= Time-integrated pulse operator has action units
For a short applied perturbation $V(t)$ in a <Quantum Hamiltonian>, define $K=\int V(t)\,dt$. Since $V$ has <energy> units, $K$ has <action> units. In the ideal impulsive limit, during which the unperturbed evolution is negligible, the <quantum state> changes by $\exp(-iK/\hbar)$ if the perturbation commutes with itself at different times. More general pulses require <time ordering>.
Writing the impulse as $V(t)=K\delta(t-t_0)$ makes the dimensions explicit: the <Dirac delta function> in time has inverse-time units. For an <off-diagonal two-state impulse> with coupling $K_{21}=k$, the transition <probability> is $\sin^2(|k|/\hbar)$. The ratio $|k|/\hbar$, rather than $|k|$ alone, is the dimensionless pulse strength. Integrating across the full pulse gives its full area; an artificial factor of one-half must not be introduced by treating the pulse as an endpoint without specifying a regularization.
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