Computational Physics

6.3 Approximations 131

It follows that the energiesE(k)are given as

E±(k)=

−(− 2 E 0 +E 1 )±

√

(− 2 E 0 +E 1 )^2 − 4 E 2 E 3

2 E 3

, (6.20)

where

E 0 =HAASAA; E 1 =SABHAB∗ +HABS∗AB;

E 2 =HAA^2 −HABH∗AB; E 3 =S^2 AA−SABSAB∗.

The matrix elementHAA(k)must be a real constant which does not depend onk
and the same holds forSAA(k). We take the first to be 0 (with a suitable shift of the
energy scale) and the second is 1 because of the normalisation of the orbitals.
You may verify that the off-diagonal elements have the form

HAB=γ 0 [exp(ik·R 1 )+exp(ik·R 2 )+exp(ik·R 3 )] (6.21)

withγ 0 a real constant independent ofk. The vectorsR 1 connectAto its three
neighbours. ForSABwe find the same form but we call the constants 0.
After some calculation, the energies are found as

E±(k)=

ε 2 p∓γ 0

√

f(k)

1 ∓s 0

√

f(k)

, (6.22)

where

f(k)= 3 +2 cos(k 1 )+2 cos(k 2 )+2 cos(k 1 −k 2 ); (6.23)

k 1 =k·a 1 , k 2 =k·a 2 etc. (6.24)

In the Brillouin zone, the pointis identified withk=(0, 0), and the pointK
is a vector of length 2π/ain the direction ofa 2 ;Mis the point 2π/

√

3 (1, 0).We
can now plot the bands (i.e. the valuesE±(k)betweenM,andK. The result is
shown inFigure 6.5. Note that we find two energy values per atom. As we know
that there should only be a single electron per pzorbital, the Fermi energy must be
the highest negative energy. We see that graphene is a metal: the bands touch each
other precisely at the Fermi energy, so infinitesimal excitations are possible which
yield nonzero momentum.
Now we turn to carbon nanotubes. These are graphene sheets rolled into a cyl-
indrical form. There are many ways in which the two long ends can be glued
together. These ways correspond to different strips which can be drawn on this
sheet such that the two sides of the strip can be connected together smoothly. This
is possible when the vector running perpendicularly across the sheet is an integer
linear combination of the basis vectorsa 1 anda 2. This is also indicated inFigure 6.4.