Quantization of electromagnetic modes and angular momentum on plasmonic nanowires

Guodong Zhu; Yangzhe Guo; Bin Dong; Yurui Fang

doi:10.1088/1674-1056/ab9698

1. Introduction

It is well known that surface plasmons polaritons (SPPs) existing at interfaces between metals and dielectrics are coherent collective oscillation of free electrons (with coupled electromagnetic field) at the surface of metal. The unique properties of plasmons on nanoscale metallic systems have produced a number of dramatic effects and interesting applications, such as molecule detection with surface-enhanced Raman scattering,[1–3] biosensing,[4,5] waveguiding,[6–8] enhanced interactions,[9,10] and switching devices below the diffraction limit.[11,12] Plasmonic nanowires have been attracting a lot of attention as SPP waveguides, analogy to optical fiber waveguides but within a hundred nanometers scale cross section, which breaks the diffraction limit. The strong confinement and small mode volume of plasmonic wires facilitate the strong coupling between quantum emitters and nanowire,[13,14] the low Q factor Fabry–Pérot resonator,[15] and nanowire-wire based plasmonic devices and chips.[11,16,17] The properties of keeping quantum information like entanglement of plasmonic waveguides[18–21] now have the potential applications in quantum information and fundamental researches.

Surface plasmons are quantized electric charge density waves, while usually a lot of the phenomena can be illustrated with classical Maxwell equations in this boosting area in the past twenty years. [22] In the meantime, quantum theory descriptions were also developed to explain the phenomena such as non-locality, [23] tunneling, [24, 25] hot electrons, [26 - 28] and so on. [29] Especially quantization of surface plasmons for common systems like plane metal interface [30] and nanoparticle [31] has been developed to deal with the interactions between the plasmons and the surrounding molecules, [32 - 34] which is, in a lot of conditions, beyond the Maxwell’s theory. Now quantum plasmonics has been a rapid growing field that involves the study of the quantum properties of SPs and their interaction with matters at the nanoscale. [35]

The quantization process was usually performed by quantizing the electric field or considering the hydrodynamics meantime. Using Hopfield’s approach, [36] Elson and Ritchie reported the quantization scheme for SPPs on a metallic surface considering the hydrodynamics. [37] Archambault et al. reported the quantization scheme of the surface wave on a plane interface without any specific model of the dielectric constant. [38] Huttner and Barnett introduced a new quantization method by extending Hopfield’s approach to polaritons in dispersive and lossy media. [39] In Archambault’s works, they redid the quantization of the plasmons on plane interface in modern fashion for explaining the spontaneous and stimulated emission of SPPs. [38] Waks have re-quantized plasmons on nanoparticles under quasi-static approximation for describing the coupling of a dipole and a particle. And in Boardman’s work, the SPPs in different typical coordinates were dealt in common way with Bloch hydrodynamical model. [40] However, quantization of SPPs on cylindrical nanowires needs more investigation to illustrate more details and phenomena.

In this paper, a quantization scheme of SPPs on a cylindrical nanowire waveguide is introduced. The canonical quantization method is used, which is to identify the generalized coordinates and conjugate momentums. After getting the expressions in Fock states, the orbital and spin angular momentums are deduced. With the results one will find that the modes on the nanowire waveguide carry both orbital and spin angular momentums, which is consistent with the classic theory result from Picardi. [41] Following that, the formulas are applied to describe the SPPs on a nanowire in the waveguide, which shows good agreement. With the properties, SPPs on nanowires can easily carry quantum information and keep the entangle properties of the incident light. [20, 42 - 44] The results will be very helpful in the quantum information and light-matter interactions research.

2. Quantization description of surface waves

ε_{2} = 1 - ω_{p}^{2} / ω^{2}

$$ \begin{eqnarray}{\nabla }^{2}\left\{\begin{array}{c}{\boldsymbol{E}}\\ {\boldsymbol{H}}\end{array}\right\}-\mu \varepsilon \displaystyle \frac{{\partial }^{2}}{\partial {t}^{2}}\left\{\begin{array}{c}{\boldsymbol{E}}\\ {\boldsymbol{H}}\end{array}\right\}=0.\end{eqnarray}$$ (1)

The solutions of E and H in the cylindrical coordinates can be deduced in standard steps (Appendix A).^[14]

E (r, t) = - \frac{1}{c} \frac{\partial A (r, t)}{\partial t}

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

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}_{k,m}(t){{\rm{e}}}^{{\rm{i}}{\boldsymbol{k}}\cdot {\boldsymbol{r}}} & = & -\displaystyle \frac{{\rm{i}}}{\omega }\{[\displaystyle \frac{{\rm{i}}m}{{k}_{j,m}\rho }{a}_{j,m}{F}_{j,m}({k}_{j,m\perp }\rho )\\ & & +\displaystyle \frac{{\rm{i}}{k}_{\parallel ,m}{k}_{j,m\perp }}{{k}_{j,m}^{2}}{b}_{j,m}{F}_{j,{m}^{^{\prime} }}({k}_{j,m\perp }\rho )]\hat{{\boldsymbol{\rho }}}\\ & & +[-\displaystyle \frac{{k}_{j\perp }}{{k}_{j}}{a}_{j,m}{F}_{j,{m}^{^{\prime} }}({k}_{j,m\perp }\rho )\\ & & -\displaystyle \frac{m{k}_{\parallel }}{{k}_{j}^{2}\rho }{b}_{j,m}{F}_{j,m}({k}_{j,m\perp }\rho )]\hat{\phi }\\ & & +\displaystyle \frac{{k}_{j,m\perp }^{2}}{{k}_{j,m}^{2}}{b}_{j,m}{F}_{j,m}({k}_{j,m\perp }\rho )\hat{{\boldsymbol{z}}}\}{{\rm{e}}}^{{\rm{i}}(m\phi +{k}_{\parallel ,m}z)}{{\rm{e}}}^{-{\rm{i}}\omega t}\\ & = & {{\boldsymbol{A}}}_{k,m}(t){{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}\cdot {\boldsymbol{r}}}={{\boldsymbol{A}}}_{k,m}(t){{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}},\end{array}\end{eqnarray}$$ (3)

where

k_{j, m ⊥} = \sqrt{k_{j}^{2} - k_{∥, m}^{2}}

k_{m} \equiv k_{∥, m}

is defined just for simple, the phase factor e^{i k_m⋅r} = e^{i(mϕ + k_∥,mz)} shows that the surface waves have both components on z and ϕ directions (Fig. 1).

k_{m} = 0 \cdot \hat{ρ} + m k_{ϕ} \hat{ϕ} + k_{z, m} \hat{z}

r = ρ \hat{ρ} + ϕ ρ \hat{ϕ} + z \hat{z}

. Here k_ϕ = 1/ρ and r_ϕ = ϕ ρ are set to keep the dimensional consistency. The subscript j = 1, 2 represents the region outside and inside of the cylinder nanowire boundary characterized by different dielectric functions. m is the azimuthal quantum number. The functions F_1,m(x) = H_m(x) and F_2,m(x) = J_m(x) are Hankel and Bessel functions. a_j,m and b_j,m are arbitrary coefficients, which can be fixed by imposing the boundary conditions of interface between the wire and surrounding dielectric. A_k,m(t) is the time-dependent amplitude and can be written as A_k,m e^{−iω t}. Because each mode in the waveguide can be represent by order m, in the following we use m to replace k_m and A_m to replace A_k,m for the modes subscript. Due to the normalization condition

$$ \begin{eqnarray}\displaystyle \int {{\rm{d}}}^{3}r{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}}=V{\delta }_{mm^{\prime} }\end{eqnarray}$$ (4)

and the orthogonal of the Bessel function, different modes are orthogonal. The amplitudes A_m are complex. The complex conjugate item is added to make the field be real

$$ \begin{eqnarray}{\boldsymbol{A}}({\boldsymbol{r}},t)=\displaystyle \frac{1}{2\sqrt{V}}\displaystyle \sum _{k,m}[{{\boldsymbol{A}}}_{m}(t){{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}+{\rm{c}}.{\rm{c}}.].\end{eqnarray}$$ (5)

The canonical momentum

p (r, t) = \frac{1}{4 π c^{2}} \dot{A}

satisfies the commutation relation^[45]

$$ \begin{eqnarray}[{{\boldsymbol{A}}}_{m}({\boldsymbol{r}},t),{{\boldsymbol{p}}}_{{m}^{^{\prime} }}({{\boldsymbol{r}}}^{^{\prime} },t)]={\rm{i}}\hslash {\delta }_{m,{m}^{^{\prime} }}\delta ({\boldsymbol{r}}-{{\boldsymbol{r}}}^{^{\prime} }).\end{eqnarray}$$ (6)

Now define the cannonical coordinates and momentums as

$$ \begin{eqnarray}{{\boldsymbol{Q}}}_{m}=\sqrt{\displaystyle \frac{1}{4\pi {c}^{2}}}({{\boldsymbol{A}}}_{m}+{{\boldsymbol{A}}}_{m}^{\ast }),\end{eqnarray}$$ (7)

$$ \begin{eqnarray}{{\boldsymbol{P}}}_{m}=-\displaystyle \frac{{\rm{i}}\omega }{\sqrt{4\pi {c}^{2}}}({{\boldsymbol{A}}}_{m}-{{\boldsymbol{A}}}_{m}^{\ast }),\end{eqnarray}$$ (8)

which make

$$ \begin{eqnarray}{\boldsymbol{A}}({\boldsymbol{r}},t)=\displaystyle \frac{\sqrt{4\pi {c}^{2}}}{2\sqrt{V}}\displaystyle \sum _{m}\left[{{\boldsymbol{Q}}}_{m}\cos {\boldsymbol{k}}\cdot {\boldsymbol{r}}-\displaystyle \frac{1}{{\omega }_{m}}{{\boldsymbol{P}}}_{m}\sin {\boldsymbol{k}}\cdot {\boldsymbol{r}}\right],\end{eqnarray}$$ (9)

$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{E}}({\boldsymbol{r}},t) & =-\displaystyle \frac{1}{c}\displaystyle \frac{\partial {\boldsymbol{A}}({\boldsymbol{r}},t)}{\partial t}\\ & =-\displaystyle \frac{\sqrt{4\pi }}{2\sqrt{V}}\displaystyle \sum _{m}{\omega }_{m}\left[{{\boldsymbol{Q}}}_{m}\sin {\boldsymbol{k}}\cdot {\boldsymbol{r}}+\displaystyle \frac{1}{{\omega }_{m}}{{\boldsymbol{P}}}_{m}\cos {\boldsymbol{k}}\cdot {\boldsymbol{r}}\right],\end{array}\end{eqnarray}$$ (10)

$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{B}}({\boldsymbol{r}},t) & =\nabla \times {\boldsymbol{A}}({\boldsymbol{r}},t)\\ & =-\displaystyle \frac{\sqrt{4\pi {c}^{2}}}{2\sqrt{V}}\displaystyle \sum _{m}{{\boldsymbol{k}}}_{m}\times \left[{{\boldsymbol{Q}}}_{m}\sin {\boldsymbol{k}}\cdot {\boldsymbol{r}}+\displaystyle \frac{1}{{\omega }_{m}}{{\boldsymbol{P}}}_{m}\cos {\boldsymbol{k}}\cdot {\boldsymbol{r}}\right].\end{array}\end{eqnarray}$$ (11)

Integrating the square of the electric and magnetic fields over the volume V and using the orthogonality relation, the total Hamiltonian is

$$ \begin{eqnarray}H=\displaystyle \frac{1}{8\pi }\displaystyle \int {{\rm{d}}}^{3}r({|{\boldsymbol{E}}|}^{2}+{|{\boldsymbol{B}}|}^{2})=\displaystyle \frac{1}{2}\displaystyle \sum _{m}({{\boldsymbol{P}}}_{m}^{2}+{\omega }_{m}^{2}{{\boldsymbol{Q}}}_{m}^{2}).\end{eqnarray}$$ (12)

Take the commutators for the canonical coordinates and canonical momentums (which can also be got from Eq. (6)) as

$$ \begin{eqnarray}[{{\boldsymbol{Q}}}_{m},{{\boldsymbol{P}}}_{m}]={\rm{i}}\hslash .\end{eqnarray}$$ (13)

Now introduce the creation and annihilation operators

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m}=\displaystyle \frac{1}{\sqrt{2\hslash \omega }}({\omega }_{m}{{\boldsymbol{Q}}}_{m}+{\rm{i}}{{\boldsymbol{P}}}_{m}),\end{eqnarray}$$ (14)

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m}^{\dagger }=\displaystyle \frac{1}{\sqrt{2\hslash \omega }}({\omega }_{m}{{\boldsymbol{Q}}}_{m}-{\rm{i}}{{\boldsymbol{P}}}_{m}).\end{eqnarray}$$ (15)

Based on the definitions, we have

$$ \begin{eqnarray}\begin{array}{ll}[{{\boldsymbol{a}}}_{m},{{\boldsymbol{a}}}_{m}^{\dagger }] & =\displaystyle \frac{1}{2\hslash {\omega }_{m}}[{\omega }_{m}{{\boldsymbol{Q}}}_{m}+{\rm{i}}{{\boldsymbol{P}}}_{m},{\omega }_{m}{{\boldsymbol{Q}}}_{m}-{\rm{i}}{{\boldsymbol{P}}}_{m}],\\ & =\displaystyle \frac{1}{2\hslash {\omega }_{m}}(-{\rm{i}}{\omega }_{m}[{{\boldsymbol{Q}}}_{m},{{\boldsymbol{P}}}_{m}]+{\rm{i}}{\omega }_{m}[{{\boldsymbol{P}}}_{m},{{\boldsymbol{Q}}}_{m}])=1.\end{array}\end{eqnarray}$$ (16)

Considering the positive and negative wave vector sums, the quantized vector potential, electromagnetic fields, and Hamiltonian are

$$ \begin{eqnarray}{\boldsymbol{A}}({\boldsymbol{r}},t)=\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle \sum _{m}\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}[{{\boldsymbol{a}}}_{m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}+{{\boldsymbol{a}}}_{m}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\end{eqnarray}$$ (17)

$$ \begin{eqnarray}{\boldsymbol{E}}({\boldsymbol{r}},t)={\rm{i}}\displaystyle \frac{\sqrt{2\pi \hslash }}{\sqrt{V}}\displaystyle \sum _{m}{\sqrt{\omega }}_{m}[{{\boldsymbol{a}}}_{m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{{\boldsymbol{a}}}_{m}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\end{eqnarray}$$ (18)

$$ \begin{eqnarray}{\boldsymbol{B}}({\boldsymbol{r}},t)={\rm{i}}\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle \sum _{m}\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}{{\boldsymbol{k}}}_{m}\times [{{\boldsymbol{a}}}_{m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{{\boldsymbol{a}}}_{m}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\end{eqnarray}$$ (19)

$$ \begin{eqnarray}H=\displaystyle \sum _{m}{H}_{m}=\displaystyle \sum _{m}\hslash {\omega }_{m}({{\boldsymbol{a}}}_{m}^{\dagger }{{\boldsymbol{a}}}_{m}+1/2)=\displaystyle \sum _{m}\hslash {\omega }_{m}({n}_{m}+1/2),\end{eqnarray}$$ (20)

with

n_{m} = a_{m}^{†} a_{m}

being the number operator. And the evolution obeys

$$ \begin{eqnarray}{\rm{i}}\hslash {\mathop{{\boldsymbol{a}}}\limits^{.}}_{m}=[{{\boldsymbol{a}}}_{m},{\boldsymbol{H}}],\end{eqnarray}$$ (21)

$$ \begin{eqnarray}{\rm{i}}\hslash {\mathop{{\boldsymbol{a}}}\limits^{.}}_{m}^{\dagger }=[{{\boldsymbol{a}}}_{m}^{\dagger },{\boldsymbol{H}}].\end{eqnarray}$$ (22)

Figure 1.Schematic illustration of SPPs propagating on a plasmonic cylindrical waveguide, $k_{m} \equiv k_{∥, m} = 0 \cdot \hat{ρ} + m k_{ϕ} \hat{ϕ} + k_{z, m} \hat{z}$ .

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3. Polarization and angular momentum

{\hat{ϵ}}_{ρ} = (1, 0, 0)

$$ \begin{eqnarray}\begin{array}{ll}\,\quad{\boldsymbol{A}}({\boldsymbol{r}},t)\\ =\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle \sum _{m,\alpha }\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}[{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}+{\hat{\epsilon }}_{\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\,\end{array}\end{eqnarray}$$ (23)

$$ \begin{eqnarray}\begin{array}{ll}\,\quad{\boldsymbol{E}}({\boldsymbol{r}},t)\\ ={\rm{i}}\sqrt{\displaystyle \frac{2\pi \hslash }{V}}\displaystyle \sum _{m,\alpha }{\sqrt{\omega }}_{m}[{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{\hat{\epsilon }}_{\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}],\end{array}\end{eqnarray}$$ (24)

$$ \begin{eqnarray}\begin{array}{ll}\,\quad{\boldsymbol{B}}({\boldsymbol{r}},t)\\ ={\rm{i}}\sqrt{\displaystyle \frac{2\pi \hslash }{V}}\displaystyle \sum _{m,\alpha }\displaystyle \frac{c{{\boldsymbol{k}}}_{m}}{{\sqrt{\omega }}_{m}}\times [{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{\hat{\epsilon }}_{\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}].\end{array}\end{eqnarray}$$ (25)

And the momentum is given by

$$ \begin{eqnarray}\begin{array}{lll}{\boldsymbol{p}} & = & \displaystyle \int {{\rm{d}}}^{3}r\displaystyle \frac{{\boldsymbol{E}}\times {\boldsymbol{B}}}{4\pi c}\\ & = & \displaystyle \frac{{{\rm{i}}}^{2}}{4\pi c}\displaystyle \frac{hc}{V}\displaystyle \sum _{m,\alpha }\displaystyle \sum _{{m}^{^{\prime} },{\alpha }^{^{\prime} }}[{\hat{\epsilon }}_{\alpha }\times ({k}^{^{\prime} }\times {\hat{\epsilon }}_{{\alpha }^{^{\prime} }})]\times \displaystyle \int {{\rm{d}}}^{3}r({{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}\\ & & -{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}})({{\boldsymbol{a}}}_{{m}^{^{\prime} },{\alpha }^{^{\prime} }}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}}-{{\boldsymbol{a}}}_{{m}^{^{\prime} },{\alpha }^{^{\prime} }}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}})\\ & = & \displaystyle \frac{h}{4\pi }\displaystyle \sum _{m,\alpha }{{\boldsymbol{k}}}_{m}[{{\boldsymbol{a}}}_{m,\alpha }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }+{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }]=\displaystyle \sum _{m}\hslash {{\boldsymbol{k}}}_{m}{n}_{m}.\end{array}\end{eqnarray}$$ (26)

The zero point term 1/2 is cancelled by the terms of + k_m and – k_m.

With the above results we will discuss the quantized field properties in the nanowire, such as spin and orbital angular momentums (SAM/OAM) and polarization. It has been known that the optical angular momentum is

$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{J}} & =\displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r{\boldsymbol{r}}\times ({\boldsymbol{E}}\times {\boldsymbol{B}})\\ & =\displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r[{E}_{\alpha }({\boldsymbol{r}}\times \nabla ){A}_{\alpha }+{\boldsymbol{E}}\times {\boldsymbol{A}}].\end{array}\end{eqnarray}$$ (27)

The first term of the right hand is the orbit part and the second is the spin part. The orbital angular momentum density on the surface is (Appendix B)

$$ \begin{eqnarray}\begin{array}{ll}{{\boldsymbol{l}}}_{{\boldsymbol{\alpha }}}= & ({\boldsymbol{r}}\times \nabla ){A}_{\alpha }{|}_{\rho =R}=\displaystyle {\sum }_{m}{\hat{r}}_{\alpha }\cdot \left(\begin{array}{c}-\displaystyle \frac{z}{R}\displaystyle \frac{\partial }{\partial \phi }\hat{{\boldsymbol{\rho }}}\\ (z\displaystyle \frac{\partial }{\partial \rho }-R\displaystyle \frac{\partial }{\partial z})\hat{\phi }\\ \displaystyle \frac{\partial }{\partial \phi }\hat{{\boldsymbol{z}}}\end{array}\right){A}_{\alpha }={\rm{i}}\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle {\sum }_{m}\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}{\hat{r}}_{\alpha }\cdot R\cdot \left(\begin{array}{c}-\displaystyle \frac{z}{R}m{k}_{\phi }\\ -{k}_{z,m}\\ m{k}_{\phi }\end{array}\right)[{{\boldsymbol{a}}}_{m\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}\\ & -{{\boldsymbol{a}}}_{m\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}]=\sqrt{\displaystyle \frac{2\pi {c}^{2}\hslash }{V}}\displaystyle {\sum }_{m}\displaystyle \frac{1}{{\sqrt{\omega }}_{m}}{\hat{r}}_{\alpha }\cdot \left(\begin{array}{c}-\displaystyle \frac{z}{R}m\\ -R{k}_{z,m}\\ m\end{array}\right)[{{\boldsymbol{a}}}_{m\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}-{{\boldsymbol{a}}}_{m\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}].\end{array}\end{eqnarray}$$ (28)

With the creation and annihilation operators, the orbital angular momentum is

$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{L}}= & \displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r{E}_{\alpha }{l}_{\alpha }=\displaystyle \frac{{\rm{i}}\hslash }{2V}\displaystyle {\sum }_{m,{m}^{^{\prime} },\alpha ,{\alpha }^{^{\prime} }}\displaystyle \frac{{\omega }_{m}}{\sqrt{{\omega }_{m}{\omega }_{{m}^{^{\prime} }}}}{\hat{\epsilon }}_{{\boldsymbol{\alpha }}}\cdot {\hat{\epsilon }}_{{{\boldsymbol{\alpha }}}^{^{\prime} }}\displaystyle \int {{\rm{d}}}^{3}rR\left(\begin{array}{c}-\displaystyle \frac{z}{R}m{k}_{\phi }\\ -{k}_{z,m}\\ m{k}_{\phi }\end{array}\right)[{{\boldsymbol{a}}}_{m\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}-{{\boldsymbol{a}}}_{m\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}}][{{\boldsymbol{a}}}_{{m}^{^{\prime} }{\alpha }^{^{\prime} }}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,{m}^{^{\prime} }}{\boldsymbol{r}}}\\ & -{{\boldsymbol{a}}}_{{m}^{^{\prime} }{\alpha }^{^{\prime} }}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,{m}^{^{\prime} }}{\boldsymbol{r}}}]=\displaystyle \frac{{\rm{i}}\hslash }{2}\displaystyle {\sum }_{m,\alpha }\left(\begin{array}{c}0\\ -R{k}_{z,m}\\ m\end{array}\right)({{\boldsymbol{a}}}_{m,\alpha }{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }+{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha })=\displaystyle {\sum }_{m}{\rm{i}}\hslash {{\boldsymbol{l}}}_{m}{{\boldsymbol{n}}}_{m},\end{array}\end{eqnarray}$$ (29)

$$ \begin{eqnarray}{{\boldsymbol{l}}}_{m}=\left(\begin{array}{c}0\\ -R{k}_{z,m}\\ m\end{array}\right),\end{eqnarray}$$ (30)

$$ \begin{eqnarray}{L}_{z}=\displaystyle \frac{{\boldsymbol{L}}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}={\rm{i}}\hslash \displaystyle {\sum }_{m}m{{\boldsymbol{n}}}_{m},\end{eqnarray}$$ (31)

where

n_{m, α} \equiv a_{m, α}^{†} a_{m, α}

and

n_{m} \equiv a_{m}^{†} a_{m}

In the above expressions, one can rewrite the phase factor

$$ \begin{eqnarray}\begin{array}{ll}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,m}{\boldsymbol{r}}} & {={\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}={{\rm{e}}}^{{\rm{i}}(m\phi +{k}_{z}z)}\\ & =[\cos (m\phi )+{\rm{i}}\sin (m\phi )]{{\rm{e}}}^{{\rm{i}}{k}_{z}z},\end{array}\end{eqnarray}$$ (32)

$$ \begin{eqnarray}\begin{array}{ll}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{\parallel ,-m}{\boldsymbol{r}}} & ={{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{-m}\cdot {\boldsymbol{r}}}={{\rm{e}}}^{{\rm{i}}(-m\phi +{k}_{z}z)}\\ & =[\cos (m\phi )-{\rm{i}}\sin (m\phi )]{{\rm{e}}}^{{\rm{i}}{k}_{z}z},\end{array}\end{eqnarray}$$ (33)

which shows that the field rotation is very similar to the circularly polarized light. Then if we absorb the phase factor e^imϕ into the creation and annihilation operators, we can directly write

$$ \begin{eqnarray}\begin{array}{ll}{{\boldsymbol{a}}}_{m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}} & =\displaystyle \sum _{\alpha }{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}\\ & =\displaystyle \sum _{\alpha }{\lambda }_{+\alpha }{\hat{\epsilon }}_{\alpha }{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{k}_{\parallel ,m}\cdot {\boldsymbol{r}}}={{\boldsymbol{a}}}_{L,m}{{\rm{e}}}^{{\rm{i}}{k}_{z}z},\end{array}\end{eqnarray}$$ (34)

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{-m}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{-m}\cdot {\boldsymbol{r}}}={{\boldsymbol{a}}}_{R,m}{{\rm{e}}}^{{\rm{i}}{k}_{z}z},\end{eqnarray}$$ (35)

where L and R sign here referring the chirality without further explanation and we will see the physics meaning later. A_−m(t) = −A_m(t), A_m(t) (m = 0, ± 1, ± 2, …) compose over-complete bases and so as a_m.

The spin angular momentum density is

$$ \begin{eqnarray}{\boldsymbol{s}}={\boldsymbol{E}}\times {\boldsymbol{A}}=\displaystyle \frac{{\rm{i}}}{\hslash }({E}_{\alpha }{({{\boldsymbol{S}}}_{\alpha })}_{\beta \gamma }{{\boldsymbol{A}}}_{\alpha }),\end{eqnarray}$$ (36)

$$ \begin{eqnarray}{({{\boldsymbol{S}}}_{\alpha })}_{\beta \gamma }=-{\rm{i}}\hslash {{\boldsymbol{\varepsilon }}}_{\alpha \beta \gamma }.\end{eqnarray}$$ (37)

The operator S_α defined here satisfies the commutation relation (Appendix B)

$$ \begin{eqnarray}[({{\boldsymbol{S}}}_{i}),({{\boldsymbol{S}}}_{j})]={\rm{i}}\hslash {{\boldsymbol{\varepsilon }}}_{ijk}({{\boldsymbol{S}}}_{k}),\end{eqnarray}$$ (38)

which shows that it is a spin operator. Calculating S², s = 1 corresponding to spin-1 (SPPs are the quasi-particles) can be obtained as

$$ \begin{eqnarray}{{\boldsymbol{S}}}^{2}={({{\boldsymbol{S}}}_{i})}_{lm}{({{\boldsymbol{S}}}_{i})}_{mn}={(-{\rm{i}}\hslash )}^{2}({{\boldsymbol{\delta }}}_{ln}{{\boldsymbol{\delta }}}_{mn}-{{\boldsymbol{\delta }}}_{ln}{{\boldsymbol{\delta }}}_{mn}),\end{eqnarray}$$ (39)

$$ \begin{eqnarray}{({{\boldsymbol{S}}}^{2})}_{ln}={(-{\rm{i}}\hslash )}^{2}({{\boldsymbol{\delta }}}_{ln}-3{{\boldsymbol{\delta }}}_{ln})=2{\hslash }^{2}{{\boldsymbol{\delta }}}_{ln},\end{eqnarray}$$ (40)

$$ \begin{eqnarray}s(s+1)=2,\,\,\,s=1.\end{eqnarray}$$ (41)

The total spin momentum is

$$ \begin{eqnarray}{\boldsymbol{S}}=\displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r({\boldsymbol{s}})=\displaystyle \frac{1}{4\pi c}\displaystyle \int {{\rm{d}}}^{3}r{E}_{\alpha }{({\boldsymbol{S}})}_{\beta \gamma }{A}_{\alpha },\end{eqnarray}$$ (42)

$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{S}}= & \displaystyle \frac{-1}{4\pi c}\displaystyle \frac{{\rm{i}}hc}{V}\displaystyle \sum _{ma}\displaystyle \sum _{{m}^{^{\prime} }{a}^{^{\prime} }}{\varepsilon }_{\alpha \beta \gamma }{\hat{\epsilon }}_{\alpha }\times {\hat{\epsilon }}_{{\alpha }^{^{\prime} }}\displaystyle \int {{\rm{d}}}^{3}r[{{\boldsymbol{a}}}_{m,\alpha }{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}\\ & +{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}}][{{\boldsymbol{a}}}_{{m}^{^{\prime} },{\alpha }^{^{\prime} }}{{\rm{e}}}^{{\rm{i}}{{\boldsymbol{k}}}_{m}\cdot {\boldsymbol{r}}}-{{\boldsymbol{a}}}_{{m}^{^{\prime} },{\alpha }^{^{\prime} }}^{\dagger }{{\rm{e}}}^{-{\rm{i}}{{\boldsymbol{k}}}_{{m}^{^{\prime} }}\cdot {\boldsymbol{r}}}],\end{array}\end{eqnarray}$$ (43)

$$ \begin{eqnarray}{\boldsymbol{S}}=\displaystyle \frac{{\rm{i}}h}{2\pi }\displaystyle \sum _{m\alpha {\alpha }^{^{\prime} }}{\varepsilon }_{\alpha \beta \gamma }{\hat{\epsilon }}_{\alpha }\times {\hat{\epsilon }}_{{\alpha }^{^{\prime} }}({{\boldsymbol{a}}}_{m,\alpha }{{\boldsymbol{a}}}_{m,{\alpha }^{^{\prime} }}^{\dagger }-{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,{\alpha }^{^{\prime} }}).\end{eqnarray}$$ (44)

Because both m = m and m = –m’ give the same results, so we put a factor 2 in the last step. Using

{\hat{ϵ}}_{α} \times {\hat{ϵ}}_{β} = {\hat{k}}_{γ}

(α, β, γ = ρ, ϕ, z), we get

$$ \begin{eqnarray}{\boldsymbol{S}}=\displaystyle \sum _{m\alpha {\alpha }^{^{\prime} }}{\rm{i}}\hslash {\hat{{\boldsymbol{k}}}}_{m,{\alpha }^{^{\prime\prime} }}({{\boldsymbol{a}}}_{m,\alpha }{{\boldsymbol{a}}}_{m,{\alpha }^{^{\prime} }}^{\dagger }-{{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,{\alpha }^{^{\prime} }}).\end{eqnarray}$$ (45)

From the expression one can clearly see that the spin in one direction (like ϕ direction) is yielded by the other two components (ρ and z) of the field. It also agrees with that there is transverse spin (

S_{⊥} = \frac{Re k \times Im k}{Re k^{2}}

)^[46] for surface waves which is represented by the elliptical trace of the field vector at local points.

{\hat{ϵ}}_{1} = \hat{x}

$$ \begin{eqnarray}{\hat{s}}_{\alpha ,L}={\hat{s}}_{\alpha ,+}=\displaystyle \frac{1}{\sqrt{2}}({\hat{\epsilon }}_{\beta }+{\rm{i}}{\hat{\epsilon }}_{\gamma }),\end{eqnarray}$$ (46)

$$ \begin{eqnarray}{\hat{s}}_{\alpha ,R}={\hat{s}}_{\alpha ,-}=\displaystyle \frac{1}{\sqrt{2}}({\hat{\epsilon }}_{\beta }-{\rm{i}}{\hat{\epsilon }}_{\gamma }).\end{eqnarray}$$ (47)

Then

$$ \begin{eqnarray}\begin{array}{ll}\,\quad{\hat{\epsilon }}_{\beta }{{\boldsymbol{a}}}_{m,\beta }+{\hat{\epsilon }}_{\gamma }{{\boldsymbol{a}}}_{m,\gamma }\\ =\displaystyle \frac{1}{\sqrt{2}}({\hat{s}}_{\alpha ,L}+{\hat{s}}_{\alpha ,R}){{\boldsymbol{a}}}_{m,\beta }+\displaystyle \frac{1}{{\rm{i}}\sqrt{2}}({\hat{s}}_{\alpha ,L}-{\hat{s}}_{\alpha ,R}){{\boldsymbol{a}}}_{m,\gamma }\\ =\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }-{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }){\hat{s}}_{\alpha ,L}+\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }+{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }){\hat{s}}_{\alpha ,R}.\end{array}\end{eqnarray}$$ (48)

So we can define new circularly polarized operators

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,L,\alpha }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }-{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }),\end{eqnarray}$$ (49)

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,R,\alpha }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }+{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }),\end{eqnarray}$$ (50)

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,L,\alpha }^{\dagger }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }^{\dagger }-{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }^{\dagger }),\end{eqnarray}$$ (51)

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,R,\alpha }^{\dagger }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,\beta }^{\dagger }+{\rm{i}}{{\boldsymbol{a}}}_{m,\gamma }^{\dagger }),\end{eqnarray}$$ (52)

with

[a_{m, L, α}, a_{m, L, α}^{†}] = 1

and

[a_{m, R, α}, a_{m, R, α}^{†}] = 1

. One can get

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,\beta }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,L,\alpha }+{{\boldsymbol{a}}}_{m,R,\alpha }),\end{eqnarray}$$ (53)

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,\gamma }=\displaystyle \frac{{\rm{i}}}{\sqrt{2}}({{\boldsymbol{a}}}_{m,L,\alpha }-{{\boldsymbol{a}}}_{m,R,\alpha }),\end{eqnarray}$$ (54)

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,\beta }^{\dagger }=\displaystyle \frac{1}{\sqrt{2}}({{\boldsymbol{a}}}_{m,L,\alpha }^{\dagger }+{{\boldsymbol{a}}}_{m,R,\alpha }^{\dagger }),\end{eqnarray}$$ (55)

$$ \begin{eqnarray}{{\boldsymbol{a}}}_{m,\gamma }^{\dagger }=\displaystyle \frac{-{\rm{i}}}{\sqrt{2}}({{\boldsymbol{a}}}_{m,L,\alpha }^{\dagger }-{{\boldsymbol{a}}}_{m,R,\alpha }^{\dagger }).\end{eqnarray}$$ (56)

Obviously,

$$ \begin{eqnarray}\displaystyle \sum _{a=\alpha ,\beta }{\hat{\epsilon }}_{a}{{\boldsymbol{a}}}_{m,a}=\displaystyle \sum _{\gamma }({\hat{s}}_{\gamma ,L}{{\boldsymbol{a}}}_{m,L,\gamma }+{\hat{s}}_{\gamma ,R}{{\boldsymbol{a}}}_{m,R,\gamma }),\end{eqnarray}$$ (57)

$$ \begin{eqnarray}\displaystyle \sum _{a=\alpha ,\beta }{\hat{\epsilon }}_{a}^{\dagger }{{\boldsymbol{a}}}_{m,a}^{\dagger }=\displaystyle \sum _{\gamma }({\hat{s}}_{\gamma ,L}^{\dagger }{{\boldsymbol{a}}}_{m,L,\gamma }^{\dagger }+{\hat{s}}_{\gamma ,R}^{\dagger }{{\boldsymbol{a}}}_{m,R,\gamma }^{\dagger }).\end{eqnarray}$$ (58)

Now the spin operator becomes

$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{S}} & =\displaystyle \sum _{m\alpha }\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{a}}}_{m,L,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,L,\alpha }-{{\boldsymbol{a}}}_{m,R,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,R,\alpha })\\ & =\displaystyle \sum _{m\alpha }\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha }),\end{array}\end{eqnarray}$$ (59)

$$ \begin{eqnarray}{\hat{S}}_{\alpha }=\displaystyle \frac{\hat{S}\cdot {{\boldsymbol{k}}}_{\alpha }}{|{{\boldsymbol{k}}}_{\alpha }|}.\end{eqnarray}$$ (60)

Then the total angular momentum can be expressed as

$$ \begin{eqnarray}\begin{array}{ll}{\boldsymbol{J}}={\boldsymbol{L}}+{\boldsymbol{S}}= & \displaystyle {\sum }_{m,\alpha }{\rm{i}}\hslash {{\boldsymbol{k}}}_{m,\alpha }{\hat{n}}_{m,\alpha }\\ & +\displaystyle \sum _{m,\alpha }\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha }).\end{array}\end{eqnarray}$$ (61)

4. SPPs on silver nanowires as waveguides

The nanowires performed as plasmonic waveguides are usually in homogeneous medium and excited with light perpendicular to the wire axis on one end. [6] The local symmetry at the terminus of the wire is broken when the SPPs are excited, so the phase factor e i mϕ becomes cos (mϕ) or sin (mϕ). For thin wires, the higher-order modes | m | ⩾ 2 are cut off. [14] And the retardation effects are significant, which will cause the excitation phase difference for | m | = 0 and | m | = 1 modes as shown in Fig. 2 . [48] The three lowest modes are excited simultaneously and propagating in a fixed phase delay. When the wire is excited by a beam of light propagating along y direction with linear polarization at 45° with the wire axis, the polarization can be decomposed into x and z directions. The z component will excite m = 0 mode and m ’ = + 1 mode (sin (mϕ)) with π /2 phase difference. The x component will excite m = + 1 mode (cos (mϕ)) with the same phase of the m = 0 mode. So the potential vector can be expressed as (only consider the outside of boundary, the inside is similar).

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}_{l=1}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & & +{{\boldsymbol{A}}}_{+1}\cos (\phi -\displaystyle \frac{\pi }{2}){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t+\displaystyle \frac{\pi }{2})}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & & +{\rm{i}}{{\boldsymbol{A}}}_{+1}\sin (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }{{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}.\end{array}\end{eqnarray}$$ (62)

It can be rewritten as

$$ \begin{eqnarray}\begin{array}{ll}{{\boldsymbol{A}}}_{l=1}^{R}({\boldsymbol{r}},t) & ={{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }{{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & ={{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}({{\boldsymbol{A}}}_{0}+{{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}(\Delta kz+\phi )}),\end{array}\end{eqnarray}$$ (63)

where Δ k = k_{z, + 1} – k_z,0,

Δ k \hat{z} + ϕ \hat{ϕ} = Δ k_{m} = k_{m, + 1} - k_{m, 0}

. When the wire is excited by a linear polarized light at –45°, it can be expressed as

$$ \begin{eqnarray}\begin{array}{ll}{{\boldsymbol{A}}}_{l=1}^{L}({\boldsymbol{r}},t) & ={{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{-1}{{\rm{e}}}^{-{\rm{i}}\phi }{{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & ={{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}({{\boldsymbol{A}}}_{0}+{{\boldsymbol{A}}}_{-1}{{\rm{e}}}^{{\rm{i}}(\Delta kz-\phi )}).\end{array}\end{eqnarray}$$ (64)

The factor e^{i(Δkz + ϕ) shows the spiral propagation of the SPPs clearly as in Ref. [48] (shown as Fig. 3), where the coherent interference of SPP waves of m = +1 and m’ = +1 gives a circularly-polarization-like wave (cos (ϕ) + i sin (ϕ) = e–iϕ}). The superposition with mode m = 0 will yield beat effect (e^iΔkz), which stretches the circular wave into a helical wave as the factor e^{i(Δkz + ϕ)} shown (Fig. 3(a)). The helical wave is very similar to the vortex wave with orbital angular momentum l = +1 (Figs. 3(a) and 3(b),

A_{1}^{R}

). Analogically, the SPP waves on cylindrical nanowires with higher modes m = ± 2,± 3,… are similar to the vortex wave of l = ± 2,± 3,…. The spiral is right handed which was mentioned in the polarization part as a_R,m. Similarly, the superposition of m = 0 and m = –1 modes will cause an l = –1 helical wave (Fig. 3(b),

A_{1}^{L}

). A special case of the condition is when the incident polarization is along the wire. Then

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}_{l=1}^{0}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\sin (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}({{\boldsymbol{A}}}_{0}+\sin (\phi ){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\Delta kz}),\end{array}\end{eqnarray}$$ (65)

which shows only a beat at the ϕ = π/2 side of the wire (Fig. 3(b),

A_{1}^{0}

Figure 2.Schematic illustration of the plasmonic cylindrical wire excited by light at one terminal for lower order modes. The light goes along y direction and the polarization is at 45° with the z axis which can be decomposed into x and z components.

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Figure 3.(a) Schematic illustration of the superposed m = 0 mode and m > 0 modes, which is similar to the vortex waves with l = 1,2,3. The red lines represent the maximum of the beats waves; the lighter lines indicate the beats behind the wire and the darker ones indicate the beats on the reader’s side. (b) The scheme of electric field distributions on plasmonic cylindrical nanowires excited by focusing light on one left terminals with polarization at + 45°(A^L), –45°(A^R) and 0°(A⁰) directions to z axis.

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With the vector potential, one can directly achieve the following results in Fock states:

$$ \begin{eqnarray}\begin{array}{ll}{H}^{L}|{n}\rangle_{m} & =\displaystyle \sum _{m=0,-1;a}{\hat{\epsilon }}_{\alpha }\hslash {\omega }_{m}({{\boldsymbol{a}}}_{m,a}^{\dagger }{{\boldsymbol{a}}}_{m,a}+1/2)|{n}\rangle_{m}\\ & =\displaystyle \sum _{m=0,-1}\hslash {\omega }_{m}({n}_{m}+1/2),\end{array}\end{eqnarray}$$ (66)

$$ \begin{eqnarray}\begin{array}{lll}{H}^{R}|{n}\rangle_{m} & = & \displaystyle \sum _{m=0,+1;a}{\hat{\epsilon }}_{\alpha }\hslash {\omega }_{m}({{\boldsymbol{a}}}_{m,a}^{\dagger }{{\boldsymbol{a}}}_{m,a}+1/2)|{n}\rangle_{m}\\ & = & \displaystyle \sum _{m=0,+1}\hslash {\omega }_{m}({n}_{m}+1/2),\end{array}\end{eqnarray}$$ (67)

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{p}}}^{L}|{n}\rangle_{m} & = & \displaystyle \frac{h}{4\pi }\displaystyle \sum _{m=0,-1}{{\boldsymbol{k}}}_{m}[{{\boldsymbol{a}}}_{m}{{\boldsymbol{a}}}_{m}^{\dagger }+{{\boldsymbol{a}}}_{m}^{\dagger }{{\boldsymbol{a}}}_{m}]|{n}\rangle_{m}\\ & = & \displaystyle \sum _{m=0,-1}\hslash {{\boldsymbol{k}}}_{m}{n}_{m},\end{array}\end{eqnarray}$$ (68)

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{p}}}^{R}|{n}\rangle_{m} & = & \displaystyle \frac{h}{4\pi }\displaystyle \sum _{m=0,+1}{{\boldsymbol{k}}}_{m}[{{\boldsymbol{a}}}_{m}{{\boldsymbol{a}}}_{m}^{\dagger }+{{\boldsymbol{a}}}_{m}^{\dagger }{{\boldsymbol{a}}}_{m}]|{n}\rangle_{m}\\ & = & \displaystyle \sum _{m=0,+1}\hslash {{\boldsymbol{k}}}_{m}{n}_{m},\end{array}\end{eqnarray}$$ (69)

$$ \begin{eqnarray}\begin{array}{lll}{L}_{z}^{L}|{n}\rangle_{m} & = & \displaystyle \frac{{\boldsymbol{L}}|{n}\rangle_{m}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}\\ & = & {\rm{i}}\hslash \displaystyle {\sum }_{m=0,-1}m\ast {n}_{m}=-{n}_{-1}\ast {\rm{i}}\hslash ,\end{array}\end{eqnarray}$$ (70)

$$ \begin{eqnarray}\begin{array}{lll}{L}_{z}^{R}|{n}\rangle_{m} & = & \displaystyle \frac{{\boldsymbol{L}}|{n}\rangle_{m}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}\\ & = & {\rm{i}}\hslash \displaystyle {\sum }_{m=0,+1}m\ast {n}_{m}={n}_{+1}\ast {\rm{i}}\hslash ,\end{array}\end{eqnarray}$$ (71)

$$ \begin{eqnarray}{{\boldsymbol{S}}}^{L}|{n}\rangle_{m}=\displaystyle \sum _{m=0,-1,a}\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha }),\end{eqnarray}$$ (72)

$$ \begin{eqnarray}{{\boldsymbol{S}}}^{R}|{n}\rangle_{m}=\displaystyle \sum _{m=0,+1,a}\hslash {\hat{{\boldsymbol{k}}}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha }).\end{eqnarray}$$ (73)

Similarly, when the nanowire is excited by right circularly polarization (RCP) light, the electric field components of RCP (

E_{z} = E_{0}, E_{x} = E_{0} e^{i \frac{π}{2}}

) illuminate on the nanowire, resulting in an analogous form of the vector potential

$$ \begin{eqnarray}\begin{array}{lll} & & {{\boldsymbol{A}}}^{{\rm{RCP}}}({\boldsymbol{r}},t)\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}{{\rm{e}}}^{{\rm{i}}\displaystyle \frac{\pi }{2}}\\ & & +{{\boldsymbol{A}}}_{+1}\cos (\phi -\displaystyle \frac{\pi }{2}){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t+\displaystyle \frac{\pi }{2})}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{\rm{i}}{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & & +{\rm{i}}{{\boldsymbol{A}}}_{+1}\sin (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{\parallel ,0}z-\omega t)}+{\rm{i}}{{\boldsymbol{A}}}_{+1}(\cos (\phi )\\ & & +\sin (\phi )){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}.\end{array}\end{eqnarray}$$ (74)

The term cos (ϕ) + sin (ϕ) reflects that the oblique distribution of the SPPs excited by the circularly polarization light is on the ϕ = + π/4 side, which is shown in Fig. 4. For symmetry, the over complete base of m can be used to describe the vector potential

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}^{{\rm{RCP}}}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+\displaystyle \frac{1}{2}(({\rm{i}}+1){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }\\ & & +{{\boldsymbol{A}}}_{+1}({\rm{i}}-1){{\rm{e}}}^{-{\rm{i}}\phi }){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+\displaystyle \frac{1}{2}(({\rm{i}}+1){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }\\ & & -{{\boldsymbol{A}}}_{-1}({\rm{i}}-1){{\rm{e}}}^{-{\rm{i}}\phi }){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}.\end{array}\end{eqnarray}$$ (75)

Figure 4.The scheme of electric field distributions on plasmonic cylindrical nanowires excited by focused LCP (A^LCP) and RCP (A^RCP) light on left ends. The insets show the cross section at the wires at positions i and ii.

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From the expression one can see that the pattern should be the superpostion of the two helical modes m = +1 and m = -1. It is also consist with the fact that the circularly polarized light can be decomposed with two orthorgnal linearly polarized light for the exciting light.

For LCP light excitation on one end, similarly we have

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}^{{\rm{LCP}}}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+{{\boldsymbol{A}}}_{+1}\cos (\phi ){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}{{\rm{e}}}^{-{\rm{i}}\displaystyle \frac{\pi }{2}}\\ & & +{{\boldsymbol{A}}}_{+1}\cos (\phi -\displaystyle \frac{\pi }{2}){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t+\displaystyle \frac{\pi }{2})}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}-{\rm{i}}{{\boldsymbol{A}}}_{+1}(\cos (\phi )\\ & & -\sin (\phi )){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\end{array}\end{eqnarray}$$ (76)

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{A}}}^{{\rm{LCP}}}({\boldsymbol{r}},t) & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+\displaystyle \frac{1}{2}((-1+{\rm{i}}){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }\\ & & +{{\boldsymbol{A}}}_{+1}(-1-{\rm{i}}){{\rm{e}}}^{-{\rm{i}}\phi }){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}\\ & = & {{\boldsymbol{A}}}_{0}{{\rm{e}}}^{{\rm{i}}({k}_{z,0}z-\omega t)}+\displaystyle \frac{1}{2}((-1+{\rm{i}}){{\boldsymbol{A}}}_{+1}{{\rm{e}}}^{{\rm{i}}\phi }\\ & & +{{\boldsymbol{A}}}_{-1}(1+{\rm{i}}){{\rm{e}}}^{-{\rm{i}}\phi }){{\rm{e}}}^{{\rm{i}}({k}_{z,+1}z-\omega t)}.\end{array}\end{eqnarray}$$ (77)

The coefficient shows that the pattern of A^LCP is the beat at the ϕ = –π/4 sides (Fig. 4). And the results in Fock states can be written as follows:

$$ \begin{eqnarray}\begin{array}{lll}{H}^{{\rm{LCP}}}{|n\rangle }_{m} & = & \displaystyle \sum _{m=0,\pm 1;\alpha }{\hat{\epsilon }}_{\alpha }\hslash {\omega }_{m}({{\boldsymbol{a}}}_{m,\alpha }^{\dagger }{{\boldsymbol{a}}}_{m,\alpha }+1/2)\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}{|n\rangle }_{m}\right]\\ & = & \displaystyle \sum _{m=0,\pm 1}{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}\hslash {\omega }_{m}({n}_{m}+1/2)\\ & = & \displaystyle \sum _{m=0,+1}\hslash {\omega }_{m}({n}_{m}+1/2),\end{array}\end{eqnarray}$$ (78)

$$ \begin{eqnarray}{H}^{{\rm{RCP}}}|{n}\rangle_{m}=\displaystyle \sum _{m=0,+1}\hslash {\omega }_{m}({n}_{m}+1/2),\end{eqnarray}$$ (79)

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{p}}}^{{\rm{LCP}}}|{n}\rangle_{m} & = & \displaystyle \frac{h}{4\pi }\displaystyle \sum _{m=0,\pm 1}{{\boldsymbol{k}}}_{m}[{{\boldsymbol{a}}}_{m}{{\boldsymbol{a}}}_{m}^{\dagger }+{{\boldsymbol{a}}}_{m}^{\dagger }{{\boldsymbol{a}}}_{m}]\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right]\\ & = & \displaystyle \sum _{m=0,\pm 1}{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}\hslash c{{\boldsymbol{k}}}_{m}{n}_{m}\\ & = & \displaystyle \sum _{m=0,+1}\hslash {k}_{m}{n}_{m},\end{array}\end{eqnarray}$$ (80)

$$ \begin{eqnarray}{{\boldsymbol{p}}}^{{\rm{RCP}}}|{n}\rangle_{m}=\displaystyle \sum _{m=0,+1}\hslash {k}_{m}{n}_{m},\end{eqnarray}$$ (81)

$$ \begin{eqnarray}\begin{array}{lll}{L}_{z}^{{\rm{LCP}}}|{n}\rangle_{m} & = & \displaystyle \frac{{\boldsymbol{L}}|{n}\rangle_{m}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}={\rm{i}}h\displaystyle {\sum }_{m=0,\pm 1}m{\hat{n}}_{m}\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right]=0,\end{array}\end{eqnarray}$$ (82)

$$ \begin{eqnarray}\begin{array}{lll}{L}_{z}^{{\rm{RCP}}}|{n}\rangle_{m} & = & \displaystyle \frac{{\boldsymbol{L}}|{n}\rangle_{m}\cdot {{\boldsymbol{k}}}_{z}}{|{{\boldsymbol{k}}}_{z}|}={\rm{i}}h\displaystyle {\sum }_{m=0,\pm 1}m{\hat{n}}_{m}\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right]=0,\end{array}\end{eqnarray}$$ (83)

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{S}}}^{{\rm{LCP}}}|{n}\rangle_{m} & = & \displaystyle \sum _{m=0,\pm 1,\alpha }\hslash {\hat{k}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha })\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right],\end{array}\end{eqnarray}$$ (84)

$$ \begin{eqnarray}\begin{array}{lll}{{\boldsymbol{S}}}^{{\rm{RCP}}}|{n}\rangle_{m} & = & \displaystyle \sum _{m=0,\pm 1,\alpha }\hslash {\hat{k}}_{m,\alpha }({{\boldsymbol{n}}}_{m,L,\alpha }-{{\boldsymbol{n}}}_{m,R,\alpha })\\ & & \times \left[{\left(+\displaystyle \frac{1}{2}\right)}^{|m|}|{n}\rangle_{m}\right].\end{array}\end{eqnarray}$$ (85)

From the expression one can find out that the orbital angular momentum of SPPs on nanowire is zero under circularly polarized light excitation.

5. Discussion

6. Conclusion

In this work, the electromagnetic modes on a plasmonic cylinder waveguide are quantized. The orbital and spin angular momentums are studied and it can be seen that similar to the vortex waves, the plasmon modes on cylinder nanowires carry both orbital angular momentum and spin angular momentum. The results may be helpful for the future quantum information applications.