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Often I hear it described that spontaneous emission is when vacuum fluctuations "kick" an atom and induce stimulated emission on its own, but I've never been able to find a convincing explanation using a quantum optics formalism.

My main question is: If I have an atom in an excited state, how do I describe the Hamiltonian or equations of motion for the effect of spontaneous emission?

Specifically I am interested in obtaining a result that includes the frequency modes $a_\omega^\dagger$ and their distribution $g(\omega)$.

If I were to guess I think that this system would have something of the form:

$\dot{\rho}_{excited} = -\gamma_{decay} \rho_{excited}\sum_\omega g(\omega)(a_\omega+a^\dagger_\omega)$

(That is over time there is decay from the excited state and production of photons with certain frequencies over some bandwidth, but I'm not sure this is correct.)

Wolpertinger
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Steven Sagona
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    Do you know the Jaynes-Cummings-model? – Norbert Schuch Mar 24 '21 at 23:29
  • @NorbertSchuch, yes but I believe that best describes an atom coupled to a cavity, and does not cover the general case where you have an unconfined excited atom that emits a photon via spontaneous emission. – Steven Sagona Mar 25 '21 at 00:12
  • It describes an atom couples to one mode (no t necessarily a cavity). If you couple to a continuum, you have to integrate over the modes. I'm just puzzled about your guess given you know the Jaynes-Cummings-model. – Norbert Schuch Mar 25 '21 at 07:41

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The minimal quantum model for spontaneous emission is given by the interaction Hamiltonian

$$ H = \int d\omega [g(\omega)a(\omega)\sigma^+ + g^*(\omega)a^\dagger(\omega)\sigma^-]$$

where $a(\omega), a^\dagger(\omega)$ are bosonic modes associated with the light field and $\sigma^\pm$ are the ladder operators of a two-level system. The coupling function $g(\omega)$ depends on the electromagnetic environment and can e.g. be evaluated for free space or for a cavity.

The sponteneous emission problem is then specifically given by the initial state where the atom is fully excited and the light field is in the vacuum state (or a thermal state at a given temperature, but I will only consider the vacuum case for simplicity).

With regards to solution methods, there are multiple levels of complexity that one can consider. The standard approach, which is probably covered in every modern quantum optics book, is Wigner-Weisskopf theory. The latter makes some approximations such as the Markov approximation, which make it applicable at weak light-matter coupling.

For strong light-matter coupling, the standard approach is to use an alternative starting Hamiltonian, namely the Jaynes-Cummings model as suggested by @NorbertSchuch in the comments. The reason is that strong coupling is often achieved by coupling the atom to a resonator where a single mode dominates.

As an advanced topic, I would like to note that the spontaneous emission problem, specifically, can also be described at strong coupling using the above Hamiltonian. There are a bunch of resources on this topic, I think one of the Cohen-Tannoudji quantum optics textbooks also contains a chapter. My personal favourite is Krimer et al., PRA 89, 033820 (2014). It essentially presents a generalization of the Wigner-Weisskopf method, where the Markov approximation is eliminated by solving certain integrals exactly using Laplace transforms. With regards to the comments, the paper also shows the transition from the Wigner-Weisskopf type emission to the Jaynes-Cummings type dynamics as you turn up the coupling by making the electromagnetic environment more confined.

Note that the central reason why this model is so useful and can be solved so accurately is the rotating wave approximation. It results in the Hamiltonian being excitation number conserving, such that only the states $|1_\mathrm{atom}\rangle|0_\omega\rangle$ and $|0_\mathrm{atom}\rangle|1_\omega\rangle$ contribute.

Even this treatment breaks down eventually. In the ultra-strong coupling regime, the minimal model given above does not work anymore, since counter-rotating terms and stuff like gauge invariance start to matter. There is a lot of literature on this topic, see e.g. What are the "strong", "ultrastrong" and "deep strong" coupling regimes of the Rabi model? and references therein.

Wolpertinger
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  • Thanks. Do you have an intuition for how this Hamiltonian can induce emission when it is in the excited state? Must I write out the full equations of motion to see that? Speaking of which, in the Hamiltonian you've written, does this show that there is decay from the excited state (transitions from excited to ground states in the atom)? I would have thought to see some "empirical" decay terms in this system, to describe this. – Steven Sagona Mar 25 '21 at 00:42
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    @StevenSagona This Hamiltonian fully contains spontaneous emission due to the continuum of bosonic modes, so there are not empirical terms needed. With regards to the intuition, I suggest to try the Wigner-Weisskopf calculation for yourself. I find it rather instructive. – Wolpertinger Mar 25 '21 at 00:45
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    Maybe to add to that: The decay terms you are referring to usually arise when one transfers to a Master equation treatment by tracing out the continuum. Note that the Wigner-Weisskopf calculation is one level above that. In fact, you can derive the decay rate in the Master equation using WW, which was indeed the historical purpose. – Wolpertinger Mar 25 '21 at 00:47
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    Thanks, I'll try that out. – Steven Sagona Mar 25 '21 at 01:03
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    I'm a bit puzzled about the strong separation between Jaynes-Cummings and this. In the end it is the same model, just once with an integral and once without. – Norbert Schuch Mar 25 '21 at 09:25
  • @NorbertSchuch hm, good question. In my view, there is no strong separation in terms of the physics. Jaynes-Cummings (with loss from the caviy mode in a Master equation treatment) can also model the decay at weak coupling (i.e. the adiabatic elimination regime). On the level of models, however, there is a difference between the continuum and the single mode description. Jaynes-Cummings (without loss) will always just oscillate, with the continuum integral you will always get some sort of decay. (cont.) – Wolpertinger Mar 25 '21 at 12:50
  • I quite like the Krimer paper in that it shows the physical transition of the dynamics while staying within the continuum model. The Jaynes-Cummings like oscillations are then obtained in the case where $g(\omega)$ is strongly peaked around the atomic frequency, as would be the case for a cavity with isolated resonances. They also have a section where they redo the calculation in a Jaynes-Cummings like formalism, based on https://doi.org/10.1103/PhysRevA.67.013805 . So that also shows the formal connection on a model level (also cf. https://doi.org/10.1103/PhysRevX.10.011008). – Wolpertinger Mar 25 '21 at 12:54
  • @NorbertSchuch I just remembered an old question of mine which might be relevant with regards to your comment https://physics.stackexchange.com/questions/330899/why-is-a-continuum-of-bath-modes-required-for-irreversible-dynamics The answer and comments below suggest that the difference is related to Poincaré recurrence. – Wolpertinger Apr 14 '21 at 18:06
  • @Wolpertinger Thanks. I didn't mean to say that the two models show similar behavior. But the question was about how to set up the model in the first place. And from that point of view, I'd say they are the same, just that you couple to a continuum of modes (so you need to integrate). That the physical behavior is different is clear. (But that's one great thing about physics: The same basic model in different scenarios can give very different behavior!) – Norbert Schuch Apr 14 '21 at 19:34
  • @NorbertSchuch ah right, agreed! I guess my long comments were unnecessary then. I’ll leave the up in case third parties are interested. – Wolpertinger Apr 15 '21 at 08:22
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If I may help, there is an alternative to Wigner-Weiskopff theory to explain spontaneous emission, assuming that the electron oscillates between upper state $\vert b >$ to lower state $\vert a>$. This theory, as far as I know introduced by Roger Boudet in his book "Relativistic Transitions in the Hydrogenic Atoms", uses classical electromagnetism with vector potential $\boldsymbol{A}$ and probability current transition $\boldsymbol{j}$:

$$ \Box \boldsymbol{A} = \mu_0 \boldsymbol{j} $$

with the transition current provided (here non-relativistic) by:

$$ \boldsymbol{j} = \frac{\hbar e}{m} \text{Re} \lbrace i \left[ \Psi_b^* \boldsymbol{\nabla} \Psi_a - \Psi_a \boldsymbol{\nabla} \Psi_b^* \right] \rbrace $$

Computing $\boldsymbol{A}$ in far field, then $\boldsymbol{E}$ and $\boldsymbol{B}$ you can have the time-averaged Poynting vector then the radiated power: you divide by $E_b - E_a$ and you obtain the Einstein spontaneous coefficient. The full derivation is available here.

fefetltl
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