1
$\begingroup$

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 $\delta_{ij}$?

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

1 Answer 1

2
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Thank you, I corrected the equation. Does it mean the $\Phi_j$ are not orthogonal although the $\phi_j$ are? $\endgroup$
    – Andrea
    Commented Jun 3, 2022 at 13:20
  • 1
    $\begingroup$ Exactly. There is no reason for the $\Phi_i$'s to be mutually orthogonal. $\endgroup$
    – mike stone
    Commented Jun 3, 2022 at 13:29
  • 1
    $\begingroup$ 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
    Commented Jun 3, 2022 at 18:07
  • $\begingroup$ Ah now I see we the $\Phi_j$ are not orthogonal, thank you very much! $\endgroup$
    – Andrea
    Commented Jun 3, 2022 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.