Atom - light field coupling and emission process

Suppose a “2-state atom” and a light field are quantized with the following Hamiltonians, respectively: $$hat{H}_A=hbaromega_{21}hat{sigma}^{dagger}hat{sigma}$$ and $$hat{H}_R=sum_{textbf{k}}hbaromega_{textbf{k}}(hat{a}^{dagger}_{textbf{k}}hat{a}_{textbf{k}} + frac{1}{2}) .$$ Where $hat{sigma}^{dagger}=left|2rightrangleleftlangle1right|$ and $hat{sigma}=left|1rightrangleleftlangle2right|$, where $left|1rightrangle,left|2rightrangle$ are the 2 states of the atom and $omega_{21}$ is for the transition from state 1 to state 2. $textbf{k}$ are the modes of the light field, and $hat{a}^{dagger}_{textbf{k}}, hat{a}_{textbf{k}}$ are the usual creation and annihilation operators.

If the interaction of the atom and the light field is modeled using a dipole moment with contribution to the total Hamiltonian of: $$hat{H}_I=sum_{textbf{k}}hbar g_{textbf{k}}(hat{a}^{dagger}_{textbf{k}} + hat{a}_{textbf{k}})(hat{sigma}^{dagger}+hat{sigma}) .$$

The interaction Hamiltonian $hat{H}_I$ shows that all the modes of the light field couple to the atom. What does that mean exactly in the case of an emission process, where the atom goes from $|2ranglerightarrow|1rangle$? In particular, are several modes of the field populated with photons at the various frequencies? Or is only a single mode populated with exactly one photon? How should I understand that a “photon is emitted” in the process, when all the light field modes couple to the atom?