Biomedical Engineering Reference
In-Depth Information
16.9 Gaussian Orbitals
None of these considerations actually broke the integrals bottleneck. There are two con-
siderations. First is the sheer number of integrals that have to be processed. If we run a
HF-LCAO calculation with
n
basis functions, we have to calculate
p
1) one-
electron integrals of each type (overlap, kinetic energy and electron-nuclear attraction).
If
n
=
1
2
n
(
n
+
100 then we have to calculate 15 150 one-electron integrals. It is the two-electron
integrals that cause the problem, for we have to calculate at most
q
=
=
1
2
p
(
p
+
1) of them; if
n
10
7
. There are approximately
8
n
4
integrals (for large
n
), although
this is to be seen as an upper limit. Many of the integrals turn out to be so tiny that they
can be neglected for practical calculations on large molecules.
Second is the mathematical intractability of the integrals, as mentioned above. The real
breakthrough came in the 1950s with the introduction of (
Cartesian) Gaussian basis func-
tions
; these are similar to Slater orbitals but they have an exponential factor that goes as
the
square
of the distance between the electron and the orbital centre:
=
100 then
q
=
1.25
×
N
G
x
l
y
m
z
n
exp
α
r
2
a
0
χ
G
=
−
(16.26)
Here
l
,
m
and
n
can have any integral values and the orbital exponent α is positive. If all of
l
,
m
and
n
are zero we talk about a 1s-type
Gaussian-type orbital
(GTO). If one of
l
,
m
or
n
is equal to 1 we have a 2p-type GTO. Expressions for normalized 1s and 2p
x
GTOs are
2α
π
a
0
3/4
exp
α
r
2
a
0
G
1s
=
−
128α
5
π
3
a
10
0
1/4
x
exp
α
r
2
a
0
G
2p
x
=
−
2) we note that there are six possibilities,
xx
,
xy
,
xz
,
yy
,
yz
,
zz
, rather than the five combinations we normally encounter for Slater-type
orbitals (STOs). The combination (
x
2
When dealing with d-type (where
l
+
m
+
n
=
z
2
) actually gives a 3s GTO, but we normally
include all six Cartesian Gaussians in calculations. Similar considerations apply to the f,
g,... orbitals.
Why use Gaussians? Let me illustrate the answer by considering the two one-dimensional
normal distribution curves (which are one-dimensional Gaussians) shown in Figure 16.4:
+
y
2
+
exp
−
x
A
)
2
=
−
G
A
(
x
)
α
A
(
x
exp
−
x
B
)
2
G
B
(
x
)
=
α
B
(
x
−
I have taken α
A
=
0.1,
x
A
=
1 (the full curve) and α
B
=
0.3 and
x
B
=−
2 (the dotted
curve). It is easily shown that the product
G
A
(
x
)
×
G
B
(
x
) is another Gaussian
G
C
whose
centre lies between
G
A
and
G
B
:
exp
2
exp
−
x
C
)
2
α
A
α
B
α
A
+
G
C
(
x
)
=
−
α
B
{
x
A
−
x
B
}
(α
A
+
α
B
)(
x
−
(16.27)
α
A
x
A
+
α
B
x
B
x
C
=
α
A
+
α
A
which is shown as the dashed curve in Figure 16.4. This is a general property of Gaussians.
