BRST operator ($Q_B$)

The BRST operator is the Grassmann-odd conserved charge that combines a gauge transformation with its ghost field. Quantum gauge consistency requires $Q_B^2=0$, and physical states are its cohomology classes.

If $L_m=\{b_m,Q\}$, nilpotence gives $[L_m,Q]=0$. The graded Jacobi identity and $[b_m,L_n]=(m-n)b_{m+n}$ then imply the displayed algebra. Thus a quantum central extension in the total generators obstructs BRST nilpotence.

For a matter state tensored with the ghost oscillator vacuum, the BRST charge produces the independent ghost states $c_0|0⟩$ and $c_{-m}|0⟩$ with coefficients $(L_0-1)|\Psi ⟩$ and $L_m|\Psi ⟩$. Thus BRST closure is equivalent to $(L_0-1)|\Psi ⟩=0$ and $L_{m>0}|\Psi ⟩=0$ in this sector. It reproduces old covariant string quantization and the open bosonic string mass spectrum.

The BRST current is the conserved worldsheet current whose closed contour integral gives the BRST operator,For the bosonic string, one conventional holomorphic representative iswhere $b,c$ are the worldsheet ghost fields, $T_m$ is the matter stress-energy tensor, and colons denote normal ordering. The last term improves the conformal transformation law without changing the closed contour integral.

Quantum consistency requires $Q_B^2=0$. With matter central charge $c_m$ and ghost central charge $-26$, the obstructing contour residue vanishes at $c_m=26$. This does not require a nonsingular operator product expansion of the currents. For the displayed representative, even at $c_m=26$ the singular terms areThere is no simple pole, so the inner contour integral vanishes. The BRST operator can therefore be nilpotent while its local current has higher-order poles.^[ref]