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I was thinking of this previous question of mine, where I was trying to implement a path-integral over the half-line: $$ Z=\int_{\varphi\ge0}\ \mathrm e^{iS[\varphi]}\mathrm d\varphi\tag1 $$

It seems to me that a possible approach is to integrate over all $\varphi$, but to regard the configurations $\varphi$ and $-\varphi$ as physically equivalent. In other words, we take the $\mathrm d\varphi$ to be unconstrained, but we introduce the gauge equivalence $\varphi\to-\varphi$; the orbits are of the form $\{\pm\varphi\}$, and a representative of each orbit is, for example, $+\varphi$. Integrating over one representative for each orbit brings us back to $(1)$.

Instead of eliminating the gauge redundancy explicitly (which would lead to the integral I don't know how to evaluate, $(1)$), we leave the gauge symmetry as is, but treat the resulting path-integral using the standard methods of gauge-theory. The problem is that I don't know how to implement the Faddeev-Popov trick for $\mathbb Z_2$-valued gauge fields or, more generally, fields over a discrete group.

Has this problem been analysed? (How) Can we implement the Faddeev-Popov trick for system with discrete gauge symmetries?

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    $\begingroup$ Sorry, I am lost, why do you think that Faddeev-Popov eliminates a redundancy, in particular a scalar field? $\endgroup$ – user178876 Jun 20 '18 at 1:41
  • $\begingroup$ @marmot It doesn't eliminate the scalar field, it eliminates the redundancy in $\varphi\sim-\varphi$. Just like in standard gauge theory, where the redundancy is $\phi(x)\sim \mathrm e^{i\lambda(x)}\phi(x)$ (in the case of $\mathrm U(1)$, with $\phi(x)$ complex). $\endgroup$ – AccidentalFourierTransform Jun 20 '18 at 1:43
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    $\begingroup$ This is completely new to me. As far as I know, the Faddeev-Popov ghosts exist already in a pure gauge theory, and are used to fix a gauge, i.e. to make the Klein-Gordon operator for gauge fields invertible. Why do you think they eliminate the phase of the scalar field? What if you do not gauge the theory? $\endgroup$ – user178876 Jun 20 '18 at 1:49
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    $\begingroup$ Well, if you are sure about this, why don't you just see what happens if you get the discrete gauge symmetry in the good old-fashioned way, namely by breaking a U(1) symmetry by giving a scalar $\Phi$ of charge 2 a VEV and look at a downstairs theory with a scalar $\phi$ that had charge 1 under the U(1). It is straightforward to see that the Faddeev-Popov ghosts do nothing of the sort you have in mind (but they also did not in the upstairs theory). $\endgroup$ – user178876 Jun 20 '18 at 2:02
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    $\begingroup$ Ah, one more thing, when you are talking about discrete gauge symmetries, you are taking them to be those of arXiv:1710.01791, right? $\endgroup$ – user178876 Jun 20 '18 at 2:04
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The answer is no. This has, however, nothing to do with the symmetry being discrete. Rather, it is simply the statement that Faddee-Popov ghosts never eliminate a redundancy of charged "matter" fields.

(There seems also to be some confusion on what gauge fixing does and what the ghosts do. From some text books one could get the impression that one has to eliminate the gauge redundancy in order to define the path integral. This is, however, incorrect. The thing that goes wrong on the gauge fields is that the Klein-Gordon operator has a nontrivial kernel and is hence not invertible. This is because if you apply a gauge transformation on a configuration in the kernel you get "another" configuration in the kernel. This is different from the situation of the matter fields, where the corresponding Klein-Gordon or Dirac operators have covariant derivatives in such that the above problem does not arise. What the gauge fixing and ghosts do for you is to make the Klein-Gordon operator for the gauge fields invertible on the set of the restricted gauge field configurations. That's why you get the gauge fixing parameter in the gauge field propagators only. That's it. They do not kill off phases of the matter fields in any way.)

Literature on Yang-Mills ghosts: I did not find any fully self-contained and clear treatment of these issues in freely available pdf files but something close to it in Timo Weigand notes. However, the discussion in Pokorski's book on "Gauge Field Theories" is IMHO clearer, yet not publicly available. Unfortunately, the otherwise great notes by Srednicki are not 100% helpful at this point.

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    $\begingroup$ Please explain. If you consider Yang-Mills with charges matter, the redundancy is “spread” across gauge fields and matter, because both participate in gauge transformations. Seems to me that it contradicts your last sentence. $\endgroup$ – Prof. Legolasov Jun 20 '18 at 16:55
  • $\begingroup$ @SolenodonParadoxus I don't think so. Please have a look at the book by Pokorski on what gauge fixing does and what the ghosts do. $\endgroup$ – user178876 Jun 20 '18 at 16:57
  • $\begingroup$ Also as a counter example consider superstring’s ghost system (whatever they call it) where there’s bosonic and fermionic ghosts canceling respectively fermionic and bosonic modes. $\endgroup$ – Prof. Legolasov Jun 20 '18 at 16:58
  • $\begingroup$ that’s not a fair argument. Either point out where I am mistaken, or at least give a reference to the exact paragraph (and if possible, a pdf link). $\endgroup$ – Prof. Legolasov Jun 20 '18 at 17:00
  • $\begingroup$ @SolenodonParadoxus I added a longer explanation on what the ghosts do and on what they don't. A very clear discussion can be found for instance in Pokorski's book on "Gauge Field Theories" (Cambridge Monographs on Mathematical Physics), for which I do not have a pdf for obvious reasons. Of course, these statements can be found in other text books as well, I just don't remember which ones. $\endgroup$ – user178876 Jun 20 '18 at 17:09

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