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Chinese Physics B
Vol. 29, Issue 8, (2020)

Quantization of electromagnetic modes and angular momentum on plasmonic nanowires

Guodong Zhu¹, Yangzhe Guo¹, Bin Dong^2、†, and Yurui Fang¹

Author Affiliations

¹Key Laboratory of Materials Modification by Laser, Electron, and Ion Beams (Ministry of Education); School of Physics, Dalian University of Technology, Dalian 6024, China

²Key Laboratory of New Energy and Rare Earth Resource Utilization of State Ethnic Affairs Commission, Key Laboratory of Photosensitive Materials & Devices of Liaoning Province, School of Physics and Materials Engineering, Dalian Nationalities University, Dalian 116600, China

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DOI: 10.1088/1674-1056/ab9698 Cite this Article

Guodong Zhu, Yangzhe Guo, Bin Dong, Yurui Fang. Quantization of electromagnetic modes and angular momentum on plasmonic nanowires[J]. Chinese Physics B, 2020, 29(8): Copy Citation Text

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Abstract

Quantum theory of surface plasmons is very important for studying the interactions between light and different metal nanostructures in nanoplasmonics. In this work, using the canonical quantization method, the SPPs on nanowires and their orbital and spin angular momentums are investigated. The results show that the SPPs on nanowire carry both orbital and spin momentums during propagation. Later, the result is applied to the plasmonic nanowire waveguide to show the agreement of the theory. The study is helpful for the nano wire based plasmonic interactions and the quantum information based optical circuit in the future.

Keywords

optical angular momentum surface plasmons surface waves

$$ \begin{eqnarray}{\nabla }^{2}\left\{\begin{array}{c}{\boldsymbol{E}}\\ {\boldsymbol{H}}\end{array}\right\}-\mu \varepsilon \displaystyle \frac{{\partial }^{2}}{\partial {t}^{2}}\left\{\begin{array}{c}{\boldsymbol{E}}\\ {\boldsymbol{H}}\end{array}\right\}=0.\end{eqnarray}$$

(1)

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$$ \begin{eqnarray}{\boldsymbol{A}}({\boldsymbol{r}},t)=\displaystyle \frac{1}{\sqrt{V}}\displaystyle \sum _{k,m}{{\boldsymbol{A}}}_{k,m}(t){{\rm{e}}}^{{\rm{i}}{\boldsymbol{k}}\cdot {\boldsymbol{r}}},\end{eqnarray}$$

(2)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}_{k,m}(t){{\rm{e}}}^{{\rm{i}}{\boldsymbol{k}}\cdot {\boldsymbol{r}}} & = & -\displaystyle \frac{{\rm{i}}}{\omega }\{[\displaystyle \frac{{\rm{i}}m}{{k}_{j,m}\rho }{a}_{j,m}{F}_{j,m}({k}_{j,m\perp }\rho )\\ & & +\displaystyle \frac{{\rm{i}}{k}_{\parallel ,m}{k}_{j,m\perp }}{{k}_{j,m}^{2}}{b}_{j,m}{F}_{j,{m}^{^{\prime} }}({k}_{j,m\perp }\rho )]\hat{{\boldsymbol{\rho }}}\\ & & +[-\displaystyle \frac{{k}_{j\perp }}{{k}_{j}}{a}_{j,m}{F}_{j,{m}^{^{\prime} }}({k}_{j,m\perp }\rho )\\ & & -\displaystyle \frac{m{k}_{\parallel }}{{k}_{j}^{2}\rho }{b}_{j,m}{F}_{j,m}({k}_{j,m\perp }\rho )]\hat{\phi }\\ & & +\displaystyle \frac{{k}_{j,m\perp }^{2}}{{k}_{j,m}^{2}}{b}_{j,m}{F}_{j,m}({k}_{j,m\perp }\rho )\hat{{\boldsymbol{z}}}\}{{\rm{e}}}^{{\rm{i}}(m\phi +{k}_{\parallel ,m}z)}{{\rm{e}}}^{-{\rm{i}}\omega t}\\ & = & {{\boldsymbol{A}}}_{k,m}(t){{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}\cdot {\boldsymbol{r}}}={{\boldsymbol{A}}}_{k,m}(t){{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}},\end{array}\end{eqnarray}$$

(3)

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$$ \begin{eqnarray}\displaystyle \int {{\rm{d}}}^{3}r{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}}=V{\delta }_{mm^{\prime} }\end{eqnarray}$$

(4)

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$$ \begin{eqnarray}{\boldsymbol{A}}({\boldsymbol{r}},t)=\displaystyle \frac{1}{2\sqrt{V}}\displaystyle \sum _{k,m}[{{\boldsymbol{A}}}_{m}(t){{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}+{\rm{c}}.{\rm{c}}.].\end{eqnarray}$$

(5)

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$$ \begin{eqnarray}[{{\boldsymbol{A}}}_{m}({\boldsymbol{r}},t),{{\boldsymbol{p}}}_{{m}^{^{\prime} }}({{\boldsymbol{r}}}^{^{\prime} },t)]={\rm{i}}\hslash {\delta }_{m,{m}^{^{\prime} }}\delta ({\boldsymbol{r}}-{{\boldsymbol{r}}}^{^{\prime} }).\end{eqnarray}$$

(6)

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$$ \begin{eqnarray}{{\boldsymbol{Q}}}_{m}=\sqrt{\displaystyle \frac{1}{4\pi {c}^{2}}}({{\boldsymbol{A}}}_{m}+{{\boldsymbol{A}}}_{m}^{\ast }),\end{eqnarray}$$

(7)

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$$ \begin{eqnarray}{{\boldsymbol{P}}}_{m}=-\displaystyle \frac{{\rm{i}}\omega }{\sqrt{4\pi {c}^{2}}}({{\boldsymbol{A}}}_{m}-{{\boldsymbol{A}}}_{m}^{\ast }),\end{eqnarray}$$

(8)

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$$ \begin{eqnarray}{\boldsymbol{A}}({\boldsymbol{r}},t)=\displaystyle \frac{\sqrt{4\pi {c}^{2}}}{2\sqrt{V}}\displaystyle \sum _{m}\left[{{\boldsymbol{Q}}}_{m}\cos {\boldsymbol{k}}\cdot {\boldsymbol{r}}-\displaystyle \frac{1}{{\omega }_{m}}{{\boldsymbol{P}}}_{m}\sin {\boldsymbol{k}}\cdot {\boldsymbol{r}}\right],\end{eqnarray}$$

(9)

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$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{E}}({\boldsymbol{r}},t) & =-\displaystyle \frac{1}{c}\displaystyle \frac{\partial {\boldsymbol{A}}({\boldsymbol{r}},t)}{\partial t}\\ & =-\displaystyle \frac{\sqrt{4\pi }}{2\sqrt{V}}\displaystyle \sum _{m}{\omega }_{m}\left[{{\boldsymbol{Q}}}_{m}\sin {\boldsymbol{k}}\cdot {\boldsymbol{r}}+\displaystyle \frac{1}{{\omega }_{m}}{{\boldsymbol{P}}}_{m}\cos {\boldsymbol{k}}\cdot {\boldsymbol{r}}\right],\end{array}\end{eqnarray}$$

(10)

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$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{B}}({\boldsymbol{r}},t) & =\nabla \times {\boldsymbol{A}}({\boldsymbol{r}},t)\\ & =-\displaystyle \frac{\sqrt{4\pi {c}^{2}}}{2\sqrt{V}}\displaystyle \sum _{m}{{\boldsymbol{k}}}_{m}\times \left[{{\boldsymbol{Q}}}_{m}\sin {\boldsymbol{k}}\cdot {\boldsymbol{r}}+\displaystyle \frac{1}{{\omega }_{m}}{{\boldsymbol{P}}}_{m}\cos {\boldsymbol{k}}\cdot {\boldsymbol{r}}\right].\end{array}\end{eqnarray}$$

(11)

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$$ \begin{eqnarray}H=\displaystyle \frac{1}{8\pi }\displaystyle \int {{\rm{d}}}^{3}r({|{\boldsymbol{E}}|}^{2}+{|{\boldsymbol{B}}|}^{2})=\displaystyle \frac{1}{2}\displaystyle \sum _{m}({{\boldsymbol{P}}}_{m}^{2}+{\omega }_{m}^{2}{{\boldsymbol{Q}}}_{m}^{2}).\end{eqnarray}$$

(12)

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$$ \begin{eqnarray}[{{\boldsymbol{Q}}}_{m},{{\boldsymbol{P}}}_{m}]={\rm{i}}\hslash .\end{eqnarray}$$

(13)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m}=\displaystyle \frac{1}{\sqrt{2\hslash \omega }}({\omega }_{m}{{\boldsymbol{Q}}}_{m}+{\rm{i}}{{\boldsymbol{P}}}_{m}),\end{eqnarray}$$

(14)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m}^{\dagger }=\displaystyle \frac{1}{\sqrt{2\hslash \omega }}({\omega }_{m}{{\boldsymbol{Q}}}_{m}-{\rm{i}}{{\boldsymbol{P}}}_{m}).\end{eqnarray}$$

(15)

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$$ \begin{eqnarray}\begin{array}{ll}[{{\boldsymbol{a}}}_{m},{{\boldsymbol{a}}}_{m}^{\dagger }] & =\displaystyle \frac{1}{2\hslash {\omega }_{m}}[{\omega }_{m}{{\boldsymbol{Q}}}_{m}+{\rm{i}}{{\boldsymbol{P}}}_{m},{\omega }_{m}{{\boldsymbol{Q}}}_{m}-{\rm{i}}{{\boldsymbol{P}}}_{m}],\\ & =\displaystyle \frac{1}{2\hslash {\omega }_{m}}(-{\rm{i}}{\omega }_{m}[{{\boldsymbol{Q}}}_{m},{{\boldsymbol{P}}}_{m}]+{\rm{i}}{\omega }_{m}[{{\boldsymbol{P}}}_{m},{{\boldsymbol{Q}}}_{m}])=1.\end{array}\end{eqnarray}$$

(16)

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$$ \begin{eqnarray}{\boldsymbol{A}}({\boldsymbol{r}},t)=\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle \sum _{m}\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}[{{\boldsymbol{a}}}_{m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}+{{\boldsymbol{a}}}_{m}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\end{eqnarray}$$

(17)

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$$ \begin{eqnarray}{\boldsymbol{E}}({\boldsymbol{r}},t)={\rm{i}}\displaystyle \frac{\sqrt{2\pi \hslash }}{\sqrt{V}}\displaystyle \sum _{m}{\sqrt{\omega }}_{m}[{{\boldsymbol{a}}}_{m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{{\boldsymbol{a}}}_{m}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\end{eqnarray}$$

(18)

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$$ \begin{eqnarray}{\boldsymbol{B}}({\boldsymbol{r}},t)={\rm{i}}\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle \sum _{m}\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}{{\boldsymbol{k}}}_{m}\times [{{\boldsymbol{a}}}_{m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{{\boldsymbol{a}}}_{m}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\end{eqnarray}$$

(19)

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$$ \begin{eqnarray}H=\displaystyle \sum _{m}{H}_{m}=\displaystyle \sum _{m}\hslash {\omega }_{m}({{\boldsymbol{a}}}_{m}^{\dagger }{{\boldsymbol{a}}}_{m}+1/2)=\displaystyle \sum _{m}\hslash {\omega }_{m}({n}_{m}+1/2),\end{eqnarray}$$

(20)

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$$ \begin{eqnarray}{\rm{i}}\hslash {\mathop{{\boldsymbol{a}}}\limits^{.}}_{m}=[{{\boldsymbol{a}}}_{m},{\boldsymbol{H}}],\end{eqnarray}$$

(21)

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$$ \begin{eqnarray}{\rm{i}}\hslash {\mathop{{\boldsymbol{a}}}\limits^{.}}_{m}^{\dagger }=[{{\boldsymbol{a}}}_{m}^{\dagger },{\boldsymbol{H}}].\end{eqnarray}$$

(22)

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$$ \begin{eqnarray}\begin{array}{ll}\,\quad{\boldsymbol{A}}({\boldsymbol{r}},t)\\ =\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle \sum _{m,\alpha }\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}[{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}+{\hat{\epsilon }}_{\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\,\end{array}\end{eqnarray}$$

(23)

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$$ \begin{eqnarray}\begin{array}{ll}\,\quad{\boldsymbol{E}}({\boldsymbol{r}},t)\\ ={\rm{i}}\sqrt{\displaystyle \frac{2\pi \hslash }{V}}\displaystyle \sum _{m,\alpha }{\sqrt{\omega }}_{m}[{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{\hat{\epsilon }}_{\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\end{array}\end{eqnarray}$$

(24)

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$$ \begin{eqnarray}\begin{array}{ll}\,\quad{\boldsymbol{B}}({\boldsymbol{r}},t)\\ ={\rm{i}}\sqrt{\displaystyle \frac{2\pi \hslash }{V}}\displaystyle \sum _{m,\alpha }\displaystyle \frac{c{{\boldsymbol{k}}}_{m}}{{\sqrt{\omega }}_{m}}\times [{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{\hat{\epsilon }}_{\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}].\end{array}\end{eqnarray}$$

(25)

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$$ \begin{eqnarray}\begin{array}{lll}{\boldsymbol{p}} & = & \displaystyle \int {{\rm{d}}}^{3}r\displaystyle \frac{{\boldsymbol{E}}\times {\boldsymbol{B}}}{4\pi c}\\ & = & \displaystyle \frac{{{\rm{i}}}^{2}}{4\pi c}\displaystyle \frac{hc}{V}\displaystyle \sum _{m,\alpha }\displaystyle \sum _{{m}^{^{\prime} },{\alpha }^{^{\prime} }}[{\hat{\epsilon }}_{\alpha }\times ({k}^{^{\prime} }\times {\hat{\epsilon }}_{{\alpha }^{^{\prime} }})]\times \displaystyle \int {{\rm{d}}}^{3}r({{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}\\ & & -{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}})({{\boldsymbol{a}}}_{{m}^{^{\prime} },{\alpha }^{^{\prime} }}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}}-{{\boldsymbol{a}}}_{{m}^{^{\prime} },{\alpha }^{^{\prime} }}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}})\\ & = & \displaystyle \frac{h}{4\pi }\displaystyle \sum _{m,\alpha }{{\boldsymbol{k}}}_{m}[{{\boldsymbol{a}}}_{m,\alpha }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }+{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }]=\displaystyle \sum _{m}\hslash {{\boldsymbol{k}}}_{m}{n}_{m}.\end{array}\end{eqnarray}$$

(26)

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$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{J}} & =\displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r{\boldsymbol{r}}\times ({\boldsymbol{E}}\times {\boldsymbol{B}})\\ & =\displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r[{E}_{\alpha }({\boldsymbol{r}}\times \nabla ){A}_{\alpha }+{\boldsymbol{E}}\times {\boldsymbol{A}}].\end{array}\end{eqnarray}$$

(27)

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$$ \begin{eqnarray}\begin{array}{ll}{{\boldsymbol{l}}}_{{\boldsymbol{\alpha }}}= & ({\boldsymbol{r}}\times \nabla ){A}_{\alpha }{|}_{\rho =R}=\displaystyle {\sum }_{m}{\hat{r}}_{\alpha }\cdot \left(\begin{array}{c}-\displaystyle \frac{z}{R}\displaystyle \frac{\partial }{\partial \phi }\hat{{\boldsymbol{\rho }}}\\ (z\displaystyle \frac{\partial }{\partial \rho }-R\displaystyle \frac{\partial }{\partial z})\hat{\phi }\\ \displaystyle \frac{\partial }{\partial \phi }\hat{{\boldsymbol{z}}}\end{array}\right){A}_{\alpha }={\rm{i}}\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle {\sum }_{m}\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}{\hat{r}}_{\alpha }\cdot R\cdot \left(\begin{array}{c}-\displaystyle \frac{z}{R}m{k}_{\phi }\\ -{k}_{z,m}\\ m{k}_{\phi }\end{array}\right)[{{\boldsymbol{a}}}_{m\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}\\ & -{{\boldsymbol{a}}}_{m\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}]=\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle {\sum }_{m}\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}{\hat{r}}_{\alpha }\cdot \left(\begin{array}{c}-\displaystyle \frac{z}{R}m\\ -R{k}_{z,m}\\ m\end{array}\right)[{{\boldsymbol{a}}}_{m\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}-{{\boldsymbol{a}}}_{m\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}].\end{array}\end{eqnarray}$$

(28)

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$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{L}}= & \displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r{E}_{\alpha }{l}_{\alpha }=\displaystyle \frac{{\rm{i}}\hslash }{2V}\displaystyle {\sum }_{m,{m}^{^{\prime} },\alpha ,{\alpha }^{^{\prime} }}\displaystyle \frac{{\omega }_{m}}{\sqrt{{\omega }_{m}{\omega }_{{m}^{^{\prime} }}}}{\hat{\epsilon }}_{{\boldsymbol{\alpha }}}\cdot {\hat{\epsilon }}_{{{\boldsymbol{\alpha }}}^{^{\prime} }}\displaystyle \int {{\rm{d}}}^{3}rR\left(\begin{array}{c}-\displaystyle \frac{z}{R}m{k}_{\phi }\\ -{k}_{z,m}\\ m{k}_{\phi }\end{array}\right)[{{\boldsymbol{a}}}_{m\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}-{{\boldsymbol{a}}}_{m\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}][{{\boldsymbol{a}}}_{{m}^{^{\prime} }{\alpha }^{^{\prime} }}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,{m}^{^{\prime} }}{\boldsymbol{r}}}\\ & -{{\boldsymbol{a}}}_{{m}^{^{\prime} }{\alpha }^{^{\prime} }}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,{m}^{^{\prime} }}{\boldsymbol{r}}}]=\displaystyle \frac{{\rm{i}}\hslash }{2}\displaystyle {\sum }_{m,\alpha }\left(\begin{array}{c}0\\ -R{k}_{z,m}\\ m\end{array}\right)({{\boldsymbol{a}}}_{m,\alpha }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }+{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha })=\displaystyle {\sum }_{m}{\rm{i}}\hslash {{\boldsymbol{l}}}_{m}{{\boldsymbol{n}}}_{m},\end{array}\end{eqnarray}$$

(29)

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$$ \begin{eqnarray}{{\boldsymbol{l}}}_{m}=\left(\begin{array}{c}0\\ -R{k}_{z,m}\\ m\end{array}\right),\end{eqnarray}$$

(30)

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$$ \begin{eqnarray}{L}_{z}=\displaystyle \frac{{\boldsymbol{L}}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}={\rm{i}}\hslash \displaystyle {\sum }_{m}m{{\boldsymbol{n}}}_{m},\end{eqnarray}$$

(31)

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$$ \begin{eqnarray}\begin{array}{ll}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}} & {={\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}={{\rm{e}}}^{{\rm{i}}(m\phi +{k}_{z}z)}\\ & =[\cos (m\phi )+{\rm{i}}\sin (m\phi )]{{\rm{e}}}^{{\rm{i}}{k}_{z}z},\end{array}\end{eqnarray}$$

(32)

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$$ \begin{eqnarray}\begin{array}{ll}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,-m}{\boldsymbol{r}}} & ={{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{-m}\cdot {\boldsymbol{r}}}={{\rm{e}}}^{{\rm{i}}(-m\phi +{k}_{z}z)}\\ & =[\cos (m\phi )-{\rm{i}}\sin (m\phi )]{{\rm{e}}}^{{\rm{i}}{k}_{z}z},\end{array}\end{eqnarray}$$

(33)

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$$ \begin{eqnarray}\begin{array}{ll}{{\boldsymbol{a}}}_{m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}} & =\displaystyle \sum _{\alpha }{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}\\ & =\displaystyle \sum _{\alpha }{\lambda }_{+\alpha }{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{k}_{\parallel ,m}\cdot {\boldsymbol{r}}}={{\boldsymbol{a}}}_{L,m}{{\rm{e}}}^{{\rm{i}}{k}_{z}z},\end{array}\end{eqnarray}$$

(34)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{-m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{-m}\cdot {\boldsymbol{r}}}={{\boldsymbol{a}}}_{R,m}{{\rm{e}}}^{{\rm{i}}{k}_{z}z},\end{eqnarray}$$

(35)

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$$ \begin{eqnarray}{\boldsymbol{s}}={\boldsymbol{E}}\times {\boldsymbol{A}}=\displaystyle \frac{{\rm{i}}}{\hslash }({E}_{\alpha }{({{\boldsymbol{S}}}_{\alpha })}_{\beta \gamma }{{\boldsymbol{A}}}_{\alpha }),\end{eqnarray}$$

(36)

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$$ \begin{eqnarray}{({{\boldsymbol{S}}}_{\alpha })}_{\beta \gamma }=-{\rm{i}}\hslash {{\boldsymbol{\varepsilon }}}_{\alpha \beta \gamma }.\end{eqnarray}$$

(37)

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$$ \begin{eqnarray}[({{\boldsymbol{S}}}_{i}),({{\boldsymbol{S}}}_{j})]={\rm{i}}\hslash {{\boldsymbol{\varepsilon }}}_{ijk}({{\boldsymbol{S}}}_{k}),\end{eqnarray}$$

(38)

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$$ \begin{eqnarray}{{\boldsymbol{S}}}^{2}={({{\boldsymbol{S}}}_{i})}_{lm}{({{\boldsymbol{S}}}_{i})}_{mn}={(-{\rm{i}}\hslash )}^{2}({{\boldsymbol{\delta }}}_{ln}{{\boldsymbol{\delta }}}_{mn}-{{\boldsymbol{\delta }}}_{ln}{{\boldsymbol{\delta }}}_{mn}),\end{eqnarray}$$

(39)

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$$ \begin{eqnarray}{({{\boldsymbol{S}}}^{2})}_{ln}={(-{\rm{i}}\hslash )}^{2}({{\boldsymbol{\delta }}}_{ln}-3{{\boldsymbol{\delta }}}_{ln})=2{\hslash }^{2}{{\boldsymbol{\delta }}}_{ln},\end{eqnarray}$$

(40)

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$$ \begin{eqnarray}s(s+1)=2,\,\,\,s=1.\end{eqnarray}$$

(41)

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$$ \begin{eqnarray}{\boldsymbol{S}}=\displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r({\boldsymbol{s}})=\displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r{E}_{\alpha }{({\boldsymbol{S}})}_{\beta \gamma }{A}_{\alpha },\end{eqnarray}$$

(42)

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$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{S}}= & \displaystyle \frac{-1}{4\pi c}\displaystyle \frac{{\rm{i}}hc}{V}\displaystyle \sum _{ma}\displaystyle \sum _{{m}^{^{\prime} }{a}^{^{\prime} }}{\varepsilon }_{\alpha \beta \gamma }{\hat{\epsilon }}_{\alpha }\times {\hat{\epsilon }}_{{\alpha }^{^{\prime} }}\displaystyle \int {{\rm{d}}}^{3}r[{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}\\ & +{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}}][{{\boldsymbol{a}}}_{{m}^{^{\prime} },{\alpha }^{^{\prime} }}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{{\boldsymbol{a}}}_{{m}^{^{\prime} },{\alpha }^{^{\prime} }}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}}],\end{array}\end{eqnarray}$$

(43)

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$$ \begin{eqnarray}{\boldsymbol{S}}=\displaystyle \frac{{\rm{i}}h}{2\pi }\displaystyle \sum _{m\alpha {\alpha }^{^{\prime} }}{\varepsilon }_{\alpha \beta \gamma }{\hat{\epsilon }}_{\alpha }\times {\hat{\epsilon }}_{{\alpha }^{^{\prime} }}({{\boldsymbol{a}}}_{m,\alpha }{{\boldsymbol{a}}}_{m,{\alpha }^{^{\prime} }}^{\dagger }-{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,{\alpha }^{^{\prime} }}).\end{eqnarray}$$

(44)

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$$ \begin{eqnarray}{\boldsymbol{S}}=\displaystyle \sum _{m\alpha {\alpha }^{^{\prime} }}{\rm{i}}\hslash {\hat{{\boldsymbol{k}}}}_{m,{\alpha }^{^{\prime\prime} }}({{\boldsymbol{a}}}_{m,\alpha }{{\boldsymbol{a}}}_{m,{\alpha }^{^{\prime} }}^{\dagger }-{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,{\alpha }^{^{\prime} }}).\end{eqnarray}$$

(45)

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$$ \begin{eqnarray}{\hat{s}}_{\alpha ,L}={\hat{s}}_{\alpha ,+}=\displaystyle \frac{1}{\sqrt{2}}({\hat{\epsilon }}_{\beta }+{\rm{i}}{\hat{\epsilon }}_{\gamma }),\end{eqnarray}$$

(46)

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$$ \begin{eqnarray}{\hat{s}}_{\alpha ,R}={\hat{s}}_{\alpha ,-}=\displaystyle \frac{1}{\sqrt{2}}({\hat{\epsilon }}_{\beta }-{\rm{i}}{\hat{\epsilon }}_{\gamma }).\end{eqnarray}$$

(47)

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$$ \begin{eqnarray}\begin{array}{ll}\,\quad{\hat{\epsilon }}_{\beta }{{\boldsymbol{a}}}_{m,\beta }+{\hat{\epsilon }}_{\gamma }{{\boldsymbol{a}}}_{m,\gamma }\\ =\displaystyle \frac{1}{\sqrt{2}}({\hat{s}}_{\alpha ,L}+{\hat{s}}_{\alpha ,R}){{\boldsymbol{a}}}_{m,\beta }+\displaystyle \frac{1}{{\rm{i}}\sqrt{2}}({\hat{s}}_{\alpha ,L}-{\hat{s}}_{\alpha ,R}){{\boldsymbol{a}}}_{m,\gamma }\\ =\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }-{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }){\hat{s}}_{\alpha ,L}+\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }+{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }){\hat{s}}_{\alpha ,R}.\end{array}\end{eqnarray}$$

(48)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,L,\alpha }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }-{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }),\end{eqnarray}$$

(49)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,R,\alpha }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }+{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }),\end{eqnarray}$$

(50)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,L,\alpha }^{\dagger }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }^{\dagger }-{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }^{\dagger }),\end{eqnarray}$$

(51)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,R,\alpha }^{\dagger }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }^{\dagger }+{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }^{\dagger }),\end{eqnarray}$$

(52)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,\beta }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,L,\alpha }+{{\boldsymbol{a}}}_{m,R,\alpha }),\end{eqnarray}$$

(53)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,\gamma }=\displaystyle \frac{{\rm{i}}}{\sqrt{2}}({{\boldsymbol{a}}}_{m,L,\alpha }-{{\boldsymbol{a}}}_{m,R,\alpha }),\end{eqnarray}$$

(54)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,\beta }^{\dagger }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,L,\alpha }^{\dagger }+{{\boldsymbol{a}}}_{m,R,\alpha }^{\dagger }),\end{eqnarray}$$

(55)

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$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,\gamma }^{\dagger }=\displaystyle \frac{-{\rm{i}}}{\sqrt{2}}({{\boldsymbol{a}}}_{m,L,\alpha }^{\dagger }-{{\boldsymbol{a}}}_{m,R,\alpha }^{\dagger }).\end{eqnarray}$$

(56)

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$$ \begin{eqnarray}\displaystyle \sum _{a=\alpha ,\beta }{\hat{\epsilon }}_{a}{{\boldsymbol{a}}}_{m,a}=\displaystyle \sum _{\gamma }({\hat{s}}_{\gamma ,L}{{\boldsymbol{a}}}_{m,L,\gamma }+{\hat{s}}_{\gamma ,R}{{\boldsymbol{a}}}_{m,R,\gamma }),\end{eqnarray}$$

(57)

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$$ \begin{eqnarray}\displaystyle \sum _{a=\alpha ,\beta }{\hat{\epsilon }}_{a}^{\dagger }{{\boldsymbol{a}}}_{m,a}^{\dagger }=\displaystyle \sum _{\gamma }({\hat{s}}_{\gamma ,L}^{\dagger }{{\boldsymbol{a}}}_{m,L,\gamma }^{\dagger }+{\hat{s}}_{\gamma ,R}^{\dagger }{{\boldsymbol{a}}}_{m,R,\gamma }^{\dagger }).\end{eqnarray}$$

(58)

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$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{S}} & =\displaystyle \sum _{m\alpha }\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{a}}}_{m,L,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,L,\alpha }-{{\boldsymbol{a}}}_{m,R,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,R,\alpha })\\ & =\displaystyle \sum _{m\alpha }\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha }),\end{array}\end{eqnarray}$$

(59)

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$$ \begin{eqnarray}{\hat{S}}_{\alpha }=\displaystyle \frac{\hat{S}\cdot {{\boldsymbol{k}}}_{\alpha }}{|{{\boldsymbol{k}}}_{\alpha }|}.\end{eqnarray}$$

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$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{J}}={\boldsymbol{L}}+{\boldsymbol{S}}= & \displaystyle {\sum }_{m,\alpha }{\rm{i}}\hslash {{\boldsymbol{k}}}_{m,\alpha }{\hat{n}}_{m,\alpha }\\ & +\displaystyle \sum _{m,\alpha }\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha }).\end{array}\end{eqnarray}$$

(61)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}_{l=1}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & & +{{\boldsymbol{A}}}_{+1}\cos (\phi -\displaystyle \frac{\pi }{2}){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t+\displaystyle \frac{\pi }{2})}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & & +{\rm{i}}{{\boldsymbol{A}}}_{+1}\sin (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }{{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}.\end{array}\end{eqnarray}$$

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$$ \begin{eqnarray}\begin{array}{ll}{{\boldsymbol{A}}}_{l=1}^{R}({\boldsymbol{r}},t) & ={{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }{{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & ={{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}({{\boldsymbol{A}}}_{0}+{{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}(\Delta kz+\phi )}),\end{array}\end{eqnarray}$$

(63)

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$$ \begin{eqnarray}\begin{array}{ll}{{\boldsymbol{A}}}_{l=1}^{L}({\boldsymbol{r}},t) & ={{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{-1}{{\rm{e}}}^{-{\rm{i}}\phi }{{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & ={{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}({{\boldsymbol{A}}}_{0}+{{\boldsymbol{A}}}_{-1}{{\rm{e}}}^{{\rm{i}}(\Delta kz-\phi )}).\end{array}\end{eqnarray}$$

(64)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}_{l=1}^{0}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\sin (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}({{\boldsymbol{A}}}_{0}+\sin (\phi ){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\Delta kz}),\end{array}\end{eqnarray}$$

(65)

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$$ \begin{eqnarray}\begin{array}{ll}{H}^{L}|{n}\rangle_{m} & =\displaystyle \sum _{m=0,-1;a}{\hat{\epsilon }}_{\alpha }\hslash {\omega }_{m}({{\boldsymbol{a}}}_{m,a}^{\dagger }{{\boldsymbol{a}}}_{m,a}+1/2)|{n}\rangle_{m}\\ & =\displaystyle \sum _{m=0,-1}\hslash {\omega }_{m}({n}_{m}+1/2),\end{array}\end{eqnarray}$$

(66)

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$$ \begin{eqnarray}\begin{array}{lll}{H}^{R}|{n}\rangle_{m} & = & \displaystyle \sum _{m=0,+1;a}{\hat{\epsilon }}_{\alpha }\hslash {\omega }_{m}({{\boldsymbol{a}}}_{m,a}^{\dagger }{{\boldsymbol{a}}}_{m,a}+1/2)|{n}\rangle_{m}\\ & = & \displaystyle \sum _{m=0,+1}\hslash {\omega }_{m}({n}_{m}+1/2),\end{array}\end{eqnarray}$$

(67)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{p}}}^{L}|{n}\rangle_{m} & = & \displaystyle \frac{h}{4\pi }\displaystyle \sum _{m=0,-1}{{\boldsymbol{k}}}_{m}[{{\boldsymbol{a}}}_{m}{{\boldsymbol{a}}}_{m}^{\dagger }+{{\boldsymbol{a}}}_{m}^{\dagger }{{\boldsymbol{a}}}_{m}]|{n}\rangle_{m}\\ & = & \displaystyle \sum _{m=0,-1}\hslash {{\boldsymbol{k}}}_{m}{n}_{m},\end{array}\end{eqnarray}$$

(68)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{p}}}^{R}|{n}\rangle_{m} & = & \displaystyle \frac{h}{4\pi }\displaystyle \sum _{m=0,+1}{{\boldsymbol{k}}}_{m}[{{\boldsymbol{a}}}_{m}{{\boldsymbol{a}}}_{m}^{\dagger }+{{\boldsymbol{a}}}_{m}^{\dagger }{{\boldsymbol{a}}}_{m}]|{n}\rangle_{m}\\ & = & \displaystyle \sum _{m=0,+1}\hslash {{\boldsymbol{k}}}_{m}{n}_{m},\end{array}\end{eqnarray}$$

(69)

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$$ \begin{eqnarray}\begin{array}{lll}{L}_{z}^{L}|{n}\rangle_{m} & = & \displaystyle \frac{{\boldsymbol{L}}|{n}\rangle_{m}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}\\ & = & {\rm{i}}\hslash \displaystyle {\sum }_{m=0,-1}m\ast {n}_{m}=-{n}_{-1}\ast {\rm{i}}\hslash ,\end{array}\end{eqnarray}$$

(70)

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$$ \begin{eqnarray}\begin{array}{lll}{L}_{z}^{R}|{n}\rangle_{m} & = & \displaystyle \frac{{\boldsymbol{L}}|{n}\rangle_{m}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}\\ & = & {\rm{i}}\hslash \displaystyle {\sum }_{m=0,+1}m\ast {n}_{m}={n}_{+1}\ast {\rm{i}}\hslash ,\end{array}\end{eqnarray}$$

(71)

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$$ \begin{eqnarray}{{\boldsymbol{S}}}^{L}|{n}\rangle_{m}=\displaystyle \sum _{m=0,-1,a}\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha }),\end{eqnarray}$$

(72)

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$$ \begin{eqnarray}{{\boldsymbol{S}}}^{R}|{n}\rangle_{m}=\displaystyle \sum _{m=0,+1,a}\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha }).\end{eqnarray}$$

(73)

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$$ \begin{eqnarray}\begin{array}{lll} & & {{\boldsymbol{A}}}^{{\rm{RCP}}}({\boldsymbol{r}},t)\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}{{\rm{e}}}^{{\rm{i}}\displaystyle \frac{\pi }{2}}\\ & & +{{\boldsymbol{A}}}_{+1}\cos (\phi -\displaystyle \frac{\pi }{2}){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t+\displaystyle \frac{\pi }{2})}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{\rm{i}}{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & & +{\rm{i}}{{\boldsymbol{A}}}_{+1}\sin (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{\parallel ,0}z-\omega t)}+{\rm{i}}{{\boldsymbol{A}}}_{+1}(\cos (\phi )\\ & & +\sin (\phi )){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}.\end{array}\end{eqnarray}$$

(74)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}^{{\rm{RCP}}}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+\displaystyle \frac{1}{2}(({\rm{i}}+1){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }\\ & & +{{\boldsymbol{A}}}_{+1}({\rm{i}}-1){{\rm{e}}}^{-{\rm{i}}\phi }){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+\displaystyle \frac{1}{2}(({\rm{i}}+1){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }\\ & & -{{\boldsymbol{A}}}_{-1}({\rm{i}}-1){{\rm{e}}}^{-{\rm{i}}\phi }){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}.\end{array}\end{eqnarray}$$

(75)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}^{{\rm{LCP}}}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}{{\rm{e}}}^{-{\rm{i}}\displaystyle \frac{\pi }{2}}\\ & & +{{\boldsymbol{A}}}_{+1}\cos (\phi -\displaystyle \frac{\pi }{2}){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t+\displaystyle \frac{\pi }{2})}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}-{\rm{i}}{{\boldsymbol{A}}}_{+1}(\cos (\phi )\\ & & -\sin (\phi )){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\end{array}\end{eqnarray}$$

(76)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}^{{\rm{LCP}}}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+\displaystyle \frac{1}{2}((-1+{\rm{i}}){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }\\ & & +{{\boldsymbol{A}}}_{+1}(-1-{\rm{i}}){{\rm{e}}}^{-{\rm{i}}\phi }){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+\displaystyle \frac{1}{2}((-1+{\rm{i}}){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }\\ & & +{{\boldsymbol{A}}}_{-1}(1+{\rm{i}}){{\rm{e}}}^{-{\rm{i}}\phi }){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}.\end{array}\end{eqnarray}$$

(77)

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$$ \begin{eqnarray}\begin{array}{lll}{H}^{{\rm{LCP}}}{|n\rangle }_{m} & = & \displaystyle \sum _{m=0,\pm 1;\alpha }{\hat{\epsilon }}_{\alpha }\hslash {\omega }_{m}({{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }+1/2)\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}{|n\rangle }_{m}\right]\\ & = & \displaystyle \sum _{m=0,\pm 1}{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}\hslash {\omega }_{m}({n}_{m}+1/2)\\ & = & \displaystyle \sum _{m=0,+1}\hslash {\omega }_{m}({n}_{m}+1/2),\end{array}\end{eqnarray}$$

(78)

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$$ \begin{eqnarray}{H}^{{\rm{RCP}}}|{n}\rangle_{m}=\displaystyle \sum _{m=0,+1}\hslash {\omega }_{m}({n}_{m}+1/2),\end{eqnarray}$$

(79)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{p}}}^{{\rm{LCP}}}|{n}\rangle_{m} & = & \displaystyle \frac{h}{4\pi }\displaystyle \sum _{m=0,\pm 1}{{\boldsymbol{k}}}_{m}[{{\boldsymbol{a}}}_{m}{{\boldsymbol{a}}}_{m}^{\dagger }+{{\boldsymbol{a}}}_{m}^{\dagger }{{\boldsymbol{a}}}_{m}]\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right]\\ & = & \displaystyle \sum _{m=0,\pm 1}{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}\hslash c{{\boldsymbol{k}}}_{m}{n}_{m}\\ & = & \displaystyle \sum _{m=0,+1}\hslash {k}_{m}{n}_{m},\end{array}\end{eqnarray}$$

(80)

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$$ \begin{eqnarray}{{\boldsymbol{p}}}^{{\rm{RCP}}}|{n}\rangle_{m}=\displaystyle \sum _{m=0,+1}\hslash {k}_{m}{n}_{m},\end{eqnarray}$$

(81)

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$$ \begin{eqnarray}\begin{array}{lll}{L}_{z}^{{\rm{LCP}}}|{n}\rangle_{m} & = & \displaystyle \frac{{\boldsymbol{L}}|{n}\rangle_{m}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}={\rm{i}}h\displaystyle {\sum }_{m=0,\pm 1}m{\hat{n}}_{m}\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right]=0,\end{array}\end{eqnarray}$$

(82)

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$$ \begin{eqnarray}\begin{array}{lll}{L}_{z}^{{\rm{RCP}}}|{n}\rangle_{m} & = & \displaystyle \frac{{\boldsymbol{L}}|{n}\rangle_{m}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}={\rm{i}}h\displaystyle {\sum }_{m=0,\pm 1}m{\hat{n}}_{m}\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right]=0,\end{array}\end{eqnarray}$$

(83)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{S}}}^{{\rm{LCP}}}|{n}\rangle_{m} & = & \displaystyle \sum _{m=0,\pm 1,\alpha }\hslash {\hat{k}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha })\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right],\end{array}\end{eqnarray}$$

(84)

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$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{S}}}^{{\rm{RCP}}}|{n}\rangle_{m} & = & \displaystyle \sum _{m=0,\pm 1,\alpha }\hslash {\hat{k}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha })\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right].\end{array}\end{eqnarray}$$

(85)

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Guodong Zhu, Yangzhe Guo, Bin Dong, Yurui Fang. Quantization of electromagnetic modes and angular momentum on plasmonic nanowires[J]. Chinese Physics B, 2020, 29(8):

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