This tutorial covers the basics for spin-polarized and open-shell calculations in QimPy, using the most basic system of all, a hydrogen atom.
First, lets set up a hydrogen atom calculation exactly as we would based on the previous tutorials. Save the following to Hatom.yaml:
lattice: system: cubic modification: face-centered a: 20. # bohrs ions: pseudopotentials: - SG15/$ID_ONCV_PBE.upf coordinates: - [H, 0., 0., 0.] electrons: basis: ke-cutoff: 30.0 checkpoint: null # disable reading checkpoint checkpoint-out: Hatom.h5 # but still create it
(qimpy) $ python -m qimpy.dft -i Hatom.yaml | tee Hatom.out
Since there is only one atom, we don’t need geometry optimization. Notice that the final energy F = -0.4601 Hartrees, which is rather different from the analytical exact energy -0.5 Hartree (= -1 Rydberg = -13.6 eV).
The reason for this disrepancy is that, by default, this DFT calculation is spin-unpolarized, that is it assumes an equal number of up and down spin electrons. This assumption is correct for the water molecule with a closed shell of 8 valence electrons that we dealt with so far, but is incorrect for the hydrogen atom which has only one electron. This electron must be either an up or down spin, so that the magnetization (Nup - Ndn) is +1 or -1. We can invoke a spin-polarized calculation and specify the magnetization by adding the following key-value pairs to Hatom.yaml and rerun QimPy:
electrons: basis: ke-cutoff: 30.0 spin-polarized: yes fillings: M: 1
Now we find F = -0.4997 Hartrees, in much better agreement with the analytical result. Check that using magnetization -1 produces exactly the same result.