How to convert a Hamiltonian for the tight-binding model in silicene into $k$-space? I am trying to convert the Hamiltonian from the paper "A topological insulator and helical zero mode in
silicene under an inhomogeneous electric field" (also on arXiv) into $k$-space.
$$
\begin{array}{rcl}
H & = &~~-t \displaystyle \sum_{\langle i,j\rangle\alpha}c_{i\alpha}^{\dagger}c_{j\alpha} \\
&&+~~i\frac{\lambda_{\mathrm{SO}}}{3\sqrt{3}} \displaystyle \sum_{\langle\langle i,j\rangle\rangle\alpha\beta}v_{ij}c_{i\alpha}^{\dagger}\sigma_{\alpha\beta}^{z}c_{j\beta} \\
&&-~~i\frac{2}{3}\lambda_{\mathrm{R}} \displaystyle \sum_{\langle\!\langle i,j\rangle\!\rangle\alpha\beta}\mu_{ij}c_{i\alpha}^{\dagger}\Bigl(\vec\sigma\times\vec{d}_{ij}^{0}\Bigr)_{\alpha\beta}^{z}c_{j\beta} \\
&&+~~\ell \displaystyle \sum_{i\alpha}\zeta_i E_z^i c_{i\alpha}^{\dagger}c_{i\alpha}
\end{array}
$$
I converted the first term into $k$-space, but I need help for the remaining 3 terms. I am not asking for the whole derivation. Hints on how to deal with the terms would be enough.
 A: Fourier transformation of the creation/annihilation operators is no different than any other Fourier transform; the spin index just comes along for the ride. For instance,
$$c^\dagger_{i \alpha} = \frac{1}{\sqrt{2\pi}} \int_\text{BZ} d^2k\,c^\dagger_{\mathbf{k} \alpha}e^{i\mathbf{k}\cdot\mathbf{x}_i}$$
With that operator in hand, your second term (ignoring the prefactor, spin index summation, and various factors of $2\pi$ throughout) becomes
$$\sum_i \sum_{\delta\in\Delta_i} \nu_{i,i+\delta} \,\sigma_{\alpha\beta}^z\int d^2k\int d^2q\, c^\dagger_{\mathbf{k}\alpha} e^{i\mathbf{k}\cdot\mathbf{x}_i}c_{\mathbf{q\beta}}e^{-i\mathbf{q}\cdot(\mathbf{x}_i+\delta)},$$
Where I have defined $\Delta_i$ as the set of next-nearest-neighbor displacement vectors associated with site $i$. Inspection of the graphene lattice will show that $\nu$ is $+1$ for three of the NNNs and $-1$ for the other three; I'll let you separate those out by hand. Once you do, you'll find that the only remaining functional dependence on the site index $i$ is in the exponentials, and since
$$ \sum_i e^{i(\mathbf{k}-\mathbf{q})\cdot\mathbf{x}_i} \propto \delta(\mathbf{k}-\mathbf{q})$$
you end up with something like
$$ \sigma_{\alpha\beta}^z\int d^2 k\,c^\dagger_{\mathbf{k}\alpha}c_{\mathbf{k}\beta}e^{-i\mathbf{k}\cdot\delta}.$$
There are no longer any spatial indices remaining; this term has been "converted into $k$-space", as you requested. Note that it's a sum of $k$-dependent operators; you're probably looking for the $k$-dependent Bloch hamiltonian, not the "true" hamiltonian operator, so take note of whether you want your hamiltonian to be $k$-dependent or not.
This is simply a rough demonstration of how to Fourier transform these operators and then reduce them using delta functions. There are several details I omitted, but you should be able to carry out manipulations like these on the remaining terms.
