How to understand operators representing observables are Hermitian?

[George Jones]

For the momentum operator a particle in a 1-dimensional infinite well, see "Self-adjoint extensions of operators and the teaching of quantum mechanics",

http://arxiv.org/abs/quant-ph/0103153

 

[R136a1]

Yes.

No, I disagree. You have as definition that hermitian immediately implies that it's bounded. I don't think this is a very common definition. All definitions I've seen is that symmetric and hermitian are equivalent, and thus hermitian is not necessarily bounded. See for example Reed & Simon, page 255. So I would say that the OP has it exactly right.

 

[George Jones]

No, I disagree. You have as definition that hermitian immediately implies that it's bounded. I don't think this is a very common definition. All definitions I've seen is that symmetric and hermitian are equivalent, and thus hermitian is not necessarily bounded. See for example Reed & Simon, page 255. So I would say that the OP has it exactly right.

Well, I have (at least) a couple of books that only use "Hermitian" with respect to bounded operators:

1) Introductory Functional Analysis by Kreyszig (the text for a course I took as a student);

2)Hilbert Space Operators in Quantum Physics by Blank, Exner, and Havilcek.

Both of these references define symmetric unbounded operators, but don not associate the term "Hermitian" with such operators.

Some references (e.g., Riesz and Nagy) do not use "Hermitian" at all. In the recent book "Quantum Theory for Mathematicians" by Hall, "Hermitian" is used only in the sentence "Physicists refer to self-adjoint operators as Hermitian." Hall does not use "Hermitian" (but does use symmetric and sel-adjoint) when doing actual functional analysis.

 

[dextercioby]

That's my mind, too. <Hermitean> is not modern terminology in functional analysis. It belongs now only in linear algebra which is done in finite dimensional spaces, of course. It stands in front of the words matrix+matrices.