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

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.

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.