10. Appendix

652 Appendix B
point group. We notice that one important difference between the E1 and E2
irreducible representation is that under a C2 operation the character of the
E1 mode is 2 while the corresponding character for the E2 mode is 2. This
suggests that both basis functions of the E2 mode remain invariant under a
C2 operation, On the other hand, the character for the E2 mode under a C6
operation is 1 while that of the E1 mode is 1.
Functions of x and y that remain invariant under a rotation by 180◦ about
the z-axis (C2 symmetry operation) are: {xy and x2 y2 }. Thus, their character
is 2 for a C2 operation. To verify that these two functions indeed form an
irreducible 2D representation of E2 symmetry, we consider a 6-fold rotation
about the z-axis. Under this symmetry operation x and y transform according
to: x ⇒ (x/2) ( √ 3y/2) while y ⇒ ( √ 3x/2) (y/2).
From these two transformation equations we find the character of the 2D
representation {x, y} to be: ( 1 1
2 ) ( 2 ) 1 consistent with the fact that the
symmetry of this representation is E1. Under the same symmetry operation
{xy, x 2 y 2 } transform according to the following equations:
xy ⇒ [(x/2) ( 3y/2)][( √ 3x/2) (y/2)]
1
2
xy ( √ 3/4)(x 2 y 2 )
while
x 2 y 2 ⇒[(x/2)( 3y/2)] 2 [( √ 3x/2)(y/2)] 2
3xy 1
(x
2
2 y 2 ).
These results show that {xy, x2 y2 } forms a 2D representation whose character
is 1 under a C6 symmetry operation. It is easy to show that the characters
of {xy, x2 y2 } under the other symmetry operations of the C6v point group
are all consistent with the E2 irreducible representation. Thus {xy, x2 y2 }
form a set of basis functions for the E2 mode.
Next we have to deduce the non-zero and linearly independent elements
of the Raman tensor of a phonon belonging to the E2 irreducible representation.
We will illustrate this process by showing that the Raman tensor component
R13 has to vanish. For simplicity, we will define f1 xy and f2 x2 y2 .
By definition
R13 ∼ (¯13/f1)f1 (¯13/f2)f2
Consider the effect of a symmetry operation Ûd (reflection in the plane containing
the y and z axes): xyz → xyz. Under this operation f1 ⇒ f1 while
f2 ⇒ f2. The second rank tensor component (¯13) ⇒ (¯13). Thus R13 ⇒ R13
under this symmetry operation and hence R13 0. By similar symmetry arguments
one can conclude that R23 0 also.
Next, we consider the diagonal term: R33 ∼ (¯33/f1)f1 (¯33/f2)f2
Under the same operation Ûd, ¯33 is invariant while f1 changes sign. Thus
(¯33/f1) changes sign and hence is zero. By considering another reflection
operation in which f2 changes sign we can show similarly that (¯33/f2) 0
and hence R33 0.
The derivation of the remaining non-zero components of the Raman tensor
for the E2 or °6 mode is left as an exercise.