Do Photons Decay? QED Predictions & Measurements

[maline]

QED predicts photon-photon scattering, via a fourth-order Feynman diagram with four external photon legs connected to an electron loop. It seems to me that the same diagram should also predict a process where a single incoming photon decays spontaneously into three photons. This is not forbidden by energy-momentum conservation, as long as the final energies add up to the original one and the momenta are all parallel.

Is this correct? If so, has this decay rate been calculated? Or at least, what is its order of magnitude? If the decay does not occur, why not?

I wonder whether this attenuation could be measurable, perhaps in the CMBR? By Lorentz invariance, or just dimensional analysis, the rate of decay (events per second per photon) should be linear in the starting frequency. Perhaps this signature could distinguish the pure photon decay from other sources of attenuation such as intergalactic dust, or processes involving more than one photon.

 

[mfb]

The rate of this process is exactly zero. There is some nice theorem about it but I forgot the name and can't find it.
To have photon-photon scattering you need a non-zero center of mass energy, something your process doesn't have.

This assumes photons are exactly massless, of course, otherwise they might be able to decay.

 

[A. Neumaier]

This is not forbidden by energy-momentum conservation, as long as the final energies add up to the original one and the momenta are all parallel.

This implies that the process is forbidden. Do the calculation, using the fact that photons are massless!

 

[maline]

This implies that the process is forbidden. Do the calculation, using the fact that photons are massless!

What do you mean? The photon is massless, so if all the directions are the same then energy and momentum are conserved together.
Is there a symmetry principle or something that I am missing?

 

[Vanadium 50]

There is some nice theorem about it but I forgot the name and can't find it.

Furry's theorem.

 

[maline]

Furry's theorem.

No, Furry's theorem is for odd n-point functions. It's basically just C symmetry. That's not an issue here; there are four external lines in total.
Remember, the diagram I am talking about is the same one that gives light-by-light scattering, just labeled with one incoming and three outgoing photons rather than two and two.

 

[Vanadium 50]

You're right. Furry blocks the 3 and 5, not the 2 and 4.

However, to make this work you need a quantum anomaly, since classically the EM wave has the same frequency as its source (and the same frequency as the wave had a moment ago). I am not an expert, but I believe for electromagnetism they all need to cancel.

 

[A. Neumaier]

What do you mean? The photon is massless, so if all the directions are the same then energy and momentum are conserved together.
Is there a symmetry principle or something that I am missing?

Oh,yes, sorry. If they are exactly parallel, conservation laws don't help. But this is a set of measure zero, which has no physical effect since momentum eigenstates are not normalizable and one needs to integrate over an open set of momenta to get something with a physical meaning.

Independent of that, the renormalization process ensures that single photon, electron, or positron states don't scatter (and hence do not decay). These requirements are part of the renormalization conditions.

 

[maline]

But this is a set of measure zero, which has no physical effect since momentum eigenstates are not normalizable and one needs to integrate over an open set of momenta to get something with a physical meaning.

Presumably the original photon has some distribution over momentum states; why shouldn't these coefficients work for the daughter photons as well?

the renormalization process ensures that single photon, electron, or positron states don't scatter (and hence do not decay). These requirements are part of the renormalization conditions.

The renormalization condition I am familiar with is that the photon propagator must have a pole at zero mass. So I guess you are saying that if decay were possible, it would shift the pole away from the origin, in the imaginary direction.
I don't know enough to be confident about that, but it sounds like fairly convincing evidence that the decay is indeed forbidden somehow. However, it cannot be eliminated as part of the renormalization itself: all of the counterterms are real-valued, and derived from loop corrections to the propagators-diagrams with only two external lines- or to the vertex, with its one external photon line.
So we still need to know what stops the decay as a process.

 

[A. Neumaier]

Presumably the original photon has some distribution over momentum states; why shouldn't these coefficients work for the daughter photons as well?

The contribution of equal momentum terms to Feynman integrals is zero because they have zero measure.

The renormalization condition I am familiar with is that the photon propagator must have a pole at zero mass. So I guess you are saying that if decay were possible, it would shift the pole away from the origin, in the imaginary direction.
I don't know enough to be confident about that, but it sounds like fairly convincing evidence that the decay is indeed forbidden somehow.

See the last axiom (S) in the five axioms required in [URL='https://www.physicsforums.com/insights/causal-perturbation-theory/']causal perturbation theory[/URL].

However, it cannot be eliminated as part of the renormalization itself: all of the counterterms are real-valued, and derived from loop corrections to the propagators-diagrams with only two external lines- or to the vertex, with its one external photon line.
So we still need to know what stops the decay as a process.

Renormalized Feynman diagrams are a poor guide to physics. Except on the tree level (representing the classical approximation), they don't indicate any processes at all.

 

[vanhees71]

Furry's theorem.

Furry's theorem is about the fact that in standard (scalar or Dirac-fermionic) QED the diagrams with only an odd number of photon lines all identically vanish.

Here the OP considers, however the decay of one photon to three photons. This process cannot happen (in vacuo!) because of the fact that energy and momentum AND the on-shell conditions for all the photon momenta on the external legs.

 

[A. Neumaier]

Here the OP considers, however the decay of one photon to three photons. This process cannot happen (in vacuo!) because of the fact that energy and momentum AND the on-shell conditions for all the photon momenta on the external legs.

Well, @maline had pointed out the exception that collinear photons are not excluded by this. But this does not contribute to physical quantities because of their vanishing measure.

 

[meopemuk]

Here is a paper that could be useful in this discussion:

G. Fiore, G. Modanese,"General properties of the decay amplitudes for massless particles", Nucl. Phys. B477 (1996), 623.

Eugene.

 

[mfb]

G. Fiore, G. Modanese,"General properties of the decay amplitudes for massless particles", Nucl. Phys. B477 (1996), 623.

On arXiv

There is a funny implication of a 1->3 photon decay. Let's say a 1 eV photon has a lifetime of 1 year before it decays to 3 photons with an average energy of 1/3 eV.
Our original photon had 1/3 eV in a reference frame moving at 80% the speed of light in the direction of the photon. In that reference frame it lived only 1/3 years. This is a general property that doesn't depend on the energy or the lifetime: If photons can decay then a photon with 1/3 the energy has 1/3 the lifetime to preserve Lorentz invariance. You can derive similar factors for other energies. Lifetime is inversely proportional to energy.
A photon decaying needs at least one photon with at most 1/3 the energy, which has 1/3 the original lifetime. If that decays we get again at least one photon with at most 1/9 the original energy, and 1/9 the original lifetime and so on. That time series converges. We get infinitely many decays in finite time.

Edit: Wrong "inversely" in the post

 

[maline]

No, lifetime will grow like the inverse of the energy.

 

[maline]

On arXiv

There is a funny implication of a 1->3 photon decay. Let's say a 1 eV photon has a lifetime of 1 year before it decays to 3 photons with an average energy of 1/3 eV.
Our original photon had 1/3 eV in a reference frame moving at 80% the speed of light in the direction of the photon. In that reference frame it lived only 1/3 years. This is a general property that doesn't depend on the energy or the lifetime: If photons can decay then a photon with 1/3 the energy has 1/3 the lifetime to preserve Lorentz invariance. You can derive similar factors for other energies. Lifetime is inversely proportional to energy.
A photon decaying needs at least one photon with at most 1/3 the energy, which has 1/3 the original lifetime. If that decays we get again at least one photon with at most 1/9 the original energy, and 1/9 the original lifetime and so on. That time series converges. We get infinitely many decays in finite time.

I apologise, you are of course correct: any 4-displacement along a straight particle path is directly proportional to the momentum 4-vector, so of course lifetime grows with energy. The avalanche result you mentioned follows. I guess it's a good the decay is disallowed...

Thank you for the Arxiv link.

 

[thephystudent]

I think another point is that photons, unlike eg. gluons, don't contain a charge themselves so they don't interact with each other.

Inside materials, similar mechanisms exist though

https://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion

 

[maline]

I think another point is that photons, unlike eg. gluons, don't contain a charge themselves so they don't interact with each other.

Inside materials, similar mechanisms exist though

https://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion

Photons in vacuum are able to interact ( very weakly) by generating a virtual electron-positron pair. This causes photon-photon scattering, an old prediction of QED which very recently was finally observed at the LHC.

My question was why the same process doesn't cause individual photons to decay, and Arnold Neumaier gave me the correct answer: since conservation of energy and momentum would force the daughter photons to be perfectly collinear (same direction of motion), the space of outcomes has measure zero. The set of possibilities has only two degrees of freedom, out of a total of nine (for the three spatial momenta). When we calculate the total rate of decay, we need to integrate over the space of outcomes, so we will end up with a rate of zero.

 

[thephystudent]

Photons in vacuum are able to interact ( very weakly) by generating a virtual electron-positron pair. This causes photon-photon scattering, an old prediction of QED which very recently was finally observed at the LHC.

You are absolutely correct in principle, when I say they don't interact I mean 'for practical purposes'. But now I see you indicated you were after this process from the beginning, my mistake of skimming too rapidly trough. Guess I just wanted to add that there's in general a name for these kind of processes.

 

[vanhees71]

Well, the process $\gamma \gamma \rightarrow \gamma \gamma$ is of course possible (fourth-order QED process). It has been observed recently by the ATLAS collaboration in ultraperipheral Pb-Pb collisions at the LHC.

 

[jfizzix]

Above the Schwinger limit of EM field strength (a limit we're getting close to attaining), the vacuum is polarizeable and behaves like a nonlinear medium (the nonlinearity being third-order due to centrosymmetry). It is conceivable to have simultaneous energy and momentum conservation (i.e. phase-matching) for the process where one high energy photon decays into an entangled photon triplet (thus not ruling it out). Without knowing what the nonlinear polarizeability of the vacuum is, I couldn't predict how efficient this process is, but looking at studies of third-order nonlinear optics in media, it's really quite unlikely (of the order of a few triplets per second for hundreds of milliwatts of laser power focused into a single (microns-wide) nonlinear waveguide).

See for example:
https://arxiv.org/abs/1307.3261https://arxiv.org/abs/physics/0203069

 

[mfb]

This is not in a vacuum anymore, however. A perfectly parallel wave in a vacuum cannot reach nonlinear effects. There is always a reference frame where it has a low power.

 

[vanhees71]

Of course, in a medium or (strong) fields, $\gamma \rightarrow \gamma \gamma \gamma$ is possible. It's called "non-linear optics". Indeed, "strong fields" implies that we are no longer in the "vacuum" (i.e., the ground state of the interacting quantum system).