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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 $\delta_{ij}$?

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

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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 $\phi_{n'}(r-R_2)$.

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  • $\begingroup$ Thank you, I corrected the equation. Does it mean the $\Phi_j$ are not orthogonal although the $\phi_j$ are? $\endgroup$
    – Andrea
    Jun 3, 2022 at 13:20
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    $\begingroup$ Exactly. There is no reason for the $\Phi_i$'s to be mutually orthogonal. $\endgroup$
    – mike stone
    Jun 3, 2022 at 13:29
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    $\begingroup$ 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: $\int d^3r \Phi_i^*(r,k)\Phi_j(r,q)\approx \delta_{k,q}\delta_{i,j}$. $\endgroup$
    – hft
    Jun 3, 2022 at 18:07
  • $\begingroup$ Ah now I see we the $\Phi_j$ are not orthogonal, thank you very much! $\endgroup$
    – Andrea
    Jun 3, 2022 at 19:04

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