One question about BRST symmetry in reading Srednicki’s book: Why should the BRST charge $Q_B$ be nilpotent? In p.453, Srednicki claims that since the BRST transformation of a BRST transformation is zero, $Q_B$, the BRST charge, must be nilpotent:
$$Q_{B}^{2}=0.\tag{74.32}$$
I don't know why.
 A: Ref. 1 claims that

$$ \delta_B^2~=~0 \qquad \Rightarrow\qquad  Q_B^2~=~0. \tag{1}$$

Here the BRST transformation $\delta_B$ and the BRST charge $Q_B$ are related by the relation
$$ \delta_B(\cdot)~=~i[Q_B,\cdot],\tag{2}$$
where $[\cdot,\cdot]$ denotes the supercommutator. OP asks why the implication (1) is true?
Well, that's a good question, and a lot less trivial than it superficially seems$^1$. Here is an argument. From the Jacobi identity for the supercommutator, we get
$$\begin{align} 0
~\stackrel{\text{nilp.}}{=}& -\delta_B^2(\cdot)\cr
~\stackrel{(2)}{=}~&[Q_B,[Q_B,\cdot]]\cr
~\stackrel{\text{Jac. id.}}{=}&\frac{1}{2}[[Q_B,Q_B],\cdot]\cr
~=~&[Q_B^2,\cdot].
\end{align} \tag{3}$$
So $Q_B^2$ is a Casimir operator. This in practice shows that
$$ \exists \text{ constant } k:~~ Q_B^2~=~k \mathbb{1} .\tag{4}$$
Arguments that the constant
$$k~=~0.\tag{5}$$
vanishes go as follows:

*

*Because the BRST charge $Q_B$ has ghost number 1, then $k$ has ghost number 2. But most physical theories don't have any ghost number carrying constants lying around, so $k$ is zero, cf. Ref. 2.

If we are not allowed to use ghost number symmetry, here's another argument:


*One would naively expect from eq. (4) that the constant $k$ is a soul-valued constant, but most physical theories don't have any soul-valued constants lying around, so $k$ is zero$^2$.

In practice (e.g. in Yang-Mills theory with matter), the constant $k$ indeed vanishes.
Altogether we get
$$  Q_B^2~\stackrel{(4)+(5)}{=}~0 .\tag{6}$$
$\Box$
References:

*

*M. Srednicki, QFT, 2007; Chapter 74. A prepublication draft PDF file is available here.


*J. Polchinski, String Theory Vol. 1, 1998; p. 128.
--
$^1$ The opposite implication ($\Leftarrow$) in eq. (1) is a triviality.
$^2$ This sentence is flawed as the following toy-example shows: $Q_B=\frac{d}{d\theta}+\theta$ leads to $k=1$.
A: If applying any  map $A$  twice gives zero, then $A^2=0$ by definition of what is meant by $A^2$.  How can it be different here?  
