# Emitting two photons in a single transition?

Tags:
1. May 13, 2015

### bcrowell

Staff Emeritus
A student asked me a question last night that stumped me. Suppose we have an atom in its first excited state A, and it's going to decay electromagnetically to the ground state B, losing energy E. We would expect this to occur through the emission of a photon with energy E. But what prevents it from emitting, say, two photons, each with energy E/2? Obviously this can't occur as a two-step process, since there isn't any atomic state mid-way between A and B. But I don't see why it can't occur as a one-step process. I certainly don't think we observe this in reality. My experience in research was in gamma-ray spectroscopy, and we never even considered such a possibility.

Is such a process possible in principle, but just very unlikely? If so, how would one estimate the probability? Are there conditions under which it might be observable?

2. May 13, 2015

### fzero

The necessary coupling would be something like $\bar{\psi}\gamma^\mu A_\mu \gamma^\nu A_\nu \psi$ which is not gauge-invariant. Two-photon processes can proceed via a virtual intermediate state $C^*$ which need not be between A and B (the 2s to 1s transition is an example).

3. May 13, 2015

### atyy

I'm not sure this is relevant, but googling suggests http://arxiv.org/abs/quant-ph/0612124. At least pictorially, their Fig.1 seems to indicate two-photon emission.

4. May 13, 2015

If a one step process can result in the emission of two photons then why why not three photons or four or five or a thousand and five etc? if this can happen there is the possibility that we could observe a continuous spectrum.This makes me think that if such multiphoton emission can occur the probability of it happening is extremely small and reduces further with number of photons.

5. May 13, 2015

### The_Duck

Sure, this is possible. You can calculate the rate in second-order perturbation theory. I found this calculating the inverse process of two-photon absorption.

6. May 13, 2015

### bcrowell

Staff Emeritus
I think I'm starting to understand this better. One thing that was misleading me was that in my field of nuclear physics, we deal with electromagnetic transitions that have lots of different multipolarities, whereas in atomic physics it's pretty much always assumed that everything is an E1.

The classic real-world example seems to be the 2s->1s transition in hydrogen, which is forbidden for an E1 transition due to parity. There are at least three ways for the 2s state to lose its immortality: (1) the atom collides with something, (2) it emits an M1 transition, and (3) it decays by the emission of two photons. Two-photon emission has a partial half-life of 1/7 s, and is observed in planetary nebulae:
http://articles.adsabs.harvard.edu/full/1951ApJ...114..407S . M1 emission has a partial half-life of 2 days: http://www.physics.umd.edu/news/News_Releases/sucher.pdf (Sucher, Rep. Prog. Physics 41 (1978) 1781).

For emission of multiple gammas, there are some nice lecture notes here http://www.tapir.caltech.edu/~chirata/ay102/ by C. Hirata ("Notes on Atomic Structure"). Paraphrasing his treatment, I get something like this. Classically, an oscillating electric dipole d radiates at a rate given by $P\sim f^4d^2c^{-3}$. If you divide by $E=hf$, you get an emission rate $R\sim P/E$, which for atoms is about 10^9 per second. You can think of 2-photon decay as an energy-nonconserving jump to some intermediate state (one that actually exists), followed by a second jump to the final state. The first jump can happen because of the energy-time form of the Heisenberg uncertainty principle, which allows you to stay in the intermediate state for a time $t\sim h/E$, which is on the order of 10^-16 s. The probability for the second photon to be emitted within this time is $Rt$. The rate of two-photon emission is $R^2t$, which comes out to be on the order of 10^2 per second. (Hirata has 10^-2 per second, which seems to be an arithmetic mistake, or else I'm having a brain fade.) Applying this to the hydrogen 2s->1s transition, the intermediate state is a 2p [*plus others, see fzero's #7*], and the result is not ridiculously bad compared to experiment, given the extremely rough nature of the estimate.

The probability of emitting n photons would be something like $R(Rt)^{n-1}$, which falls off very quickly with n.

Last edited: May 13, 2015
7. May 13, 2015

### fzero

The semiclassical picture is qualitatively decent enough. As The_Duck said, the QM calculation involves 2nd order perturbation theory. I found a source for the full calculation as the 2nd problem in this solution set. The 2p state makes the largest contribution to the process, but you have to remember that this is QM and we have to sum over all possible intermediate states. These will satisfy $|\Delta L|=1$ and the contribution from large quantum numbers will be small, but the contribution from say 3p will not be much smaller than that from the 2p intermediate state.

Last edited: May 13, 2015