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An impulse integrated across its full support produces the impulsive quantum kick $K=e^{-i\hat{U}/\hslash }$. Thus the leading transition amplitude is $-i{U}_{21}/\hslash$ andHere $U_{21}$ has units of action. The off-diagonal two-state impulse is exactly $P={\sin }^2(|U_{21}|/\hslash )$ for a Hermitian two-state impulse with zero diagonal. No energy denominator occurs: the impulse is instantaneous.

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.