# Topology and Polarization of Optical Vortex Fields from Atomic Phased Arrays
## 4 Polarization Parameters of Atomic-Array Radiation
 \tag{32}\]  
To evaluate the orientation and alignment parameters, \(p_{z}\) and \(p_{zz}\), we use the definitions from Equations (30) and (31) and the electric field components from the above equation,  
\[p_{z}=\frac{|E^{\mu}_{+1}|^{2}-|E^{\mu}_{-1}|^{2}}{|E^{\mu}_{+1}|^{2}+|E^{\mu} _{-1}|^{2}+|E^{\mu}_{0}|^{2}} \tag{33}\]  
\[p_{zz}=\frac{|E^{\mu}_{+1}|^{2}+|E^{\mu}_{-1}|^{2}-2|E^{\mu}_{0}|^{2}}{|E^{\mu }_{+1}|^{2}+|E^{\mu}_{-1}|^{2}+|E^{\mu}_{0}|^{2}}. \tag{34}\]  
We are interested in particular in the far-field polarization properties of the array's radiation in the vicinity of phase singularity, where \(\theta_{k}\approx\frac{R}{z}\ll 1\); therefore, [PERSON] expansion in \(\theta_{k}\) is justified. Keeping only the leading terms in \(\theta_{k}\), we obtain for the positive-helicity component (\(\eta_{1}\)),  
\[ \begin{split}&(1+m_{z}\cos^{2}\theta_{k})(-i)^{-1}J_{l+m_{z}-1}( \kappa\rho)\approx\\ &(1+m_{z}-m_{z}\theta_{k}^{2})(-i)^{-1}\frac{1}{(l+m_{z}-1)!}( \frac{x\theta_{k}}{2})^{|l+m_{z}-1|},\end{split}  \tag{35}\]  
where \(x=k\rho=\frac{2\pi}{\lambda}\rho\). For the negative-helicity component (\(\eta_{-1}\)), we have  
\[ \begin{split}&(1-m_{z}\cos^{2}\theta_{k})(-i)J_{l+m_{z}+1}( \kappa\rho)\approx\\ &(1-m_{z}+m_{z}\theta_{k}^{2})(-i)\frac{1}{(l+m_{z}+1)!}(\frac{x \theta_{k}}{2})^{|l+m_{z}+1|},\end{split}  \tag{36}\]  
and for the longitudinal field component (\(\eta_{0}\)),  
\[ \begin{split}&\frac{1}{\sqrt{2}}\sin 2\theta_{k})_{l+m_{z}}( \kappa\rho)\approx\\ &\sqrt{2}\theta_{k}\frac{1}{(l+m_{z})!}(\frac{x\theta_{k}}{2})^{ |l+m_{z}|},\end{split}  \tag{37}\]  
where we left out a common factor that can be restored from Equation (32). Substituting \(m_{z}=+1\) in the above expressions, we obtain  
\[ \begin{split}|E_{-1}^{\mu}|^{2}&\sim\theta_{k}^{2(l+4) }\frac{1}{(l+2)!^{2}}(\frac{x}{2})^{2|l+2|}\\ |E_{+1}^{\mu}|^{2}&\sim\frac{4}{l!^{2}}(\frac{x}{2} )^{2l}\theta_{k}^{2|l|}\\ |E_{0}^{\mu}|^{2}&\sim\frac{2\theta_{k}^{2(l+2)}}{( l+1)!^{2}}(\frac{x}{2})^{2|l+1|},\end{split}  \tag{38}\]  
where a common factor is omitted since it does not affect polarization parameters. The term \(|E_{+1}^{\mu}|^{2}\) contains the lowest order of \(\theta_{k}^{2l}\). Therefore, in the leading order of \(\theta_{k}\) expansion  
\[ \begin{split} p_{z}&\approx\frac{|E_{+1}^{\mu}|^{2 }}{|E_{+1}^{\mu}|^{2}}=1\\ p_{zz}&\approx\frac{|E_{+1}^{\mu}|^{2}}{|E_{+1}^{ \mu}|^{2}}=1.\end{split}  \tag{39}\]  
Therefore, both \(p_{z}\) (orientation) and \(p_{zz}\) (alignment) are equal to unity in the leading order of \(\theta_{k}\), i.e., in the far field case, provided that the \(l\) and \(m_{z}\) parameters have the same sign. The same result, \(p_{z}=p_{zz}=1\), is obtained for \(l\)=0. When \(m_{z}=-1\), we have for the field components  
\[ \begin{split}|E_{-1}^{\mu}|^{2}&\sim\frac{4\theta_ {k}^{2l}}{l!^{2}}(\frac{x}{2})^{2|l|}\\ |E_{+1}^{\mu}|^{2}&\sim\frac{\theta_{k}^{2l}}{(l-2)!^{2}}(\frac{x}{2})^{2|l-2|}\\ |E_{0}^{\mu}|^{2}&\sim\frac{2\theta_{k}^{2l}}{(l-1)!^{2}}(\frac{x}{2})^{2|l-1|},\end{split}  \tag{40}\]  
where \(|E_{-1}^{\mu}|^{2}\) and \(|E_{0}^{\mu}|^{2}\) have the same orders \(\theta_{k}^{2(l+m_{z})+2}\) and \(|E_{+1}^{\mu}|^{2}\) remains the order \(\theta_{k}^{2(l+m_{z})+2}\). \[p_{z}=\frac{\frac{1}{(l-2)!^{2}}(\frac{x}{2})^{2|l-2|}-\frac{4}{l!^{2}}(\frac {x}{2})^{2|l|}}{\frac{1}{(l-2)!^{2}}(\frac{x}{2})^{2|l-2|}+\frac{4}{l!^{2}}( \frac{x}{2})^{2|l|}+\frac{2}{(l-1)!^{2}}(\frac{x}{2})^{2|l-1|}}. \tag{41}\]  
\[p_{zz}=\frac{\frac{1}{(l-2)!^{2}}(\frac{x}{2})^{2|l-2|}+\frac{4}{l!^{2}}( \frac{x}{2})^{2|l|}-\frac{4}{(l-1)!^{2}}(\frac{x}{2})^{2|l-1|}}{\frac{1}{(l-2 )!^{2}}(\frac{x}{2})^{2|l-2|}+\frac{4}{l!^{2}}(\frac{x}{2})^{2|l|}+\frac{2}{( l-1)!^{2}}(\frac{x}{2})^{2|l-1|}}. \tag{42}\]  
Specifically for \(l=0\), \(m_{z}=-1\),  
\[ \begin{split} p_{z}&=-1,\\ p_{zz}&=1.\end{split}  \tag{43}\]  
For \(l=1\), \(m_{z}=-1\),  
\[ \begin{split} p_{z}&=\frac{-(\frac{x}{2})^{2}}{( \frac{x}{2})^{2}+0.5},\\ p_{zz}&=\frac{(\frac{x}{2})^{2}-1}{(\frac{x}{2})^{2 }+0.5}.\end{split}  \tag{44}\]For \(l=2\), \(m_{z}=-1\),  
\[