Let $H := -\partial^2_x : L^2(\mathbb{R}) \to L^2(\mathbb{R})$ be the free Hamiltonian, self adjoint with respect to the Sobolev space $H^2(\mathbb{R})$. We have $H \ge 0$, and the spectral theorem for self-adjoint operators furnishes us a resolution of the identity associated to $H$, from which we can define operators $f(H)$, for Borel functions $f : \mathbb{R} \to \mathbb{C}$. In particular, we consider the function $f(x) = |x|^{1/2}$, with associated operator which we denote by $H^{1/2}$.
Denote by $\langle x \rangle$ the usual Japanese bracket $\langle x \rangle := (1 + |x|^2)^{1/2}$.
I would like to know if it's the case that for each $\varphi \in C^\infty_0(\mathbb{R})$, it holds that $\langle x \rangle^{s} H^{1/2} \langle x \rangle^{-s} \varphi \in L^2(\mathbb{R})$ and
$$\| \langle x \rangle^{s} H^{1/2} \langle x \rangle^{-s} \varphi \|_{L^2(\mathbb{R})} \lesssim \| \varphi \|_{H^1(\mathbb{R})}.$$
The estimate is clearly true if we replace $H^{1/2}$ on the left side by $\partial_x$, but the difficulty in our case seems to be that it is not so clear how to "commute" the Japanese bracket factors, when the operator in the middle is defined in more abstract way, in this case, defined via the functional calculus.
Since nonnegative self-adjoint operators have unique nonnegative self-adjoint square roots, we can just as well take $H^{1/2}$ to be the usual Fourier multiplier: $H^{1/2} \varphi = \mathcal{F}^{-1} | \xi | \mathcal{F} \varphi$, where $\mathcal{F}$ denotes Fourier transform.
I hope the above estimate is "well-known" and can be proved using minimal machinery, e.g., perhaps some Fourier analysis tools. In particular, I would like to avoiding using any pseudodifferential operator methods, but maybe there is no way to get around them.
It's worth mentioning we have $\| H^{1/2} \varphi \|^2_{L^2} = \| \partial_x \varphi \|^2_{L^2}$, but it's not clear to me how to leverage this yet.