I've so far always been told, that the symmetrization requirement is an axiom on the level of the Schrödinger equation and the statistical interpretation of the wave function (or it's absolute value). Some time ago, however, I found the following little calculation (which I modified a little bit, but it is hopefully still correct):
Let $\Psi \left(\vec{n_1},\vec{n_2}\right)$ be the wave function of a two particle system and $\vec{n_1}$ and $\vec{n_2}$ be the quantum numbers of the particles. Now if the two particles are identical (i.e. indistinguishable), we shouldn't be able to observe any changes when exchanging their quantum numbers, which leaves us with:
$$
{\left|\Psi \left(\vec{n_1},\vec{n_2}\right)\right|}^2={\left|\Psi \left(\vec{n_2},\vec{n_1}\right)\right|}^2
$$
Now we can conclude:
$$
\Psi \left(\vec{n_1},\vec{n_2}\right)=\text{e}^{i\delta}\Psi \left(\vec{n_2},\vec{n_1}\right)
$$
I.e. the wave function acquires a factor $\text{e}^{i\delta}$ when we exchange its arguments. Exchanging the arguments again, leaves us with:
$$
\text{e}^{i 2\delta}=1\ \therefore\ \text{e}^{i\delta}=\pm1
$$
Which basically is what the Pauli principle states.
If this calculation is correct, shouldn't the Pauli principle be regarded as a consequence of the indistinguishability of identical particles and the statistical interpretation, rather than an axiom?
Best Answer
This argument just replaces one axiom by another.
It assumes that if a quantum system consists of identical particles, then the state of the system should not change (it get's multiplied by a phase) under exchange of quantum numbers.
Although this is (perhaps) a more intuitive way of thinking about states of identical particles, it's still a strong assumption in the model that doesn't follow from the other axioms.
The fact is that no matter what you do, you're going to need some extra logical input to deal with systems of identical particles.