Hubbard Hamiltonian and alternating potential I am trying to diagonalize the Hubbard Hamiltonian (for a 1d fermionic chain) with an alternating potential and find its eigenenergies. The given Hamiltonian is the following:
$$
H=t∑_{j=1}^N(c^†_{j+1}c_j+c^†_{j}c_{j+1})+v∑_j(-1)^jc^†_{j}c_j
$$
As a first step, I have replaced the Fourier transform of the raising\lowering operators: 
$$
c_j=\frac{1}{\sqrt{N}}\Sigma_{k\epsilon[-\pi;\pi]}c_ke^{ijk}
$$
And I obtained: 
$$
H=t∑_{k}2\cos(k)(c^†_{k}c_k)+v∑_kc^†_{k}c_{k+\pi}
$$
Then, I must restrict the values of k to $$[-\pi/2;\pi/2]$$
I tried to separate into three intervals and change the index of the sum, but it does not work. Does anyone have any idea?
Thank you in advance
 A: What if you try treating the system as an explicitly bipartite lattice? That is:
$$
H = \nu \sum_j \left(c_j^\dagger c_j - d_j^\dagger d_j\right) + t\sum_{\langle ij\rangle} \left(c_i^\dagger d_j + d_j^\dagger c_i\right)
$$
Now, using what you wrote
$$
c_j = \frac{1}{\sqrt{N}}\sum_k c_k e^{ijk}\,,
$$
we for the on-site part of the Hamiltonian
$$
H_S = \frac{1}{N}\nu\sum_j\left(\sum_{pk} c^\dagger_p c_k e^{i\left(k-p\right)j} - \sum_{pk} d^\dagger_p d_k e^{i\left(k-p\right)j}\right) = \nu\sum_p \left(c_p^\dagger c_p - d_p^\dagger d_p\right)\,.
$$
Next, we look at the hopping part:
$$
H_h = \frac{1}{N}t\sum_{\langle nm\rangle}\sum_{pk}\left(c_p^\dagger d_k e^{i\left(km - pn\right)}+ d_k^\dagger c_pe^{-i\left(km - pn\right)}\right)
\\
= \frac{1}{N}t\sum_{n}\sum_{pk}\left(c_p^\dagger d_k e^{i\left(k\left(n+1\right) - pn\right)} + c_p^\dagger d_k e^{i\left(k\left(n-1\right) - pn\right)} + d_k^\dagger c_pe^{-i\left(k\left(n+1\right) - pn\right)} + d_k^\dagger c_pe^{-i\left(k\left(n-1\right) - pn\right)} \right)
\\
= \frac{1}{N}t\sum_{n}\sum_{pk}\left(e^{ik}c_p^\dagger d_k e^{i\left(kn - pn\right)} + e^{-ik}c_p^\dagger d_k e^{i\left(kn - pn\right)} + e^{-ik}d_k^\dagger c_pe^{-i\left(kn - pn\right)} + e^{ik}d_k^\dagger c_pe^{-i\left(kn - pn\right)} \right)
\\
= t\sum_{k}\left(e^{ik}c_k^\dagger d_k  + e^{-ik}c_k^\dagger d_k  + e^{-ik}d_k^\dagger c_k + e^{ik}d_k^\dagger c_k \right) = 2t\sum_k\left[\cos\left(k\right)c_k^\dagger d_k + \cos\left(k\right) d_k^\dagger c_k\right]\,.
$$
This leads us to
$$
H = 
\sum_k
\begin{pmatrix}
c_k^\dagger, d_k^\dagger
\end{pmatrix}
\begin{pmatrix}
\nu&2t\cos k
\\
2t\cos k&-\nu
\end{pmatrix}
\begin{pmatrix}
c_k \\ d_k
\end{pmatrix}\,,
$$
which is easily diagonalizable!
