
4.5 General observations from the Brillouin zone 87
K
in Figure 4.7a) leading to the conclusion that the (3, 3) CNT is metallic. Indeed,
it is straightforward to show that arbitrary (n, n) armchair CNTs are metallic. To
derive this, let us recall the vector from the -point to the other high-symmetry
points of the hexagonal Brillouin zone of graphene:
M =
b
1
+ b
2
2
=
2π
√
3a
,0
, K =
2π
√
3a
,
2π
3a
. (4.24)
The length of the 1D bands of an CNT is 2π /a, which is greater than the lengths
of the sides of the hexagonal Brillouin zone of graphene (length = 4π/3a). As a
result, if any of the 1D bands of armchair nanotubes intersect an M-point of the
hexagon, it will also simultaneously intersect a K-point. Therefore, in order for a
1D band of an armchair CNT to be metallic, the M vector has to be an integer
multiple of K
c
. Mathematically, this condition is equivalent to
10
j|K
c
|=j
2π
√
3an
≡|M|, ∴ j = n. (4.25)
This condition is satisfied by all armchair nanotubes by the j = nth band or
Brillouin zone line at k =±2π/3a. Hence, in general, armchair CNTs are metallic.
Likewise, we can apply similar reasoning to zigzag nanotubes to determine the
conditions in which they are metallic. The Brillouin zone of a (10, 0) zigzag CNT
was previously shown in Figure 4.8a revealing that the 1D bands are parallel to
the K vector. Hence, for a zigzag nanotube to be metallic, the K vector has to
be an integer multiple of K
c
:
j|K
c
|=j
2π
an
≡|K|, ∴ j =
2
3
n, (4.26)
which is satisfied when j = 2n/3atk = 0. However, since j is restricted to integer
values (see Eq. (4.21)), only a zigzag CNT with a chirality that is an integer multiple
of 3 (i.e. n/3 is aninteger) is metallic, otherwise thezigzag CNT is semiconducting.
For example, (12, 0), (15, 0), (18, 0) are metallic CNTs, whereas (10, 0), (11, 0),
and (13, 0) are semiconducting nanotubes. In general, an arbitrary chiral CNT is
metallic if the angle between jK
c
and the K vector is the chiral angle:
j|K
c
|=j
2π
a
√
n
2
+ nm + m
2
≡|K|cos θ =
2π(2n + m)
3a
√
n
2
+ nm + m
2
, (4.27)
which is satisfied only when j = (2n+m)/3. This leads to the celebrated condition
that a CNT is metallic if (2n + m) or equivalently (n − m) is an integer multiple
of3or0,
11
otherwise the CNT is semiconducting. Invariably, a very important
10
The equivalence symbol ≡ is used to enforce the equivalence of the LHS expression and the RHS
expression.
11
j = (2n +m)/3 is equivalent to j =[(n −m)/3]+[(n +2m)/3], which results in an integer value
for j when n − m is a multiple of 3.