Emitting two photons in a single transition?

In summary, two-photon emission is possible, but very unlikely. It would require a coupling between the atom and a virtual state, which is not always present.
  • #1
bcrowell
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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?
 
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  • #2
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).
 
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  • #3
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
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
bcrowell said:
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?

Sure, this is possible. You can calculate the rate in second-order perturbation theory. I found http://cua.mit.edu/8.421_S06/Chapter9.pdf calculating the inverse process of two-photon absorption.
 
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  • #6
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.
 
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  • #7
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 http://isites.harvard.edu/fs/docs/icb.topic893193.files/Solutions5.pdf. 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.
 
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Related to Emitting two photons in a single transition?

1. What is the process of emitting two photons in a single transition?

Emitting two photons in a single transition, also known as a two-photon emission, is a quantum mechanical process in which an excited atom or molecule releases two photons at the same time. This process occurs when the energy levels of the atom or molecule are such that the energy required for the transition corresponds to the sum of the energies of two photons.

2. How is two-photon emission different from single-photon emission?

Two-photon emission differs from single-photon emission in that it involves the simultaneous release of two photons, while single-photon emission involves the release of only one photon. Two-photon emission is also a less common process compared to single-photon emission, as it requires specific energy level configurations in the emitting atom or molecule.

3. What is the significance of two-photon emission in scientific research?

The significance of two-photon emission lies in its unique properties and potential applications. Two-photon emission is a nonlinear process, which means that the energy of the two photons combined is greater than the energy of the initial excited state. This can be useful in various fields such as microscopy, spectroscopy, and quantum computing.

4. Can two-photon emission occur in any type of atom or molecule?

Two-photon emission can occur in any atom or molecule that has the appropriate energy level configuration. However, it is more commonly observed in atoms or molecules with large transition dipole moments, such as rare earth elements and certain organic molecules.

5. How is two-photon emission related to the concept of entanglement?

Two-photon emission is closely related to the concept of entanglement, which is a phenomenon in quantum mechanics where two particles become connected in such a way that the state of one particle affects the state of the other, regardless of the distance between them. In two-photon emission, the two photons are entangled, meaning that their polarization, momentum, and other properties are correlated, even after they are separated.

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