Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The property of hermitian is the sufficient condition for eigenvalue being real. Is there any non-hermitian operator on Hilbert Space with all real eigenvalues? If there exist, then can all eigenstates be orthogonal to each other? And these operators have any application in Quantum mechanics?

share|improve this question

1 Answer 1

It is indeed possible to have a non-hermitian operator with all real eigenvalues. However, in that case at least two of its eigenstates must be non-orthogonal: having real eigenvalues and orthogonal eigenstates us sufficient for hermiticity.

For an example of the former, try e.g. $$ \begin{pmatrix} 2 & - 1 \\ 0 & 1 \end{pmatrix}, $$ whose second eigenvector is $(1,1)$.

share|improve this answer

Your Answer

 
discard

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.