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Question paper on graphene, Assignments of Physics

Good assignment sheet on tight binding on bipartite lattice and plane wave basis

Typology: Assignments

2024/2025

Uploaded on 03/08/2025

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Indian Association for the Cultivation of Science
Advanced Condensed Matter Physics (PHS-5203,
PHD-222)
Assignment-3
Due on 26-2-2025
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 ~a1=a(3/2,1/2) and ~a2=a(0,1) ,where a=0.246 nm is the lattice constant.
An isolated carbon atom has the electronic configuration 1s22s22p2. To form graphene,
one electron is excited from 2s to 2p, and the new configuration 1s22s12p1
x2p1
y2p1
z. The 2s
, 2px, and 2pyorbitals get hybridized(mixed) and form three sp2orbitals that are oriented
in the x-y plane at 120with each other. The sp2orbitals on nearby atoms form strong
bonds in the plane, giving rise to the honeycomb structure, and they broaden into the
σbands that lie very low in energy. The pzorbital on each atom is perpendicular to
the graphene plane and is occupied by one electron. The pzorbitals broaden into two
πbands (there are two atoms, hence two pzorbitals, per unit cell).
Figure: Graphene
Let us take atom A to sit at the origin of coordinates. The pzorbital on this atom is
φ(~r), and that on any atom type A, that can be reached from A by a lattice vector ~
Rn,
is φ(~r ~
Rn). The pzorbital on B is φ(~r ~
δ), where ~
δis 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,
Zφ(~r ~
Rn)(~r ~
Rn)d3r=Zφ(~r ~
δ~
Rn)(~r ~
δ~
Rn)d3r=.
1
pf3
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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

  • The overlap between p z

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 =