Non-abelian Gauge invariance (chapter 15.1 in Peskin/Schroeder)

[Stalafin]

I am trying to understand the derivation of the covariant derivative in Peskin/Schroeder (chapter 15.1, page 483).

This is the important stuff:
n^\mu\partial_\mu\psi=\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon}\left[\psi(x+\epsilon n)-\psi(x)\right]

Scalar quantity: U(y,x):
U(y,x) \rightarrow e^{i\alpha(y)} U(y,x) e^{-i\alpha(x)}

Conditions for zero separation and pure phase:
U(y,y)=1
U(y,x)=exp[i\phi(y,x)]

Covariant derivative:
n^\mu D_\mu \psi = \lim_{\epsilon\rightarrow 0} \frac{1}{\epsilon}\left[\psi(x+\epsilon n)-U(x+\epsilon n,x)\psi(x)\right]

Expansion of U(y,x) in the separation of the two points:
U(x+\epsilon n,x) = 1 - i e\epsilon n^\mu A_\mu(x) + \mathcal{O}(\epsilon^2)

A_\mu(x) is a new vector field and is the coefficient of the displacement \epsilon n^\mu. Why?! I don't see how the author gets there.

 

[Bill_K]

Isn't this just by definition. You expand U in powers of ε n^μ, and the coefficient in the linear term, you define that to be A_μ.

 

[Stalafin]

I thought about making an expansion... This is what I came up with (Peskin expands for epsilon):
\left. U(x+\epsilon n,x) \right|_{\epsilon=0} = U(x,x) + U'(x,x) n \epsilon + \mathcal{O}(\epsilon^2)

So, U(x,x) becomes one from the zero separation condition; what I have trouble understanding is the next part. How does that work out?

U'(x,x) n \epsilon = - ie\epsilon n^\mu A_\mu(x)

Do I use the pure phase condition for that? And how am I supposed to read the n? As a four-vector itself? Or is that:
n=n^\mu A_\mu(x)

Where does the rest come from?

 

[Avodyne]

The Taylor expansion is
U(x+\epsilon n,x)=U(x,x)+\epsilon n^\mu\partial_\mu U(x,x) + O(\epsilon^2)
Then we define
A_\mu(x)\equiv (i/e) \partial_\mu U(x,x)
Note that, with the factor of i, unitarity of U implies that A is hermitian. And to be precise, the partial derivative actually acts only on the first x.