Документ взят из кэша поисковой машины. Адрес оригинального документа : http://classic.chem.msu.su/gran/gamess/eri_ineqs.pdf Дата изменения: Tue Nov 21 14:33:28 2006 Дата индексирования: Mon Oct 1 19:48:35 2012 Кодировка:

Searching efficient inequalities for 2-e AO integrals screening

Alexander A. Granovsky* Vladimir I. Pupyshev+
+

*Laboratory of Chemical Cybernetics Laboratory of Molecular Structure and Quantum Mechanics M.V. Lomonosov Moscow State University, Moscow, Russia
November 15, 2006


Notations
p, q, r, s - atomic orbitals (pq|rs) - 2-e integral in AO basis Chemical notation: p(1) q(1), r(2), s(2)


Simple known inequalities
Schwarz inequalities:
Most powerful and most useful: |(pq|rs)| |(pq|pq)(rs|rs)|1/2 (1)
Simple sequences (weaker inequalities):
· · · · |(pq|rs)| |(pq|rs)| |(pq|rs)| |(pq|rs)| |(pq|pq)(rs|rs)|1/2 |(pp|qq)(rs|rs)|1/2 |(pq|pq)(rs|rs)|1/2 |(pq|pq)(rr|ss)|1/2 |(pq|pq)(rs|rs)|1/2 |(pp|qq)(rr|ss)|1/2 |(pp|pp)(qq|qq)(rr|rr)(ss|ss)|1/4

Less useful but different: |(pq|rs)| |(pp|rr)(qq|ss)|1/2 (2) |(pq|rs)| |(pp|ss)(qq|rr)|1/2 (3)


Much less trivial
Hardy-Littlewood-Sobolev inequality:
p,r>1, 0< < n,
11 + + =2 pnr

,fLp(Rn),hLr(Rn)

Rn R



n

f ( x )h ( y ) n n d x d y C ( n, , p ) || f ||p || h ||r |x- y|
p | f ( x) | d n x || f ||p = Rn
/3
1/ p

where

3 4 () C ( n, , p ) 3- 3

1 pr



/3 1 - 1/ p

/3

+ 1 - 1/ r
/2

/3

/3


/ n -1

p = r = 2n/(2n­):

C ( n, , p ) =

( n / 2 - / 2 ) ( n / 2 ) ( n ) ( n - / 2)


Simple sequences, symmetric case
For n = 3, = 1, f, h L6/5(R3):

R3 R



3

f ( x )h ( y ) 3 3 8 2 d xd y |x- y| 3

1/ 3

|| f ||6 / 5 || h ||6

/5

(

pq | rs ) =

R3 R



3

p ( x )q( x )r ( y )s ( y ) 3 3 8 2 d xd y 3 |x- y|

1/ 3

|| pq ||6 / 5 || rs ||6 / 5 ( 4)


Great disappointment
Inequality (4) is weaker than much simpler Schwarz inequality (1): Statement (1): Let |(pq|rs)| |G(p,q)G(r,s)|1/2 Then: |(pq|rs)| |(pq|pq)(rs|rs)|1/2
= ||G(p,q)G(p,q)|1/2 |G(r,s)G(r,s)|1/2|1/2 = |G(p,q)G(r,s)|1/2

Sequence: the only way to improve (1) is to use non-symmetric Sobolev inequalities


Fully non-symmetric case
p = 1, r = n/(n­) = 3/2: C ( n, , p = 1)
Finite C: does it exist at all?

Our numerical experiments show that at least for GTOs:
8 2 C (3,1,1) C (3,1,6 / 5) = 3 Promising result!
1/ 3


Applying this to estimate (pq|rs)
(
pq | rs ) =
R3 R



3

8 2 p ( x )q( x )r ( y )s ( y ) 3 3 d xd y |x- y| 3

1/ 3

|| pq ||1|| rs ||3 / 2 (5)

(

pq | rs ) =

R3 R



3

p ( x )q( x )r ( y )s ( y ) 3 3 8 d xd y 3 |x- y|

2

1/ 3

|| pq ||3 / 2 || rs ||1 (6)


Combining them together
|(pq|rs)| G1(p,q)G2(r,s) (5) |(pq|rs)| G2(p,q)G1(r,s) (6)
|(pq|rs)| min(G1(p,q)G2(r,s),G2(p,q)G1(r,s)) (7)

The statement (1) above is no longer applicable!


Results of numerical experiments
Inequality (7) is not identical to (1) Inequality (7) seems to be slightly more accurate in most cases The best way is to use both (1) and (7) at the same time Considerable CPU time required for evaluation of non-trivial integrals entering inequality (7) makes its practical applications senseless!
The only acceptable case: fully uncontracted basis sets


Conclusions: what are the inequalities we actually need?
Inequality (1) does not include 1/r dependence Inequalities (2) and (3) include 1/r dependence but do not decay exponentially We need the set of inequalities combining properties of both (1) and (2)-(3)! Our hope: strengthened Schwarz inequality


Thank you for your attention!