Introduction to the Tightbinding (LCAO) Method

Tightbinding: 1 Dimensional Model #1 Consider an Infinite Linear Chain of identical atoms, with 1 s-orbital valence 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 thes orbital on site n interacts with the s orbitals on sites n – 1 & n + 1 only! n = Atomic Label  a   n =

The periodic potential V(x) for this Monatomic Linear Chain of atoms looks qualitatively like this: n =  a   V(x) = V(x + a)

The localized atomic orbitals on each site for this Monatomic Linear Chain of atoms look qualitatively like this: n = The spherically symmetric s orbitals on each site overlap slightly with those of their neighbors, as shown. This allows the electron on site n to interact with its nearest-neighbors on sites n – 1 & n + 1!  a  

The True Hamiltonian in the solid is: H = (p) 2 /(2m o ) + V(x), with V(x) = V(x + a). Instead, approximate it as H  ∑ n H at (n) + ∑ n,n U(n,n) where, H at (n)  Atomic Hamiltonian for atom n. U(n,n)  Interaction Energy between atoms n & n. Use the assumption of only nearest-neighbor interactions:  U(n,n) = 0 unless n = n -1 or n = n +1 With this assumption, the Approximate Hamiltonian is H  ∑ n [H at (n) + U(n,n -1) + U(n,n + 1)]

Goal: Calculate the bandstructure E k by solving the Schrödinger Equation: HΨ k (x) = E k Ψ k (x) Use the LCAO (Tightbinding) Assumptions: 1. H is as above. 2. Solutions to the atomic Schrödinger Equation are known: H at (n)ψ n (x) = E n ψ n (x) 3. In our simple case of 1 s-orbital/atom: E n = ε = the energy of the atomic e - (known) 4. ψ n (x) is very localized around atom n 5. The Crucial (LCAO) assumption is: Ψ k (x)  ∑ n e ikna ψ n (x) That is, the Bloch Functions are linear combinations of atomic orbitals!

Dirac notation: E k   Ψ k |H|Ψ k  (This Matrix Element is shorthand for a spatial integral!) Using the assumptions for H & Ψ k (x) already listed:  E k =  Ψ k |∑ n H at (n) |Ψ k  +  Ψ k |[∑ n U(n,n-1) + U(n,n-1)]|Ψ k  also note that H at (n)|ψ n  = ε|ψ n  The LCAO assumption is: |Ψ k   ∑ n e ikna |ψ n  Assume orthogonality of the atomic orbitals:   ψ n |ψ n  = δ n,n (= 1, n = n; = 0, n  n) Nearest-neighbor interaction assumption:  There is nearest-neighbor overlap energy only! (α = constant)  ψ n  |U(n,n  1)|ψ n   - α; (n  = n, & n = n  1)  ψ n  |U(n,n  1)|ψ n  = 0, otherwise It can be shown that for α > 0, this must be negative!

As a student exercise, show that the “energy band” of this model is: E k = ε - 2αcos(ka) or E k = ε - 2α + 4α sin 2 [(½)ka] A trig identity was used to get last form. ε & α are usually taken as parameters in the theory, instead of actually calculating them from the atomic ψ n  The “Bandstructure” for this monatomic chain with nearest-neighbor interactions only looks like (assuming 2α < ε ): (E T  E k - ε + 2α) It’s interesting to note that: The form E k = ε - 2αcos(ka) is similar to Krönig-Penney model results in the linear approximation for the messy transcendental function! There, we got: E k = A - Bcos(ka) where A & B were constants. ETET 4α4α

n = Tightbinding: 1 Dimensional Model #2  A 1-dimensional “semiconductor”! Consider an Infinite Linear Chain consisting of 2 atom types, A & B (a crystal with 2-atom unit cells), 1 s-orbital valence e - per atom & unit cell repeat distance = a. Approximation: Only Nearest-Neighbor interactions. (Interactions between atoms further apart than ~ (½ )a are ~ 0). This model is called the “Diatomic Chain”. A B A B A B A   a

The True Hamiltonian in the solid is: H = (p) 2 /(2m o ) + V(x), with V(x) = V(x + a). Instead, approximate it (with γ = A or =B) as H  ∑ γn H at (γ,n) + ∑ γn,γn U(γ,n;γ,n) where, H at (γ,n)  Atomic Hamiltonian for atom γ in cell n. U(γ,n;γ,n)  Interaction Energy between atom of type γ in cell n & atom of type γ in cell n. Use the assumption of only nearest-neighbor interactions:  The only non-zero U(γ,n;γ,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 Hamiltonian is: H  ∑ γn H at (γ,n) + ∑ n [U(n,n -1) + U(n,n + 1)]

Goal: Calculate the bandstructure E k by solving the Schrödinger Equation: HΨ k (x) = E k Ψ k (x) Use the LCAO (Tightbinding) Assumptions: 1. H is as above. 2. Solutions to the atomic Schrödinger Equation are known: H at (γ,n)ψ γn (x) = E γn ψ γn (x) 3. In our simple case of 1 s-orbital/atom: E An = ε A = the energy of the atomic e - on atom A E Bn = ε B = the energy of the atomic e - on atom B 4. ψ γn (x) is very localized near cell n 5. The Crucial (LCAO) assumption is: Ψ k (x)  ∑ n e ikna ∑ γ C γ ψ γn (x) That is, the Bloch Functions are linear combinations of atomic orbitals! Note!! The C γ ’s are unknown

Dirac Notation: Schrödinger Equation: E k   Ψ k |H|Ψ k   ψ An |H|Ψ k  = E k  ψ An |H|Ψ k  (1) Manipulation of (1), using LCAO assumptions, gives (student exercise): ε A e ikna C A + μ[e ik(n-1)a + e ik(n+1)a ]C B = E k e ikna C A (1a) Similarly:  ψ Bn |H|Ψ k  = E k  ψ Bn |H|Ψ k  (2) Manipulation of (2), using LCAO assumptions, gives (student exercise): ε B e ikna C B + μ[e ik(n-1)a + e ik(n+1)a ]C A = E k e ikna C A (2a) Here, μ   ψ An |U(n,n-1)|ψ B,n-1    ψ Bn |U(n,n+1)|ψ A,n+1  = constant (nearest-neighbor overlap energy) analogous to α in the previous 1d model

Student exercise to show that these simplify to: 0 = (ε A - E k )C A + 2μcos(ka)C B, (3) and 0 = 2μcos(ka)C A + (ε B - E k )C B, (4) ε A, ε B, μ are usually taken as parameters in the theory, instead of computing them from the atomic ψ γn (3) & (4) are linear, homogeneous algebraic equations for C A & C B  2  2 determinant of coefficients = 0 This gives: (ε A - E k )(ε B - E k ) - 4 μ 2 [cos(ka)] 2 = 0  A quadratic equation for E k !  2 solutions: a “valence” band & a “conduction” band!

Results: “Bandstructure” of the Diatomic Linear Chain (2 bands) : E  (k) = (½)(ε A + ε B )  [(¼)(ε A - ε B ) 2 + 4μ 2 {cos(ka)} 2 ] This gives a k = 0 bandgap of E G = E + (0) - E - (0) = 2[(¼)(ε A - ε B ) 2 + 4μ 2 ] ½ For simplicity, plot in the case 4μ 2 ε A  Expand the [ ….] ½ part of E  (k) & keep the lowest order term  E + (k)  ε B + A[cos(ka)] 2, E - (k)  ε A - A[cos(ka)] 2 E G (0)  ε A – ε B + 2A, where A  (4μ 2 )/|ε A - ε B |

“Bandstructure” of a 1-dimensional “semiconductor”:

Tightbinding Method: 3 Dimensional Model Model: Consider a monatomic solid, 3d, with only nearest-neighbor interactions. Hamiltonian: H = (p) 2 /(2m o ) + V(r) V(r) = crystal potential, with the full lattice symmetry & periodicity. Assume (R,R = lattice sites): H  ∑ R H at (R) + ∑ R,R U(R,R) H at (R)  Atomic Hamiltonian for atom at R U(R,R)  Interaction Potential between atoms at R & R Near-neighbor interactions only!  U(R,R) = 0 unless R & R are nearest-neighbors

Goal: Calculate the bandstructure E k by solving the Schrödinger Equation: HΨ k (r) = E k Ψ k (r) Use the LCAO (Tightbinding) Assumptions: 1. H is as on previous page. 2. Solutions to the atomic Schrödinger Equation are known: H at (R)ψ n (R) = E n ψ n (R), n = Orbital Label (s, p, d,..), E n = Atomic energy of the e - in orbital n 3. ψ n (R) is very localized around R 4. The Crucial (LCAO) assumption is: Ψ k (r) = ∑ R e ik  R ∑ n b n ψ n (r-R) (b n to be determined) ψ n (R): The atomic functions are orthogonal for different n & R That is, the Bloch Functions are linear combinations of atomic orbitals!

Dirac Notation: Solve the Schrödinger Equation: E k   Ψ k |H|Ψ k  The LCAO assumption: |Ψ k  = ∑ R e ik  R ∑ n b n |ψ n  (1) (b n to be determined) Consider a particular orbital with label m:  ψ m |H|Ψ k  = E k  ψ m |H|Ψ k  (2) Use (1) in (2). Then use 1. The orthogonality of the atomic orbitals 2. The assumed form of H 3. The fact that ψ n (R) is very localized around R 4. That we know the atomic solutions to H at |ψ n  = E n |ψ n  5. The nearest neighbor assumption that U(R,R) = 0 unless R & R are nearest-neighors.

Manipulate (several pages of algebra) to get: (E k - E m )b m + ∑ n ∑ R  0 e ik  R γ mn (R)b n = 0, (I) where: γ mn (R)   ψ m |U(0,R)|ψ n   “Overlap Energy Integral” The γ mn (R) are analogous to the α & μ in the 1d models. They are similar to V ssσ, etc. in real materials, discussed next! The integrals are horrendous to do for real atomic ψ m ! In practice, they are treated as parameters to fit to experimental data. Equation (I): Is a system of N homogeneous, linear, algebraic equations for the coefficients b n. N = number of atomic orbitals. Equation (I) for N atomic states  The solution is obtained by taking an N  N determinant! This results in N bands which have their roots in the atomic orbitals! If the γ mn (R) are “small”, each band can be thought of as E k ~ E n + k dependent corrections That is, the bands are ~ atomic levels + corrections

Equation (I): A system of homogeneous, linear, algebraic equations for the b n N atomic states  Solve an N  N determinant! N bands Note: We’ve 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-like) band Artificial Special Case #2: Three p levels per atom  3 (p-like) bands Artificial Special Case #3: One s and three p levels per atom & sp 3 bonding  4 bands NOTE that For n atoms /unit cell, multiply by n to get the number of bands!

Back to: (E k - E m )b m + ∑ n ∑ R  0 e ik  R γ mn (R)b n = 0, (I) where: γ mn (R)   ψ m |U(0,R)|ψ n   “Overlap Energy Integral” Also: Assume nearest neighbor interactions only  ∑ R  0 is ONLY over nearest neighbors! Artificial Special Case #1: One s level per atom  1 (s-like) band: E k = E s - ∑ R=nn e ik  R γ(R) But γ(R) = γ is the same for all neighbors so: E k = E s - γ∑ R=nn e ik  R Assume, for example, a simple cubic lattice: E k = E s -2γ[cos(k x a) + cos(k y a) +cos(k z a)]

Artificial Special Case #2: Three p levels per atom. Gives a 3  3 determinant to solve.  3 (p-like) bands Student exercise!! Artificial Special Case #3: One s and three p levels per atom & sp 3 bonding Gives a 3  3 determinant to solve.  4 bands Student exercise!!