In the case of the Hermitian system, there are only real eigenvalues. We find the ground state by finding the minimum eigenvalue of the Hermitian matrix. However, for a non-Hermitian case, we get real as well as imaginary eigenvalues. In that case, how one can find the ground state as there are real and imaginary eigenvalues?
For any physically realizable system, the Hamiltonian should be Hermitian, and hence give real Energy eigenvalues.
This can be taken as a statement to as why Hamiltonian should always be Hermitian and not otherwise.
A system with loss and gain can be described by PT-symmetric non-Hermitian Hamiltonian. See the reference:
Article Making Sense of Non-Hermitian Hamiltonians
For non-hermitian matrix you can work with the singular values instead, take M×M^H or M^H×M, where M^H is the hermitian conjugate of the matrix M.
The spectral decomposition provides an ordered eigenspectrum for eigenvalues of a hermitian matrix, since they are all real,similar thing happens with singular value decomposition because the singular values are real and non negative for an arbitrary complex matrix, if you use any of the above two matrices in the Rayleigh Quotient form instead of M itself, this form is bounded both below and above for a finite complex matrix M, square or rectangular.
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