10. Appendix

Solution to Problem 7.5 651
erties as (em) we expect that its linearly-independent and non-zero elements
are: c (¯/Q)15, a (¯/Q)31, and b (¯/Q)33. The number of non-zero
elements for the °1 optical phonon will be further reduced since we can put
Q1 Q2 0 and Q3 z. Thus the second rank Raman tensor is obtained by
the contraction of (¯/Q) with (0, 0, z):
⎛ ⎞
az
⎛
⎞ ⎜ az ⎟ ⎛ ⎞
0 0 0 0 c 0 ⎜ ⎟ a 0 0
(0 0 z) ⎝ 0 0 0 c 0 0⎠
⎜ bz ⎟
⎜ ⎟ ⇔ ⎝ 0 a 0 ⎠ z
⎜ 0 ⎟
a a b 0 0 0 ⎝
0
⎠ 0 0 b
0
The final form of the Raman tensor for the °1 mode, after dividing by the
phonon amplitude z, is therefore:
⎛ ⎞
a 0 0
Rij(°1) ⎝ 0 a 0 ⎠
0 0 b
The Raman tensor for the 2D °5 modes can be obtained similarly by assuming
that Q (x, 0, 0) for one of the two modes and (0, y, 0) for the remaining
mode. The second rank Raman tensor obtained by the contraction of (¯/Q)
with (x, 0, 0) is:
⎛ ⎞
0
⎛
⎞ ⎜ 0 ⎟ ⎛ ⎞
0 0 0 0 c 0 ⎜ ⎟ 0 0 c
(x 00) ⎝ 0 0 0 c 0 0⎠
⎜ 0 ⎟
⎜ ⎟ ⇔ ⎝ 0 0 0⎠x
.
⎜ 0 ⎟
a a b 0 0 0 ⎝
cx
⎠ c 0 0
0
The final form of the Raman tensor for the °5(x) mode, after dividing by the
phonon amplitude x, is therefore:
⎛ ⎞
0 0 c
Rij(°5(x)) ⎝ 0 0 0⎠
.
c 0 0
Similarly, the Raman tensor for the °5(y) mode is:
⎛ ⎞
0 0 0
Rij(°5(y)) ⎝ 0 0 c ⎠ .
0 c 0
Finally, the °6 mode is also doubly degenerate. However, there is an important
difference between this mode and the °5 mode. While the °5 mode is
infrared-active and can be represented by the components of a vector, the °6
mode is not infrared-active and it cannot be represented by a vector (it can
be represented instead by a pseudovector). As a result, we cannot deduce the
symmetry of the Raman tensor for the °6 mode by taking advantage of the
known symmetry of the third rank tensor in the wurtzite structure. Instead,
we have to derive the symmetry of the Raman tensor by first finding some
basis functions to represent the °6 modes. It is easier to do this for the C6v