\overrightarrow{S}_{I,2}-\frac 14n_{I,1}^an_{I,2}^a) \\ 
&  \\ 
& +2(b_{I,1,\uparrow }^{+}a_{I,1,\downarrow }^{+}a_{I,2,\downarrow
}b_{I,2,\uparrow }+b_{I,2,\uparrow }^{+}a_{I,2,\downarrow
}^{+}a_{I,1,\downarrow }b_{I,1,\uparrow }) \\ 
&  \\ 
& +(1-n_{I,1}^a)n_{I,2}^b+(1-n_{I,2}^a)n_{I,1}^b-n_{I,1}^bn_{I,2}^b]
\end{array}
\\ 
&  \\ 
& +\frac t{U\sqrt{1+(\frac{4t}U)^2}}\sum\limits_{\sigma} \sum\limits_{\alpha
=1,2}(a_{I,\alpha ,\sigma }^{+}b_{I,\overline{\alpha },\sigma }+b_{I,\alpha
,\sigma }^{+}a_{I,\overline{\alpha },\sigma }).
\end{array}$$

Now, the Hamiltonian (14) can also be obtained from (5) (or (7)) with the use of the projection operator (16) or (17) using the relation

$$\overline{H}_I^d=P_IH_I^dP_I$$

and applying a straightforward but laborious algebraic calculation.

Hubbard model for large $U$
===========================

The Hamiltonian of the whole crystal (4) can be expressed (similar to (7)) in the form $$H=\sum\limits_{\gamma} \overline{E}_{\gamma} |\overline{E}_{\gamma} \rangle
\langle \overline{E}_{\gamma} |$$ with unknown energies $\overline{E}_{\gamma} $ and eigenvectors $|\overline{E%
}_{\gamma} \rangle $. We can, however, expand the eigenvectors $|\overline{E}%
_{\gamma} \rangle $ in the series of the dimer eigenvectors (see (A.1))

$$|\overline{E}_{\gamma} \rangle =\sum\limits_{\gamma _1,..,\gamma
_M}c_{\gamma _{1,}.._{,}\gamma _M}^{\gamma} |E_{\gamma _1}\rangle
..|E_{\gamma _M}\rangle$$

assuming that the crystal consists of $M$ dimers. Using (19) and (20) we obtain

$$H=\sum\limits_{\gamma} \overline{E}_{\gamma} \sum\limits_{\gamma
_1,..,\gamma _M}\sum\limits_{\gamma _1^{^{,}},..,\gamma _M^{,}}c_{\gamma
_{1,}.._{,}\gamma _M}^{\gamma} c_{\gamma _{1,}^{,}.._{,}\gamma
_M^{,}}^{\gamma *}|E_{\gamma _1}\rangle ..|E_{\gamma _M}\rangle \langle
E_{\gamma _1^{,}}|..\langle E_{\gamma _M^{,}}|.$$

It is clear that to obtain the Hubbard Hamiltonian for large $U$ we have to project (21) onto the subspace of the lowest 