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I'm aware that, according to the Pauli principle, identical fermions obey the Fermi-Dirac statistics, so they don't occupy the same state because they just can't, it's simply how they behave. I'm wondering, are there any different interpretation about this that don't postulate it as a principle but describe it in terms of exchanged effective particles, or some other device?

Possibly unrelated, but I thought about this after learning that, in the context of Path Integral formulation, the commutation relations aren't postulated, they are indeed derived from prime principles, so an answer with a similar example would be great.

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    $\begingroup$ For the bosonic case, you might be able to make an argument that Bose-Einstein condensation can't be modeled as distinguishable particles interacting via a force. But I'm probably not the right person to write up such an argument as an answer (if in fact it's possible to make it.) $\endgroup$ Jul 30, 2020 at 13:49

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The Pauli principle is not simply a prohibition on the two particles to occupy the same state, but a symmetry requirement in respect to interchanging two particles. For example, the wave function of two identical fermions should satisfy: $$\psi(x_1, x_2) = -\psi(x_2, x_1).$$ Although it follows from this that the function is zero when $x_1 = x_2$, this condition is more stringent than just requiring that $$\psi(x, x) = 0.$$

Having said that, it is worth noting that Pauli principle does play a role similar to a repulsive force, preventing collapse of materials. Also, some genuinely repulsive forces can be represented as spin. For example, neutrons and protons are made of different combinations of quarks and are not really the states of the same particle. The strong force between them is described as exchange of mesons. Yet, one can consider them as two isospin states of the same particle and describe their interaction by an effective (spin-dependent) force.

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    $\begingroup$ I've removed some comments that seemed to be going in a not-constructive direction. Be kind, friends. $\endgroup$
    – rob
    Jul 30, 2020 at 13:43
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Wavefunctions of identical fermions have to be antisymmetric, and this antisymmetry leads to an “exchange integral” that is purely quantum in nature, with no classical analogue.

This term contributes to the energy but is not a “force” in the usual sense as there is no carrier of this force.

The spin-statistics theorem is unavoidable to explain the origin of this exchange term but also in bosonic systems to explain bunching. There is no known alternative explanation for the exchange terms in bosonic or fermionic systems.

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  • $\begingroup$ "The Pauli principle forces the wavefunction of identical fermions to be antisymmetric." Shouldn't this be the other way round? I thought the Pauli Exclusion Principle was a corollary of the wavefunction (for a system of identical fermions) being antisymmetric under particle exchange. In the spin-statistics theorem Wikipedia article you have linked, it also states: "The spin–statistics theorem implies that half-integer–spin particles are subject to the Pauli exclusion principle, while integer-spin particles are not." This to me seems like the antisymmetry is more fundamental. $\endgroup$
    – Shrey
    Jul 30, 2020 at 14:11
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    $\begingroup$ @Shrey interesting point. Historically, the Pauli principle predates antisymmetry and led to antisymmetric wavefunctions, not the other way around. It is true that, if you start from the spin-statistics theorem, then the Pauli principle "follows" from that. I edited my answer to (I hope) remove any confusion. Thanks. $\endgroup$ Jul 30, 2020 at 14:18
  • $\begingroup$ The "exchange integral" is an electromagnetic consequence of antisymmetry and not related to any cause of it. $\endgroup$
    – my2cts
    Jul 30, 2020 at 14:43
  • $\begingroup$ @my2cts I'm not sure I understand your comment. The exchange integral is certainly a consequence of antisymmetry, it certainly exists for any potential (and thus interaction). What do you mean by "it" in "not related to the cause of it"? $\endgroup$ Jul 30, 2020 at 14:49
  • $\begingroup$ If there is no interaction potential there is no "exchange integral, but there still is antisymmetry. The OP, quite correctly, does not talk about a potential. "it", unambiguously, refers to "antisymmetry". $\endgroup$
    – my2cts
    Jul 30, 2020 at 16:01

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