# Spin-Orbital Transformation in a Tight Focus of an Optical Vortex with Circular Polarization
## 3 The Total Lengthwise OAM and SAM Averaged over the Cross-Section of the Beam  
In reference [36], an expression was obtained for the partial power of the angular harmonics included in (3), (4), and (7) in the form:  
\[W_{\xi}=4\pi f^{2}\int\limits_{0}^{\beta}\sin^{2\xi+1}\left(\frac{\psi}{2} \right)\cos^{5-2\xi}\left(\frac{\psi}{2}\right)|A(\psi)|^{2}d\psi. \tag{8}\]  
Integrals (8) can be analytically calculated only in ordinary cases, but, nevertheless, it is possible to estimate the contribution of each angular harmonic. For instance, if a uniform field with a constant amplitude \(A(\psi)\equiv 1\) is focused, then [36]:  
\[\begin{array}{l}W_{0}=4\pi f^{2}\int\limits_{0}^{\beta}\sin\left(\frac{\psi}{ 2}\right)\cos^{5}\left(\frac{\psi}{2}\right)d\psi=\frac{4}{3}\pi f^{2}\left[1 -\cos^{6}\left(\frac{\beta}{2}\right)\right],\\ W_{1}=4\pi f^{2}\int\limits_{0}^{\beta}\sin^{3}\left(\frac{\psi}{2}\right) \cos^{3}\left(\frac{\psi}{2}\right)d\psi=\frac{2}{3}\pi f^{2}\sin^{4}\left( \frac{\beta}{2}\right)\left[1+2\cos^{2}\left(\frac{\beta}{2}\right)\right],\\ W_{2}=4\pi f^{2}\int\limits_{0}^{\beta}\sin^{5}\left(\frac{\psi}{2}\right) \cos\left(\frac{\psi}{2}\right)d\psi=\frac{4}{3}\pi f^{2}\sin^{6}\left(\frac{ \beta}{2}\right),\\ W=2\pi f^{2}[1-\cos(\beta)].\end{array} \tag{9}\]  
In the limiting case, when the numerical aperture is close to unity, i.e., \(\beta\approx\pi/2\), it can be obtained: \(W_{0}=(7/6)\pi f^{2}\), \(W_{1}=(1/3)\pi f^{2}\), \(W_{2}=(1/6)\pi f^{2}\). All energy coincides with the area of the hemisphere: \(W=W_{0}+W_{2}+2W_{1}=2\pi f^{2}\). It also should be noted that \(W_{0}-W_{2}=\)_W/2_.  
Based on (3), (4), (7), and (8), we find the total (averaged over the cross-section of the whole beam) intensity (total beam power), the OAM, and the lengthwise SAM in the focal plane:  
\[\hat{I}=W=\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi}I(\rho,\theta)\rho d \rho d\theta=\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi}\rho d\rho d \theta\left(\gamma_{0,m}^{2}+\gamma_{2,m+2}^{2}+2\gamma_{1,m+1}^{2}\right)=W_ {0}+W_{2}+2W_{1}. \tag{10}\]  
\[\hat{S}_{z}=\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi}S_{z}\rho d\rho d \theta=\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi}\rho d\rho d\theta\left( \gamma_{0,m}^{2}-\gamma_{2,m+2}^{2}\right)=W_{0}-W_{2}. \tag{11}\]  
\[\begin{array}{l}\hat{L}_{z}=\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi}L _{z}\rho d\rho d\theta=\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi}\rho d \rho d\theta\left(m\gamma_{0,m}^{2}+(m+2)\gamma_{2,m+2}^{2}+2(m+1)\gamma_{1,m +1}^{2}\right)=\\ =mW_{0}+(m+2)W_{2}+2(m+1)W_{1}.\end{array} \tag{12}\]  
Expressions (10)-(12) are the main result of this study. It follows from them that the sum of the full longitudinal OAM and SAM is equal to:  
\[\hat{S}_{z}+\hat{L}_{z}=mW_{0}+(m+2)W_{2}+2(m+1)W_{1}+W_{0}-W_{2}=(m+1)(W_{0}+ W_{2}+2W_{1})=(m+1)W. \tag{13}\]Equation (13) should be supplemented with a similar sum in the initial beam plane (1). To do this, we obtain the denseness of the lengthwise OAM component and the total longitudinal OAM in the launch plane from (1):  
\[\begin{array}{l}L_{z,0}=\text{Im}\left(E_{x}^{*}\frac{\partial E_{x}}{\partial q \varphi}+E_{y}^{*}\frac{\partial E_{y}}{\partial q\varphi}\right)=\ m|A(\psi)|^{ 2},\\ \hat{L}_{z,0}=\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi}L_{z,0}\rho d\rho d \vartheta=mW.\end{array} \tag{14}\]  
Adding (2) and (14), we obtain the sum of the OAM and SAM in the initial plane of the beam (1):  
\[\hat{S}_{z,0}+\hat{L}_{z,0}=W+mW=(m+1)W. \tag{15}\]