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I've been working my way through L. Baulieu's excellent paper [Perturbative gauge theories, Physics Reports, Volume 129, Issue 1, December 1985, Pages 1-74]. Towards the end, he goes on to prove that renormalized Yang-Mills theories that are BRST invariant exhibit gauge independence and unitary. At the end of the proof he states:

One should note that we have implicitly assumed in this demonstration the existence of the S-matrix, i.e., the applicability of LSZ reduction formula [...] the proof is only valid when all gauge bosons are massive. Whenever massless gauge bosons are present in theory, all our arguments become formal.

In the case of QCD, the gluons are massless, so does the LSZ formula not apply to QCD? Is there a formal argument that proves gauge-independence as he suggests?

Qmechanic
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user45213
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1 Answers1

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The LSZ formula definitely applies to theories with massless gauge bosons, like QED and QCD.

The S-matrix is given by the LSZ formula, which relates the former to correlation functions, which are in turn given by a path integral. The LSZ formula assumes asymptotic in- and out-states for particles of interest. In the path integral formalism, gauges can be conveniently implemented through functional determinants. The most general S-matrix that arises from this so-called Fadeev-Popov procedure is unitary, but not gauge invariant. If we, however, restrict ourselves to gauge-invariant asymptotic states, like transversely polarized photons or gluons (as opposed to unphysical, longitudinal ones), gauge invariance is recovered. This can be formulated mathematically as a projection of the S-matrix to the relevant subspace of gauge-invariant states.

The question is now whether this projection yields a unitary S-matrix. Non-unitarity in this context means that transverse polarizations at the far past could evolve into other polarizations in the far future. This is obviously something we do not want to happen. In QED, this problem is easy to handle, but QCD requires a more subtle solution. It turns out that BRST symmetry makes sure that no unphysical polarizations can arise. This is due to the fact that the BRST operator $Q$ commutes with the Hamiltonian, which guarantees that physical states (states annihilated by $Q$) can only evolve into physical states. Therefore, unitarity is preserved.

A more detailed discussion can be found in Peskin and Schroeder.

Frederic Brünner
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