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Should the presence of a d block element in the unit cell mandate the need for performing spin-polarized calculations? I wanted to find out the DOS, PDOS, and the band structure. I'm new to DFT and I'm using the Quantum ESPRESSO package (where spin polarized calculations correspond to nspin = 2. It would be nice if someone could guide me through doing it.

Thanks in advance.

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  • $\begingroup$ When your system is magnetic, you should do the spin-polarized calculation. $\endgroup$ – Jack Sep 20 at 10:38
  • $\begingroup$ +1. Welcome to our community, and thank you for contributing your excellent question here! We hope to see much more of you !!! $\endgroup$ – Nike Dattani Sep 20 at 16:21
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I agree with Tristan that the fool proof method is to do a spin-polarized self consistent calculation and look at the final magnetic moment. If the final magnetic moment is non-zero you should include spin polarization in your calculations.

However, from a practical point of view you can have a look at the existing databases. In particular if you are working on bulk materials, you can look at the materials project database. For example according to the database $\ce{BaTiO_3}$ is non magentic and $\ce{BaFeO_3}$ is magnetic. This will give you an idea prior to doing calculations and might be useful if you are screening for materials.

PS: As suggested by Andrew, you should not rely on DFT databases too much with regards to magnetic behavior. They almost all use a high-spin magnetic initialization, which is entirely arbitrary. It is possible that the magnetic ground state is something entirely different (potentially even a state with no magnetic moments at all).

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    $\begingroup$ You should not rely on DFT databases too much with regards to magnetic behavior. They almost all use a high-spin magnetic initialization, which is entirely arbitrary. It is possible that the magnetic ground state is something entirely different (potentially even a state with no magnetic moments at all). $\endgroup$ – Andrew Rosen Sep 21 at 2:26
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    $\begingroup$ Yes. I am not suggesting to take it for granted. Might give an idea, what to expect. I will add your comment to the answer since its important $\endgroup$ – Thomas Sep 21 at 2:31
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    $\begingroup$ Agreed. It's a quick and easy thing to check, which is always nice. $\endgroup$ – Andrew Rosen Sep 21 at 4:51
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    $\begingroup$ As a representative of Materials Project, I just want to add that while we have not predicted magnetic orderings for all materials yet, we are working through these slowly (to @Andrew Rosen's point, the orderings are not always arbitrary). This is following the method in High-throughput prediction of the ground-state collinear magnetic order of inorganic materials using density functional theory. However, this is indeed a difficult thing, and I agree that materials databases are no substitute for in-depth exploration of a given material! $\endgroup$ – Matt Horton Sep 24 at 18:30
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    $\begingroup$ Thanks for chiming in, @MattHorton! Yes, this is one of the few examples of a really nice, improved approach! My philosophy with regards to magnetic behavior is generally: check out the Materials Project (or OQMD or whatever you like) for some solid insight. And if you need to be sure of the behavior, then there's only one way to find out! Great paper, by the way :) As an aside, there's also a scheme published in the AMP$^2$ paper for identifying antiferromagnetic behavior in an automated fashion (I have only skimmed this). $\endgroup$ – Andrew Rosen Sep 24 at 20:44
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Unfortunately, without performing spin polarized calculations you cannot know if it will matter. Further, you must perform different magnetic configurations (AFM, FM, PM) to know which is stable.

If you can experimentally show if its magnetic or not this will help. Certain elements tend to be magnetic as well. If you have Ni, Co, or Fe then you almost definitely need spin polarization.

In short, you cannot know without checking. You can save time by starting from non-spin polarized calculations though.

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