130 Solving the Schrödinger equation in periodic solids
Figure 6.4. Hexagonal lattice of the graphene sheet with basis vectorsa 1 anda 2
indicated. The zigzag nanotube is also indicated.basis consisting of two atoms,AandB(see Figure 6.4). The two basis vectors are
a 1 =a(
1
2
√
3,
1
2
)
; a 2 =a(
1
2
√
3,−
1
2
)
. (6.18)
where the lattice constanta = 2.461 Å. We only include nearest neighbour
interactions.
The most relevant part of the band structure is the valence band – it turns out that
this is formed by theπ-orbitals which are built from the pz-atomic orbitals. This
leads us to taking only a single orbital per atom into account. Furthermore we keep
only nearest neighbour matrix elements in the tight-binding matrices. We see from
Figure 6.4 that anA-atom has nearest neighbours of the typeBonly.
The essential idea of using Bloch’s theorem in calculating the band structure
is to reduce the entire problem to that of the unit cell, which contains only two
orbitals: the pzorbitals ofAandB. We must therefore calculateHAA(k)=HBB(k),
SAA(k)=SBB(k),HAB(k)=HBA∗ (k)andSAB(k)=S∗BA(k)for each Bloch vector
k. The spectrum is then given by the equation:
(
HAA(k) HAB(k)
HBA(k) HBB(k))(
ψA(k)
ψB(k))
=E(k)(
SAA(k) SAB(k)
SBA(k) SBB(k))(
ψA(k)
ψB(k)