D:\282216493.doc
05/28/16
Page 1 of 5
Abstract
By completing squares Gaussian integrals over all space can be reduced to the same basic form.
This enables the analytic evaluation of forms containing rij. These will then be compared with MonteCarlo estimates of the same quantities in order to test the accuracy of various Monte-Carlo estimates.
Introduction
The integral form considered here is
F
I (1)
d
i
G
J
H
K
The basic procedure of completing the squares to integrate

FR I
I
  F
d
A, Ri  z
d r expd
 Ar  2r  Ri  GJ expG J (2)
HAK HA K
g zr1 r1 d r d r
b
  ,  , N , m, n 
3
m
1
3
1
n
12
N
N
N N
 
exp   ri 2     ri  r j
i 1
i 1 j 1
3/ 2
3
2
2
2
is in Intemp.doc .htm.
Expressing the integration in terms of an auxiliary function H
Expanding the double sum in equation (1) yields
N
N
N
N
 
 N 
2
2
2
r

2
r

r

r

2
N
r

2
r
 i
 i  i   rj
i
j
j
i 1 j 1
i 1
i 1
(3)
j 1
so that
Fb g r  2  r  r I
G
J
H
K
or defining
F A r  2  r  r I (5)
1 1
Hb
A,  , N , m, ng
z
d r d r expG
J
r r
H
K
b
 ,  , N , m, ng
 Hb
  2 N ,  , N , m, ng (6)
b
g zr1 r1 d r d r
  ,  , N , m, n 
3
m
1
3
1
n
12
N
exp    2 N
m
1
i
N
i
i 1
N
3
1
n
12
N
2
i 1
N
3
N
j
j 1
N
2
N
i
i
i 1
i 1
j
j 1
The N electron term, H, integrated over d3rN
Sorting out the last integration variable
b
g zr1 r1 d r d r expF
G
H A r
FbA  2 gr  4r   r I
expG
J
H
K
H A,  , N , m, n 
z
d 3 rN
N 1
3
m
1
n
12
3
1
N 1
i
i 1
N 1
2
N
N
j
j 1
Then using the integration template the last integral becomes
2
I
J
K (7)
N 1 N 1
 
 2  ri  r j 
i 1
j 1
(4)
D:\282216493.doc
05/28/16
z
 3/ 2 exp

 
 A, R  d 3 r exp  Ar 2  2r  R 
d i
d
i
Page 2 of 5
F
R I
G
HA J
K
2
A 3/ 2
F4  r  r I (8)
G
J

I
J
  r J
expG
A  2 g J
b
K bA  2 g G
G
J
H
K
N 1
z
3
d rN
FbA  2 gr  4r
expG
H
2
N
N 1
N
i
3/ 2
j
i 1
3/ 3
j 1
Putting the exponential with the similar term in the integral over r1 yields
b
F A r
expG
H
g bA  2 g zr1 r1 d r d r
F2  4 I r  r I
G
H bA  2 gJ
K J
K
3/ 2
H A,  , N , m, n 
N 1
2
i
3
3/ 3
m
1
2
N 1
3
1
n
12
N 1

(9)
N 1
i
i 1
j
i 1
j 1
Simplify the second exponential
b
g b g zr1 r1 d r d r 
(10)
F
F
2 A I   I
expG
HbA  2 gJ
K r  r J
H A r  G
K
And now recognize the integral as
FA, A , N  1, m, nIJ (11)
HG
H A  2
K
so that we have
FA, A , N  1, m, nIJ (12)

Hb
A,  , N , m, ng

HG
K
bA  2 g H A  2
 3/ 2
H A,  , N , m, n 
A  2
N 1
3
3/ 2
m
1
N 1
N 1
n
12
3
1
N 1
2
i
i
i 1
i 1
j
j 1
3/ 2
3/ 2
FUNCTION H(A,BETA0,N)
BETA=BETA0 -- INPUT VALUE
FACT=1
DO NM=N,3,-1
FACT=FACT*
 3/ 2
A  2
b
g
3/ 2
BETA=A*BETA/(A-2*BETA)
ENDDO
For N>2, the sequence is independent of m and n.
H=FACT*H(A,beta,2,m,n)
The bottom layer for m and n = 0 is found to be
b
g
H A,  ,2,0,0 
3
cAbA  4 gh
3/ 2
N 1
2
(13)
in H200.doc .htm. The Fortran code is in testing\hfuns.for.
j
j 1
D:\282216493.doc
05/28/16
Page 3 of 5
Inverse powers of r.
The worry comes when evaluating terms that could be singular. The first of these is the nuclear
electron potential 1/R1 and its square (part of p4) which goes as 1/R12. The change needed in equation (1)is
only to the r1 integral
F
G
H
I
J
K
The r integration is exactly the same as in H200.doc .htm leading to equation 6. Thus
FF AbA  4 gIr I (15)

1
Hb
A,  ,2, m,0g

d
r
exp
HbA  2 gJ
KJ
HG
K
bA  2 g z r G
Change to spherical coordinates to take advantage of the lack of angular dependency in the integrand.
FF AbA  4 gIr I (16)
 4
Hb
A,  ,2, m,0g

r dr expG
z
HbA  2 gJ
KJ
HG
K
bA  2 g
The cases of interest are m=0,1 and 2
b
A  2 g
 4

Hb
A,  ,2, m,0g


bA  2 g 4cAbA  4 gh cAbA  4 gh m  0
bA  2 g 
 4
2

(17)
bA  2 g 2 AbA  4 g AbA  4 gbA  2 g m  1
b
A  2 g
 4
2


bA  2 g 2cAbA  4 gh bA  2 gcAbA  4 gh m  2
gz
b
2
2
1 3
 2 
2
d
r
exp

A
r

2

r


2
i
i  r j (14)
r1m
i 1
i 1
j 1
H A,  ,2, m,0  d 3 r1
2
3/ 2
3
1
3/ 3
3/ 2

3/ 3
2
1
m
1
2m
2
0
3/ 2
3/ 2
3/ 2
3
3/ 2
3/ 2
3/ 2
5/ 2
3/ 2
1/ 2
1/ 2
3/ 2
3/ 2
3
1/ 2
1/ 2
In practice, both of these will be calculated on each test run and they will be summed over all N electrons.
NMC= 1048576
THE NUMBER OF ELECTRONS IS
2
INTEGRAL IS 4.448057117553909E-001 +- 1.676313541383681E-003
AINTEGRAL IS 4.423603938062348E-001 RES=
1.458747357691631
RI=
1.376489907610462 +- 3.932147188571423E-003
ARI=
1.372032810009793 RES=
1.133502228406804
RI2=
3.345312960514719 +- 7.052165793538759E-003
ARI2=
3.342278530980440 RES= 4.302833516844589E-001
Inverse powers of r12 .
H
The change to r1, r2 and r12 in the exponent is detailed in Htor12.doc .htm. The result is
 ,  ,2,0, n     4 ,  , n (18)
b
g b
where
b gz
g
z
 A,  , n  d 3 r1 d 3 r2
The variable change from
b g
 A,  , n  8 2
z

0
ec
h
1
exp  A r12  r22  2 r122 (19)
n
r12
j
d 3r1d 3r2 to r1dr1r2 dr2 r12 dr12 is detailed in r12b.doc .htm.
c h zr expc Ar hdr z x
r1 exp  Ar12 dr1

0
2
2
2
r1  r2
2
r1  r2
1 n
c
h
exp 2 x 2 dx (20)
The case n=0, the normalization which is equal to the results derived without this variable change is
detailed in r1r2x.doc .htm to arrive at
D:\282216493.doc
b g
 A,  ,0 
05/28/16
3
(21)
cAbA  4 gh
3/ 2
b
Page 4 of 5
g b
g
H A,  ,2,0,0   A  4  ,  ,0 
3
cAbA  4 gh
3/ 2
(22)
with equation (22) in explicit agreement with equation (13).
At this point note that
zc

1
2

x

h
exp  x 2 t 2 dt
(23)
0
This means that
b g

zz

2
 A,  ,1 


8 2
0
0
zc
g z
ec

2

c h zr expc Ar hdr z x expe2c  t / 2hx jdxdt
r1 exp  Ar12 dr1

0
r1  r2
2
2
2
2
1
2
2
r1  r2
(24)
h
 A,   t 2 / 2,0 dt
0
Using the value of (A,B,0) from (21)
b
 A,  ,1 

2

3
A A  4   2t
0
2
hj
3/ 2
dt 
2 5/ 2
A 3/ 2
zc

0
1
A  4   2t 2
h
3/ 2
dt
(25)
1
This integral is given by Jefferey as
zc
dx
a  cx
h
2 3/ 2

1
x
a a  cx 2
c
(26)
h
1/ 2
In our integral the value at the lower limit is zero, while the infinity at the upper limit gives
zc

0
dx
a  cx
h
2 3/ 2

1
(27)
ac 1/ 2
so that
b g A bA 24 g
 A,  ,1 
5/ 2
(28)
3/ 2
For n=2, we use the relationship
z
c h
zc
zz
c h zr expc Ar hdr z x expe2c  t / 2hx jdxdt


h
1
 2 t exp  x 2 t 2 dt  exp  x 2 y dy
2
x
0
0
(29)
This leads to
b g

 A,  ,2  2 t 8 2
0
zc


0
r1 exp  Ar12 dr1

0
2
2
2
r1  r2
2
r1  r2
1
2
2
(30)
h
 2 t A,   t 2 / 2,0 dt
0
Again using the value of (A,B,0) from (21)
1
Alan Jefferey, Handbook of Mathematical Formulas and Integrals, Academic Press 1995, p.158.
D:\282216493.doc
b g
 A,  ,2  2
zec

0

3
A
3/ 2
05/28/16
 3t
A A  4   2t 2
hj

1
bA  4  2 yg

1/ 2
0
3/ 2
dt 
3
A
3/ 2
1
A 3/ 2
zb

0
Page 5 of 5
3
A  4  2 y
g dy (31)
3/ 2
1
bA  4 g
1/ 2
Equation (18) can be used to put the n=1 and n=2 results into the H form needed by the code.
b
g b
g b  4 g2 
H  ,  ,2,0,1     4 ,  ,1 
5/ 2
3/ 2
(32)
and
3
H  ,  ,2,0,2     4 ,  ,2 
3/ 2
  4

b
g b
gb g
(33)
The numerical integrals are in testing\testing.mdp – (This assumes Power Station Fortran)
The list below is of the main codes. These will run in watfor or watcom. They have includes to the
necessary subroutines.
testing\moltest.for
testing\moltest2\moltest.for
testing\rijtest.for