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Thinking about some quantum mechanics issues, I stumbled across the following functional analysis problem which confuses me a lot.

Let $A$, $B$ be self-adjoint, unbounded and positive operators with domains $\rm{dom}(A) \subset H_1$ and $\rm{dom}(B) \subset H_2$, where $H_1, H_2$ are some separable Hilbert spaces ($L^2$ of something, in fact).

What is the domain of $A + B$, which denotes the closure of the operator $A \otimes 1 + 1 \otimes B$ definable on $\rm{dom}(A) \otimes \rm{dom}(B) \subset H_1 \otimes H_2$?

I know that the closure can in principle lead to complications here. But I thought that the answer should be something like $\rm{dom}(A) \otimes H_2 \cap H_1 \otimes \rm{dom}(B)$ because both operators are positive and there cannot be any strange cancellations that would further enlarge the domain.

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  • $\begingroup$ any reason why the answer is not cl(dom(A)$\otimes$dom(B))? $\endgroup$ – LinAlg Jul 25 '18 at 21:03
  • $\begingroup$ @LinAlg: Well, this is obviously true, but also not totally helpful if you want to work with it. (Notice that this closure is taken in the graph norm of A + B) $\endgroup$ – Luke Jul 27 '18 at 9:01
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Too long for a comment:

In Brian C. Hall's book 'Quantum Theory for Mathematicians' the problem is discussed in Definition 19.15 (Page 423) as follows:

It is not, however, obvious how to select a domain for $A \otimes I + I \otimes B$ in such a way that this operator will be self-adjoint. The easiest way to deal with this issue is to use Stones's theorem, as in the following definition.

Definition 19.15 If $A$ and $B$ are self-adjoint operators on $\mathbf H_1$ and $\mathbf H_2$, define the operator $A$$\otimes$$I$$+$$I$$\otimes$$B$ to be the infinitesimal generator of the strongly continuous one-parameter unitary group $\mathrm{e}^{itA} \otimes \mathrm{e}^{itB}$. Thus, by Stone's theorem, $A \otimes I + I \otimes B$ is self-adjoint.

It is not hard to check that $\mathrm{e}^{itA} \otimes \mathrm{e}^{itB}$ is indeed strongly continuous. [...] If $A$ and $B$ happen to be bounded, then $A$$\otimes$$I$$+$$I$$\otimes $$B$ defined by Definition 19.15 coincides with $A \otimes I + I \otimes B$ defined as the sum of tensor products of bounded operators, as in Sect. A.4.5.

This quote does not provide a direct answer to your question, but might help at least in the physical context. (I have not jet found a counterexample, which shows how the naive approach fails.)

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I found a good source myself after some research: In the book by K. Schmüdgen, "Unbounded Self-Adjoint Operators on Hilbert Space", he states on p. 160 that the operator $A \otimes 1 + B \otimes 1$ is closed on the natural domain if $A$ and $B$ are semibounded from below. This answers my question: For positive operators, the closure of the domain is actually the natural domain of $A +B$, i.e. $\rm{dom} A \otimes H_2 \cap H_1 \otimes \rm{dom} B$.

The proof is left as Exercise 17 in Chapter 7, so I don't really have a proof at hand, but I think that my heuristic idea that there cannot be cancellations if both operators are positive goes through (so you can show that if $(A + B)\psi_n$ is a Cauchy sequence, also $A \psi_n$ and $B \psi_n$ are, and from there closedness on the domain $\rm{dom} A \otimes H_2 \cap H_1 \otimes \rm{dom} B$ follows).

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