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Indian Association for the Cultivation of Science
Advanced Condensed Matter Physics (PHS-5203,
PHD-222)
Assignment-
Due on 26-2-
1. Tight-Binding Model for Graphene
1(i) Graphene has two atoms per unit cell,denoted by A and B. The x and y-axes are chosen
such that a~ 1
= a(
3 / 2 , − 1 /2) and a~ 2
= a(0, 1) ,where a=0.246 nm is the lattice constant.
An isolated carbon atom has the electronic configuration 1s
2 2 s
2 2 p
2
. To form graphene,
one electron is excited from 2s to 2p, and the new configuration 1s
2 2 s
1 2 p
1
x
2 p
1
y
2 p
1
z
. The 2s
, 2p x
, and 2p y
orbitals get hybridized(mixed) and form three sp
2 orbitals that are oriented
in the x-y plane at 120
◦
with each other. The sp
2
orbitals on nearby atoms form strong
bonds in the plane, giving rise to the honeycomb structure, and they broaden into the
σ−bands that lie very low in energy. The p z
orbital on each atom is perpendicular to
the graphene plane and is occupied by one electron. The p z
orbitals broaden into two
π−bands (there are two atoms, hence two pz orbitals, per unit cell).
Figure: Graphene
Let us take atom A to sit at the origin of coordinates. The p z
orbital on this atom is
φ(~r), and that on any atom type A, that can be reached from A by a lattice vector
R
n
is φ(~r −
R
n
). The p z
orbital on B is φ(~r −
δ), where
δ is the vector from A to B, and that
on any atom of type B, separated from B by
Rn, is φ(~r −
δ −
Rn).We have,
φ
∗
(~r −
R
n
)Hφ(~r −
R
n
)d
3
r =
φ
∗
(~r −
δ −
R
n
)Hφ(~r −
δ −
R
n
)d
3
r = .
We may shift the zero of energy and set =0. The equality of the matrix elements in the
above equation is due to the symmetry of the graphene under reflection on a plane that
is perpendicular bisector of the bond connecting atoms A and B. For simplicity, we make
two assumptions:
- Only nearest neighbor atoms interact;the matrix element of the Hamiltonian be-
tween orbitals on neighboring atoms is −t(t ≈ 3 eV ):
∫
φ
∗ (~r −
δ)Hφ(~r −
δ)d
3 r= −t
orbitals on different sites is ignored, i.e., we assume that
∫
φ
∗
(~r −
δ)φ(~r)d
3
r= 0
From the p z
orbitals on atoms of type A and B, the Bloch functions
ψ
A
k
(~r) =
N
n
e
i
~ k·
~ Rn
φ(~r −
R
n
ψ
B
k
(~r) =
N
n
e
i
~ k·
~ Rn
φ(~r −
δ −
R
n
are constructed. To solve the Schrodinger equation Hψ k
(~r) = E k
ψ k
(~r), we try a solution
of the form ψ k
(~r) = aψ
A
k
(~r) + bψ
B
k
(~r).
(a) Find the primitive reciprocal lattice vectors
b 1 and
b 2 and draw the first Brillouin
zone.
(b) Show that E k
= ±t|g k
|, where
g k
= 1 + exp[i(−
3 k x
a/2 + k y
a/2)] + exp[−i(
3 k x
a/ + k y
a/2)]
(c) Reduce E k
to the form :
Ek = ±t[3 + 4 cos(
3 kxa/2) cos(kya/2) + 2cos(kya)]
1 / 2
(d) In the vicinity of the points K(2π/
3 a, 2 π/ 3 a) and K
′ (0, 4 π/
3 a, ) in the first
Brillouin zone, show that E k
= ±¯hv F
k where v F
3 ta/2¯h is the Fermi velocity,
k = |
k − (2π/
3 a, 2 π/ 3 a)| (near point K), or k = |
k − (0, 4 π/ 3 a)| (near point K
′ ).
The energy dispersion is thus linear in the vicinity of K and K
′
.
(e) Assuming linear dispersion, show that the density of states per unit area is d(E) =
8 |E|/(3πa
2 t
2 ).
(ii) More on Graphene. Assume that the p z
orbital on each site in graphene is described by
the wave function φ(~r) = Ar cos(θ) exp(−Zr/ 2 a 0
), where A is a normalization constant,a 0
is the Bohr radius, θ is the angle ~r makes with the c-axis (the axis perpendicular to the
graphene plane), and Z is the effective charge on the nucleus ( the nuclear charge is
screened by the two core electrons in the 1s orbital, and to a lesser extent by the valence
electrons; Z ≈ 3). Show that
φ
∗
(~r) exp (−i~q · ~r)φ(~r)d
3
r = [1 + (qa 0
/Z)
2
]
− 3
the number of k-points within in the shell because of spin degeneracy. Therefore,
D()d = 2
V
(2π)
3
shell
d
3
k = 2
V
(2π)
3
dSdk⊥
where dS
is an area element on the inner surface and dk ⊥
is the perpendicular distance
between the two surfaces of the shell. Show that
D()d = 2
V
(2π)
3
dS
k
E
k
Ek =
