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CHAPTER 10Tight-Binding Model

Linear Combination of Atomic Orbitals (LCAO)

Application to Bands from s-LevelsGeneral Features of Tight-Binding Levels

Wannier Functions

2 2 6 1Na : 1S 2S 2P 3S

Core

NFE����Semicore

TB ☺☺☺☺ Valence

NFE ☺☺☺☺

Neither NFE nor TB is suitable for the study of transition metals and f-electron systems.In fact, 3d, 4f and 5f electrons are not so weak that can be treated using NFE model, and are not so strong that can be considered as valence electrons.

This electron is so low that can be considered as a free electron.

Indeed it is localized, tightly bonded to a nucleus, and belongs to a specific atom.

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This electron is so high that can be considered as a free electron.

Indeed it is not localized and belongs to all the atoms.

In chemistry, sigma bonds (σ bonds) are a type of covalent chemical bond.

Sigma bonding is most clearly defined for diatomic molecules using the language

and tools of symmetry groups. In this formal approach, a σ-bond is symmetrical

with respect to rotation about the bond axis. By this definition, common forms of

sigma bonds are s+s, pz+pz, and s+pz, and dz2+dz2 (where z is defined as the

bond axis). Quantum theory also indicates that molecular orbitals (MO) of identical symmetry mix. As a practical consequence of mixing in diatomic molecules, the

wavefunctions s+s and pz+pz molecular orbitals become blended. The extent of

mixing (or blending) depends on the relative energies of the like-symmetry MO's.

In chemistry, pi bonds (π bonds) are covalent chemical bonds where two lobes

of one involved electron orbital overlap two lobes of the other involved electron orbital. Only one of the orbital's nodal planes passes through both of the involved

nuclei.

Pi bonds are usually weaker than sigma bonds because their (negatively charged)

electron density is farther from the positive charge of the atomic nucleus, which requires more energy. From the perspective of quantum mechanics, this bond's

weakness is explained by significantly less overlap between the component p-

orbitals due to their parallel orientation.

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In TB model we consider the crystal as an extremely huge molecule,

and use atomic orbitals rather than plane waves to expand the wave function of the crystal.

MODEL

FE NFE TB Isolated atom

Suppose we have two atoms that constitutes a molecule.

Let these atoms to be in their ground states, and the electron be in the 1S

orbital.

1 1 1 1H φ ε φ=2 2 2 2H φ ε φ=

Now Let they come closer to each other. In this case two different situations may occur:

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( )1

11S 2S

2ψ = + ( )1

11S 2S

2ψ = −

While the bonding state (red) has an energy minimum at the

internuclear distance where the reduction in energy due to the

delocalisation of the electrons is balanced by the repulsive Coulomb

interaction between the nuclei. The anti-bonding state has no such

minimum; its energy is always higher than that of two separate atoms

at infinite distance.

When the two atoms, represented as Coulomb wells, approach each

other, the potential barrier between the two wells is gradually

reduced. This allows electrons to occupy shared (delocalised) states

with probability density at both atoms rather than only at the

originating atom. For these electrons, the well size increases. The

energy eigenvalues decrease with increased well size (cf. inverse-

square relationship for the particle in a square well). This energetic

advantage is what stabilises the chemical bond.

Both the bonding and anti-bonding states are generated at the same

time - as always, the number of states obtained by linear

combination is the same as the number of states put in. The energy

advantage of the bonding state matches exactly the disadvantage of

the anti-bonding state. Each state can be occupied by two electrons

(of opposite spin). If both atomic orbitals are fully occupied, then

both molecular states will be occupied and there is no net energy

gain. If on the other hand (as in the Fig.) each atomic state

contributes only one electron, then the electrons from both atoms

can share the bonding molecular state. The result is a stable, i.e.

energetically favourable, chemical bond.

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λ →∞

a

2aλ =

2aλ =

k

ε

0a

π

First point of view: The atoms are

considered as a set of isolated atoms:

i i i iH φ ε φ=

Isolated atomic potential

22 U(r) (r) (r)

2mφ εφ

− ∇ + =

h

Second point of view: The set atoms is

considered in a bcc crystalline environment

with the an infinite lattice parameter to

avoid any interactions.

bcc Lattice Parametera →∞

( )1 2 N...φ φ φ

For the case of first point of view we have N localized electrons:

For the case of second point of view we have a wave function. This

wave function has to satisfy crystalline electron properties:

k :ψ

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One expects to be these two points of view identical: k jk j

j

C ( )ψ φ=∑ r - r

The coefficients should be

chosen such that the wave

function becomes Bloch’s function:

ji

k j

j

1e ( )

Nψ φ= ∑

k.r

r - r

Claim: The above wave function is Bloch’s function. We will get

back to this point and prove it soon.

Since the distance between atoms are so large, therefore

depends only on one atom and the summation is performed over independent terms:

j( )φ r - r

2

jφ2

j 1 φ +

Despite there is over all the crystal space,

we (as our second claim) can prove that:kψ 2

k j

j

1

Nψ φ= ∑

λ →∞

2

jφ2

j 1 φ +

2

k j

j

1

Nψ φ= ∑

2aλ =

k

ε

0 / aπ

j2 i

k j

j

1e ( )

Nψ φ= ∑

k.r

r - r

If the atoms are brought close then “k” will be

important, and thus we

cannot do summation over

independent terms. Indeed

in the summation, k vector

varies from one term to

another term, hence phase factor will not be vanished.

PROOF OF THE FIRST CLAIM

ji

k j

j

1e ( )

Nψ φ= ∑

k.r

r - r

ji ( )ik j

j

1e e ( )

Nψ φ

−= ∑

k. r rk.rr - r

ji ( )

j

j

U( ) e ( )φ−

=∑k. r r

r r - r

ik

1e U( )

Nψ = k.r

r

jr

R

ir i j= +r r R

ji ( )

j

j

U( ) e ( )φ+ −

+ = +∑k. r R r

r R r R - r

ii ( )i

i

e ( )φ−=∑ k. r rr - r

U( )= r ☺☺☺☺

PROOF OF THE SECOND CLAIM

ji

k j

j

1e ( )

Nψ φ= ∑

k.r

r - r

ji* *k j

j

1e ( )

Nψ φ

−= ∑

k.r

r - r

2 *k k kψ ψ ψ=

( )

( )

1 1

1 1

i i1 2

i i* *1 2

1e ( ) e ( ) ...

N

e ( ) e ( ) ...

φ φ

φ φ− −

= + + ×

+ +

k.r k.r

k.r k.r

r - r r - r

r - r r - r

( )2 2

1 2

1( ) ( ) ...

Nφ φ= + +r - r r - r

2

j

j

1( )

Nφ= ∑ r - r

Since everywhere is zero,

is nonzero and vice versa, then

are orthogonal. i.e.,

1( )φ r - r 1( )φ r -r'sφ

1 2( ) ( ) 0φ φ =r - r r -r

22

k j

j

1( )

Nψ φ= ∑ r - r

22

k j

j

N ( )ψ φ=∑ r - r

Charge distribution from

crystalline point of view

Charge distribution from

atomic point of view

Which are identical

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This is the quantum state where n=6, l=4, and m=1:

This is an equal superposition of the |3,2,1> and |3,1,-1> eigenstates:

This is an equal superposition of the |3,2,2> and |3,1,-1> eigenstates:

This is an equal superposition of the |4,3,3> and |4,1,0> eigenstates: