Someone studying the exchange hole in a Slater determinant argues:"Hartree-Fock…
PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Atomic, Molecular & Optical Physics · Many-electron atoms and the periodic table
Deriving the Spectrum, Ch Self-consistent fields and correlation
Problem
Someone studying the exchange hole in a Slater determinant argues:"Hartree-Fock exchange makes each electron dig a hole in the density of the other electrons, and this hole lowers the energy. The two electrons of H− (one spin ↑, one spin ↓) are both present in the Hartree-Fock determinant, so each should dig a hole around the other, exactly as the same-spin electrons do in helium's triplet state. Hartree-Fock's energy for H−, −0.4879Eh, should therefore already include as much mutual avoidance between the two electrons as is physically possible. The fact that this lies above the −0.5Eh needed for binding (the energy of a free H atom plus a free electron at rest) must be some other failure of the method, not a missing hole."
((a)) Identify the error in this reasoning.
((b)) State which pairs of electrons in a Slater determinant actually carry an exchange hole, and why.
((c)) Explain what physically distinguishes the avoidance between H−'s two electrons that Hartree-Fock does include from the avoidance that is missing.