Gauge-fixing fermion ($\Psi$)

A gauge-fixing fermion is a Grassmann-odd functional whose BRST transformation $s\Psi$ supplies the gauge-fixing and ghost terms. Nilpotence gives $s(s\Psi )=0$ immediately.

For a left left-acting BRST differential with $sA=Dc$, $s\bar{c}=b$, $sb=0$, choose $\Psi ={\bar{c}}^a(\partial \cdot A^a+\xi b^a/2)$. The graded Leibniz rule gives $s\Psi =b\partial \cdot A+\xi b^2/2-\bar{c}\partial \cdot Dc$. Integrating the auxiliary Gaussian yields $-(\partial \cdot A{)}^2/(2\xi )-\bar{c}\partial \cdot Dc$. At $\xi =1$ this is Feynman gauge. Opposite antighost or exact-action sign conventions require a corresponding change of fermion.

// Target: relativistic-quantum-field.bigb

For the left graded Leibniz rule and $s\bar{c}=-b$, $sb=0$, the displayed gauge-fixing fermion gives $b\partial \cdot A+\xi b^2/2+\bar{c}\partial \cdot Dc$. The ghost signs follow from odd parity. BRST nilpotence implies invariance of this density without using field equations. Eliminating the Nakanishi-Lautrup field gives $-(\partial \cdot A{)}^2/(2\xi )$ for nonzero $\xi$.

For the convention $s\bar{c}=-b$, the action $S_{\mathrm{YM}}-s\Psi$ contains $-b\partial \cdot A+\xi b^2/2-\bar{c}\partial \cdot Dc$ after integration by parts. Integrating the Nakanishi-Lautrup field gives $-(\partial \cdot A{)}^2/(2\xi )$. The Grassmann Gaussian integral represents the Faddeev-Popov determinant.