I am wondering, in patricle physics, when we have an annihilation vertex, we always have a particle and an antiparticle annihilating. Why is that ?

Let me make an example to be clearer. Let's take a weak interaction vertex. We know an electron and an electronic antineutrino can annihilate each other on a weak interaction vertex to give a W- boson. Why is this not possible with an electron and an electronic neutrino ?

I understand in some cases it is forbidden by conservation of charge, that is why an electron and positron can annihilate but not 2 electrons. But in the example above, I fail to see which conservation is violated, if there is one. Or is this annihilation rule an experimental postulate ?

  • $\begingroup$ Lepton number would be violated. $\endgroup$ Dec 27, 2016 at 3:27
  • $\begingroup$ Lepton number conservation would be violated. $\endgroup$ Dec 27, 2016 at 3:27
  • $\begingroup$ But it seems to me that this lepton number is introduced somewhat artificially ? Is there a deeper reason ? Electric charge is for example observed by its effect on the QED interaction. Does lepton number arise anywhere else than this conservation ? $\endgroup$
    – Frotaur
    Dec 27, 2016 at 3:45
  • $\begingroup$ @Frotaur you can't write down a Lorentz-invariant and gauge-invariant QFT Lagrangian which would give you such an interaction vertex. This is the deeper reason. $\endgroup$ Dec 27, 2016 at 3:51
  • $\begingroup$ Here's the usual problem with why questions: we can only answer them if we have deeper theory than the one you are asking. In point of fact it is an observation about the world that matter does not self-annihilate and that matter/anti-matter combinations do. We encode this fact in the notion of conserved quantum numbers for various families of particles: lepton number, baryon number (and some approximately-but-not-quite-conserved numbers like strangeness number and lepton flavor numbers). For the moment that is the bottom answer in the chain of "Why?" questions. $\endgroup$ Dec 27, 2016 at 20:11

2 Answers 2


One has to realize that words in physics have a definite meaning. This meaning, for words used in physics, is dependent on the specific mathematical model that uses it.

The mathematical model for particle physics is called the standard model


All matter and energy emerges from interactions and composites of these particle. The model encapsulates the large number of particle data measured over the last half century.

In all particle interactions , particles may transmute at the vertices changing quantum numbers and identity as defined in the table. Changing identity is not called annihilation. The basic definition of annihilation as used in particle physics is that after the interaction all the quantum numbers of the initiating particles add up to zero. The particles coming out of the vertex add up to zero, and thus new pairs can appear with different quantum numbers.

Thus we get, as an example, e+e- annihilating into a quark antiquark pair, which have completely different quantum numbers than the electron positron pair, but their addition adds up to zero.

In your example an electron_antineutrino scattering on an electron will make the lepton number zero but not the charge, so it is not within the definition of "annihilation". It is merely an interaction.

Of course an electron_neutrino on an electron gives a lepton number of 2, and it can only be a scattering.

Searching for Feynman diagrams I did find this:


A Feynman diagram representing the annihilation of an electron neutrino and a positron to a muon neutrino and a muon.

So the author does not include charge zero within the definition of annihilation so one might find such a usage of the word, but it is not mainstream and it should be noted that it is from an astrophysics course, not a particle physics :).


Lepton number (which is negative for ant-leptons) - if they still use that, what with neutrino oscilations. Chirality is also largely conserved.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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