Non-hermitian Hamiltonian $H^1(t)=H^1e^{−i\omega t}$ on Wikipedia page for Rabi problem

Question

I read about the Rabi problem at Wikipedia, but there is something I don't understand. They introduce a periodic perturbation $$H^1(t)=H^1e^{−i\omega t}.$$ But isn't this hamiltonian non-hermitian? Specifically, the hermitian adjoint would look like $$H^1(t)^{\dagger}=(H^1)^{\dagger} e^{i\omega t}.$$ This is different from $H^1(t)$ whenever $\omega\neq0$.

Answer 1

This is only one half of a physical Hamiltonian. The correct Hamiltonian should be $$ H_1(e^{-i\omega t}+e^{i\omega t}) $$ but the treatment for periodic perturbation has an integration where the only significant contribution is determined by the energy conservation condition $E_i-E_f=\pm \hbar \omega$. In other words, once you do away with the integration over time, only one of $\pm \omega$ leads to a significant contribution, while the other is neglected.

The advantage of using the truncated form with one frequency is to simplify the algebra; the truncated form is very convenient to derive Fermi's Golden rule for instance. The cost (as you realized) is some confusion as to the hermiticity of $H$.