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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – CuriousOne
    Feb 4, 2016 at 19:07
  • $\begingroup$ @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. $\endgroup$
    – Frank
    Feb 4, 2016 at 19:24
  • $\begingroup$ 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. $\endgroup$
    – CuriousOne
    Feb 4, 2016 at 19:31
  • 2
    $\begingroup$ 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. $\endgroup$ Feb 4, 2016 at 19:48
  • 2
    $\begingroup$ Does this answer your question? Does spontaneous emission actually emit in a random direction, or is it measured in a random direction? $\endgroup$
    – Roger V.
    Oct 7, 2021 at 9:09

1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.