Adjoint of the direct sum of operators vs. direct sum of their adjoints Let $\mathcal{H}$ be an infinite-dimensional complex Hilbert space which decomposes as the direct sum of a countable family of Hilbert spaces $\{\mathcal{H}_n\}_{n\in\mathbb{N}}$, namely
\begin{equation}
\mathcal{H}=\bigoplus_{n\in\mathbb{N}}\mathcal{H}_n.
\end{equation}
For all $n$, let $A_n$ be a (possibly unbounded) densely defined operator on $\mathcal{H}_n$, with domain $\mathcal{D}(A_n)$. Define
\begin{equation}
A=\bigoplus_{n\in\mathbb{N}}A_n,
\end{equation}
that is, $A$ is the operator on $\mathcal{H}$ with domain
\begin{equation}
\mathcal{D}(A)=\left\{\Psi=\{\psi_n\}_{n\in\mathbb{N}}:\,\psi_n\in\mathcal{D}(A_n),\,\sum_n\|A_n\psi_n\|^2<\infty\right\}.
\end{equation}
acting as $A\psi_n=A_n\psi_n$.
Here is my question. Let $A^*$, $A^*_n$ be the adjoints of $A$ and $A_n$ respectively. Under which conditions does the equality
\begin{equation}
A^*=\bigoplus_{n\in\mathbb{N}}A^*_n
\end{equation}
hold? This is certainly true for self-adjoint operators, as stated e.g. by Reed and Simon (Methods of Modern Mathematical Physics vol 4: Analysis of Operators, Theorem XIII.85): if each $A_n$ is self-adjoint, then so is $A$. However, here I'm interested in what happens in the non-self-adjoint case; the proof in the reference only works for self-adjoint operators.
I have the "feeling" that the equation above should hold as well, possibly up to a closure, at least for sufficiently well-behaved operators (e.g. normal operators), but right now I have not found any reference about this case, nor I have managed to show that by directly applying the definition of adjoint. Indeed, by definition, $A^*$ has domain
\begin{equation}
\mathcal{D}(A^*)=\left\{\Psi=\{\psi_n\}_n\in\mathcal{H}:\;\forall\Phi=\{\phi_n\}_n\in\mathcal{D}(A)\,\exists\tilde{\Psi}=\{\tilde{\psi}_n\}_n\in\mathcal{H}\,\text{s.t.}\sum_n\left\langle\psi_n,A_n\phi_n\right\rangle_{\mathcal{H}_n}=\sum_n\left\langle\tilde{\psi}_n,\phi_n\right\rangle_{\mathcal{H}_n}\right\}
\end{equation}
and $A^*\psi_n=\tilde{\psi}_n$, with $\langle\cdot,\cdot\rangle_{\mathcal{H}_n}$ being the scalar product on $\mathcal{H}_n$. Whether this operator does correspond to the direct sum of the adjoints $A^*_n$ is not clear to me, since the equality on the two sums does not imply, in general, the equality between each term of the sums.
 A: The situation can be more complex. For example, consider $A$ to be the momentum operator on the line, with a  potential in the origin. Such operator has a family of self-adjoint extensions, take one of them. It can be seen as the direct sum of two momentum operators on half-lines $A^-$, $A^+$, with appropriate transmission conditions between the two half-lines. Each operator is not self-adjoint. Indeed, their minimal operators have defect indeces (0,1) and (1,0)) but the minimal operator of $A$ has defect indices (1,1). So, the starting operator is self-adjoint, $A=A^*$ but $A^\pm \neq A^{\pm *}$.
A: I think this is true without any further assumptions.
If every $A_n$ is densely defined, then $A$ is densely defined: The set of vectors with finitely many non-zero components in the direct sum decomposition is dense in $\mathcal H$, and every vector with finitely many non-zero components can clearly be approximated by elements in the domain of $A$.
If $\xi\in D\left(\bigoplus_n A_n^\ast\right)$ and $\eta\in D(A)$, then
$$
\langle\xi,A\eta\rangle=\sum_n \langle \xi_n,A_n \eta_n\rangle=\sum_n\langle A_n^\ast\xi_n,\eta_n\rangle=\langle(\bigoplus_nA_n^\ast)\xi,\eta\rangle.
$$
Thus $\xi\in D(A^\ast)$ and $A^\ast \xi=(\bigoplus_n A_n^\ast)\xi$.
It remains to prove $D(A^\ast)\subset D(\bigoplus_n A_n^\ast)$. Let $\xi \in D(A^\ast)$ and $\eta\in D(A)$ such that $\eta_n=0$ for $n\neq k$. Then
$$
\langle (A^\ast\xi)_k,\eta_k\rangle=\langle A^\ast\xi,\eta\rangle=\langle \xi,A\eta\rangle=\langle \xi_k,A_k\eta_k\rangle.
$$
Thus $\xi_k\in D(A_k^\ast)$ and $(A^\ast \xi)_k=A_k^\ast\xi_k$. In particular,
$$
\sum_k \|A_k^\ast\xi_k\|^2=\sum_k\|(A^\ast \xi)_k\|^2=\|A^\ast \xi\|^2<\infty.
$$
Hence $\xi\in D(\bigoplus_n A_n^\ast)$.
