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?