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Molecular Integral Evaluation

Trygve Helgaker

Centre for Theoretical and Computational ChemistryDepartment of Chemistry, University of Oslo, Norway

The 11th Sostrup Summer SchoolQuantum Chemistry and Molecular Properties

July 416 2010

Trygve Helgaker (CTCC, University of Oslo) Molecular Integral Evaluation 11th Sostrup Summer School (2010) 1 / 34

Molecular integral evaluation

I one-electron interactions

I overlap, multipole-moment, and kinetic-energy integralsI Coulomb attraction integrals

Oab =

a(r)O(r)b(r) dr

I two-electron interactions

I Coulomb integrals

gabcd =

a(r1)b(r1)c (r2)d (r2)

r12dr1 dr2

I Coulomb and exchange contributions to Fock/KS matrix

Fab =cd

(2gabcdDcd gacbdDcd

)I basis functions

I primitive Cartesian GTOs

I integration schemes

I McMurchieDavidson, ObaraSaika and Rys schemes

Trygve Helgaker (CTCC, University of Oslo) Introduction 11th Sostrup Summer School (2010) 2 / 34

Overview

1 Cartesian and Hermite Gaussians

I properties of Cartesian GaussiansI properties of Hermite Gaussians

2 Simple one-electron integrals

I Gaussian product rule and overlap distributionsI overlap distributions expanded in Hermite GaussiansI overlap and kinetic-energy integrals

3 Coulomb integrals

I Gaussian electrostatics and the Boys functionI one- and two-electron Coulomb integrals over Cartesian Gaussians

4 Sparsity and screening

I CauchySchwarz screening

5 Coulomb potential

I early density matrix contractionI density fitting

Trygve Helgaker (CTCC, University of Oslo) Introduction 11th Sostrup Summer School (2010) 3 / 34

Cartesian Gaussian-type orbitals (GTOs)

I We shall consider integration over primitive Cartesian Gaussians centered at A:

Gijk (r, a,A) = xiAy

jAz

kA exp(ar

2A),

a > 0 orbital exponent

rA = r A electronic coordinatesi 0, j 0, k 0 quantum numbers

I total angular-momentum quantum number ` = i + j + k 0I Gaussians of a given ` constitutes a shell:

s shell: G000, p shell: G100, G010, G001, d shell: G200, G110, G101, G020, G011, G002

I Each Gaussian factorizes in the Cartesian directions:

Gijk (a, rA) = Gi (a, xA)Gj (a, yA)Gk (a, zA), Gi (a, xA) = xiA exp(ax

2A)

I note: this is not true for spherical-harmonic Gaussians nor for Slater-type orbitals

I Gaussians satisfy a simple recurrence relation:

xAGi (a, xA) = Gi+1(a, xA)

I The differentiation of a Gaussian yields a linear combination of two Gaussians

Gi (a, xA)

Ax=

Gi (a, xA)

x= 2aGi+1(a, xA) iGi1(a, xA)

Trygve Helgaker (CTCC, University of Oslo) Cartesian and Hermite Gaussians 11th Sostrup Summer School (2010) 4 / 34

Hermite GaussiansI The Hermite Gaussians are defined as

tuv (r, p,P) =

(

Px

)t ( Py

)u ( Pz

)vexp

(pr2P

), rP = r P

I Like Cartesian Gaussians, they also factorize in the Cartesian directions:

t(xP) = (/Px )t exp

(px2P

) a Gaussian times a polynomial of degree t

I Hermite Gaussians yields the same spherical-harmonic functions as do Cartesian Gaussians

I they may therefore be used as basis functions in place of Cartesian functions

I We shall only consider their use as intermediates in the evaluation of Gaussian integrals

Cartesian product Gi (xA)Gj (xB) =i+jt=0

E ijt t(xP) Hermite expansion

I such expansions are useful because of simple integration properties such as

t(x)dx = t0

p

I McMurchie and Davidson (1978)

I We shall make extensive use of the McMurchieDavidson scheme for integration


Integration over Hermite Gaussians

I From the definition of Hermite Gaussians, we have

t(x) dx =

(

Px

)texp

(px2P

)dx

I We now change the order of differentiation and integration by Leibniz rule:

t(x) dx =

(

Px

)t

exp(px2P

)dx

I The basic Gaussian integral is given by

exp(px2P

)dx =

p

I Since the integral is independent of P, differentiation with respect to P gives zero:

t(x) dx = t0

p

I only integrals over Hermite s functions do not vanish


Hermite recurrence relationI Cartesian Gaussians satisfy the simple recurrence relation

xAGi = Gi+1

I The corresponding Hermite recurrence relation is slightly more complicated:

xPt =1

2pt+1 + tt1

I The proof is simple:

I from the definition of Hermite Gaussians, we have

t+1 =

(

Px

)t Px

exp(px2P) = 2p(

Px

)txP0

I inserting the identity(

Px

)txP = xP

(

Px

)t t

(

Px

)t1we then obtain

t+1 = 2p[xP

( Px

)t t( Px

)t1]0 = 2p (xPt tt1)


Comparison of Cartesian and Hermite Gaussians

Cartesian Gaussians Hermite Gaussians

definition Gi = xiA exp

(ax2A

)t =

t

Ptxexp

(px2P

)recurrence xAGi = Gi+1 xPt =

12p

t+1 + tt1

differentiation GiAx

= 2aGi+1 iGi1 tPx = t+1

!3 !2 !1 1 2 3

!2

!1

1

2

!3 !2 !1 1 2 3

!2

!1

1

2


The Gaussian product rule

I The most important property of Gaussians is the Gaussian product rule

I The product of two Gaussians is another Gaussian centered somewhere on the lineconnecting the original Gaussians; its exponent is the sum of the original exponents:

exp(ax2A

)exp

(bx2B

)= exp

(X 2AB

) exponential prefactor

exp(px2P

) product Gaussian

where

Px =aAx + bBx

p center of mass

p = a + b total exponentXAB = Ax Bx , relative separation

=ab

a + b reduced exponent

-3 -2 -1 1 2 3

0.2

0.4

0.6

0.8

1

I The Gaussian product rule greatly simplifies integral evaluation

I two-center integrals are reduced to one-center integralsI four-center integrals are reduced to two-center integrals

Trygve Helgaker (CTCC, University of Oslo) Gaussian product rule and Hermite expansions 11th Sostrup Summer School (2010) 9 / 34

Overlap distributions

I The product of two Cartesian Gaussians is known as an overlap distribution:

ij (x) = Gi (x , a,Ax )Gj (x , b,Bx )

I The Gaussian product rule reduces two-center integrals to one-center integralsij (x) dx = KAB

x iAx

jB exp

(px2P

)dx

I the Cartesian monomials still make the integration awkwardI we would like to utilize the simple integration properties of Hermite Gaussians

I We therefore expand Cartesian overlap distributions in Hermite Gaussians

overlap distribution ij (x) =i+j

t=0Eijt t(xP) Hermite Gaussians

I note: ij (x) is a single Gaussian times a polynomial in x of degree i + jI it may be exactly represented as a linear combination of t with 0 t i + j

I The expansion coefficients may be evaluated recursively:

E i+1,jt =1

2pE ijt1 + XPAE

ijt + (t + 1)E

ijt+1

I we shall now prove these recurrence relations


Recurrence relations for Hermite expansion coefficients

I A straighforward Hermite expansion of i+1,j

i+1,j = KABxi+1A x

jB exp

(pr2P

)=t

E i+1,jt t

I An alternative Hermite expansion of i+1,j

i+1,j = xAij = (x Ax )ij= (x Px )ij + (Px Ax )ij = xPij + XPAij

=t

E ijt xPt + XPAt

E ijt t

=t

E ijt

(1

2pt+1 + tt1 + XPAt

)=t

[1

2pE ijt1 + (t + 1)E

ijt+1 + XPAE

ijt

]t

I we have here used the recurrence xPt =1

2pt+1 + tt1

I A comparison of the two expansions yields the recurrence relations

E i+1,jt =1

2pE ijt1 + XPAE

ijt + (t + 1)E

ijt+1


Overlap integrals

I We now have everything we need to evaluate overlap integrals:

Sab = Ga |Gb Ga = Gikm (r, a,A) = Gi (xA)Gk (yA)Gm(zA)

Gb = Gjln (r, b,B) = Gj (xB)Gl (yB)Gn(zB)

I The overlap integral factorizes in the Cartesian directions:

Sab = SijSklSmn, Sij =Gi (xA)

Gj (xB)I Use the Gaussian product rule and Hermite expansion:

Sij =

ij (x) dx =

i+jt=0

E ijt

t(xP) dx =

i+jt=0

E ijt t0p

= E ij0

p

I only one term survives!

I The total overlap integral is therefore given by

Sab = Eij0 E

kl0 E

mn0

(

p

)3/2

Trygve Helgaker (CTCC, University of Oslo) Simple one-electron integrals 11th Sostrup Summer School (2010) 12 / 34

Dipole-moment integrals

I Many other integrals may be evaluated in the same manner

I For example, dipole-moment integrals are obtained as

Dij =Gi (xA) |xC |Gj (xB)

=

ij (xP)xCdx

=

i+jt=0

E ijt

t(xP)xcdx (expand in Hermite Gaussians)

=

i+jt=0

E ijt

[xPt(xP) + XPCt(xP)] dx (use xC = xP + XPC )

=

i+jt=0

E ijt

[1

2pt+1(xP) + XPCt(xP) + tt1(xP)

]dx (use xPt = 12p t+1 + tt1)

= E ij0 XPC

p+ E ij1

p(s functions integration)

I The final dipole-moment integral is given by

Dij =(E ij1 + XPCE

ij0

)p


Kinetic-energy integrals

I As a final example of non-Coulomb integrals, we consider the kinetic-energy integrals:

Tab = 12Ga

2x2

+ 2

y2+

2

z2

GbTab = TijSklSmn + SijTklSmn + SijSklTmn

Sij =Gi (xA)

Gj (xB)Tij = 12

Gi (xA)

2x2

Gj (xB)I Differentiation of Cartesian Gaussians gives

d

dxGj (xB) = 2bGj+1 + jGj1

d2

dx2Gj (xB) = 4b

2Gj+2 2b(2j + 1)Gj + j(j 1)Gj2

I the first derivative is linear combination of two undifferentiated GaussiansI the second derivative is linear combination of three undifferentiated Gaussians

I Kinetic-energy integrals are linear combinations of up to three overlap integrals:

Tij = 2b2Si,j+2 + b(2j + 1)Si,j 12 j(j 1)Si,j2


Coulomb integrals

I We shall consider two types of Coulomb integrals:

I one-electron nuclear-attraction integralsGa(rA)

r1C Gb(rB)I two-electron repulsion integrals

Ga(r1A)Gb(r1B)r112 Gc (r2C )Gd (r2D)

I Coulomb integrals are not separable in the Cartesian directions

I however, they may be reduced to a one-dimensional integral on a finite interval:

Fn(x) =

10

exp(xt2

)t2ndt the Boys function

I this makes the evaluation of Gaussian Coulomb integrals relatively simple

I We begin by evaluating the one-electron spherical Coulomb integral:

Vp =

exp

(pr2P

)rC

dr =2

pF0(pR2PC

)I we shall next go on to consider more general Coulomb integrals

Trygve Helgaker (CTCC, University of Oslo) Coulomb integrals 11th Sostrup Summer School (2010) 15 / 34

Coulomb integral over a spherical Gaussian I

I We would like to evaluate the three-dimensional Coulomb-potential integral

Vp =

exp

(pr2P

)rC

dr

1 The presence of r1C is awkward and is avoided by the substitution

1

rC=

1

exp(r2C t

2)dt Laplace transform

to yield the four-dimensional integral

Vp =1

exp

(pr2P

)

exp(t2r2C

)dt dr

2 To prepare for integration over r, we invoke the Gaussian product rule

exp(pr2P

)exp

(t2r2C

)= exp

(

pt2

p + t2R2CP

)exp

[(p + t2

)r2S]

to obtain (where the exact value of S in rS does not matter):

Vp =1

exp[(p + t2)r2S

]dr exp

(

pt2

p + t2R2CP

)dt

to be continued. . .

Trygve Helgaker (CTCC, University of Oslo) Spherical Coulomb integrals 11th Sostrup Summer School (2010) 16 / 34

Coulomb integral over a spherical Gaussian II

carried over from previous slide:

Vp =1

[(p + t2)r2S

]dr exp

(

pt2

p + t2R2CP

)dt

3 Integration over all space r now yields a one-dimensional integral

Vp =2

0

(

p + t2

)3/2exp(pR2CP

t2

p + t2

)dt

4 To introduce a finite integration range, we perform the substitution

u2 =t2

p + t2 1 + pt2 = u2 pt3 dt = u3du dt = p1

(t2

u2

)3/2du

and obtain

Vp =2

p

10

exp(pR2CPu

2)du =

2

pF0(pR2PC

)I We have here introduced the Boys function

Fn(x) =

10

exp(xt2

)t2ndt

a 3D integral over all space has been reduced to a 1D integral over [0, 1]


Gaussian electrostaticsI The Boys function is related to the error function

erf(x) =2

x0

exp(t2

)dt =

2xF0(x2)

I Introducing unit charge distributions and the reduced exponent as

p(rP) =( p

)3/2exp

(pr2P

), =

pq

p + q

we obtain formulas similar to those of point-charge electrostatics (but damped)

p (rP)

rCdr =

erf(

pRPC)

RPCp (r1P) q (r2Q)

r12dr1dr2

=erf(RPQ

)RPQ

2 4 6 8 10

0.5

1.0

1.5


The Boys function Fn(x)I The Boys function is central to molecular integral evaluation

I it is evaluated by a combination of numerical techniques

I Fn(x) > 0 since the integrand is positive:

Fn(x) =

10

exp(xt2

)t2n dt > 0

I Fn(x) is convex and decreasing:

F n(x) = Fn+1(x) < 0F n (x) = Fn+2(x) > 0

2 4 6 8 10 12

0.2

0.4

0.6

0.8

1.0

I Evaluation for small and large values:

Fn(x) =1

2n + 1+k=1

(x)k

k!(2n + 2k + 1)(x small)

Fn(x)

0exp

(xt2

)t2ndt =

(2n 1)!!2n+1

x2n+1(x large)

I Different n-values are related downward recurrence relations:

Fn1(x) =2xFn(x) + exp(x)

2n 1


Cartesian Coulomb integrals

I We have obtained a simple result for one-center spherical Gaussians:exp

(pr2P

)rC

dr =2

pF0(pR2PC

)I We shall now consider general two-center one-electron Coulomb integrals:

Vab =Ga

r1C Gb = ab(r)rC drI The overlap distribution is expanded in Hermite Gaussians

ab(r) = ij (x)kl (y)mn(z), ij (x) =i+j

t=0 Eijt t(xP)

I The total overlap distribution may therefore be written in the form

ab(r) =tuv

E ijt Eklu E

mnv tuv (rP) =

tuv

E abtuvtuv (rP)

I We now expand two-center Coulomb integrals in one-center Hermite integrals:

two-center integral Vab =tuv

E abtuv

tuv (rP)

rCdr one-center integrals

I Our next task is therefore to evaluate Hermite Coulomb integrals

Trygve Helgaker (CTCC, University of Oslo) Cartesian Coulomb integrals 11th Sostrup Summer School (2010) 20 / 34

Coulomb integrals over Hermite Gaussians

I Expansion of two-center Coulomb integrals in one-center Hermite integrals

Vab =tuv

E abtuv

tuv (rP)

rCdr,

exp

(pr2P

)rC

dr =2

pF0(pR2PC

)I Changing the order of integration and differentiation by Leibniz rule, we obtain

tuv (rP)

rCdr =

t+u+v

PtxPuy P

vz

exp(pR2P)

rCdr

=2

p

t+u+vF0(pR2PC )

PtxPuy P

vz

=2

pRtuv (p,RPC )

where the one-center Hermite Coulomb integral is given by

Rtuv (p,RPC ) =t+u+vF0

(pR2PC

)PtxP

uy P

vz

I The Hermite Coulomb integrals are derivatives of the Boys function

I can be obtained by repeated differentiation, using F n(x) = Fn+1(x)I recursion is simpler and more efficient


Evaluation of Hermite Coulomb integrals

I To set up recursion for the Hermite integrals, we introduce the auxiliary Hermite integrals

Rntuv (p,P) = (2p)nt+u+vFn

(pR2PC

)PtxP

uy P

vz

which include as special cases the source and target integrals

Rn000 = (2p)Fn R0tuv =t

Ptx

u

Puy

v

PvzF0

I The Hermite integrals can now be generated from the recurrence relations:

Rnt+1,u,v = tRn+1t1,u,v + XPCR

n+1tuv

Rnt,u+1,v = uRn+1t,u1,v + YPCR

n+1tuv

Rnt,u,v+1 = vRn+1t,u,v1 + ZPCR

n+1tuv


Summary one-electron Coulomb integral evaluation

1 Calculate Hermite expansion coefficients for overlap distributions by recurrence

E000 = exp(X 2AB

)E i+1,jt =

12pE ijt1 + XPAE

ijt + (t + 1)E

ijt+1

2 Calculate the Boys function by a variety of numerical techniques

Fn(x) =

10

exp(xt2

)t2ndt (x = pR2PC )

Fn(x) =2xFn+1(x) + exp(x)

2n + 1

3 Calculate Hermite Coulomb integrals by recurrence

Rn000 = (2p)nFnRnt+1,u,v = XPCR

n+1tuv + tR

n+1t1,u,v

4 Obtain the Cartesian Coulomb integrals by expansion in Hermite integralsGa

r1C Gb = tuv

E ijt Eklu E

mnv R

0tuv


Two-electron Coulomb integrals

I Two-electron integrals are treated in the same way as one-electron integrals

I Evaluation of one-electron integrals:Ga

r1C Gb = abrC dr=tuv

E abtuv

tuv

rCdr =

2

p

tuv

E abtuvRtuv (p,RPC )

I Evaluation of two-electron integrals:Ga(a)Gb(1)

r112 Gc (2)Gd (2) = ab(1)cd (2)r12 dr1dr2=tuv

E abtuv

E cd

tuv (1)(2)

r12dr1dr2

=25/2

pqp + q

tuv

E abtuv

E cd(1)++Rt+,u+,v+(,RPQ)

I Hermit expansion for both electrons, around centers P and QI integrals depend on RPQ and = pq/(p + q), with p = a + b and q = c + d


Direct SCF and other techniques

I The general expression for two-electron integrals is given by

gabcd =25/2

pqp + q

tuv

E abtuv

E cd(1)++Rt+,u+,v+(,RPQ)

I In SCF theories, these integrals make contributions to the Fock/KS matrix:

Fab =cd

(2gabcdDcd Coulomb

gacbdDcd exchange

)I in early days, one would write all integrals to disk and read back as requiredI in direct SCF theories, integrals are calculated as neededI this development (1980) made much larger calculations possible

I Many developments have since improved the efficiency of direct SCF

I screening of integralsI early contraction with density matricesI density fittingI multipole methods


Screening of overlap integrals

I The product of two s functions is by the Gaussian product rule

exp(ar2A

)exp

(br2B

)= exp( ab

a+bR2AB) exp

((a + b)r2P

)I This gives a simple expression for the overlap integral between two such orbitals:

Sab =

(

a + b

)3/2exp

(

ab

a + bR2AB

)I the number of such integrals scales quadratically with system sizeI however, Sab decreases rapidly with the separation RAB

I Let us assume that we may neglect all integrals smaller than 10k :

|Sab| < 10k insignificant integrals

I We may then neglect integrals separated by more than

RAB >

a1minln

[( 2amin

)3102k

]I in a large system, most integrals becomes small and may be neglectedI the number of significant integrals increases linearly with system size

Trygve Helgaker (CTCC, University of Oslo) Screening of integrals 11th Sostrup Summer School (2010) 26 / 34

Screening of two-electron integrals

I A two-electron ssss integral may be written in the form:

gabcd = erf(RPQ

) SabScdRPQ

I the total number of integrals scales quartically with system sizeI the number of significant integrals scales quadratically

Sab, Scd 0 rapidlyR1PQ 0 very slowly

I Let us decompose the integral into classical and nonclassical parts:

gabcd =SabScd

RPQ classical

erfc(RPQ

) SabScdRPQ

nonclassical

I quadratic scaling of classical part, can be treated by multipole methodsI linear scaling of nonclassical part since Sab 0, Scd 0, erfc 0 rapidly

erfc(RPQ)

exp(R2PQ)RPO


The scaling properties of molecular integrals

I linear system of up to 16 1s GTOs of unit exponent, separated by 1a0

3 5 7 9 11 13 15

1!103

3!103 nuclear attraction n3

n2

n1

3 5 7 9 11 13 15

1!103

3!103

3 5 7 9 11 13 15

2!104

5!104 electron repulsion

n4

n2

n13 5 7 9 11 13 15

2!104

5!104

3 5 7 9 11 13 15

1!102

2!102 overlap n2

n1

3 5 7 9 11 13 15

1!102

2!102

3 5 7 9 11 13 15

1!102

2!102 kinetic energy n2

n1

3 5 7 9 11 13 15

1!102

2!102


Integral prescreening

I Small integrals (< 1010) are not needed and should be avoided by prescreening

I The two-electron integrals

gabcd =

ab(1)cd (2)

r12dr1dr2

are the elements of a positive definite matrix with diagonal elements

gab,ab 0

I The conditions for an inner product are thus satisfied

I The CauchySchwarz inequality yieldsgab,ab gab,abgcd,cdI Precalculate

Gab =gab,ab

and prescreen gab,cd GabGcd


Early contraction with density matrixI In direct SCF, two-electron integrals are only used for Fock/KS-matrix construction

Jab =25/2

pqp + q

tuv

E abtuv

cd

E cd(1)++Rt+,u+,v+(,RPQ)Dcd

I only integrals that contribute significantly are evaluatedI screening is made based on the size of the density matrix

I It is not necessary to evaluate significant integrals fully before contraction with Dcd

I An early contraction of Dcd with the coefficients Ecd is more efficient:

KQ = (1)++

cdQ

DcdEcd

Jab =25/2

pqp + q

tuv

E abtuvQ

KQRt+,u+,v+(,RPQ)

I Timings (seconds) for benzene:

cc-pVDZ cc-pVTZ cc-pVQZlate contraction 27 374 3823early contraction 10 90 687

Trygve Helgaker (CTCC, University of Oslo) Early density-matrix contraction 11th Sostrup Summer School (2010) 30 / 34

Density fitting

I Traditionally, the electron density is expanded in orbital products

(r) =

abDab ab(r)

I the number of terms is n2, where n is the number of AOs

I The Coulomb contribution to the Fock/KS matrix is evaluated as

Jab =

ab(r1)r112 (r2)

I the formal cost of this evaluation is therefore quartic (n4)

I Consider now an approximate density (r), expanded in an auxiliary basis (r):

(r) = c(r), (r) (r)

I the size of the auxiliary basis N increases linearly with system size

I We may now evaluate the Coulomb contribution approximately as

Jab =

ab(r1)r112 (r2)

I the cost of the evaluation is therefore cubic (n2N)I if the is sufficiently accurate and easy to obtain, this may give large savings

Trygve Helgaker (CTCC, University of Oslo) Density fitting 11th Sostrup Summer School (2010) 31 / 34

Coulomb density fitting

I Consider the evaluation of the Coulomb potential

Jab =

cd (ab |cd )Dcd

I We now introduce an auxiliary basis and invoke the resolution of identity:

(ab |cd ) (ab | ) ( | )

1 ( |cd )

I in a complete auxiliary basis, the two expressions are identical

I We may now calculate the Coulomb potential in two different ways:

Jab =

cd (ab |cd )Dcd = (ab | )

cd

(ab | ) ( | )

1 ( |cd )Dcd = (ab | ) c = (ab | ) = Jab

where the coefficients c are obtained from the linear sets of equations ( | ) c =

cd ( |cd )Dcd ( | ) = ( | )

I no four-center integrals, the formal cost is cubic or smallerI note: the solution of the linear equations scales cubically

I This particular approach to density fitting is called Coulomb density fitting


Robust density fitting

I In Coulomb density fitting, the exact and fitted Coulomb matrices are given by

Jab = (ab | ) , Jab = (ab | ) , ( | ) = ( | )

I We may then calculate the exact and density-fitted Coulomb energies as

E =

ab JabDab = ( | ) , E =

ab JabDab = ( | )

I The true Coulomb energy is an upper bound to the density-fitted energy:

E E = ( | ) ( | ) = ( | ) ( | ) ( | ) + ( | ) = ( | )

where we have used the relation (|) = (|) which follows from ( | ) = ( | )I the error in the density-fitted energy is quadratic in the error in the densityI Coulomb fitting is thus equivalent to minimization of

E E = ( | ) 0

I It is possible to determine the approximate density in other ways

I quadratic energy error is ensured by using the robust formula

E = ( | ) + ( | ) ( | ) = E ( | )I however, the error in the energy is no longer minimized


Density fitting: sample calculations

I Calculation on benzene

cc-pVDZ cc-pVTZ cc-pVQZexact density late 27 374 3823exact density early 10 90 687fitted density 1 7 34exact energy 230.09671 230.17927 230.19463density-fitted energy 230.09733 230.17933 230.19466

I Clearly, very large gains can be achieved with density fitting

I Formal scaling is cubical

I for large systems, screening yields quadratic scaling for integral evaluationI the cubic cost of solving linear equations remains for large systems

I Linear scaling is achieved by boxed density-fitting and fast multipole methods

I the density is partitioned into boxes, which are fitted one at a timeI fast multipole methods for matrix elements


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