Let $A$ be an unbounded self-adjoint operator acting on a Hilbert space $H$ (typically $L^2(\mathbb{R}^d)$).

Then, using Stone's theorem, the operator $A^{\otimes 2}:=A\otimes 1+1\otimes A$ defines a self-adjoint operator on $H\otimes H$, the domain of which might be difficult to determine.

Is it however true that if $A$ is furthermore essentially self-adjoint on $\mathcal{C}$, then $A^{\otimes 2}$ is essentially self-adjoint on $\mathcal{C}\otimes \mathcal{C}$ ?


1 Answer 1


This is true and in fact even more is true: it is proved in

Theorem VIII.33 (a) of Reed & Simon, Methods of Modern Mathematical Physics I: Functional Analysis, 1980


If $\{A_k : \mathcal{H}_k \to \mathcal{H}_k \}$, $1 \leq k \leq N$, is a collection of self-adjoint operators with $D^e_k \subset \mathcal{H}_k$ the domain of essential self-adjointness for $A_k$ and $P \in \mathbb{R}[x_1, \ldots, x_N]$ is of degree $n_k$ in the variable $x_k$, then $P(A_1^{(1)}, \ldots, A_N^{(N)})$ is essentially self-adjoint on $D^e := \bigotimes_{k=1}^N D^e_k$,

wherein I've used the notation $A_k^{(k)} := I \otimes \cdots \otimes A_k \otimes \cdots I$ (which in the text is written just as $A_k$, overloading the notation). The specialization to $\sum_{k=1}^N A_k^{(k)}$, which is only slightly more general than your question, is then

Corollary (a) to Theorem VIII.33, ibid.

The proof to your specific form would involve, I believe, essentially the same ingredients as that to Theorem VIII.33 (a). The natural consequence for the closure of the spectrum is also dealt with in the same place.

  • $\begingroup$ Thank you very much for your help. R&S contains so much information, i will include this reference. $\endgroup$
    – Chr
    Apr 19, 2020 at 8:32

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