# Transverse Spin Hall Effect and Twisted Polarization Ribbons at the Sharp Focus
## 6 Numerical Modeling
Such a strip with an even number of half-turns is called a polarization-twisted ribbon [6]. In addition, it is seen from Figure 1 that upon moving along the ellipse around its center, approximately half-way in the yellow intensity pattern, the direction of the major axis of polarization ellipse changes to the opposite six times at uniform 60-degree intervals. Thus, farther from the optical axis in the focal plane, the polarization singularity index of field (1) equals \(lc=3\), being equal to the topological charge (\(n=3\)) of the original field in (1). Making note of the four blue arrows at the top of the internal red intensity ellipse in Figure 1, we can see that they twice change their direction to the opposite. Considering that the four blue-arrow polarization ellipses are narrow, we infer that the polarization ellipses are nearly perpendicular to the focal plane. Their blue color means that the minor axis of the polarization ellipse is opposite to the optical axis, whereas the SAM vector is near vertical and almost in the focal plane (being slightly negatively tilted relative to the optical axis). Hence, for two of the four blue arrows, the spin vector \(\mathbf{S}\) is directed nearly vertically upwards, being directed downwards for the other two. Such a spatial separation of focal regions with the oppositely directed transverse spin \(\mathbf{S}\) (upwards and downwards in Figure 1) may be called a transverse spin Hall effect. Such reasoning based on Figure 1 agrees well with the theoretical conclusions from Equation (13). Figure 2 depicts an intensity pattern (squeezed white-yellow-red ring) in the focal plane from the original cylindrical second-order vector beam of Equation (17) at \(n=2\). From (18), the intensity at the focus is seen to depend on \(\cos(2\varphi)\), leading to two intensity maxima on the horizontal axis (marked white in Figure 2). Note that at the center, instead of an intensity null, there is an intensity minimum proportional to the term \(I_{2,0}\) in (18). From (18), it follows that there are two zero-intensity points on the vertical axis, because at \(\varphi=\pi/2\) and \(\varphi=3\pi/2\), field (18) has the only non-zero projection: \(E_{x}=i(I_{2,0}-I_{0,2})\), which becomes equal to zero at \(I_{0,2}-I_{2,0}=0\). At these points of field (18), there are V-point polarization singularities where polarization is uncertain. From Figure 2, it can be seen that when making a full circle around these two points, the linear polarization vector rotates by \(2\pi\), similar to azimuthal polarization. That is, the said two V-points have the indices +1 (upper) and -1 (lower). At \(n=2\), the initial field (17) has an on-axis V-point with the index 2, which spits at the focus into two V-points, with the sum of their indices being equal to zero. It would seem that the polarization singularity index should not conserve upon tight focusing. However, when making a circle of a larger radius, the seeming contradiction is eliminated. When making a full circle along the outside (red) intensity ellipse, the linear polarization vectors make two full turns (four semi-turns); because of this, the field index in Figure 2 equals 2, coinciding with the original field index. From Figure 2, the polarization vectors are seen to lie in the longitudinal planes and perpendicularly to the focal plane (forming photonic wheels), meaning that only linear polarization vectors (the major axes of polarization ellipses) lie in the focal plane. In Figure 2, blue arrows are seen to prevail in the upper right and lower left quadrants, with red arrows dominating the upper left and lower right quadrants. Red arrows mark the right-handed circular polarization, with the spin vector \(\mathbf{S}\) in the focal plane being directed predominantly downwards. Blue arrows mark the left-handed circular polarization, with the spin vector directed mainly upwards.