# Driving $\sigma$ transition with light in superposition of $\pi_x$ and $\pi_y$ polarization of slightly different frequencies

Lets assume the following experiment.

Circularly polarized laser light is sent through a Mach-Zender interferometer $\left(l_1 = l_2 \sim \,\mathrm{cm}\right)$ made up of polarizing beam splitters instead of beam splitters. Second, light in one of the arms is frequency shifted $(\mbox{a few }\delta f\sim \,\mathrm{MHz})$. Light at the output is used to excite atoms.

Say that in some moment in the region where atoms are, two light modes are roughly $\pi/2$ shifted. Can light now drive circular light transition $\left(m_f= \pm1\right)$, even though two linear polarizations don't have the exact same frequency? If yes, how does this influence excitation and how to determine detuning from atomic transition in this case? Some tips about how to treat this exactly would be appreciated.

An additional assumption that might be important is that Rabi freq. is in $\mathrm{GHz}$ regime for proper circular polarized light.

The two frequency components have polarization $|V\rangle$ and $|H\rangle$, respectively. Linear polarization propagating along the quantization axis of the atoms will drive $\sigma^+$ and $\sigma^-$ transitions, so we should decompose the linear polarizations into circular: up to normalization and sign conventions, we can say $|V\rangle = |R\rangle + |L\rangle$ and $|H\rangle = -i(|R\rangle - |L\rangle)$ (where $|R\rangle$ and $|L\rangle$ are right handed and left handed). The $|R\rangle$ components drive $\sigma^+$ transitions and the $|L\rangle$ components drive $\sigma^-$ transitions.
Now, suppose both components have the same frequency but one component has phase $e^{i\pi/2} = i$. The first component $|V\rangle$ drives $\sigma^+$ and $\sigma^-$. The second component $i |H\rangle = |R\rangle - |L\rangle$ drives $\sigma^+$ and $-\sigma^-$ transitions. The $\sigma^+$ driving constructively interferes and the $\sigma^-$ driving cancels out - we are left with just $\sigma^+$ driving as if the combined field were purely circularly polarized (which it is!)