To quantize gauge field, one usually use gauge-fixing procedure and then plus ghost field, my question is what the relation between BRST quantization and gauge fixing quantization is? Because it seems the latter is a special case of the former. Thanks!
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1$\begingroup$ Don't forget BV and BFV! $\endgroup$– AHusainCommented Jul 16, 2017 at 7:41
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1$\begingroup$ Roughly speaking you want to work with M/G one way is to make qoutient directly - gauge fixing; another way is to use RESOLUTION i.e. to find homological complex such that its say zero cohomology describe the quotient that is BRST, BV , Faddev-Popov ghosts.... $\endgroup$– Alexander ChervovCommented Jul 16, 2017 at 9:28
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2$\begingroup$ I'm voting to close this question as off-topic because it belongs on Physics StackExchange and is introductory material in the field. $\endgroup$– Chris GerigCommented Jul 16, 2017 at 18:00
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4$\begingroup$ @ChrisGerig really? It's actually a serious mathematical issue. Urs Schreiber, for instance, is working to understand quantisation at a deep mathematical level, and how the various versions (deformation quantisation, geometric quantisation, and all the variants) relate. $\endgroup$– David Roberts ♦Commented Aug 14, 2017 at 3:40
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I do not think that it makes sense to say that the gauge-fixing is a special case of BRST quantization:
$\rightarrow$ The gauge-fixing procedure is actually a normalization technique and it is utilized in order to eliminate unphysical degrees of freedom (gauge directions) in the path integral quantization of the various gauge theories. This -rather naturally- destroys the gauge invariance. A standard method to restore the lost gauge invariance -and thus to recover the physics- is the Batalin-Vilkovisky (BV) or antifield formalism:
$\rightarrow$ after introducing some extra fields in the theory (ghost fields and conjugate antifields), an extended action is constructed involving all these variables. This action is invariant under the BRST symmetry operator $s$ (usually called BRST-differential), which now replaces the original gauge symmetry. Its cohomology essentially contains the physics, in the following sense:
$\rightarrow$ The BRST-differential is nilpotent $s^2=0$ and thus its cohomological groups $H^n(s)$ can be constructed. $H^0(s)$ consists of the gauge invariant functions (that is: the observables).
You can find more details in these notes and the references therein.
Edit: The question: What is the BRST-anti-BRST formalism? and the answer there, are also related and may be of some further help.
answered Aug 14, 2017 at 0:41