Does it make sense to introduce Faddeev–Popov ghost fields for abelian gauge field theories?

Wikipedia says the coupling term in the Lagrangian "doesn't have any effect", but I don't really know what that means. If it doesn't work at all (probably because structure constants are zero?) then why doesn't it work, physically? I mean you still have to do things like getting rid of unphysical degrees of freedom/gauge fixing.

Since it isn't done in QED afaik, I guess its not a reasonable way of doing things like computation of self energy/computing corrections of propagators. I wondered, because the gauge field-ghost-ghost vertices for Yang-Mills basically look just like the gauge field-fermion-fermion vertex in electrodynamics.

  • $\begingroup$ You can make photons and ghosts "interact" if you want to, e.g. by choosing a non-linear gauge fixing function like $\partial_\mu A^\mu+A_\mu A^\mu$. The standard choice $\partial_\mu A^\mu$ just produces a $\bar c\square c$ term, which is just a normalisation constant after the Grassman integration. $\endgroup$ Jul 3, 2021 at 4:34

1 Answer 1


I think you are misinterpreting the statement that "it doesn't have any effect". This statement doesn't mean that the Faddeev-Popov methodology "doesn't work", as you wrote later. Instead, it means that it is completely unnecessary.

If you look at the Faddeev-Popov ghosts' Lagrangian, you will see that for Abelian groups, the structure constants $f_{abc}$ vanish and we're left with $$ {\mathcal L}_{\rm ghost} = \partial_\mu \bar c^a \partial^\mu c^a $$ which means that the ghosts are completely decoupled. They don't interact with the gauge fields (photons). You may still use the Faddeev-Popov machinery and the BRST formalism based upon it to identify the physical states as the cohomologies of $Q$, the BRST operator.

But what this BRST machinery tells you is something you may easily describe without any Faddeev-Popov ghosts, too. It just tells you that the excitations of $\bar c,c$ are unphysical much like the excitations of time-like and longitudinal photons. That's why the BRST problem in the case of Abelinan gauge groups is "solvable" in such a way that you may simply eliminate the ghosts completely, together with 2 unphysical polarizations of the photon. And that's why QED may be taught without any Faddeev-Popov ghosts and one may still construct nice Feynman rules for any multiloop diagrams.

For non-Abelian theories, the counting still works – ghosts, antighosts, and two polarizations of gluons etc. are unphysical. However, because there are interactions of ghosts with the gluons in that case, there's no easy way to describe the physical states without the Faddeev-Popov ghosts.

  • $\begingroup$ Okay, so if you do introduce these unnecessary ghosts, I guess you still have to do the gauge fixing then? $\endgroup$
    – Nikolaj-K
    Jan 29, 2012 at 18:22
  • $\begingroup$ If you want specific states that one encounters in gauge-fixing, you have to be gauge-fixing. But the simplification in Abelian theories is that there won't actually be any loop diagrams with FP ghosts, no nontrivial Jacobians from the gauge fixing etc. Those things are non-issues for U(1) which is why the FP ghosts are redundant in this case. $\endgroup$ Jan 29, 2012 at 18:48
  • $\begingroup$ Let me also mention that it is not true that gauge-fixing is "necessary", not even in non-Abelian gauge theories. The physical states upon the vacuum are cohomologies of the BRST operators. These cohomologies form a specific Hilbert space and to describe this Hilbert space, one doesn't have to gauge fix in any specific way. Gauge fixing is more "paramount" to describe classical backgrounds etc. but not physical multiparticle states. $\endgroup$ Jan 30, 2012 at 17:03
  • $\begingroup$ @Motl: Okay, but I guess constructing this state of cohomologies, which sounds like building a factor space of the naive Hilbert space, might require some work. $\endgroup$
    – Nikolaj-K
    Jan 30, 2012 at 17:07

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