How does commutator of $Q_B$ with change in $H$ results in moving around in gauge space

Question

In Polchinski I, section 4.2, BRST quantization, the author states the following:

In order to move around in the space of gauge choices, the BRST charge [$Q_B$] must remain conserved. Thus $Q_B$ must commute with the change in the Hamiltonian [under variation of the gauge-fixing parameter $F^A[\phi] \to F^A[\phi] +\delta F^A[\phi]$], $$[Q_B, \{Q_B, b_A \delta F^A\}] =[Q_B^2, b_A \delta F^A] = 0 \tag{4.2.12}.$$

Commutation with $H$ implies some sort of conservation in time, does this picture fit in above scenario?

Also, the (super)-commutator with a Noether charge $[Q_B, G\}$ encodes the change in some G under the symmetry variation, so does it mean that the change in Hamiltonian send us to different gauge space?

The Hamiltonian is just one component of $T^{ab}$, while in the gauge transformation the missing equation of motion is $\langle\psi|T^{ab}|\psi'\rangle=0$, so we are leaving 2 components of this relation - do they not affect $Q_B$ and gauge space?

Answer 1

We have gauge fixed. Therefore if we now change gauge fixing parameter $F^A(\phi)$ infinitesimally, say $F_2 = F_1 + \epsilon\delta F$, then correlators $\langle f|i\rangle$ will change, according to:

$$\epsilon \delta_\text{gauge} \langle f|i\rangle =\langle f| e^{-\epsilon\delta_\text{gauge}S} |i\rangle - \langle f|i\rangle=\langle f| -\epsilon\delta_\text{gauge}S|i\rangle + O(\epsilon^2).$$

As Polchinski (I, 4.2.7) shows, the change in the action $\epsilon\delta_\text{gauge}S \equiv S[F_2]-S[F_1]$ is precisely equal to $\frac{1}{i}\delta_\text{BRST} (b_A \delta F^A)$, which means that we can write

$$\epsilon \delta_\text{gauge} \langle f|i\rangle = i\langle f|\delta_\text{BRST} (b_A \delta F^A) |i\rangle = -\epsilon \langle f|\{Q_\text{BRST}, b_A \delta F^A\}|i\rangle \tag{1}.$$

We can equally rewrite the BRST variation as an anticommutator with the BRST charge $Q_B$, which is conserved: $[Q_B,H_1]=0$. But now we reinterpret eq. (1) in Hamiltonian formulation: it says that the integrated Hamiltonian operator has changed,

$$\int H_1 dt \to \int H_2 dt = \int H_1 dt + \epsilon \{Q_\text{BRST}, b_A \delta F^A\}. \tag{2}$$

What Polchinski means by "In order to move around in space of gauge choices, the BRST charge must remain conserved" is that we knew that $[Q_B, H_1]=0$ in our original gauge-fixing, but we must also have for consistency in the new gauge-fixing that $[Q_B, H_2]=0$. This integrates to give via eq. (2) that $[Q_B, \{Q_\text{BRST}, b_A \delta F^A\}]=0$, as required.

---

For more on this, see Weinberg II, section 5.7.