The correct way to gauge-fix in a path integral is to insert the Faddeev-Popov determinant, and add a delta-functional constraint. The final action contains three contributions: a Yang-Mills (I'm dealing with a Yang-Mills field for now), a ghost part and a gauge-fixed part.

So the partition function is then $$Z[J] = \int \mathcal{D} A_{\mu}^{a} \mathcal{D} c_a \mathcal{D} \overline{c}_a \exp \left( i ( S_{YM} + S_{gh} + S_{gf} ) \right)$$ where $$S_{gf} = - \int d^4 x \frac{1}{2 \xi} (\partial^{\mu} A_{\mu}^{a}) (\partial^{\nu} A_{\nu}^{a})$$ is the gauge-fixed action.

Now I was wondering whether this doesn't make the partition function depend explicitly on the choice of gauge? I mean: we calculate stuff like correlation functions from the partition function. How are these results then independent of the specific gauge fixing condition we had? Or is this actually not a problem?

• Making sure the partition function does not depend on the choice of gauge-fixing function is precisely the goal of the Faddeev-Popov trick. This should be thoroughly emphasised in any textbook on the subject. Jun 13, 2018 at 17:39
• @AccidentalFourierTransform. Do you mean the trick where you separate out the integration over the gauge transformations $\int \mathcal{D} U$? This contributes to like an overall normalization constant. Is this the step that ensures that the final result is independent of gauge-fixing? Jun 13, 2018 at 17:49

The path integral and observables are independent$^1$ of the gauge-fixing condition. Perhaps a simple toy example is in order:

• Toy example: Imagine an action $S_0$ that doesn't depend on the variable $x$. In other words, $x$ is a gauge variable. Let $f(x)\approx 0$ be a gauge-fixing condition. Here the gauge-fixing function $f$ is assumed to belong to the class of differentiable, monotonically increasing functions with a simple zero.

Consider the full gauge-fixed action $$S~=~S_0 + S_{FP} + S_{gf}, \qquad S_{FP} ~=~ c f^{\prime}(x) \bar{c}, \qquad S_{gf} ~=~ \lambda f(x), \tag{1}$$ where $c$ and $\bar{c}$ are a Grassmann-odd Faddeev-Popov ghost and antighost, respectively, and where $\lambda$ is a Lagrange multiplier.

The toy path integral \begin{align} Z_f&~=~ \int \! dx ~d\bar{c}~dc~d\lambda~\exp\left(\frac{i}{\hbar}S \right) \cr &~\stackrel{(1)}{=}~\int \! dx ~\exp\left(\frac{i}{\hbar}S_0 \right)~\cdot~\frac{i}{\hbar}f^{\prime}(x)~\cdot~2\pi\hbar\delta(f(x))\cr &~=~\int \! dx ~\exp\left(\frac{i}{\hbar}S_0 \right)~2\pi i~\delta(x-x_0)\cr &~=~ 2\pi i\exp\left(\frac{i}{\hbar}S_0 \right)\end{align}\tag{2} does not depend on the gauge-fixing function $f$ within the above-mentioned class! In eq. (2) we used the Berezin integral $\int dc~c=1$ and the Fourier representation of the Dirac delta distribution.

See e.g. this Phys.SE post for another simple toy example.

For a more systematic discussion of independence of gauge-fixing choice from a BRST perspective, see e.g. this related Phys.SE post.

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$^1$ In this answer, we are skipping various technical fineprints, such as, e.g. topological obstructions, etc.

• Since the path integral integrates over distinct physical states, the theory is defined by a path integral over gauge-inequivalent configurations of the gauge field, $$Z = \int \mathcal{D}A' e^{iS}$$
• In the Faddeev-Popov procedure, we "multiply by 1" in order to convert the path integral to $$Z = \int \mathcal{D} A\, e^{iS'}$$, where the measure $$\mathcal{D} A$$ behaves formally like that of a non-gauge field, integrating redundantly over gauge-equivalent configurations, and hence is easier to handle.
• In the process, the action picks up extra "ghost" and "gauge-fixing" terms, $$S' = S + S_{\text{gf}} + S_{\text{gh}}$$.