# Transverse Spin Hall Effect and Twisted Polarization Ribbons at the Sharp Focus
## 4 Polarization-Twisted Ribbons in the Focal Plane
### An Azimuthally Polarized Optical Vortex  
Below, we analyze another example of an original vector light field that allows obtaining analytical relationships for projections of the major and minor axes of the polarization ellipse at the tight focus. An original azimuthally polarized optical vortex of topological charge \(n\) is described by the Jones vector:  
\[\textbf{E}(\varphi)=\exp(in\varphi)\bigg{(} \begin{array}{c}-\sin\varphi\\ \cos\varphi\end{array} \bigg{)}. \tag{21}\]  
A specific feature of light field (21) is that at its tight focus, the longitudinal projection of the **E**-vector equals zero. At the center of a linearly polarized light field (21), there is a V-point polarization singularity where the linear polarization vector is uncertain. For the V-point polarization singularity, the Poincare-Hopf index is \(\eta=1\), which means that when making a full circle about the beam center in the original plane, the linear polarization vectors make a full turn. According to Richards-Wolf theory [12], all projections of the **E**-vector onto the focal plane take the form [16]:  
\[ \begin{array}{l}E_{x}=\frac{\varphi^{+1}}{\varphi^{2}}e^{in\varphi}\big{[}e ^{i\varphi}R(r)+e^{-i\varphi}T(r)\big{]},\\ E_{y}=\frac{\varphi}{\varphi}e^{in\varphi}\big{[}e^{i\varphi}R(r)-e^{-i\varphi }T(r)\big{]},\\ E_{z}=0,\\ R(r)=(I_{0,n+1}+I_{2,n+1}),\\ T(r)=(I_{0,n-1}+I_{2,n-1}).\end{array}  \tag{22}\]The first two equations in (22) can be rearranged to Equation (7) by explicitly expressing projections of vectors of the major and minor axes of a polarization ellipse in the focal plane:  
\[ \begin{array}{l}\mathbf{E}=e^{i\gamma}(\mathbf{A}+i\mathbf{B}),\\ A_{x}=\frac{1}{2}[R(r)\cos(n+1)\varphi+T(r)\cos(n-1)\varphi],\\ A_{y}=\frac{1}{2}[R(r)\sin(n+1)\varphi-T(r)\sin(n-1)\varphi],\\ A_{z}=0,\\ B_{x}=\frac{1}{2}[R(r)\sin(n+1)\varphi+T(r)\sin(n-1)\varphi],\\ B_{y}=\frac{1}{2}[-R(r)\cos(n+1)\varphi+T(r)\cos(n-1)\varphi],\\ B_{z}=0,\\ \gamma=(n+1)\frac{\pi}{2}.\end{array}  \tag{23}\]  
Putting \(n=1\), (23) is rearranged to:  
\[ \begin{array}{l}A_{x,1}=\frac{1}{2}[R(r)\cos 2\varphi+T(r)],\\ A_{y,1}=\frac{1}{2}R(r)\sin 2\varphi,\\ A_{z}=0,\\ B_{x,1}=\frac{1}{2}R(r)\sin 2\varphi,\\ B_{y}=\frac{1}{2}[-R(r)\cos 2\varphi+T],\\ B_{z}=0.\end{array}  \tag{24}\]  
The dot product of the vectors \(\mathbf{A}\) and \(\mathbf{B}\) equals: \(\mathbf{AB}=RT\sin(2n\varphi)/2\), suggesting that these vectors form the polarization ellipse axes only at \(n=0\) (azimuthal polarization). At other values of \(n\), although lying in the focal plane, the \(\mathbf{A}\) and \(\mathbf{B}\) vectors are mutually orthogonal only on circles of certain radius when either \(R(r)=0\), or \(T(r)=0\), or at angles \(\varphi=\pi p/(2n)\), \(p=0\),\(1\),\(2\),... From (24) we find that a C-point of circular polarization at the tight focus center has the index \(lc=2\), because upon making a circle around the optical axis, the polarization ellipse major axis makes four half-turns, as projections of the \(\mathbf{A}\) vector in (24) depend on the double azimuthal angle. At the tight focus, field (21) has just a longitudinal SAM vector projection:  
\[S_{z}=2\text{Im}\big{(}E_{x}^{*}E_{y}\big{)}=2\big{(}A_{x}B_{y}-A_{y}B_{x} \big{)}=\Big{(}T^{2}(r)-R^{2}(r)\Big{)}/2. \tag{25}\]  
With the longitudinal SAM vector projection in (25) being \(\varphi\)-independent, it changes sign or equals zero on circles of certain radius. So, the intensity distribution at the focus has radial symmetry: \(I(r)=|\mathbf{A}|^{2}+|\mathbf{B}|^{2}=\big{(}R^{2}(r)+T^{2}(r)\big{)}/2\). In the focal plane, on a circle of radius \(r_{2}\) on which an equality \(R(r_{2})=T(r_{2})\) holds, field (21) is linearly polarized because the longitudinal SAM component in (25) equals zero, \(S_{z}=0\).