Why don't atoms emit more than one photon during an energy level transition?

If you look at the emission spectrum of an atom, there are sharp lines corresponding to the different energy level transitions. That's because the single photon emitted during each transition carries the entire energy of the transition.

My question is: how does it happen that just a single photon carries all the energy? Why isn't the energy sometimes split among two or more photons?

I understand how the rules of quantum mechanics constrain the energy states of the atom to a discrete set of levels, but I don't understand how they also constrain the number of produced photons to be equal to one.

• I have a feeling that quantum electrodynamics, rather than quantum mechanics itself, imposes this constraint. – probably_someone Oct 29 '19 at 18:24
• A high-quality answer to this question might address the second-order process of two-photon absorption, which is the time-reversed partner of two-photon emission. – rob Oct 29 '19 at 18:33
• As @rob alludes multi-photon processes are a thing, just a highly surpressed thing. The factor of $1/\alpha$ from the extra vertex is just the beginning. – dmckee --- ex-moderator kitten Oct 29 '19 at 18:50
• – Dvij Mankad Oct 29 '19 at 19:11
• Here is a comment related to two-photon emission. I also wrote an answer that examines the mechanism of two-photon excitation. – dmckee --- ex-moderator kitten Oct 29 '19 at 22:35

Two-photon emission does exist, or else the 2s state of hydrogen would be stable. You can get a pretty decent estimate for this kind of rate without any fancy math or physics, just using the energy-time uncertainty principle. The typical rate of emission for a photon, when not forbidden by parity, is $$R \sim 10^9\ \text{s}^{-1}$$. We can think of the two-photon decay as an energy-nonconserving jump up to some higher-energy state, with the emission of a photon, followed by the emission of a second photon leading down to the ground state. The first jump can happen because of the energy-time uncertainty relation, which allows the electron to stay in the intermediate state for a time t ∼ h/E, which is on the order of $$10^{−15}$$ s. The probability for the second photon to be emitted within this time is $$Rt$$, so the rate for the whole two-photon process is $$R^ 2 t \sim 10\ \text{s}^{-1}$$. Considering the extremely crude nature of this calculation, the result is in good agreement with the observed rate of about $$0.1\ \text{s}^{-1}$$ for two-photon decay of the 2s state in hydrogen.

• That makes sense, but why do we have to think of two jumps when we want to get two photons? Why can't each jump itself emit two or more photons as long as the energies all add up properly? – Ian H Oct 29 '19 at 19:30
• "We can think of the two-photon decay as an energy-nonconserving jump up to some higher-energy state, with the emission of a photon, [...]" Huh? Jump 'up', yet emit a photon while doing so??? – Gert Oct 29 '19 at 20:17
• @Gert Higher up with respect to the ground state. – my2cts Oct 29 '19 at 20:50
• That makes sense, but why do we have to think of two jumps when we want to get two photons? Why can't each jump itself emit two or more photons as long as the energies all add up properly? QED only has vertices with three lines, not four. – user4552 Oct 29 '19 at 21:02
• QED only has vertices with three lines, not four. This is more the kind of answer I was looking for. Would you mind going into more detail? I get that this is a statement about Feynman diagrams, but I don't know how you would model something like an atomic transition using Feynman diagrams. – Ian H Oct 29 '19 at 21:41

Another form of two photon emission is two photon excited fluorescence.

Two photon absorption is the absorption of two photons by a molecule to excite from ground state to excited state.

Two-photon absorption can lead to two-photon-excited fluorescence where the excited state produced by TPA decays by spontaneous emission to a lower energy state.

First, there is a two photon absorption, then a non-radiative deexitation, and a fluorescent emission. The electron returns to ground state by another non-radiative deexitation.

https://en.wikipedia.org/wiki/Two-photon_absorption

Now you are asking about two photon emission, and phosphorescence is a type of multi-photon emission, where the absorbed energy is released by the emission of multiple photons.

Phosphorescence is a type of photoluminescence related to fluorescence. Unlike fluorescence, a phosphorescent material does not immediately re-emit the radiation it absorbs. The slower time scales of the re-emission are associated with "forbidden" energy state transitions in quantum mechanics. As these transitions occur very slowly in certain materials, absorbed radiation is re-emitted at a lower intensity for up to several hours after the original excitation.

https://en.wikipedia.org/wiki/Phosphorescence

Two photon emission (stimulated) is possible by semiconductors.

We report the first experimental observations of two-photon emission from semiconductors, to the best of our knowledge, and develop a corresponding theory for the room-temperature process. Spontaneous two-photon emission is demonstrated in optically-pumped bulk GaAs and in electrically-driven GaInP/AlGaInP quantum wells.

https://arxiv.org/abs/quant-ph/0701114

• Stimulated emission would also be an example of multi-photon emission if we include the cases in which one or more photons are absorbed as well, right? – Dvij Mankad Oct 29 '19 at 21:01
• @correct, like singly stimulated two photon emission in semiconductors. – Árpád Szendrei Oct 29 '19 at 21:03
• Yes, just saw the edit after posting the comment :P – Dvij Mankad Oct 29 '19 at 21:04
• I think this question is about simultaneous emission of two photons in a single transition, rather than about a cascade of single-photon transitions through well-defined states. Phosphorescence is the latter; the delay comes when one state in the cascade had a longer lifetime than the others. – rob Oct 29 '19 at 23:49
• Minor comment: please link to arXiv abstract pages, not to PDFs. See physics.meta.stackexchange.com/q/11400/44126 – rob Oct 30 '19 at 0:09