Let $\Psi \in \mathcal{L}^{2}(\mathbb{R}^{6}; \mathbb{C}^{2}\otimes\mathbb{C}^{2})$ so that $\Psi$ represents the state of two interacting fermions.
Traditionally, we define the exchange operator $P_{12}$ via \begin{align*} P_{12}\Psi(\mathbf{r}_1, \mathbf{r}_2) = \Psi(\mathbf{r}_2, \mathbf{r}_1) \end{align*} but pretty soon we realize this is incorrect, as we know that we need to interchange the spin state as well. This is often achieved by writing \begin{align*} P_{12}\Psi(\mathbf{r}_1, \sigma_1, \mathbf{r}_2, \sigma_2) = \Psi(\mathbf{r}_2, \sigma_2, \mathbf{r}_1, \sigma_1) \end{align*} but this is still awkward, as the spin state is a dependent variable in the theory.
To avoid these problems I would like to define the exchange operator without any reference to the dummy variables $\mathbf{r}_1$ and $\mathbf{r}_2$.
This is what I have so far: Suppose $\{\psi_{n}\}$ separates $\mathcal{L}^{2}(\mathbb{R}^{3}, \mathbb{C})$. Then any function $\Psi \in \mathcal{L}^{2}(\mathbb{R}^{6}; \mathbb{C}^{2}\otimes\mathbb{C}^{2})$ can be written as \begin{align}\label{separate} \Psi &= \sum_{n} a_{nm} \begin{pmatrix} \psi_{n} \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ \psi_{m} \end{pmatrix} + b_{nm} \begin{pmatrix} \psi_{n} \\ 0 \end{pmatrix} \otimes \begin{pmatrix} \psi_{m} \\ 0 \end{pmatrix} + c_{nm} \begin{pmatrix} 0 \\ \psi_{n} \end{pmatrix} \otimes \begin{pmatrix} 0 \\ \psi_{m} \end{pmatrix} + d_{nm} \begin{pmatrix} 0 \\ \psi_{n} \end{pmatrix} \otimes \begin{pmatrix} \psi_{m} \\ 0 \end{pmatrix} \end{align} where the first part of the tensor product is assumed to act on $\mathbf{r}_1$ and the second assumed to act of $\mathbf{r}_2$. Now we define $P_{12}$ on the basis, interchanging the functions rather than the dummy variables. Exchange of the spin-coordinate is now trivial, defined by \begin{align*} P_{12}^{\mathrm{spin}} \begin{pmatrix} \psi_{n} \\ \psi_{m} \end{pmatrix} \otimes \begin{pmatrix} \psi_{j} \\ \psi_{k} \end{pmatrix} = \begin{pmatrix} \psi_{m} \\ \psi_{n} \end{pmatrix} \otimes \begin{pmatrix} \psi_{k} \\ \psi_{j} \end{pmatrix} \end{align*} However, what I would consider the natural space exchange definition \begin{align*} P_{12}^{\mathrm{space}} \begin{pmatrix} \psi_{n} \\ \psi_{m} \end{pmatrix} \otimes \begin{pmatrix} \psi_{j} \\ \psi_{k} \end{pmatrix} = \begin{pmatrix} \psi_{j} \\ \psi_{k} \end{pmatrix} \otimes \begin{pmatrix} \psi_{n} \\ \psi_{m} \end{pmatrix} \end{align*} gives results which are in disagreement with the notation found in standard quantum mechanics texts. For instance, take $\Psi(\mathbf{r}_1, \mathbf{r}_2) = \psi(\mathbf{r}_1)\psi(\mathbf{r}_2)(\uparrow \downarrow - \downarrow \uparrow)$. This is clearly antisymmetric, but is symmetric under $P_{12}^{\mathrm{space}}P_{12}^{\mathrm{spin}}$. This can be patched up by defining $P_{12}^{\mathrm{space}}$ differently on each of the four types of terms is \ref{separate}, but that seems so unnatural that I feel like I'm on the wrong track.
Is there a natural definition of $P_{12}$ which doesn't make reference to the dummy variables $\mathbf{r}_1$ and $\mathbf{r}_2$?