Chemistry Reference
In-Depth Information
reverses both vectors, a character of
2, and each of the reflections leaves the vector it
contains unchanged but reverses the other one, 1
−
0. So we have found the trace of
the matrices by inspection. This is the approach we will adopt in the remaining chapters
of this topic. What we need to do now is show that the trace is sufficient to identify the
appropriate irreducible representations.
For the
C
2v
point group, we have already found all the standard labels, and the characters
for the irreducible representations are given in the character table shown in Table 4.6. None
of these standard representations have a 2 under the
E
column; in fact, they all have 1. This
is not a surprise, since we have already shown that
x
belongs to the
B
1
representation and
y
to the
B
2
representation by inspecting each basis vector in turn in Section 4.2. So the
totalled character for the basis obtained above could be reduced to those for
B
1
and
B
2
, i.e.
the characters for the individual vectors. Table 4.7 shows that the sum of the characters
B
1
+
−
1
=
B
2
agrees with the totals laid out in Equation (4.23). The reducible representation is
usually given the symbol
, and so in this case we have shown that
=
B
1
+
B
2
(4.24)
In general, it will be possible to simplify the total representation for a basis of our choos-
ing into a sum of the standard irreducible representations from the point group character
table. The reducible and irreducible representations are linked by the fact that the sum of
characters of the irreducible representations for a basis must give the characters of their
reducible representation:
χ
(
C
)
=
n
i
χ
i
(
C
)
(4.25)
i
Here, we use the symbol
and
i
indicating that the
character is taken from the reducible representation and the
i
th irreducible representation
respectively, for a given class of operations
C
. This formula simply says that the diagonal
χ
for a character, the subscripts
Tab l e 4 . 6
The C
2v
character table.
C
2v
E
2
σ
ν
σ
v
A
1
1
1
1
1
z
x
2
,
y
2
,
z
2
A
2
1
1
−
1
−
1
R
z
xy
B
1
1
−
1
1
−
1
x
,
R
y
xz
B
2
1
−
1
−
1
1
y
,
R
x
yz
Tab l e 4 . 7
The sum of the characters
from the B
1
and B
2
representations in
the C
2v
point group.
C
2v
E
2
σ
ν
σ
v
B
1
1
−
1
1
−
1
B
2
1
−
1
−
1
1
B
1
+
B
2
2
−
2
0
0
