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If I do two calculations for some energy difference, one with spin polarization and one without, is it always true that the spin polarized one is more "accurate" in some sense? That is, will it always either give the same as the non-spin polarized calculation or something that is physically better, or are there situations where the spin polarized calculation can give a worse physical description of some system?

no, the accuracy is the same on both levels. Which solution is more reasonable
only depends on whether the system really is magnetic or not:
-- if it is (either FM of AFM), the calculation has to be done spin-polarized, and the total
energy of the cell will be more negative than for the non-polarized calculation
-- if it is not, the magnetic moments will be 0 and the energy should be the same
as for the non-polarized cell, even is ISPIN=2 is set.

Thanks. So, spin polarized calculations should always give the lowest energy configuration, regardless of the type of system. It will be the same as the non-spin polarized calculation if the system is non-magnetic. The only reason not to do spin-polarized calculations on every system, then, is probably due to speed and efficiency of the code being worse for such calculations?

I am calculating a saddle point energy for a vacancy migration process and am finding that with ISPIN=2, the energy for the saddle is actually higher than for a non-spin polarized calculation. For every minimum I've calculated, the energy is either the same or lower with ISPIN=2. I'm not quite sure what this means. Is the ISPIN=2 calculation not converging correctly (though the forces went to "zero" as expected)? Any toughts would be greatly appreciated.

I've seen a similar issue and it was related to the high initial spin state that is the default for vasp. The solution for me was to start with no spin on any atom, by setting MAGMON=0 0 0 0 ... in the INCAR files (one 0 for each atom in the system). Then using ISIN=2 relaxes the spin from the non spin polarized density, and the energy should go down or be the same if the net spin is optimally zero. We were looking at Pd on MgO. Joel Kress saw this problem for (I think) FeO2, and pointed me to the solution.