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So we know that in Quantum Mechanics we require the operators to be Hermitian, so that their eigenvalues are real ($\in \mathbb{R}$) because they correspond to observables.

What about a non-Hermitian operator which, among the others, also has real ($\mathbb{R}$) eigenvalues? Would they correspond to observables? If no, why not?

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Essentially a duplicate of this and this Phys.SE questions. –  Qmechanic Mar 29 '14 at 0:24
They don't say anything about whether or not a measurement of that quantity can be performed –  SuperCiocia Mar 29 '14 at 0:27
Yes they do. The answer given there says that there will in general be non-zero overlap between the eigenstates that are not orthogonal. Thus measuring an eigenvalue would not be a guarantee that the system is in the corresponding eigenstate. In the Copenhagen interpretation, the collapse of the wave function is no longer a well-defined procedure. So a non-Hermitian operator is not a well-defined observable. –  Qmechanic Mar 29 '14 at 0:34
Right I meant there is no yes/no answer in there, which is what I was looking for. You say that "measuring an eigenvalue would not be a guarantee that the system is in the corresponding eigenstate", but can that eigenvalue (real, but of a non-Hermitian matrix) be measured? –  SuperCiocia Mar 29 '14 at 0:40

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For Hermitian matrices eigenvectors corresponding to different eigenvalues are orthogonal. This guarantees that not only are the eigenvalues real, expectation values are too.

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so for a non-Hermitian operator the eigenvectors corresponding to the real eigenvalues are not orthogonal? I realise this means that one cannot know which eigenstate the wavefunction collapsed to, but do these eigenvalues correspond to observables? Can you measure them? –  SuperCiocia Mar 30 '14 at 15:39

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