Strongly continuous group and generator commute, what about square roots? Let $A$ be a positive self-adjoint operator, then $iA$ generates a unitary strongly continuous semigroup (Stone's theorem) $T$. Then from basic semigroup theory we know that $T$ and $A$ commute, but does this imply that $T$ and $\sqrt{A}$ commute, too?  
By the way: I was already able to show that $T(D(\sqrt{A})) \subset D(\sqrt{A})$.
 A: Using the Spectral Theorem,
$$
               Ax=\int_{0}^{\infty}\lambda dE(\lambda)x \\
     \mathcal{D}(A) = \left\{ x : \int_{0}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty \right\}.
$$
Then the positive square root $\sqrt{A}$ is
$$
              \sqrt{A}x = \int_{0}^{\infty}\sqrt{\lambda}dE(\lambda)x \\
      \mathcal{D}(\sqrt{A}) = \left\{ x : \int_{0}^{\infty}\lambda d\|E(\lambda)x\|^2 < \infty \right\}.
$$
The Unitary group is $e^{itA}x=\int_{0}^{\infty}e^{it\lambda}dE(\lambda)x$. By the functional calculus,
$$
               e^{itA}\int_{0}^{R}\sqrt{\lambda} dE(\lambda)x = \int_{0}^{R}e^{it\lambda}\sqrt{\lambda} dE(\lambda)x = \int_{0}^{R}\sqrt{\lambda} dE(\lambda)e^{itA}x \\
      \left\|e^{itA}\int_{0}^{R}\sqrt{\lambda} dE(\lambda)x\right\|^2
   = \int_{0}^{R}\lambda d\|E(\lambda)x\|^2 = \int_{0}^{R}\lambda d\|E(\lambda)e^{itA}x\|^2
$$
Therefore $x \in \mathcal{D}(\sqrt{A})$ iff $e^{itA}x \in \mathcal{D}(\sqrt{A})$. For such $x$, the following limits exist and are equal:
$$
     e^{itA}\sqrt{A}x = \lim_{R\rightarrow\infty}e^{itA}\int_{0}^{R}\sqrt{\lambda}dE(\lambda) x =\lim_{R\rightarrow\infty}\int_{0}^{R}\sqrt{\lambda}dE(\lambda)e^{itA}x = \sqrt{A}e^{itA}x.
$$
