Dyson series

Remarks

If $\norm{V(s)} \le K$ for $s \in [t_0, t]$, the $n$-th term has norm at most $\frac{1}{n!}\left( \frac{K (t - t_0)}{\hbar} \right)^n$, so the series converges in operator norm. Its sum is the unitary solution of $i\hbar \, \partial_t U(t, t_0) = V(t) U(t, t_0)$ with $U(t_0, t_0) = I$. Writing the $n$-fold integral over the cube $[t_0, t]^n$ with the factors ordered by decreasing time gives the notation $U(t, t_0) = \mathcal{T} \exp\left( -\frac{i}{\hbar} \int_{t_0}^{t} V(s)\, ds \right)$, the time-ordered exponential.

Example: if $V(t) = V$ does not depend on $t$, the $n$-fold integral equals $\frac{(t - t_0)^n}{n!} V^n$, and the series is $e^{-iV(t - t_0)/\hbar}$.

Non-example of the hypothesis: let $V(t)$ be unbounded, as for $V = x^4$ on $L^2(\R)$. Then the terms need not be defined on a common domain, and the series need not converge. It is then only a formal expansion.