The gauge transfomation of the Wilson line

Question

I have a question in Capter 15 of Peskin & Schroeder.

The gauge transformation here in its infinitesimal form: \begin{cases} \psi(x) \to V(x)\psi(x) \quad \quad \quad \quad \quad \quad \,\,\,\, \text{(15.41)} \\ V(x)=1+i\alpha^a(x)t^a+ \mathcal{O}(\alpha^2) \quad \quad \text{(15.42)} \\ A_{\mu}^a \to A_{\mu}^a +\frac{1}{g} \partial_{\mu}\alpha^a +f^{abc}A_{\mu}^b \alpha^c \quad \,\,\,\, \text{(15.46)} \end{cases} When $\tilde{s}$ is a parameter of the path $P$, running from 0 at $x=y$ to s at $x=z$ and $P{}$ denotes path-ordering, the Wilson line is written $$ U_p(z(s),y)=P \biggl\{ \exp \left[ ig\int_0^s d\tilde{s}\frac{dx^{\mu}}{d\tilde{s}}A_{\mu}^a(x(\tilde{s}))\,t^a \right] \biggl\} $$

By an analogy with the propagator (4.23) of the time-ordered exponential, $U_P$ is the solution of a differential equation $$ \begin{align}\frac{d}{ds}U_p(x(s),y) &=\left( ig\frac{dx^{\mu}}{ds}A_{\mu}^a(x(s))\,t^a \right) U_p(x(s),y). \tag{15.57} \\ \Leftrightarrow ~~~~\frac{dx^{\mu}}{ds} D_{\mu} U_p(x,y) &=0 \tag{15.58} \end{align}$$

In the following this book is going to show $$ U_p\left(z,y,A^V\right)=V(z)U_p(z,y,A)V^{\dagger}(y) \tag{15.59} $$ where $A^V$ is the gauge transform of $A$.

And $$ D_{\mu}\left(A^V\right)V(x)=V(x)D_{\mu}(A) \tag{15.60} $$ is proved in its infinitesimal version before.

This relation implies that the right-hand side of (15.59) satisfies (15.58) for the gauge field $A^V$ if $U_P(z,y,A)$ satisfies this equation for the gauge field $A$. But the solution of a first-order differential equation with a fixed boundary condition is unique. Thus, if $U_P(z,y)$ is defined to be the solution of (15.57) or (15.58), it indeed has the transformation law (15.59).

I think that the evolution along the path $x^{\mu}$ (or the parameter $s$)
and the evolution in terms of the gauge transfomation are completely different issues.
Why does this book claim as the last line of this paragraph?

Answer 1

I will try to make it clear what the logic of the argument is. We know that $U_P(x,y,A)$ satisfies a particular differential equation, namely $\frac{dx^{\mu}}{dt}(\partial_{\mu}-igA_{\mu})U_P(x,y,A)=0$, for any gauge field values $A$. We wish to know how $U_P(x,y,A^V)$ is related to $U_P(x,y,A)$, where $A^V$ is related to $A$ by a gauge transformation. We consider the proposal, $$ U_P(x,y,A^V)=V(x)U_P(x,y,A)V^{\dagger}(y)\,. $$ It is then easy to show that if $U_P(x,y,A)$ satisfies the differential equation $\frac{dx^{\mu}}{dt}(\partial_{\mu}-igA_{\mu})U_P(x,y,A)=0$, then with this proposal $U_P(x,y,A^V)$ satisfies $\frac{dx^{\mu}}{dt}(\partial_{\mu}-igA^V_{\mu})U_P(x,y,A^V)=0$.

However we also know that whatever is the true relation between $U_P(x,y,A^V)$ and $U_P(x,y,A)$, it always must be the case that $U_P(x,y,A^V)$ satisfies $\frac{dx^{\mu}}{dt}(\partial_{\mu}-igA^V_{\mu})U_P(x,y,A^V)=0$ because we know this equation holds for any gauge field values. And indeed our proposed relation does have this necessary property, as we have just seen. But as Peskin and Schroeder state, this differential equation is first order, and if there is in addition a boundary condition that must be satisfied, then the solution is unique. Hence the proposed relation must be the unique correct relation.

(You say in a comment that $V(z)U_P(z,y,A)$ would also satisfy the differential equation, so why should we say that $V(z)U_P(z,y,A)V^{\dagger}(y)$ is the correct relation? The reason is that there is in addition a boundary condition: we require that $U_P(x,x,A)=1$.)