Does an observable have to be represented by a self-adjoint operator?

[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)

 

[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.

 

[Bill_K]

¿ Is it the same self-adjoint operator that hermitian operator If it is not the same, what is the difference?

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.

 

[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?

 

[dextercioby]

That's right.

 

[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.

 

[Ravi Mohan]

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

No.

If it is not the same, what is the difference?

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.

And an observable is an operator whose eigenvectors form basis in the Hilbert space, and it is hermitian, or self-adjoint?

After studying it you will see that observable must be represented by self-adjoint operators.