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?
[Physics] ny non-hermitian operator on Hilbert Space with all real eigenvalues
hilbert-spaceobservablesoperatorsquantum mechanics
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Best Answer
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 is sufficient for hermiticity.
For an example of the former, try e.g. $$ \begin{pmatrix} 2 & - 1 \\ 0 & 1 \end{pmatrix}, $$ whose second eigenvector is $(1,1)$.
Note, however, that in a case like this (non-hermitian diagonalizable operator with all real eigenvalues) you can always find a second inner product with respect to which the operator is hermitian, essentially by declaring its eigenbasis to be orthogonal, by decree, on the new inner product.