Path Integrals and Non-Abelian Gauge Theories

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SUMMARY

The discussion focuses on the Faddeev-Popov procedure for quantizing non-Abelian gauge theories using path integrals. It clarifies that the function Δ, which acts as a Jacobian, is defined as Δ[𝑨] = det(∂f/∂θ). The conversation emphasizes the complexity of delta functions in the context of field spaces and highlights the challenges posed by non-Abelian gauge theories, particularly regarding the density of gauge orbits. The use of a non-physical Fermionic field is proposed as a solution to maintain the integrity of perturbation theory despite the complications introduced by non-polynomial terms.

PREREQUISITES
  • Understanding of path integrals in quantum field theory
  • Familiarity with non-Abelian gauge theories
  • Knowledge of Faddeev-Popov ghosts and their role in quantization
  • Basic concepts of differential geometry and fiber bundles
NEXT STEPS
  • Study the derivation and implications of the Faddeev-Popov procedure in detail
  • Explore the role of Jacobians in path integrals, particularly in non-Abelian contexts
  • Investigate the use of Fermionic path integrals for computing operator determinants
  • Learn about the structure of gauge orbits and their impact on perturbation theory
USEFUL FOR

The discussion is beneficial for theoretical physicists, particularly those specializing in quantum field theory, gauge theories, and anyone involved in advanced studies of path integrals and their applications in particle physics.

BenTheMan
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Ok, I have a question about this Fade'ev Popov procedure of teasing out the ghosts when one quantizes a non-Abelian gauge theory with path integrals.

The factor of 1 that people insert, for some gauge fixing function f, and some non-Abelian symmetry [tex]\mathcal{G}[/tex] is:

[tex]1=\int \mathcal{D}U \delta[f(\mathbf{A})] \Delta[\mathbf{A}][/tex],

where

[tex]\mathcal{D}U = \Pi d\theta[/tex],

and

[tex]U \in \mathcal{G}[/tex].

This is probably a stupid question, but the function [tex]\Delta[/tex] works out just to be a Jacobian of some sort over the manifold [tex]\mathcal{G}[/tex], right?

[tex]\Delta[\mathbf{A}] = det \left(\frac{\delta f}{\delta \theta}\right)[/tex]

I am confused because no one actually says this. Am I completely off base?

Thanks in advance for helping me, and tolerating a (possibly) stupid question!
 
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Yes it is the Jacobian typical to multi-dimensional delta functions. Though consider how abstract an object a delta function on the space of Fields is. The space of fields is:

$$\bigcup_{E}\Gamma(E),\quad \pi(E) = \mathcal{M}$$

that is the space of sections of a Fiber Bundle, ##\Gamma(E)##, taken over all Fiber Bundles with a common base space ##\mathcal{M}##. Typically this base space can be taken to be four-dimensional Euclidean space. However since fields which don't die off at infinity have zero weight in the path intgeral one can reduce it to ##\mathcal{M} = S^{4}##.

So then we have the space of functions of compact support on this space:
$$\mathcal{C}_{0}\left(\bigcup_{E}\Gamma(E)\right)$$

And then the delta function used for the Faddeev-Popov procedure is a map from this space to the reals:
$$\delta_{FP} : \mathcal{C}_{0}\left(\bigcup_{E}\Gamma(E)\right) \rightarrow \mathbb{R}$$

This is because delta functions (or any distribution) on any space are defined as maps from compactly supported functions to the reals.

What's really happening with Faddeev-Popov ghosts is that you are restricting the integral to a single element from each gauge equivalent orbit.

In the case of QED this is no problem, but in non-Abelian theories any such surface you select out with your Gauge condition, e.g. ##\partial_{\mu}A^{\mu} = 0##, will not intersect the gauge orbits perpendicularly. Thus the gauge orbits will be denser on some parts of the surface than others. ##det\left(D^{\mu}\partial_{\mu}\right)## then measures this density. Unfortunately this spoils perturbation theory as you now have a term which is not polynomial in the fields other than the exponential term with the action ##e^{-S}##. The combinatorics and methods of perturbation theory are based on only integrating polynomials in the field.

Fortunately we use the fact that any operator determinant can be computed from a Fermionic path integral and invent a non-physical (not part of the actual scattering spectrum or physical states) Fermionic field to compute this term. The Fermionic field's action will be added to the path integral to compute this determinant, but does so in a way that we once again only have a term like ##e^{-S}## as our sole non-polynomial term and thus perturbation theory can continue.
 
Last edited:
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