1. Jun 17, 2013

StarsRuler

¿ Is it the same self-adjoint operator that hermitian operator

If it is not the same, what is the difference? And an observable is an operator whose eigenvectors form basis in the Hilbert space, and it is hermitian, or self-adjoint?

I always considered both terms like sinonynms, in the textbook use both terms, but with the same definition, hermitian and self-adjoint ( the last term is obvious) : it is an operator that it is the same that his adjoint (transpose conjugate)

2. Jun 17, 2013

dextercioby

The difference among the 2 terms is given by the difference between the books which use hand-waving mathematics versus (instead of) real functional analysis. I would advice for the use of <self-adjoint> in all possible (suitable) contexts.

3. Jun 17, 2013

Bill_K

StarsRuler, I believe that some books use the two terms to disitnguish between the adjoint of an operator in Hilbert space and the matrix adjoint of a Dirac matrix.

4. Jun 17, 2013

StarsRuler

Ok. Then an observable in QM is represented by an self-adjoint operator which eigenvectors form basis in Hilbert Space ( therefore if we use function representation or bra and kets representation), is not it?

5. Jun 17, 2013

dextercioby

That's right.

6. Jun 17, 2013

DrDu

At least in mathematical physics, a Hermitian or synonymously symmetric mean that the operator and it's adjoint have the same operational form (i.e. d/^2dx^2). However, for a symmetric operator to be self-adjoint, the (dense) domains of the two operators have to be the same. The later condition is non-trivial for unbounded operators which can't be defined on all the Hilbert space.

7. Jun 17, 2013

Ravi Mohan

No.
https://www.physicsforums.com/showpost.php?p=4401816&postcount=13
The difference is given on page 13, however you can read whole paper. It is interesting.
After studying it you will see that observable must be represented by self-adjoint operators.