Relationship between a non-Hermitian Hamiltonian and its solution

[SeM]

Hello, I Have a non-Hermitian Hamiltonian, which is defined as an ill-condition numbered complex matrix, with non-orthogonal elements and linearily independent vectors spanning an open subspace.

However, when accurate initial conditions are given to the ODE of the Hamiltoanian, it appears to give an orthogonal solution. Does that make sense? I would have thought that a non-Hermitian Hamiltonian gives only non-orthogonal solutions?

Thanks

 

[DrDu]

I would have thought that a non-Hermitian Hamiltonian gives only non-orthogonal solutions?

Thanks

No, for example, normal operators are non-Hermitian. They have a complete set of orthogonal eigenvectors but the corresponding eigenvalues are complex and not real, in general.
In general, you can always do a singular value decomposition. It turns out that a matrix always has a set of orthogonal eigenvectors, but in general, these eigenvectors are not also eigenvectors of the transposed (or adjoint) matrix.

 

[SeM]

Thanks Dr. Du! In my case, I have a complex matrix, which is not unitary and not normal, and is very ill-conditioned. However, its ODE are solved using some initial conditions, and orthogonal solutions are found. However the orthogonal solutions have both complex as well as real expectation values in their integrals:

\begin{equation}
\int_0^{2\pi} \psi \Omega \psi^*
\end{equation}