# Transverse Spin Hall Effect and Twisted Polarization Ribbons at the Sharp Focus
## 4 Polarization-Twisted Ribbons in the Focal Plane
### Cylindrical Vector Fields  
The Jones vector in the original plane of a cylindrical vector beam of \(n\)-th order (\(n\) is an integer) is given by:  
\[\mathbf{E}(\varphi)=\bigg{(} \begin{array}{c}\cos n\varphi\\ \sin n\varphi\end{array} \bigg{)}. \tag{17}\]  
Relationships for the \(\mathbf{E}\)-vector projections onto the focal plane from the original field in (17) have been known [8,12]:  
\[ \begin{array}{l}E_{x}=i^{n-1}(I_{0,n}\cos n\varphi+I_{2,n-2}\cos(n-2)\varphi ),\\ E_{y}=i^{n-1}(I_{0,n}\sin n\varphi-I_{2,n-2}\sin(n-2)\varphi),\\ E_{z}=2^{n}I_{1,n-1}\cos(n-1)\varphi.\end{array}  \tag{18}\]  
In (18), the integrals \(I_{\mu\
u}\) are defined by (3). Field projections in (18) can be expressed via projections of vectors \(\mathbf{A}\) and \(\mathbf{B}\) using Equation (8), then:  
\[ \begin{array}{l}\mathbf{E}=e^{i\gamma}(\mathbf{A}+i\mathbf{B}),\\ A_{x}=I_{0,n}\cos n\varphi+I_{2,n-2}\cos(n-2)\varphi,\\ A_{y}=I_{0,n}\sin n\varphi-I_{2,n-2}\sin(n-2)\varphi,\\ A_{z}=0,\\ B_{x}=0,\\ B_{y}=0,\\ B_{z}=2I_{1,n-1}\cos(n-1)\varphi,\\ \gamma=(n-1)\frac{\pi}{2}.\end{array}  \tag{19}\]  
From (19), the vectors \(\mathbf{A}\) and \(\mathbf{B}\) lying in the polarization ellipse plane are seen to be mutually orthogonal: \(\mathbf{AB}=0\). Considering that \(I_{0,n}>I_{1,n-1}\), the \(\mathbf{A}\) vector corresponds to the polarization ellipse major axis, with the \(\mathbf{B}\) vector corresponding to the minor axis. Considering that the \(\mathbf{B}\) vector has just a longitudinal projection, it is perpendicular to the focal plane. Meanwhile, with the \(\mathbf{A}\) vector having just transverse projections, it lies in the focal plane. At the same time, the spin density vector \(\mathbf{S}\) perpendicular to the two above-said vectors is described by the projections:  
\[ \begin{array}{l}S_{x}=4I_{1,n-1}\cos((n-1)\varphi)[I_{0,n}\sin(n\varphi)-I_ {2,n-2}\sin((n-2)\varphi)],\\ S_{y}=-4I_{1,n-1}\cos((n-1)\varphi)[I_{0,n}\cos(n\varphi)+I_{2,n-2}\cos((n-2) \varphi)],\\ S_{z}=0.\end{array}  \tag{20}\]  
Hence, we can see that the situation that we observed for field (1) just on a circle of radius \(r_{0}\), for field (17), occurs in the whole focal plane. What is more, the field at the focus is linearly polarized when \(\cos(n-1)\varphi=0\), i.e., at angles \(\varphi_{p}=\pi(1/2+p)/(n-1)\), where \(p=0\), 1, 2,... At angles somewhat smaller than \(\varphi_{p}\), the projection of the vector **B** is positive if \(I_{1,n-1}>0\), whereas at angles somewhat larger than \(\varphi_{p}\), the projection of the vector **B** is negative. At the same time, the **A** vector lying in the focal plane hardly changes. We can infer that the spin density vector **S**, which also lies in the focal plane, on passing the angles \(\varphi_{p}\) changes the direction to the opposite. Thus, in the focal plane of field (17), there is no longitudinal spin Hall effect, but there is a transverse spin Hall effect. In other words, on an axis-centered circle of arbitrary radius in the focal plane, there are \(4(n-1)\) regions such that the transverse spin components have an alternating sign in the neighboring regions. Besides, lying on this circle are \(4(n-1)\) points at which light is linearly polarized (the **B** vector equals zero) and the transverse spin also equals zero, \(\textbf{S}_{\perp}=0\). So, when making a full axis-centered circle of any radius lying in the focal plane, the polarization ellipse always remains perpendicular to the focal plane, with the polarization vectors rotating in the ellipse plane like spokes of a wheel. Thus, field (17) produces at the tight focus a \"photonic wheel\" effect [15]. A remarkable finding is that while traversing the focal plane, the light wave is not transversely polarized, as the wave vector and polarization ellipse lie in the same plane.