Coulomb Integral

For convenience'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>\langle ii|ii\rangle_{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 consistent with the wavelengths of the species involved? (And this isn't the case.)

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

Solution

Suppose we define <math>\rho'</math> such that:

<math>\rho'(r)=\rho(r)</math> if <math>r</math> lies within the primitive cell

<math>\rho'(r)=0</math> otherwise.

We can then calculate the above integral over the maximum displacement cell, dc:

<math>\langle ii|ii\rangle_{cell}=\int_{dc} \frac{\rho'(r_1) \rho'(r_2)}{r_{12}} dr_1 dr_2.</math>

That is, we could calculate the integral over a cell 8 times the volume of the primitive cell.

We note that we never access the values of <math>r_{12}^{-1}</math> unless both <math>r_1</math> and <math>r_2</math> are within the primitive cell. Hence, despite the range of integration, we can perform the Fourier transform of both functions over the effective displacement cell of vectors within the primitive cell:

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

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

Hence:

<math>\langle ii|ii\rangle_{cell}=\sum_{G_d} \rho'[G_d] \rho'[-G_d] \tilde v[G_d].</math>

Thoughts? I'm concerned about the Fourier transform of the <math>r_{12}^{-1}</math> function. Admittedly, I could just code it up...

--james 18:58, 29 December 2006 (GMT)

The Fourier method above calculates the integral above for a Gaussian distribution to within 0.2% (such an integral is analytic). --james 16:41, 8 January 2007 (GMT)