Tight Binding for Graphene

7/30/2019 Tight Binding for Graphene

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Tight Binding for Grapheneby Sowmya Sathyanarayana Murthy

Tight-binding method - A brief introduction and bands of Graphene

TB method with

First nearest neighbours

Second nearest neighbours

Third nearest neighbours

1 Tight-binding method - A brief introduc-

tion:

The basic idea behind the tight-binding method is that the crystal Hamil-tonian can be approximated by an atomic Hamiltonian when the basis or-bitals are chosen to be localised. By considering the hopping of the electrons

between the nearest neighbours (atmost the third nearest neighbours), thebands corresponding to the localised orbitals could be obtained.

Statement of Blochs theorem:

The eigenstates of the one-electron Hamiltonian H= h22/2m + U(r),where U(r + R) = U(r) for all R in a Bravais lattice, can be chosen to havethe form of a plane wave times a function with the periodicity of the Bravaislattice:

nk(r) = eik.runk(r) (1)

Where unk(r + R) = unk(r) for all R in the Bravais lattice.

In other words, the tight binding method involves the linear combinationof the atomic orbitals located on the various atoms of the crystal, the coef-

ficients being the values of the planewaves e(ikR) at various positions R at

which the atoms are located. The atomic states are used as basis orbitals.If the states are denoted by up(r R), where p refers to the atomic levels ofatom located at R, the Bloch functions are given as,

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p,k(r) =1N

R

eik Rup(r R) (2)

and a general Bloch state is a linear combination of the Bloch basis functionsgiven by

k(r) = pCp(k)p,k(r) (3)

The coefficients Cp(k) could be obtained by solving the Schrodinger equa-tion in matrix form:

HC(k) = ESC(k) (4)

The Hamiltonian and the overlap matrix elements could be written as:

Hpq = p,k|H|q,k (5)Spq =

p,k

|q,k

(6)

Where

p,k(r) =1N

R

eik Rup(r R)

and

Hpq = R

eik R d3rup(r

R)Huq(r)

Spq =R

eik R

d3rup(r R)Suq(r)

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2 Structural details of Graphene:

Graphene is a 2D sheet of carbon atoms arranged in a hexagonal pattern, asshown in Fig. 1. Each carbon atom has 3 nearest neighbours, 6 next-nearestneighbours and 3 third nearest neighbours.

The unit cell consists of 2 Carbon atoms, labelled as A and B.

1a

a2

A

3R

1R

2R B

Figure 1: Sketch of graphene showing the unit cell (light-blue region) de-limitated by the two dimensional lattice vectors a1 = a(

3/2, 1/2) and

a2 = a(

3/2,1/2). Lattice constant = a0 and a = a03 Vectors Ri referto the first nearest-neighbour distances.

Carbon atom has 6 electrons and the bonding of the atoms is due to theoverlapping of the orbitals. sp2 hybridization due to the overlapping of the 2sorbital with the 2px and 2py orbitals and this is attributed to the structureof Graphene. The overlap between the pz orbitals (that are perpendicularto the plane) is responsible for the electronic properties of Graphene. Thepz orbitals could be almost treated as localized and the overlapping betweenthe pz orbitals gives rise to the bonding and antibonding

bands.

3 and bands of GrapheneHAA(k) HAB(k)

HBA(k) HBB(k)

A(k)

B(k)

= E(k)

SAA(k) SAB(k)

SBA(k) SBB(k)

A(k)

B(k)

(7)

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Figure 2: Nearest neighbours B1i i=1-3, second nn A2i i=1-6, Thirdnn B3i i=1-3

det

HAA(k) E(k)SAA(k) HAB(k) E(k)SAB(k)HBA(k) E(k)SBA(k) HBB(k) E(k)SBB(k)

= 0 (8)

In order to obtain the energy eigenvalues, the above secular has to be

solved.

Evaluation of the matrix elements

3.1 Taking nearest neighbours only:

For the A Carbon atom, the first nearest neighbours (or simply the near-est neighbours) are B atoms, which could be labelled as B1,B2 and B3.The

vectors R1, R2, R3 connect A atoms to B1, B2 and B3 respectively. Since

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Figure 3: Brillouin zone of 2D graphene showing the high symmetry points,= (0,0),K = (2/

3,-2/3),M=(2/

3,0)

both the atoms of the unit cell are Carbon atoms described by one single pzorbital, the following conditions apply:

HAA(k) = HBB(k)

SAA(k) = SBB(k)

HAB(k) = HBA(k)

SAB(k) = SBA(

k)

From Eq. 8 it follows:

[HAA(k)E(k)(SAA(k)][HBB(k))E(k)SBB(k)][HAB(k)E(k)SAB(k)][HBA(k)E(k)SBA(k)] = 0By suppressing the dependency on (k) one obtains:

H2AAHABHAB2EHAASAA+E2(S2AASABSAB)+E(SABHAB+HABSAB) = 0 (9)Given the definition:

HAASAA = E0 (10)

SABHAB + HABS

AB = E1 (11)

H2AA HABHAB = E2 (12)S2AA SABSAB = E3 (13)

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Substituting E0, E1, E2, E3 in 9, a quadratic equation in E is obtained

E2 2EE0 + E2E3 + EE1 = 0 (14)whose solution yields the following dispersion relation:

E(k) =(E1 2E0) [(E1 2E0)2 4E2E3]1/2

2E3(15)

By considering nearest-neigbhor interactions only, the matrix elementsrequired to evaluate E0, E1, E2 and E3 are obtained in the following way:

HAA =1

N

RA

RA

eik( R

A RA)A(r RA|H|A(r RA ) (16)

where A and A

refer to the same atom-type and N is the number of unitcells in the crystal.

HAA =1

N

RA

A(r RA|H|A(r RA) = 2p (17)

HAB =1

N

RA

RB

eik( RB RA)A(r RA|H|B(r RB) = 0(eik R1+eik R2+eik R3)

(18)

with

0 = A(r RA|H|B(r RA Ri) with (i = 1, 2, 3)

HAB = HBA

The same treatment yields the overlap matrix element SAA & SAB:

SAA = 1 (normalization) (19)

SAB =1

N

RA

RB

eik( RB RA)A(r RA|B(r RB) = s0(eik R1+eik R2+eik R3)

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(20)

with

s0 = A(r RA)|B(r RA Ri) with (i = 1, 2, 3)

Where Ri is the vector pointing from atom A to its nearest neighbours Bi(see Fig. 1)

Summing up we have:

HAB = 0(eik R1 + ei

k R2 + eik R3)

SAB = s0(eik R1 + ei

k R2 + eik R3)

HAA = 2p

SAA = 1

HAB = 0(eik R1 + ei

k R2 + eik R3)

SAB = s0(eik R1 + ei

k R2 + eik R3)

Now we insert the Hamiltonian and the overlap matrix elements into theexpression for E(k) that was previously obtained (Eq. 37). E0,E1,E2,E3become:

E0 = 2p

E1 = SABHAB + HABS

AB = 2s00(e

ik R1 + eik R2 + ei

k R3)eik R1 + ei

k R2 + eik R3)

E2 = H2AA HABHAB = 22p 20(eik R1 + eik R2 + eik R3)(eik R1 + eik R2 + eik R3)

E3 = S2AA SABSAB = 1 s20(eik R1 + eik R2 + eik R3)(eik R1 + eik R2 + eik R3)

To evaluate (eik R1 + eik R2 + eik R3) (eik R1 + eik R2 + eik R3) we use the

trigonometric relationship ei = cos + isin from which

(eik R1 + ei

k R2 + eik R3)(ei

k R1 + eik R2 + ei

k R3) =

= [(cosk R1 + isink R1) + (cosk R2 + isink R2) + (cosk R3 + isink R3)][(cosk R1 isink R1) + (cosk R2 isink R2)(cosk R3 isink R3)]

= (cosk R1 + cosk R2 + cosk R3)2 + (sink R1 + sink R2 + sink R3)2

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= [cos2k

R1+cos

2k

R2+cos

2k

R3+2(cosk

R1cosk

R2+cosk

R2cosk

R3cosk

R1cosk

R3)]

[sin2k R1+sin2k R2+sin2k R3+2(sink R1sink R2+sink R2sink R3+sink R1sink R3)]= 3 + 2cosk ( R1 R2) + 2cosk ( R2 R3) + 2cosk ( R1 R3)= 3 + 2cosk1 + 2cosk2 + 2cos(k1 k2)

where k1 = k a1, k2 = k a2.The value of (ei

k R1 + eik R2 + eik R3) (eik R1 + eik R2 + eik R3) can thusbe represented as a function f(k)

f(k) = 3 + 2cosk1 + 2cosk2 + 2cos(k1 k2) (21)

and the corresponding E0,E1,E2,E3 values finally reads:

E0 = 2p (22)

E1 = SABHAB + HABS

AB = 2s00(f(

k)) (23)

E2 = H2AA HABHAB = 22p 20(f(k)) (24)

E3 = S2AA SABSAB = 1 s20(f(k)) (25)

(26)

Substituting the above values in E(k), we obtain:

E(k) =(E1 2E0) [(E1 2E0)2 4E2E3]1/2

2E3

=(2s00f(k) 22p) [(2s00(f(k)) 22p)2 4(22p 20(f(k)))(1 s20(f(k)))]1/2

2(1 s20(f(k)))

=

(

2s00f(k) + 22p)

[(4s20

20f

2(k)

8s002pf(k) + 4

22p

(4

4s20f(

k))(22p

20f(

k))]1/2

2(1 s20(f(k)))Expanding the terms inside the square root,

=(2s00f(k) + 22p) [4s2020f2(k) 8s002pf(k) + 422p 422p + 420f(k) + 4s2022pf(k) 4s2020f2(k)]1/2

2(1 s20(f(k)))

=(2s00f(k) + 22p) [f(k)(20 2s02p)2]1/2

2(1 s20(f(k)))

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=(

2s00f(k) + 2

2p)

(20

2s02p

)f(k)2(1 s20(f(k)))

=(s00f(k) + 2p) (0 s02p)

f(k)

(1 s20(f(k)))We obtain two values for E(k), namely

E(k) =

(s00f(k)+2p)+(0s02p)

f(k)

(1s20(f(k)))

(s00f(k)+2p)(0s02p)

f(k)

(1s2

0(f(k)))

E(k) =

2p(1s0

f(k))+0

f(k)(1s0

f(k))

(1s0

f(k))(1s0

f(k))

2p(1s0

f(k))0

f(k)(1s0

f(k))

(1s0

f(k))(1+s0

f(k))

E(k) =

(2p+0

f(k))(1s0

f(k))

(1s0

f(k))(1s0

f(k))

(2p0

f(k))(1+s0

f(k))

(1s0

f(k))(1+s0

f(k))

The final formula for E(k) is:

E(k) =2p 0

f(k)

1s0

f(k)

The values of2p, 0, and s0 are usually found by fitting experimental orfirst principles data. The values that are taken are 2p = 0 eV,0 between-2.5 and -3 eV, s0 below 0.1.[1] Since s0 is small, it is usually neglected. By

using 2p = 0 eV, 0 = 2.78 and s0 = 0.006 we obtain the band structureshown in Fig. 4, where the E(k) values are plotted against k along the high-symmetry lines M--K-M. The values E > 0 correspond to the antibonding states and the values E < 0 correspond to the bonding states.

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-20

-10

0

10

20

Energy(eV)

VASPTB - first nn

VASP and TB bandsTB bands-with first nearest neighbours

M K M

Figure 4: TB band structure of graphene-with nearest neighbours only

3.2 With second nearest neighbours:

For the A carbon atoms, the second nearest neighbours are A atoms present

at A21

a3

2, a2

, A22

a3

2,a

2

, A23 (0,a), A24

a

3

2,a

2

, A25

a

3

2, a2

,

A26 (0, a)

As per the definition,

HAASAA = E0 (27)

SABHAB + HABS

AB = E1 (28)

H2AA HABHAB = E2 (29)S2AA SABSAB = E3 (30)

Except HAA and SAA, all the other values remain the same ( second near-est neighbours are A atoms).

General definition for HAA is,

HAA =1

N

RA

RA

A(r RA|H|A(r RA) (31)

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RA = RA + R2i Where i goes from 1 to 6.

HAA = 2p + 1 u(kx, ky) Where,

1 =1N

AA(r RA|H|A(r RA

SAA = 1 + s1 u(kx, ky) Where,

s1 =1N

AA(r RA|A(r RA

Where u(kx, ky) = 2 cos(kya) + 4 cos3a2

kx cos a2kyE0 = (2p + 1u(kx, ky)) (1 + s1u(kx, ky)) (32)

E1 = SABHAB + HABS

AB = 2s00(f(

k)) (33)

E2 = H2AA HABHAB = (2p + 1u(kx, ky))2 20(f(k)) (34)

E3 = S2AA SABSAB = (1 + 1u(kx, ky))2 s20(f(k)) (35)

(36)

Where u(kx, ky) = 2 cos(kya) + 4 cos

3a2

kx

cos

a2ky

Substituting the above values, the energy eigen values are given by theexpression,

E(k) =(E1 2E0) [(E1 2E0)2 4E2E3]1/2

2E3(37)

The parameters used for the TB bands are given in Table:1

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-20

-10

0

10

20

Energy(eV)

VASPTB-with second nn

VASP and TB bandsTB - with second nearest neighbours

M K M

Figure 5: TB band structure of graphene-with second nearest neighbours

3.3 With third nearest neighbours

For the A carbon atom, the third nearest neighbours are B atoms present at

B31

a

23, a2

, B32

a

23,a

2

, B31

a

3, 0

So the elements HAA and SAA remain the same but the elements HABand SAB include the additional terms due to the third nearest neighbouringatoms.

E0 = (2p + 1u(kx, ky)) (1 + s1u(kx, ky))

E1 = SABHAB + HABS

AB = 2s00(f(

k)) + (2s0 + 0s2)v(kx, ky) + 2s22(f(2k))

E2 = H2AA HABHAB = (2p + 1u(kx, ky))2

20(f(

k)) + 202v(kx, ky) + 22(f(2

k))

E3 = S2AA SABSAB = (1 + 1u(kx, ky)2

s20(f(k)) + 2s0s2v(kx, ky) + s22(f(2k))

Where v(kx, ky) = 4 cos (aky) + 8 cos

32akx

cosa2ky

+ 4 cos

32akx

cos

3a2ky

+ 2 cos (

3akx)Substituting the above values, the energy eigen values are given by the

expression,

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E(k) =(E1 2E0) [(E1 2E0)2 4E2E3]1/2

2E3

-20

-10

0

10

20

Energy(eV)

VASPTB -Third nn

VASP and TB bandsTB bands - With third nearest neighbours

M K M

Figure 6: TB band structure of graphene-with third nearest neighbours

Table 1: Parameters used for the TB modelParameters 1stnn 2ndnn 3rdnn

2p 0 -0.28 eV -0.280 -2.78 eV -2.78 -2.781 -0.073 eV -0.0732 -0.3 eVs0 0.06 0.06 0.06s1 0.002 0.002

s1 0.001

It is to be noted that the parameters used for the first, second and thirdnearest neighbour bands are not varied i.e., the values are kept the same foreach case inorder to show the improvement of the fitting with the inclusion ofthe second and third nearest neighbour hoppings as compared to that of thefirst nearest neighbours. But for each case, the parameters could be chosenfreely inorder to obtain the best fit.

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Tight Binding for Graphene