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

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Discussion Overview

The discussion centers around the relationship between self-adjoint operators and Hermitian operators in the context of quantum mechanics, specifically whether an observable must be represented by a self-adjoint operator. Participants explore definitions, distinctions, and implications of these terms within the framework of functional analysis and quantum theory.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question whether self-adjoint and Hermitian operators are synonymous, suggesting that definitions may vary between texts.
  • One participant emphasizes the importance of using the term "self-adjoint" in appropriate contexts, implying a distinction in mathematical rigor.
  • Another participant notes that some literature distinguishes between the adjoint of an operator in Hilbert space and the matrix adjoint of a Dirac matrix.
  • It is proposed that in quantum mechanics, an observable is represented by a self-adjoint operator whose eigenvectors form a basis in Hilbert space.
  • A later reply clarifies that for a symmetric operator to be self-adjoint, the domains of the operator and its adjoint must be the same, highlighting complexities with unbounded operators.
  • One participant asserts that observables must be represented by self-adjoint operators after reviewing relevant literature.

Areas of Agreement / Disagreement

Participants express differing views on the equivalence of self-adjoint and Hermitian operators, with no consensus reached on the definitions or implications of these terms. The discussion remains unresolved regarding the nuances of these concepts.

Contextual Notes

Limitations include potential ambiguities in definitions of self-adjoint and Hermitian operators, as well as the implications of these definitions for observables in quantum mechanics. The discussion also touches on the complexities introduced by unbounded operators.

StarsRuler
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¿ 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)
 
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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.
 
StarsRuler said:
¿ 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.
 
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?
 
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.
 
StarsRuler said:
¿ Is it the same self-adjoint operator that hermitian operator
No.
StarsRuler said:
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.
StarsRuler said:
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.
 
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