• A charged particle (e.g. proton) meeting with its own charged antiparticle (e.g., antiproton) annihilates and energy is given off in the form of radiation. Do we have any clue why this happens?

  • Does a neutral particle (e.g. neutron, neutrino etc) meeting with its neutral antiparticle (e.g. antineutron, antineutrino) also annihilate and produce radiation? If not, why?

  • What does a proton meeting with an antineutron do? Since the system has a net nonzero charge, this upon contact cannot annihilate to radiation.

  • 1
    $\begingroup$ I've answered some of your questions here: physics.stackexchange.com/a/451337/123208 $\endgroup$
    – PM 2Ring
    Commented Feb 3, 2020 at 12:45
  • $\begingroup$ @PM2Ring Thanks for the answer. I did give a look at it. But I was looking for a more field-theory kind of explanation as given by yu-v. $\endgroup$ Commented Feb 3, 2020 at 12:57
  • $\begingroup$ No worries. As I said in that answer, the cross-section for neutrino + antineutrino annihilation is really tiny, since they annihilate to Z bosons, which have a huge mass. The Z boson has a half-life of around $10^{-25}$ seconds, and it can decay into all sorts of things, as explained here: en.wikipedia.org/wiki/W_and_Z_bosons#Z_bosons_2 $\endgroup$
    – PM 2Ring
    Commented Feb 3, 2020 at 13:03
  • $\begingroup$ note that your 3rd bullet point gives insight into the 1st two: "because they can"...if no conservation laws are violated. $\endgroup$
    – JEB
    Commented Feb 3, 2020 at 15:28

2 Answers 2


They annihilate because in the Lagrangian that describes their behavior there is a term $\propto -i e A \Psi \bar{\Psi}$ where $\Psi$ is the matter field, $\bar{\Psi}$ its conjugate, which describes the anti-matter field, and $A$ the gauge field, for example photons in quantum electrodynamics. This means that there is a term which allows for a particle and an anti-particle to "come in" and a photon to "come out". In reality, two massive (anti/)matter particles cannot annihilate to a single massless gauge particle as you cannot place this process on the mass-energy shell, and two massive (anti/)matter particles will annihilate to at least two massless gauge particles.

If you consider neutrons as elementary particles, that is - you write a field theory that contains them as fundamental excitations, and only consider the electromagentic field as a gauge field, then neutrons will not annihilate as they do not couple to it. If, however, you consider them as composite particles composed of quarcks, then anti-neutron can annihilate with a neutron, via the coupling to the color gauge field (the strong interaction). Then, of course, the neutron is no longer a neutral particle, as its individual quark components will have charge under the strong interaction.

  • $\begingroup$ If we have a case where a neutral elementary fermion and its antiparticle meet, wouldn't they annihilate? For example, a neutrino and an antineutrino? $\endgroup$ Commented Feb 3, 2020 at 12:46
  • $\begingroup$ neutrinos have a charge under the weak interaction (what's called the "weak isospin") $\endgroup$
    – user245141
    Commented Feb 3, 2020 at 12:51
  • $\begingroup$ Sure. But to couple to photons, they need to have the electric charge. Since neutrinos are electrically neutral, the meeting of a neutrino-antineutrino pair does not annihilation to photons. Is it wrong? $\endgroup$ Commented Feb 3, 2020 at 12:54
  • $\begingroup$ @mithusengupta123 You can get some photons eventually (plus other stuff), from neutrino + antineutrino annihilation, after the Z boson(s) have decayed. But you're correct that simple annihilation to photons doesn't happen. If it did, we'd get low energy (visible) photons from the antineutrinos of beta minus decay annihilating with solar & cosmic neutrinos, but the probability of that happening is ridiculously tiny. $\endgroup$
    – PM 2Ring
    Commented Feb 3, 2020 at 13:15
  • $\begingroup$ they cannot annihilate to photons directly. only through some more complex higher-order process $\endgroup$
    – user245141
    Commented Feb 3, 2020 at 13:16

The concept of annihilation in particle physics depends on quantum mechanics, special relativity and the quantum numbers axiomatically assigned to the particles in the standard model of particle physics, in addition to energy conservation.

Special relativity assures that all particles have intrinsic energy, and are described by four vectors . The antiparticles have the same mass as the particles, and when they meet their quantum numbers add up to zero ( that is what annihilation means) and depending on their kinetic energy there is a high probability that new particle antiparticle pairs will come out, in addition to photons, i.e.electromagnetic radiation.

Particle meeting an antiparticle is a quantum mechanical interaction and there are calculable probabilities of the output. Depending on the energy carried by the input pair, they might scatter off, might create more particles remaining themselves intact, or annihilate into a multitude of different particle-antiparticle pairs .

A good example is electron positron scattering. For low energy either they are elastically scattered or the quantum numbers (+1 lepton nunber for electron-1 for positron) add to zero and two photons ( for momentum conservation in the center of mass) come out. For high energy in the incoming particles, as at LEP a large number of particles can come out, as long as the quantum numbers add up to zero. The standard model was checked and validated by the data coming from LEP.

When a proton meets and antineutron, one has to conserve charge, the other quantum numbers add up to zero. The possible output particles must have charges summing up to +1. As protons and neutrons are composed of triplets of quarks , the particle antiparticle pairs will have a probability to rearrange so that the total charge is +1. An antiup with a down quark make a pi+, and the rest a pi0 , if there is more energy a lot more pairs.

The important thing to retain is that it is quantum numbers that annihilate and the"radiation" depending on probabilities of scattering and the energies involved can be a lot more particle pairs.


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