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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"?

Qmechanic
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Simplyorange
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1 Answers1

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  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.
Qmechanic
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