Dyson Series

Formula

U(t,t_0) =\mathcal T\exp\!\left[-\frac{i}{\hbar} \int_{t_0}^{t}H(s)\,ds\right]

means

\begin{aligned} U(t,t_0) &=I+\sum_{n=1}^{\infty} \left(-\frac{i}{\hbar}\right)^n \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2\cdots\\ &\qquad\times \int_{t_0}^{t_{n-1}}dt_n\, H(t_1)H(t_2)\cdots H(t_n). \end{aligned}

The first terms are

U^{(1)}=-\frac{i}{\hbar}\int_{t_0}^{t}H(t_1)\,dt_1,

U^{(2)}=-\frac{1}{\hbar^2} \int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\, H(t_1)H(t_2).

Later-time operators stand to the left.
In the interaction picture replace $H$ by the interaction $V_I(t)$ .
If $[H(t_1),H(t_2)]=0$ for all times, time ordering drops out and the ordinary exponential of the integral is exact.

Symbol	Meaning
$U(t,t_0)$	evolution operator
$\mathcal T$	time-ordering operator
$H(t)$	time-dependent Hamiltonian or generator
$t_1\gt\cdots\gt t_n$	ordered integration region

The full series is a formal solution; it becomes an approximation only after truncation.
A finite truncation is generally unitary only through its retained order.
For bounded $H$ , with $M=\hbar^{-1}\int\|H\|dt$ , the $n$ th term is bounded by $M^n/n!$ ; unbounded operators require domain analysis.
A first-order transition amplitude usually produces a second-order probability when the zeroth-order amplitude vanishes.
Do not reverse operator order or omit the sign $(-i)^2=-1$ .

The integral-equation derivation, convergence estimate, unitarity bookkeeping, exercises, and references are at Dyson Expansion as Formal Evolution.

Detailed transition amplitudes belong at Dyson Expansion for Transition Amplitudes.