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To quantize the non-Abelien gauge theory. We multiply the path integral by:

$f[A]=\int \mathcal{D}\pi exp[-i\int d^4 x {1\over \xi}(\partial_\mu D_\mu \pi^a)^2]$

then we can shift the argument in the round parentheses by $\partial_\mu A^\mu$ and perform the Stueckelberg trick to eliminate $\pi$s. These are the steps before we introduce Faddeev-Popov ghosts to deal with f[A].

However, all these steps only work at the linear order of gauge transformation(first order in $\pi$), where gauge transformations look like:

$A_\mu^a \rightarrow A_\mu^a+ {1\over g }D_\mu \pi^a$

while it should be:

$A_{\mu}\rightarrow UA_{\mu}U^\dagger-{i\over g}(\partial_\mu U)U^\dagger$

where $U=e^{i\pi^{a}(x)T_a}$ , $A_\mu =A_\mu^a T^a$ and $T^a$s are the generators of $SU(N)$

So it doesn't really make sense to work in the linear order.

Why is it acceptable?

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  • $\begingroup$ It is acceptable because the gauge transformation is inside a delta function(al), so you can expand it to lowest order on the fields. I answered essentially the same question here physics.stackexchange.com/a/784046/348916. Ignore the last part (about BRST). $\endgroup$
    – Bairrao
    Commented Mar 8 at 13:31
  • $\begingroup$ @BairraoB dunno about the context in string theory, In quantization of gauge field, the argument in the delta function has a zeroth order term. Your argument only applies if the argument in delta function is zero when $\pi=0$ $\endgroup$
    – Bababeluma
    Commented Mar 8 at 15:20
  • $\begingroup$ @BairraoB, also the method of quantization discussed in this post doesn't involve any delta function. I know of the method using delta function to fix gauge. $\endgroup$
    – Bababeluma
    Commented Mar 8 at 15:22
  • $\begingroup$ it’s because you are computing a jacobian for a change of coordinates, and that always only involves the first derivative. recall also that the jacobian for a change of variables in the base space is the same as the jacobian for a change of variables in the tangent space. $\endgroup$
    – Wakabaloola
    Commented Mar 10 at 8:53
  • $\begingroup$ @Wakabaloola yes, only the first derivative. However, suppose you're making a change of coordinate y=e^x, you should have dy=e^x dx instead of dy=dx(expanding e^x to linear order), this is exactly what is wrong with the quantization of non-abelian theory. $\endgroup$
    – Bababeluma
    Commented Mar 12 at 4:05

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