How to derive Pauli Exclusion Principle without assuming anti-symmetry?

Question

So, it appears that the statement of the Pauli Exclusion Principle is equivalent to the statement that fermions are anti-symmetric. That is, if you assume that fermions are anti-symmetric, then you can derive the Pauli Exclusion Principle. If you assume the Pauli Exclusion Principle, you can derive anti-symmetry in the wavefunction.

At some point, one of these must have been proved independently of the other. I can't seem to find such a proof anywhere. I think this confusion also comes from the fact that spin is a difficult thing to conceptualize, let alone measure. Does some experimental data to necessarily imply one of the two? I guess, some insight on how spin was discovered would probably help.

Answer 1

The Pauli Exclusion principle is determined from experimental facts alone. First, its consequences were observed and then it was incorporated in the theory. See Kaplan´s article for a more erudite answer. Also, see Towler´s exposition. Although, he dwells in foundational issues.

Important: they are not equivalent unless you consider only totally symmtric/antisymmetric functions! The antisymmetry is assumed, it cannot be derived rigorously from particle independence nor tested with experiments.

As for getting to the conclusion that the wavefunction of fermions has to be antisymmetric here is a weak argument without requiring the Pauli Exclusion principle. You could also say that since you cannot follow the trajectory of particles therefore they are indistinguishable -- at least, if they are of the same kind. That is, the permutation of any two of them $$ \Psi(\pmb x_1, \pmb x_2, ...) \to \Psi(\pmb x_2, \pmb x_1, ...) $$ maintains the same physical description $$ |\Psi(\pmb x_1, \pmb x_2, ...)|^2 = |\Psi(\pmb x_2, \pmb x_1, ...)|^2.$$ The permutation can be described as the action of an operator that satisfies $P^2 = 1$. There are only two ways to acomplish this. Either the wavefunction is symmetric (the eigenvalue of $P$ is $-1$) under the exchange or antisymmetric (eigenvalue $-1$). $$ \Psi(\pmb x_1, \pmb x_2, ...) = \Psi(\pmb x_2, \pmb x_1, ...) \quad \text{ Symmetric: bosons}$$ $$ \Psi(\pmb x_1, \pmb x_2, ...) = -\Psi(\pmb x_2, \pmb x_1, ...)\quad \text{Antisymmetric: fermions}$$

Therefore, if $\pmb x_1 = \pmb x_2$ then $\Psi(\pmb x_1, \pmb x_1, ...) = -\Psi(\pmb x_1, \pmb x_1, ...) = 0$. That is, Pauli Exclusion principle.

The reason as to why this argument, that is the one that was elaborated to justify Pauli Exclusion principle and is seen nowadays in most textbooks, is weak is neatly explained in the reference mentioned above.