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 $\exp (-iK/\hslash )$ 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 ${\sin }^2(|k|/\hslash )$. The ratio $|k|/\hslash$, 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.