10. Appendix

Solution to Problem 8.9 661
of 19.5 ◦ with the surface. Show that the former is energetically favorable when
cleaving a crystal parallel to one of the equivalent (111) surfaces.
The primitive translation vectors of a (100) surface are (a/2)(011) and
(a/2)(01-1).
The two corresponding reciprocal lattice vectors can be obtained from
(2.9) by defining a3 a(001). Using these two vectors it is easy to draw the
corresponding Brillouin zone.
The groups of the k-vector at the center of the Brillouin.zone. are also C2v
and C3v.
Solution to Problem 8.9
Let us start with the simplest case, that of a zincblende crystal (point group at
°: Td).
We will discuss the hexagonal faces first and then the tetragonal faces.
The point in a tetragonal face with the highest symmetry is the X-point:
(2°/a0)(100). The corresponding group of the k-vector is D2d (Table 2.15). The
slope of a given band perpendicular to this face is proportional to the expectation
value of px according to (2.35). It is easy to check with Table 2.15 that
px has X3 symmetry. All bands are non-degenerate (we are neglecting spin at
this point) at the X point except those of X5 symmetry which are twofold degenerate
(see Fig. 2.14. In this figure the X5 bands are split into X6 and X7
by spin-orbit interaction. In the absence of spin-orbit interaction these bands
remain doubly degenerate along the ¢ direction because of time reversal invariance).
Any nondegenerate bands will approach the X-point along ¢ with
zero slope because the product of two nondegenerate representations is the
identity representation, X1 in our case. Since px belongs to X3, its expectation
value for a nondegenerate representation must vanish. Since the X5 degeneracy
does not split along ¢, both bands belonging to it act as a single band
which also end at the X point with zero slope. The only other high symmetry
point on the (100) face is the W-point, (1, 0, 1
2 ). Its point group is S4 [2.4]. It is
easy to see that the matrix elements of px do not vanish by symmetry at this
point and thus no one-dimensional van Hove singularities occur there.
Let us consider now a face perpendicular to the [111] axis. The highest
symmetry point of k-space on this surface is L. The corresponding group is
C3v (see Table 2.12). The component of p along the [111] direction is invariant
under all operations of C3v and, therefore, we cannot use an argument
similar to that used for the (100) face to prove that the bands have zero slope
perpendicular to the (111) face. An examination of Figs. 2.14 and 2.15, however,
suggests that this is still the case. In order to prove it, we use the following
arguments. Let us move along the § axis, perpendicular to the (111)
face, by an infinitesimally small amount ‰. The resulting point we label as
(apple/a0)(1 ‰)(111). Using time reversal symmetry we can show that the energy
of a given band at this point is equal to that at (apple/a0)(1 ‰)(111). Adding to