(2.2.13) Baerends : So in what sense are Kohn-Sham orbitals better than Hartree-Fock orbitals? When the energy of the determinant of Kohn-Sham orbitals is calculated with the full Hamiltonian, its energy is of course higher than the Hartree-Fock energy, but actually by only a tiny amount. On the other hand, the Kohn-Sham density, being equal to the exact one and not so diffuse as the Hartree-Fock one (in molecules), leads to much improved (more negative) electron-nuclear energy. Also the orbital shapes are "better" than the too diffuse Hartree-Fock orbitals (in molecules), so the kinetic energy is also considerably better (higher). The errors of the Hartree-Fock model for these two energy terms are large, in molecules often larger than the bond energy, and they rapidly increase upon bond lengthening. They cancel to some extent, which is why they are not so readily recognized. The Hartree term is of course also better (exact) with Kohn-Sham orbitals and density.