When considering a bilattice you need to distinguish two type of sites.

```
A B A B
-------o-------o---------------o-------o----
|<--a-->|<------b------>|
```

For instance you can denote the two kind of sites with letters $A$ and $B$ as it is shown above. Then you have now two different creation and destruction operators . $c_{iA}^{\dagger}$ and $c_{iA}$ in order to create or annihilate a particle in $A$ sites and $c_{iB}^{\dagger}$ and $c_{iB}$ for $B$ sites.

You have to be careful with indices $i$. With a simple lattice the sites had indices like that:

```
A A A A A
-------o-------o-------o-------o-------o
i-1 i i+1 i+2 i+3
|<--a-->|<--a-->|<--a-->|
```

But know the indices are different since there is two kinds of sites:

```
A B A B
-------o-------o---------------o-------o----
i i i+1 i+1
|<--a-->|<------b------>|
```

All this changes expressions of Fourier transforms and hamiltonian:
$$
c_{kA}=\frac{1}{\sqrt{V/2}}\sum_{i}{c_{iA} e^{i k r_{i}}} \quad \text{where} \quad
r_i=i*(a+b)
$$
$$
c_{kB}=\frac{1}{\sqrt{V/2}}\sum_{i}{c_{iB} e^{i k r_{i}}} \quad \text{where} \quad
r_i=i*(a+b)+a
$$
**EDIT**:The volume of the system has to be divided by two in Fourier transforms since there is now two sites in each primitive cell (there were only one before). **END EDIT**

The hamiltonian now takes the form:
$$
H=-\sum_{i}{t_s (c^{\dagger}_{iA} c_{iB} + c^{\dagger}_{iB} c_{iA})+t_l (c^{\dagger}_{iB} c_{(i+1)A} + c^{\dagger}_{(i+1)A} c_{iB})}
$$
where I have considered a one-dimensional problem. Since the distance between sites is not always the same, you might want to consider two different hopping parameters: $t_s$ for short jumps and $t_l$ for long ones.

**EDIT:** I had forgotten terms in the Hamiltonian, nearest neighbor hopping terms must be present in both ways $(A,i)\rightarrow (B,i)$ and $(B,i) \rightarrow(A,i)$. The *long* jumps are $(A,i+1)\rightarrow (B,i)$ and $(B,i)\rightarrow (A,i+1)$.
**END EDIT**

If you want to obtain a diagonal form for your Hamiltonian, you can try to find a $2\times2$ matrix $M$ such that:
$$
H=-\sum_{k}{(c^{\dagger}_{kA} c^{\dagger}_{kB})M\binom{c_{kA}}{c_{kB}}}
$$
$M$ will contain hopping parameters $t_s$ and $t_l$, once you have diagonalized $M$ your problem is solved.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user ChocoPouce