= Solution
Let $\Delta=\Omega_0-\omega$ and $R(t)=e^{-i\omega tS_z/\hbar}$. Differentiating $|\Psi\rangle=R|\Psi_R\rangle$ gives the rotating-frame <Hamiltonian operator> $R^\dagger HR-i\hbar R^\dagger\dot R$. The <angular momentum> <commutator> yields $R^\dagger S_\pm R=e^{\pm i\omega t}S_\pm$, while $-i\hbar R^\dagger\dot R=-\omega S_z$. Thus the <rotating-frame Rabi Hamiltonian> is
$$
\boxed{H_R=\Delta S_z+\Omega_RS_x
=\frac\hbar2\begin{pmatrix}\Delta&\Omega_R\\\Omega_R&-\Delta\end{pmatrix}},\qquad
i\hbar\partial_t|\Psi_R\rangle=H_R|\Psi_R\rangle.
$$
This transformation is exact for the circular drive; no <rotating-wave approximation> has been used.
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