BRST charge action on fields

Question

1. How can we prove that the BRST charge $Q$ acts on bosonic fields as $$\delta A(x) = \epsilon[Q,A(x)]\tag{1}$$ and on fermionic fields as $$\delta\psi(x) = \epsilon\{Q, \psi(x)\}\tag{2}.$$

2. And how should we consider the BRST charge? Is it a generator or a conserved charge?

Answer 1

1. Eqs. (1) & (2) are the standard way a Noether charge $Q$ generates a symmetry transformation, cf. e.g. this & this related Phys.SE posts.

   More generally, an operator $\hat{F}$ is transformed as $$\begin{align}\hat{F}~\longrightarrow~e^{\epsilon\hat{Q}}\hat{F}e^{-\epsilon\hat{Q}}~=~&e^{[\epsilon\hat{Q},~\cdot~]_{C}}\hat{F}\cr ~=~&e^{\epsilon[\hat{Q},~\cdot~]_{SC}}\hat{F}\cr ~=~&\hat{F}+\epsilon[\hat{Q},\hat{F}]_{SC}+{\cal O}(\epsilon^2),\end{align}\tag{A}$$ where $[\cdot,\cdot]_{C}$ and $[\cdot,\cdot]_{SC}$ are the commutator and supercommutator, respectively. In eq. (A) we have assumed that $\hat{Q}$ and $\epsilon$ carry the same Grassmann parity.

   See also e.g. this related Phys.SE post.

2. In case of the BRST symmetry, the charge $Q$ is Grassmann-odd.

   In terms of value, a Grassmann-odd charge $Q$ is soul-valued, i.e. a definite non-zero value does not make sense, cf. e.g. this related Phys.SE post.