Given a C*-algebra $\mathcal{A}$
Consider a free generator $\delta_0:\mathcal{D}_0\to\mathcal{A}$ with $\overline{\mathcal{D}_0}=\mathcal{A}$.
Introduce a perturbation $\delta_V:\mathcal{A}\to\mathcal{A}:A\mapsto\imath[V,A]$ with $V=V^*\in\mathcal{A}$.
Denote to full generator by: $\delta:=\delta_0+\delta_V$
Investigate its dynamics: $\tau^{t}(A):=e^{it\delta}(A)$
How to derive the Schwinder-Dyon-expansion: $$\tau^t(A)=\tau_0^t(A)+\sum_{n=1}^\infty\int_{0\leq s_1\leq\ldots\leq s_n\leq t}\imath[\tau^{s_1}(V),\ldots,\imath[\tau^{s_n}(V),\tau^{t}(A)]\ldots]\mathrm{d}s_1\ldots\mathrm{d}s_n$$ (I'm so stuck on this since weeks and still no idea how to unfold this relation. Can you help me?)
Also I still struggle with the rigorous construction of the dynamics: $$A\in\mathcal{D}_0^\omega:\quad\tau^{t}(A):=\sum_{k=0}^\infty\delta^k(A)$$ (The problem here is that the analytic elements may be to few in several senses, or?)
Moreover, can't it happen that it defines no dynamics at all?
Hahaaa, I got it thanks to my supervisor. =D
All formal calculations...
Consider the abstract cauchy problem: $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\tau^t\tau_0^{-t}\right)(Z)=\tau^t(\delta-\delta_0)\tau_0^{-t}(Z)=\tau^t\imath\left[V,\tau_0^{-t}(Z)\right]=\left(\tau^t\tau_0^{-t}\right)\imath\left[\tau_0^{t}(V),Z\right]$$ so one gets the recursion relation: $$\left(\tau^t\tau_0^{-t}\right)(Z)=Z+\int_0^t\mathrm{d}s\left(\tau^t\tau_0^{-t}\right)\imath\left[\tau_0^{s}(V),Z\right]$$ and therefore the expansion: $$\left(\tau^t\tau_0^{-t}\right)(Z)=Z+\imath\int_0^t\mathrm{d}s\left[\tau_0^{s}(V),Z\right]+\imath^2\int_0^t\mathrm{d}s\int_0^s\mathrm{d}r\left[\tau_0^r(V),\left[\tau_0^{s}(V),Z\right]\right]+\ldots$$ That is after inserting the original argument: $$Z:=\tau_0^t(X):\quad\tau^t(X)=\sum_{n=0}^\infty\imath^n\int_{0\leq s_n\leq\ldots\leq s_1\leq s_0=t}\mathrm{d}s_1\ldots\mathrm{d}s_n\left[\tau_0^{s_n}(V),\ldots\left[\tau_0^{s_1}(V),\tau_0^t(X)\right]\ldots\right]$$ (Mind the order of integral bounds w.r.t. the order of commutators.)
(Note that starting with the reverse expression leads to the same result after suitable substitutions.)