1
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Qmechanic Jan 30 '15 at 16:55
  • $\begingroup$ @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) $\endgroup$ – Xin Wang Jan 30 '15 at 16:59
  • $\begingroup$ 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. $\endgroup$ – Phoenix87 Jan 30 '15 at 17:03
  • $\begingroup$ @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 $\endgroup$ – Xin Wang Jan 30 '15 at 17:11
  • $\begingroup$ see arXiv:1008.4680 for a short overview and arXiv:0810.5643 for the full package $\endgroup$ – Christoph Jan 30 '15 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.