# Non Hermitian Quantum Mechanics

I was just reading about Non-Hermitian Quantum Mechanics dealing with Hamiltonians $H$ that are not Hermitian operators.

Then it is unclear that we get orthonormal eigenstates. Now, I was reading a paper where they suggested such a Hamiltonian for their model and used a so-called C-product to describe expectation values and so on. The idea was that they showed that their Hamiltonian $H$ ($n \times n$) matrix is diagonalisable as there are $n-$ distinct eigenvalues. Thus, we get $n$ eigenvectors $|i \rangle$ and another set of $n$ eigenvectors to the adjoint matrix $H^{\dagger}$ that I want to call $|i*\rangle$. Now these eigenvectors satisfy $\langle j* | i \rangle = \delta_{ji}$ and then they called $\langle j* |H|i \rangle$ the C-product which defines in this setting expectation values and so on. My question would be now: Is this a meainingful definition or are there some problems with this (on a fundamental level)?

• Comments to the post (v1): 1. Which paper? 2. Rather than asking about general non-Hermitian QM, are you really just asking about PT symmetric QM? – Qmechanic Jan 30 '15 at 16:55
• @Qmechanic if you could give me a clue what I need to check to find out if a matrix describes a PT symmetric Hamiltonian, I could maybe answer your quesiton. (They were dealing with an upper triangular matrix with real entries on the diagonal, which ensured the existence of real eigenvalues) – Xin Wang Jan 30 '15 at 16:59
• my feeling is that $|i*\rangle$ is just the dual basis of $|i\rangle$? I presume it is then used to somehow make up for the fact that the basis $|i\rangle$ is not orthonormal in general. – Phoenix87 Jan 30 '15 at 17:03
• @Phoenix87 that is exactl it, but my question is: Is this also meaningful from a physical perspective or are there some problems with this definition – Xin Wang Jan 30 '15 at 17:11
• see arXiv:1008.4680 for a short overview and arXiv:0810.5643 for the full package – Christoph Jan 30 '15 at 17:49