question: WHY BRST formulation works? In more details:

  1. What are the conditions we need to impose on QFT to find the BRST (global) symmetry?
  2. Why can we demand the BRST parameter $\epsilon$ directly that relates the gauge symmetry parameter $\alpha^a(x)$ to the ghost field?
  3. And how to determine how many BRST parameters $$\epsilon_1,\epsilon_2,\epsilon_3,...$$ can we introduce? Does each continuous BRST parameter $\epsilon_j$ introduce a $U(1)$ or a supergroup (?) global symmetry? (But note that $\epsilon_j$ is an anticommuting Grassman number.) and correspond to a ghost field one by one?

(You can follow discussions on Peskin and Schroeder (PS) Chap 16.4 if you wish, also with this understanding.)

Given a gauge theory such as nonabelian Yang-Mills gauge theory, we know there is a gauge symmetry transformation on the 1-form gauge field as (in the standard QFT notation of spacetime index $\nu$ and the gauge Lie algebra (adjoint) index $a, b, c$) $$ A^a_\nu \to A^a_\nu +\delta A^a_\nu = A^a_\nu +\frac{1}{g}D_\nu^{ac} \alpha^c =A^a_\nu +\frac{1}{g} (\partial_\nu \delta^{ac} + g f^{abc} A^b_\nu ) \alpha^c $$ with a 0-form gauge parameter $\alpha^a$.

However the BRST formulation declares that we can introduce the global symmetry parameter $\epsilon$ and a $C$ ghost field: $$ A^a_\nu \to A^a_\nu +\delta A^a_\nu = A^a_\nu +\epsilon D_\nu^{ac} C^c $$ such that we need a relation between the gauge symmetry and BRST global symmetry: $$ \boxed{\alpha^a = g \epsilon C^a}. $$ (Pardon me here that I capitalize the ghost field $C$ to distinguish it from the gauge index $c$.)

  • But how do we know such a "single" BRST parameter can be introduced? (What conditions do we need to impose and declare such global symmetry $\epsilon$?)

  • Follow the question 3 above, If we have more ghost fields (such as Chap 2.5 Polchinski) with $b$ and $c$ two ghost fields for $bc$ CFT, is it possible to introduce more BRST parameters $\epsilon_1,\epsilon_2,..$? Is the number of the BRST symmetry parameters the same as the number of ghost fields?


Only 1 Grassmann-odd global parameter $\epsilon\in \mathbb{R}^{0|1}$ is needed in the BRST formulation $\delta=\epsilon {\bf s}$ even if the underlying gauge theory contains several gauge parameters.

The formal proof of the existence of a BRST formulation for an arbitrary Hamiltonian & Lagrangian gauge theory with possibly reducible and open gauge algebra was given in a series of articles by Batalin, Fradkin & Vilkovisky, cf. e.g. Ref. 1 and references therein.

Roughly speaking, the main tool in the existence proof of a Grassmann-odd nilpotent BRST symmetry ${\bf s}$ of ghost number 1 is deformation of a cohomological complex of fields.


  1. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994; chapters 9 + 10 + 17.
| cite | improve this answer | |
  • $\begingroup$ thanks so much, I am looking at this 500+ ref book. Can I regard the BRST global symmetry in terms of generating some continuous group? since $\epsilon$ is a continuous parameter? What is that group? It is not easy to think in terms of the usual group because the $\epsilon$ is a Grassmann. (Does it have anything to do with SUSY or super group?) $\endgroup$ – annie marie heart Oct 7 at 1:29
  • $\begingroup$ (I meant this book is 500 page +) $\endgroup$ – annie marie heart Oct 7 at 1:39
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Oct 7 at 13:51

If you want to understand the quantization of gauge theories and the BRST procedure in particular in detail, the best reference is probably "Quantization of Gauge Systems" (QoGS) by Henneaux and Teitelboim. Requirements for a BRST symmetry to exist is that we have a gauge theory - or equivalently a constrained Hamiltonian theory - and that this gauge theory fulfills regularity constrains I discuss in this answer of mine (but which is also taken from QoGS).

There is no notion of multiple "BRST parameters". The BRST symmetry generator $\Omega$ is the generator of the BRST differential $s$ such that $sF = \{F,\Omega\}$ where $\{-,-\}$ is the Poisson bracket on extended phase space (i.e. including ghosts, and taking into account the grading w.r.t bosonic and fermionic variables). I explain the construction of one part of it in this answer of mine and its rough connection to the Lie algebra in a generic Yang-Mills theory in this answer of mine

The proof that the BRST differential exists for a general gauge theory obeying certain regularity constraints and has a generator is rather technical. It requires proving the "main theorem of homological perturbation theory" (theorem 8.3 in QoGS) and then using the differential $\delta$ on the ghosts discussed in my answers linked above. Once you have the generator $\Omega$, the infinitesimal BRST symmetry is just $F\mapsto F + \epsilon \{F,\Omega\}$, with a single parameter $\epsilon$. The number of ghosts depends on the number of constraints in the Hamiltonian formulation of the gauge theory, and the BRST operator in the simplest cases is given by $\eta^a G_a$, where the $\eta^a$ are the ghosts associated with the first-class constraints $G_a$.

Note that the BRST symmetry acts on a very different space than the gauge symmetry - the BRST symmetry acts on the extended phase space with ghosts (and ghosts-of-ghosts, etc.) and only BRST-invariant quantites are physically meaningful, while the original gauge symmetry (with potentially multiple parameters) acts on the original phase space. It is subtle and dangerous to try to compare their actions directly.

| cite | improve this answer | |
  1. To learn about BRST properly, you need the „Bible”.

  2. As to the main question ("why does the BRST work?"), the answer is very simple: simply by replacing it, it "cures" the "illness" of gauge symmetry, in the sense that the path integral (Lagrangian or Hamiltonian) is no longer infinite, but finite, thus allowing one to compute the QFT observables, which are typically scattering cross-sections (tree-level Feynman diagrams) or corrections stemming from loop-level diagrams.

  3. How does it work? There are a number of needed extra variables which are added to the theory (called ghosts and antighosts or antifields) and the gauge symmetry, typically exhibited by a differential operator in some principal fiber bundle over space-time ("the gauge-covariant" derivative), is replaced by the BRST symmetry, which is exhibited through a novel (nilpotent in degree 2) differential operator which is usually denoted by "s" and the "symmetry" itself is the requirement that the BRST/gauge-fixed Lagrangian/Hamiltonian action up to s-exact terms is invariant under "s".

If you allow me some personal remarks, the "BRST Bible" of Henneaux and Teitelboim is not standard (=non-PhD) material, and some QFT books as the one you quote do a miserable job to offer an incomplete view on this huge topic, probably with the notable exception of Weinberg's 3 volume QFT book. The rest of the QFT books, thankfully, do not mention it all (they typically give a sketch of Faddeev & Popov's calculation of the gauge-fixed QCD action), which is to be desired. The school where I learnt about BRST 15-16 years ago had proper full-courses in BRST. I do not recall of the „BRST parameters” terminology.

| cite | improve this answer | |
  • $\begingroup$ „BRST parameters” I meant 𝜖, thanks voted up +1 $\endgroup$ – annie marie heart Oct 7 at 1:44
  • $\begingroup$ thanks so much, I am looking at this Bible. Can I regard the BRST global symmetry in terms of generating some continuous group? since 𝜖 is a continuous parameter? What is that group? It is not easy to think in terms of the usual group because the 𝜖 is a Grassmann. (Does it have anything to do with SUSY or super group?) $\endgroup$ – annie marie heart Oct 7 at 1:47
  • $\begingroup$ (I post a new question feel free to comment more. Still digesting the book.) $\endgroup$ – annie marie heart Oct 7 at 3:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.