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Faddeev-Popov Ghosts

+ 3 like - 0 dislike
3439 views

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: $$ \begin{cases} \psi_{c_{i}} & \mapsto\left(1+i\alpha^{a}t^{a}\right)_{c_{i}c_{j}}\psi_{c_{j}}+\mathcal{O}\left(\left(\alpha^{a}\right)^{2}\right)\\ A_{\mu}\,^{a} & \mapsto A_{\mu}\,^{a}+\frac{1}{g}D_{\mu}\,^{ab}\alpha^{b}+\mathcal{O}\left(\left(\alpha^{a}\right)^{2}\right) \end{cases} $$

This post imported from StackExchange Physics at 2014-08-10 20:08 (UCT), posted by SE-user PPR
asked Aug 10, 2014 in Theoretical Physics by PPR (130 points) [ no revision ]

2 Answers

+ 4 like - 0 dislike

Once you uniquely fix the gauge for the gauge field, any two mathematically distinct configurations of the matter field are also physically distinct, they aren't gauge related. Changing a gauge changes both matter and gauge field, so that making sure that the gauge field is constrained to a surface which intersects each gauge orbit once and only once implies that any matter field configuration is different from any other, because you no longer have freedom to change the gauge.

What that means is that you can fix the gauge using a constraint on the gauge field or a constraint on the matter field, but not both. The thing that is going on can be explained using coordinate transformations. If you have three points in the plane, you can describe their relative configuration using one of the points as an origin, and another point on the x-axis, then the location of the third point is arbitrary pair of real number cartesian coordinates, and each mathematically different pair of points is a different configuration (ignoring reflection symmetry). You could also fix the coordinates by putting the third particle on the y-axis, but then you can't simultaneously insist that the second particle is on the x-axis.

When you fix the using a condition on the matter field, there is an issue, at least for linear representations: a zero matter field doesn't transform under gauge transformations at all, so the usual vacuum for matter fields doesn't work to fix a gauge! But in the case where there is a nonzero VEV, the VEV can be used to fix a gauge, the unique gauge where it is real and pointing in one particular direction in the representation space.

When you choose this matter gauge fixing, any two mathematical configurations of the gauge field are now physically distinct, and the action is now a physical action path-integrated over all the configurations of $A$. The matter field kinetic term contributes a part quadratic in $A^2$, and this action describes a collection of massive vector mesons interacting according to the gauge interaction, in other words, because of the VEV, you have a Higgs mechanism. This is the usual treatment of the Higgs mechanism, using the matter field to define the gauge condition.

answered Aug 10, 2014 by Ron Maimon (7,535 points) [ no revision ]

One question: I thought only scalar fields can have non-zero VeV, for otherwise there would be a violation of Lorentz invariance. Is that really the case, and if so, how does it connect to your post?

The field with a VEV is a Lorentz scalar, it's just not a gauge singlet. The standard Higgs-mechanism gauge fixing uses the scalar VEV to fix the gauge, so that the vector parts are a full unconstrained path integral.

+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-08-10 20:08 (UCT), posted by SE-user ACuriousMind
answered Aug 10, 2014 by ACuriousMind (730 points) [ no revision ]

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