Vert_{L^2} + \alpha_2 \Vert \Delta f \Vert_{L^2} \text{ for some }\alpha_i \ge 0?$$

I would appreciate insights on any of these questions. My main interest is in small $c>0.$

One way to think about it may be to follow Example 4 in Reed-Simon II on page 160.

There the authors prove, going via a 1D reduction, that if

$c \ge 1,$ then $H$ is essentially self-adjoint (there is a unique self-adjoint extension) on $C_c^{\infty}(\mathbb R^2 \setminus \{0\})$ and if

$1\ge c \ge \frac{1}{4},$ then $H$ is not essentially self-adjoint on $C_c^{\infty}(\mathbb R^2 \setminus \{0\}).$

The case $c \in (0,\frac{1}{4})$ seems to be missing, but I think one quickly sees by analysing the argument that it is the same case as the last one above (not essentially self-adjoint).

In a separate question Iosif Pinelis has shown that the quadratic form generated by the potential is not relatively form bounded with respect to the Dirichlet form.

The reason I asked this question was that I suspected that $H$ was self-adjoint on the Dirichlet domain of $-\Delta$, i.e. $H_0^1(\mathbb R^2\setminus \{0\}) \cap H^2(\mathbb R^2\setminus \{0\})$, but I am now not so sure about my conjecture anymore.

Consider the Schroedinger operator

$$H = -\Delta + \frac{c}{\vert x \vert^2}$$ in two dimensions with $c >0$

This operator has a self-adjoint realization, since it is a positive symmetric operator on $C_c^{\infty}( \mathbb R^2 \setminus \{0\}).$ I would like to understand three things:

0.) What are possible domains $D(H)$ on which $H$ is self-adjoint?

1.) Does there exist a domain $D(H)$ of $H$ on which $-\Delta$ and $H$ are both self-adjoint.

2.) On these domains $D(H)$, can we expect $$\Vert \frac{f}{\vert x \vert^2}\Vert_{L^2} \le \alpha_1 \Vert f \Vert_{L^2} + \alpha_2 \Vert \Delta f \Vert_{L^2} \text{ for some }\alpha_i \ge 0?$$

I would appreciate insights on any of these questions. My main interest is in small $c>0.$

One way to think about it may be to follow Example 4 in Reed-Simon II on page 160.

There the authors prove, going via a 1D reduction, that if

$c \ge 1,$ then $H$ is essentially self-adjoint (there is a unique self-adjoint extension) on $C_c^{\infty}(\mathbb R^2 \setminus \{0\})$ and if

$c \in [ \frac{1}{4},1),$ then $H$ is not essentially self-adjoint on $C_c^{\infty}(\mathbb R^2 \setminus \{0\}).$

The case $c \in (0,\frac{1}{4})$ seems to be missing, but I think one quickly sees by analysing the argument that it is the same case as the last one above (not essentially self-adjoint).

In a separate question Iosif Pinelis has shown that the quadratic form generated by the potential is not relatively form bounded with respect to the Dirichlet form.

The reason I asked this question was that I suspected that $H$ was self-adjoint on the Dirichlet domain of $-\Delta$, i.e. $H_0^1(\mathbb R^2\setminus \{0\}) \cap H^2(\mathbb R^2\setminus \{0\})$, but I am now not so sure about my conjecture anymore.

Consider the Schroedinger operator

$$H = -\Delta + \frac{c}{\vert x \vert^2}$$ in two dimensions with $c >0$

This operator has a self-adjoint realization, since it is a positive symmetric operator on $C_c^{\infty}( \mathbb R^2 \setminus \{0\}).$ I would like to understand three things:

0.) What are possible domains $D(H)$ on which $H$ is self-adjoint?

1.) Does there exist a domain $D(H)$ of $H$ on which $-\Delta$ and $H$ are both self-adjoint.

2.) On these domains $D(H)$, can we expect $$\Vert \frac{f}{\vert x \vert^2}\Vert_{L^2} \le \alpha_1 \Vert f \Vert_{L^2} + \alpha_2 \Vert \Delta f \Vert_{L^2} \text{ for some }\alpha_i \ge 0?$$

I would appreciate insights on any of these questions.

One way to think about it may be to follow Example 4 in Reed-Simon II on page 160.

There the authors prove, going via a 1D reduction, that if

$c \ge 1,$ then $H$ is essentially self-adjoint (there is a unique self-adjoint extension) on $C_c^{\infty}(\mathbb R^2 \setminus \{0\})$ and if

$1\ge c \ge \frac{1}{4},$ then $H$ is not essentially self-adjoint on $C_c^{\infty}(\mathbb R^2 \setminus \{0\}).$

The case $c \in (0,\frac{1}{4})$ seems to be missing, but I think one quickly sees by analysing the argument that it is the same case as the last one above (not essentially self-adjoint).

In a separate question Iosif Pinelis has shown that the quadratic form generated by the potential is not relatively form bounded with respect to the Dirichlet form.

Consider the Schroedinger operator

$$H = -\Delta + \frac{c}{\vert x \vert^2}$$ in two dimensions with $c >0$

This operator has a self-adjoint realization, since it is a positive symmetric operator on $C_c^{\infty}( \mathbb R^2 \setminus \{0\}).$ I would like to understand three things:

0.) What are possible domains $D(H)$ on which $H$ is self-adjoint?

1.) Does there exist a domain $D(H)$ of $H$ on which $-\Delta$ and $H$ are both self-adjoint.

2.) On these domains $D(H)$, can we expect $$\Vert \frac{f}{\vert x \vert^2}\Vert_{L^2} \le \alpha_1 \Vert f \Vert_{L^2} + \alpha_2 \Vert \Delta f \Vert_{L^2} \text{ for some }\alpha_i \ge 0?$$

I would appreciate insights on any of these questions. My main interest is in small $c>0.$

One way to think about it may be to follow Example 4 in Reed-Simon II on page 160.

There the authors prove, going via a 1D reduction, that if

$c \ge 1,$ then $H$ is essentially self-adjoint (there is a unique self-adjoint extension) on $C_c^{\infty}(\mathbb R^2 \setminus \{0\})$ and if

$1\ge c \ge \frac{1}{4},$ then $H$ is not essentially self-adjoint on $C_c^{\infty}(\mathbb R^2 \setminus \{0\}).$

The case $c \in (0,\frac{1}{4})$ seems to be missing, but I think one quickly sees by analysing the argument that it is the same case as the last one above (not essentially self-adjoint).

In a separate question Iosif Pinelis has shown that the quadratic form generated by the potential is not relatively form bounded with respect to the Dirichlet form.

The reason I asked this question was that I suspected that $H$ was self-adjoint on the Dirichlet domain of $-\Delta$, i.e. $H_0^1(\mathbb R^2\setminus \{0\}) \cap H^2(\mathbb R^2\setminus \{0\})$, but I am now not so sure about my conjecture anymore.