# Momentum Space Representation of the Tight Binding Hamiltonian

I am trying to represent the tight-binding Hamiltonian $$$$\hat{H}_{TB} = \sum_{\sigma} \sum_{\alpha,\beta} \sum_{\mathbf{R}_1,\mathbf{R}_2} t^{\alpha,\beta}_{\mathbf{R}_1,\mathbf{R}_2} \hat{c}^{\dagger}_{\alpha,\mathbf{R}_1,\sigma} \hat{c}_{\beta,\mathbf{R}_2,\sigma} \label{eq:Htb}\tag{1}$$$$

in the momentum space, and it is not clear this relation $$$$\sum_{\mathbf{R}_1,\mathbf{R}_2} e^{-i\mathbf{k}_1 \cdot \mathbf{R}_1} e^{i\mathbf{k}_2 \cdot \mathbf{R}_2} t_{\mathbf{R}_1,\mathbf{R}_2}^{\alpha,\beta} = \frac{1}{M} \sum_{\mathbf{R}_0} \sum_{\mathbf{R}_1,\mathbf{R}_2} e^{-i\mathbf{k}_1 \cdot \mathbf{R}_1} e^{i\mathbf{k}_2 \cdot \mathbf{R}_2} t_{\mathbf{R}_1 - \mathbf{R}_0,\mathbf{R}_2 - \mathbf{R}_1 - \mathbf{R}_0}^{\alpha,\beta} \label{eq:pass2}\tag{2}$$$$ where $$M$$ is the number of lattice sites and the exponentials come out of the Fourier transform of the operators in the real space to those in the momentum space $$$$\hat{c}_{n,\mathbf{R},\sigma} = \frac{1}{\sqrt{M}} \sum_{\mathbf{k}} e^{i\mathbf{k} \cdot \mathbf{R}} \hat{c}_{n,\mathbf{k},\sigma} \label{eq:c_R}\tag{3}$$$$ Moreover the translational invariance of the lattice imply $$$$t_{\mathbf{R}_1,\mathbf{R}_2}^{\alpha,\beta} = t_{\mathbf{R}_1 - \mathbf{R}_0,\mathbf{R}_2 - \mathbf{R}_0}^{\alpha,\beta} \quad \forall \mathbf{R}_0 \label{eq:hopping_transl} \tag{4}$$$$

In (2) we can substitute $$t^{\alpha,\beta}_{\mathbf{R}_{1}-\mathbf{R}_{0},\mathbf{R}_{2}-\mathbf{R}_{0}}$$. Then since the left hand side of (2) does not depend on $$\mathbf{R}_{0}$$, if we sum on it we have M times the same thing, so if we divide by M, we have a relation equivalent to the previous one