I find the formulation confusing, for the following reason: If the ``unperturbed'' Hamiltonian is not Hermitian, this means that, generically, its eigenvalues are complex. This, in turn, means that the ground state risks being unstable-that's what complex eigenvalues mean. Now you add the perturbation, that's, also, not Hermitian. I don't, immediately, see how a sensible perturbation expansion can be set up at all. A lot will depend on very special properties of the Hamiltonians.

In this case, I'd guess that a much better approach would be studying a reduced model, where explicit diagonalization (numerical, of course, included) could be attempted, in order to acquire some intuition.

@ Stam: The unperturbed Hamiltonian is not Hermitian but has real eigenvalues. The eigenvalues are 0 and -1 and so are not positive.

The issue that's relevant is that they're finite. Therefore the ground state has energy -1 and it's a two-level system. This *would* imply that the ground state is stable, were the Hamiltonian Hermitian, i.e. were the evolution operator unitary.

That the Hamiltonian is non-Hermitian, however, would mean that the evolution operator isn't unitary, therefore the computation of transition probabilities requires care. This is even before the perturbation is considered. For, if you write this non-Hermitian Hamiltonian as a sum of a Hermitian part and an anti-Hermitian part, then the contribution of the latter will be to destabilize the ground state, defined by the former.

Therefore perturbation theory risks having problems, since it's not obvious that a ground state could be defined about which to perturb. But this isn't an artifact of perturbation theory: the anti-Hermitian part of the full Hamiltonian will destabilize the ground state, defined by the Hermitian part. Therefore the system is sensitive to the details that define the anti-Hermitian part and asymptotic states will be sensitive to this prescription.

In dealing with non-Hermitian Hamiltonians, one has to deal with two sets of eigenstates, the left- and the right-eigenstates, which form a bi-orthogonal / bi-orthonormal set (the left-eigenstates are orthogonal / orthonormal on the right-eigenstates). The subject matter is briefly discussed in Vol. 1 of Morse & Feshbach (Chapter 7, Section 5). For perturbation expansion, the basic projectors are of the form |R_i> denotes the i-th right-eigenket and

Since the eigenvalues of the ``unperturbed'' Hamiltonian are only -1 and 0, then the Hamiltonian in question is a projector onto the subspace of eigenvalue -1 and its complement projects on the other subspace. So it suffices to study how the perturbation acts on the two subspaces. If it doesn't commute with the ``unperturbed'' Hamiltonian, then it will induce transitions between the two subspaces.

I wonder whether one should look at other mathematical properties of both perturbed and unperturbed operators. It could well be that the requirements of a meaningful time-dependent perturbation theory can only be met for operators in a Hilbert space, say, that have rather specific mathematical properties.

Some interesting points here may concern

   * the dimensionality of the Hilbert space

   * the nature of the spectrum of the unperturbed operator (discrete, continuous, real, essential, ...)

   * properties of the operators like bounded, unbounded, ... or normal, Fredholm, ...

Thus, it could help to reverse the question: If there _is_ a well-defined time-dependent perturbation theory, how does this restrict the operator.

Of course, one of the essential questions then is the exact meaning of "well-defined" here.

 Up to now, I did not come across a thorough mathematical investigation to answer the reverse question.

A neat way to set up problems like this is the following: look at the eigenvectors of the unperturbed Hamiltonian. Since the eigenvalues are real, if you now define a new scalar product such that these eigenvectors are orthogonal, the unperturbed Hamiltonian becomes Hermitian---at least at a formal level---in this scalar product. Thus time evolution will conserve the norm defined in the new scalar product. Thus you reduce the problem at least to a non-Hermitian perturbation of a Hermitian Hamiltonian, which is a  mild improvement. See the work by Bender and others on non Hermitian Hamiltonians