If the term $V(t)$ in the Hamiltonian changes suddenly ("impulsively"') between time $t$ and $t+\Delta t$, in a time $\Delta t$ short compared with all relevant periods, and assuming only that $\left[V\left(t^{\prime}\right), V\left(t^{\prime \prime}\right)\right]=0$ during the impulse, show that the time development operator is given by
$$
T(t+\Delta t, t)=\exp \left[-\frac{i}{\hbar} \int_t^{t+\Delta t} V\left(t^{\prime}\right) d t^{\prime}\right]
$$
Note especially that the state vector remains unchanged during a sudden change of $V$ by a finite amount.