I am permanently confused about the distinction between Hermitian and self-adjoint operators in an infinite-dimensional space. The preceding statement may even be ill-defined. My confusion is due to consulting Wikipedia, upon which action I have the following notion.

Let $H$ be a pre-Hilbert space equipped with an inner product ${\langle}.,.{\rangle}$ and $T:D(T){\subset}H{\longmapsto}H$ a linear operator. Then


If ${\langle}Tx,y{\rangle}$=${\langle}x,Ty{\rangle}$ for all $x,y{\in}D(T)$ then $T$ is symmetric.
If $T$ is symmetric and also bounded then it is Hermitian.
If $T$ is symmetric and $D(T)=H$ then $T$ is self-adjoint.


As a corollary, if the above is true then a symmetric and self-adjoint operator must be Hermitian since a symmetric operator defined on all of $H$ must be bounded. On the other hand, a Hermitian operator need not be self-adjoint: it would not be if its domain were a strict subset of $H$.

Would people agree with this?  I always see the second and third of these treated as equivalent, hence my confusion.

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:


$T$ 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 $D(T)$.)
$T$ 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.
$T$ 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.
