Past exam of the theoretical physics II course of the University of Cambridge / 2013 / Paper 1 / 1 / a / Solution

Let $\Delta ={\Omega }_0-\omega$ and $R(t)=e^{-i\omega t{S}_z/\hslash }$. Differentiating $|\Psi ⟩=R|{\Psi }_R⟩$ gives the rotating-frame Hamiltonian operator $R^{†}HR-i\hslash R^{†}\dot{R}$. The angular momentum commutator yields $R^{†}{S}_{\pm }R=e^{\pm i\omega t}{S}_{\pm }$, while $-i\hslash R^{†}\dot{R}=-\omega S_z$. Thus the rotating-frame Rabi Hamiltonian isThis transformation is exact for the circular drive; no rotating-wave approximation has been used.