Polarization vectors with inverted momentum Given a $2\rightarrow2$ body scattering problem in CM frame, i.e., $\gamma_1\gamma_2\rightarrow\gamma_3\gamma_4$ process, and the coordinate is chosen such that $\gamma_1$ is moving along $+z$-axis, with scattering angle $\theta$ of $\gamma_3$ in the $xz$-plane.
Then the polarization vectors for for $\gamma_1$ is by convention:
\begin{align}
\hat{e}_+&=-\frac{1}{\sqrt{2}}(\hat{x}+i\hat{y})\\
\hat{e}_-&=\frac{1}{\sqrt{2}}(\hat{x}-i\hat{y})
\end{align}
and for $\gamma_3$, one just needs to apply $R_y(\theta)$ on the vectors above.
However, how do I find the polarization vectors for $\gamma_2$ and $\gamma_4$, where they have momentum $\vec{k}_2=-\vec{k}_1$ and $\vec{k}_3=-\vec{k}_4$?
My idea is to apply a rotation in $y$-direction with angle $\pi$ and $\pi-\theta$ on polarization vectors of $\gamma_1$ and $\gamma_3$ correspondingly.
In short, the question would be how to find the polarization vectors with inverted momentum $\vec{k}$?
 A: The polarization vectors for any photon in the $xz$-plane* whose momentum vector makes an angle $\vartheta$ with the $z$-axis are those obtained from rotating $\hat{e}_{+}=\hat{e}^{(1)}_{+}$ and $\hat{e}_{-}=\hat{e}^{(1)}_{-}$ with $R_{y}(\vartheta)$. Since the momentum $\vec{k}_{2}$ of the second incoming photon points in the direction opposite ($180^{\circ}$ away from) $\vec{k}_{1}$,
the right- and left-circular polarization vectors for $\gamma_{2}$ are
\begin{align}
\hat{e}^{(2)}_{+}=R_{y}(\pi)\hat{e}^{(1)}_{+} & =-\frac{1}{\sqrt{2}}(-\hat{x}+i\hat{y})=\hat{e}^{(1)}_{-} \\
\hat{e}^{(2)}_{-}=R_{y}(\pi)\hat{e}^{(1)}_{-} & =\frac{1}{\sqrt{2}}(-\hat{x}-i\hat{y})=\hat{e}^{(1)}_{+}.
\end{align}
So for a photon traveling the opposite direction, the roles of the two circular polarization vectors are interchanged. This uses, that $R_{y}(\pi)\hat{x}=-\hat{x}$ and
$R_{y}(\pi)\hat{y}=\hat{y}$.
The relationship between the helicity basis polarization vectors for $\gamma_{3}$ and $\gamma_{4}$ is the same:
\begin{align}
\hat{e}^{(4)}_{+}=R_{y}(\pi)\hat{e}^{(3)}_{+} & =R_{y}(\pi)R_{y}(\theta)\hat{e}^{(1)}_{+}=R_{y}(\pi+\theta)\hat{e}^{(1)}_{+}=\hat{e}^{(3)}_{-} \\
\hat{e}^{(4)}_{-}=R_{y}(\pi)\hat{e}^{(3)}_{-} & =R_{y}(\pi)R_{y}(\theta)\hat{e}^{(1)}_{-}=R_{y}(\pi+\theta)\hat{e}^{(1)}_{-}=\hat{e}^{(3)}_{+}.
\end{align}
*You can always choose coordinates for the $2\rightarrow2$ process so it likes in the $xz$-plane.
