Suppose $A :D(A) \subset \mathcal{H} \to \mathcal{H}$ is self-adjoint over the separable Hilbert space $\mathcal{H}$. By Stone's Theorem, $A$ generates a unique one-parameter strongly continuous group $U(t) = \exp(-itA)$ of unitary operators. It is a standard fact that $U(t)$ preserves the domain of $A$, i.e., $U(t)[D(A)] \subset D(A)$.
Suppose now $A$ is also semibounded from below, i.e., $A + \mu\mathrm{Id} \geq 0$ for some $\mu \in \mathbb{R}$, so that $A$ it is associated with a unique bounded sesquilinear form $a : Q(A) \times Q(A) \to \mathcal{H}$, where $Q(A) = D((A+\mu \mathrm{Id})^{1/2}) \supset D(A)$ is the form domain of $A$. (On $D(A)$, $a(u,v) = \langle u,Av\rangle$.)
Question: Does $U(t)$ also preserve the form domain, i.e.,
$$ U(t)[Q(A)] \subset Q(A) \quad ? $$
By replacing $A$ with $A+\mu I$ we may assume that $A\ge 0.$ Let $E(x)$ denote the resolution of the identity associated with $A.$ Then $$D(A^{1/2})=\left\{v\in \mathcal{H}\,:\, \int\limits_{0^-}^\infty x\,d\langle E(x)v,v\rangle <\infty\right \}$$ The operators $e^{itA}$ commute with $E(x),$ hence $$\langle E(x)e^{itA}v,e^{itA}v\rangle =\langle e^{-itA}E(x)e^{itA}v,v\rangle =\langle E(x)v,v\rangle\quad (*) $$
Let $v\in D(A^{1/2}).$ Then, in view of $(*),$ for $w=e^{itA}v$ we have $$\int\limits_{0^-}^\infty x\,d\langle E(x)w,w\rangle= \int\limits_{0^-}^\infty x\,d\langle E(x)v,v\rangle <\infty $$ Hence $w\in D(A^{1/2}).$