Mathematical Methods for Physics and Engineering - Matematica.NET

29.8 CONSTRUCTION OF A CHARACTER TABLEm d ′m xm dm yFigure 29.3 The mirror planes associated with 4mm, the group of twodimensionalsymmetries of a square.that at least one irrep has dimension 2 or greater. However, there can be no irrep withdimension 3 or greater, since 3 2 > 8, nor can there be more than one two-dimensionalirrep, since 2 2 +2 2 = 8 would rule out a contribution to the sum in (iii) of 1 2 from theidentity irrep, and this must be present. Thus the only possibility is one two-dimensionalirrep and, to make the sum in (iii) correct, four one-dimensional irreps.Therefore using (i) we can now deduce that there are five classes. This same conclusioncan be reached by evaluating X −1 YX for every pair of elements in G, as in the descriptionof conjugacy classes given in the previous chapter. However, it is tedious to do so andcertainly much longer than the above. The five classes are I, Q, {R, R ′ }, {m x , m y }, {m d , m d ′}.It is straightforward to show that only I and Q commute with every element of thegroup, so they are the only elements in classes of their own. Each other class must haveat least 2 members, but, as there are three classes to accommodate 8 − 2 = 6 elements,there must be exactly 2 in each class. This does not pair up the remaining 6 elements, butdoes say that the five classes have 1, 1, 2, 2, and 2 elements. Of course, if we had startedby dividing the group into classes, we would know the number of elements in each classdirectly.We cannot entirely ignore the group structure (though it sometimes happens that theresults are independent of the group structure – for example, all non-Abelian groups oforder 8 have the same character table!); thus we need to note in the present case thatm 2 i = I for i = x, y, d or d ′ and, as can be proved directly, Rm i = m i R ′ for the same fourvalues of label i. We also recall that for any pair of elements X and Y , D(XY )=D(X)D(Y ).We may conclude the following for the one-dimensional irreps.(a) In view of result (vi), χ(m i )=D(m i )=±1.(b) Since R 4 = I, result (vi) requires that χ(R) is one of 1, i, −1, −i. But, sinceD(R)D(m i )=D(m i )D(R ′ ), and the D(m i ) are just numbers, D(R) =D(R ′ ). FurtherD(R)D(R) =D(R)D(R ′ )=D(RR ′ )=D(I) =1,and so D(R) =±1 =D(R ′ ).(c) D(Q) =D(RR) =D(R)D(R) =1.If we add this to the fact that the characters of the identity irrep A 1 are all unity then wecan fill in those entries in character table 29.4 shown in bold.Suppose now that the three missing entries in a one-dimensional irrep are p, q and r,where each can only be ±1. Then, allowing for the numbers in each class, orthogonality1101