Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.