January 26, 2004 16:26 WSPC/Book Trim Size for 9in x 6in b ook2
Periodic Lattice of Atoms 43
For a sufficiently large momentum matrix element, the effective mass of the
lower (valence) band can become negative, as is the case for the example
shown in Fig. 3.1, while the effective mass of the conduction band becomes
much smaller than the free electron mass.
Eq. (3.70) shows that the effective masses are determined by the inter-
band matrix element of the momentum operator and by the energy gap. Be-
cause the effective masses can be measured experimentally, one frequently
uses this relation to express the interband momentum matrix element in
terms of the effective electron mass m
e
= m
c
and the hole mass m
h
= −m
v
.
Eq. (3.70) allows one to express p
2
in terms of the reduced electron–hole
mass m
r
:
1
m
r
=
1
m
e
+
1
m
h
=
4p
2
m
2
0
E
g
. (3.71)
This result is often used to estimate the value of p
2
.
Next, we have to consider the case that some of the bands are degener-
ate. In the element semiconductors of group IV, the four electrons in the
outer shell of the atoms populate the sp
3
orbitals. The same is true for the
isoelectronic compound semiconductors of the groups III–V and II–VI. In
a cubic symmetry, the valence band states at k =0are made up of three
degenerate p-like states. The conduction band at k =0consists of an s-like
state. Thus, at the center of the Brillouin zone we can approximate the
cubic symmetry, which is the only case we consider here, by a spherical
one and use in the following the eigenfunctions of the angular momentum
operator as basis states. These four states are |l =0,m
l
=0 = |0, 0, |l =
1,m
l
= ±1 = |1, ±1 and |l =1,m
l
=0 = |1, 0. Following Kane, we
diagonalize the Hamiltonian H = H
0
+ k · p/m
0
in the basis of these four
states. A linear combination of these states at a finite wave vector k is
|ψ(k) =
l
=0,1,|m
l
|≤l
c
l
,m
l
(k)|l
,m
l
. (3.72)
From the stationary Schrödinger equation (3.62) we get
l, m
l
|H
0
+
m
0
(k · p) − E(k)|ψ(k) =0 . (3.73)
With the selection rules (see problem 3.4)
l, m
l
|k · p|l
,m
l
= kpδ
l,l
±1
δ
m
l
,m
l
(3.74)