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I'm trying to understand why the Faddeev-Popov ghost that appear in the quantization of non-abelian gauge theories are anti-commuting fields.

I've seen a number of books (chapters), lecture notes and tutorials regarding the topic but they all say something in the lines of: as is well known these fields are anti-commuting or these fields are unphysical because they violate the spin statistic theorem, but never actually prove that they are Grassmann fields.

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    $\begingroup$ Very short answer: they are anticommuting so that when doing the path integral over them you get the determinant you want instead of the inverse determinant. $\endgroup$ – Javier Aug 3 '16 at 19:12
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    $\begingroup$ Really what @Javier said; not much more to add. $\endgroup$ – gented Aug 3 '16 at 20:28
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They are fermionic by construction, so it is something that you can't prove, just accept. It is constructed in that way because is convenient for various reasons. The first one is that we can represent a path integral Jacobian for the gauge fixing in terms of a Gaussian integration. This is so because fermionic fields obeys:

$$ \int \prod_{\alpha} dc^{\alpha}\prod_{\beta}db_{\beta} \exp\left(b_{\beta\,}\Delta^{\beta}\,_{\alpha}c^{\alpha}\right)=\det\left(\Delta\right) $$

assuming no zero-modes for the fermionic fields.

The second one is that we can use the BRST construction where we have a nilpotent fermionic conserved charge $Q^2=0$ generating a symmetry of the path integral (assuming no anomaly), given by

$$ Q=c^{\alpha}G_{\alpha} - \frac{i}{2}b_{\alpha}f^{\alpha}\,_{\beta\gamma}c^{\beta}c^{\gamma} $$

where $G_{\alpha}$ is the generator of the gauge invariances and $f^{\alpha}\,_{\beta\gamma}$ is the structure constant of the Gauge Group. This BRST charge is useful to explore new "gauge fixings" as well as to obtain the gauge invariant spectrum in a covariant manner. New "gauge fixings" comes from adding terms in the action of the form $S\rightarrow S+Q(...)$. The physical spectrum could be worked out by looking at the representations of the cohomology of $Q$ that it is also covariant. The opposite statistic of the ghosts will lead to an erasing of the wrong polarizations that you introduce in order to work in a covariant manner.

All this holds because the ghosts has the opposite statistics of the gauge parameters.

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  • $\begingroup$ Can you briefly explain the proof that ghosts have the opposite statistics of the gauge parameters? For example, can you illustrate with Fadeev-Popov ghosts? $\endgroup$ – wilsonw Apr 17 at 10:32

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