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Dear VASP community,

We are trying to wrap our head around the question if and how a DFT+U calculation (utilizing the Hubbard model via LDAU) in VASP is generating meaningful outcomes for non-spin-polarized calculations (ISPIN=1). (Assuming that sensible U parameters were chosen and the systems have d- or f-valence shells)

Looking at the different types and work equations of LDAU, it appears that for at least for LDAUTYPE=2 and LDAUTYPE=3 spin-polarization would need to be enabled by setting ISPIN=2, but nothing is preventing the calculation to be performed with ISPIN=1. (albeit the resulting energies change a bit)

Thus, VASP apparently has all necessary information about the electrons for performing the +U correction even without explicit spin-polariziation?

For context: Other periodic DFT codes do enforce spin-polarization when performing DFT+U. From this observation our discussion has started

Hi,

Technically, with LDAUTPYE=1 and 2, choosing ISPIN=1 seems ok, because the results are as expected identical to what is obtained with ISPIN=2 for non-magnetic systems (this is a sanity check). LDAUTPYE=3 should probably not be used because it is not mentioned at LDAUTYPE, and in addition the results are (for a non-magnetic system) completely different depending on ISPIN, which is of course not logical. Concerning LDAUTYPE=4, calculations crash with ISPIN=1.

From the physical point of view, it may sometimes make sense to apply DFT+U to non-magnetic systems (e.g., TiO2, ZnO); I recommend to check the literature.

A good review on DFT+U is http://dx.doi.org/10.1103/PhysRevB.79.035103.

Thank you for the insight! That's very helpful to us :-)

Hi,

What I wrote about LDAUTYPE=3 in my previous answer is wrong. The use of LDAUTYPE=3 is in fact explained at wiki/index.php/Calculate_U_for_LSDA%2BU, where the following is mentioned:
"Note that for LDAUTYPE=3 the LDAUU and LDAUJ tags specify the strength (in eV) of the spherical potential acting on the spin-up and spin-down d-manifolds, respectively.".

I have tested for LDAUTYPE=3, that if LDAUU and LDAUJ are chosen equal, then the results obtained with ISPIN=1 and 2 are identical. That should mean that using LDAUTYPE=3 with ISPIN=1 is ok, at least technically.