Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2009
Laura Gagliardi
Lecture 29, November 20, 2009
Solved Homework
Water, H
2
O, involves 2 hydrogen atoms and an oxygen atom. To minimally
represent hydrogen is simple—one needs only a 1s function. As there are two H atoms,
we will need two such functions, one centered on each atom. Oxygen has electrons in the
second principal quantum level, so we will need one 1s, one 2s, and three 2p functions
(one each of p
x
, p
y
, and p
z
). So, that's five functions total on O, plus one each on the H
atoms which is then 7 total.
The total number of one-electron integrals we need to solve is determined by
considering that we need kinetic-energy integrals
T
μ
!
"
1
2
#
2
!
and potential energy integrals
V
μ
"
=
μ
#
Z
k
r
k
k
nuclei
$
"
Since both
μ
and
ν
can be any one of the 7 basis functions, there are 7
x
7 possible
kinetic energy integrals (49) and the same number of potential energy integrals
for each
nucleus
. As there are 3 nuclei, that is 7
x
7
x
3 = 147 integrals. The grand total of one-
electron integrals is thus 196.
The two-electron integrals are
μ
! "#
( ) =
$
μ
1
( )
$
!
1
( )
1
r
12
$
"
2
( )
$
#
2
( )
dr
1
dr
2
%
where now
μ
,
ν
,
λ
, and
σ
can be any one of the seven basis functions (note that letting
them be
any
function covers both all Coulomb and all exchange integrals). There are thus
7
x
7
x
7
x
7 = 2401 two-electron integrals.
The numbers computed above involve the contracted basis functions, each of
which, since the basis is STO-3G, is composed of 3 primitive functions. Thus, for any
individual one-electron integral, there will be 3
x
3 = 9 separate integrals involving the
primitives. There are thus 9
x
196 = 1764 individual primitive one-electron integrals.