# Nilpotency of the BRST operator

I'm styding chapter 16 of Peskin and Schroeder, in section 16.4 on the BRST symmetry, Peskin and Schroeder first checks (on page 518) that if $$Q$$ is the BRST symmetry operator, then $$Q^2\phi=0\tag{16.48}$$ for any field operator $$\phi$$, then they claim on page 519:

Because the Lagrangian has the continuous symmetry (the BRST symmetry), the theory will have a conserved current, and the integral of the time component of this current will be a conserved charge $$Q$$ that commutes with $$H.$$ The action of $$Q$$ on field configurations will be just that described in the previous paragraph.

From here Peskin and Schroeder says if $$Q$$ now denotes the integral of the time component of the conserved charge, then $$Q^2=0\tag{16.50}.$$ My question is: why is it that the BRST symmetry operator itself being nilpotent implies that the conserved charge is also nilpotent? Why is it true that " the action of $$Q$$ on field configurations will be just that described in the previous paragraph"?

## 1 Answer

1. The situation is easiest to explain in a Hamiltonian formulation. (P&S start in a Lagrangian formulation, but state on top of p. 519 that one should transcribe their formulas into the Hamiltonian language.) Then the infinitesimal BRST variation $$\delta$$ is given by $$\delta ~=~ \epsilon [Q,\cdot],$$ where $$Q$$ is the Grassmann-odd BRST operator, $$\epsilon$$ is an infinitesimal Grassmann-odd parameter, and $$[\cdot,\cdot]$$ is a super-commutator.

2. The BRST operator $$Q$$ is also the Noether charge for the BRST symmetry, i.e. $$Q$$ is the BRST charge, cf. e.g. my Phys.SE answer here.

3. Ref. 1 shows that given a gauge symmetry, there exists a Grassmann-odd Hermitian nilpotent BRST operator $$Q^2=0$$ (of ghost-number 1), and a (possibly BRST-improved) Grassmann-even Hermitian Hamiltonian operator $$H$$ (of ghost-number 0) that commutes with $$Q$$.

References:

1. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994; Chapter 9.