Nilpotency of the BRST operator

Question

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"?

Answer 1

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.