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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?

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    $\begingroup$ see the answers to physics.stackexchange.com/q/73670 $\endgroup$ – Valter Moretti Dec 16 '13 at 19:26
  • $\begingroup$ Combining your answer and joshphysics', I now conclude, that in the case of a two particle system, the axioms (Pauli principle and the assumption, that the state does not change when exchanging quantum numbers) are equivalent. But in the case of a system with more than two identical particles, the Pauli principle is "stronger" and must therefore be used in order for the theory to describe nature correctly. $\endgroup$ – user35915 Dec 16 '13 at 22:13
  • $\begingroup$ @user35915 Even in the case of more than two particles, I think you might be able use a similar argument to show that they're basically equivalent because any permutation can be decomposed into a product of transpositions, but I haven't thought this through completely. $\endgroup$ – joshphysics Dec 16 '13 at 22:41
  • $\begingroup$ In the case of more than two particles you can still show, that the wave function will simply change signs under the exchange of any pair of its arguments, but, if I got VM9's explanation, the problem is that you cannot show that it will always change signs the same way. I.e. if I exchange arguments 1 and 2 I might get a minus, but if I exchange 1 and 3 I might get a plus. In other words: one cannot show, that the wave function has to be FULLY symmetric or antisymmetric as is stated by the Pauli principle. Which is, why I called it "stronger". $\endgroup$ – user35915 Dec 17 '13 at 6:46
  • $\begingroup$ @user35915: Yes you have understood well. Following that way it impossible to prove that wavefunctions of more than two entries must be either completely symmetric or completely antisymmetric. The modern version of Pauli's principle requires completely antisymmetry of that state of $n$ fermions, so the argument in the question is not equivalent to Pauli principle but is weaker. In particular it cannot distinguish between statistics and parastatistics. $\endgroup$ – Valter Moretti Dec 17 '13 at 7:40
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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.

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The modern version of Pauli's principle requires completely antisymmetry of a state of $n$ fermions. Instead, the argument discussed in the body of the question only implies that a state of $n$ fermions has to be either symmetric or antisymmetric under interchange of a pair of particle. It is impossible, following this way, to prove that the full state is either completely symmetric or completely antisymmetric, since different pairs of particles in the state could have different character, giving rise to a so called parastatistics. Actually, in 1+3 dimension, parastatistics are forbidden as a consequence of the Poincaré covariance of the theory (it is the celebrated spin statistics theorem).
See also my answer to Is the symmetrisation postulate unnecessary according to Landau Lifshitz?

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The axiom that probability density is symmetric: $$ |\psi(\mathbf r_1,\mathbf r_2)|^2 = |\psi(\mathbf r_2,\mathbf r_1)|^2 $$ only implies $$ \psi(\mathbf{r}_1,\mathbf{r}_2) = e^{i\delta(\mathbf{r}_2,\mathbf{r}_1)}\psi(\mathbf{r}_2,\mathbf{r}_1)~~~(*) $$

where $\delta(\mathbf{a},\mathbf{b})$ is arbitrary real function of $\mathbf{a},\mathbf{b}$. It does not imply that $\delta$ is constant throughout the whole configuration space $\mathbf{r}_1,\mathbf{r}_2 \in \mathbb{R}^6$ and so it is not possible generally to obtain

$$ e^{i2\delta} \psi $$ after application of the transposition operator to (*). What is obtained instead is $$ \psi(\mathbf{r}_2,\mathbf{r}_1) = e^{i\delta(\mathbf{r}_1,\mathbf{r}_2)}\psi(\mathbf{r}_1,\mathbf{r}_2) $$ From these two relations it follows $$ e^{i\delta(\mathbf{r}_1,\mathbf{r}_2)}.e^{i\delta(\mathbf{r}_2,\mathbf{r}_1)} = 1 $$ so $$ \delta(\mathbf{r}_1,\mathbf{r}_2) + \delta(\mathbf{r}_2,\mathbf{r}_1) = k.2\pi $$ but it does not follow that $e^{i\delta}$ has definite value for all $\mathbf r_1,\mathbf r_2$.

To obtain symmetric and antisymmetric functions, one has to make stronger assumption. For example, if we assume that all multiples of wave function represent the same state, and postulate that transposition does not change the state, then we have $$ \psi(\mathbf{r}_1,\mathbf{r}_2) = e^{i\delta}\psi(\mathbf{r}_2,\mathbf{r}_1)~~~(*) $$ with $\delta$ constant and the rest of the usual argument can be used to derive $e^{i\delta} = \pm 1$.

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