In a Feynman diagram, the particles behave like free particles everywhere except at one of the vertex of the diagram. The vertex is the site for an interaction to occur.

So, if we have an electron which emits and later absorbs the same photon, we already have 2 vertex in the diagram. But i think that a third vertex is necessary for the electron to change its direction of motion and catch up with the photon.

This is because being free particles, after emission, the electron and the photon will keep on moving away from each other. So, we need a third vertex to scatter the electron and help it catch up with the virtual photon. If the photon returns on its own, this will violate the law of conservation of momentum.

But the funny thing is- I have seen Feynman himself draw the Feynman diagram i describe with just 2 vertex in a QED lecture video.

So, do i have some misconceptions regarding Feynman diagrams? Please help!

  • 2
    $\begingroup$ Internal lines (those that run from vertex to vertex) are not the same as free particles even when they are labeled as a particle. You can't reason about that by saying "that's light so ..." because it is 'off-shell' or 'virtual'. $\endgroup$ May 12, 2017 at 18:33
  • $\begingroup$ So, two vertex Feynman diagram for an electron emitting and absorbing the same photon is possible then, and would not violate the momentum conservation law? $\endgroup$ May 12, 2017 at 18:52
  • $\begingroup$ It is a standard diagram leading to one of the first big mathematical difficulties of QFTs. Search terms: "bare mass" and "dressed mass". $\endgroup$ May 12, 2017 at 18:55

1 Answer 1


Feynman diagrams are an iconal representation of the terms entering a perturbative expansion when one wants to calculate a cross section or a decay .

They are not simple but a summation of terms with smaller and smaller contribution, for a calculation that makes sense. The usualy seen Feynman diagrams are the first order ones. Look at the complexity of the higher order diagrams that have to be included for accurate predictions.

The DYSON equation can be achieved by classifying the various contributions in arbitrary FEYNMAN diagrams. DYSON's equation summarizes the FEYNMAN-DYSON perturbation theory in a particularly compact form.


Figure 3.5: The GREEN's function expanded in terms of connected diagrams

If you read the link you will see that the diagram you describe with two vertices is included in the possible diagrams. For a single electron and a photon loop it is the self energy that is taken into account by renormalization. The energy and momentum are conserved at the two vertices because the photon is virtual, i.e. off mass shell, so it does not "go away" as you think.


This first order diagram of e- e- scattering clears the meaning of " virtual" :


  • $\begingroup$ Thank you. But I have a doubt. According to the Feynman rules, each internal line for a virtual photon contributes a factor $\frac{ig_{uv}}{q^{2}}$ in the amplitude, where $q$ is the momentum of the virtual photon. But if a virtual photon is able to return to the source electron, then what do we mean by a momentum associated with it? Surely this momentum is a variable and for some reason, the virtual particles can experience forces and hence are not free. Or alternatively, does this means that Feynman rules are a set of abstract mathematical procedures with no simple physical interpretation. $\endgroup$ May 12, 2017 at 19:59
  • $\begingroup$ Of course the momentum is variable in internal lines. Internal lines are under integration always, and the variables are connected with the four vector of the internal line. That is why the denoted particle, photon in your case, is off mass shell. They are a set of rules which make sense in quantum number conservation as far as internal lines go. That is why they are identified with a particle, they have the charge, strangeness baryon number etc of the named particle. $\endgroup$
    – anna v
    May 13, 2017 at 3:08
  • $\begingroup$ By the momentum of the internal line being a variable, i meant them being a variable with respect to time for a free particle. I think that the simplicity of the Feynman diagram lies in the fact that any change in the momentum of a particle occurs only at the vertex of the diagram, in which momentum is conserved. After a particle leaves the vertex, it's momentum becomes constant until it approaches another vertex. In other words, The particle behaves as if it is free. I still think that this is the case. The real reason why i think i am having trouble with this concept is that i am taking... $\endgroup$ May 13, 2017 at 3:20
  • $\begingroup$ (Cont.) the off mass shell momentum of the virtual particles too literally. I now think that a virtual particle with $x$ momentum along a direction is not moving in that direction in a literal sense. $\endgroup$ May 13, 2017 at 3:24
  • $\begingroup$ even if mathematically it can be considered to be moving, the value changes depending on the integral denoted by the diagram. $\endgroup$
    – anna v
    May 13, 2017 at 5:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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