TITLE: C*-Algebras: Dynamics vs. Derivations
QUESTION [5 upvotes]: Problem
Given a C*-algebra $\mathcal{A}$.
Consider dynamics $\tau:\mathbb{R}\to\mathrm{Aut}(\mathcal{A})$ and $\tau':\mathbb{R}\to\mathrm{Aut}(\mathcal{A})$.
(More precisely, strongly continuous one-parameter groups.)
Denote their derivations by $\delta:\mathcal{D}\to\mathcal{A}$ and $\delta':\mathcal{D}'\to\mathcal{A}$.
Then one has:
$$\delta=\delta'\implies\tau=\tau'$$
(Here, equality is meant in terms of operators resp. maps.)
How do I check this?
For dynamics over Hilbert spaces I would proceed by:
$$i\frac{\mathrm{d}}{\mathrm{d}t}\|\varphi(t)\|^2=\langle H\varphi(t),\varphi(t)\rangle-\langle\varphi(t),H\varphi(t)\rangle=0$$
But for the C*-algebra case this path is not directly available.
Disclaimer
I hope to get a hint from here.
(I haven't got any respond yet from stack exchange.)

REPLY [6 votes]: As was discussed in the comments, it suffices to see that $\delta$ determines $\tau$ uniquely on $\mathrm{dom}(\delta)$ which is dense in $\mathcal{A}$.  Suppose that $x_0 \in \mathrm{dom}(\delta)$.  Check that $t \mapsto \tau^t(x_0)$ is a solution to the initial value problem
\begin{align*}
\frac{d}{dt} x(t)  = \delta( x(t)) && x(0) = x_0.
\end{align*}
If $x$ is any solution then
$$\frac{d}{dt} \left( \tau^{-t}(x(t)) \right) = -\tau^{-t}\delta(x(t)) + \tau^{-t}\left(\delta(x(t))\right) = 0$$
and it follows that $\tau^{-t}(x(t)) \equiv x_0$ so that $x(t) = \tau^t (x_0)$ is the unique solution.
Basically this all works for Banach space flows too. See the answers to my own question here.