0
$\begingroup$

I am trying to represent the tight-binding Hamiltonian \begin{equation} \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} \end{equation}

in the momentum space, and it is not clear this relation \begin{equation} \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} \end{equation} 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 \begin{equation} \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} \end{equation} Moreover the translational invariance of the lattice imply \begin{equation} 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} \end{equation}

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.