To describe a quantum mechanical system, that consist out of a fixed number of particles we take advantage of the tensor product:
$\vert ab \rangle = \vert a \rangle \otimes \vert b \rangle \in H_1 \otimes H_2$.
When dealing with identical particles, the corresponding quantum mechanical states of these systems either have to be symmetric or antisymmetric.
Lets assume that we describe a system of two electrons and the spin-dependent part of the state is given by
$ \vert \Phi\rangle = \frac{1}{\sqrt{2}} ( \vert \uparrow_1 \downarrow_2 \rangle - \vert \downarrow_1 \uparrow_2 \rangle)$
By exchanging both particles, we get:
$ \hat{P}_{1,2} \vert \Phi\rangle = \frac{1}{\sqrt{2}} ( \vert \uparrow_2 \downarrow_1 \rangle - \vert \downarrow_2 \uparrow_1 \rangle) = - \vert \Phi\rangle$
What makes me wonder is, why the second expression is equal to the first one (except for the minus sign). Apparently $\vert \uparrow_1 \downarrow_2 \rangle$ equals to $\vert \downarrow_2 \uparrow_1 \rangle$.
Does this mean, that the tensor product is always symmetric in the sense of
$\vert ab \rangle = \vert a \rangle \otimes \vert b \rangle = \vert b \rangle \otimes \vert a \rangle = \vert b a \rangle$
while $H_1 \otimes H_2$ and $H_2 \otimes H_1$ are basically the same vector space?