In $\mathcal{N} = 4$ Super-Yang mills there are only massless particles. If one wishes to obtain a heavy quark one can see the SYM theory as a stack of (N+1)-branes in AdS$_5 \times$S$^5$ where one brane has been higssed, i.e we separete one brane from the stack and take it to infinity. Once we have the massive quark we can find the Wilson-Loop operator for the SYM theory, wich is given by App. A:

$W(\mathcal{C}) = \frac{1}{dim(\mathcal{R})}Tr_{\mathcal{R}} \mathcal{P} exp(\oint (iA_{\mu}\dot{x}^{\mu}+\Phi_i\dot{y}^i)ds)$

Where $\mathcal{R}$ is the dimension of the representation of the gauge group, $A_{\mu}$ is the gauge field and $\Phi_i$ are the fields of the R-symmetry of the theory.

The vev of the wilson loop operator:

$<W(\mathcal{C})> = 1 - g_{YM}^2 \frac{Tr(t^at^b)}{dim(\mathcal{R})} \oint ds \oint ds' [\dot{x}^{\mu}(s) \dot{x}^{\nu}(s')G_{\mu \nu}(x(s) - x(s')) - \dot{y}^i(s) \dot{y}^j(s')G_{ij}(x(s) - x(s'))] $

Where one chooses the branc $<A_{\mu}>=0$. And $G_{\mu \nu}$, $G_{ij}$ are the gauge and scalar propagator. So in order to get the vev for the wilson loop one has to obtain the propagators:

$<A_{\mu}^a(s) A_{\nu}^b(s')> = \int \mathcal{D}S e^{-S_{SYM}} A_{\mu}^a(s) A_{\nu}^b(s')$

And the same for the scalar fields $\Phi_i$. The lagrangian for SYM (doing $\Phi_i \mapsto X_i$):

$L = \operatorname{tr} \left\{-\frac{1}{2g^2}F_{\mu\nu}F^{\mu\nu}+\frac{\theta_I}{8\pi^2}F_{\mu\nu}\bar{F}^{\mu\nu}- i \overline{\lambda}^a\overline{\sigma}^\mu D_\mu \lambda_a -D_\mu X^i D^\mu X^i +g C^{ab}_i \lambda_a[X^i,\lambda_b] + g \overline{C}_{iab}\overline{\lambda}^a[X^i,\overline{\lambda}^b]+\frac{g^2}{2}[X^i,X^j]^2 \right\}$

Wich terms of the lagrangian are the ones that matter and why? (ie. I only need the free terms of the lagrangian?, why?). How is the method for attacking this problem (getting correlation/propagators) in QFT?.


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