correspond to the negative of the transverse component of the chromo-electric field, $\bm{\Pi}^a(\x) = - 
 \E^a_T(\x)$, $\na \cdot \E^a_T = 0$. The chromo-magnetic field, $\B$ contains linear and quadratic terms in $\A$. It will also be convenient to transform to the momentum space components of the fields by $$\A^a(\k) = \int d\x\A^a(\x) e^{-i\k\cdot x},$$ and similarly for $\bm{\Pi}^a(\k)$. The Coulomb potential ${\hat V}_C$ may be expressed in terms of the longitudinal component of the chromo-electric field, $${\hat  V}_C = \frac{1}{2}\int d\x {\bf E}^a(\x) {\bf E}^a(\x),$$ with $${\bf E}^a(\x) = \int d\y d\z { \frac{\x - \y}{4\pi|\x - \y|^3}} 
 \left[ \frac{g}{1 - \lambda}\right]^{ab}_{\y,\z} \rho^b(\z). \label{e}$$ Here $(1-\lambda)$ is the Faddeev-Popov (FP) operator which in the configuration-color space is determined by, $$[\lambda]^{ab}_{\x,\y} = \int {\frac{d\p}{(2\pi)^3}} {\frac{d\q}{(2\pi)^3}} 
   e^{i\p\cdot\x} e^{-i\q\cdot\y} 
\lambda_{ab}(\p,\q),$$ where $$\lambda_{ab}(\p,\q) = ig f_{acb} {\frac{\A^c(\p-\q)\cdot \q}{\q^2}},$$ $f$ are the $SU(N_c)$ structure constants, and $g$ is the bare coupling. In [Eq. (\[e\])]{}, $\rho$ is the color charge density given by $$\rho^a(\x) = \psi^{\dag}(\x)T^a \psi(\x) + f_{abc} \A^b(\x)\cdot 
\bm{\Pi}^c(\x),$$ with the two terms representing the quark and the gluon contribution, respectively; the former is replaced by a $c$-number for static quarks. Without light flavors there is no other dependence on the quark degrees of 