This question is motivated by my recent foray into Quantum Field Theory. Just to make this clear straight off the bat; I am not suggesting in any way shape or form that the photon has a charge. I am well aware that the photon does not have a charge! I am not some sort of Crank trying to promote some hilarious 'not even wrong' theory. This is a purely theoretical question motivated by Quantum Chromodynamics.

With that dealt with I'll briefly discuss what led me to think about this question. I was reading about Quantum Chromodynamics. Part of what I found particularly interesting was that the force-carrier of the color-charge interaction in QCD is the gluon, a massless particle with zero electric charge. Unlike the photon however, which is the force-carrier of the electromagnetic interaction, the gluon itself has 'color' charge, whereas the photon itself lacks 'electric' charge. (I remember reading somewhere that this implies that the gluon participates in the strong interaction in addition to mediating it, which complicates matters compared to QCD where the photon mediates the electromagnetic interaction, but it does not directly participate in it).

So, with all that mind here's my question. How would the electromagnetic interaction as described by quantum-electrodynamics change if the photon itself was charged? I know that this is a pretty broad question so I'll try to narrow it down to two principle questions. 1.) How would the feynman diagrams for tree-level processes change in quantum-electrodynamics if the photon itself was charged? 2.) How would the effective coulomb potential change if the photon itself was charged (to tree-level)?

I apologize if this question is broad, I've really tried to narrow it down as much as I can. Thanks in advance for your answers!

  • 3
    $\begingroup$ QED is a $U(1)$ Abelian gauge theory, which means that the photon isn't charged. There's no natural way of making it charged. Simply writing a term in the Lagrangian will break gauge invariance and consequentially invalidate Ward-Takahashi identities and result in nonrenormalizability. $\endgroup$ May 3, 2017 at 6:15
  • $\begingroup$ Fair enough. Looks like the questions don't make sense in the context of quantum electrodynamics. Thanks for your response. $\endgroup$ May 4, 2017 at 3:33
  • $\begingroup$ Also see arxiv.org/abs/1209.2052, introduction, 3rd para. $\endgroup$ May 4, 2017 at 6:51
  • $\begingroup$ What would the charge of that photon be? +1 or -1? $\endgroup$
    – QuantumDot
    May 12, 2017 at 13:31
  • $\begingroup$ arxiv.org/abs/hep-ph/0505250 $\endgroup$ Jun 30, 2017 at 22:03

1 Answer 1


There is no way for the photon to be charged in QED since this would violate Unitarity. However, if you want to play around with photon-photon interactions a nice exercise is to use a non-linear gauge-fixing function. This results in a gauge-fixed action that appear to contain photon self-interaction terms, but which of course cancel out in the end.


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