We all know Faddeev-Popov ghosts are needed in manifestly Lorentz covariant nonabelian quantum gauge theories. We also all know they decouple from the rest of matter asymptotically, although they "superficially" interact over finite time periods.

  1. So, what is their ontological status?

  2. If they don't really exist, can we just formulate a local theory without them?

  3. OK, we can using spin foams?

  4. So are they not really needed?

  5. If they do "exist" out there, then say a state with an electron and no ghosts is "actually" different in "reality" from the state with same electron with the same momentum, but with a ghost added?

  6. According to the formalism, they are different states. But observationally, we can never tell the difference. Both states will always give the same observable results for any physical observation. So are they the same state or not?

  7. Do we just take the partial trace over the ghost sector?

  8. But that's not gauge invariant except asymptotically, is it?

  • 10
    $\begingroup$ The ghosts are not present in all gauges, you don't have an invariant meaning to the statement "an electron plus a ghost added". You need to say "an electron plus a ghost in such and so gauge" and then this is an intermediate state which always comes with "an electron plus a gauge particle in the same gauge" and the ghost just cancels out the unphysical part of the gauge field. $\endgroup$
    – Ron Maimon
    Jul 5, 2012 at 15:45
  • $\begingroup$ In the lattice approach to QFT, Fadeev-Popov ghosts never appear. They are an artifact of a particular quantization method, and as such have no ontological status. $\endgroup$
    – user1504
    Jun 4, 2020 at 16:02

2 Answers 2


1,3&4. If you fixed a gauge and got rid of unphysical degrees of freedom (sacrificing manifest gauge invariance), you would not have to muck around with ghosts. You need to come up with them in order to get rid of the unphysical modes and still preserve manifest gauge invariance. You take a call on what that means ontologically; as far as physics is concerned, that doesn't matter.

5&6. As @RonMaimon says, ghosts are gauge dependent. It doesn't make sense to compare a state in one gauge (without ghost) to the same state in another gauge (with ghost). Remember, a gauge is kind of like the setup in which you "measure" the gauge field. And no physical observable will depend on your gauge choice.

7&8. We don't need to take any partial trace. The ghosts decouple honest-to-god. So you might as well not bother with them. If they did not decouple completely, qft would start behaving very weirdly, to put it mildly.

q2. Huh! Where did spin foams enter this discussion?


I'd say BRST ghosts have more elements of reality compared to longitudinal gauge bosons which are pure gauge. Look at the inner product structure. Physical states belong to the BRST cohomology of BRST-closed modulo BRST-exact. pure gauge longitudinal gauge bosons are BRST-exact and have zero inner product with any other BRST-closed state. BRST ghosts have nonzero inner products with themselves and other BRST-closed states. BRST ghost states are merely BRST-closed. BRST-closed states with different ghost numbers have zero inner product between themselves, even while having nonzero norms.

The BRST ghost structure is independent of the choice of gauge-fixing. Different choices of gauge fixings correspond to different extended Hamiltonians but these Hamiltonians can only differ by BRST-exact quantities. So, different gauges only differ by BRST-exact differences to be quotiented over.

  • $\begingroup$ To the last paragraph: so is it possible to get rid of ghosts completely by some gauge fixing? $\endgroup$
    – firtree
    May 9, 2013 at 5:53

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