$\sim \hat 1$ & {\sc yes} &
{\sc no} & {\sc no} &\cr
\noalign{\vskip6pt}
& $G_M^{T=1}$ & $A_k$ & $\Omega^0 + \Omega^1$ &
$\sim \tau^3$ & {\sc no} + {\sc yes} &
{\sc no} & {\sc yes} + {\sc no} &\cr
\noalign{\vskip6pt}
}
\hrule height1pt  }$$

The electric scalar form factor can be immediately extracted from () and (). Since only the isoscalar part of the operator $\hat Q$ contributes the rotational matrices $R(t)$ vanish from the action. Hence, the $D{\cal R}$ integration in () becomes trivial. Also, in this case the contribution from the terms linear in $\Omega$ is exactly equal to zero. Finally we arrive at:

$$\openup3\jot
\eqalign{
G_E^{T=0} (q^2) = &\int d^3 x\ e^{-i \vec q \vec x}
\ {N_c\over 3}
\ \Bigg\lbrace \sum_{n\in VAL} \psi_n^{\dagger}(\vec x) \,
\psi_n(\vec x)\ - \cr
&- \ \sum_{n\in ALL} \ {1\over 2}\ {\rm sign}(\epsilon_n)
\ \psi_n^{\dagger}(\vec x) \, \psi_n(\vec x)
\ \Bigg\rbrace \ , \cr }
\eqno(\ELSCAL)$$

where [val]{} denotes the valence whereas [all]{} stands for all eigenstates of the Hamiltonian including the valence one as well. The first term in eq.() comes from the valence quarks and the second term is due to the quark vacuum polarization. As has 