(28)). The total Hamiltonian $\overline{H}$ (43), we are interested in, is much more complicated than $\overline{%
\overline{H}}^{(1)}$ and $\overline{\overline{H}}^{(2)}$alone (see (41)-(43)) because of the presence of the projection operators $P^{(1)}$ and $P^{(2)}$ (see (34)-(37)) in the expression (43).

It is also interesting to see what happens in the special case when taking the limit $U\rightarrow \infty $. The second term in the parentheses of Eq.(43) vanishes in this limit (cf (35) and (42)). Besides, each lattice site cannot be at the same time occupied by two electrons what is equivalent to the assumption ($\alpha =1,2)$

$$n_{I,\alpha ,\sigma }^b=b_{I,\alpha ,\sigma }^{+}b_{I,\alpha ,\sigma
}=n_{I,\alpha ,\sigma }n_{I,\alpha ,-\sigma }=0,$$

$$n_{I,\alpha }^b=\sum\limits_{\sigma} n_{I,\alpha ,\sigma }^b=0$$

and (cf (38))

$$\widetilde{a}_{I,\alpha ,\sigma }=a_{I,\alpha ,\sigma }.$$

The total Hamiltonian (43) is thus given by a simple formula

$$\begin{array}{ll}
\overline{H}=P^{(1)}\overline{\overline{H}}= & -t\sum\limits_{I,\sigma
}\left[ a_{I,1,\sigma }^{+}a_{I,2,\sigma }+a_{I,2,\sigma }^{+}a_{I,1,\sigma
}\right] \\
&  \\
& -t\sum\limits_{I,\sigma }\left[ a_{I,2,\sigma }^{+}a_{I+1,1,\sigma
}+a_{I+1,1,\sigma }^{+}a_{I,2,\sigma }\right] \\
&  \\
& -t\sum\limits_{I\neq J,\sigma }\sum\limits_{\alpha =1}^2a_{I,\alpha
,\sigma }^{+}a_{J,\alpha ,\sigma }.
\end{array}$$

Going back to the original lattice (cf (4),(5) and (1)) it is easy to see that the exact Hubbard Hamiltonian in the limit $U\rightarrow \infty $ takes on the form

$$\overline{H}=-t\sum\limits_{i,j,\sigma }a_{i,\sigma }^{+}a_{j,\sigma }$$

where $i$ 