10. Appendix

Solution to Problem 2.6 585
transform like {xyz 2 , xy 2 z, x 2 yz} xyz{z, y, x}. Since {xyz} transforms like
°1 and {x, y, z} transform like °4 these three functions transform like °4.
(d) The derivation of the symmetry of the three functions {sin(2applex/a)
cos(2appley/a) cos(2applez/a), cos(2applex/a) sin(2appley/a) cos(2applez/a), cos(2applex/a) cos(2appley/a)
sin(2applez/a)} is left as an exercise.
Solution to Problem 2.6
In this solution we demonstrate how to derive the compatibility relation between
° and ¢. The reader should repeat the calculation for the remaining
compatibility relations in this problem. First, we need to find the symmetry
operations of the group of ° which are also symmetry operations of ¢. From
Table 2.13 we find the following correspondence between the symmetry operations
in those two groups.
Operation in ° Corresponding Operation in ¢
{E} {E}
{C2} {C2 4 }
{Û} {md}
{Û ′ } {m ′ d }
According to the definition on page 45, two representations are compatible
if they have the same characters for the corresponding classes in the above
table. Based on this definition °1 is compatible with ¢1 while °2 is compatible
with ¢2.
For °3 the characters of the above 4 classes are:
{E} {C2} {Û} {Û ′ }
°3 2 2 0 0
From Table 2.14 we see that the characters of ¢1 ¢2 are exactly the same
for the corresponding classes of °:
{E} {C 2 4 } {md} {m ′ d }
¢1 ¢2 2 2 0 0
Hence °3 is compatible with ¢1 ¢2.