introduction to the tight binding (lcao) method

7/30/2019 Introduction to the Tight binding (LCAO) Method

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Introductionto the

Tightbinding

(LCAO)Method


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Tightbinding: 1 Dimensional Model #1

Consider anInfinite Linear Chainof identical atoms, with 1s-orbitalvalence e- per atom & interatomic spacing = a

Approximation:Only Nearest-Neighbor interactions.(Interactions between atoms further apart than a are ~ 0).

This model is called the Monatomic Chain.

Each atom has s electron orbitals only!

Near-neighbor interaction only means that thesorbital on site

ninteracts with the sorbitals on sites n1 & n + 1 only!

n =Atomic Label a n = -3 -2 -1 0 1 2 3 4


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The periodic potential V(x) for this Monatomic Linear

Chainof atoms looks qualitatively like this:

n = -4 -3 -2 -1 0 1 2 3

a

V(x) = V(x + a)


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The localized atomic orbitals on each site for this Monatomic

Linear Chainof atoms look qualitatively like this:

n = -4 -3 -2 -1 0 1 2 3

The spherically symmetric sorbitals on each site overlap

slightly with those of their neighbors, as shown. Thisallows the electron on site n to interact with its

nearest-neighbors on sites n1 & n + 1!

a


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The True Hamiltonian in the solid is:

H = (p)2/(2mo) + V(x), with V(x) = V(x + a).

I nstead, approximate it as H n Hat(n) + n,nU(n,n)where, Hat(n) Atomic Hamiltonian for atom n.U(n,n) Interaction Energy between atoms n & n.

Use the assumption ofonly nearest-neighbor interactions:U(n,n) = 0 unless n = n -1 orn = n +1

With this assumption, the Approximate Hamil tonianisH n [Hat(n) + U(n,n -1) + U(n,n + 1)]


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H n [Hat(n) + U(n,n -1) + U(n,n + 1)] Goal:Calculate the bandstructure Ekby solving the

Schrdinger Equation:

Hk(x) = Ekk(x)

Use the LCAO(Tightbinding) Assumptions:

1. H is as above.2. Solutions to the atomic Schrdinger Equation are known:

Hat(n)n(x) = Enn(x)

3. In our simple case of1 s-orbital/atom:

En= = the energy of the atomic e- (known)

4.n(x) is very localized around atom n5. The Crucial(LCAO) assumption is:

k(x) neiknan(x)That is, the Bloch Functions are linear combinations of atomic orbitals!


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As a student exercise, show that the energy band of this model is:

Ek= - 2cos(ka) or Ek= - 2 + 4 sin2[()ka]

A trig identity was used to get last form.&are usually taken as parameters

in the theory, instead of actually calculating them from the atomic nThe Bandstructure for this monatomic chain with nearest-neighborinteractions only looks like (assuming 2< ): (ET Ek- + 2)

Its interesting to note that:The form Ek= - 2cos(ka)is similar to Krnig-Penney model

results in the linear approximation

for the messy transcendental

function! There, we got:Ek= A - Bcos(ka)

where A & B were constants.

ET

4


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n = -1 0 1

Tightbinding: 1 Dimensional Model #2A1-dimensional semiconductor! Consider anInfinite Linear Chainconsisting of 2 atom types, A

& B (a crystal with 2-atom unit cells), 1 s-orbitalvalence e- per atom &unit cell repeat distance = a.

Approximation:Only Nearest-Neighbor interactions.(Interactions between atoms further apart than ~ ( )a are ~ 0).

This model is called the Diatomic Chain.

A B A B A B A

a


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The True Hamiltonian in the solid is:

H = (p)2/(2mo) + V(x), with V(x) = V(x + a).

I nstead, approximate it (with = A or =B) asH nHat(,n) + n,nU(,n;,n)

where, Hat(,n) Atomic Hamiltonian for atom in cell n.U(,n;,n) Interaction Energy between atom of type

in cell n & atom of type in cell n.Use the assumption ofonly nearest-neighbor interactions:The only non-zero U(,n;,n)

are U(A,n;B,n-1) = U(B,n;A,n+1) U(n,n-1) U(n,n+1) With this assumption, the Approximate Hamil tonianis:

H nHat(,n) + n[U(n,n -1) + U(n,n + 1)]


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H nHat(,n) + n[U(n,n -1) + U(n,n + 1)] Goal: Calculate the bandstructure Ekby solving the Schrdinger Equation:

Hk(x) = Ekk(x)

Use the LCAO(Tightbinding) Assumptions:1. H is as above.

2. Solutions to the atomic Schrdinger Equation are known:

Hat

(,n)n

(x) = En

n

(x)

3. In our simple case of1 s-orbital/atom:EAn= A = the energy of the atomic e

- on atom A

EBn= B = the energy of the atomic e- on atom B

4.n

(x) is very localized near cell n

5. The Crucial(LCAO) assumption is:

k(x) neiknaCn(x)That is, the Bloch Functions are linear combinations

of atomic orbitals!

Note!!

The Cs are unknown


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Dirac Notation:Schrdinger Equation: Ekk|H|kAn|H|k = EkAn|H|k (1) Manipulation of (1), using LCAO assumptions, gives (studentexercise):

AeiknaCA+ [e

ik(n-1)a + eik(n+1)a]CB = EkeiknaCA (1a)

Similarly: Bn|H|k = EkBn|H|k (2) Manipulation of (2), usingLCAO assumptions,gives (student exercise):

BeiknaCB+ [e

ik(n-1)a + eik(n+1)a]CA = EkeiknaCA (2a)

Here, An

|U(n,n-1)|B,n-1

Bn

|U(n,n+1)|A,n+1

= constant (nearest-neighbor overlap energy) analogous

to in the previous 1d model


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Student exercise to show that these simplify to:

0 = (A - Ek)CA + 2cos(ka)CB, (3)

and0 = 2cos(ka)CA + (B - Ek)CB, (4)

A,B , are usually taken as parameters in the theory, instead of

computing them from the atomic n

(3) & (4) are linear, homogeneous algebraic equations forCA& CB

22 determinant of coefficients = 0 This gives: (A - Ek)(B - Ek) - 4

2[cos(ka)]2 = 0

A quadratic equation forEk!

2 solutions: a valence band

& a conduction band!


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Results:

Bandstructure of the Diatomic Linear Chain(2 bands):

E(k) = ()(A + B) [()(A - B)2 + 42 {cos(ka)}2] This gives a k = 0 bandgap of

EG= E+(0) - E-(0) = 2[()(A - B)2 + 42]

For simplicity, plot in the case

42 A

Expand the [ .]part ofE(k) & keep the lowest order termE+(k) B + A[cos(ka)]2, E-(k) A - A[cos(ka)]2EG(0) AB + 2A, where A (42)/|A - B|


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Bandstructure of a 1-dimensional semiconductor:


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Tightbinding Method: 3 Dimensional Model

Model: Consider a monatomic solid, 3d, with only

nearest-neighbor interactions. Hamiltonian:H = (p)2/(2mo) + V(r)

V(r) = crystal potential,

with the full lattice symmetry & periodicity.

Assume (R,R = lattice sites):H RHat(R) + R,RU(R,R)

Hat

(R) Atomic Hamiltonian for atom at RU(R,R) Interaction Potential between atoms at R & R

Near-neighbor interactions only!

U(R,R) = 0 unless R & Rare nearest-neighbors


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Goal:Calculate the bandstructure Ekby solving the Schrdinger Equation:

Hk(r) = Ekk(r)

Use the LCAO(Tightbinding) Assumptions:1. H is as on previous page.

2. Solutions to the atomic Schrdinger Equation are known:

Hat(R)n(R) = Enn(R), n = Orbital Label (s, p, d,..),

En= Atomic energy of the e- in orbital n

3.n(R) is very localized around R

4. The Crucial(LCAO) assumption is:

k(r) = ReikRnbnn(r-R) (bnto be determined)n(R): The atomic functions are orthogonal for different n & R

That is, the Bloch Functions are linear combinations of

atomic orbitals!


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Dirac Notation:Solve the Schrdinger Equation: Ekk|H|kThe LCAOassumption:

|k= ReikRnbn|n (1)(bnto be determined)

Consider a particular orbital with label m:

m|H|k = Ekm|H|k(2) Use (1) in (2). Then use

1. The orthogonality of the atomic orbitals

2. The assumed form ofH3. The fact that n(R) is very localized around R

4. That we know the atomic solutions to Hat|n = En|n5. The nearest neighbor assumption that U(R,R) = 0

unless R& R are nearest-neighors.


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Manipulate (several pages of algebra) to get:

(Ek- Em)bm + nR0 eikRmn(R)bn = 0 , (I)where: mn(R) m|U(0,R)|n Over lap Energy I ntegral

The mn(R) are analogous to the &in the 1d models. They are

similar to Vss, etc. in real materials, discussed next! The integrals

are horrendous to do for real atomic m! In practice, they are

treated as parameters to fit to experimental data.

Equation (I): Is a system ofN homogeneous, linear, algebraicequations for the coefficients bn. N = number of atomic orbitals.

Equation (I) forN atomic states

The solution is obtained by taking an N N determinant!This results in N bandswhich have their roots in the atomic orbitals!

If the mn(R) are small, each band can be thought of as

Ek~ En + kdependent corrections

That is, the bands are ~atomic levels + corrections


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Equation (I): A system of homogeneous, linear, algebraic

equations for the bn

N atomic states Solve an N N determinant!N bands

Note:Weve implicitly assumed 1 atom/unit cell.If there are n atoms/unit cell, we get nN equations &nN bands!

Artificial Special Case #1: One s level per atom

1 (s-l ike) band Artificial Special Case #2:Three p levels per atom3 (p-like) bands Artificial Special Case #3:

One s and three p levels per atom & sp3 bonding 4 bandsNOTE that

For n atoms /unit cel l, mul tiply by n to get the number of bands!


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Back to:

(Ek- Em)bm + nR0 eikRmn(R)bn = 0 , (I)where: mn(R)

m|U(0,R)|n

Overlap Energy I ntegral

Also:Assume nearest neighbor interactions onlyR0is ONLYover nearest neighbors! Artificial Special Case #1: One s level per atom

1 (s-l ike) band: Ek= Es - R=nneikR(R)But (R) = is the same for all neighbors so:

Ek= Es - R=nneikR Assume, for example, a simple cubic lattice:

Ek= Es -2[cos(kxa) + cos(kya) +cos(kza)]


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Artificial Special Case #2:Three p levels per atom.

Gives a 3 3 determinant to solve.3 (p-l ike) bandsStudent exercise!!

Artificial Special Case #3:One s and three p levels per atom & sp3 bonding

Gives a 3 3 determinant to solve. 4 bandsStudent exercise!!

introduction to the tight binding (lcao) method

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