$d^{\dag}$ an anti-quark. From we see that $b$ and $d^{\dag}$ annihilate the drained state, which means that this state is empty of quarks, and filled with anti-quarks. The local baryon density operator is $$B_x \equiv \frac1{N}\sum_{a=1}^{N} \left[ b^{\dag a} b^{a} - d^{\dag a} d^{a} \right]_x =\frac1{N} \sum_a\psi^{\dag\,a}_x \psi^a_x - 1. \label{B}$$ According to , the baryon number of the drained state is $-1$, corresponding to filling the site with anti-quarks. The vacuum $|0\rangle$ is the state with no quarks and no anti-quarks. This state is the filled Dirac sea on a single site and obeys, $$b|0\rangle=d|0\rangle=0.$$ The baryon number of this state is $B=0$. Because of the Pauli exclusion principle we cannot put too many fermions on a single site. For a single flavor theory the maximum number of local baryon number will be $1$, and is found only in the state $|\textrm{filled} \rangle$ $$b^{\dag}|\textrm{filled}\rangle=d|\textrm{filled}\rangle=0.$$ Below we will see that gauge invariance puts more restrictions on the single site Hilbert space in order that it be color neutral.

Moving to the gauge Hilbert space, we also write it as a direct product of the form $$|\Omega\>_G = \prod_{\otimes x} \left( |\Omega\>_G \right)_{x}.$$ Next, we denote the state with no electric 