In order for a photon to decay into a pair of $e^+ e^-$, it must have at least $E_{\gamma}=1.022$ MeV and must be near a nucleus in order to satisfy the conservation of energy-momentum.

But would this happen even if the photon is near a neutron and not necesarily a nucleus? Does the fact that the nucleus is charged have anything to do with this decay? Who acts upon the photon to induce the interaction?


3 Answers 3


Quantum mechanics says that everything that is not forbidden is compulsory. Any process that doesn't violate a conservation law will happen, with some rate or cross-section. However, this general principle doesn't tell you what the rate is. For example, it's theoretically possible for 124Te to decay into two 62Ni nuclei plus four electrons and four antineutrinos, but to predict the (very small) rate, you need to know the relevant nuclear physics.

In your example, the process probably would go at some rate determined by electromagnetic interactions, because the neutron has a magnetic field. But the rate would presumably be small because the magnetic field of a dipole falls off like $1/r^3$, and magnetic effects are usually down by $\sim v/c$ compared to electric effects.


Yes, pair production can occur even near a lone neutron. The presence of a mass for the photon to interact with is required for conservation of momentum (further explanation can be found here). To my knowledge, the charge of the nucleus is not significant to the process of pair production, though the probability of pair production increases approximately with atomic number squared. Experimentally, this would be hard to demonstrate as neutrons are hard to control and have a relatively short half life (~10.3 minutes).

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    $\begingroup$ To my knowledge, the charge of the nucleus is not significant to the process of pair production This is not true, and it contradicts your later statement about the $Z^2$ dependence. The charge is relevant because it's a QED process, and the $Z^2$ dependence tells us that this vanishes to first order in QED for the neutron. However, one would expect some higher-order Feynman diagram to provide some nonvanishing rate. $\endgroup$
    – user4552
    May 23, 2019 at 19:02
  • $\begingroup$ Good point. I tried to find the relevant higher order correction, but I could only find corrections for large mass elements, and not for $Z=0$. I'll have to keep looking. Thanks $\endgroup$ May 23, 2019 at 23:09

If you think of pair production as the exchange of a photon between a recoil nucleus and either the electron or the positron in the final state, then that is not going to happen with a neutron which doesn't have electric charge for the photon to couple to.

There are two ways to convince oneself that pair production off neutrons nevertheless happens. The first stays close to the picture you probably have of pair production, the second goes a bit deeper into what these pictures mean or express:

1) the neutron is not an elementary particle, it is composed of quarks which are charged. The photon that balances the recoil momentum with the mass-shell requirement can interact with those quarks instead of the neutron as a whole. It is just very unlikely: the electrical field of the neutron is limited to its interior, whereas the field of a proton (in the usual case) extends to the whole atom whose nucleus it is contained in.

2) taking a step back, and more in line with the previous answers arguing based on unitarity (i.e. "anything that can happen, happens"): what is pair production? Above we imagined a possible Feynman diagram for pair production. But that is not what nature does. We don't know what nature does. All we do is observe the following: photon and neutron go in, positron, electron and neutron go out. What happened in between, we can't know. In Quantum Field Theory we use a so-called on-shell renormalization scheme which makes the objects that we use for calculation very similar to the objects we observe in the lab, and thus allows us to describe processes with great accuracy with little calculatory effort -- but nevertheless, we are still only looking at part of could happen in the process "photon + neutron in; e-, e+, n out". So in that sense the mental picture that I gave in 1) doesn't describe what happens in nature, not even in the case where the recoil is a proton. It's just a convenient approximation.

Now given that, we can let our phantasy roam and invent all kinds of intermediate processes that would lead to the observation, while being able to balance energy and momentum (i.e. the photon mass). E.g., the photon could split into an intermediate electron and positron, and the positron could then be absorbed by the neutron (or rather its constituents, but let's ignore that, hadronization is hard), which would now now in some state of charge 1, say a proton, but a $\Delta^ +$ resonance also fits the bill. Conservation of Lepton number (a fundamental law) requires simultaneous emission of a neutrino. The electron and the neutrino meet, become a $W^-$ boson, and the $W^-$ boson could radiate off a $Z$ boson before being absorbed by the proton, turning the proton back into a neutron. Finally the Z could decay into two electrons.

This process is infinitely unlikely. But since we only observed $\gamma, n$ going in, and $e^+, e^-, n$ going out, who is going to tell nature that this isn't what she did?

The intermediate processes that we imagine are just tools to guide us through the calculation, they do not describe what nature does.

  • $\begingroup$ While the neutron's electric field is on the inside, so can be the photon. We're still talking about excitation in one quantum field producing excitations in other fields. Mind, this would still make this less likely than the same thing with a proton, all other things equal, but I definitely don't have the apparatus to quantify this in any way :D $\endgroup$
    – Luaan
    May 24, 2019 at 7:14
  • $\begingroup$ Without wanting to think too much about this, very coarsely, the proton radius is 1/1000 of the hydrogen radius, so the cross-sectional area scales as 1/1000000. Thus the area where a photon would see the electrical field of a neutron (which has more or less the same size as a proton) is a millionth of that where it would see the EM field of the proton in a hydrogen atom. This estimate neglect just about everything, but that way it also avoids a lot of potential sources of errors. $\endgroup$
    – tobi_s
    May 24, 2019 at 9:10

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