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Could the Fermi statistic be a force?

Physics Asked on October 1, 2021

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.

2 Answers

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.

Answered by Roger Vadim on October 1, 2021

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.

Answered by ZeroTheHero on October 1, 2021

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