# Why does the exchange interaction in Hartree-Fock theory lower the total energy?

In Hartree-Fock theory, the expectation value of the total energy can be written as

$$E = \langle\Psi| H |\Psi\rangle = \sum_{a} \langle a| h |a \rangle + \frac{1}{2}\sum_{ab} \big( [aa |bb] - [ab|ba] \big),$$

where the first summation contains the single electron kinetic energy and the interaction energy of the external field. The second summation contains a classical Coulombic term plus an exchange term. The exchange term is introduced by the antisymmetrization of the wavefunction (through a Slater determinant) and takes the form of

$$[ab|ba] = \int \mathrm { d } \mathbf { r } _ { 1 } \mathrm { d } \mathbf { r } _ { 2 } \psi _ { a } ^ { * } \left( \mathbf { r } _ { 1 } \right) \psi _ { b } \left( \mathbf { r } _ { 1 } \right) \frac { 1 } { r _ { 12 } } \psi _ { a } ^ { * } \left( \mathbf { r } _ { 2 } \right) \psi _ { b } \left( \mathbf { r } _ { 2 } \right).$$

This integral is always positive, and so the exchange contribution to the total energy is always negative. Can anyone explain why the exchange contribution to the total energy is negative? In Fermi-Dirac statistics, we can see that the effect of Pauli exclusion of indistinguishable Fermions is an effective outward pressure. Hartree-Fock theory seems to suggest that the exchange interaction lowers the energy of the system, and therefore stabilizes it. Shouldn't the effect of exchange, therefore, be an effective inward pressure?

• Would you check if this is answered by chemistry.stackexchange.com/a/61185/39153 ? It's quite a bit to read, but the relevant bit is in there: the $[aa|bb]$ term is valid for a Hartree product - it intuitively cannot be valid on its own for a Slater determinant. The latter disallows some combinations of orbitals/orbital forms because it does account for Pauli exclusion. $[aa|bb]$ does not accurately calculate the resulting energy by itself, so $[ab|ba]$ must be included to correctly lower the energy. – TAR86 Apr 10 at 5:38
• I understand how the mathematics of a Slater determinant give rise to the form of the exchange integrals. I am having difficulty reconciling the fact that if we used a Hartree product, which is equivalent to removing exchange antisymmetry/Pauli exclusion, then the total energy is higher. This suggests bonding would be weaker if Fermion exchange antisymmetry did not exist. How could the same principle that is responsible for stronger bonds also cause an effective outward pressure in Fermi statistics? – CuriousChemStudent Apr 10 at 19:24
• @CuriousChemStudent I believe the common explanation of why exchange lowers the energy is that it allows the electron motion to be somewhat more correlated. Same spin electrons will avoid each other due to the exchange interaction, reducing the unfavorable electron-electron repulsion and thus lowering the energy. – Tyberius Apr 13 at 1:55
• Thank you, I get it now! There is an effective outward pressure created by a reduction in effective volume cause by Pauli exchange. The volume that is excluded is entirely the highest energy configurations of phase space. Therefore, the electrons correlated by Pauli exchange do not on average exist in locations of high energy with respect to each other. This causes an effective reduction in energy. – CuriousChemStudent Apr 17 at 6:20
• @CuriousChemStudent that's how I've understood it. You could convert your comment to an answer with a bit more detail and I think it would be well received. – Tyberius May 15 at 13:36