q+1}\hat{b}_{p, q}^k  \right) + {\rm H.c.}
                   \Bigr ],
\label{ham_ss_h1}\end{aligned}$$ with the SF order parameter $\phi^{k}_{p,q}$ as the perturbation parameter. Then, from the first-order perturbative correction to the ground state wavefunction (details given in Appendix \[appendix\_a\]), we have $$\!\!\!\!\!\phi^{k}_{p,q} = J\bar{\phi}^{k}_{p,q} 
    \left(\frac{n^{k}_{p,q} +1}{n^{k}_{p,q}U -\bar{\mu}^{k}_{p,q}}
        -\frac{n^k_{p,q}}{(n^k_{p,q} -1)U -\bar{\mu}^{k}_{p,q}}\right),
 \label{order_par_ana}$$ with $$\begin{aligned}
   \bar{\mu}^{k}_{p,q} &= &\tilde{\mu}^{k}_{p,q}-U_{12}n^{3-k}_{p,q},
                          \nonumber \\
   \bar{\phi}^{k}_{p,q} &= &\left(\phi^{k}_{p+1,q}+\phi^{k}_{p-1,q}
                           +\phi^{k}_{p,q+1}+\phi^{k}_{p,q-1}\right).
                           \nonumber\end{aligned}$$ For a homogeneous lattice system $\varepsilon^{k}_{p,q} = 0$. Then, in the MI phase the total density $\rho = \rho^{1}+\rho^{2}$ is integer commensurate and $\phi^{k}_{p,q}=0$. In the SF phase, the order parameter is non-zero and uniform, say $\phi^{k}_{p,q}=\varphi^{k}_{0}$. With these considerations, $\bar{\phi}^{k}_{p,q} = \bar{\phi}^{k} = 4\varphi^{k}_{0}$. Starting from the SF phase, at the SF-MI phase boundary $\varphi^{k}_{0}\rightarrow 0^{+}$. Considering this limit in Eq. (\[order\_par\_ana\]), we obtain the equation which defines the MI-SF phase boundary in terms of $J$ for a particular value of $\mu$.

### TBHM

For the MI phase with $\rho = 2$,in the miscible domain, atoms of both the species fill all the lattice sites. That is, $n^1_{p,q} =n^2_{p,q} =1$. The MI-SF phase boundary is, then, defined by $$\begin{aligned}
   \frac{1}{4J}= \frac{2}{U -\mu +U_{12}} +\,\frac{1}{\mu -U_{12}}.
   \label{tbhm_even}\end{aligned}$$ On the other hand for finite $U_{12}$, the system is in the immiscible domain 