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Fix the current sign convention by defining the localized variation as ${\delta }_{ϵ}S=-\int d^{d}x({\partial }_{\mu }{ϵ}^a)j_a^{\mu }$. For a first-derivative Lagrangian invariant without a boundary term under constant parameters, this means $j_a^{\mu }=-\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )t_a\varphi$; include the usual improvement term if the constant variation is a total derivative. On solutions, arbitrary compactly supported parameters imply ${\partial }_{\mu }{j}_a^{\mu }=0$. The Noether charge is $Q_a=\int d^{d-1}x{j}_a^0$ and is conserved if the spatial flux vanishes. This sign convention matches the printed Ward identity; reversing the current also reverses the corresponding generator convention.

Assume an invariant regulated functional measure, invariant vacuum boundary conditions and no quantum anomaly. Changing variables in the normalized path integral gives $0=⟨{\delta }_{ϵ}X⟩+i⟨X{\delta }_{ϵ}S⟩$. Integration by parts then yieldsFor a product of scalar fields, the local Ward identity contact terms areAway from the insertions this is current conservation. Time-ordering or the distributional functional identity supplies the contact terms.

For the gauge theory write $[X,Y{]}^a=f^{abc}{X}^b{Y}^c$ and use the left-acting BRST differential $s$, with ${\delta }_{ϵ}=ϵs$. It obeys the graded Leibniz rule andAssume the structure constants satisfy the Jacobi identity and the dot product is invariant; antisymmetry alone would not be enough. Since $c$ and $D_{\mu }c$ are odd, their Lie brackets are symmetric in these two arguments. ConsequentlyAlso $s{F}_{\mu \nu }=D_{\mu }{D}_{\nu }c-D_{\nu }{D}_{\mu }c=[F_{\mu \nu },c]$. The graded Jacobi identity gives $s[c,c]=0$, so $s^{2}c=0$; the other three fields have zero second variation immediately. For independent odd parameters, ${\delta }_{{ϵ}^{\prime }}{\delta }_{ϵ}=-{ϵ}^{\prime }ϵs^2=0$. These are off-shell BRST nilpotence identities because the Nakanishi-Lautrup field $b$ is retained.

The Yang-Mills theory variation is proportional to $F^{\mu \nu }\cdot [F_{\mu \nu },c]=0$. The gauge-fixing variation is $({\partial }^{\mu }b)\cdot D_{\mu }c$, while the ghost variation is its negative; the $b^2$ term does not vary. Equivalently these terms are $s\Psi$ for the gauge-fixing fermion $\Psi =({\partial }^{\mu }\bar{c})\cdot A_{\mu }+\xi \bar{c}\cdot b/2$. BRST nilpotence makes this expression invariant. Ghost-number scaling also leaves every term invariant: the ghost and antighost factors carry opposite weights.

Here are the two explicit currents in the same sign convention as the Ward identity. Localizing the even ghost parameter gives coefficient $({\partial }_{\mu }\theta )[\bar{c}\cdot D^{\mu }c-({\partial }^{\mu }\bar{c})\cdot c]$. Localizing the odd parameter, keeping it on the left, gives coefficientThe last sign comes from moving the odd parameter through the odd antighost derivative. Since the current was defined as minus this coefficient,These are the ghost-number Noether current and the BRST current in derivative-b gauge fixing. If the opposite Noether sign is used, both displayed currents acquire an overall minus sign. Integrating the gauge-fixing term by parts changes the Noether representative by the associated boundary improvement; mixing the two Lagrangian conventions without that improvement gives incorrect signs.

With no BRST anomaly, the conserved odd BRST charge has $Q_B^2=0$. The BRST cohomology identifies closed states $Q_B|\psi ⟩=0$ modulo exact states $Q_B|\chi ⟩$. Nilpotence puts every exact state in the closed space. In the usual indefinite gauge-fixed state space, a Hermitian BRST charge makes exact states orthogonal to closed states; the standard no-ghost/positivity assumptions then give a physical inner product on the quotient. The physical sector is its ghost-number-zero component,The ghost number assigns $+1$ to $c$, $-1$ to $\bar{c}$ and zero to gauge and auxiliary fields. Gauge-invariant observables and the chosen vacuum have ghost number zero; unphysical ghost excitations are removed in BRST pairs. Thus physical representatives are expected to satisfy $Q_G|\psi ⟩=0$. This zero-grading selection is part of the physical-state prescription, not a consequence of nilpotence alone.