29.13 HINTS AND ANSWERS
(a) Make an initial sketch showing an arbitrary small mass displacement from,
say, vertexC. Draw the results of operating on this initial sketch with each
of the symmetry elements of the group 3m(C 3 v).
(b) Superimpose the results, weighting them according to the characters of irrep
A 1 (table 29.1 in section 29.6) and verify that the resultant is a symmetrical
arrangement in which all three masses move symmetrically towards (or away
from) the centroid of the triangle. The mode is illustrated in figure 29.7(a).
(c) Start again, this time considering a displacementδofCparallel to thex-axis.
Form a similar superposition of sketches weighted according to the characters
of irrep E (note that the reflections are not needed). The resultant contains
some bodily displacement of the triangle, since this also transforms according
to E. Show that the displacement of the centre of mass is ̄x=δ, ̄y=0.
Subtract this out, and verify that the remainder is of the form shown in
figure 29.7(c).
(d) Using an initial displacement parallel to they-axis, and an analogous proce-
dure, generate the remaining normal mode, degenerate with that in (c)and
shown in figure 29.7(b).29.13 Further investigation of the crystalline compound considered in exercise 29.7
shows that the octahedron is not quite perfect but is elongated along the (1, 1 ,1)
direction with the sulphur atoms at positions±(a+δ, δ, δ),±(δ, a+δ, δ),±(δ, δ, a+
δ), whereδa. This structure is invariant under the (crystallographic) symmetry
group 32 with three two-fold axes along directions typified by (1,− 1 ,0). The
latter axes, which are perpendicular to the (1, 1 ,1) direction, are axes of two-
fold symmetry for the perfect octahedron. The group 32 is really the three-
dimensional version of the group 3mand has the same character table as table 29.1
(section 29.6). Use this to show that, when the distortion of the octahedron is
included, the doublet found in exercise 29.7 is unsplit but the triplet breaks up
into a singlet and a doublet.
29.13 Hints and answers29.1 There are four classes and hence four one-dimensional irreps, which must have
entriesasfollows:1,1,1,1; 1,1,−1,−1; 1,−1, 1,−1; 1,−1,−1, 1. The
characters ofDare 2,−2, 0, 0 and so the irreps present are the last two of these.
29.3 There are five classes{ 1 },{− 1 },{±i},{±j},{±k}; there are four one-dimensional
irreps and one two-dimensional irrep. Show thatab=ba. The homomorphism
is± 1 →I,±i→a,±j→b,±k→ab.Vis Abelian and hence has four
one-dimensional irreps.
In the class order given above, the characters forQare as follows:
Dˆ(1), 1 , 1 , 1 , 1 ,1;Dˆ(2), 1 , 1 , 1 ,− 1 ,−1;Dˆ(3), 1 , 1 ,− 1 , 1 ,−1;
Dˆ(4), 1 , 1 ,− 1 ,− 1 ,1;Dˆ(5), 2 ,− 2 , 0 , 0 ,0.
29.5 Note that the fourth and fifth classes each have 6 members.
29.7 The five basis functions of the representation are multiplied by 1,e−iφ′,e+iφ′,
e−^2 iφ
′
,e+2iφ′
as a result of the rotation. The character is the sum of these for
rotations of 0, 2π/3,π,π/2,π;Drep=E+T 2.
29.9 (b) The bodily translation has irrep T 2 and the rotation has irrep T 1. The irreps
of the internal vibrations are A 1 ,E,T 2 , with respective degeneracies 1, 2, 3,
making six internal coordinates (12 in total, minus three translational, minus
three rotational).
29.11 There are four classes and hence four irreps, which can only be the identity
irrep, one other one-dimensional irrep, and two two-dimensional irreps. In the
class order{I},{R, R^4 },{R^2 ,R^3 },{mi}the second one-dimensional irrep must