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Is it necessary for an operator to be Hermitian in order to be a physical observable or is it just sufficient that the operator obeys the eigenvalue equation? If I were to check whether an operator is a physical observable, must I also check for Hermiticity?

Qmechanic
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Artemisia
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  • Hermiticity is necessary for the numerical range of an operator to be a subset of the reals. It means that it is a necessary condition to guarantee that the expectation value of the observable would yield a real number in any quantum state. – yuggib Dec 04 '14 at 19:48
  • Related: http://physics.stackexchange.com/q/39602/2451 , http://physics.stackexchange.com/q/27038/2451 , http://physics.stackexchange.com/q/75401/2451 , http://physics.stackexchange.com/q/82613/2451 , http://physics.stackexchange.com/q/87551/2451 and links therein. – Qmechanic Dec 04 '14 at 19:53
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    note that in the infinite-dimensional case, observables need to be self-adjoint, not necessarily Hermitian (eg position and momentum operators of the free particle aren't Hermitian); operators with continous or unbounded spectrum are a bit more complicated than most introductory courses on quantum mechanics suggest – Christoph Dec 05 '14 at 08:35

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Hermitian observables yield real eigenvalues. Also, for an observable to be a good observable, it must be diagonal in some basis, in which case it has real diagonal elements. So your observable is related to a real, diagonal matrix by a unitary transformation. That pretty much restricts what you can have. This isn't axiomatic, of course. It's more from a practical point of view.

lionelbrits
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  • Oh ok. But if I have an operator that yields an eigenvalue when I insert it into the eigenvalue equation, isn't it a physical observable? Not all Hermitian operators work, do they? – Artemisia Dec 04 '14 at 19:48
  • Well, axiomatically, all Hermitian operators correspond to observables, because they have real eigenvalues. But whether or not you can practically measure them is a different story. – lionelbrits Dec 04 '14 at 19:52
  • Hmm alright. What I meant is that it's not an if and only if situation, is it? As in a Hermitian operator corresponds to an observable but it doesn't go the other way-so the test for Hermiticity is insufficient. – Artemisia Dec 04 '14 at 19:54
  • In QM, for all practical purposes, it goes both ways, although sometimes an operator is said to be self adjoint instead. – lionelbrits Dec 04 '14 at 20:19
  • Oh ok. So the eigenvalue equation isn't really relevant? – Artemisia Dec 04 '14 at 20:42
  • Well, the eigenvalue is what you measure. You couldn't measure, say, an observable quantity corresponding to a a matrix with random complex elements. You measure real numbers. – lionelbrits Dec 04 '14 at 20:56