Chemistry Reference
In-Depth Information
4.1 Symmetry-Adapted Linear Combinations of Hydrogen
Orbitals in Ammonia
From now on we shall no longer be working with the nuclei, but with electronic
orbital functions that are anchored on the nuclei. As an example, we take the 1
s
orbitals on the hydrogen atoms in ammonia. The
is defined in Eq. (
4.1
), where
R
A
denotes the position vector of atom A with respect to the Cartesian origin;
a
0
is
the Bohr radius (0.529 Å), which is the atomic unit of length.
|
1
s
A
1
a
0
3
/
2
exp
1
√
π
−
|
r
−
R
A
|
a
0
|
1
s
A
=
(4.1)
Following Sect.
1.2
, the transformation of this function by the threefold axis is given
by
1
a
0
3
/
2
exp
−
|[
C
−
1
]−
R
A
|
1
√
π
r
C
3
|
3
1
s
A
=
(4.2)
a
0
The distance between the electron at position
C
−
3
r
and nucleus A is equal to the
distance between the electron at position
r
and nucleus B. This can be established by
working out the distances as functions of the Cartesian coordinates, but a straight-
forward demonstration is based on the fact that the distance does not change if we
rotate
both
nuclei and electrons:
C
−
1
3
r
−
R
A
=
Q
3
C
−
1
r
−
R
A
=
Q
3
C
−
1
r
−
Q
3
R
A
=|
−
R
B
|
r
(4.3)
3
3
Here, we have denoted the bodily rotation of the entire molecule as
Q
3
in order to
indicate that its action also involves the nuclei, as opposed to
C
3
, which is reserved
for electrons. In the expression of Eq. (
4.1
),
R
A
is replaced by
R
B
,or
C
3
|
1
s
A
=|
1
s
B
(4.4)
This confirms the active view, propagated from the beginning, as applied to the
functions. We rotate the
orbital itself counterclockwise over 120
◦
.Theresult
|
1
s
A
is equal to the
orbital. We now put the three components of the function space
together in a row vector:
|
1
s
B
=
|
1
s
C
|
f
1
s
A
|
1
s
B
|
(4.5)
The action of the operator in the function space now reads as follows:
C
3
|
f
=|
f
D
(C
3
)
(4.6)
where the representation matrix is given by
⎛
⎞
001
100
010
⎝
⎠
D
(C
3
)
=
(4.7)
