I'm wondering if someone could explain me what happens to charge during electron-positron annihilation. For example if weak interaction (like electron + positron forming neutrino and antineutrino) occurs W boson transfers negative charge so charge conservation is not being violated. But in electron-positron annihilation with photon production (talking about low energies only) there's electromagnetic interaction (is it?) which means a photon being a virtual particle. So there's an exchange of a photon between electron and positron which results into production of two photons? How? In weak interaction there would be a W boson emission which resulted into charge change but due to what interaction does it happen in the case shown here? enter image description here

I understand that there's no violation because total charge is the same but I don't get how it happens?

Also in pair production as I could understand from here enter image description here there's an electron being a virtual particle that transfers momentum to the nucleus to absorb (if a wavy line going downwards really indicates this or I am completely off the right way; I'm wondering why is it presented as a photon?) and the system results into electron and positron as a charge was carried by a virtual electron. Can it be somehow nearly true? If so can I use the same principle with the annihilation and write on the first Feynman diagram that it's an electron which acts as a virtual particle?

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    $\begingroup$ Virtual particles are not real. (Those are links to answers of mine where I say the same things in more words) $\endgroup$
    – ACuriousMind
    Aug 9, 2015 at 12:32
  • $\begingroup$ Yes, the red line still is an electron going to the right (or a positron going to the left) by construction of feynman diagrams. Lepton number has to be annihilated too. They just have not labeled the virtual with an electron tag $\endgroup$
    – anna v
    Aug 9, 2015 at 12:36
  • $\begingroup$ @anna v So there's not only a photon that can be a virtual particle in electromagnetic interaction? Depending on whether charge changes or not a virtual particle can be represented by an electron? $\endgroup$
    – spfortray
    Aug 9, 2015 at 13:04
  • $\begingroup$ @ACuriousMind I do understand that but we imagine the exchange of those in purpose to represent the system changes. I know that they don't have to obey rules themselves being off mass shell however the point of them to be considered is to show particles interactions that do obey conservation rules. I'm sorry if I'm wrong, just trying to understand $\endgroup$
    – spfortray
    Aug 9, 2015 at 13:12
  • $\begingroup$ A virtual particle has all the attributes of a particle, i.e. the quantum numbers except they are off mass shell. They are a calculational symbol, very useful in translating feynman diagrams into integrals for calculating crossections etc. $\endgroup$
    – anna v
    Aug 9, 2015 at 19:28

2 Answers 2


Any particle can be a virtual particle, even an electron.

The incoming electron and positron are real, so they must be on shell. The virtual electron-positron can be off shell, which means different frames might even disagree about whether it is a virtual electron or a virtual positron .

But it is not really an electron or a positron. It is a stand in for a calculation. You write down a diagram and then consider all the ways energy and momentum can balance at a vertex and have the charge and other quantum numbers to be carried off to another vertex.

So in general you have to consider every possible energy and momentum for the internal lines that can be off shell and integrate the contribution from each.

And in general you also include every diagram to some number of vertexes. If you have just two vertices like in your picture above the answer won't be completely correct. If you have 3,4,5, or even 10 vertices you would still be incorrect but the idea is to be closer to the real value.

So the virtual electron isn't a real thing, it is a stand in on how how to do a perturbative expansion. But anything can be virtual, even electrons and positrons because anything can show up in the middle of a perturbative expansion.


For myself I created a model where exist two quanta e and p which form electrons and positrons too. The number of e-quanta equals the number of p-quanta. Unfortunately the quanta are so tiny that from the light spectra until now it is not possible to conclude about the energy of such quanta. The only difference between electrons and positrons is, that at the end of electrons field lines are sitting e-quanta and at the end of positrons field lines are sitting p-quanta. Because the amount of e- and p-quanta is equal in both particles the get transformed in photons completely.

Protons and anti-protons are made from this e-and p-quanta too, but in bigger amounts as in electrons and positrons. When an electron falls into nucleus the transformation of some e- and p- quanta into photons happens too, but due to the bigger amount in nucleus particles it stops in defined levels (electron orbitals).


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