*} \right)   \times \frac{{\bf K}^-}{2}} \frac{d^2 {\bf k}^{\prime}_\bot }
{2\pi } \frac{d^2 {\bf k}_\bot }{2\pi },$$ where ${\bf K}^{\pm}={\mathbf{k}} \pm {\mathbf{k'}}$. Note that in the decomposition (\[eqn:33\])–(\[eqn:35\]) the two helicity components are exactly separated without an interference term [@Berry2009]. Also, the “electro-magnetic democracy” discussed by Berry [@Berry2009] is accomodated, because switching to magnetic-field plane-wave helicity amplitudes, ${{\tilde E}^\sigma } \rightarrow {{\tilde H}^\sigma } =  -i\sigma {{\tilde E}^\sigma }$, keeps Eqs. (\[eqn:33\])–(\[eqn:35\]) invariant.

The linear momentum of the beam (per unit $z$-length) is given by the 2D space integration of the momentum densities (\[eqn:33\])–(\[eqn:35\]): $\mathbf{P} = \int {\bm{\pi }\,d^2 \mathbf{r}_ \bot  }$. In doing so, we find that the spin momentum density makes no contribution to the linear momentum [@Li]: ${\bf P}^{\rm s} = \int {\bm{\pi }^{\rm s}  \,d^2 \mathbf{r}_ \bot = 0}$, while the orbital contribution yields $$\label{eqn:36}
\mathbf{P}^{\rm o}   = \mathbf{P} = \int {\bm{\pi }^{\rm o}  d^2 \mathbf{r}_ \bot   = \int {\mathbf{k}\left| {\tilde E^\sigma  } \right|^2 } d^2 \mathbf{k}_ \bot  }~,$$ which obviously coincides with Eq. (\[eqn:13\]).

The SAM and OAM of the beam (per unit $z$-length) can be obtained by the 2D space integration of their densities, i.e.: $$\label{eqn:37}
\mathbf{L} = \int {\mathbf{r} \times \bm{\pi }^{\rm o}  d^2 