Permutation of Identical Fermions - Spatial and Spin Decomposition

Question

As far as I know, fermions are the particles which exhibit antisymmetric states: $\hat{P}\left|n_1\right>\otimes\left|n_2\right> = -\left|n_2\right>\otimes\left|n_1\right>$. Often times, we decompose the state $\left|n_1\right>\otimes\left|n_2\right>$ into its spatial and spin parts: $$\psi(x_1,x_2) \chi_{1,2} = \Psi = \sum_{m_{s1}}\sum_{m_{s2}} C(m_{s1},m_{s2}) \left< x_1,m_{s1} |n_1\right>\otimes\left< x_2,m_{s2} |n_2\right> $$ (see Kardar eq. 7.31) since then we can conclude that, if $\psi$ is symmetric, then $\chi$ must be antisymmetric and vice-versa.

What prohibits us from having neither $\psi$ nor $\chi$ symmetric/antisymmetric. For example, I believe that the following would maintain our desired antisymmetry of $\left|n_1\right>\otimes\left|n_2\right>$: $$ \psi(x_1,x_2)=i\,\psi(x_2,x_1) \quad \text{and} \quad \chi_{1,2}=i\, \chi_{2,1}$$ since then $$ \sum_{m_{s1}}\sum_{m_{s2}} C(m_{s1},m_{s2}) \bigg(\left< x_1,m_{s1} \right|\otimes\left< x_2,m_{s2}\right|\bigg) \hat{P}\bigg(\left| n_1\right> \otimes\left|n_2\right>\bigg) = \sum_{m_{s1}}\sum_{m_{s2}} C(m_{s1},m_{s2}) \left< x_1,m_{s1} |n_2\right>\otimes\left< x_2,m_{s2} |n_1\right> = \psi(x_2,x_1)\chi_{2,1} = -\psi(x_1,x_2)\chi_{1,2} $$ as desired.

Is this just allowed mathematically but doesn't occur in reality? Is there a process prohibiting it? Did I make a mistake?

Answer 1

The permutation $P_{12}$ satisfies $P_{12}^2=\hat{\mathbb{I}}$ and your transformation does not satisfy this since it multiples the spatial or the spin part (which are separate objects) by $i$. In other words you have $P_{12}^2\psi(x_1,x_2)=-\psi(x_1,x_2)$ which is not possible for a transposition.

A more sophisticated answer is that there are only irreducible representations of $S_2$; one multiplies by $-1$ the other by $+1$; note that repeating the permutation twice then gives the identity, as expected.