# Why don't photons split up into multiple lower energy versions of themselves?

A photon could spontaneously split up into two or more versions of itself and all the conservation laws I'm aware of would not be violated by this process. (I think.) I've given this some thought, and a system consisting of multiple lower energy photons would have a significantly higher number of micro-states (and consequentially higher entropy) than one consisting of a single photon with that much energy. This would make the process more favorable.

Why does this not happen?

• Can you make a $\gamma\to2\gamma$ vertex (interaction term in the Lagrangian) that is gauge invariant? – Ryan Unger Aug 28 '15 at 14:04
• No (because I haven't learnt that much yet!) Could you explain it in a simpler way, possibly? – Hritik Narayan Aug 28 '15 at 14:10
• To tree level in Feynman diagrams, you have to have a term in the Lagrangian like $A^3$ for $\gamma\to 2\gamma$, where $A$ is the photon vector field. I don't know how to make this either Lorentz or gauge invariant, which are both fundamental symmetries of nature. Note that this does not mean this decay can't happen via higher order diagrams, hence why this is not an answer :) – Ryan Unger Aug 28 '15 at 14:26
• A photon can split after interacting with a charge, but this process has a very small amplitude. It's a suggested exercise in the QFT book by Itzykson and Zuber in the section about photon-photon scattering. – Count Iblis Aug 28 '15 at 17:17
• A single photon has momentum in a very specific direction. For the sake of the argument, assume that it is travelling down the positive x axis, in the positive direction. For this photon to split in two, there would have to be a small momentum in the vertical (or depth - "z") direction. If this vertical momentum suddenly appears from nowhere, you would have violated conservation of momentum, as total momentum AND each component of total momentum is conserved in any interaction. This means that the first sentence in your problem statement is probably incorrect. – David White Sep 2 '15 at 2:07

After the hypothetical split, 2 photons with the same energy would be propagating at an angle ok with momentum conservation. Then there would be a rest frame where the angle is 180 degrees. Now if you stay in this restframe and go back in time before the split, your single photon would be at rest. However, that is not possible: According to relativity, speed of light is constant for all frames. Thus, there can be not split of a single photon into two in vacuum (i.e. without momentum transfer during split). Mathematically, the reason is that the Lorentz group is non-compact, which means that the parameter gamma can take any value from [1, infinity) but not infinity itself which would correspond to a coordinate frame moving at lightspeed with all massive particles having infinite kinetic energy.

• It is the argument I am using also but +1 for the colinear case; – anna v Aug 29 '15 at 3:09
• This seems like a circular argument: for physical process X, examine it in a frame which does not exist according to special relativity. Then, since the frame can't exist, the process must be impossible. To put it another way, you could equally well say that a single photon, before splitting, cannot exist. Because in some frame it must be at rest, but such a frame is impossible. Circular. – Paco Jain Sep 3 '15 at 21:26
• rdjain1, the argument is neither circular nor generalizable to other physical processes. In fact, the argument does not go into a forbidden frame per se, but rather the rest frame. Now the only physical process which does not have a rest frame is the single photon propagating in vacuum. This fact leads to the fact that single photons cannot split even though they might be allowed to do so when one only looks at energy and momentum conservation. – user3042711 Sep 4 '15 at 15:57
• This explanation is wrong: two collinear photons with the same energy simply don't have a rest frame: the rest frame for the photon pair is the same as for the rest frame of a single photon. From a purely kinematic standpoint, one photon going to two photons in the same direction with the same energy (half the total) is totally allowed. – Chris Mar 1 '16 at 1:40
• @Chris I altogether agree. An $N$-fold split into $N$ momentum states collinear with the original (and with the same sense) and each with $N^{-1}$ times the original energy / 3-momentum is the unique solution consistent with conservation 4-momentum. In neither case is there a center of mass frame. There's definitely more to all this. – WetSavannaAnimal Jun 26 '16 at 5:55

A photon is an elementary particle. As much elementary and as much particle as the electron .

A single elementary particle has a fixed mass and cannot emit another particle without violating energy conservation, because its mass is fixed. In the center of mass of a massive elementary particle, electron, there is no energy for an emission , for a radiating electron in a field the energy is supplied by the field.

If a zero mass elementary particle like the photon could split into two, suddenly an invariant mass will appear and the before the split has zero invariant mass, after the split a measurable invariant mass. This means both momentum and energy conservation are violated, as the invariant mass is the measure of the four vector, before and after the split. A photon can also interact with a field in higher order diagrams , but cannot split in the sense you envisage.

Assume a photon could decay into two photons.

These photons will have four vectors. There are two situations: their three momenta are parallel in the laboratory to the original photon, or there exists an angle of the three momenta with the original photon and also between them. In the latter case the two decay photons define a center of mass ( similar to a pi0 at rest). In this system the two momenta add up to zero, but there will be energy giving an invariant mass to the system, which violates energy conservation as the original photon had 0 invariant mass, i.e. cannot supply this energy. The original photon in the center of mass of the decayed photons will still be moving with velocity c, and so have a momentum different than zero, thus momentum conservation is also violated.

In the case of two collinear photons in the lab , their invariant mass will be zero at the limit of the angle between them being exactly 0, otherwise the above argument holds. If it is exactly 0 no center of mass can be defined because a zero mass system moves with the velocity of light.

So the question becomes: why a photon of frequency nu does not turn into two exactly collinear in the lab photons of lower frequency. Experimentally this has not been observed so if it can happen it is a very very low probability process. In the comments Lubos Motl gives this statement :"For photons, this amplitude is 0 due to the Abelian gauge symmetry and other symmetries." I am still looking for a link on this.

In the next answer the collinear case is excluded by special relativity,

Mathematically, the reason is that the Lorentz group is non-compact, which means that the parameter gamma can take any value from [1, infinity) but not infinity itself which would correspond to a coordinate frame moving at lightspeed with all massive particles having infinite kinetic energy.

Photons come with chirality, so you should consider angular momentum conservation as well. For $1\gamma \to 2\gamma$ scattering, this will not be possible. (I'm assuming production of collinear photons only; it's obvious when two are not collinear, energy and momentum conservation will be violated)

• I'm not a particle physicist, so I'm going to ask why couldn't the initial photon be linearly polarized, i.e. in a superposition of two opposite AM circular polarizations so that the expected value of the spin observable were nought? – WetSavannaAnimal Jun 26 '16 at 3:48
• To be more precise, let the two photons state be: \alpha c^{\dagger } _{p \uparrow}c^{\dagger } _{p \uparrow} – pppqqq Jun 26 '16 at 9:26
• Sorry, my previous comment was wrong and I'm unable to delete the one above, I'm going to give a answer. – pppqqq Jun 26 '16 at 9:43

Each photon has intrinsic angular momentum, spin, which is non-zero. It's 1 for a photon. That's why the selection rule for an atom emitting a photon is Delta j = +/- 1. Therefore, it's a case of angular momentum violation if a photon is going to split into two or more, while the conservation of energy still holds. A system containing one photon has the total angular momentum 1, but if this photon splits into 2 photons the total angular momentum will be different.

• I'm not a particle physicist, so I'm going to ask why couldn't the initial photon be linearly polarized, i.e. in a superposition of two opposite AM circular polarizations so that the expected value of the spin observable were nought? – WetSavannaAnimal Jun 26 '16 at 3:48
• A superposition with no angular momentum is TWO photons. The described reaction, two photons going in, and two different photons coming out, is not observed. We can imagine the reverse of an electron/positron annihilation (photons creating antiparticles), but the rate of such a reaction is TINY (photons are big, and electron/positron pairs are small, so the matrix element for the reaction would vanish unless those photons are focused to an electron-radius spot). – Whit3rd Jun 26 '16 at 8:03
• @Whit3rd no, that's not right, I'm talking about the one-particle superposition of the form $(e^{i\,\phi} |\psi_\ell\rangle + e^{i\,\theta}|\psi_r\rangle)/\sqrt{2}$, where $\psi_\ell$ and $\psi_r$ are the left and right handed eigenstates, respectively and $\phi,\,\theta \in\mathbb{R}$. – WetSavannaAnimal Jun 26 '16 at 9:21

It looks like the reaction $\gamma \to 2\gamma$ is not only dynamically forbidden (Furry's theorem), but also kinematically forbidden.

As Dexter Kim points out, the only way to conserve energy and momentum is that the two photons are emitted at $0°$, in which case the angular momentum along the direction of motion is given by the coupling of the two photon spins.

The photon's spin can assume only the values $m=\pm 1$. Looking at the $1+1\to 1$ Clebsch Gordan table, we realize that the only possible coupling of two photon's spins with $j=1$ has $m=0$. But, again, the initial photon has $m=\pm 1$. Therefore, angular momentum cannot be conserved together with four-momentum in $\gamma \to 2 \gamma$.

The are two factors going into the decay of any particle. There is the matrix element, which comes from fundamental physics, and the density of states, which comes from kinematics, mostly how heavy the mother particle is relative to the daughter particles. Either of these being very small slows down the decay (or being zero keeps the decay from happening at all).

An example:

The free neutron has a matrix element roughly equal to the free muon. But the muon decays into particles much lighter than itself (the muon has a mass of 106 MeV, and the decay products combined only have a mass of about .5 MeV). The neutron on the other hand is barely heavier than its products: The neutron has a mass of 939.6 MeV and its products add up to 938.8 MeV.

The result? The muon has a lifetime of .0000022 seconds, while the neutron has a lifetime of 890 seconds. The neutron lives 400 million times longer! In particle physics jargon, we call an interaction like this phase-space suppressed.

An interaction like $\gamma \rightarrow \gamma + \gamma$ is even more phase-space suppressed. So phase-space suppressed, in fact, that the decay rate of this interaction formally computes to zero even if the matrix element is not zero. (In the standard model, the matrix element for $\gamma \rightarrow 2\gamma$ is zero anyway, and we know from experiment that it is very small regardless)

This same argument works for one photon splitting into any number of photons. So it forbids one photon splitting into three or four or more photons as well.

• That's your problem. We don't even know for sure why particle decays let alone saying how many factors for a particle to decay like it is an axiom. In addition, putting specific numbers into your argument only convince a weak mind better. Let me tell you about this fundamental: time has been defined to be based on particle decay. And you say that decay bases on density and density bases on kinematic which bases on motion which bases on distance and time. See the circular argument now??? – user121963 Jun 26 '16 at 4:59

It is certainly thermodynamically possible for a high energy photon to vanish and a multiplicity of lower energy photons to be created. This is observable as a cascade of events (photoelectric absorption of a photon, followed by multiple fluorescence photons) in thermalization of a high energy photon interacting with matter. It isn't a simple photon-in, two-photons-out reaction, because that doesn't balance as a particle reaction (can't conserve energy and momentum and angular momentum).

The usual cascade that thermalizes energy, in matter, from an X-ray photon, might generate some other photons, but mainly generates phonons or unstable atomic states (excited electrons). There may be photons MUCH later; thermoluminescent devices accumulate X-ray exposure for weeks, to create IR photons when the radiation badge is in the reader.