Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When quantizing Yang-Mills theory, we introduce the ghosts as a way to gauge-fix the path integral and make sure that we "count" only one contribution from each gauge-orbit of the gauge field $A_\mu\,^a$, because physically only the orbits themselves correspond to distinct physical configurations whereas the motion within the gauge-orbit should not contribute to the path-integral.

How come we don't run into this problem when we quantize the Fermions, which also have gauge transformations, and also have a gauge orbit? Shouldn't we include a gauge-fixing term for the Fermions as well, or does the term introduced for the Boson fields already pick out the gauge orbit for the Fermions as well? How does this technically come to be?

So far I introduce a gauge fixing term into the Lagrangian as $$ 1 = \int d\left[\alpha\right]\det\left(\frac{\delta G\left[A_{\mu}\left[\alpha\right]\right]}{\delta\alpha}\right)\delta\left(G\left[A\left[\alpha\right]\right]\right) $$ where $\alpha(x)$ are the gauge functions, and $G[]$ is a functional which is non-zero only for a unique gauge-representative in each gauge-orbit, where we have the transformations as: $$ \begin{cases} \psi_{c_{i}} & \mapsto\left(1+i\alpha^{a}t^{a}\right)_{c_{i}c_{j}}\psi_{c_{j}}+\mathcal{O}\left(\left(\alpha^{a}\right)^{2}\right)\\ A_{\mu}\,^{a} & \mapsto A_{\mu}\,^{a}+\frac{1}{g}D_{\mu}\,^{ab}\alpha^{b}+\mathcal{O}\left(\left(\alpha^{a}\right)^{2}\right) \end{cases} $$

share|cite|improve this question
up vote 6 down vote accepted

Why do we gauge-fix the path integral in the first place? If we were doing lattice gauge theory, we didn't need to gauge-fix. But in the continuum case, (the Hessian of) the action for a generalized$^1$ gauge theory has zero-directions that lead to infinite factors when performing the path integral over gauge orbits. In a BRST formulation (such as, e.g., the Batalin-Vilkovisky formulation) of a generalized gauge theory, the gauge-fixing conditions can in principle depend on gauge fields, matter fields, ghost fields, anti-ghost fields, Lagrange multipliers, etc. Perturbatively, a necessary condition for a good gauge-fixing procedure is that the gauge-fixed Hessian is non-degenerate (in the extended field-configuration space). Generically, the number of gauge-fixing conditions should match the number of gauge symmetries.

For Yang-Mills theory with Lie group $G$, one needs ${\rm dim}(G)$ gauge-fixing conditions. One may check that for various standard gauges that only involve the gauge fields, it is not necessary to gauge-fix matter fields to achieve a non-degenerate Hessian.


$^1$ By the word generalized gauge theories, we mean gauge theories that are not necessarily of Yang-Mills type.

share|cite|improve this answer

We only have one contribution from each gauge-equivalent matter field configuration:

Let $P$ be the principal $G$-bundle associated to our gauge theory on the spacetime $\mathcal{M}$ (for simplicity, assume it is $\mathcal{M} \times G$. The matter fields are constructed as sections of an associated vector bundle $P \times_G V_\rho$, where $V_\rho$ is a vector space on which a representation $\rho$ of the gauge group exists.

Now, the associated bundle is constructed from $P \times V_\rho$ by dividing out the equivalence relation $$(p,v) \sim (q,w) \iff \exists g\in G \; : \; (p,v) = (qg,\rho(g^{-1})w)$$

Thereby, points which differ only by a gauge transformation are identified, and so are matter field configurations that differ only by a gauge transformation, since they correspond to the exact same section. Therefore, if we model the matter fields by the right kind of functions right away, we can formally take the path integral over the space of sections of the associated bundle, counting each matter gauge configuration exactly once.

I'm not 100% sure if this is done in the standard approach to the path integral, though.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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