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.

  • $\begingroup$ It should be clear from your question what you are asking. Please post the Hamiltonian and explain how you have dealt with the 1st term. Also, why are the remaining terms causing difficulty? Please be more specific. Thanks. $\endgroup$ Jul 9, 2016 at 20:30
  • $\begingroup$ Sorry for any inconvenience ,this is my first time posting here so i am not used to it.I am about to go into my 3rd year of college and i just completed my quantum mechanics course. My professor introduced me to topological insulators and wanted me to understand this paper.Actually more or less he wants me to understand how the calculations are going on . The first term was pretty easy with 3 nearest neighbour atoms and no spin distinction but in 2nd term due the spin and next nearest neighbour atoms i don't know how to fourier transform it let alone simplify it. $\endgroup$
    – Student
    Jul 9, 2016 at 21:15
  • $\begingroup$ Some more information would be desirable. ..e.g. what do the terms in your expression represent? What does <<i,j>> and <i,j> mean? Some of it looks familiar but much of it isnt obvious what it means (at least to me). If you could elaborate (and even rewrite using LaTEX) thatd be most excellent $\endgroup$
    – user122066
    Jul 9, 2016 at 22:48
  • $\begingroup$ I haven't looked through your picture in great detail but for next nearest neighbours the two sublattices of the honeycomb decouples. So this you should be able to treat as two ordinary decoupled nearest neighbour fourier transforms $\endgroup$ Jul 9, 2016 at 23:08
  • $\begingroup$ Actually i don't get what you are trying to say 'cause next nearest neighbours belong to the same sub lattice(A or B). It's not a combination of both. $\endgroup$
    – Student
    Jul 9, 2016 at 23:22

1 Answer 1


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.


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