Does "non-interacting" (fermions) really mean "no interactions other than Pauli exclusion"?

Question

When one speaks of non-interacting elections (or other ferimons), doesn't one technically mean non-interacting but with the exception of Pauli exclusion? I wonder if it is appropriate to view Pauli exclusion as essentially an infinitely strong short ranged interaction, the same as the condition one imposes to create a model of "hard core bosons".

Answer 1

Not really. The concept of 'non-interacting fermions' refers to the fact that a many-body Hamiltonian quadratic in fermion operators can be reduced to a single-particle Hamiltonian, where the wavefunction and energy of a single fermion is unaffected by the others. To see this, consider Hamiltonian $$ H = \sum_{ab} H_{ab} \psi^\dagger_a \psi_b, $$ then instead of analyzing the entire Hilbert space ($2^N$ dimensional, with $N$ flavors of fermions), one can diagonalize the matrix $H_{ab}$ ($N$ dimensional) to get $$ H = \sum_{\alpha} E_\alpha \gamma^\dagger_\alpha \gamma_\alpha, $$ with $E_\alpha$ the eigenvalues. The spectrum is very simple now. Occupying a fermion $\gamma_a^\dagger$ will increase the energy of the state by $E_a$, regardless of whether other flavors $\gamma_\beta$ are occupied or unoccupied. This is what we mean by non-interacting; the presence/absence of a fermion operator $\gamma_\alpha$ will change the energy by $E_\alpha$: there is no effect of other fermions on this fact. Pauli exclusion is built into the fermions, and the property that the many-body Hamiltonian reduces to a single particle (noninteracting) Hamiltonian is unaffected by this property. You could make the same argument for bosons, with the minor change that you can have more than one boson per flavor.

When we refer to interactions, we mean terms of the form such as $$ \sum_{ab} V_{ab} \psi^\dagger_a \psi_a \psi^\dagger_b \psi_b,$$ which ruin this single-particle decomposition.