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Plotly: Tight-binding Model for Graphene

Honeycomb lattice of graphene

The unit cell of graphene’s lattice consists of two different types of sites, which we will refer to as A and B sites (see Figure 1).

Figure 1. Honeycomb lattice of graphene where different colors are used to denote the two sublattices. The basis vectors of the unit cell are shown with black arrows. This figure is generated by TikZ/LaTeX.

With the basis vectors, the cell can be defined by the cell vector

Rn=ja1+ka2

Below we will used (j,k) to denote the cell index.

Nearest-neighbour Tighting Binding Model for Graphene

The Hamiltonian of the graphene lattice

ˆH=22m2+U(r)

where U(r) is the periodic potential of the crystal. The Schrödinger equation follows

[22m2+U(r)]ψk(r)=E(k)ψk(r)

where ψk(r) is the wavefunction of the electron, and E(r) is the corresponding energy.

Assuming that wα(rRn) are orthonormal wavefunctions located at site α of the cell Rn, i.e.

Vwα(rRn)wβ(rRm)dr=δαβδmn

According to the Bloch’s theorem, the wavefunction of the electron in a periodic potential can be written as linear combination of the Bloch sum, i.e.

ψk(r)=1NαCα(k)neikRnwα(rRn)

By expanding the Hamiltonian in the Bloch sum basis, we have

H(k)=[H11H12H21H22]

where Hαβ is given by

Hαβ=1NnmVeik(RmRn)wα(rRn)H(r)wβ(rRm)dr=nVeikRnwα(rRn)H(r)wβ(rR0)dr

where we used the fact that the integrand only depends on the distance between the site Rn and Rm.

In the nearest-neighbours tight-binding model, we only considter those terms with |n|1.

The diagonal matrix elements

Apparently, only n=0 contributes to the matrix elements H11 and H22, therefore

H11=H22=Vw1(rR0)H(r)w1(rR0)dr=Vw2(rR0)H(r)w2(rR0)dr=ε0

The off-diagonal matrix elements

Let’s first inspect the H12 term. One can see from Figure 1 that the nearest-neighbours of the A site in cell (j,k) located in the cell (j,k), (j1,k) and (j,k1), respectively. As a result, the H12 term can be readily written.

H12=Vw1(rR0)H(r)w2(rR0)dr+eika1Vw1(rR(1,0))H(r)w2(rR0)dr+eika2Vw1(rR(0,1))H(r)w2(rR0)dr

By further assuming that the integral of the last three terms are the same, i.e.

t=Vw1(rR0)H(r)w2(rR0)dr

we have

H12=t[1+eika1+eika2]

Similarly, we have

H21=t[1+eika1+eika2]

The complete matrix

Putting together all the matrix element, we have

H(k)=[ε0t[1+eika1+eika2]t[1+eika1+eika2]ε0]

The eigenvalues E(k) can then be easily obtained

E(k)=ε0±t1+eika1+eika1+eika2+eika2+(eika1+eika2)(eika1+eika2)=ε0±t3+2cos(ka1)+2cos(ka2)+2cos(k(a1a2))

One of the definitions for the hexagonal lattice is

a1=a2(3,3)a2=a2(3,3)

where a is the bond-length of the graphene. With this definition, the energy band of graphene follows.

E(kx,ky)=±t3+2cos(3kya)+4cos(32kxa)cos(32kya)

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