Distinguishing between symmetric, Hermitian and self-adjoint operators

Solution 1:

These are not the usual definitions as I know them.$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

First, I am only familiar with the situation that $H$ is a Hilbert space and $D(T)$ is dense in $H$ (which entails no loss of generality, as we can replace $H$ with the completion of $D(T)$.)

I would say:

1. $T$ is symmetric if $\inner{Tx}{y} = \inner{x}{Ty}$ for all $x,y \in D(T)$. (Note your definition doesn't make sense, because you are applying $T$ to vectors that may not be in $D(T)$.)

2. $T$ is Hermitian if it is symmetric and bounded. (If $T$ is bounded then it has a unique bounded extension to all of $H$, so we may as well assume $D(T) = H$ in this case.) Since a symmetric operator is always closable, the closed graph theorem implies that a symmetric operator with $D(T) = H$ is automatically bounded.

3. $T$ is self-adjoint if the following, more complicated condition holds. Let $D(T^*)$ be the set of all $y \in H$ such that $|\inner{Tx}{y}| \le C_y ||x||$ for all $x \in D(T)$, where $C_y$ is some constant depending on $y$. If $T$ is symmetric, one can show that $D(T) \subset D(T^*)$; $T$ is said to be self-adjoint if it is symmetric and $D(T) = D(T^*)$.

With these definitions, we have Hermitian implies self-adjoint implies symmetric, but all converse implications are false.

The definition of self-adjoint is rather subtle and this may not be the place for an extended discussion. However, I'd recommend a textbook such as Reed and Simon Vol. I. Perhaps I'll just say that symmetric operators, although the definition is simple, turn out not to be good for much, per se. One needs at least self-adjointness to prove useful theorems.

Solution 2:

The definition is quite simple when you realize it. But it takes some time to realize the difference. There are some contradictions with Nate answer, but this just a matter of terminology.

• $\mathrm T$ is Hermitian if $\forall x,y \in D(\mathrm T) (\mathrm Tx,y) = (x,\mathrm T y)$
• $\mathrm T$ is symmetric if $\mathrm T$ is Hermitian and densely defined. As far as i understand the only advantage of symmetric op over Hermitian is guaranted exitance of $\mathrm T$'s closure.

• $\mathrm T$ is self-adjoint if $\mathrm T^* == \mathrm T$, where $\mathrm T^*$ defined as from following relation$\forall x \in D(\mathrm T) \exists y,z \in \mathbb H: (\mathrm Tx,y) = (x,z)$. The operator $\mathrm T^*: z = T^*y$ and is called adjoint.

  For finite-dimensional spaces all this definitions turn to be the same. Bounded symmetric operators are essentially self-adjoint (closure is self-adjoint).