Recently, I have got to learn that if time-reversal symmetry and inversion symmetry are present simultaneously in the system we have the following conditions on energy of Bloch's states:

$$E_{n,\chi }(\overrightarrow{k})=E_{n,-\chi }(\overrightarrow{-k})\:\:\:\:\:\:(time\: reversal)$$ $$E_{n,\chi }(\overrightarrow{k})=E_{n,\chi }(\overrightarrow{-k})\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(inversion)$$ $$\Rightarrow E_{n,-\chi }(\overrightarrow{k})=E_{n,\chi }(\overrightarrow{k})\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$

Where, $\chi$ represents the spin-up, and $-\chi$ represents the spin-down electronic states. The third equation is the consequence of the first two equations. This implies that for each band-index($n$), the energy of spin-up and spin-down electronic states are degenerate at a particular $\overrightarrow{k}$ in momentum space.

Does, this mean if we perform the spin polarised DFT calculation on any system, in which we account for up and down electronics states differently, it breaks the time-reversal symmetry of the system?

If the system is non-magnetic, do we get the same band structure for spin-up and spin down electrons if spin polarised DFT calculation is performed? Hence the TRS is not broken?

Please clarify my doubts!
Thank you!


1 Answer 1


Yes, broken symmetry solutions do break time-reversal symmetry, and that's one of the reasons why they are unphysical when the total magnetic moment of the system is 0 (although they are physical when the total magnetic moment is not 0, due to spin polarization). The reason is that, suppose you have a spin polarized $M_S=0$ state, then if you flip the spin of every electron, you get another spin polarized solution with the same energy. But the new state has the same symmetry as the original state, so the two states will generally mix, and since they have the same energy, they will mix in a 1:1 ratio, giving an in-phase combination and an out-of-phase combination. In both linear combinations, the spin densities of the two states cancel out, and you obtain two spin-unpolarized, yet multiconfigurational states.

The very reason that one uses spin-polarized DFT is to get correct energetics, e.g. correct $J$ parameters. This is related to the size inconsistency of spin-unpolarized DFT. E.g. consider an infinitely stretched $H_2$ molecule. If you do a spin-unpolarized DFT calculation, you retain time-reversal symmetry but get an energy that is higher than the sum of two $H$ atoms; yet if you do a spin-polarized DFT calculation, you get the same energy as the sum of two $H$ atoms, but you break time-reversal symmetry. Usually energies are more important than time-reversal symmetry, so people usually prefer to do spin-polarized calculations (whenever the system is so large that beyond-DFT approaches are not feasible), and if necessary, qualitatively or semiquantitatively discuss the changes of the result if we could restore the time-reversal symmetry. There is a nice review by Weitao Yang et al. on this topic.


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