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