The Faddeev-Popov approach to make the generating functional corresponding to free gluon fields well defined, introduces two independent Grassmann fields. Since these are scalar, their quanta can be interpreted as spin-0 fermions, which means they cannot correspond to physical states.

The complete derivation of the method is at the moment a bit outside of my mathematical ability, but I am still curious about the motivation of using Grassmann fields in this procedure. Is a similar approach with commuting fields (as opposed to the anti-commuting properties of the Grassmann fields) impossible or very difficult? On the surface of it, such an approach would make more sense to me.

Are results that incorporates fields such as these annoying to theorists? Or is there no reason to care as long as it does not impact the model's ability to make predictions of observables?


2 Answers 2


Well, one argument goes as follows:

  1. The path integral $Z$ for a gauge theory needs gauge-fixing, typically a product of Dirac delta distributions: $\prod_{x,a} \delta(G^a(x))$.

  2. To make the path integral independent of gauge-fixing, there must also be a Faddeev-Popov (FP) determinant.

  3. We would like the FP determinant to be the result of Gaussian integrations.

  4. Unfortunately Grassmann-even Gaussian integrations produce inverse/reciprocal determinants. However Grassmann-odd Gaussian integrations do the job.


  1. M. Srednicki, QFT, 2007; Chapter 71. A prepublication draft PDF file is available here.
  • $\begingroup$ The first three steps are usually well motivated and explained in the texts I have read. I suppose what seems strange is the choice of Grassmann-odd Gaussian integrations over the even ones, 'before we know' they do not work. It seems a more instructive approach would be to demonstrate the failure of the even case, before resorting to the odd one. $\endgroup$
    – Tjommen
    Sep 22, 2022 at 14:14
  • 1
    $\begingroup$ Hi @Tjommen. Thanks for the feedback. Point 4 argues that the even case fails. $\endgroup$
    – Qmechanic
    Sep 22, 2022 at 15:41
Main idea

Indeed as @JeanbaptisteRoux pointed out in a comment, the motivation is that you can think of the determinant of an operator to a positive power as coming out of a fermionic integral. To illustrate this without the necessity of all the technology of non-abelian gauge theories let me give you a toy example that could be treated with Faddeev-Popov ghosts; a massless particle in one-dimension.

Toy example

Consider then a free massless field: $$\newcommand{\d}{\mathrm{d}}\newcommand{\pd}{\partial}\newcommand{\D}{\mathrm{D}}S[\phi] := \frac12 \int_0^1\d{\tau}\ \big(\pd_\tau\phi(\tau)\big)^2.$$ For concreteness in the calculations I will be in the Euclidean signature, so this is the Euclidean action and $\tau$ is Euclidean time.

This action has a shift symmetry $\phi\mapsto \phi+\lambda$ where $\alpha$ is a constant. You can gauge the symmetry by considering a gauge field $A_\tau$ that transforms as $A_\tau\mapsto A_\tau^\lambda :=A_\tau + \pd_\tau \lambda$ under the symmetry, so the gauged action is $$S[\phi, A] = \frac12\int_0^1 \d{\tau}\ \big(\pd_\tau \phi(\tau) - A_\tau\big)^2.$$ To compute the partition function of the theory you need to integrate over $\phi$ and $A$ and divide by the volume of the gauge group — in this case this is $\mathcal{G}=\operatorname{Maps}(\mathbb{S}^1\to\mathbb{R})$ — to avoid overcounting gauge configurations. All in all you have $$ Z_\text{gauge} = \int \frac{\D{\phi}\D{A}}{\operatorname{vol}(\mathcal{G})}\ \exp(-S[\phi,A]).$$ Now comes the first Faddeev-Popov trick. Choose a specific gauge-field that you like, $\hat{A}_\tau$ (this could be for example $\hat{A}_\tau=0$ in this case) and note that $$\DeclareMathOperator{\det}{det} \det(\pd_\tau) \int\D{\lambda}\ \delta\!\left(A-\hat{A}^\lambda\right) = 1, \tag{FP1}$$ where $\delta(\cdot)$ is a functional generalisation of the delta-function. To convince yourself that this is true, remember the property of the Dirac delta-function $$\delta\big(f(x)) = \left\vert\frac{1}{f'(x_0)}\right\vert\delta(x-x_0),$$ where $x_0$ is the only zero of $f(x)$; then generalise for a multidimensional delta-function, to finally generalise for a functional delta-function and land on (FP1). In the above I have been very cavalier with zero-modes; the correct equation needs to deal with zero-modes separately, but that is a story for another day (question).

Cool, now insert $1$ in the guise of (FP1) in your partition function to get $$Z_\text{gauge} \overset{\text{(FP1)}}{=\!\!=\!\!=} \int \frac{\D{\phi}\D{A}}{\operatorname{vol}(\mathcal{G})}\ \det(\pd_\tau) \int\D{\lambda}\ \delta\!\left(A-\hat{A}^\lambda\right)\exp(-S[\phi,A]).$$ Next thing to notice is that you can do the integral over $A$ (let's choose for simplicity $\hat{A}=0$) because of the delta-function and then everything that remains is gauge-invariant so you can do the $\lambda$-integral too. What you get is $$\int\D\lambda = \operatorname{vol}(\mathcal{G}),$$ cancelling exactly the volume in the denominator. This was probably a bit too fast but if you take a bit of time at every non-trivial step you can convince yourself it is true.

All in all, what you're left with is $$Z_\text{gauge} = \det(\pd_\tau)\int\D{\phi} \exp(-S[\phi]).$$ And here comes the second Faddeev-Popov trick. Note the following $$\det(\pd_\tau) = \int \D{\bar{c}}\D{c}\ \exp\!\left(-\int_0^1\d{\tau}\ \bar{c}(\tau)\pd_\tau c(\tau)\right),\tag{FP2}$$ where $c$ is a complex Grassmann-valued field; a ghost; a spin-0 fermion. This is simply a property of Grasmannian integration. Again, I have sacrificed (important) subtleties of zero-modes in order to give the big picture, without many digressions.

So in total you can think of the partition function of your gauge theory as coming from a theory containing a weird bunch of spin-0 fermions, i.e. \begin{align}Z_\text{gauge} &= \int \D{\phi}\D{\bar{c}}\D{c}\ \exp(-S_\text{tot}[\phi,c]),\qquad \text{where}\\ S_\text{tot}[\phi,c] &= \int_0^1\d{\tau}\ \left[\frac12\big(\pd_\tau\phi(t)\big)^2+\bar{c}(\tau)\pd_\tau c(\tau)\right]. \end{align}


The exact same steps lead to the introduction of ghosts in QCD. What differs is that the gauge transformation is, instead that the gluons (the analogues of $A_\tau$) have $\mathrm{SU}(N_\text{colour})$ non-abelian gauge transformations, and that the gauge-fixing condition (the analogue of the choice $\hat{A}=0$) is different. This leads to more complicated descriptions for the total action $S_\text{tot}$, but conceptually it is the same procedure.

Some other comments

To answer some of your other questions:

  1. As you saw above, in the toy example, the description in terms of ghosts is just a choice. It is a convenient choice, because it is the only way to express $\det[\text{operator}]$ in a local way and so it helps with doing perturbative calculations (which is more or less the only thing we can do with interacting theories). We can refrain from using anticommuting variables by simply keeping the determinant as it is, without expressing it locally. The price to pay is that to calculate the determinant, one must solve a, typically very difficult PDE (which is usually impossible).

  2. Theorists aren't afraid of ghosts. They understand where they come from and what their role in life is, and they know the ghosts must die when calculating physical observables.


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