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The standard textbook for second quantization is Reed & Simon. However, I am a bit confused with their notation. They write:

Let $\mathscr{H}$ be a Hilbert space, $\mathcal{F}(\mathscr{H})$ the associated Fock space over $\mathscr{H}$ with domain of essential self-adjointness $D$. Corresponding to such $A$ we can define and operator $d\Gamma(A)$ on $\mathcal{F}(\mathscr{H})$ as follows. Let $A^{(n)} = A\otimes I \otimes \cdots \otimes I + I \otimes A \otimes \cdots \otimes I+ \cdots + I\otimes I \otimes \cdots \otimes A$ on $\bigotimes_{k=1}^{n}D$. Let $D_{A} \subset \mathcal{F}(\mathscr{H})$ be the set of $\psi =\{\psi_{0},\psi_{1},...\}$ such that $\psi_{n} = 0$ for $n$ large enough and $\psi_{n} \in \bigotimes_{k=1}^{n}D$ for each $n$. $D_{A}$ is dense in $\mathcal{F}(\mathscr{H})$ because $D$ is dense in $\mathscr{H}$. Define $A^{(0)} = 0$ and $d\Gamma(A) = \sum_{n=0}^{\infty}A^{(n)}$.

My problem is that they use the same notation for different things when it comes to tensor products and densely-defined operators. So, I would like to know:

  • Should $\bigotimes_{k=1}^{n}D$ be taken to be the algebraic tensor product of $D$, that is, the vector subspace of all finite linear combinations of elements of $\psi_{1}\otimes \cdots \otimes \psi_{n}$, $\psi_{i} \in D$ for every $i = 1,...,n$? Or should it be taken to be its closure, that is, the "actual" tensor product?
  • What do they exactly mean by $A^{(n)} = A\otimes I \otimes \cdots + I \otimes A \otimes \cdots \otimes I+ \cdots + I\otimes I \otimes \cdots \otimes A$? Is it the linear operator $A^{(n)} (\psi_{1}\otimes \cdots \psi_{n}) = A\psi_{1}\otimes \psi_{2}\otimes \cdots \otimes \psi_{n} + \cdots +\psi_{1}\otimes \cdots \otimes A\psi_{n}$ defined on $\bigotimes_{k=1}^{n}D$ or its closure?
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    $\begingroup$ If $\bigotimes_{k = 1}^n D$ were closed up, then wouldn't it be the same as the $n$-fold tensor power of $\mathscr H$, making it pointless to refer to $D$? For your second question, I think so—what else would the notation mean? $\endgroup$
    – LSpice
    Commented Feb 6 at 20:39
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    $\begingroup$ For example, given $A$ and $B$ two densely defined operators with domains $D(A)$ and $D(B)$ they also write $A \otimes B$ as the closure on $D(A)\otimes D(B)$. $\endgroup$
    – MathMath
    Commented Feb 6 at 20:41
  • $\begingroup$ (1) algebraic tensor product, (2) defined on $\bigotimes D$ (though the operator can then be closed up, but that doesn't mean it extends to the closure of $\bigotimes D$) $\endgroup$
    – Nik Weaver
    Commented Feb 6 at 21:47

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