10. Appendix

612 Appendix B
The calculation for V ′ xx (2) is essentially similar except that the angle ı′ 2x
equal to ı2x. Instead
cos ı ′ 2x
1 2‰
√ 3 4‰ .
is not
If we expand cos ı ′ 2x to the lowest order in ‰ we get: cos 2ı′ 2x ∼ (1/3)[1
(8‰/3)]. Using this result we can show that:
V ′
xx (2) 1 2¢d2
VppÛ
1
d2 3
8‰
3
2Vppapple
1
3
4‰
3
≈ VppÛ
1
3
4‰
3
2Vppapple
1
3
8‰
3
Similarly V ′ xx (3) V′ xx (4) can be shown to be given by:
V ′
VppÛ
xx (3) ≈ 1
3
8‰
3
2Vppapple
1
3
2‰
.
3
Finally
V ′ xx
4
i1
V ′ VppÛ 8Vppapple
xx (i)
3 3 O(‰2 ) ≈ Vxx .
This proves that the deformation changes V ′ xx only by amounts that are of
second order in ‰. This is not true for V ′ xy . To calculate the corresponding
results for V ′ xy we can first show that: V′ xy (1) cos ı′ 1x cos ı′ 1y (V′ ppÛ V′ ppapple )
where cos ı ′ 1x cos ı′ 1y 1/3. Thus we can show that V′ 1y (l) ∼ (l 4‰)(VppÛ
Vppapple)/3 to the lowest order in ‰. The other matrix elements V ′ xy (2), V′ xy (3) and
(4) can be calculated similarly. In particular the reader should show that:
V ′ xy
cos ı ′ 2y
1
√
3 4‰
so that cos ı ′ 2x cos ı′ 2y (1/3)[1 (2‰/3)] cos ı′ 3x cos ı′ 3y and
cos ı ′ 4x cos ı′ 4y
(1/3)[1 (4‰/3)] all to the lowest order in ‰. Combining these
results one obtains
V ′ xy ≈ 8‰
9 (VppÛ Vppapple).
In this way one can obtain the 6 × 6 determinant on p. 152. The solutions to
the rest of this problem can be obtained by following the directions given in
the problem.
Solution to Problem 3.15
We note first that the third rank electromechanical tensor (em)ijk is symmetric
with respect to interchange of the indices i and j.
(a) By applying the symmetry operations of the zincblende crystal to this third
rank tensor as in Problem 3.5 we can show that: