Wilson loop as path integral of parallel transport action I am trying to get that the path integral of the parallel transport action is the Wilson loop. Here is the setting:
Let $w$ be a complex vector dimension $N$, and $A_{\mu}$ a fixed Yang-Mills connection. We will work with $G=SU(N)$. Using the parallel transport equation and the constraint:
$$i\frac{dw}{d\tau} = \frac{dx^{\mu}}{d\tau} A_{\mu}w$$
$$w^{\dagger}w=1 $$
We construct the following action with a Lagrange multiplier $\lambda$ enforcing the constraint above. Evidently the equation of motion of this action is the parallel transport equation.
$$S_w = \int  \left(i w^{\dagger}\frac{dw}{dt} + \lambda(w^{\dagger}w-1)+w^{\dagger}A(x(\tau))w \right)d\tau$$
This vector satisfies:
$$[w_i,w_j^{\dagger}]=\delta_{ij}$$
Now let $\tau \in \mathbf{S}$ to allow for large gauge transformations. I finally arrive to the following path integral, where I had to insert the $w_i$ factors for it to not vanish:
$$ Z_w[A]= \int e^{iS_w (w;\lambda;A)} w_i(\tau=\infty)w_i^{\dagger}(\tau=-\infty) \mathcal{D}\lambda\mathcal{D}w \mathcal{D}w^{\dagger}$$
I am supposed to get that 
$$Z_w[A] = tr \mathcal{P}e^{i\int A d\tau}$$
How do I compute that specific path integral? 
 A: I will follow Monopoles and Wilson Lines, by David Tong, Kenny Wong.
Let's work in perturbation theory. Action:
$$
S_w = \int  \left(i w^{\dagger}\frac{dw}{dt} + \alpha(w^{\dagger}w-\kappa)+w^{\dagger}A(x(\tau))w \right)d\tau
$$
Propogator for free (non-interacting) part of action, in following we use $\langle \dots \rangle$ for averaging by free theory:
$$
\langle w_i^\dagger(\tau_1) w_j(\tau_2)\rangle = \theta(\tau_1-\tau_2) \delta_{ij}
$$


*

*Vacuum bubbles
As usual in perturbation theory, we can factorise vacuum diagrams. You will obtain up to notations:

All $n ≥ 2$ factors vanish because the product of the step functions vanishes everywhere except on a set of measure zero.
So we have only one contribution ($\theta(0)=1/2$):
$$
\exp\left(i (N/2 - \kappa )\int dt\; \alpha(t)\right)
=
\exp\left(-i \kappa_{eff} \int dt\; \alpha(t)\right)
$$


*Path integral with insertions
$$
Z_w[A]= \int\mathcal{D}\lambda\mathcal{D}w \mathcal{D}w^{\dagger}\; e^{iS_w (w;\lambda;A)} w_i(\tau=\infty)w_i^{\dagger}(\tau=-\infty)
$$
This integral correspond to following series of diagrams (times to vacuum bubbles factor):

This series correspondence to:

Including vacuum bubbles we left with:
$$
Z_\omega[A] = W[A]\int D\alpha e^{-i\int dt \;\alpha(t) (\kappa_{eff}-1)}
$$
If $\kappa_{eff}= 1$, we obtain:
$$
Z_\omega[A] = W[A]
$$
