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How can we determine the exact number of photons produced in a decay or other event? This has puzzled me because photons can have arbitrarily low energy and momentum, so how do we tell if two photons are produced or three photons one of which has arbitrarily low energy? (or infinitely many soft photons for that matter)

For instance we hear the singlet state of positronium decays mostly to two photons but the triplet has to decay to three photons or higher odd numbers due to charge parity or the Landau Yang theorem.

What if one of those odd photons had arbitrarily small energy, wouldn't that look like a decay to even photons? Don't we expect infinitely many soft photons in any process in any case?

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  • $\begingroup$ You may be interested in reading Chapter 6 of Peskin and Schroeder. It gives a treatment of radiative corrections and infrared divergences, and there it is presented as crucial that final states differing by emission of additional photons with arbitrarily small energies cannot be physically distinguished. $\endgroup$
    – diracula
    Commented Mar 11, 2017 at 22:22
  • $\begingroup$ When we say positronium decays to two photons, we ignore the soft photons. We then compute the energy distribution of the two resulting hard photons. Then we can add the soft photons back in, which does not significantly change the energy distribution of the hard photons. The ultimate physical prediction, which is not ambiguous at all, is that most of the time you get two hard photons, along with the usual infinity of soft photons. $\endgroup$
    – knzhou
    Commented Oct 31, 2019 at 23:04
  • $\begingroup$ If you're unlucky, a "hard" photon as stated above can come out with too low energy to be detected. But the phase space for this to happen is negligibly small, so we don't talk about it. $\endgroup$
    – knzhou
    Commented Oct 31, 2019 at 23:06
  • $\begingroup$ @knzhou, So in the cases where a particle can not decay to two photons for symmetry reasons, we can still in principle observe a decay to two photons with low probability? $\endgroup$
    – octonion
    Commented Nov 2, 2019 at 3:41
  • $\begingroup$ Yes, but the probability of this is so incredibly small that it’s not relevant to experimentalists (try estimating it!) — things like detector inefficiency or pileup or whatever are many orders of magnitude worse. $\endgroup$
    – knzhou
    Commented Nov 2, 2019 at 5:24

2 Answers 2

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When you ask "how can we determine...", that sounds like an experimental question. But when you ask "don't we expect infinitely many soft photons?", that sounds like a theoretical question. I'll focus on the the experimental question.

The facetious answer to how you determine the number of photons in a decay is that you measure them. That's obviously not very satisfying, so I'll go into a little more detail, but that's still the crux of the answer.

Two points to keep in mind:

First, a high-energy physics detector has almost 4$\pi$ coverage, so you can measure the total energy of all the decay products in the event. You can also compare it to the total energy of the collision, which (depending on the event) may give you an idea of whether you have all the decay products that were produced. There are of course some complications:

  • Collision energy: In proton facilities like LHC, you don't actually know the collision energy because you don't know the fraction of the energy in the colliding partons. However, you do know what the distribution is, and you also know the maximum energy. On the other hand, in lepton colliders like LEP, the collision energy is known very precisely. (In fact, corrections were made in LEP to account for the level of water in Lake Geneva, and also for the operating schedule of the French TGV trains, which produced an even more subtle effect). That's why lepton colliders are so good for precision measurements.

  • Detector coverage: It's not really 4$\pi$ coverage; you lose coverage around the beam pipe (and in LHC experiments you always lose something down the beam pipe), and traditionally the endcaps and detectors at low angles are not as precise as the barrels. So people (especially in LEP analyses) sometimes require that the measured event energy be contained in the barrel portion of the detector. And of course there's all the subdetector calibration efforts, which I won't go into.

  • Energy Loss: It is not possible to detect all decay products in proton collisions like LHC. Lepton collisions like LEP are much cleaner; some events (such as $e^+ e^- \to e^+ e^-$ Bhabha scattering) contain the full collision energy (and thus are useful for calibration, for example). On the other hand, hadronic decays end up as jets; in such cases there will always be energy losses due to neutrinos even if you could precisely measure everything else. However, you can still get an idea of what decayed by looking at vertex information (i.e., where exactly the decay originated) and things like the invariant mass of the detected decay products.

The second point is that all high energy physics analyses are statistical. You never draw a conclusion from a single event; you always use a lot of events (the actual number is highly variable) and look at the distribution.

That second point is the real answer to the question. Especially in clean decays (such as those from lepton colliders), you can measure the invariant mass of the decay products such as the photons that you assume are from the particle that decayed. For example, suppose you suspect a given particle decays into three photons. You select events consistent with the decay of that particle by looking for events with two or three photons with an invariant mass equal to the particle's mass (plus whatever other event characteristics would apply). You calculate the invariant mass by looking at the photon energies and assuming they originated at a decay point which you estimate based on the particle's lifetime. With enough such events, you can then look at the energy distribution of (say) the photon with the lowest energy, and estimate/extrapolate how many events would appear to be two-photon events because the 3rd photon had an energy too small to measure. You can then compare that number with the actual observed number of two-photon events with that invariant mass.

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  • $\begingroup$ Thanks, the statistical point of view makes sense and is a satisfying answer. So I guess observing whether there are more 2 photon events than would be expected from pure phase space grounds would be a test of the theory that it only decays to odd photons. $\endgroup$
    – octonion
    Commented Nov 6, 2019 at 0:34
  • $\begingroup$ Yes, that's correct. Of course, a big part of any experimental analysis is understanding the background and the systematic uncertainties, so that when you subtract the background you can quantify the uncertainty of the remaining signal. This sort of estimation is often what takes up the bulk of an experimental analysis. But given that, you're exactly right: you compare the observed number of signal events (to include the uncertainty) the phenomenological prediction. $\endgroup$
    – Richter65
    Commented Nov 6, 2019 at 14:10
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how do experimentalists distinguish between the number of photons produced in a decay or other event?

By designing an experiment that can detect photon direction and energies of the photon , and using previously established conservation laws , energy, momentum, and quantum numbers to interpret the data.

because photons can have arbitrarily low energy and momentum,

Not if coming from a specific quantum mechanical state, as is electron positron annihilation. Energy must be conserved, so the two photons detected must have at least the energy of the mass of two electrons.

This is used in astrophysics, as an example:

The production of positrons and their annihilation in the galactic interstellar medium (ISM) is one of the pioneering topics of γ-ray astronomy. Since the detection of the 0.511-MeV line

With high enough energies electrons and positrons can annihilate to many other particles, the whole LEP experiment studied these interactions.

Every photon added in a decay reduces the probability of happening because the feynman diagrams will be depressed by 1/137, the electromagnetic coupling constant. There are some publications on this.

The paper presents results of experimental imaging of point-like sources using the 3-photon annihilations registered by a system of three high energy resolution detectors in coincidence. After filtering out the irrelevant random coincidences images of activity distributions are reconstructed. The positions of the sources are reproduced with good accuracy. The influence of random triple coincidences arising from the predominant 2-gamma annihilations, which may contribute to image noise is discussed. The analysis of experimental results is reinforced by computer simulations.

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  • $\begingroup$ The total energy of the photons must be greater than the mass of the electron and positron. But if there needs to be more than two due to charge parity or the Landau Yang theorem, what stops one of them from being so soft as to be undetectable? That's my question. $\endgroup$
    – octonion
    Commented Mar 11, 2017 at 6:16
  • $\begingroup$ nothing stops them except probabilities there do exist three gamma annihilations ( btw postironium annihilation just gives those two .511 mev gamma, if it is at rest $\endgroup$
    – anna v
    Commented Mar 11, 2017 at 6:56
  • $\begingroup$ I was specifically wondering about the case where we expect three or more photons to avoid the simple kinematic argument. Positronium can annihilate to more than two photons, and depending on the exact state sometimes it must annihilate to more than two. But there is a region of phase space where one of those has very low energy, and naively I would think that would look like charge parity violation. But given there are well defined values in data tables for decays to 2 or 3 photons, someone must understand how to deal with this. $\endgroup$
    – octonion
    Commented Mar 12, 2017 at 0:32
  • $\begingroup$ annihilation is not like a decay, the incoming state can be spin one or zero and the outgoing can have the appropriate combinations . One would have to design a complicated experiment to check incoming spins, outgoing spins and accumulate the numbers to observe differences not accounted by the CP conservation. All the calculations for positronium are confirmed by the data. en.wikipedia.org/wiki/Positronium#Measurements $\endgroup$
    – anna v
    Commented Mar 12, 2017 at 4:50

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