1. Introduction to the Tightbinding (LCAO) Method

2. 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”.
n = Atomic Label
 a 
 
n =
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!

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

4. The spherically symmetric s orbitals on each site overlap
The localized atomic orbitals on each site for this Monatomic Linear Chain of atoms look qualitatively like this:
 a 
 
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!

5. The True Hamiltonian in the solid is:
H = (p)2/(2mo) + V(x), with V(x) = V(x + a).
Instead, approximate it as H  ∑n Hat(n) + ∑n,nU(n,n)
where, Hat(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 [Hat(n) + U(n,n -1) + U(n,n + 1)]

6. H  ∑n [Hat(n) + U(n,n -1) + U(n,n + 1)]
Goal: Calculate the bandstructure Ek by solving the Schrödinger Equation:
HΨk(x) = Ek Ψk(x)
Use the LCAO (Tightbinding) Assumptions:
1. H is as above.
2. Solutions to the atomic Schrödinger Equation are known:
Hat(n)ψn(x) = Enψn(x)
3. In our simple case of 1 s-orbital/atom:
En= ε = the energy of the atomic e- (known)
4. ψn(x) is very localized around atom n
5. The Crucial (LCAO) assumption is:
Ψk(x)  ∑neiknaψn(x)
That is, the Bloch Functions are linear combinations of atomic orbitals!

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

8. Ek= ε - 2αcos(ka) or Ek = ε - 2α + 4α sin2[(½)ka]
As a student exercise, show that the “energy band” of this model is:
Ek= ε - 2αcos(ka) or Ek = ε - 2α + 4α sin2[(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 ψn
 The “Bandstructure” for this monatomic chain with nearest-neighbor interactions only looks like (assuming 2α < ε ): (ET  Ek - ε + 2α)
It’s interesting to note that:
The form Ek= ε - 2αcos(ka) is
similar to Krönig-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 4α

9. This model is called the “Diatomic Chain”.
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
n =
  a

10. The True Hamiltonian in the solid is:
H = (p)2/(2mo) + V(x), with V(x) = V(x + a).
Instead, approximate it (with γ = A or =B) as
H  ∑γnHat(γ,n) + ∑γn,γnU(γ,n;γ,n)
where, Hat(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  ∑γnHat(γ,n) + ∑n[U(n,n -1) + U(n,n + 1)]

11. H  ∑γnHat(γ,n) + ∑n[U(n,n -1) + U(n,n + 1)] HΨk(x) = EkΨk(x)
Goal: Calculate the bandstructure Ek by solving the Schrödinger Equation:
HΨk(x) = EkΨk(x)
Use the LCAO (Tightbinding) Assumptions:
1. H is as above.
2. Solutions to the atomic Schrödinger Equation are known:
Hat(γ,n)ψγn(x) = Eγnψγn(x)
3. In our simple case of 1 s-orbital/atom:
EAn= εA = the energy of the atomic e- on atom A
EBn= εB = the energy of the atomic e- on atom B
4. ψγn(x) is very localized near cell n
5. The Crucial (LCAO) assumption is:
Ψk(x)  ∑neikna∑γCγψγn(x)
That is, the Bloch Functions are linear combinations of atomic orbitals!
Note!!
The Cγ’s are unknown

12. Dirac Notation: Schrödinger Equation: Ek  Ψk|H|Ψk
ψAn|H|Ψk = EkψAn|H|Ψk (1)
Manipulation of (1), using LCAO assumptions, gives (student exercise):
εAeiknaCA+ μ[eik(n-1)a + eik(n+1)a]CB = EkeiknaCA (1a)
Similarly: ψBn|H|Ψk = EkψBn|H|Ψk (2)
Manipulation of (2), using LCAO assumptions, gives (student exercise):
εBeiknaCB+ μ[eik(n-1)a + eik(n+1)a]CA = EkeiknaCA (2a)
Here, μ  ψAn|U(n,n-1)|ψB,n-1  ψBn|U(n,n+1)|ψA,n+1 = constant (nearest-neighbor overlap energy) analogous to α in previous 1d model

13.  A quadratic equation for Ek!  2 solutions: a “valence” band
Student exercise to show that these simplify to:
0 = (εA - Ek)CA + 2μcos(ka)CB , (3)
and
0 = 2μcos(ka)CA + (εB - Ek)CB, (4)
εA, εB , μ are usually taken as parameters in the theory, instead of computing them from the atomic ψγn
(3) & (4) are linear, homogeneous algebraic equations for CA & CB
 22 determinant of coefficients = 0
This gives: (εA - Ek)(εB - Ek) - 4 μ2[cos(ka)]2 = 0
 A quadratic equation for Ek!
 2 solutions: a “valence” band
& a “conduction” band!

14. EG= E+(0) - E-(0) = 2[(¼)(εA - εB)2 + 4μ2]½
Results:
“Bandstructure” of the Diatomic Linear Chain (2 bands):
E(k) = (½)(εA + εB) [(¼)(εA - εB)2 + 4μ2 {cos(ka)}2]½
This gives a k = 0 bandgap of
EG= E+(0) - E-(0) = 2[(¼)(εA - εB)2 + 4μ2]½
For simplicity, plot in the case
4μ2 << (¼)(εA - εB)2 & εB > εA
 Expand the [ ….]½ part of E(k) & keep the lowest order term
 E+(k)  εB + A[cos(ka)]2, E-(k)  εA - A[cos(ka)]2
EG(0)  εA – εB + 2A , where A  (4μ2)/|εA - εB|

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

16. 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,RU(R,R)
Hat(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

17. HΨk(r) = EkΨk(r) Use the LCAO (Tightbinding) Assumptions:
Goal: Calculate the bandstructure Ek by solving the Schrödinger Equation:
HΨk(r) = EkΨk(r)
Use the LCAO (Tightbinding) Assumptions:
1. H is as on previous page.
2. Solutions to the atomic Schrödinger Equation are known:
Hat(R)ψn(R) = Enψn(R), n = Orbital Label (s, p, d,..),
En= Atomic energy of the e- in orbital n
3. ψn(R) is very localized around R
4. The Crucial (LCAO) assumption is:
Ψk(r) = ∑ReikR∑nbnψn(r-R) (bn to be determined)
ψn(R): The atomic functions are orthogonal for different n & R
That is, the Bloch Functions are linear combinations of atomic orbitals!

18. Dirac Notation: Solve the Schrödinger Equation: Ek  Ψk|H|Ψk
The LCAO assumption:
|Ψk = ∑ReikR ∑nbn|ψn (1)
(bn to be determined)
Consider a particular orbital with label m:
ψm|H|Ψk = Ekψm|H|Ψk (2)
Use (1) in (2).
Then use 1. The orthogonality of the atomic orbitals
2. The assumed form of H
3. The fact that ψn(R) is very localized around R
4. That we know the atomic solutions to Hat|ψn = En|ψn
5. The nearest neighbor assumption that U(R,R) = 0
unless R & R are nearest-neighors.

19. (Ek - Em)bm + ∑n∑R0 eikR γmn(R)bn = 0 , (I)
Manipulate (several pages of algebra) to get:
(Ek - Em)bm + ∑n∑R0 eikR γmn(R)bn = 0 , (I)
where: γmn(R)  ψm|U(0,R)|ψn  “Overlap Energy Integral”
The γmn(R) are analogous to the α & μ in the 1d models. They are similar to Vssσ, etc. in real materials, discussed next! The integrals are horrendous to do for real atomic ψm! 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 bn. 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 γmn(R) are “small”, each band can be thought of as
Ek ~ En + k dependent corrections
That is, the bands are ~ atomic levels + corrections

20. For n atoms /unit cell, multiply by n to get the number of bands!
Equation (I): A system of homogeneous, linear, algebraic eqtns for the bn
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 & sp3 bonding
 4 bands
NOTE that
For n atoms /unit cell, multiply by n to get the number of bands!

21. Ek = Es -2γ[cos(kxa) + cos(kya) +cos(kza)]
Back to:
(Ek - Em)bm + ∑n∑R0 eikR γmn(R)bn = 0 , (I)
where: γmn(R)  ψm|U(0,R)|ψn  “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: Ek = Es - ∑R=nneikR γ(R)
But γ(R) = γ is the same for all neighbors so:
Ek = Es - γ∑R=nneikR
Assume, for example, a simple cubic lattice:
Ek = Es -2γ[cos(kxa) + cos(kya) +cos(kza)]

22. 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 & sp3 bonding
 4 bands