Coulomb Integral

For convenvience's sake, we consider a <math>\Gamma</math>-point calculation, so need not think about effective supercells.

We need to subtract terms of the type:

<math><ii|ii>_{cell}=\int_{cell} \frac{\rho(r_1) \rho(r_2)}{r_{12}} dr_1 dr_2,</math>

where cell refers to the integral over the primitive cell.

As usual, we can express the functions involved as Fourier series:

<math>\rho(r_1)=\sum_{G_p} e^{-iG_p.r_1}\tilde \rho [G_p]</math>

<math>\frac{1}{r_{12}} = \sum_{G_d} e^{-iG_d.(r_1-r_2)} \tilde v[G_d],</math>

where <math>G_p</math> are G-vectors of the primitive cell and <math>G_d</math> are G-vectors of the maximum displacement cell. [Note that the reciprocal lattice vectors of the maximum displacement cell are half those of the primitive cell.]

We get:

<math>\int_{cell} \frac{\rho(r_1) \rho(r_2)}{r_{12}} dr_1 dr_2 = \sum_{G_p}\tilde\rho[G_p] \sum_{G'_p} \tilde\rho [G'_p] \sum_{G_d} \tilde v[G_d]

\int_{cell} e^{-i(G_p+G_d).r_1} dr_1 \int_{cell} e^{-i(G'_p-G_d).r_2} dr_2</math>

For the integrals to vanish or equal 1, don't we require the integrals to be performed over a cell consumerate with the wavelengths of the species involved? (And this isn't the case.)

--james 20:02, 20 December 2006 (GMT)