Why is there no exchange interaction when there is no coulomb interaction? The Hartree-Fock energy may be written as
$$
E_{HF}= \langle\Psi| H |\Psi\rangle = \sum_{a} \langle a| h |a \rangle + \frac{1}{2}\sum_{ab} \big( [aa |bb] - [ab|ba] \big),$$
with the exchange term
$$[ab|ba] = \int \mathrm { d } \mathbf { r } _ { 1 } \mathrm { d } \mathbf { r } _ { 2 } \psi _ { a } ^ { * } \left( \mathbf { r } _ { 1 } \right) \psi _ { b } \left( \mathbf { r } _ { 1 } \right) \frac { 1 } { r _ { 12 } } \psi _ { a } ^ { * } \left( \mathbf { r } _ { 2 } \right) \psi _ { b } \left( \mathbf { r } _ { 2 } \right).$$
So if we neglect the electron-electron interaction given by $$1/r_{12}$$ the exchange interaction [ab|ba] would be zero.
My question is:
The exchange interaction is said to be due to the fact that we have indistinguishable particles i.e. that we use a Slater determinant (Pauli exclusion principle).

*

*Why is it then, that the exchange interaction is zero, if we neglect the coulomb interaction? We said the exchange is due to the Pauli principle, not due to the Coulomb repulsion.

*Is it always the case that the exchange is zero, if there is no interaction between particles?

*Is this the reason why it is often stated that „Pauli exclusion is not a fundamental interaction (like electromagnetism or the weak nuclear force)“?

 A: 


*Is this the reason why it is often stated that „Pauli exclusion is not a fundamental interaction (like electromagnetism or the weak nuclear force)“?


Exchange interaction is not in itself an interaction, but a contribution to other interactions (like Coulomb interaction) that arises from indistinguishably of particles.
Note also that the particular form of the interaction, Fock term, arises when the Coulomb term is expressed in terms of single-particle wave functions (either as a Slater determinant or when writing the Hamiltonian in second quantization, in terms of single particle creation/annihilation operators.)


*

*Why is it then, that the exchange interaction is zero if we neglect the coulomb interaction? We said the exchange is due to the Pauli principle, not due to the coulomb repulsion.

*Is it always the case that the exchange is zero if there is no interaction between particles?


Echange does not disappear when we neglect the Coulomb interaction - the many-particle wave function is still antisymmetric in respect to exchanging any two particles. However, now it can be exactly expressed in terms of single-particle wave functions (as a Slater determinant), which can be found by solving single-particle Schrödinger equation.
