# The gauge transfomation of the Wilson line

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?

• In over the ‎first half I've rewriten the sentences to shorten them. But if I quote all the sentences literally as written in the book, does it violate the Copyright Act (because the quotation is too long) ? – GotchaP Nov 3 '16 at 4:28
• This probably doesn't answer your question completely, but you could substitute the transformed holonomy (Wilson line, whatever) in the defining differential equation for the transformed gauge connection. The fact that it still holds is the proof for the transformation law. – Prof. Legolasov Nov 3 '16 at 11:56
• @Solenodon: Umm..... if (15.59) were $U_p\left(z,y,A^V\right)=V(z)U_p(z,y,A)$ , the RHS of this would satisfy (15.58) too. But that's mentioned by diracula's answer. – GotchaP Feb 19 '17 at 23:15
• yes, but it won't satisfy the initial conditions. You have a first-order differential equation, which means that there could be only one solution for each initial condition. – Prof. Legolasov Feb 20 '17 at 1:29
• once you agree with that, it is easy to see that the parallel-transport equation, along with the initial condition $U(0) = 1$ completely determines the holonomy. You can gauge-transform this equation and study the holonomy defined by its gauge-transformed version. This is naturally the gauge-transformed holonomy, which turns out to be $v(0)U(\tau)v^{-1}(\tau)$. The choice $U(\tau)v^{-1}(\tau)$ doesn't suffice, because it violates the initial condition $U(0) = 1$, which is a part of the definition of holonomy. – Prof. Legolasov Feb 21 '17 at 11:32

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 $U_P(x,x,A)=1$.)
• In a first-order differential equation, we can get the numerically-solved solution from discrete values of the derivatives with small spacing. For clearness, I label $A_{\mu}^a$ along the motion of the gauge transformation as $A_{\mu}^a(i)\,, i\in \mathbb{Z}$ in a discrete fashion. Then $A_{\mu}^a(i)$ is transformed sequentially $$A_{\mu}^a(i=0) \to A_{\mu}^a(i=1) \to A_{\mu}^a(i=2) \to A_{\mu}^a(i=3) \cdots$$ and $U_P(x,y,A)$ is also transformed sequentially $$U_P(x,y,A)(i=0) \to U_P(x,y,A)(i=1) \to U_P(x,y,A)(i=2) \to U_P(x,y,A) \cdots$$ – GotchaP Feb 19 '17 at 23:06
• $U_p\left(z,y,A^V\right)=V(z)U_p(z,y,A)V^{\dagger}(y)$(15.59) is about the evolution of $U_P(x,y,A)$ with regard to $A$. So the differential equation were differentiated by $A_{\mu}^a$(or $\alpha^a$), I could understand. Does your answer resolve my question? – GotchaP Feb 19 '17 at 23:14
• @GotchaP The gauge field $A$ has some fixed values everywhere, and then given those values $U_P(x,y,A)$ is some function of that field configuration (in particular we need to integrate those gauge field values over some path). The differential equation for $U_P$ just reformulates the problem of working out $U_P$ (an integral is traded for a differential equation), and this differential equation can be used to determine $U_P$ numerically if you like. However $A$ is not being evolved in that differential equation, except in that as we move along the path we sample different gauge field values. – diracula Feb 20 '17 at 15:24
• $U_P$ is a function of the field configuration $A$, and hence it has dependence on those field values, in particular the field values in this loop that we're integrating around. But we have never written a differential equation like $\frac{\partial U_P}{\partial A}=\ldots$, by which we could figure out the dependence of $U_P$ on $A$. (The differential equation we do have tells us about the dependence of $U_P$ on a space-time coordinate $x$.) Essentially we already know the dependence on $A$, from the expression for $U_P$. But the logic in the book is just a shortcut to a neat relation. – diracula Feb 20 '17 at 15:29
• From $\frac{dx^{\mu}}{ds} D_{\mu}(A^V) U_p(x,y,A^V)=0$, $\frac{dx^{\mu}}{ds} D_{\mu}(A^V) V(x)U_p(x,y,A)V^{\dagger}(y)=0$ and tha uniqueness of the solution of a first-order differential equation, we can state that $U_p\left(x,y,A^V\right)=V(x)U_p(x,y,A)V^{\dagger}(y)$. Now I understand. Thank you so much! – GotchaP Feb 22 '17 at 1:57