# 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 $$\vec{k}$$?

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.