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