= Solution
{c}
Choose the relative phases of the two <energy eigenstates> so that $d=\langle\psi_2|\widehat D_0|\psi_1\rangle$ is real and nonnegative. Define the resonant <Rabi angular frequency> $\Omega=d/\hbar$, the <Rabi detuning> $\delta=\omega_0-\omega$, and $\Sigma=\omega_0+\omega$. Substitution into the <Schrödinger equation>, followed by projection onto the two <eigenstates>, gives the exact <interaction picture> equations
$$
i\dot c_1=\frac{\Omega}{2}(e^{-i\delta t}+e^{-i\Sigma t})c_2,\qquad
i\dot c_2=\frac{\Omega}{2}(e^{i\delta t}+e^{i\Sigma t})c_1.
$$
Keep both <probability amplitudes> dynamical: replacing $c_1$ by one would incorrectly impose weak excitation and miss complete population transfer.
For $|\delta|,\Omega\ll\omega_0$, apply the <rotating-wave approximation> to discard the rapidly oscillating terms at $\Sigma$. Differentiating the second remaining equation and using the first gives
$$
\ddot c_2-i\delta\dot c_2+\frac{\Omega^2}{4}c_2=0.
$$
The <initial conditions> are $c_2(0)=0$ and $\dot c_2(0)=-i\Omega/2$. With $\Omega_R=\sqrt{\Omega^2+\delta^2}$, their solution is
$$
c_2(t)=-i\frac{\Omega}{\Omega_R}e^{i\delta t/2}\sin\frac{\Omega_Rt}{2},\qquad
c_1(t)=e^{-i\delta t/2}\left(\cos\frac{\Omega_Rt}{2}+i\frac{\delta}{\Omega_R}\sin\frac{\Omega_Rt}{2}\right).
$$
Consequently,
$$
\boxed{|c_2(t)|^2=U\sin^2(\alpha t),\qquad
U=\frac{\Omega^2}{\Omega^2+\delta^2},\qquad
\alpha=\tfrac12\sqrt{\Omega^2+\delta^2}.}
$$
This result permits order-one excitation, including full population transfer on <resonance>, even though the drive is small compared with the optical <angular frequency>.
For an average from switch-on to time $\tau$, integrate the squared sine:
$$
\boxed{\overline P_2(\tau)=\frac1\tau\int_0^\tau|c_2(t)|^2dt
=\frac U2\left[1-\frac{\sin(\Omega_R\tau)}{\Omega_R\tau}\right],\qquad
\lim_{\tau\to\infty}\overline P_2=\frac U2.}
$$
On <resonance> this limiting average is $1/2$, although the instantaneous <probability> repeatedly reaches one. The zero-coupling, zero-<Rabi detuning> case is understood separately as $P_2=0$.
\b[Near <resonance> alone does not make this an exact result for arbitrarily strong driving.] The stated sinusoidal <Hamiltonian> also needs the <rotating-wave approximation>. A concrete counterexample is $\omega=\omega_0$ and $\Omega=\omega_0$: the exact equations give $P_2(t)=\Omega^2t^2+O(t^4)$ at switch-on, while the claimed rotating-wave sinusoid gives $\Omega^2t^2/4+O(t^4)$. Thus “without assuming weak radiation” can mean without assuming small excitation, but cannot remove the drive-to-carrier scale separation. The infinite-time average above belongs to the rotating-wave model; omitted corrections can accumulate over sufficiently long times.
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