What happens to atoms in far detuned states?

[Malamala]

Hello! Assuming we have a 2 level system interacting with an EM field in the RWA and dipole approximation, we would have in the basis of the 2 unperturbed atomic states a 2x2 Hamiltonian matrix with off-diagonal terms. By diagonalizing this Hamiltonian we obtain dressed states which are the eigenstates of the atom-EM field interacting system and are shifted by a given amount relative to the unperturbed levels. Mathematically it makes sense and it is straightforward, but I am not sure what is happening physically, especially in the case of far detuning. On resonance for example, the atom undergoes Rabi oscillations and there is a clear interaction with the EM field, but in the far detuned region, how does the atom knows about the existence of the EM? Is it simply because there is a very small probability (small Rabi frequency) of Rabi oscillations due to the fact that the transition between the 2 levels has a line width and even if the laser frequency is far detuned it will still excite the transition once in a while? Or is there some other mechanism by which the atom feels the EM field?

 

[Twigg]

The atom feels the EM field through the electric dipole interaction (the coupling of the electric dipole moment of the atom to the electric field), whether it's near detuned or far detuned. Now there are two kinds of "far detuned", there's $\omega - \omega_0 \gg \Gamma$ and there's $\omega - \omega_0$ being on the same order as $\omega_0$. In the former case, the dressed state picture is still valid, and in the latter case the whole RWA fails and there be monsters. Which do you mean when you say "far detuned"?