1. Sep 25, 2008

Heirot

Hello,

what's the difference between Hermitian and self-adjoint operators? Our professor in Group Theory made a comment once that the two are very similar, but with a subtle distinction (which, of course, he failed to mention )

Thanks!

2. Sep 25, 2008

Heirot

Oops... I found the answer. Sorry

3. Sep 25, 2008

morphism

Well, what's this subtle distinction?

4. Sep 25, 2008

Heirot

5. Sep 25, 2008

morphism

Just make sure you're aware that it's a matter of convention. Most people use the two words to mean the same thing.

6. Sep 25, 2008

George Jones

Staff Emeritus
While this is true for bounded operators, all the books I have checked on my shelf make a distinction for unbounded operators. According to the books that I have checked, a Hermitian operator is a bounded self-adjoint operator, i.e., all Hermitian operators are self-adjoint, but not all self-adjoint operators are Hermitian.

7. Sep 25, 2008

atyy

Is this an example of the difference you are talking about, or is it yet another difference?
http://mathworld.wolfram.com/BerryConjecture.html:"...physicists define this operator to be Hermitian, mathematicians do not."

8. Sep 26, 2008

George Jones

Staff Emeritus
Let $A$ be a densely defined linear operator on a Hilbert space. One of the defining properties, *, for all cases is

$$\left<Ax,y\right> = \left<x,Ay\right>, \ *$$

where $x$ and $y$ are elements of the Hilbert space.

The text that was used for the functional analysis course that I took as a student makes the following definitions:

$A$ is Hermitian if $A$ is bounded and * is true for every $x$ and $y$ in the Hilbert space;

$A$ is symmetric if * holds for for every $x$ and $y$ in the domain of $A$;

$A$ is self-adjoint if $A$ is symmetric and the domain of $A$ equals the domain of $A^\dagger$.

According to these defintions, every Hermitian operator is and self-adjoint, but not all self-adjoint operators are Hemitian. Some books leave off the first definition and call symmetric operators Hermitian. Then, every self-adjoint operator is Hermitian, but not all Hermitian operators are self-adjoint.
A symmetric (or Hemitian, depending on the terminology used) operator is self-adjoint iff it has trivial deficiency indices. Some physicists and physics references do not distinguish between the above defintions, and use the terms Hermitian and self-adjoint interchangeably to refer to all cases.

If $A$ is unbounded, then the Hilbert space is necessarily infinite-dimensional, and the domain of $A$ need not be all of the Hilbert space. If this is the case and if $A$ is symmetric, then the domain of $A$ is a subset of the domain of $A^\dagger$, which I have not defined.

In physics, the canonical commutation relation is important. If self-adjoint operators $A$ and $B$ satisfy such a relation, then it is easy to show that at least one of the operators must be unbounded. Suppose it is $A$. The Hellinger-Toeplitz theorem states any linear operator that satisfies * for all elements of the Hilbert space must bounded. Since $A$ is self-adjoint and unbounded, the domain of physical observable $A$ cannot be all of Hilbert space!

For one consequence of these concepts, see

Last edited: Dec 17, 2013
9. Sep 26, 2008

morphism

That's a very interesting thread, George. Thanks for sharing!

By the way, in your references, are unbounded operators ever referred to as (unqualified) operators?

10. Sep 27, 2008

atyy

11. Dec 16, 2008