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How is the Pauli Exclusion Principle a consequence of antisymmetric wavefunction?

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    $\begingroup$ I would rather change the order of the implications. Pauli's principle forces the wavefunction to be totally antisymmetric $\endgroup$
    – Phoenix87
    Jan 5, 2015 at 15:04
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    $\begingroup$ @Phoenix87 it's not Pauli principle which would force the wavefunction to be antisymmetric. It's rather the principle AND requirement of differentiability of the wavefunction in the absense of singularities in potential. Otherwise you could just symmetrically reflect the wavefunction from the $\vec r_1=\vec r_2$ hyperplane, setting the wavefunction zero at that hyperplane. $\endgroup$
    – Ruslan
    Jan 5, 2015 at 17:58

2 Answers 2

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Take for example a 2 particle fermion system that are not interacting. Because they're not interacting we can assume the two-particle wavefunction can be written as a product of the single particle wavefunctions. Let's label the two single particles with $a_1$ and $a_2$, we have:

$$ \psi(\mathbf{r}_1,\mathbf{r}_2)=\psi_{a_1}(\mathbf{r}_1)\psi_{a_2}(\mathbf{r}_2)$$

Since we can't disntinguish between the two particles, we can also write the above wavefunction as: $$\psi'(\mathbf{r}_1,\mathbf{r}_2)=\psi_{a_1}(\mathbf{r}_2)\psi_{a_2}(\mathbf{r}_1)$$

All we can say is that the system must be in a linear superposition of $\psi$ and $\psi'.$ Mathematically we can only combine the two in only two correctly normalised way: First the symmetric case (Bosons, e.g. photons): $$\Psi_s(\mathbf{r}_1,\mathbf{r}_2)=\frac{1}{\sqrt{2}}[\psi_{a_1}(\mathbf{r}_1)\psi_{a_2}(\mathbf{r}_2)+\psi_{a_1}(\mathbf{r}_2)\psi_{a_2}(\mathbf{r}_1)]$$

And second case being the antisymmetric combination (fermions, e.g. electrons): $$\Psi_{anti}(\mathbf{r}_1,\mathbf{r}_2)=\frac{1}{\sqrt{2}}[\psi_{a_1}(\mathbf{r}_1)\psi_{a_2}(\mathbf{r}_2)-\psi_{a_1}(\mathbf{r}_2)\psi_{a_2}(\mathbf{r}_1)]$$

Now since we're considering fermions, for identical single particles i.e. $a_1=a_2$, then $\Psi_{anti}=0$, which means the probability amplitude of two fermions occupying the same state is $0.$ So you see that just by considering the form of the wavefunction for a system of identical elements we managed to arrive at Pauli's exclusion principle. Finally, as you can tell from $\Psi_s\neq 0$ for identical particles, simply implies that bosons do not follow such exclusion principle and nothing prohibits them from occupying the same state.

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Or could we just say, if we exchange coordinates once, it leads to a wave function proportional to the original because they are indistinguishable and it can be proportional with some constant c. If we do it back, we get c to the power of 2, but since we have to have the same function we conclude that c=1 or c=-1. Both cases are found in nature. If the c=-1 we call a function antisymmetric. But now, in this case if we have two particles in the same state and then interchange them, we get the same function multiplied by -1. That can only be true for the case in which our function is identically equal to zero.

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