Question about the Faddeev-Popov method for gauge-fixing

Question

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?

Answer 1

The logic in your answer states the Faddeev-Popov procedure a bit backwards. It should be:

• 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, $Z = \int \mathcal{D}A' e^{iS}$
• This integral is hard, because the space of gauge-inequivalent configurations is complicated.
• 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, $S' = S + S_{\text{gf}} + S_{\text{gh}}$.

No matter what the gauge fixing condition is, we are really computing the exact same thing from the start, so the result has to be independent of the gauge fixing procedure.

However, some confusion is understandable because, in a first QFT class, the logic often isn't there. Typically, one would note that canonical quantization doesn't work for the QED Lagrangian, then artificially add a "gauge fixing" term to the Lagrangian and move on without comment. In this case, it indeed needs to be justified that the results are independent of the gauge fixing. The Faddeev-Popov procedure is precisely that justification.

Answer 2

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.

--

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