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?


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.

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  • $\begingroup$ 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? $\endgroup$ – Artemisia Dec 4 '14 at 19:48
  • $\begingroup$ 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. $\endgroup$ – lionelbrits Dec 4 '14 at 19:52
  • $\begingroup$ 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. $\endgroup$ – Artemisia Dec 4 '14 at 19:54
  • $\begingroup$ In QM, for all practical purposes, it goes both ways, although sometimes an operator is said to be self adjoint instead. $\endgroup$ – lionelbrits Dec 4 '14 at 20:19
  • $\begingroup$ Oh ok. So the eigenvalue equation isn't really relevant? $\endgroup$ – Artemisia Dec 4 '14 at 20:42

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