Hermitian vs. self-adjoint operators

[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!

 

[Heirot]

Oops... I found the answer. Sorry

 

[morphism]

Well, what's this subtle distinction?

 

[Heirot]

https://www.physicsforums.com/showthread.php?t=202864

 

[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.

 

[George Jones]

Most people use the two words to mean the same thing.

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.

 

[atyy]

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.

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."

 

[George Jones]

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."

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.

Note that $A$ is symmetric but might have nontrivial deficiency indices, so while physicists define this operator to be Hermitian, mathematicians do not.

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

https://www.physicsforums.com/showthread.php?t=122063&highlight=dirac.

 

[morphism]

https://www.physicsforums.com/showthread.php?t=122063&highlight=dirac.

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?

 

[atyy]

For one consequence of these concepts, see

https://www.physicsforums.com/showthread.php?t=122063&highlight=dirac.

Thanks! That was a wickedly good puzzle (at least for those of us whose professors told them functional analysis is useless)

 

[Adam....]

I know, George you have to be the most interesting person i know, gosh gosh

"what are you going to do today napoleon?"
"whatever the frick i want to do gosh! gosh"