# Tight-Binding method and orthogonality of Bloch functions

Tight binding summary

When computing the electronic bands of a crystal in the tight-Binding approximation, the standard way to do it is to construct Bloch solution as

$$\Psi_n(\textbf{r}, \textbf{k}) = \sum_{\textbf{R}}e^{i\textbf{k}\cdot \textbf{R}}\psi_n(\textbf{r}-\textbf{R})$$

where in general $$\psi_n(\textbf{r}-\textbf{R})$$ is a LCAO, namely

$$\psi_n(\textbf{r}-\textbf{R}) = \sum_{n'= 1}^N C_{nn'} \phi_{n'}(\textbf{r}-\textbf{R})\;\;\; n = 1, ..., N$$

and $$\phi_n(\vec{r})$$ is the $$n^{th}$$ atomic orbital. To compute the bands, we use

$$E_i(\textbf{k}) = \frac{\langle\Psi_i(\textbf{r}, \textbf{k})|\hat{H}|\Psi_i(\textbf{r}, \textbf{k})\rangle}{\langle\Psi_i(\textbf{r}, \textbf{k})|\Psi_i(\textbf{r}, \textbf{k})\rangle}$$

if we now introduce the functions

$$\Phi_n(\textbf{r}, \textbf{k}) = \frac{1}{\sqrt{N}}\sum_{\textbf{R}}e^{i\textbf{k}\textbf{R}}\phi_{n}(\textbf{r}-\textbf{R})$$

we can define the matrices $$H_{ij} = \langle\Phi_i|\hat{H}|\Phi_j\rangle$$ and $$S_{ij} = \langle\Phi_i|\Phi_j\rangle$$ and the eigenvalue equation gives the secular system

$$E_i = \frac{\sum_{n, n'}H_{nn'}C^*_{in}C_{in'}}{\sum_{n, n'}S_{nn'}C^*_{in}C_{in'}}$$

by minimizing the $$E_i$$ we an find the eigenvalues, namely the electronic bands.

My questions

• Why are we taking the number of LCAO combinations the same as the number of atomic orbitals we are taking into account? Is it because we are looking for $$N$$ linearly independent functions, therefore we can't make more than that, having $$N$$ linearly independent atomic orbitals?

• If $$\phi_n(\textbf{r})$$ are the atomic orbitals, namely eigenvectors of the atomic Hamiltonian, they are an orthogonal basis. Then I expect the functions $$\Phi_n(\textbf{r}, \textbf{k})$$ to be orthogonal as well. If this is true, why do we bother computing $$S_{ij}$$ when it is just $$\delta_{ij}$$?

• Your 2nd to last equation has an $n$ on the LHS, but and $n'$ on the RHS.
– hft
Commented Jun 3, 2022 at 18:02
• thanks I fixed it Commented Jun 3, 2022 at 19:04

Your second equation should be $$\psi_n(\vec{r}-\vec{R}) = \sum_{n'= 1}^N C_{nn'} \phi_{n'}(\vec{r}-\vec{R})\;\;\; n = 1, ..., N,$$ and while the $$\phi_{n'}$$ and $$\phi_{n}$$ at the same site are orthogonal, there is no reason for $$\phi_{n}(r-R_1)$$ to be orthogonal to $$\phi_{n'}(r-R_2)$$.
• Thank you, I corrected the equation. Does it mean the $\Phi_j$ are not orthogonal although the $\phi_j$ are? Commented Jun 3, 2022 at 13:20
• Exactly. There is no reason for the $\Phi_i$'s to be mutually orthogonal. Commented Jun 3, 2022 at 13:29
• The $\phi_n(r-R)$ are not orthogonal. The $\Phi_i$ are not orthogonal, but they are approximately orthogonal if the $\vec R$ are well separated compared to the orbital size. In that case: $\int d^3r \Phi_i^*(r,k)\Phi_j(r,q)\approx \delta_{k,q}\delta_{i,j}$.
• Ah now I see we the $\Phi_j$ are not orthogonal, thank you very much! Commented Jun 3, 2022 at 19:04