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Following some suggestions on my previous posts, I'm trying to reformulate my question in a more specific way. This is a continuation of my original post. Since the mentioned post, I think I've learned some things, so let me elaborate a little more on the issue.

Let $\mathcal{H}$ be a Hilbert space and $\mathcal{F}_{f}(\mathcal{H}) = \bigoplus_{n=0}^{\infty}\mathcal{H}^{-}_{n}$ the fermionic Fock space associated to $\mathcal{H}$. Here, $\mathcal{H}^{-}_{n}$ is the space of all anti-symmetric tensors $\mathcal{H}_{n} := \overbrace{\mathcal{H}\otimes\cdots\otimes\mathcal{H}}^{\text{$n$ times}}$. Let $\varphi \in \mathcal{H}$ be fixed and $a(\varphi): \mathcal{H}_{n}^{-}\to \mathcal{H}_{n-1}^{-}$ and $a^{\dagger}(\varphi):\mathcal{H}_{n}^{-}\to \mathcal{H}_{n+1}^{-}$ be the usual annihilation and creation operators, respectivelly. These operators are bounded operators and we can extend them to $\mathcal{F}_{f}(\mathcal{H})$ (actually to a dense subset of it).

Now, let $\mathcal{H} = L^{2}(\mathbb{R}^{3})$. Suppose we identify $\psi \in \mathcal{H}_{n}^{-}$ with $\psi \in L^{2}_{-}(\mathbb{R}^{3n})$ the space of all anti-symmetric $f\in L^{2}(\mathbb{R}^{3n})$. In other words: $$f \in L^{2}_{-}(\mathbb{R}^{3n}) \iff f(x_{\sigma(1)},...,x_{\sigma(n)}) = \epsilon_{\sigma}f(x_{1},...,x_{n})$$ for every permutation $\sigma$ of $\{1,...,n\}$. Then $a(\varphi)\psi \in H_{n-1}^{-}$ is identified with: \begin{eqnarray} (a(\varphi)\psi)(x_{1},...,x_{n-1}) = \sqrt{n}\int \overline{h(x)}\psi(x,x_{1},...x_{n-1})dx \tag{1}\label{1} \end{eqnarray} If $\psi \in L^{2}_{-}(\mathbb{R}^{3n})$ is a continuous function, then we can take $\varphi$ to be a delta distribution, so we can define an operator: \begin{eqnarray} (a(x)\psi)(x_{1},...,x_{n-1}) = \sqrt{n}\psi(x,x_{1},...,x_{n-1}) \tag{2}\label{2} \end{eqnarray} This operator can be extended to a suitable dense domain on the Fock space.

My original question was to understand the creation and annihilation operators $\varphi(x,\sigma)$ and $\varphi^{\dagger}(x,\sigma)$ introduced on page 18 of Feldman, Trubowitz and Knörrer's notes. I believe the mentioned extension of my $a(x)$'s and $a^{\dagger}(x)$'s is precisely what FTK's notes are talking about if the spins $\sigma$ were not to be considered. However, I'd like to consider spins in my above description, so to define $\varphi(x,\sigma)$ and $\varphi^{\dagger}(x,\sigma)$.

Question: How to introduce spin variables within the above theory?

My guess is as follows: a particle now is described not only by not only its wave function $\psi \in L^{2}(\mathbb{R}^{3})$ but also by its spin $\sigma$, so I should probably consider, say, $L^{2}(\mathbb{R}^{3})\otimes \mathbb{C}^{2}$, or something like this. Then, maybe use $L^{2}(\mathbb{R}^{3})\otimes \mathbb{C}^{2} \cong L^{2}(\mathbb{R}^{3}; \mathbb{C}^{2})$? But then, if $\mathcal{H} = L^{2}(\mathbb{R}^{3};\mathbb{C}^{2})$, an element $\psi \in H_{n}^{-}$ should be identified to what? An element $\psi \in L^{2}_{-}(\mathbb{R}^{3n};\mathbb{C}^{2})$?

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    $\begingroup$ Your guess is correct. $\endgroup$
    – gmvh
    Commented Sep 12, 2020 at 16:01
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    $\begingroup$ Your very last question is essentially this: what is $\mathcal{H}_n^-$? Not unexpectantly, it is always $\mathcal{H}_n^- = \bigwedge^n \mathcal{H}$, denoting the fully anti-symmetrized Hilbert tensor product. The further isomorphism $L^2(\mathbb{R}^3; \mathbb{C}^2) \cong L^2(\mathbb{R}^3\times\{+1,-1\})$ should make it clear what $\mathcal{H}_n^-$ is. Exercise: fill in the blank in $\mathcal{H}_n^- \subset L^2(\mathbb{R}^{3n}; ?)$. $\endgroup$ Commented Sep 12, 2020 at 17:48
  • $\begingroup$ @IgorKhavkine Maybe I could use the following: if $\mathcal{H} = L^{2}(\mathbb{R}^{3};\mathbb{C}^{2}) \cong L^{2}(\mathbb{R}^{3}\times \{+1,-1\})$ it seems that $\mathcal{H}_{n} \cong L^{2}(\mathbb{R}^{3}\times \{+1,1\})\otimes \cdots \otimes L^{2}(\mathbb{R}^{3}\times \{+1,-1\}) \cong L^{2}(\mathbb{R}^{3n}\times \{+1,-1\}^{n})$? (Don't know for sure if the last isomorphism holds tho).The latter seems more useful to define $\varphi(x,\sigma)$ and $\varphi^{\dagger}(x,\sigma)$ since $x \in \mathbb{R}^{3}$ and $\sigma$ is a spin variable, probably taking values $\{+1,-1\}$. $\endgroup$ Commented Sep 12, 2020 at 20:01
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    $\begingroup$ @IamWill All correct. Don't see any reason for you to doubt yourself there. $\endgroup$ Commented Sep 13, 2020 at 5:08

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