 Solution

Define $\Omega =\sqrt{{\Delta }^2+{\Omega }_R^2}$. Squaring the two-by-two matrix gives $H_R^2={\hslash }^2{\Omega }^{2}I/4$, soHere $|\pm {⟩}_R$ is the corresponding normalized eigenstate at $t=0$. In the rotating frame only its quantum phase changes. In the laboratory frame its spin direction also rotates about $z$ at angular velocity $\omega$: $⟨\mathbf{S}{⟩}_{\pm }=\pm \hslash ({\Omega }_R\cos \omega t,{\Omega }_R\sin \omega t,\Delta )/(2\Omega )$. An arbitrary initial state is the same fixed linear combination of these two solutions. If $\Omega =0$, the rotating-frame Hamiltonian operator vanishes and the entire doublet is degenerate; evolution is just $R(t)$.