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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".

Tobias Fünke
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Aqualone
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    Related: https://physics.stackexchange.com/q/44712/50583 – ACuriousMind Oct 11 '22 at 19:00
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    The Pauli exclusion principle, in effect (at least for non-relativistic theories), constraints the Hilbert space of $N$ identical fermions to the $N$-fold anti-symmetrized tensor product of the one-particle Hilbert space; this is independent of the Hamiltonian, i.e. whether or not the particles interact. – Tobias Fünke Oct 11 '22 at 19:01

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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.

AureySteader
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  • What is a flavor of fermions? Can you explain how you obtain the dimensions of your Hilbert space? Do you mean that the single-particle Hilbert space is of dimension $N$? Then you're considering the Fock space, no? Because this has dimension $2^N$ then. – Tobias Fünke Oct 11 '22 at 20:26
  • @AureySteader So are you saying that one can also reduce a hard-core-boson model to a single particle problem? – Aqualone Oct 11 '22 at 20:42
  • @Aqualone Unfortunately, in general, no. Enforcing the hard-core constraint on bosons can be performed by adding a term $U \sum_{a} (\psi^\dagger_a \psi_a - \frac{1}{2})^2$, $U \rightarrow \infty$, which penalizes occupying more than $1$ boson per flavor. This term is an interaction. You can do it in some cases in one dimension, via the Jordan-Wigner transformation that maps spins (hard-core bosons) into electrons (fermions). – AureySteader Oct 13 '22 at 23:06
  • @JasonFunderberker Flavor = type of fermion. That index $a$ I call flavor can label spin, momentum, position, band index, etc. There are $N$ total fermions, so the Hilbert space is dimension $2^N$. But the matrix $H_{ab}$ is still $N \times N$. – AureySteader Oct 13 '22 at 23:08
  • I don't see that the Hilbert space of $N$ fermions is $2^N$ dimensional. If the single-particle Hilbert space has dimension $d$, then the dimension of the $N$-particle space is $\binom{d}{N}$, where $d\geq N$. IMHO your definitions are a bit weird, but this could be only due to a different terminology than I am used to. Perhaps you could provide a resource? – Tobias Fünke Oct 14 '22 at 05:54
  • @TobiasFünke $2^N$ is the dimension of the Fock space, with all possible particles numbers. You have $N$ single particle states, each of which can either be occupied or unoccupied – By Symmetry Aug 04 '23 at 08:33
  • @BySymmetry Yes. Have I said anything contrary? See the first comment... And see the previous comment by the OP of the answer. They seem to label by $N$ the particle number. – Tobias Fünke Aug 04 '23 at 08:49
  • @TobiasFünke $N$ total flavors of fermions, not particle number. Consider this: let's say there are $N=2$ flavors, a fermion of spin up and one of spin down: $\psi_\uparrow$ and $\psi_\downarrow$. Then there are $2^N = 4$ states: the vacuum: $|0 \rangle$, the two single particle states $\psi_\uparrow^\dagger |0 \rangle$ and $\psi_\downarrow^\dagger |0 \rangle$, and the doubly occupied state $\psi_\uparrow^\dagger \psi_\downarrow^\dagger |0 \rangle$. $2^2 = 4$ total states, from $2$ flavors of fermions. – AureySteader Aug 08 '23 at 19:07
  • I still don't understand, but fine, this is not a big problem and probably only due to a different terminology. Anyway in one of your previous comments you said: "There are $N$ total fermions, so the Hilbert space is dimension $2^N$."- and I disagreed here. – Tobias Fünke Aug 08 '23 at 19:23