Finding energy bands of trimerized 1D periodic crystal in second quantization formalism I am trying to figure out the energy bands of a trimerized 1D periodic crystal model. Suppose we have three atom types per unit cell $A,B,C$, and we have the Hamiltonian
$$H = - \sum_j c_{j+1, A}^\dagger c_{j,C} + c_{j, C}^\dagger c_{j+1,A}$$
where $c_{j,\sigma}^\dagger$ and $c_{j,\sigma}$ are the creation and annihilation operators (respectively) of electrons at cell-position $x_j$ in atoms of type $\sigma$. In this case, I just use a Fourier transformation to obtain 
$$H = -2 \sum_k \cos(k \Delta) c_{k,A}^\dagger c_{k,C} = \sum_k \epsilon_k c_{k,A}^\dagger c_{k,C}$$
where $\Delta$ is the distance between the atom of type $A$ and the neighboring atom of type $C$.
But, I get one band only. Should I get more bands like with the dimerized model?
 A: First of all, there is an issue with the Hamiltonian, since it contains couplings only between two types of atoms (A and C). A more general Hamiltonian would be
$$H = - \sum_j\sum_{\sigma,\sigma'}\left[t_{\sigma\sigma'}c_{j+1,\sigma}^\dagger c_{j,\sigma'} + t_{\sigma\sigma'}^*c_{j,\sigma'}^\dagger c_{j+1,\sigma}\right].$$
The hopping constant $t_{\sigma\sigma'}$ can be taken real, if there is no magnetic field.
Then, in order to diagnonalize this Hamiltonian we could use conjecture
$$c_{k,\alpha} \propto \sum_j e^{ikj}\sum_\sigma x_{\alpha\sigma}c_{j,\sigma}.$$
Exponent $e^{ikj}$ should take care of the spatial dependence, leaving us to diagonalize a 3-by-3 matrix, whose eigenvectors I labeled by $\alpha$.
Alternative approach
One could try to solve this problem in two steps:


*

*One performs the Fourier transform, obtaining
$$H = - \sum_k\sum_{\sigma,\sigma'}2\cos(k\Delta)t_{\sigma\sigma'}c_{k,\sigma}^\dagger c_{k,\sigma'}.$$

*One then has to diagonalize the 3-by-3 Hamiltonian for each $k$.

