This question is prompted by Exercise 25.6 of Schwartz's Quantum Field Theory and the Standard Model. The first part of the exercise asks you to show that the path-ordering in the definition of a Wilson line is necessary for the gauge transformation property at leading non-trivial order in perturbation theory. I am having trouble getting rid of one term in the calculation, and I don't know if I have a physical misunderstanding or a mathematical one.
For simplicity, I'll consider the case where the starting potential vanishes. So I want to show that for any gauge transformation $U$ and any path $P$ from $y$ to $x$ we have \begin{equation} \tag{$Q$} P\operatorname{exp}\left( ig\int_{y}^{x} dz^{\mu}\, (\partial_{\mu}\alpha^{a})(z)\, T^{a} \right) = U(x)\, U(y)^{-1} \end{equation} where $$ U(x) = \exp\left(ig\alpha^{a}(x)\, T^{a}\right) $$ To zeroth order in $g$ both sides are just $1$, and at order $g$ they are both $ig(\alpha^{a}(x) - \alpha^{a}(y))T^{a}$.
My difficulty arises at order $g^{2}$. For the right hand side, I have \begin{equation} \tag{$R$} - \frac{1}{2} g^{2} \alpha^{a}(x)\, \alpha^{b}(x)\, T^{a} T^{b} + g^{2} \alpha^{a}(x)\, \alpha^{b}(y)\, T^{a} T^{b} - \frac{1}{2} g^{2} \alpha^{a}(y)\, \alpha^{b}(y)\, T^{a}T^{b} \end{equation} where the first and last terms come from the $\mathcal{O}(g^{2})$ parts of $U(x)$ and $U(y)^{-1}$, respectively, and the middle term from their $\mathcal{O}(g^{1})$ parts. However, for the path-ordered exponential I calculate as follows. By definition of the path-ordered exponential, the $\mathcal{O}(g^{2})$ part is $$ - \frac{1}{2} g^{2} \int_{0}^{1} d\lambda \int_{0}^{1} d\tau \frac{dz^{\mu}(\lambda)}{d\lambda} \frac{dz^{\nu}(\tau)}{d\tau} (\partial_{\mu}\alpha^{a})(z(\lambda))\, (\partial_{\nu}\alpha^{b})(z(\tau)) \left[ T^{a} T^{b}\, \theta(\lambda - \tau) + T^{b} T^{a}\, \theta(\tau - \lambda) \right] $$ At this point, I would use the $\theta$ functions to set the bounds of integration on one of the two integrals. Taking the integral over $\tau$ in both cases, the expression becomes $$ - \frac{1}{2} g^{2} \int_{0}^{1} d\lambda \int_{0}^{\lambda} d\tau \frac{dz^{\mu}(\lambda)}{d\lambda} \frac{dz^{\nu}(\tau)}{d\tau} (\partial_{\mu}\alpha^{a})(z(\lambda))\, (\partial_{\nu}\alpha^{b})(z(\tau))\, T^{a} T^{b}\\ - \frac{1}{2} g^{2} \int_{0}^{1} d\lambda \int_{\lambda}^{1} d\tau \frac{dz^{\mu}(\lambda)}{d\lambda} \frac{dz^{\nu}(\tau)}{d\tau} (\partial_{\mu}\alpha^{a})(z(\lambda))\, (\partial_{\nu}\alpha^{b})(z(\tau))\, T^{b} T^{a} $$ Applying Stokes' theorem to the $\tau$ integrals, we obtain $$ \frac{1}{2} g^{2} \alpha^{b}(y) \int_{0}^{1} d\lambda \frac{dz^{\mu}(\lambda)}{d\lambda} (\partial_{\mu}\alpha^{a})(z(\lambda))\, T^{a} T^{b} - \frac{1}{2} g^{2} \alpha^{b}(x) \int_{0}^{1} d\lambda \frac{dz^{\mu}(\lambda)}{d\lambda} (\partial_{\mu}\alpha^{a})(z(\lambda))\, T^{b} T^{a} \\ - \frac{1}{2} g^{2} \int_{0}^{1} d\lambda \frac{dz^{\mu}(\lambda)}{d\lambda} (\partial_{\mu}\alpha^{a})(z(\lambda))\, \alpha^{b}(z(\lambda))\, [T^{a}, T^{b}] $$ Applying Stokes' theorem again to the top line and shuffling some indices, we obtain \begin{equation} \tag{$L$} - \frac{1}{2} g^{2} \alpha^{a}(x)\, \alpha^{b}(x)\, T^{b} T^{a} + g^{2} \alpha^{a}(x)\, \alpha^{b}(y)\, T^{a} T^{b} - \frac{1}{2} g^{2} \alpha^{a}(y)\, \alpha^{b}(y)\, T^{a} T^{b} \\ - \frac{1}{2} g^{2} \int_{0}^{1} d\lambda \frac{dz^{\mu}(\lambda)}{d\lambda} (\partial_{\mu}\alpha^{a})(z(\lambda))\, \alpha^{b}(z(\lambda))\, [T^{a}, T^{b}] \end{equation} This differs from Eq. ($R$) in that it has an extra commutator term. So it seems that I need to show \begin{equation} \tag{$\star$} \int_{y}^{x} dz^{\mu}\, (\alpha^{a}\partial_{\mu}\alpha^{b})(z)\, [T^{a}, T^{b}] = 0 \end{equation} However, I don't see how to proceed.
Therefore, my question(s) are:
- Are Eqs. ($L$) and ($R$) indeed correct expressions for the left and right hand sides of Eq. ($Q$)? I.e., did I expand the exponentials in $U(x)$ correctly and apply the definition of $P\operatorname{exp}$ and Stokes' theorem properly?
- How can I show ($\star$)? Since $[T^{a}, T^{b}]$ is anti-symmetric in $a$ and $b$, a natural strategy would be to try to show that the integral is symmetric in $a$ and $b$. But it's not: integration by parts and the anti-symmetry of $[T^{a}, T^{b}]$ make the integral anti-symmetric in $a$ and $b$ as well. I tried looking at simple example gauge transformations and seemed to find that ($\star$) is false in those examples, so I'm making a mistake somewhere.
