# Can someone give a simple mathematical explanation of the tight binding method?

I'm trying to understand the tight binding method but I'm struggling with a lot of the mathematical formalism. A lot of the mathematical formalism I read jumps into explaining it a few too many steps ahead of where my understanding is.

I understand it's an approach to calculating the band structure in solids.

[(-ħ2/2m)∇2 + V(r)]Ψ = EΨ

Coulomb potential for a hydrogen atom:

V(r) = -e2/4πϵr

Right now I'm imagining a 2D case where hydrogen atoms are lined up in a row. The electron in question experiences a coulomb potential from other atoms in the crystal.

i V(r - Ri)

This will tell us what all the other coulomb potentials are. When we expand it out we get V(r) [the coulomb potential the electron experiences from it's own nucleus] and V(Ri) - [the potential the electron experiences from the nucleus of nearby atoms]

[(-ħ2/2m)∇2 + ∑i V(r - Ri)] = EΨ

This only describes what the energy of 1 electron is. From here I get a bit confused with it all.

I'm not enough of an expert that I could, say, write a TB code, but here's what I get. In order to describe a wavefunction you need to write it down as a linear combination of some set of 'basis' functions:

$$\psi = \sum_ic_i\psi_i$$

In general, you can then write the Hamiltonian as a matrix whose elements correspond to the kinetic and potential energy terms for each of these basis functions:

$$H_{ij} = \langle\psi_j|H|\psi_i\rangle$$

and then diagonalise said matrix to find the eigenvalues (energies) and eigenvectors (states). Tight binding uses as basis functions the hydrogen-like orbitals on each of the atoms of the system. This is not an orthogonal basis: two nearby atoms will have slightly overlapping orbitals. If you carry out:

$$S = \langle\psi_{1s}(\mathbf{r})|\psi_{1s}(\mathbf{r}-\mathbf{r}_0)\rangle$$

this overlapping integral isn't zero. What tight binding does is parametrise these sort of integrals for a number of different distances and orientations for each possible pair of elements by using lower level simulation methods (like plane waves or gaussian basis DFT), and then use these fitted parametrisations to compute the Hamiltonian for your specific system, diagonalise it, find the states, and fill them in - from which the rest (like band structure) follows naturally.

• Why are the wavefunctions indexed with i and j? – mcsquared Aug 8 '18 at 10:28
• Because those are the generic indices I gave to $H_{ij}$, aka the matrix element of the Hamiltonian. I just mean that the matrix contains all possible such combinations, for $i$ and $j$ running over all your basis set. – Okarin Aug 8 '18 at 13:06