Atom - light field coupling and emission process Suppose a "2-state atom" and a light field are quantized with the following Hamiltonians, respectively: $$\hat{H}_A=\hbar\omega_{21}\hat{\sigma}^{\dagger}\hat{\sigma}$$ and $$\hat{H}_R=\sum_{\textbf{k}}\hbar\omega_{\textbf{k}}(\hat{a}^{\dagger}_{\textbf{k}}\hat{a}_{\textbf{k}} + \frac{1}{2})\ .$$ Where $\hat{\sigma}^{\dagger}=\left|2\right\rangle\left\langle1\right|$ and $\hat{\sigma}=\left|1\right\rangle\left\langle2\right|$, where $\left|1\right\rangle,\left|2\right\rangle$ 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 $|2\rangle\rightarrow|1\rangle$? 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?
 A: In the process of spontaneous decay, a single photon is eventually emitted from the atom (assuming the atom is initially in the pure state $\left|\psi\right\rangle = \left|2\right\rangle$). If there are many modes that the photon can be emitted into (e.g. multiple values of $g_{\mathbf k}\ne 0$), then the state of the emitted photon will be in a quantum/coherent superposition of possible EM modes (with the weight in each mode determined by $g_{\mathbf k}\ne 0$). In fact, if you aren't sure a photon has been emitted yet, then you are actually in a quantum superposition of an excited atom and no photons, and an atom in the ground state plus a single photon in many EM modes.
If you don't measured which state the photon is emitted into (or when), then you have some uncertainty in the total state of the atom. This type of ignorance is the origin of quantum decoherence, and if you properly average over all possible EM states, will give you a mixed state representation for the state of the atom (characterized by non-zero entropy).
