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Let $\mathscr{H}_{1}$ and $\mathscr{H}_{2}$ be Hilbert spaces. If $\psi_{1}\in \mathscr{H}_{1}$ and $\psi_{2}\in \mathscr{H}_{2}$, define $\psi_{1}\otimes \psi_{2}$ to be a function on $\mathscr{H}_{1}\times \mathscr{H}_{2}$ given by: $$(\psi_{1}\otimes \psi_{2})(\psi,\varphi) := \langle \psi, \psi_{1}\rangle \langle \varphi, \psi_{2}\rangle $$ Let $\mathscr{H}_{1}\hat{\otimes}\mathscr{H}_{2}$ be the set of all finite linear combinations of functions of the form $\psi_{1}\otimes \psi_{2}$. Then $\mathscr{H}_{1}\hat{\otimes}\mathscr{H}_{2}$ is a vector space and: $$\langle \psi \otimes \varphi, \eta \otimes \nu\rangle := \langle \psi, \eta \rangle \langle \varphi, \nu \rangle $$ is an inner product on this space. Its completion, denoted by $\mathscr{H}_{1}\otimes \mathscr{H}_{2}$ is called the tensor product of $\mathscr{H}_{1}$ and $\mathscr{H}_{2}$.

The above definitions are discussed in many books and it extends by finite induction to $\mathscr{H}_{1}\otimes \cdots \otimes \mathscr{H}_{n}$.

In many textbooks, however, tensor products of dense subspaces are mentioned but usually not defined. To be more precise, let $(\forall j = 1,...,n)$ $D_{j}\subset \mathscr{H}_{j}$ be a dense subspace. What is the definition of $D_{1}\otimes \cdots \otimes D_{n}$? Is it just the set of all finite linear combinations of elements of the form $\psi_{1}\otimes \cdots \otimes \psi_{n}$ (as defined above) with $\psi_{j} \in D_{j}$, $j=1,2,...,n$?

Remark: I don't know if there is a general definition of tensor product of dense subspaces or not, so it might be useful to mention the motivation of my question. The above scenario is commonly found in the process of second quantization, as discussed in chapter $X$ of Reed & Simon, vol 2. Usually, many interesting linear operators in quantum mechanics are unbounded, so densely defined and to construct, say, its second quantizations on subspaces of Fock spaces. Say $A$ is self-adjoint (not necessarily bounded) with domain of essential self-adjointness $D\subset D(A)$. Let $F_{0}$ the set of all vectors $\psi = \{\psi^{(n)}\}_{n=0}^{\infty}$ of the symmetric Fock space $\mathcal{F}_{s}(\mathscr{H})$ with $\psi^{(n)}=0$ for all but finitely many entries. Define: $$D_{A}:=\{\psi \in F_{0}: \hspace{0.1cm} \psi^{(n)} \in \bigotimes_{k=1}^{n}D \hspace{0.1cm} \mbox{for each $n$}\}$$ Then, the second quantization of $A$ is the operator $d\Gamma(A)$ on $D_{A}$ defined by: $$d\Gamma(A):= \bigoplus_{n=0}^{\infty}A^{(n)}$$ where: $$A^{(n)} := A\otimes I \otimes \cdots I + I\otimes A \otimes \cdots \otimes I + \cdots + I\otimes I \otimes \cdots \otimes A $$ This is an example where tensor products of dense subspaces arrise: $\bigotimes_{k=1}^{n}D$. More generally, if $A$ is densely defined, with dense domain $D(A)$, what does it mean to consider $D(A) \otimes \cdots \otimes D(A)$ $n$ times?

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  • $\begingroup$ I don't think there's any standard way of making such a definition. Anyway it's not what's going on in your illustration about second quantization. In that setting you define $D_A$ as you've stated using the algebraic (uncompleted) tensor product, then you take the closure of $d\Gamma(A)$ as defined on $D_A$. $\endgroup$
    – Nik Weaver
    Commented Feb 27, 2021 at 13:42
  • $\begingroup$ @NikWeaver thanks for the comment! So, $D\otimes\cdots\otimes D$ should be the set of all linear combinations of elements $\psi_{1}\otimes\cdots \otimes \psi_{n}$, with $\psi_{i} \in D_{i}$? $\endgroup$
    – MathMath
    Commented Feb 27, 2021 at 14:19
  • $\begingroup$ The was my guess since if we take $A = I$ the identity map, this should lead to the number operator. So I think $D\otimes \cdots \otimes D$ is defined as the algebraic tensor product. $\endgroup$
    – MathMath
    Commented Feb 27, 2021 at 14:20

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