Why does the S squared value have no meaning in Density Functional Theory?

Question

It is not the first time that I stumble upon the affirmation that spin and the $S^2$ value, under DFT, has no "real physical meaning". Why?

If that is true, why are we using it all the time with UKS?

Answer 1

The Kohn-Sham wavefunction is the wavefunction of a hypothetical system where:

(1) there is no interaction between the electrons,

(2) the electrons are subject to a repulsive potential (the Hartree-exchange-correlation potential) in addition to the nuclear attraction potential, and

(3) the effect of (2) effectively counterbalances the effect of (1), in the sense that the total electronic density of the hypothetical system is exactly the same with the density of the corresponding real electronic system. Usually we are talking about spin density functional theory, which not only requires the total density to be equal, but also that the alpha electronic density and the beta electronic density are both equal to their exact counterparts.

As we can see, there is no requirement that the $S^2$ value of the hypothetical system be the same as that of the real system. The corollary is that a spin-pure electronic state may correspond to a Kohn-Sham wavefunction that is spin-contaminated. And this does happen frequently, as can be shown from the following argument: imagine an exact electronic eigenstate whose $S=1/2$, and we plot its spin density. The spin density will have positive values in most parts of the space, but can adopt small negative values in certain regions (e.g. near atomic nuclei) due to spin polarization. A spin-adapted (restricted open-shell) Kohn-Sham determinant however can only give an everywhere non-negative spin density. This proves that the exact Kohn-Sham wavefunction must be spin-contaminated, so that a ROKS wavefunction is less theoretically justified than a UKS wavefunction.

That being said, the $\langle S^2 \rangle$ value can still be used to characterize the reliability of a DFT calculation. This is because, while the exact DFT functional performs equally well (by definition, always exactly) for all electronic systems, essentially all approximate functionals only perform well when the exact wavefunction is mostly single-determinantal, and heuristically, the exact wavefunction is usually manifestly multi-determinantal when the corresponding Kohn-Sham wavefunction is heavily spin-contaminated. Thus, while the $\langle S^2 \rangle$ value in DFT is not rigorously tied to any physically observable quantity, one can still arrive at meaningful conclusions about the reliability of a DFT calculation using the $\langle S^2 \rangle$ value.