Let $\delta$ be a positive self-adjoint unbounded operator and let $u$ be a unitary operator on an infinite-dimensional Hilbert space.
Is it true that $(u \delta u)^{it} = u \delta^{it} u^{*}$ for all $t \in \mathbb{R}$?
I have a computation that would indicate this is true, but I'm not sure if all the steps are actually valid. First, note that there exists an unbounded self-adjoint operator $A$ such that $\delta^{it} = e^{itA}$. We then have \begin{equation*} -i\frac{d}{dt}\bigg|_{t=0} (u \delta u^{*})^{it} =-i\frac{d}{dt}\bigg|_{t=0} (u e^{A} u^{*})^{it} = -i\frac{d}{dt}\bigg|_{t=0} (e^{uAu^{*}})^{it} = uAu^{*}. \end{equation*} On the other hand, we have \begin{equation*} -i\frac{d}{dt}\bigg|_{t=0} u \delta^{it} u^{*} = -i\frac{d}{dt}\bigg|_{t=0} u e^{itA} u^{*} = uAu^{*}. \end{equation*} By Stone's theorem on one-parameter unitary groups we would then conclude that indeed $(u\delta u)^{it} = u \delta^{it}u^{*}$.
I'm not sure if I'm actually allowed to conclude that $\delta = e^{A}$, and I'm also unsure about $(u e^{A} u^{*}) = e^{uAu^{*}}$.