Faraday rotation from circular phases ($\theta =({\varphi }_{+}-{\varphi }_{-})/2$)

The factor of one half follows from superposition, without assigning ambiguous right/left names to the circular polarizations. Use the complex amplitude convention $\mathrm{Re}[\mathbf{E}(z)e^{-i\omega t}]$ and fixed transverse unit vectorsFor equal transmitted wave amplitudes, let the accumulated wave phases be ${\varphi }_{+}$ and ${\varphi }_{-}$. Their sum isThe common wave phase $\bar{\varphi }$ sets the timing of the electric field oscillation; the real direction in parentheses sets its linear polarization. Opposite circular polarizations have changed their relative wave phase, yet their sum still oscillates along a line. Unequal transmitted wave amplitudes would generally give elliptical polarization instead.

In a uniform sample, with refractive indices $n_{+},n_{-}$ and vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the phase difference is $(k_{+}-k_{-})L=k_0(n_{+}-n_{-})L$. ThusThe second relation identifies the small signed circular birefringence responsible for the rotation in this basis. Unlike a waveplate, whose separated modes have orthogonal linear polarizations, this ideal rotator has modes with opposite circular polarizations.