at $V=\pm 0.519$ (metal) and at $V=0$ (insulator). $\epsilon_c=0$ and $\epsilon_a=2$. []{data-label="fig:e1"}](fig2.eps){width="65mm"}

Consider the Hubbard model $$H = \sum_{ij\sigma} t_{ij} c_{i\sigma}^\dagger c_{j\sigma} 
  + \frac{U}{2} \sum_{i\sigma} n_{i\sigma} n_{i-\sigma}$$ at temperature $T=0$ and chemical potential $\mu = U/2$. For symmetric conditions this implies half-filling. The Hamiltonian of the reference system is given by $H'= \sum_i H'(i)$ with $$\begin{aligned}
  H'(i) &=& \sum_{\sigma} \epsilon^{(c)}_{i} c_{i\sigma}^\dagger c_{i\sigma} 
            + \frac{U}{2} \sum_{\sigma} n_{i\sigma} n_{i-\sigma} 
        \nonumber \\
    &+& \sum_{\sigma,k=2}^{n_{\rm s}} \epsilon^{(a)}_{i,k} a^\dagger_{ik\sigma} a_{ik\sigma}
     +  \sum_{\sigma,k} \left( V_{i,k} c_{i\sigma}^\dagger a_{ik\sigma}  + \mbox{h.c.} \right) \: .
     \nonumber \\
\label{eq:siam2}\end{aligned}$$ For the sake of simplicity we consider a homogeneous paramagnetic phase and the most simple case $n_{\rm s}=2$ where one is left with three independent variational parameters only, namely $V \equiv V_{i,k=2}$, $\epsilon_a \equiv \epsilon^{(a)}_{i,k=2}$, and $\epsilon_c \equiv \epsilon^{(c)}_{i}$. The computation of the different contributions to the grand potential, Eq. (\[eq:vvv\]), is straightforward: Diagonalization of $H'$ yields the ground-state energy $E_0'$ and $\Omega' = E_0' -\mu \langle N' \rangle$ as well as the excitation energies, the ground state and the excited states. The Green’s function ${\bf G}'$ and the free Green’s function ${\bf G}'_0$ can be computed from their respective Lehmann representations. The self-energy of the reference system is obtained as 