Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I read that a particle will meet its antiparticle and annihilate to generate a photon. Is it important for the pairs to be of the same type? What will happen when for example a neutron meets an antiproton or a proton meets a positron? Are there any rules to determine what happens when such particles meet?

share|improve this question
More on matter and anti-matter –  voix Sep 11 '11 at 12:28
add comment

3 Answers

up vote 3 down vote accepted

Yes, there are rules that depend on the quantum numbers carried by the particles under question and the energy available for the interaction.

In general we label as annihilation when particle meets antiparticle because all the characterising quantum numbers are equal and opposite in sign and add and become 0, allowing for the decay into two photons, two because you need momentum conservation.

A positron meeting a proton will be repulsed by the electromagnetic interaction, unless it has very high energy and can interact with the quarks inside the proton, according to the rules of the standard model interactions.

When a neutron meets an antiproton the only quantum number that is not equal and opposite is the charge, so we cannot have annihilation to just photons, but the constituent antiquarks of the antiproton will annihilate with some of the quarks in the neutron there will no longer be any baryons, just mesons and photons, and all these interactions are given by the rules and crossections of the standard model.

share|improve this answer
I'm having trouble seeing why a reaction like $\mu^{+}e^{-}\rightarrow \gamma \gamma$ is disallowed by an argument like this, but I"m also having trouble generating a diagram that does this that doesn't have a ton of vertices involving annhilating neutrinos. –  Jerry Schirmer Sep 11 '11 at 17:06
@Jerry that violates the lepton flavor conservation, so you need those weak verticies to make it go ahead. It's got to be massively suppressed. –  dmckee Sep 11 '11 at 17:24
@Jerry that is why we have nu_tau , nu_mu and nu_e neutrinos to keep the flavour count. There are oscillations though, but that is another story. –  anna v Sep 11 '11 at 18:58
add comment

The major problem you run into with this question is the protons and neutrons are not fundamental particles.

The short answer is that $x + \bar{y}$ does not result in any annihilation at the vertex level (but sometimes other reactions are possible), but $n + \bar{p}$ is not expressed at the vertex level.

Instead a neutron is a composite object made up of two (matter) down quarks and one (matter) up quark and a seething mess of virtual particles (often called "the sea") that pop into and out-of existence all the time.

A anti-proton is made up of two anti-up quarks and one anti-down quark and another seething mess of virtual particles.

So if a neutron meets and anti-neutron ($(udd+ \text{sea}) + (\bar{u}\bar{u}\bar{d} + \text{sea}) \to \text{??}$) you get reactions at the quark level, and the two composite particles can be destroyed leaving zero baryons but a lot of hadronic spray.

share|improve this answer
add comment

The only thing that a particle and an antiparticle can do always for sure in theory is annihilate into gravitons. Everything else depends on the particle.

The reason people talk about particle/antiparticle annhilation is to convey that antiparticles are particles going backward in time, in an S-matrix particle-path picture, like Feynman diagrams. If you have interactions that can knock a particle sideways, then these interactions can also knock a particle back in time, so that any external interaction can produce particle/antiparticle pairs, and can lead to creation and annihilation of pairs. If the external potential can absorb the particles, if they aren't protected by conservation laws, then it can produce the particle and antiparticle singly.

There are no conservation laws which can forbid a particle from annihilating with its antiparticle, because gravity can always knock a particle back in time.

But some particles are hardly interacting, externally or internally. These particles don't annihilate with their antiparticle, they just ignore it. A neutrino is its own antiparticle, but two stopped neutrinos will just sit there next to each other, wavefunctions spreading, doing nothing much at all. Their cross section for annihilating into anything is negligible. Same with two photons, or two gravitons.

Charged particle/antiparticle pairs can annihilate into photons, usually two or three depending on the conservation laws, but not all particles are charged. neutrons and antineutrons have a hard time finding each other, and I don't think its fair to say that they annihilate. When they collide, they can in theory turn into photons, but they almost always will just go past each other, or, if they collide, they turn into mesons.

The most extreme example of annihilation is at the highest energies, a massive charged spinning black hole. The antiparticle has opposite charge. Two such black holes collide to make a big neutral probably spinning black hole, which then slowly decays.

share|improve this answer
"A neutrino is its own antiparticle" This may be true in the favored theories, but it is not settled. We'll get back to you in a few years... –  dmckee Sep 12 '11 at 17:36
@dmckee: the fact that there are irresponsible theorists in the world should not change the story we tell ourselves. That neutrinos are one chirality only was established by Feynman,Gell-Mann, Sudarshan, and Marshak in the 1950s! This predicts the neutrino masses order of magnitude right. It is impossible that neutrinos are Dirac and somebody has to say it. That does not mean that experimentalists should not verify that it is Majorana, not Dirac, but the theorists should not be dithering on this obvious fact. To have a Dirac neutrino you need a light sterile neutrino, absurd fine-tuning. –  Ron Maimon Sep 12 '11 at 20:32
Ok--- now I know it's not just Georg. What's the problem here? The answer is correct. –  Ron Maimon Oct 23 '11 at 18:57
+1: You are correct but I think people are downvoting because you are leaving out any discussion of the iconic particle-antiparticle annihilation of an electron-positron annihilation into 2 photons. This has very high cross section and extremely low cross section into anything else ( such as neutrino + antineutrino ). You also put down weak annihilation but graviton annihilation would have a lifetime of much greater than the age of the universe ( I'm guessing ). –  FrankH Oct 23 '11 at 21:56
@FrankH: yes, you are right, but one downvote is just reflexive. The point of this is to answer the original question about why people talk about annihilation with regards to particle/antiparticle at all. The reason is to explain that any particle path that can go forward in time can double-back to a particle antiparticle annihilation, but the only universal thing that can go at the turnaround point is a graviton. –  Ron Maimon Oct 24 '11 at 4:01
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.