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?

• What the light field is populated with depends on your choice of "environmental conditions". Are you putting your atom into a cavity with thermal radiation, are there electromagnetic fields, etc.. That's your choice. QFT takes the classical potentials away from you and replaces them with a population of states of the light field. If all you want is to describe the emission process, then, yes, start with an empty initial condition and end on a final with one photon. – CuriousOne Feb 4 '16 at 19:07
• @CuriousOne - how about we couple the atom with the vacuum state of each mode? That is, we would be looking at spontaneous emission. I am trying to understand what is emitted: a photon in each mode, or a single photon in a single mode? There has to be conservation of energy, for starters. – Frank Feb 4 '16 at 19:24
• There is a conservation of energy (and more importantly) angular momentum. You can only emit one photon per transition, but then, when you consider a thermal environment with sufficient temperature, you do get high frequencies of absorption, stimulated emission and emission processes. – CuriousOne Feb 4 '16 at 19:31
• Some comments: (i) There's not really a difference between "a single photon in a superposition mode" and "a superposition of states which have a single photon in different modes", by linearity. (ii) Single-photon states need not be eigenstates of the radiation hamiltonian, since this includes superpositions of single-photon states of modes with different frequency and therefore different energy. (iii) Energy does need to be conserved, but this only means that $H=H_A+H_R+H_I$ is conserved, which says much less about the dynamics than one would like to think. – Emilio Pisanty Feb 4 '16 at 19:48
• @Emilio - so would the emitted photon be representable as a wave packet, involving the frequencies of all the modes? Further, is that equivalent to a linear combination of Fock states, one for each mode, or am I off base here (I am a beginner in a quantum optics class). – Frank Feb 4 '16 at 20:05

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.