equivalent to Eq. (\[eqn:12\]).

To derive the linear and angular momenta of the field, we use the Poynting vector which determines the momentum density (energy flow) [@Born]: $$\label{eqn:32}
{\bm{\pi}} = \frac{1}{g^2} \operatorname{Im} \left[ {{{\mathbf{E}}^*} \times \left( {\nabla  \times {\mathbf{E}}} \right)} \right]~.$$ Substituting here the Fourier decomposition (\[eqn:A6\]), we obtain $$\label{eqn:33}
\bm{\pi} = {\mathop{\rm Re}\nolimits} \iint {e^{i\left( {{\bf k} - {\bf k'}} \right) \cdot {\bf r}} \tilde E^{\sigma\prime *}  \tilde E^\sigma  {\bf e}^{\sigma\prime *} \times \left( {{\bf k} \times {\bf e}^\sigma  } \right)} \frac{d^2 {\bf k}^{\prime}_\bot}{2\pi} {\frac{d^2 {\bf k}_\bot}{2\pi} },$$ where $\tilde E^{\sigma\prime}   = \tilde E^\sigma  \left( {{\bf k'}} \right)$, ${\bf e}^{\sigma\prime}   = {\bf e}^\sigma  \left( {{\bf k'}} \right)$, and summation over $\sigma =\pm 1$ is implied hereinafter. After some calculations the total momentum density (\[eqn:33\]) can be decomposed into the orbital and spin parts as suggested in [@Berry2009; @Li], $\bm{\pi}=\bm{\pi}^{\rm o}+\bm{\pi}^{\rm s}$: $$\label{eqn:34}
{\bm \pi}^{\rm o} = \iint {e^{i {\bf K}^{-} \cdot {\bf r}} {\tilde E}^{\sigma} {\tilde E}^{\sigma\prime *} \left( {\bf e}^{\sigma} \cdot {\bf e}^{\sigma\prime *} \right) \frac{{\bf K}^+}{2}} 
\frac{{d^2 {\mathbf{k}}^{\prime}_\bot }}
{{2\pi }} \frac{{d^2{\mathbf{k}}_\bot}}
{{2\pi }},$$ $$\label{eqn:35}
{\bm \pi}^{\rm s} = \iint {
e^{i {\bf K}^{-} \cdot {\bf r}}{\tilde E}^{\sigma} {\tilde E}^{\sigma\prime *} \left( {\bf e}^{\sigma} \times {\bf e}^{\sigma\prime 