Hi Nicola, I would like to express my opinion on this very interesting topic that you mentioned. These are just my very arguable ideas or what I think I have understood from a little thinking and studying of this problem.
Let's start from B and H2+ or 1 e- TM atoms with only 1 d state occupied (e.g. Sc or Y): I think that the "real" ground state is cylindrical. So for example in the H2+ molecule at large interionic distance one should get the localization of the e- on one of the protons. The symmetry of the problem (and of the potential) is re-established on a statistical level because the physical situation can be described by a statistical mixture of states with 1 e- on one proton or the other one that have the same probability. So in the case of B it would be a misture of px, py, pz with weights of 0.333 each (BTW this case is a bit like the Scroedinger cat because any cartesian coordinate system you fix is arbitrary; you would probably need a filed to break this equivalence but in that case you would perturb the system....). Now problem is: would should we expect from DFT? I think this is at the core of the old problem that DFT "prefers" the solution where the charge density has the same symmetry of the potential (a "spherical" one for B) instead one of the possible and equivalent materializations of the statistical mixture for which symmetry is broken. Why? I think because a) you would need to break the symmetry "by hand"; b) there is a finite, negative contribution to the energy (coming from an approximate Hxc potential) even for one electron systems that shouldn't be there and gives lower energy for the symmetrical solution (e.g. the electron delocalized over the H2+ molecule at any distance). In fact corrections as "simple" as SIC can solve these problems (I guess/hope). Situation is more involved with multi-electron systems like, for example H2. If you pull the two nuclei apart one gets (in real life) 1 e- localized on each side. What is the spin of these two e- (of course they were in a singlet state at short distances)? Any relative orientation is good. So again I think the infinite distance solution is a statistical mixture of 4 states with equal probability. How this solution connects to the one at finite distances? I think it continuously connects to a multi-reference ground state described by a combination of slater determinants. What to expect from DFT? I think that to get the best out of DFT one has to impose a broken symmetry solution (e.g. an antiferromagnetic one for H2); this corresponds in fact to one of the states in the statistical mixture of the infinite distance case and to one of the slater determinants in the real ground state wavefunction. How about the charge density? If we sum over the spin it still has the symmetry of the problem. If we had the perfect DFT I think that, with a broken symmetry solution (that has the advantage also that the two body density matrix rho(x,x')-->0 for large interatomic distances), we would get the right energy and density (not wfc that is a fake one) because that is what DFT is meant to do. if we have a symmetrical solution with each electron split between both sites I think this won't come out right even with a perfect xc functional (which nobody has). For multi-electron, multi-states systems of course the situation is more complicated than this. But I would prefer a broken symmetry solution in any case (and hopefully a better xc functional). hope this helps, Matteo Nicola Marzari wrote: >> I agree that symmetrization (due to the periodization) induces a >> degeneracy but the >> "real" system that I want to simulate (a dimer alone in space) has even more >> symmetries. In particular it has a rotational symmetry that will never >> be there >> > > > Well, this is the part I've never understood - that's why I was > mentioning the boron atom. I wouldn't necessarily agree that your > dimer charge density should have the same symmetry of the ionic > potential - as much as a dissociated H2+ molecular ion doesn't have > mirror symmetry on the plan bisecting the bond, as you dissociate the > molecule, or antiferromagnetic MnO has a lower electronic-structure > symmetry, since the Mn are inequivalent from the point of view of spin > density. > > Also, the potential in atoms is spherical, but I'm sure there are > solutions lower in energy that are cylindrical. Think at a transition > metal with only one d electron: do you fill up 1 orbital 1.0, 2 orbitals > 0.5, 3 orbitals 0.33333, 5 orbitals 0.2 ? Only the last is spherical, > all others are cylindrical, although of course an ensemble of > measurements on atoms will always converge on the spherical limit. > > Am I the only one worrying about this ? It seems a key problem > that I've never seen addressed - Englisch and Englisch in 1983 > had a paper on the fact that fractional occupations are somewhat > not compatible with v-representability, but I don't recall it very well. > > > nicola > > > --------------------------------------------------------------------- > Prof Nicola Marzari Department of Materials Science and Engineering > 13-5066 MIT 77 Massachusetts Avenue Cambridge MA 02139-4307 USA > tel 617.4522758 fax 2586534 marzari at mit.edu http://quasiamore.mit.edu > _______________________________________________ > Pw_forum mailing list > Pw_forum at pwscf.org > http://www.democritos.it/mailman/listinfo/pw_forum > -------------- next part -------------- A non-text attachment was scrubbed... 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