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In this paper by Golovach et al.: http://journals.aps.org/prb/abstract/10.1103/PhysRevB.74.165319 there is the following equation for spin evolution: $$\langle \dot{\bf{S}} \rangle=({\boldsymbol \omega_z}+\delta{\boldsymbol \omega}(t))\times \langle {\bf S} \rangle,$$ where ${\boldsymbol \omega}_z=g\mu_BB{\bf n}/\hbar$. When they consider a general driving field: $$\delta{\boldsymbol \omega}(t) = \delta{\boldsymbol \omega}_a\sin(\omega_{ac} t)+\delta{\boldsymbol \omega}_b\cos(\omega_{ac} t)$$ they obtain the following expression for the Rabi frequency: $${\boldsymbol \omega}_R(t)=\frac{1}{2} \left( \delta{\boldsymbol \omega}_a\times{\bf n}-\left[\delta{\boldsymbol \omega}_b \times {\bf n} \right]\times {\bf n} \right)$$ by using the rotating wave approximation. This last equation is what I really want to understand. The case for ${\bf n}={\bf k}$ is quite easy to prove but how do we obtain the relation for $\bf n$ at an arbitrary direction?

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    $\begingroup$ I'm not entirely sure what part you want to know about so I'll just leave this other post here and see if you have questions :D $\endgroup$ – DanielSank Jan 19 '16 at 7:49
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I guess I will take my own answer as the possible solution: taking the Hamiltonian in the interaction picture and applying the rotating wave approximation. The transformed Hamiltonian includes a term with the Rabi vector in the form: ${\boldsymbol \omega}_R(t)\cdot {\bf{S}}$. The only problem is that I also obtain an additional time dependent term of the form: $(\delta{\boldsymbol \omega}(t)\cdot {\bf n})({\bf n}\cdot{\bf{S}})$. This term could be obviated if both $\delta{\boldsymbol \omega}_a$ and $\delta{\boldsymbol \omega}_b$ were perpendicular to ${\bf n}$, this last condition, however, is not mentioned in the paper.

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