How do I go about determining the tight-binding Hamiltonian for the crystal structure below? I have identified the primitive lattice vectors $\mathbf{a}_1=(a,0)$ and $\mathbf{a}_2=(0,a)$ for lattice constant $a$ and labelled the three basis atoms. All the atoms are identical and have one s-orbital each. All nearest-neighbour distances are equal.

It is assumed that the diagonal elements of the Hamiltonian are zero. Also, the parameter for nearest-neighbour hopping is given as $t=\langle \phi|H|\phi\rangle$ where $|\phi\rangle$ is an s-orbital. I understand the derivation for the square lattice, but the addition of a basis has confused me.

  • 1
    $\begingroup$ What's your question? Do you want to write out a Hamiltonian or what? $\endgroup$ Nov 22, 2021 at 3:17
  • $\begingroup$ @RoderickLee Exactly yes $\endgroup$
    – soupdragon
    Nov 22, 2021 at 3:20
  • $\begingroup$ isn't that just hoping terms with every pair of connected sites? And since there are three basis atoms, you can write the hopping term individually. $\endgroup$ Nov 22, 2021 at 3:26
  • $\begingroup$ see my answer below $\endgroup$ Nov 22, 2021 at 3:28

1 Answer 1


$$H=-t\sum_{i,j}c_{(i,j),1}^\dagger c_{(i,j),2}+c_{(i,j),1}^\dagger c_{(i,j),3}+c_{(i,j),2}^\dagger c_{(i+1,j),1}+c_{(i,j),3}^\dagger c_{(i,j+1),1}+\mathrm{h.c.}$$ $i$ for $x$-index and $j$ for $y$-index.


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