The exact domain on which the Euclidean Dirac operator is self-adjoint I use the convention of the Weinberg QFT textbooks, that is, $(-,+,+,+)$.
According to Weinberg QFT vol 2 p. 369, he says the Euclidean Dirac operator
\begin{equation}
{D}:=[i\partial_i +t_\alpha A_{i \alpha}]\gamma_i
\end{equation}
is Hermitian. Here, $\partial_4:=-i\partial_0$, $A_{4\alpha}:= iA^0_\alpha$ and $\gamma_4:=i \gamma^0$. $\alpha$ is the gauge index.
Let us simplify and focus on the case of $U(1)$ gauge theory so that there is no $\alpha$. Also, assume that $A_0, A_1, A_2, A_3$ are real-valued Schwartz functions on $\mathbb{R}^4$.
Then, the above $D$ must be a densely defined unbounded operator on the Hilbert space $[L^2(\mathbb{R}^4)]^4$.
My question is, what would be the maximal domain of $D$ that makes it self-adjoint? Physicists would not care about this kind of subtlety but, I feel uncomfortable about not distinguishing between Hermitian and self-adjoint.
Could anyone please clarify?
 A: Recall that a densely defined operator $T$ on a Hilbert space $H$ is essentially self-adjoint if and only if its minimal closure $\overline{T}$ is self-adjoint, if and only if $T$ is symmetric and has a unique extension to a self-adjoint operator (i.e., the minimal closure $\overline{T}$).
In general, you can consider the spin Dirac operator $D$ on a complete Riemannian spin manifold $(M,g)$ with spinor bundle $S$ (e.g., the Dirac operator you consider on Euclidean $\mathbb{R}^4$). By a theorem of Wolf, $D$ (as well as its square $D^2$) is essentially self-adjoint on the subspace $C_c^\infty(M,S)$ of compactly supported smooth spinor fields. More generally, it follows from a theorem of Chernoff that $D^k$ is essentially self-adjoint on $C_c^\infty(M,S)$ for every $k \in \mathbb{N}$.
What about the domain of the minimal closure $\overline{D}$? By a theorem of Gromov and Lawson, if the curvature of the spinor Levi-Civita connection is uniformly bounded in the appropriate sense, then the domain of $\overline{D}$ is the first $L^2$ Sobolev space $H^1(M,S) = L^{1,2}(M,S)$ of square-integrable spinor fields with square-integrable weak covariant derivatives (w.r.t. the spinor Levi-Civita connection). In your case, this means that the domain of the unique self-adjoint extension of the Dirac operator on Euclidean $\mathbb{R}^4$ is precisely the space $H^1(\mathbb{R}^4) \otimes \mathbb{C}^4$, where $H^1(\mathbb{R}^4)$ is the Sobolev space of square integrable functions on $\mathbb{R}^4$ with square-integrable weak first-order partial derivatives (or equivalently square-integrable weak gradient).
