\Big> \right)_{ \bf j} \right\},$$ where the upper and lower positions are the two ($xz/yz$) orbitals, the arrows show spin and the two particle states form the standard triplet. The main task now is to consider the matrix elements of (\[hamt\]-\[hamt2\]) in the basis (\[site basis\]).

Because the Hilbert space is 12-dimensional there are 144 different matrix elements. However, it is easy to see that only a small number of them is nonzero. In order to find them one can use a graphical technique ilustrated in Table \[tab:hop\].

The matrix elements listed above and their complex conjugates constitute all nonzero matrix elements. In total there is 32 of them. There are, however, some symmetries that help us write the Hamiltonian in a simpler form.

First, $(M_1,M_2,M_5,M_6,M_9,M_{10},M_{13},M_{14})$ and conjugates come from $(H_t^{xz})_{{\bf i},{\bf i}+{\bf e}_{xz}}$, the rest comes from $(H_t^{yz})_{{\bf i},{\bf i}+{\bf e}_{yz}}$. We only need to consider one of these Hamiltonians as the matrix elements of other one can be obtained in an analogous way. From now on we will only consider $(H_t^{xz})_{{\bf i},{\bf i}+{\bf e}_{xz}}$.

Second, let us look at $M_4$ and $M_{11}^*$ or $M_{11}$ and $M_{4}^*$, where a star denotes complex conjugation. They represent exactly the same process 