species at each lattice sites are the same $n^1_{p,q}=n^2_{p,q}$. From this the $\Delta \rho^1=-\Delta\rho^2$ states are obtained after translation of one of the species by one lattice constant either along $x$ or $y$ direction. Thus, in the latter we have $n^{1,A}=n^{2,B}$ and $n^{1,B}=n^{2,A}$. It is to be noted that the $\rho=1$ phase of the TBHM has the same average density as the DW(1,0). However, the two have different symmetries. The $\rho=1$ phase of the TBHM has atoms from the two species with random occupancies and has no diagonal long range order. But, the DW(1,0) has diagonal order arising from the non-zero $\Delta\rho^k$. As an example consider the DW(1,0) phase, the two degenerate states correspond to $\Delta\rho^1=\Delta\rho^2=1$ and $\Delta\rho^1=-\Delta\rho^2=1$. At higher $\mu$, the DW(2,1) intervenes the transition from MI(1,1) to MI(2,2) phase.

To study the effect of the inter-species interaction we increase $U_{12}$ retaining $V_{12}$ and $V_k$ fixed at $0$ and $0.05U$, respectively. The phase diagram corresponding to $U_{12} = 0.4U$ is shown in Fig. \[ebhm\_v12\_0\] (b). At finite $U_{12}$ the MI phase is energetically costly due to repulsion between atoms of the two-species co-existing on the same lattice site. So MI phase shifts to higher $\mu/U$ values with increasing $U_{12}$ 