\left(-{1\over \na^2}\right)
       {g\over {1 - \lambda}}\right]_{{\R\over 2},-{\R\over 2}}
 |\Psi  \rangle }
/{ \langle \Psi|\Psi \rangle } 
}, \nonumber \\ 
&\hspace{-5mm} = -C_F \int {{d\p}\over {(2\pi)^3}} {{d^2(p) f(p)}\over {p^2}}
 e^{i\p\cdot \R},
\label{vc} \end{aligned}$$ and the net self-energy contribution is, $$\begin{aligned}
\Sigma &=  C_F {{\langle \Psi | 
\left[ {g\over {1 - \lambda}} \left(-{1\over \na^2}\right)
       {g\over {1 - \lambda}}\right]_{\pm {\R\over 2},\pm {\R\over 2}}
 |\Psi  \rangle }
/{ \langle \Psi|\Psi \rangle } }, \nonumber \\
 &=  C_F \int {{d\p} \over {(2\pi)^3}} {{d^2(p) f(p)}\over {p^2}}.
 \label{sigma} \end{aligned}$$ In lattice simulations it has been shown [@Greensite] that the Coulomb energy and the phenomenological static $Q\bar Q$ potential obtained from the Wilson loop are different. In particular it was found that the Coulomb potential string tension is about three times larger than the phenomenological string tension. This is in agreement with the “no confinement without Coulomb Confinement” scenario discussed by Zwanziger [@zwancon]. It is simple to understand the origin of the difference. Even if $|\Psi[\A]\rangle $ were the true vacuum state (without sources) of the Coulomb gauge QCD Hamiltonian (here we approximate it by a variational ansatz) the state $|Q{\bar Q},R\rangle \equiv  Q(\R/2){\bar Q}(-\R/2) |\Psi[A]\rangle$ is no longer an eigenstate. For example ${\hat V}_C$ acting on $|Q{\bar
  