When calculating the lowest order self energy corrections for an electron for example, feynman diagrams involving the emission and re-absorption of a (virtual) photon need to be considered, (as here for example: http://quantummechanics.ucsd.edu/ph130a/130_notes/node475.html )

but how is it possible for the emitted photon to be re-absorbed by the same electron? why hasn't the photon immediately raced away into the distance leaving the electron far behind before it can be re-absorbed? is this just a case of not taking these diagrams too literally?

Edit: After the electron emits the photon, both the particles should be considered as free particles and hence they cannot meet again unless the photon collides with some other particle and races back to be absorbed by the source electron. In other words, we need atleast a third vertex in the Feynman diagram for it to be possible for an electron to emit and absorb the same photon.

So, how can such a process be possible with just two vertex?


1 Answer 1


Indeed, do not take Feynman diagrams as literal representations of what is happening in a particle picture. Only the external lines of a diagram correspond to real particles - the internal lines, though called virtual particles, are little more than artifacts of the perturbative expansion we do to calculate QFT amplitudes, and there is little reason to ascribe to these virtual particles any kind of physical reality.

Instead of saying "the correction to the self-energy comes from the electron emitting and reabsorbing a virtual photon", it is perfectly fine to say that the self energy correction comes from interaction with the electromagnetic field, and is, to lowest order in the canonical perturbative expansion, represented by the Feynman graph you descibe. The talk about emitting/absorbing virtual particles is wholly superfluous.

  • $\begingroup$ But in perturbation theory, we have to assume the electrons and photons to be free particles (except, of course, at the vertex of the Feynman diagrams, where an interaction with other particles happen). So, if an electron emits and, after a while, absorbs the same photon, with no other interaction happening for both of them in between, then this must violate conservation of momentum, because we need a vertex in the Feynman diagram for the escaped photon to head back to the source electron. $\endgroup$
    – Prem
    Commented May 12, 2017 at 17:38
  • 3
    $\begingroup$ @Raja a virtual photon doesn't go anywhere so there's no need for it to "turn around and come back." $\endgroup$
    – Asher
    Commented May 12, 2017 at 21:25

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