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For any molecule with open or closed shells considering the electronic state to be the ground state:

  1. Is the exact electron density totally symmetric?
  2. is the Kohn-Sham electron density totally symmetric?

(spatially upon symmetry operations of the molecular point group). I refer to "exact" as the density built using the wave function of the molecular Hamiltonian in the Born-Oppenheimer approximation. And I refer as the KS electron density as the one built by a single Slater determinant solution of the KS equations with the exact exchange-correlation functional or approximations. In principle due to the Hogenberg-Kohn theorems these two electron densities should be the same, nevertheless it seems that they cannot be in certain cases. This is why I distinguish the two densities.

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    $\begingroup$ Symmetric with respect to what (i.e. symmetric with respect to permuting electrons or spatially symmetric with respect to a group operation)? Or do you mean the density matrix itself? $\endgroup$
    – Bertram
    Jan 7, 2021 at 3:56
  • $\begingroup$ I mean, totally symmetric with respect to symmetry operations of the point group of the molecule $\endgroup$
    – Horse time
    Jan 7, 2021 at 4:11
  • $\begingroup$ Yes, if the molecular geometry is in a point group, the electron density will transform as one of the irreducible representations of the point group, with a very few exceptions. For the ground state, it will almost always be the simplest irreducible representation (the A1 representation for C2V, for instance.) This representation is all 1's, so it retains the symmetry under group operations without multiplication by -1. $\endgroup$
    – Bertram
    Jan 7, 2021 at 4:53

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