Let $A: H\to H$ be an unbounded operator on a Hilbert space $H$. An unbounded operator $A^*$ is its adjoint if
An adjoint does not need to exist in general.
An unbounded operator is symmetric if $dom(A)\subset dom(A^*)$ and $A x = A^*x$ for all $x\in dom(A)$ (one also writes $A\subset A^*$).
The domain of $A^*$ is the set of all vectors $y\in H$ such that the linear functional $x\mapsto (Ax|y)$ is bounded on $dom(A)$.
The graph $\Gamma_A\subset H\oplus H$ satisfies $\Gamma_{A^*} = \tau(\Gamma_A)^\perp$ where $\perp$ denotes the orthogonal complement and $\tau$ denotes the transposition of the direct summands changing the sign of one of the factors, i.e. $x\oplus y\mapsto -y\oplus x$. An unbounded operator $A$ is closed if $\Gamma_A$ is closed subspace of $H\oplus H$. An operator $B$ is a closure of an operator $A$ if $\Gamma_B$ is a closure of operator $\Gamma_A$. It is said that $B$ is an extension of $A$ and one writes $B\supset A$ if $\Gamma_B\supset \Gamma_A$. The closure of an unbounded operator does not need to exist.
For any unbounded operator $A$ with a dense $dom(A)\subset H$, if the adjoint operator $A^*$ exists, then $A^*$ is closed, and if $(A^*)^*$ exists then it coincides with a closure of $A$.
An unbounded operator $A : H\to H$ on a Hilbert space $H$ is self-adjoint if
An (unbounded) operator is essentially self-adjoint if it is symmetric and its spectrum (as a subspace of the complex plane) is contained in the real line. Alternatively, it is symmetric if its closure is self-adjoint.
A Hermitean (or hermitian) operator is a bounded symmetric operator (which is necessarily self-adjoint), although some authors use the term for any self-adjoint operator.
For a bounded operator $A: H\to K$ between Hilbert spaces, define the Hermitean conjugate operator $A^*: K\to H$ by $(Ax|y)_H = (x|A^*y)_K$, for all $x\in K$, $y\in H$. Distinguish it from the concept of the transposed operator? $A^T: K^*\to H^*$ between the dual spaces.
In an arbitrary $*$-algebra, a self-adjoint or hermitian element is any element $A$ such that $A^* = A$.
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