Mathematical Methods for Physics and Engineering : A Comprehensive Guide

29.7 COUNTING IRREPS USING CHARACTERS

whilst forDˆ

(λ)

=Dˆ

(μ)

= E, it gives

1(2^2 ) + 2(1) + 3(0) = 6.

(ii) ForDˆ

(λ)

=A 2 andDˆ

(μ)

= E, say, (29.15) reads

1(1)(2) + 2(1)(−1) + 3(−1)(0) = 0.

(iii) ForX 1 =AandX 2 =D, say, (29.16) reads

1(1) + 1(−1) + (−1)(0) = 0,

whilst forX 1 =CandX 2 =E, both of which belong to classC 3 for which

c 3 =3,

1(1) + (−1)(−1) + (0)(0) = 2 =

6

3

.

29.7 Counting irreps using characters

The expression of a general representationD={D(X)}in terms of irreps, as

given in (29.11), can be simplified by going from the full matrix form to that of

characters. Thus

D(X)=m 1 Dˆ

(1)

(X)⊕m 2 Dˆ

(2)

(X)⊕···⊕mNDˆ

(N)

(X)

becomes, on taking the trace of both sides,

χ(X)=

∑N

λ=1

mλχ(λ)(X). (29.17)

Given the characters of the irreps of the groupGto which the elementsXbelong,

and the characters of the representationD={D(X)},thegequations (29.17)

can be solved as simultaneous equations in themλ, either by inspection or by

multiplying both sides by

[

χ(μ)(X)

]∗

and summing overX, making use of (29.14)

and (29.15), to obtain

mμ=

1

g

∑

X

[

χ(μ)(X)

]∗

χ(X)=

1

g

∑

i

ci

[

χ(μ)(Xi)

]∗

χ(Xi). (29.18)

That an unambiguous formula can be given for eachmλ, once thecharacter

set(the set of characters of each of the group elements or, equivalently, of

each of the conjugacy classes) ofDis known, shows that, for any particular

group, two representations with the same characters are equivalent. This strongly

suggests something that can be shown, namely,the number of irreps = the number

of conjugacy classes.The argument is as follows. Equation (29.17) is a set of

simultaneous equations forNunknowns, themλ, some of which may be zero. The

value ofNis equal to the number of irreps ofG. There aregdifferent values of

X, but the number ofdifferentequations is only equal to the number of distinct