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I am looking for a proof or some interpretation around why the domain of the new extension $D(A_U)$ in the Theorem below is given by its specific formula.

I have already searched in papers and here but I could find only applications of the Theorem. Does anybody know a reference with it or can give me an idea of the behavior of extension domain $D(A_U)$?

Theorem: Let $A$ be a Hermitian, densely defined, and closed operator, $A^\dagger$ its conjugate transpose, $K_\pm = \{ \psi_\pm \in D(A^\dagger) : A^\dagger \psi_\pm = \pm i\eta\psi_\pm \}$ and $n_\pm = dim(K_\pm)$. Then:

1) if $n_+ = n_- = 0$, A is self-adjoint

2) if $n_+ \neq n_-$, A is not self-adjoint and it has no self-adjoint extensions

3) if $n_+ = n_- > 0$, A has infinitely many self-adjoint extensions $A_U$ parametrized by $n \times n$ unitary matrix maps $U: K_+ \rightarrow K_-$. For a choice of $U$, $D(A) \subset D(A_U) \subset D(A^\dagger)$, and $D(A_U) = \{\psi = \phi + \psi_+ + U\psi_+ : \phi \in D(A), \psi_+ \in K_+ \}$

I am looking for proof or some interpretation around why the domain of the new extension $D(A_U)$ in the Theorem below is given by its specific formula.

I have already searched in papers and here but I could find only applications of the Theorem. Does anybody know a reference with it or can give me an idea of the behavior of extension domain $D(A_U)$?

Theorem: Let $A$ be a Hermitian, densely defined, and closed operator, $A^\dagger$ its conjugate transpose, $K_\pm = \{ \psi_\pm \in D(A^\dagger) : A^\dagger \psi_\pm = \pm i\eta\psi_\pm \}$ and $n_\pm = \dim (K_\pm)$. Then:

1) if $n_+ = n_- = 0$, A is self-adjoint;

2) if $n_+ \neq n_-$, A is not self-adjoint and it has no self-adjoint extensions;

3) if $n_+ = n_- > 0$, A has infinitely many self-adjoint extensions $A_U$ parametrized by $n \times n$ unitary matrix maps $U: K_+ \rightarrow K_-$. For a choice of $U$, $D(A) \subset D(A_U) \subset D(A^\dagger)$, and $D(A_U) = \{\psi = \phi + \psi_+ + U\psi_+ : \phi \in D(A), \psi_+ \in K_+ \}$.

Theorem: Let $A$ be a Hermitian, densely defined, and closed operator, $A^\dagger$ its conjugate transpose, $K_\pm = \{ \psi_\pm \in D(A^\dagger) : A^\dagger \psi_\pm = \pm i\eta\psi_\pm \}$ and $n_\pm = dim(K_\pm)$. Then:

1) if $n_+ = n_- = 0$, A is self-adjoint

2) if $n_+ \neq n_-$, A is not self-adjoint and it has no self-adjoint extensions

3) if $n_+ = n_- > 0$, A has infinitely many self-adjoint extensions $A_U$ parametrized by $n \times n$ unitary matrix maps $U: K_+ \rightarrow K_-$. For a choice of $U$, $D(A) \subset D(A_U) \subset D(A^\dagger)$, and $D(A_U) = \{\psi = \phi + \psi_+ + U\psi_+ : \phi \in D(A), \psi_+ \in K_+ \}$

I am looking for a proof or some interpretation around why the domain of the new extension $D(A_U)$ is given by the formula above.

I have already searched in google and here in stack exchange but with no success. Does anybody know a reference/book with it or can give me an idea of the behavior of the domain of the extension $D(A_U)$ ?

I am looking for a proof or some interpretation around why the domain of the new extension $D(A_U)$ in the Theorem below is given by its specific formula.

I have already searched in papers and here but I could find only applications of the Theorem. Does anybody know a reference with it or can give me an idea of the behavior of extension domain $D(A_U)$?

Theorem: Let $A$ be a Hermitian, densely defined, and closed operator, $A^\dagger$ its conjugate transpose, $K_\pm = \{ \psi_\pm \in D(A^\dagger) : A^\dagger \psi_\pm = \pm i\eta\psi_\pm \}$ and $n_\pm = dim(K_\pm)$. Then:

1) if $n_+ = n_- = 0$, A is self-adjoint

2) if $n_+ \neq n_-$, A is not self-adjoint and it has no self-adjoint extensions

3) if $n_+ = n_- > 0$, A has infinitely many self-adjoint extensions $A_U$ parametrized by $n \times n$ unitary matrix maps $U: K_+ \rightarrow K_-$. For a choice of $U$, $D(A) \subset D(A_U) \subset D(A^\dagger)$, and $D(A_U) = \{\psi = \phi + \psi_+ + U\psi_+ : \phi \in D(A), \psi_+ \in K_+ \}$