Mathematical Methods for Physics and Engineering - Matematica.NET

REPRESENTATION THEORYy 3x 3y 1y2x 1x 2Figure 29.5An equilateral array of masses and springs.29.11.3 Degeneracy of normal modesAs our final area for illustrating the usefulness of group theoretical results weconsider the normal modes of a vibrating system (see chapter 9). This analysishas far-reaching applications in physics, chemistry and engineering. For a givensystem, normal modes that are related by some symmetry operation have the samefrequency of vibration; the modes are said to be degenerate. It can be shown thatsuch modes span a vector space that transforms according to some irrep of thegroup G of symmetry operations of the system. Moreover, the degeneracy ofthe modes equals the dimension of the irrep. As an illustration, we consider thefollowing example.◮Investigate the possible vibrational modes of the equilateral triangular arrangement ofequal masses and springs shown in figure 29.5. Demonstrate that two are degenerate.Clearly the symmetry group is that of the symmetry operations on an equilateral triangle,namely 3m (or C 3v ), whose character table is table 29.1. As on a previous occasion, it ismost convenient to use the natural representation D nat of this group (it almost alwayssaves having to write out matrices explicitly) acting on the six-dimensional vector space(x 1 , y 1 , x 2 , y 2 , x 3 , y 3 ). In this example the natural and regular representations coincide, butthis is not usually the case.We note that in table 29.1 the second class contains the rotations A (by π/3) and B (by2π/3), also known as R and R ′ . This class is known as 3 z in crystallographic notation, orC 3 in chemical notation, as explained in section 29.9. The third class contains C, D, E, thethree mirror reflections.Clearly χ(I) = 6. Since all position labels are changed by a rotation, χ(3 z )=0.Forthemirror reflections the simplest representative class member to choose is the reflection m y inthe plane containing the y 3 -axis, since then only label 3 is unchanged; under m y , x 3 →−x 3and y 3 → y 3 , leading to the conclusion that χ(m y ) = 0. Thus the character set is 6, 0, 0.Using (29.18) and the character table 29.1 shows thatD nat =A 1 ⊕ A 2 ⊕ 2E.1110