Let $\mathcal{H}$ be a (complex) Hilbert space.
Let $H$ be a self-adjoint operator on $\mathcal{H}$ with dense domain $\mathcal{D}(H) \subset \mathcal{H}$, generating the unitary one-parameter semigroup $U_t = e^{-iHt}$ for $t \in \mathbb{R}$. Let also $A$ be a self-adjoint operator on $\mathcal{H}$ with dense domain $\mathcal{D}(A) \subset \mathcal{H}$ and such that $\mathcal{D}(A)$ is invariant under $U_t$, i.e. $U_t \mathcal{D}(A) \subset \mathcal{D}(A)$. Assume also that the space $G = \mathcal{D}(A) \cap \mathcal{D}(H)$ is dense in $\mathcal{H}$.
I would like to make sense of $$U_t A U_{-t} - A = \int_0^t U_s [iA, H] U_s ds.$$
This should be an unbounded operator on $\mathcal{H}$ with domain $G$? It is somehow a statement of the function $t \mapsto U_t A U_{-t}$ being $C^1$. I used the fundamental theorem of calculus in a formal sense to derive it. Am I missing something here or is it as simple as the fact that $U_s$ leaves $G$ invariant?
What troubles me is the integrand. In general, how does one define the commutator $[A, H]$? I can define the sesquilinear form $$\alpha(f, g) = \langle Hf, Ag \rangle - \langle A f, Hg \rangle$$ acting on $G$ and equip it with the intersection norm ($||f||^2_{A \cap H} = ||f||^2 + ||Af||^2 + ||Hf||^2$) and inner product and it becomes a Hilbert space. But how do I go from this to the continuity of $t \mapsto U_t A U_{-t}$? The definition of the commutator $[A,H]$ also feels very closely related to differentiating $U_t A U_{-t}$ at $t = 0$.
Would appreciate any input/explanations/pointers. Thanks.
