with respect to the $a$-axis, whereas $\hat V$ is the constant transformation from linear- to circular-polarization basis. Making the transformation of operators (\[eqn:2\]) and (\[eqn:3\]) to the helicity basis, ${\bf \hat O'} \to \hat U^\dag  {\bf \hat O'}\hat U$, we obtain: $$\begin{aligned}
\label{eqn:6}
{\bf \hat S'} = {\bm \kappa }\hat \sigma~,~~{\bf \hat L'} = 
- i{\bf k} \times \partial _{\bf k}  - {\bf \hat A}_B  \times {\bf k}~,\\
\label{eqn:7}
{\bf \hat r'} = i\partial _{\bf k}  - {\bf \hat A}_B~,~~{\bf \hat p} = {\bf k}~,~~{\hat w}=\omega\end{aligned}$$ Here we included the momentum and energy operators, ${\bf \hat p}$ and ${\hat w}$ (which are unaffected by the transformations), $\omega$ is the frequency, the helicity is diagonal: $\hat \sigma  = {\mathop{\rm diag}\nolimits} \left( {1, - 1,0} \right)$, and $$\label{eqn:8}
{\bf \hat A}_B  =  - \frac{{{\bf k} \times {\bf \hat S}}}{{k^2 }} - i\hat U^\dag  \partial _{\bf k} \hat U = \frac{{m - \cos \theta }}{{k\sin \theta }}\hat \sigma {\bf e}_\phi$$ is the Berry gauge field (connection) which corresponds to the monopole curvature ${\bf \hat F}_B  = \partial _{\bf k}  \times {\bf \hat A}_B  = \hat \sigma \,{\bf k}/k^3$ [@Photon; @Hawton]. Hereafter we choose the gauge $m=1$, which corresponds to the absence of the 