Usually the Coulomb potential (electron-electron interaction) can be Fourier transformed (aside from prefactors) like that: $$ \frac{1}{|\vec r_1 -\vec r_2|} = \int \frac{\text d ^3 k}{(2\pi)^3} \frac{\text e^{\text i \vec k (\vec r_1 - \vec r_2)}}{\vec k^2}\;. $$ But sometimes, when dealing with periodic crystals, I see things that: $$ \frac{1}{|\vec r_1 -\vec r_2|} = \sum_{\vec G} \frac{\text e^{\text i \vec G (\vec r_1 - \vec r_2)}}{\vec G^2}\;, $$ where the sum goes only over the reciprocal lattice vectors $\vec G$.

But I don't understand why? The electron-electron interaction is not periodic! Does it have something to do with the finite size of the system?

  • $\begingroup$ Is the electron potential the only electric potential to consider ? $\endgroup$
    – Declan
    Commented Oct 20, 2015 at 16:14
  • $\begingroup$ In the case of my question yes. I don't speak of the potential of the nuclei... $\endgroup$
    – thyme
    Commented Oct 20, 2015 at 16:16
  • $\begingroup$ Within prefactors, Fourier transforming the crystal results in the reciprocal lattice vectors G. So, any solution that has the crystal symmetry will be represented as a sum across all reciprocal lattice vectors. $\endgroup$
    – Jon Custer
    Commented Oct 20, 2015 at 16:32
  • $\begingroup$ Hm, I don't understand what you mean. The electron-electron interaction does not have the crystal symmetry. That's why I wrote this question. $\endgroup$
    – thyme
    Commented Oct 20, 2015 at 16:34
  • $\begingroup$ I think it assumes system is periodically added to itself such as wherever you go, you pass same observation window again. Something like whole space is made of your observation window (infinite copies) of particles. $\endgroup$ Commented Jun 14, 2018 at 10:53

2 Answers 2


Much as Okarin wrote, in the crystal there is a periodic arrangement of Coulomb potentials. The Fourier transform of this periodic arrangement of Coulomb potentials is discrete; it is zero except for the reciprocal lattice vectors.

Like the Fourier transform of a single Dirac delta function is constant, but the Fourier transform of an infinite comb of Dirac delta functions is an infinite comb of Dirac delta functions. Using the convolution theorem you can see that the Fourier transform of the infinite comb of Coulomb potentials is the infinite comb of Dirac delta functions times the Fourier transform of the Coulomb potential.


I'm very late at this but I'll answer. The reason is that for a periodic crystal the Fourier transform and the Fourier series are equivalent. Instead of thinking of a point-like charge, consider a generic charge density $\rho(r_2)$. The case of a point-like electron is one where $\rho(r_2) = \delta(r_2-r_e)$, using the Dirac delta. Then the Fourier transform of it is the constant function, $\rho(k)=1$ for any $k$, and you retrieve the first formula.
However, if you have a periodic system with lattice parameters $a,b,c$, then for each electron at $r_e$ you will also have infinite more electrons at $r_e+ia+jb+lc$, with $i,j,l$ integers. Then your density isn't a Dirac delta any more, but a Dirac comb: $\rho(r_2) = \sum_{ijl}\delta(r_2-r_e-ia-jb-lc)$. And the Fourier transform of a Dirac comb is another Dirac comb! Hence, you now have a Fourier transform that is itself defined on a lattice of sort - non-zero at the various reciprocal lattice vectors $G$, and zero everywhere else. And that's why the integral above becomes, effectively a sum. It's still an integral, really: it's just that the integrand is zero everywhere other than at those discrete points.


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