Coulomb Integral - Revision history

import>Ajwt3: PeriodicImages moved to Coulomb Integral

2007-01-17T18:20:53Z

PeriodicImages moved to Coulomb Integral

← Older revision	Revision as of 18:20, 17 January 2007
(No difference)

import>Ajwt3 at 17:26, 10 January 2007

2007-01-10T17:26:09Z

← Older revision		Revision as of 17:26, 10 January 2007
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	We need to subtract terms of the type:		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>		:<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.		where ''cell'' refers to the integral over the ''primitive'' cell.

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	We can then calculate the above integral over the maximum displacement cell, dc:		We can then calculate the above integral over the maximum displacement cell, dc:
	:<math><ii\|ii>_{cell}=\int_{dc} \frac{\rho'(r_1) \rho'(r_2)}{r_{12}} dr_1 dr_2.</math>		:<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.		That is, we could calculate the integral over a cell 8 times the volume of the primitive cell.

Line 33:		Line 33:
	:<math>\frac{1}{r_{12}} = \sum_{G_d} e^{-iG_d.(r_1-r_2)} \tilde v[G_d],</math>		:<math>\frac{1}{r_{12}} = \sum_{G_d} e^{-iG_d.(r_1-r_2)} \tilde v[G_d],</math>
	Hence:		Hence:
	:<math><ii\|ii>_{cell}=\sum_{G_d} \rho'[G_d] \rho'[-G_d] \tilde v[G_d].</math>		:<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...		Thoughts? I'm concerned about the Fourier transform of the <math>r_{12}^{-1}</math> function. Admittedly, I could just code it up...

import>Jss43: /* Solution */

2007-01-08T16:41:20Z

Solution

← Older revision		Revision as of 16:41, 8 January 2007
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	--[[User:jss43\|james]] 18:58, 29 December 2006 (GMT)		--[[User:jss43\|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).
			--[[User:jss43\|james]] 16:41, 8 January 2007 (GMT)

import>Jss43 at 18:58, 29 December 2006

2006-12-29T18:58:38Z

← Older revision		Revision as of 18:58, 29 December 2006
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	--[[User:jss43\|james]] 20:02, 20 December 2006 (GMT)		--[[User:jss43\|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><ii\|ii>_{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><ii\|ii>_{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...

			--[[User:jss43\|james]] 18:58, 29 December 2006 (GMT)

import>Jss43 at 10:34, 21 December 2006

2006-12-21T10:34:34Z

← Older revision		Revision as of 10:34, 21 December 2006
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	For ~~convenvience~~'s sake, we consider a <math>\Gamma</math>-point calculation, so need not think about effective supercells.		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:		We need to subtract terms of the type:
Line 14:		Line 14:
	\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>		\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.)		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.)

	--[[User:jss43\|james]] 20:02, 20 December 2006 (GMT)		--[[User:jss43\|james]] 20:02, 20 December 2006 (GMT)

import>Jss43 at 20:19, 20 December 2006

2006-12-20T20:19:51Z

← Older revision		Revision as of 20:19, 20 December 2006
Line 2:		Line 2:

	We need to subtract terms of the type:		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>		:<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:		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>\rho(r_1)=\sum_{G_p} e^{-iG_p.r_1}\tilde \rho [G_p]</math>
Line 12:		Line 14:
	\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>		\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?		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.)

	--[[User:jss43\|james]] 20:02, 20 December 2006 (GMT)		--[[User:jss43\|james]] 20:02, 20 December 2006 (GMT)

import>Jss43 at 20:06, 20 December 2006

2006-12-20T20:06:51Z

← Older revision		Revision as of 20:06, 20 December 2006
Line 3:		Line 3:
	We need to subtract terms of the type:		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>		:<math><ii\|ii>_{cell}=\int_{cell} \frac{\rho(r_1) \rho(r_2)}{r_{12}} dr_1 dr_2.</math>
	~~Expressing~~ the functions involved as Fourier series:		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>\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>		:<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.		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:		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]		:<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>		\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>

	~~Surely to for~~ the integrals to vanish or equal 1, we require the integrals to be performed over a cell consumerate with the wavelengths of the species involved?		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?

	--[[User:jss43\|james]] 20:02, 20 December 2006 (GMT)		--[[User:jss43\|james]] 20:02, 20 December 2006 (GMT)

import>Jss43 at 20:02, 20 December 2006

2006-12-20T20:02:41Z

← Older revision		Revision as of 20:02, 20 December 2006
Line 12:		Line 12:

	Surely to for the integrals to vanish or equal 1, we require the integrals to be performed over a cell consumerate with the wavelengths of the species involved?		Surely to for the integrals to vanish or equal 1, we require the integrals to be performed over a cell consumerate with the wavelengths of the species involved?

			--[[User:jss43\|james]] 20:02, 20 December 2006 (GMT)

import>Jss43 at 20:00, 20 December 2006

2006-12-20T20:00:07Z

← Older revision		Revision as of 20:00, 20 December 2006
Line 1:		Line 1:
	For convenvience's sake, we consider a <math>\Gamma</math>-point calculation.		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:		We need to subtract terms of the type:
Line 11:		Line 11:
	\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>		\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>

			Surely to for the integrals to vanish or equal 1, we require the integrals to be performed over a cell consumerate with the wavelengths of the species involved?
	Assuming that <math>G_p-G_d</math> vanishes when <math></math>

import>Jss43 at 19:51, 20 December 2006

2006-12-20T19:51:37Z

New page

For convenvience's sake, we consider a <math>\Gamma</math>-point calculation.

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>
Expressing 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>

Assuming that <math>G_p-G_d</math> vanishes when <math></math>