Identical particles and the tensor product

Question

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?

Answer 1

I think the notation $\left|\uparrow_1\downarrow_2\right>$ is not particularly helpful because it makes it seem like there is a difference between $\left|\uparrow_1\downarrow_2\right>$ and $\left|\uparrow_2\downarrow_1\right>$. Presumably the point is that in the state $\left|\uparrow_1\downarrow_2\right>$, the first particle is in the state $\left|\uparrow\right>$ while the second is in the state $\left|\downarrow\right>$, but that is already obvious because of the ordering, which renders the subscripts redundant at best and actively harmful at worst (which appears to be the case here). Getting rid of the subscripts and simply writing $\left|\uparrow\downarrow\right> \equiv \left|\uparrow\right>\otimes \left|\downarrow\right>$ is a better idea. Exchanging the particles then yields $$\hat P_{1,2}\left|\uparrow\downarrow\right> = \left|\downarrow\uparrow\right>$$ which renders your original question rather straightforward.

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?

No, the tensor product is emphatically not always symmetric (otherwise the symmetry (resp. antisymmetry) of the composite state would be trivial (resp. impossible). However, a system of e.g. two indistinguishable particles is modeled using a tensor product space of the form $H\otimes H$ with $H$ being the single-particle Hilbert space; if $H_1\neq H_2$, then $H_1 \otimes H_2 \neq H_2\otimes H_1$ but it also wouldn't make much sense to write $$|\psi\rangle \otimes |\phi\rangle \overset{?}{\leftrightarrow} |\phi \rangle\otimes |\psi\rangle$$ since the two vectors $|\psi\rangle$ and $|\phi\rangle$ don't even belong to the same space.