10. Appendix

Solution to Problem 3.8 (a, c and d) 609
(d) In case the spin orbit-coupling is zero, the Pikus-Bir Hamiltonian can be
written as:
L 2 L2
x
HPB a(exx eyy ezz) 3b ∗
exx c.p.
3
6d∗
1
√
3 2 (LxLy
Lylx)exy c.p. .
By repeating the calculation in (a) we obtain for [100] stress:
HPB(X) a(S11 2S12)X 3b ∗ (S11 S12)X[L 2 x (L2 /3)] .
The matrix elements corresponding to those in part (c) are:
while
〈1, 1|L 2 x|1, 1〉 〈1, 1|L 2 x|1, 1〉 1 and 〈1, 0|L 2 x|1, 0〉 0
〈1, 1|L 2 |1, 1〉 〈1, 1|L 2 |1, 1〉 〈1, 0|L 2 |1, 0〉 2.
Clearly, the mL ±1 states remain degenerate while these two states are split
from the mL 0 state by the uniaxial stress. Calculating the matrix elements
for the Pikus-Bir Hamiltonian we obtain:
〈1, 1|3b ∗ (S11 S12)X[L 2 x (L2 /3)]|1, 1〉 3b ∗ (S11 S12)X[1 (2/3)]
b ∗ (S11 S12)X
and
〈1, 0|3b ∗ (S11 S12)X[L 2 x (L2 /3)]|1, 0〉 3b ∗ (S11 S12)X[0 (2/3)]
2b ∗ (S11 S12)X .
Thus the three L 1 states split into a doublet (mL ±1 states) and a singlet
(mL 0) with the energy of the splitting equal to 3b∗ (S11 S12)X.
The reason why the stress induced splitting depends on the angular momentum
can be traced to the degeneracy of the initial states. In case of semiconductors
with a large spin-orbit coupling we can assume that the spin-orbit
splitting ¢ will be much larger than the stress-induced splitting. Thus the initial
state J 3/2 before stress is applied is 4-fold degenerate. [100] uniaxial stress
splits these states into two doublets with a splitting equal to 2b(S11S12)X. On
the other hand, if we neglect spin-orbit coupling then the initial state L 1is
six-fold degenerate (including spin). [100] uniaxial stress splits them into two
pairs of degenerate states (one is a spin doublet while the other is a fourfold
degenerate state when spin degeneracy is included) with total splitting
3b∗ (S11 S12)X. Thus, the splitting is larger than that for the J 3/2 states by
a factor of (3/2) which is equal to the ratio of the degeneracy of their initial
states.
We have assumed in this book the general case where b and b∗ , d and d∗ may be different. Usually we can assume that the spin-orbit coupling is not dependent
on strain. In this case, within this approximation b b∗ and d d∗ .
It is with this approximation in mind that the strain Hamiltonians with and
without the spin-orbit coupling are defined with a difference of a factor of 3
in the shear terms.