Physical observables and hermiticity

Question

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?

Answer 1

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.