Talk:BRST formalism

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This article is both extremely technical and utterly confusing. Bodera 21:33, 3 November 2005 (UTC)

However, it is far better than most mathematical or physical articles, sinse it can be useful for professional :)

Not really -- I am a professional theoretical physicist and I know what BRST symmetry is in the case of gauge theories, but this article is not written in a particularly useful way. 142.3.164.195 17:33, 2 March 2006 (UTC)

Yes, this article needs a total re-write. linas 00:43, 3 March 2006 (UTC)
There's a draft of a new exposition at BRST Quantization. I'm not a professional theoretical physicist, so it may contain howlers; but I'm at least attempting to make connections to functional quantization and differential geometry on fiber bundles. Michael K. Edwards 20:41, 30 August 2006 (UTC)

[edit] Page should be totally re-written

This page is - even to an expert and Ph.D. in theoretical physics - completely confusing and quite misleading. Instead, I would organize the text as follows: 1. to understand the BV formalism you need to introduce the concept of cohomology. An operator Q with Q^2=0 has an image Im(Q) which is a subset of the kernel ker(Q), since if a = Q(b) then Q(a) =Q^2(b)=0. In this case you can define a cohomology of Q, or H(Q). An example is the de Rham cohomology of the operator d. 2. Now, the central idea of the BRST construction is to replace the original gauge symmetry by the BRST symmetry s, which is still present even after one has fixed the gauge. The BRST symmetry is nilpotent, s^2=0 and you can define the cohomology H(s). 3. The BRST transformation of any field, F, is then written in terms of an anti-bracket, sF =(S,F) - here S is the generating function for the BRST transformation, or the generalized action. 4. Introduce the master equation, (S,S) =0; observables of the theory are identified with the zero'th cohomology of s: H^0(s) = {observables}.

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