Lecture 6 Tight-binding model In Lecture 3 we discussed the Krönig-Penny model and we have seen that, depending on the strength of the potential barrier inside the unit cell, the electrons can behave like in a FEG (in the limit µ = 0) or like electrons in atoms (in the limit µ = ). In Lecture 4 we introduced the nearly-free electron model, which describes reasonably situations where the electrons feel a weak tal potential. In this lecture we want to look at the other end of the spectrum, where µ is large but not infinite. This corresponds to a situation where electrons are tightly bound to their nuclei. While in the FEG the starting point are travelling electron waves, in the tight binding (TB) model the starting point are atomic wavefunctions. Figure 6.1 illustrates schematically this conceptual difference. In order to understand TB model we start with the H 2 molecule and then we move on to the simplest 3D model of a solid, the cubium. Figure 6.1: Schematic view of the theoretical models that we can use to describe solids depending on the strength of the tal potential. For very weak potentials the electrons are essentially travelling waves and we can use the NFEG approximation. For very strong potentials it could be more convenient, at least conceptually, to start from an atomic-like description as given by the tight-binding approximation. While in the NFEG approximation we write the wavefunctions as linear combinations of planewaves, in TB we write the wavefunctions as linear combinations of atomic wavefunctions.
48 6 Tight-binding model 6.1 Hydrogen molecule We start by considering the H atom. The Schrödinger equation for the ground state 1s wavefunction is: 2 φ 1s (r) + v(r)φ 1s (r) = E 1s φ 1s (r), (6.1) with E 1s = 13.6 ev, e2 v(r) = 4πɛ 0 r, (6.2) φ 1s (r) = 1 π a 3/2 0 e r /a 0, (6.3) and a 0 = 0.53 A. The hydrogen molecule is formed by putting together two H atoms. If we have atom A and atom B in the positions A and B, respectively, the total potential seen by the electrons is: V (r) = v(r A ) + v(r B ), (6.4) and the Schrödinger equation becomes: 2 ψ(r) + V (r)ψ(r) = Eψ(r), (6.5) 2 ψ(r) + [v(r A ) + v(r B )]ψ(r) = Eψ(r). (6.6) The tight-binding approximation consists of assuming that the electrons in this molecule remain tightly bound to their nuclei. If this is the case, then we can find a solution which is very similar to the two original 1s wavefunctions. The simplest guess is to look for a solution such as: ψ(r) = c A φ 1s (r A ) + c B φ 1s (r B ), (6.7) i.e. a linear combination of atomic orbitals (LCAO). In practice, while in the NFEG we were looking for a solution which is a linear combination of planewaves, here we use a linear combination of local orbitals: we expand the wavefunctions in a localized basis set. In order to find the wavefunctions of the H 2 molecule we now replace Eq. (6.7) inside Eq. (6.5). Before doing this, however, it is convenient to simplify the notation. We use the following definitions: φ A (r) = φ 1s (r A ), (6.8) φ B (r) = φ 1s (r B ), (6.9) v A (r) = v(r A ), (6.10) v B (r) = v(r B ). (6.11)
Hydrogen molecule 49 We can now perform the substitution and obtain: 2 (c A φ A + c B φ B ) + (v A + v B )(c A φ A + c B φ B ) = E(c A φ A + c B φ B ). (6.12) From Eq. (6.1) we already know that: 2 φ A + v A φ A = E 1s φ A, (6.13) 2 φ B + v B φ B = E 1s φ B, (6.14) therefore we can use these equalities to simplify Eq. (6.12): c A (E 1s E + v B )φ A + c B (E 1s E + v A )φ B = 0. (6.15) In order to determine the coefficients c A and c B we need two equations. These equations can be found as follows: we multiply both sides of Eq. (6.15) by φ A and integrate over the whole space. This gives us the first equation relating c A and c B. The second equation can be found by multiplying both sides of Eq. (6.12) by φ B and integrating over the whole space. The integration leads to the following equations: c A [(E 1s E)S AA + γ] + c B [(E 1s E)S AB + β] = 0, (6.16) having defined: S AA = S AB = γ = β = dr φ A (r) 2, (6.17) dr φ A (r)φ B (r), (6.18) dr φ A 2 v B (r), (6.19) dr φ A (r)v A (r)φ B (r). (6.20) Now S AA = 1 since the H wavefunction is normalized. S AB is called the overlap integral, and can be neglected because the H wavefunctions decay exponentially fast. The term γ is called the tal field shift, and we are going to neglect it because the 1s wavefunction of the atom A is negligibly small on the nuclear site of atom B. Finally the term β is called the bond integral, and cannot be neglected since the 1/r potential on site A diverges, hence the overlap between v A φ A and φ B is not small. We note also that β < 0 because both φ A and φ B are positive while the potential v A is negative. Taking into account the above considerations we can simplify Eq. (6.16) as follows: c A (E 1s E) + β c B = 0. (6.21)
50 6 Tight-binding model The second equation that we need for determining the coefficients c A and c B can be obtained similarly and reads: β c A + (E 1s E)c B = 0. (6.22) Be rewriting the previous two equations in matrix notation we obatin: [ ] [ ] E1s E β ca = 0. (6.23) β E 1s E The solutions of the secular equation (E 1s E) 2 = β 2 are therefore: and the final wavefunctions can be written as: c B E b = E 1s β, E a = E 1s + β, (6.24) ψ b (r) = 1 2 [φ 1s (r A ) + φ 1s (r B )], (6.25) ψ a (r) = 1 2 [φ 1s (r A ) φ 1s (r B )], (6.26) where the prefactor comes from the normalization condition. The lowest energy solution is the bonding combination of the two hydrogenic wavefunctions, while the highest energy solution is the antibonding combination. The difference between these two solutions is illustrated in Fig. 6.2. Figure 6.2: Electron eigenstates of the H 2 molecule within the tight-binding approximation. The bonding state piles up electronic charge in the middle of the H-H bond and lowers the total energy of the system. An electron in this bonding state stabilizes the molecule through the formation of a covalent bond. The antibonding state piles up electronic charge around the nuclear sites, thereby increasing the total energy of the system.
Cubium 51 6.2 Cubium The cubium is a toy-model of a material which consists of a simple cubic lattice with one s orbital per site. For instance we can imagine an ideal sc lattice of H atoms. The Schrödinger equation for this system is: 2 ψ(r) + V (r)ψ(r) = Eψ(r). (6.27) In analogy with the H 2 molecule the potential is given by the sum of the individual atomic potentials: V (r) = v(r ), (6.28) where the sum runs over all the direct lattice vectors = a(n x u x + n y u y + n z u z ). As in the case of the H 2 molecule we look for a solution given by a linear combination of atomic orbitals: ψ(r) = c() φ 1s (r ), (6.29) with φ 1s (r) given by Eq. (6.3). In the case of the H 2 molecule we found the eigenstates and eigenvalues by performing a diagonalization. In this case the matrix would be huge, because we have as many coefficients c() as the number of lattice sites. It is therefore convenient to take an alternative approach, and start by requiring that our solution for a given wavevector k satisfies the Bloch theorem: ψ k (r + ) = ψ k (r) e ik. (6.30) We now replace Eq. (6.29) in the last equation in order to determine a condition for the coefficients c k () (the following steps are very tedious): c k ( ) φ 1s (r + ) = c k ( ) φ 1s (r ) e ik. [c k ( + ) c k ( ) e ik ] φ 1s (r ) = 0. (6.31) c k ( + ) φ 1s (r ) = c k ( ) φ 1s (r )e ik. The last equality must be satisfied for every r, therefore the only possibility is that the term between square brackets vanishes identically: If we rewrite this for the particular case = 0 we obtain: c k ( + ) = c k ( ) e ik. (6.32) c k () = c k (0) e ik. (6.33)
52 6 Tight-binding model We can now rewrite our Bloch wavefunction Eq. (6.29) as follows: ψ k (r) = c k (0) e ik φ 1s (r ), = e ik φ 1s (r ), (6.34) having set c k (0) = 1 in order to have the wavefunction normalized in the unit cell. At this point we have found the eigenstates ψ k (r) and we only need to find the associated eigenvalues. In order to simplify the notation it is convenient to introduce the Hamiltonian operator as follows: Ĥ = 2 + v(r ). (6.35) Using this notation Eq. (6.27) can be rewritten in a more compact form as: Ĥ(r)ψ k (r) = E k ψ k (r). (6.36) In order to find the eigenvalues E k we multiply both sides of the last equation by ψk (r) and integrate over the tal volume, obtaining: dr ψ k(r)ĥ(r)ψ k(r) = E k dr ψk(r)ψ k (r) = N c E k, (6.37) N c being the number of unit cells in the tal. The last equality follows from the normalization of the wavefunctions. We can now replace Eq. (6.34) inside Eq. (6.37) (the following steps are also very tedious): E k = 1 dr ψ N k(r)ĥ(r)ψ k(r) c = 1 e ik ( ) dr φ 1s (r N )Ĥ(r) φ 1s(r ) c = 1 N c = 1 N c = e ik e ik ( ) e ik ( ) dr φ 1s (r )Ĥ(r ) φ 1s [r + ( )] dr φ 1s (r)ĥ(r) φ 1s[r ( )] dr φ 1s (r)ĥ(r) φ 1s(r ). (6.38) In the second and third steps we made use of the fact that the Hamiltonian operator is translationally invariant (because the kinetic energy and the potential are both translationally invariant). The integral in the last equation needs to be computed by writing
Cubium 53 explicitly the Hamiltonian operator: dr φ 1s (r)ĥ(r) φ 1s(r ) = [ = dr φ 1s (r) 2 + ] v(r ) φ 1s (r ) ] = dr φ 1s (r) [ 2 + v(r) φ 1s (r ) + dr φ 1s (r)v(r ) φ 1s (r ) 0 = E 1s dr φ 1s (r)φ 1s (r ) + dr φ 1s (r)v(r ) φ 1s (r ). 0 = E 1s δ,0 + dr φ 1s (r)v(r ) φ 1s (r ), (6.39) 0 with δ,0 = 1 if = 0, and 0 otherwise. At this stage we need to introduce the tight-binding approximations, in analogy with the case of the H 2 molecule. When = 0 the integral dr φ 1s(r)v(r ) φ 1s (r ) corresponds to the tal field shift (cf. Sec. 6.1). Since 0 in the sum, we can neglect this term because φ 1s (r) is localized around = 0. Now we are left with the case 0 and 0. If = we have a bond integral and we call it β(). If and both are nonzero, then φ 1s (r), v(r ), and φ 1s (r ) are located at three different lattice sites. This term is called a three-centre integral. Since the atomic orbitals are localized, the product of these three functions is always very small and we can neglect this term. By combining these results we have: dr φ 1s (r)ĥ(r) φ 1s(r ) = E 1s δ,0 + β()(1 δ,0 ). (6.40) We can finally rewrite Eq. (6.38) using this result: E k = e ik [E 1s δ,0 + β()(1 δ,0 )]. = E 1s + 0 e ik β(). (6.41) With reference to Fig. 6.3 we now make the approximation that only the bond integrals β() between nearest neighbor sites are non negligible. This means that the only sites we need to consider are ±au x, ±au y, and ±au z. The bond integrals between nearest neighbor sites are all the same by symmetry, and we will call them simply β. Within
54 6 Tight-binding model Figure 6.3: 2D representation of a simple cubic lattice. In the tight-binding approximation we keep only the bond integrals between nearest-neighbor lattice sites (NN in the figure). The rationale for this choice is that atomic wavefunctions are localized around their nuclei and decay exponentially with the distance. Sometimes it may be necessary to include interactions up to the next-nearest neighbors (NNN in the figure), and even possibly further neighbors, depending on the application. such approximation the energy vs momentum dispersion relations in Eq. (6.41) simplifies to: E k = E 1s + β[e i(k ux)a + e i(k ux)a + e i(k uy)a + e i(k uy)a + e i(k uz)a + e i(k uz)a ] = E 1s + β[e ikxa + e ikxa + e ikya + e ikya + e ikza + e ikza ] = E 1s + 2β(cos k x a + cos k y a + cos k z a). (6.42) It is useful to remember that β < 0 and rewrite this equation as E k = E 1s 2 β (cos k x a + cos k y a + cos k z a). (6.43) Equation (6.43) gives the tight-binding s-band for a simple cubic lattice. Figure 6.4 shows the band structure of cubium and the associated density of states.
Cubium 55 Figure 6.4: Band structure and DOS of cubium. The wavevector k is in units of π/a. The energy is given in dimensionless units (E E 1s )/ β. The highest energy state is found for k = (1, 1, 1)π/a and corresponds to the wavefunction ψ 111 (r) = eiπ(nx+ny+nz ) φ 1s (r ). The phase factor exp[iπ(n x +n y +n z )] alternates in sign when we move across nearest-neighbor lattice sites, therefore the state ψ 111 corresponds to a checkerboard configuration of φ 1s (r) and φ 1s (r) functions on each lattice site. The lowest energy state is found for k = (0, 0, 0) and corresponds to the wavefunction ψ 000 (r) = φ 1s(r ). Therefore in the state ψ 000 all the φ 1s (r) functions carry the same sign. The width of the band is 12 β and gives a measure of the bond integral β.