Faddeev-Popov ghosts are introduced in the quantization of Yang-Mills theory to absorb the Faddeev-Popov determinant into the action, $$\det \Delta_{\text{FP}} = \int \mathcal{D} \bar{c} \mathcal{D} c \, e^{i \int dx \, \bar{c}^a(x) (\Delta_{\text{FP}} c(x))^a}.$$ Here, $c$ is a Lorentz scalar, and it must be fermionic, as if we used a bosonic variable $\psi$ instead, we would get the determinant in the denominator rather than the numerator, $$\frac{1}{\det \Delta_{\text{FP}}} \propto \int \mathcal{D} \bar{\psi} \mathcal{D} \psi \, e^{i \int dx \, \bar{\psi}^a(x) (\Delta_{\text{FP}} \psi(x))^a}.$$ So we have to accept a violation of spin-statistics to quantize Yang-Mills.

But a friend of mine came up with a simple alternative: simply note that $$\det \Delta_{\text{FP}} = \frac{1}{\det \Delta_{\text{FP}}^{-1}} \propto \int \mathcal{D} \bar{\psi} \mathcal{D} \psi \, e^{i \int dx \, \bar{\psi}^a(x) (\Delta_{\text{FP}}^{-1} \psi(x))^a}.$$ Note that the inverse exists because the determinant is nonzero; if the determinant were zero the whole path integral would be zero and we wouldn't be able to do anything, ghosts or not.

Is there anything wrong with this method? Does it lead to some complications down the line? If not, why are the spin-statistics violating Faddeev-Popov ghosts typically used instead, when this setup looks much nicer?

  • 2
    $\begingroup$ Formally, this is correct (a simliar method, called pseudo-fermions, is sometimes used to treat fermions in lattice gauge theory). However, you end up with a non-local action, which is difficult to treat in perturbation theory. $\endgroup$
    – Thomas
    Apr 30, 2018 at 17:50
  • 4
    $\begingroup$ Also please note (and I cannot overemphasize this) that ghosts do not violate spin-statistics... Spin-statistics is a statement about the physical Hilbert space. Ghosts have no effect on this space whatsoever (the no-ghost theorem), as can be shown by computing the asymptotic BRST operator cohomology. Ghosts are mathematical tools, and the only reason we even consider them is that they prove convenient in practical computations. $\endgroup$ May 2, 2018 at 1:58

1 Answer 1


Note that the inverse of $\Delta_{FP}$ is the Green function $G_{FP}(x,y)$ obeying:

$$ \Delta_{FP}|_x G_{FP}(x,y)=\delta(x-y) $$

so, the action that you wrote down is non-local, should be of the form:

$$ S_{FP}^{bosonic}=i\int dx \int dy\, \bar{\psi}^{a}(x)G_{FP}(x,y) \psi^{a}(y) $$

this action is non-local, since we cannot get away with just one integral in space-time. Since we are dealing with local theories, the presence of non-local interactions make the computations very complicated with lots of cancellations that are obscure in this formalism. Those cancellations should happen since the theory is local, so all the non-local terms must conspire to produce a local theory at the very end.

Also, it is better to have the ghosts as fermions since they will shows up in loops with the opposite sign of the pure gauge loops, erasing them. In this formalism of bosonic ghosts is not clear how this cancellation will be happening, but of course will be happen since there is no inconsistency over here. There will be extra cancellations that are obscure in this formalism.

  • $\begingroup$ what is the trouble with nonlocality? I think nowadays nonlocality is fairly accepted as a part of a consistent quantum unitary theories. Maybe the trouble is that it is hard to observe? $\endgroup$
    – lurscher
    Apr 30, 2018 at 18:51
  • $\begingroup$ @lurscher well, the main trouble is that it is cumbersome to deal with, in practical terms. Calculating Feynman diagrams with non-local interactions is a pain in the neck; and proving renormalisability much more. $\endgroup$ Apr 30, 2018 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.