10. Appendix

644 Appendix B
Solution to Problem 6.21
The ratio ¢0/¢1 in Table 6.2 is not close to 3/2 for the two compounds GaN
and InP. To understand why the ratio of the spin-orbit (S-O) splittings deviates
from 3/2 for these compounds we have to find the reason why this ratio should
be equal to 3/2 in the first place. Quantum mechanics teaches us that the S-O
coupling is a relativistic effect described by the Hamiltonian: (see (2.45a)):
HSO [/4c 2 m 2 ][(∇Vxp) · Û]
Where V is the potential seen by the electron, p and Û are, respectively, the
electron momentum operators and the Pauli spin matrices. In atoms the nuclear
potential has spherical symmetry so one can express the S-O coupling in
terms of the electron orbital angular momentum operator L and spin operator
S as:
HSO ÏL · S
where Ï is known as the S-O coupling constant. In cubic semiconductors with
the zincblende and diamond structure we find that the top valence band wave
functions at k 0 are “p-like”. As a result we can “treat” these wave functions
as if they were eigenfunction of L with eigenvalue L 1. Within this
model, we can define a total angular momentum operator J L S. Following
the results of atomic physics we symmetrize the k 0 valence band
wave functions to correspond to J 3/2 and J 1/2. Using the relation that
L · S (1/2)[J 2 L 2 S 2 ] we can show that:
〈J 3/2|HSO|J 3/2〉 Ï/2 while 〉J 1/2|HSO|J 1/2〉 Ï.
The S-O splitting ¢0 given by the separation between the J 3/2 and 1/2
states is, therefore, 3Ï/2. In atoms the parameter Ï depends on the atomic orbitals
involved. In crystals, we have pointed out on p. 59 that the conduction
and valence band wave functions contain two parts: a smooth plane-wave like
part which is called the pseudo-wave function and an oscillatory part which
is localized mainly in the core region. Since HSO depends on ∇V most of
the contribution to Ï comes from the oscillatory part of the wave function.
Hence we expect that HSO in a semiconductor will depend on the S-O coupling
of the core states (p and d states only since s states do not have S-O
coupling) of its constituent atoms. In the cases of atoms with several p and
d core states, the outermost occupied states are expected to make the biggest
contribution to Ï of the conduction and valence bands. This is because the
deeper core states tend to be screened more and hence contribute less to
the valence and conduction electrons. For example, in Ge atoms the deep 2p,
3d and 3p cores states all have larger S-O constants than the outermost 4p
state. However, most of the S-O interaction in the valence band of Ge crystal
at k 0 comes from the 4p atomic state. For the diamond-type semiconductors,
such as Si and Ge, one may expect that S-O coupling in the crystal
is related to the S-O coupling of the outermost occupied atomic p states