Up till now we have been assuming that we always had some set of bands which could each fit two identical electrons. Essentially we have been ignoring the electron spin. If you want to examine, for example, a magnetic system then the spin of the electrons is important. It can also be important in modelling atomic or molecular systems. We’ll cover several examples of this in this lab.

## The Hydrogen Atom and electron spin

As a first example, let’s look again at hydrogen. In the molecule we have two hydrogen atoms bonded together, and two electrons in total, and we can do this with a standard calculation.

But if we instead want to find the energy of an isolated H atom accurately it’s a little more tricky. A hydrogen atom has one electron, which means if we treat it with doubly-degenerate bands, we would have a single half occupied band. To calculate it like this we can treat it as a metal, and use some small smearing, so that we allow partial occupation of bands. This is equivalent to assuming that we have half an electron in each spin state. If we however restrict it to being in one spin state or another, we may find a slightly different energy (coming primarily from differences in how the DFT exchange term is calculated in each case, with the latter being more physical).

To perform a spin-polarized calculation in quantum espresso with the `pw.x`

code, there are two additional variables you’ll need to set.

`nspin`

: this is 1 by default so no spin polarization is taken into account. To perform a spin polarized calculation it should be set to 2.`tot_magnetization`

: this is difference between the number of spin-up and spin-down electrons in the cell. If we want a single spin up electron we can set this to`1.0`

.

The directory `01_H1_metal`

has an input file for a single
H atom using a small smearing, while the directory `01_H1_spin`

has the same calculation, but with no smearing, and we have used the two
input variables above to enable a spin polarized calculation.

- Run the input files in these two folders.
- Compare the total energy obtained in each case. In which case is the energy lower?
- Compare the energies of the lowest energy calculated bands.
- Enabling smearing for the “metal” calculation will automatically add extra bands, but only the lowest energy band will be occupied in this calculation.
- In the spin polarized calculation if you check the output you will see two sections for the band energies. One listing the energies of the “spin up” bands, and the other listing “spin down” bands. Since we have said we want 1 spin up electron in the calculation, the spin up band will be occupied and should be lower in energy than the unoccupied spin down band.

- Slightly above the band energies in the output of the spin polarized calculation, you’ll see that quantum espresso also outputs the magnetic moment associated with each atom in the system. And slightly below the final total energy output, it will list the total magnetization of the system in Bohr magnetons per cell. The measured value for hydrogen is 1. How close are you here?

## The Oxygen Molecule

Even if a system is not necessarily magnetic and we might imagine that representing it with some set of fully occupied, doubly degenerate bands will work, including spin polarization can lead to important differences. One example of this is the O2 molecule.

In this case, we have a system with two interacting oxygen atoms. Each oxygen has 8 electrons in total, with the configuration 1s2 2s2 2p4 (the 1s orbital will be contained within the pseudopotential for the DFT calculations done here, so you will have 6 electrons from each oxygen atom). For a single oxygen, from Hund’s rule the three p orbitals should be filled singly before being filled in pairs, so that one of the p-orbitals will have two electrons, and the other two should have one each. However, if we assume doubly occupied orbitals, we’ll have the two p-orbitals with two electrons and one that is empty. This means a calculation where we assume a set of doubly occupied bands will have trouble converging to the ground state of the system. For the molecule the situation is similar, but the s and p orbitals from each atom combine to form bonding and anti-bonding sigma and pi orbitals.

The directory `02_O2`

contains an input file to calculate the total
energy of the system at the measured bond length. Here the calculation has
been set up exactly as you’ve seen in the past.

- Try running this calculation and see what happens.

While it’s possible that the system may randomly meet the convergence criteria in the self-consistent cycle, this calculation will most likely not converge. If you look at the estimate accuracy at the end of each iteration in the output, it will likely vary from step to step, rather than steadily decreasing as in a well-behaved calculation.

The situation we have is similar to a metal: we have two bands and the ground state of the system should be when there is one electron in each of them.

- Create a copy of the
`02_O2`

directory called`02_O2_metal`

. Modify the input file in it to use a metallic occupation scheme with a small smearing width and run the calculation. The relevant input variables have been used in the H atom example discussed above.- Does the calculation now converge?
- Take a look at the file
`pwscf.xml`

in the calculation directory, and try to find the occupations of each band at each k-point. Are these as expected?

As with the hydrogen case above however, treating this system as a metal may not get the physics of the system right.

- Create another copy of
`02_O2`

called`02_O2_spin`

. This time modify the input file in it to turn on spin polarization, as was done in the hydrogen example. This time try setting the total magnetization to 0, which would be the case if we don’t have any net magnetization in the molecule, as both spins point in opposite directions. Then try setting the total magnetization to 2.0, which corresponds to both spins pointing in the same direction.- How does the total energy compare for each case compare to the metallic scheme? Which is the more energetically favourable configuration? How do the orbital energies vary?

## Magnetic Systems - Iron

Now that you’ve seen how including spin polarization can allow us a correctly describe the ground state of our system in your calculation, the next step is to use it to describe a magnetic system.

In a magnetic system there is a net spin polarization in the unit cell. This means that we’ll probably have an odd number of electrons, and the energy of the system when we include a net spin polarization is lower than the energy when we don’t.

One of the most common magnetic systems is iron, so we’ll examine this.
The directory `03_Fe`

contains an input file for iron. Note this is
a BCC structure (as set by `ibrav = 3`

in the input file), whereas most of the
crystals structures you have examined previously were FCC. The calculation
has been set up in the usual way for a metallic system.

- Run this calculation and check everything worked as expected.
- Now make a copy of the calculation directory and in this, modify the
calculation to turn on spin polarization. Try running the calculation
with
`tot_magnetization = 0.0`

first, and compare your total energy to that obtained using doubly degenerate bands.**Note**while in the case of the atoms and molecules above, we were able to get our calculations to at least converge my using a metallic occupation instead of using spin polarization, in the case of iron, it will still be a metal when you use spin polarization, so you should not remove the input variables associated with this.- The total energies should agree within the accuracy of the calculation.

- Now try setting the total magnetization to 1.0 and see how total energy
changes.
- Which is the more energetically favourable configuration?

- Try setting the total magnetization to 2.0.
- How does this compare to the previous value?

From this we could test many guesses for the total magnetization, and find
the value which gives the lowest overall total energy. However, we can instead
pass an option that tells quantum espresso to automatically find the best
value. This is done by setting the `starting_magnetization`

input variable.

Make another copy of the

`03_Fe`

directory, and this time set`nspin = 2`

, and`starting_magnetizaton = 1.0`

(do not include the`tot_magnetization`

variable as this fixes a value). Run the calculation and see what the final total magnetization per cell is. See if you can find a measured value for iron to compare to.See if you can use what we covered in previous labs to calculate and make a plot of the electronic band structure of BCC Fe.

- Plot the spin-up and spin-down bands in different colours.
- There are different Fermi energies for the spin-up and spin-down bands: indicate these on your plot in some sensible way.
- As the Brillouin zone is different to the one’s you have calculated so far you’ll need to select a few sensible high-symmetry points yourself to plot with.

## Summary

In this lab you have seen:

- how to do a DFT calculation including spin polarization.
- how some systems need to be done with spin polarization to converge to the correct ground state.
- how to use spin polarized calculations to find the correct magnetization of a magnetic system by letting the code find the total magnetization that produces the lowest overall total energy.