I'm working out some calculations with Wilson lines, defined as path-ordered exponential integrals of a gauge field:
$$U = \mathcal{P}\exp\biggl(ig\int_{-\infty}^{\infty}\mathrm{d}x^\mu T^c A_{c\mu}\biggr)$$
Here $g$ is the vertex factor for strong interactions, $A_{c\mu}$ is the gluon field which carries an $\mathrm{SU}(3)$ index $c\in\{1,\ldots,8\}$ and a Lorentz index $\mu$, and $T^c$ is an $\mathrm{SU}(3)$ generator in the fundamental representation. (Repeated indices are summed, as usual.) I can also write the equivalent in the adjoint representation,
$$W = \mathcal{P}\exp\biggl(ig\int_{-\infty}^{\infty}\mathrm{d}x^\mu \tilde{T^c} A_{c\mu}\biggr)$$
where now $\tilde{T^c}$ is an $\mathrm{SU}(3)$ generator in the adjoint rep.
I'm trying to prove the identity
$$\boxed{W^{ab} = 2\operatorname{Tr}\bigl[T^a U T^b U^\dagger\bigr]}$$
but I'm kind of stuck on how to handle the generators in the exponentials.
So far I see two possible ways to go about this: I can use the identity $(\tilde{T^c})^{jk} = 2i\operatorname{Tr}\bigl(T^c\bigl[T^j,T^k\bigr]\bigr)$ to write
$$W^{ab} = \mathcal{P}\exp\biggl(ig\int_{-\infty}^{\infty}\mathrm{d}x^\mu (2i)\operatorname{Tr}\bigl(T^c\bigl[T^a,T^b\bigr]\bigr) A_{c\mu}\biggr)$$
and then try to separate that into two factors of the form $TU$, but I think I need at least some identity like
$$T^a \mathcal{P}\exp\bigl(i T^c X_c\bigr) \overset ?= \mathcal{P}\exp\bigl(i (\text{something}) X_c\bigr)$$
and I can't find one.
Alternatively, I can start from the other side and expand to first order,
$$\begin{align}
2\operatorname{Tr}\bigl[T^a U T^b U^\dagger\bigr]
&\approx 2\operatorname{Tr}\biggl[T^a \biggl(1 + ig\int\mathrm{d}x^\mu T^c A_{c\mu}\biggr)T^b \biggl(1 – ig\int\mathrm{d}x^\mu T^c A_{c\mu}\biggr)\biggr] \\
&= 2\operatorname{Tr}\biggl[T^a T^b + ig\int\mathrm{d}x^\mu \bigl(T^c \bigl[T^b, T^a\bigr]\bigr) A_{c\mu}\biggr]\quad\text{(cyclic property in trace)} \\
&= \delta^{ab} + g\int\mathrm{d}x^\mu (\tilde{T^c})^{ab} A_{c\mu}
\end{align}$$
which gets close, but this leaves me without a needed factor of $i$, and also it's only approximate, since there's no reason to assume higher order terms will completely vanish (e.g. no $g\to 0$ limit or anything like that).
Can anyone suggest another method, or fill in the gaps?
Best Answer
For SU(N), the adjoint representation can be obtained from a tensor product of the fundamental representation with its dual and projecting out the scalar. Thus we can replace the adjoint representation indices $a, ...$ by double indices $i \bar{j}$ and write this relation as:
$$W^{i \bar{j}}_{k \bar{l}} = U^i_k U^{\dagger j}_l - \frac{1}{N} \delta^i_k \delta^j_l$$
(The factor $\frac{1}{N}$ can be obtained by taking the traces in the case $U=\mathbb{1}$).
On the other hand, the generators of the Lie algebra are the traceless Hermitian matrices, which have the following form in the double index notation:
$$ T^{i\bar{j}} = \frac{1}{\sqrt{2}}(E^{ij} - \frac{\mathbf{1}}{N} \delta^{ij})$$
The prefactor consists of a standard normalization and $\mathbb{1}$ is the unit matrix in the fundamental representation.
Substituting the second equation into the required identity gives you the first equation.