# functional analysis – Extension of essentially self-adjoint operators on tensor products The content of my question is based on Reed & Simon. Let $$A$$ be a given densely-defined self-adjoint operator on a Hilbert space $$mathscr{H}$$. Let $$mathcal{D}_{A}$$ be a domain of self-adjointness of $$A$$. Thus, the restriction of $$A$$ to $$mathcal{D}_{A}$$ is essentially self-adjoint, which means that its closure $$bar{A}$$ is self-adjoint. Here, we could take $$mathcal{D}_{A}$$ to be the domain $$D(A)$$ of $$A$$ itself, in which case $$bar{A} = A$$, since $$A=A^{*}$$ is closed.

Now, let $$A_{1},…,A_{n}$$ be (densely-defined) self-adjoint operators on $$mathscr{H}_{1},…mathscr{H}_{n}$$ respectivelly. For each $$A_{i}$$, let $$mathcal{D}_{A_{i}}$$ be a domain of self-adjointness of $$A_{i}$$ and set $$mathcal{D}_{A_{1}}otimes cdots otimes mathcal{D}_{A_{n}}$$ to be the set of all finite linear combinations of elements of the form $$varphi_{1}otimes cdots otimes varphi_{n}$$, with $$varphi_{i}in D_{A_{i}}$$. In addition, let $$A_{pi} := A_{1}otimes cdots otimes A_{n}$$ be defined on $$mathcal{D}_{A_{1}}otimes cdots otimes mathcal{D}_{A_{n}}$$ as follows. For each $$varphi_{1}otimes cdots otimes varphi_{n}$$, set:
$$(A_{1}otimes cdots otimes A_{n})(varphi_{1}otimes cdots otimes varphi_{n}) := A_{1}varphi_{1}otimes cdots otimes A_{n}varphi_{n}$$
and extend it to $$mathcal{D}_{A_{1}}otimes cdots otimes mathcal{D}_{A_{n}}$$ by linearity. Finally, if $$I_{i}$$ denotes the identity operator on $$mathscr{H}_{i}$$, let $$A_{Sigma}$$ be defined on $$mathcal{D}(A_{1})otimes cdots otimes mathcal{D}(A_{n})$$ by:
$$A_{Sigma}:= A_{1}otimes I_{2}otimes cdots otimes I_{n} + I_{1}otimes A_{2}otimes cdots otimes I_{n}+ cdots + I_{1}otimes I_{2}otimes cdots otimes A_{n}$$
As proved in Reed & Simon, $$A_{Sigma}$$ is essentially self-adjoint on its domain. This implies that $$A_{Sigma}$$ has a self-adjoint extension. But, as before, we could take $$D_{A_{i}}= D(A_{i})$$ in which case I’d expect that the self-adjoint extension of $$A_{Sigma}$$ would be itself. Is this correct? And, if so, is there a easy way to prove it? Posted on Categories Articles