Molecular Symmetry and Group Theory Workshop 3 — Character tables

• Exercise 2

The C3v character table is given opposite:
C3v E 2C3 v
A1 1 1 1
A2 1 1 -1
E 2 -1 0

Note that the numbers in a character table are not necessarily restricted to 1 or -1. We shall consider the meaning of the character 2, which describes the effect of the identity operation for the irreducible representation labelled E, in Workshop 4.

Suppose we have the following reducible representation (reducible representations are normally given the label Γ):

E 2C3 3σv 4 1 -2

This may be reduced to its constituent irreducible representations by applying the reduction formula:

n(A1) = 1/h Σ(χR × χI × N)

n(A1) h — is the number of times the irreducible representation A1 occurs in the reducible representation, — is the order of the group i.e. the total number of operations in the group, — the sum is taken over all classes, — is the character of the reducible representation, — is the character of the irreducible representation, — number of symmetry operations in the class.

Fortunately this formula is much easier to use than to remember! Consider the reducible representation given above. For A1 we apply the following procedure:

1. group the operations in classes and write Γ and A1 as shown:

E 2C3 3σv 4 1 -2 1 1 1

2. N×χR×χI for each column, Multiply together the two characters and the number of operations in the class:

 1×4×1 1×1×2 1×(-2)×3 4 2 -6

3. Add: 4 + 2 + (-6) = 0
4. Divide by the order (6) of the point group C3v: 0/6 = 0

number of A1 = 0

Full Example of A1:

E 2C3 v
 Order = ?
Γ 4 1 -2
A1 1 1 1
N×χR×χI
?
?
?
 Σ(N×χR×χI) = ?

Therefore, number of A1 is:
 Σ(N×χR×χI)/Order = ?

Repeat the procedure above for A2 and E, and thus determine the composition of Γ.

E 2C3 v
 Order =
Γ 4 1 -2
A2 1 1 -1
N×χR×χI
 Σ(N×χR×χI) =

Therefore, number of A2 is:
 Σ(N×χR×χI)/Order =

E 2C3 v
 Order =
Γ 4 1 -2
E 2 -1 0
N×χR×χI
 Σ(N×χR×χI) =

Therefore, number of E is:
 Σ(N×χR×χI)/Order =

Therefore, the composition of Γ is:   Γ = A2 + E