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When we think about non-hermitian systems, the physical meaning of the eigenvalue of $H$ seems hard to be understood directly as the energy of the system.

In other words, if we experimentally measure a non-hermitian system, can we still measure energy levels? Is the energy level at this time related to the real part of the eigenvalue of $H$?

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  • $\begingroup$ Generally, for quantum system, this means that time reversal invariance does not hold. Anyway, there are some non-hermitian Hamiltonian admitting real eigenvalues and this is an active area of research. $\endgroup$ – Jon Nov 21 '19 at 15:02
  • $\begingroup$ @Jon Give me one example such kind hamiltonian. $\endgroup$ – baponkar Nov 21 '19 at 16:58
  • $\begingroup$ Check e.g. arxiv.org/abs/1203.6590 but a search by Carl Bender on arxiv will give you a lot of results about this matter. These are the so called ${\cal PT}$ invariant Hamiltonians. $\endgroup$ – Jon Nov 21 '19 at 17:24
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If a Hamiltonian is hermitian then its eigenvalues are real (and we deduce they are observable) and its eigenstates are complete (form a basis in the relevant Hilbert space).

In general, a non-hermitian Hamiltonian will not have real eigenvalues and as such we would not propose being able to observe / measure a "complex energy." However, as Jon points out, there are some non-Hermitian Hamiltonians that have real eigenvalues in which case there is nothing to stop us from interpreting the eigenvalues as measurable energies. On the other hand, time evolution would be non-unitary, which is usually taken as a desirable property of quantum mechanics.

However, the spectral theorem that informs us that the eigenstates of self-adjoint operators are complete would no longer apply, so it's not clear to me that such Hamiltonians are "useful" -- how would you decompose an arbitrary state into a linear combination (superposition) of eigenstates of the Hamiltonian? In any case, it is an ongoing line of research; in my institute we had two talks by external visitors on this work in the past 18 months....

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  • $\begingroup$ Thank you very much for your answer. What I'm most curious about is, in a dissipated system, the dissipated part is reflected in the imaginary part of the energy, but can the dissipated system also have PT symmetry? $\endgroup$ – xingsh xu Jan 6 at 7:57
  • $\begingroup$ That's a good question, but I think it should be submitted as a separate question here - it wasn't mentioned in your initial post! One thing I will say is that the probability current (and even the normalisation of states) may not be conserved with an imaginary part. If you do post a new question please don't forget to accept my answer here before moving on, assuming you're happy with it :-) $\endgroup$ – lux Jan 6 at 14:29

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