Hamiltonian-commutation, hermiticity and non-hermiticity (QM)

Question

When we have a QM system in an energy eigenstate (say after a measurement of energy) then we can measure any time another quantity that is described by an hermitian operator that commutes with the Hamiltonian and expect to get a precisely predictable result, namely an eigenvalue. Now when I measure a quantity that is associated to an operator that does not commute with the one from the last measurement than the result will be "unpredictable" but one can exactly determine the specific expectation value of the quantity, OK.

What about non-hermitian "operators" like when I have a perturbation of the system e.g. an external magnetic field) such that say no component of the angular momentum (nor its square) is conserved "in the state". That means $\hat{l}_z$ is a non-hermitian operator (question 1: Is that correct?) but when I nevertheless compute $\langle\psi|\hat{l}_z|\psi\rangle$ I think I would get a complex value as an "expectation value" (question 2: Is that correct?).

Question 3: Is there any sensical interpretation to the real and the imaginary component of that?

At least the size of the imaginary component would indicate the deviation from hermiticity I suppose.

As a side note:

In the context of some paper (that I cannot find right now again) on $\mathcal{P,T}$ reversal symmetry I came about the statement that the imaginary component would indicate a "flow" of the quantity under regard into and out of the system and in case of $\mathcal{P,T}$ reversal symmetry they are just balanced, such that we have purely real eigenvalues but can have non-hermitian operators. Maybe someone can also comment on that.

Answer 1

In your side note I think you may be referring to this paper. (1)

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Question 1:

What about non-hermitian "operators" like when I have a perturbation of the system e.g. an external magnetic field) such that say no component of the angular momentum (nor its square) is conserved "in the state". That means $\hat{L}_z$ is a non-hermitian operator.

I'm not familiar with the situation you're referring to, but this is not in general correct. Even in a system where $\hat{L}_z$ is not conserved, that doesn't mean that $\hat{L}_z$ is not hermitian.

Question 2: Yes, that's correct. If $\hat{O}$ is a non-hermitian operator, there is no reason for $\langle \hat{O} \rangle$ to in general be real.

Question 3: It appears that, yes, that there is a way of making sense of non-hermitian hamiltonians under additional assumptions (cf. 1). As you pointed out in your side note this requires additional assumptions. Namely,

$$[\hat{H}, \hat{P}] = [\hat{H},\hat{T}] =0 $$

and that the eigenstates of $\hat{H}$ be $\hat{P}\hat{T}$ invariant. The exact details of this are outlines in (1), though it should be known that this is still an ongoing field of research.