as $| A \rangle \langle A |$ and $| B \rangle \langle B |$ in the covalent limit. On the other hand, both electrons occupy $ | A \rangle$ in the ionic limit and thus these two matrices should equal $| A \rangle \langle A |$ as $| \Psi _{i} \rangle$ is dominated. By taking [@Huang1] $$\begin{aligned}
\rho ^{(1)} = | A \rangle \langle A | \text{ and } \rho ^{(2)} = | L \rangle \langle L |              \end{aligned}$$ with $ | L \rangle = \alpha _ {i} | A \rangle + \alpha _{c} | B \rangle$ in $\mathbb{C} ^{2} _{ \Omega }$, the matrix $\rho ^{(1)}$ just represents the quasielectron occupying $| A \rangle$ under both limits. In addition, the matrix $\rho ^{(2)}$ corresponds to the other one jumping to $| A \rangle$ from $| B \rangle$ as $| \Psi _{i} \rangle$ becomes significant. The operator $\rho _{sb} = \frac{1}{2} (\rho ^{(1)} + \rho ^{(2)} + d ^{(1)} )$ satisfies $\langle {\bf r} ^{\prime} | \rho _{sb} | {\bf r} ^{ \prime \prime } \rangle = \langle \Psi _{b} | \psi _{ \uparrow } ^{ \dagger } ({\bf r} ^{ \prime } ) \psi _{ \uparrow } ({\bf r} 