Proposition: Gauge-fixing normalization

The auxiliary equation is $b=C[h]/2$. Eliminating it gives

and therefore reproduces the frozen gauge-fixing action $\frac{1}{4}\int C_{\mu }{C}^{\mu }$. In the notation $\Psi =\int \bar{c}(C+\alpha b)$, the required value is $\alpha =-1$.

Proof. Variation in $b$ gives $C-2b=0$. Completing the square proves the result.