On P & S'QFT section 16.2, the book discussed the Faddeev-Popov Lagrangian. The idea is to insert an identity to the function integral: $$ 1=\int \mathcal{D} \alpha(x) \delta\left(G\left(A^\alpha\right)\right) \operatorname{det}\left(\frac{\delta G\left(A^\alpha\right)}{\delta \alpha}\right) \tag{16.23} $$ where $A^\alpha$ is a gauge field under finite gauge transformation:

$$\left(A^\alpha\right)_\mu^a t^a=e^{i \alpha^a t^a}\left[A_\mu^b t^b+\frac{i}{g} \partial_\mu\right] e^{-i \alpha^c t^c} \tag{16.24}$$

also the infinitesimal form: $$\left(A^\alpha\right)_\mu^a=A_\mu^a+\frac{1}{g} \partial_\mu \alpha^a+f^{a b c} A_\mu^b \alpha^c=A_\mu^a+\frac{1}{g} D_\mu \alpha^a \tag{16.25} $$

The book said as long as the gauge-fixing function $G(A)$ is linear, the functional derivative $\delta G(A^\alpha)/\delta \alpha$ is independent of $\alpha$.

I can notice this independent of $\alpha$ from the infinitesimal in (16.25), but cannot from finite form (16.24). So I am troubled if the $\delta G(A^\alpha)/\delta \alpha$ only works for infinitesimal form, why? For me, it seems that from (16.24) $\delta G(A^\alpha)/\delta \alpha$ still depend on $\alpha$.


1 Answer 1


Because of the delta function, you are evaluating the functional derivative at $\alpha=0$. The derivative of the finite transformation at $\alpha=0$ will agree with the coefficient of the linear term in $\alpha$ for the infinitesimal transformation, essentially by definition.

  • $\begingroup$ Thank you very much, that's helpful! May I add one comment? So "evaluating the functional derivative at $\alpha=0$" is the definition of it? I don't see this kind of description in P&S's book. Thanks! $\endgroup$
    – Daren
    Nov 28, 2022 at 6:00
  • $\begingroup$ @Daren It follows because of the delta function setting $G=0$. The idea of the path integral is that you set $A=U^\dagger(\alpha) A_g U(\alpha) + U^\dagger(\alpha) \partial U(\alpha)$, where $A_g$ is $A$ in a gauge where the gauge condition is satisfied (ie, $G[A_g]=0$), and $U(\alpha)$ is associated with a gauge transformation, with $U(0)=1$. The delta function forces $A=A_g$, so $\alpha=0$. $\endgroup$
    – Andrew
    Nov 28, 2022 at 12:42

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