Permutation of Identical Fermions - Spatial and Spin Decomposition

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?

• can you write down a function of 2 variables, $f(x,y)$, that satisfies $f(y,x) = if(x,y)$? I have been unable to get past $f(x,y)=0$.
– JEB
Commented Nov 3, 2020 at 20:15
• I believe that I can give you the general form of such an $f$! If we want $f(x,y)=if(y,x)$, then all of the information in the function occurs in the half plane $y<x$ (or one could choose $y>x$); the other half plane will be given by symmetry. If $\theta(x)$ is the standard Heaviside theta function, then $1-\theta(y-x)$ is equal to $1$ for $y<x$ and $0$ otherwise. Similarly, $1-\theta(x-y)$ is equal to $1$ for $y>x$ and $0$ otherwise. Thus, we can construct $f$ from any function $g(x,y)$ as $f(x,y)=[(1-\theta(y-x))+i(\theta(x-y)]g(x,y)$. Commented Nov 3, 2020 at 21:30

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.
• How is it proper to apply $\hat{P}$ to $\psi(x_1,x_2)$? I (implicitly) defined $\hat{P}$ as an endomorphism of $\mathcal{H}_1\otimes\mathcal{H}_2$ where $\mathcal{H}_1$ is the vector space for particle $1$ and $\mathcal{H}_2$ is the vector space for particle $2$ (the vector spaces can be identical). I do not believe that function $\psi$ lives in $\mathcal{H}_1\otimes\mathcal{H}_2$ so I don't see how I can apply $\hat{P}$ to it. Commented Nov 3, 2020 at 18:18
• Edit to above comment: I misspoke. Not only can $\mathcal{H}_1=\mathcal{H}_2$, but I believe that we must have that. Commented Nov 3, 2020 at 18:42
• If you’re thinking of endomorphism this is complicated. Of course you need the 2 copies of the Hilbert spaces to be identical else the particles cannot be identical. $\psi(x_1,x_2)$ is already a tensor product space so $P_{12}[\phi_a(x_1)\otimes \phi_b(x_2)]= \phi_a(x_2)\otimes \phi_b(x_1)$. Commented Nov 3, 2020 at 21:39
• Yes - I mis-implied that $\mathcal{H}_1$ could be different than $\mathcal{H}_2$. I do not understand what you mean by $\psi(x_1,x_2)$ is already a tensor product space, however. As far as I know, $\psi$ is just a function of two real variables $x_1,x_2\in\mathbb{R}^3$ and thus $\psi$ is not an element of $\mathcal{H}_1\otimes\mathcal{H}_1$ and thus not in the domain of $\hat{P}$. While I can see a natural definition of a permutation operator on $\psi$, I do not believe that $\hat{P}$ is said operator. Commented Nov 3, 2020 at 21:47
• $\psi(x_1,x_2)$ can be expanded in a basis of product wave functions since the set of all such products is complete; by linearity you may thus work on the product states. Moreover non-interacting particles have product states they serve as basis set. Commented Nov 3, 2020 at 22:49