down-spin density $\langle {\bf r} | \rho _{sb} | {\bf r} \rangle $ at position ${\bf r}$ includes the factor $\langle {\bf r} | d ^{(1)} | {\bf r} \rangle$, which follows $\int _{ {\bf r} \in \Omega}  d ^{3} r  \langle {\bf r} | d ^{(1)} | {\bf r} \rangle=tr d ^{(1)} =  ( \sqrt{2} -1) tr [ ( I - \rho ^{(1)}) \rho ^{(1)} \rho ^{(2)}  +  \rho ^{(2)} \rho ^{(1)}  ( I - \rho ^{(1)}) ]=0 $ because $( I - \rho ^{(1)}) \rho ^{(1)}= \rho ^{(1)}  ( I - \rho ^{(1)})=0$ under Eq. (11). Therefore, $\langle {\bf r} | d ^{(1)} | {\bf r} \rangle$ has no contribution to the total charge $2 \int _{ {\bf r} \in \Omega}  d ^{3} r \langle {\bf r} | \rho _{sb} | {\bf r} \rangle $, and represents the fluctuating charge density due to the coexistence of the ionic and covalent components. I note that $\delta \rho$, the deviation from the average density [@Zhang] due to the correlation in the quantum Hall effect [@DHLee; @You], also has no contribution to the total charge.

To include the correlation energy due to $\hat{U}$ in the quasielectron space, we can further introduce 