REPRESENTATION THEORY
x 1x 2x 3y (^1) y 2
y 3
Figure 29.5 An equilateral array of masses and springs.
29.11.3 Degeneracy of normal modes
As our final area for illustrating the usefulness of group theoretical results we
consider the normal modes of a vibrating system (see chapter 9). This analysis
has far-reaching applications in physics, chemistry and engineering. For a given
system, normal modes that are related by some symmetry operation have the same
frequency of vibration; the modes are said to bedegenerate. It can be shown that
such modes span a vector space that transforms according to some irrep of the
groupGof symmetry operations of the system. Moreover, the degeneracy of
the modes equals the dimension of the irrep. As an illustration, we consider the
following example.
Investigate the possible vibrational modes of the equilateral triangular arrangement of
equal 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(orC 3 v), whose character table is table 29.1. As on a previous occasion, it is
most convenient to use the natural representationDnatof this group (it almost always
saves 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, but
this is not usually the case.
We note that in table 29.1 the second class contains the rotationsA(byπ/3) andB(by
2 π/3), also known asRandR′. This class is known as 3zin crystallographic notation, or
C 3 in chemical notation, as explained in section 29.9. The third class containsC,D,E,the
three mirror reflections.
Clearlyχ(I) = 6. Since all position labels are changed by a rotation,χ(3z)=0.Forthe
mirror reflections the simplest representative class member to choose is the reflectionmyin
the plane containing they 3 -axis, since then only label 3 is unchanged; undermy,x 3 →−x 3
andy 3 →y 3 , leading to the conclusion thatχ(my) = 0. Thus the character set is 6, 0, 0.
Using (29.18) and the character table 29.1 shows that
Dnat=A 1 ⊕A 2 ⊕2E.