Advanced Solid State Physics

Figure 38: Lattice, primitive lattice vectors and reciprocal lattice of of graphene

the energy shift (in a covalent bond) we describe the state of the whole system in terms of these two
wavefunctions and construct the hamiltonian matrix:∣
∣∣
∣∣
〈ψa|H|ψa〉−E 〈ψa|H|ψb〉
〈ψb|H|ψa〉 〈ψb|H|ψb〉−E

∣∣

∣∣

∣=

∣∣

∣∣

∣

ε−E −t

−t∗ ε−E

∣∣

∣∣

∣= 0 → E=ε±|t| (60)

It is only a two-by-two matrix, so it is easy to solve. εis the energy eigenvalue without interaction
andtis the overlap integral between the atomsaandb(so the nearest neighbour matrix elements).
For the eigenenergyEwe get two solutions, one for the bonding state and one for the anti-bonding
state.

Now we do the same thing for two carbon sublattices. We also get the same kind of answer like in
eqn. (60), but this is just the calculation fork= 0. So there are also two eigenvalues fork= 0. For
k 6 = 0we have to redo the calculation multiplied by a phase factor. So we get
∣∣
∣∣
∣

ε−E −tf(k)

−tf∗(k) ε−E

∣∣

∣∣

∣

= 0

with the phase factor

f(k) = exp

(

ikxa

2

)(

1 + exp

(

−i

(√

3 kxa

2

−

kya

2

))

+ exp

(

−i

(√

3 kxa

2

+

kya

2

)))

,

which depends onk(and is the sum over the three nearest neighbours). We plug the phase factor in
the matrix equation
∣∣
∣∣ ε−E t

(

1 + exp

(

−i

(√ 3 kxa

2 −

kya

2

))

+ exp

(

−i

(√ 3 kxa

2 +

kya

2

)))

t

(

1 + exp

(

+i

(√ 3 kxa

2 −

kya

2

))

+ exp

(

+i

(√ 3 kxa

2 +

kya

2

)))

ε−E

∣∣

∣∣= 0

and after some calculation (using the trigonometric identitiescos(a±b) = cos(a) cos(b)∓sin(a) sin(b)
andcos(2a) = 2 cos^2 (a)− 1 ) we get

E=ε±t

√√

√√

1 + 4 cos

(√

3 kxa

2

)

cos

(

kya

2

)

+ 4 cos^2

(

kya

2

)

.

So again we get two solutions (eigenenergies) for everykfrom this matrix. The energyEis the atomic
energyεplus the overlap integralttimes some function ofk. When we plot that function atk= 0
it starts to rise like a parabola (see fig. 39). The two solutions (for the bonding and the anti-bonding
state) are just mirror images of each other. In the upper direction there is the energy. The solutions
touch at six points and the fermi energy also cuts through this point. So we have a fermi surface
which consists of six points. At the fermi surface the dispersion relationship is linear (usually it has
some curvature like a parabola).