8
$\begingroup$

I'd like to use the tight-binding model with a linear combination of atomic orbitals.

It should show the full energy bands, as well as the spatial probability distribution for each eigenfunction of the Hamiltonian. Ideally it would be easy to use with many tunable parameters.

Can anyone recommend a software for this?

$\endgroup$
3
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Nov 24, 2021 at 18:39
  • $\begingroup$ +1 and welcome to our new community! Thank you for contributing your question here and we hope to see much more of you! The two downvotes you received here were likely an error, but typically that doesn't happen here. $\endgroup$ Nov 24, 2021 at 20:24
  • $\begingroup$ If you have the orbital coefficients, you can generate orbital-like distributions, and cube files from them. Such can be visualized with many free software, eg VESTA $\endgroup$
    – Greg
    Nov 25, 2021 at 19:01

1 Answer 1

9
$\begingroup$

The sisl python package might be useful to you. It has:

  • Tools to build hamiltonians with user-defined basis orbitals.
  • A very high level API to visualize the most common analysis from your hamiltonian (bands, PDOS, wavefunctions, etc...).
  • Lower level APIs to customize your analysis as you wish.

Here's an example of how you might define a tight binding Hamiltonian for graphene:

import sisl
import sisl.viz
import numpy as np

# Define a basis orbital, with a radial function that is defined by us
r = np.linspace(0, 3.5, 50)
f = np.exp(-r)
orb = sisl.AtomicOrbital('2pzZ', (r, f))

# Build a graphene structure using our orbital as basis for the C atoms.
geom = sisl.geom.graphene(atoms=sisl.Atom(6, orb))

# Initialize a hamiltonian matrix for the geometry
H = sisl.Hamiltonian(geom)
# And set the non zero values.
# In this case, matrix elements of atoms that are closer than 0.1 Ang
# (i.e. the same atom) are assigned 0 eV and connections of less than
# 1.44 Ang (first neighbours) are assigned -2.7 eV. 
H.construct([(0.1, 1.44), (0, -2.7)], )
# To make it more interesting, we will change the on-site energies of
# the two atoms
H[0,0] = -1
H[1,1] = 1

Now, you can go on and use any of the methods of the Hamiltonian (see docs).

However, if you want a quick analysis you can use the high level API of the visualization module (docs). Following I show you some of the quick analysis you can do. I tweak some parameters of each analysis, but there are more parameters for which I'm using the defaults.

  • PDOS:
pdos_plot = H.plot.pdos(
    kgrid=[90,90,1], nE=500, Erange=[-4, 4],
    distribution={"method": "gaussian", "smearing": 0.1}
)

pdos_plot.split_DOS(on="atoms", name="Atom $atoms")

enter image description here

  • Wavefunctions (given your defined basis):
H.plot.wavefunction(
    i=1, grid_prec=0.05, axes="xy", 
    represent="mod", nsc=[2,2,1], zsmooth="best"
)

enter image description here

  • Bands:
# We need to define a path for the band structure
bz = sisl.BandStructure(
    H, points=[[0, 0, 0], [2/3, 1/3, 0], [1/2, 0, 0]],
    divisions=30, names=["Gamma", "M", "K"]
)

# Then just plot it
bz.plot()

enter image description here

  • Fatbands:
fatbands = bz.plot.fatbands()

fatbands.split_groups(on="atoms", name="Atom $atoms")

enter image description here

If you have doubts you can go through the documentation or ask specific questions on the discord server.

$\endgroup$
3
  • 1
    $\begingroup$ Welcome to the site and a great first couple of answers! $\endgroup$
    – Tyberius
    Dec 14, 2021 at 21:34
  • 2
    $\begingroup$ +1 - I've been looking for something like this for teaching - thanks! $\endgroup$ Dec 21, 2021 at 15:46
  • $\begingroup$ Yes, exactly! I believe it is a great tool for students to play around since they can quickly see the consequences of changing parameters, etc... I myself have learned a lot by using it :) $\endgroup$
    – Pol Febrer
    Dec 21, 2021 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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