Chemistry Reference
In-Depth Information
All these irreps occur only once; hence, when we choose a trigonal symmetry adap-
tation for constructing representation matrices, for all these matrices, there will be
exactly one index
k
that labels the component that is totally symmetric in
C
3
v
.For
this value, the matrix elements read
Φ
ik
|
H
|
Φ
ik
=
β
D
kk
(C
4
)
+
D
kk
C
−
4
+
D
kk
(C
2
)
(4.146)
To obtain these eigenvalues, we thus have to mould the representation matrices for
O
h
in a trigonal setting. For the one-dimensional irreps,
A
1
g
and
A
2
u
, this is trivial
since the matrix elements in Eq. (
4.146
) are simply the characters. We thus obtain:
β
χ
A
1
g
(C
4
)
χ
A
1
g
C
−
4
+
χ
A
1
g
(C
2
)
=
=
+
E(A
1
g
)
3
β
(4.147)
β
χ
A
2
u
(C
4
)
χ
A
2
u
C
−
4
+
χ
A
2
u
(C
2
)
=−
=
+
E(A
2
u
)
3
β
For the
T
1
u
irrep, we can use the set of the
p
-orbitals. In the standard Cartesian ori-
entation, this set is adapted to the tetragonal site symmetry. In Fig.
4.9
we illustrate
an alternative trigonal basis (see also Fig.
3.6
(d)). The transformation between the
two sets is given by
√
3
|
p
z
1
p
z
=
|
p
x
+|
p
y
+|
√
2
|
p
y
1
p
x
=
|
p
x
−|
(4.148)
√
6
|
p
z
1
p
y
=
|
p
x
+|
p
y
−
2
|
Here, the first component points in the threefold direction and thus is adapted to
the
C
3
v
site symmetry. We require only the diagonal matrix elements for this com-
ponent. They can easily be obtained by expressing these elements in the standard
canonical set:
=
p
z
|
C
4
|
p
z
D
z
z
(C
4
)
3
p
x
|+
p
y
|+
p
z
|
C
4
|
p
x
+|
p
y
+|
p
z
1
=
3
p
x
|+
p
y
|+
p
z
|
|
p
y
−|
p
x
+|
p
z
=
1
1
3
=
D
z
z
C
−
4
=
p
z
|
C
−
4
|
p
z
3
p
x
|+
p
z
C
−
4
|
p
x
+|
p
z
1
=
p
y
|+|
p
y
+|
3
p
x
|+
p
y
|+
p
z
|
−|
p
x
+|
p
z
=
1
1
3
=
p
y
+|
