CHAPTER 10 Tight-Binding Model

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1 CHAPTER 0 Tight-Binding Model Linear Combination of Atomic Orbitals (LCAO) Application to Bands from s-levels General Features of Tight-Binding Levels Wannier Functions 6 a : S S P 3S Core FE Semicore TB Valence FE either FE nor TB is suitable for the study of transition metals and f-electron systems. In fact, 3d, 4f and 5f electrons are not so wea that can be treated using FE 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.

2 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 dz+dz (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 lie-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 weaer 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 weaness is explained by significantly less overlap between the component p- orbitals due to their parallel orientation.

3 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. FE FE TB Isolated atom MODEL Suppose we have two atoms that constitutes a molecule. Let these atoms to be in their ground states, and the electron be in the S orbital. H φ = εφ Hφ = εφ ow Let they come closer to each other. In this case two different situations may occur: 3

4 ψ = ( S + S ) ψ = ( S S ) 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. 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. inversesquare relationship for the particle in a square well). This energetic advantage is what stabilises the chemical bond. 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. 4

5 λ ε a λ = a 0 π a λ = a First point of view: The atoms are considered as a set of isolated atoms: H φ = ε φ i i i i h + U(r) φ(r) = εφ(r) m Isolated atomic potential For the case of first point of view we have localized electrons: ( φ φ... φ ) For the case of second point of view we have a wave function. This wave function has to satisfy crystalline electron properties: bcc Lattice Parameter a Second point of view: The set atoms is considered in a bcc crystalline environment with the an infinite lattice parameter to avoid any interactions. ψ : 5

6 ψ = C φ( r -r ) One expects to be these two points of view identical: The coefficients should be i chosen such that the wave ψ φ function becomes Bloch s function: = e.r ( r - r ) Claim: The above wave function is Bloch s function. We will get bac to this point and prove it soon. = Despite ψ there is over all the crystal space, we (as our second claim) can prove that: ψ φ λ φ φ + Since the distance between atoms are so large, therefore φ( r - r) depends only on one atom and the summation is performed over independent terms: φ φ + ψ = φ λ = a PROOF OF THE FIRST CLAIM ψ =.r r - r i e φ( ) i ( + ). r R r U( r + R) = e φ( r + R - r ) ε ψ i = e.r φ( r - r) If the atoms are brought close then will be important, and thus we cannot do summation over independent terms. Indeed in the summation, vector varies from one term to another term, hence phase factor will not be vanished. i i ( ) e.r e φ ( r - r ) ψ =. r r ψ = e i.r U( r ) i ( ) =. r r U( r) e φ( r - r ) R i ( i ) =. r r φ r - ri i e ( ) = U( r) 0 π / a r r = r + R ri i PROOF OF THE SECOD CLAIM ψ =.r r - r i e φ( ) ψ = e.r ( r - r ) ψ * i * φ * = ψ ψ ( e i.r ( i ) e.r φ r - r φ ( r - r )... ) i.r * i.r * ( e φ ( r - r ) + e φ ( r - r ) +...) = + + Since everywhere φ( r - r is zero, ) φ( r -r ) is nonzero and vice versa, then φ 's are orthogonal. i.e., φ( r - r ) φ ( r -r ) = 0 ( ( r - r ) ) φ ( r - r )... = φ + + = φ( r - r ) ψ Charge distribution from crystalline point of view = φ( r - r ) ψ = φ( r - r ) Which are identical Charge distribution from atomic point of view 6

7 This is the quantum state where n=6, l=4, and m=: This is an equal superposition of the 3,,> and 3,,-> eigenstates: This is an equal superposition of the 3,,> and 3,,-> eigenstates: This is an equal superposition of the 4,3,3> and 4,,0> eigenstates: 7