Circularly polarized Hankel vortices

Victor V. Kotlyar; Alexey A. Kovalev

doi:10.1364/OE.25.007778

1. Introduction

There have been few vector laser beams that satisfy Maxwell's equations and for which explicit analytic relations to describe projections of the E-and H-field vectors have been derived. For such beams, exact calculation of energy flow and angular momentum is possible at any point in space. In this group, most widely known are Bessel beams. For a symmetric zero-order Bessel beam, relations to express all six projections of the E-and H-field vectors in each point of space have been reported [1]. In [2–5] rather than considering the proper Bessel beam, projections of the E- and H-field vectors of its TE- and TM-modes were obtained. Expressions for an arbitrary-order vector Bessel beam with linear and circular polarization were deduced in [6]. A vector Bessel beam was derived in [6] using Hertz vector potentials. Along with the Bessel beam there is well know exact solution of Maxwell's equation – spherical wave. Spherical waves are highly diverging or converging, and they do not have the angular momentum. Bessel beams are, contrary, neither diverging nor converging, but they do have angular momentum. To the best of our knowledge, there are no known optical fields, that are simultaneously: (i) described by an exact and relatively simple solution of the Maxwell equations, (ii) having angular momentum, and (iii) diverging or converging like spherical wave. Most recently, we have introduced a novel vector beam (alongside the familiar Bessel beams) that can be fully defined by analytical relations. Nonparaxial scalar [7] and vector [8,9] Hankel beams with linear polarization were discussed. The beams were derived by using an expansion in terms of plane waves [10]. Vector cylindrical beams with radial and azimuthal polarization have also been proposed [11]. Note, however, that no analytical relations to express projections of the E- and H-field vectors of such beams have been derived so far. The cylindrical beams are analyzed using Richards-Wolf formulae and used for tightly focusing laser light [12].

Explicit formulae for projections of the E- and H-field vectors of a laser beam enable obtaining analytical relations to describe the Poynting vector and angular momentum (AM) of the light field [6, 9, 13, 14].

In this work, we expand the results we reported in [9] by deriving explicit relations to define the projections of the E-and H-field vectors of Hankel beams with clockwise and anticlockwise circular polarization. The relations derived suggest that in a Hankel beam with topological charge n, the amplitude of the longitudinal E- and H-field components is characterized by the topological charge of n + 1 and n – 1, respectively, for clockwise and anticlockwise polarization. It is consistent with previous works on nonparaxial circularly polarized optical vortices [15–17]. We show that for non-negative topological charges Hankel beams with clockwise and anticlockwise circular polarization propagate in free space differently. The Hankel beam with the clockwise circular polarization has radial divergence (ratio between the radial and longitudinal projections of the Poynting vector) similar to that of the spherical wave, while the beam with the anticlockwise circular polarization has greater radial divergence. The beam with anticlockwise circular polarization has stronger longitudinal component of the electric vector than the beam with clockwise circular polarization. Laser beams under analysis can be generated by using high-resolution diffractive optical elements illuminated by circularly polarized light, showing promise for applications in optically trapping and rotating dielectric microobjects due to their nonparaxiality and absence of sidelobes.

2. Projections of the electromagnetic field vectors for clockwise and anticlockwise circular polarization

In this paper, we start with the Maxwell's equations for monochromatic light propagating in free space:

{\begin{cases} rot H = - i ω ε_{0} E, \\ rot E = i ω μ_{0} H, \\ div E = 0, \\ div H = 0, \end{cases}

where ω is the frequency of light, E and H are respectively electric and magnetic field strengths, ε₀ and μ₀ are the permittivity and permeability of free space. We suppose that the functions E and H depend only on spatial coordinates, while the time dependence factor exp(–iωt) is omitted for brevity. From Eq. (1), the scalar Helmholtz equation follows

\nabla^{2} P + k^{2} P = 0,

where P is any Cartesian component of the vectors E or H, k is the wavenumber of light with a wavelength of λ. It is known [18] that if some function P is an exact solution of the Helmholtz Eq. (2), then its derivative with respect to the longitudinal coordinate ∂P/∂z is also a solution of this equation. So, if the transverse components of the electric strength read as ∂P/∂z, i.e. E_x = α_x∂P/∂z and E_y = α_y∂P/∂z, where α_x and α_y are arbitrary constants, then the other components can be obtained by the Maxwell's Eqs. (1):

E_{z} = - α_{x} \frac{\partial P}{\partial x} - α_{y} \frac{\partial P}{\partial y},

H = - \frac{i}{k} \sqrt{\frac{ε_{0}}{μ_{0}}} rot E .

From now on, for obtaining the vector H by using Eq. (4) we will omit the factor (ε₀/μ₀)^1/2 for brevity.

In [9], an exact solution of the scalar Helmholtz equation was obtained (up to constant multipliers):

P (r, φ, z) = \frac{1}{2} i^{n + 1} λ^{3 / 2} {(r e^{i φ})}^{n} ψ_{n + 1 / 2} (R),

where (r, φ, z) are cylindrical coordinates, R = (r² + z²)^1/2 and

ψ_{ν} (R) = \frac{H_{ν}^{(1)} (k R)}{R^{ν}},

where

H_{ν}^{(1)} (x)

is the ν-th order Hankel function of the 1-st kind. It can be verified by direct substitution of the function P into the Helmholtz Eq. (2), which leads to the following identity:

{\frac{d^{2}}{d ξ^{2}} + \frac{1}{ξ} \frac{d}{d ξ} + [1 - {(n + \frac{1}{2})}^{2} \frac{1}{ξ^{2}}]} H_{n + 1 / 2}^{(1)} (ξ) = 0,

where ξ = kR. This identity is obviously true, since it is the Bessel differential equation. Earlier, we supposed that n ≥_□0, but it can be shown that if a function E(r, φ, z) = A(r, z)exp(imφ) (A(r, z) is an arbitrary complex function) is a solution of the Helmholtz Eq. (2), then a function E(r, φ, z) = A(r, z)exp(–imφ) is also a solution of this equation. Thus, the solution (5) can be generalized for the negative topological charges:

P (r, φ, z) = \frac{1}{2} i^{n + 1} λ^{3 / 2} r^{| n |} e^{i n φ} ψ_{| n | + 1 / 2} (R) .

Using the procedure (3), (4), and the function P of Eq. (5), linearly polarized Hankel beams were studied in [9]. Following the same calculation procedure, we can derive all projections of the electric E = (E_x, E_y, E_z) and magnetic H = (H_x, H_y, H_z) vectors of a Hankel beam with clockwise and anticlockwise circular polarization. To distinguish between the clockwise and anticlockwise circular polarizations, the amplitudes of the projections of the E- and H-field components will be marked with superscript ' + ' and “–”. Then, the amplitudes of transverse components of electric vector of the Hankel beam with clockwise polarization are:

E_{n y}^{+} (r, φ, z) = i E_{n x} (r, φ, z) = i \frac{\partial P}{\partial z} = i^{n} π \sqrt{λ} z r^{| n |} e^{i n φ} ψ_{| n | + 3 / 2} (R) .

Therefore, other components of the electromagnetic field read as (α_x = 1 and α_y = i in (3))

\begin{array}{l} E_{n z}^{+} (r, φ, z) = - \frac{\partial P}{\partial x} - i \frac{\partial P}{\partial y} = - e^{i φ} (\frac{\partial P}{\partial r} + \frac{i}{r} \frac{\partial P}{\partial φ}) = \\ = \frac{1}{2} i^{n - 1} λ^{3 / 2} r^{| n | - 1} e^{i (n + 1) φ} [(| n | - n) ψ_{| n | + 1 / 2} (R) - k r^{2} ψ_{| n | + 3 / 2} (R)], \end{array}

\begin{array}{l} H_{n x}^{+} (r, φ, z) = \frac{- i}{k} (\frac{\partial E_{n z}^{+}}{\partial y} - i \frac{\partial E_{n x}}{\partial z}) = \\ = \frac{i^{n + 1} λ^{3 / 2}}{2} r^{| n |} e^{i n φ} {- (| n | - n) (n + 1) k^{- 1} r^{- 2} e^{2 i φ} ψ_{| n | + 1 / 2} (R) + \\ + [| n | + 2 - (| n | - n) e^{2 i φ}] ψ_{| n | + 3 / 2} (R) - k (z^{2} - i r^{2} e^{i φ} \sin φ) ψ_{| n | + 5 / 2} (R)}, \end{array}

\begin{array}{l} H_{n y}^{+} (r, φ, z) = \frac{- i}{k} (\frac{\partial E_{n x}}{\partial z} - \frac{\partial E_{n z}^{+}}{\partial x}) = \\ = \frac{i^{n} λ^{3 / 2}}{2} r^{| n |} e^{i n φ} {- (| n | - n) (n + 1) k^{- 1} r^{- 2} e^{2 i φ} ψ_{| n | + 1 / 2} (R) - \\ - [| n | + 2 + (| n | - n) e^{2 i φ}] ψ_{| n | + 3 / 2} (R) + k (z^{2} + r^{2} e^{i φ} \cos φ) ψ_{| n | + 5 / 2} (R)}, \end{array}

\begin{array}{l} H_{n z}^{+} (r, φ, z) = \frac{- i}{k} (i \frac{\partial}{\partial x} - \frac{\partial}{\partial y}) E_{n x} = \frac{- i}{k} e^{i φ} (i \frac{\partial}{\partial r} - \frac{1}{r} \frac{\partial}{\partial φ}) E_{n x} = \\ = i^{n + 1} π λ^{1 / 2} z r^{| n | + 1} e^{i (n + 1) φ} [ψ_{| n | + 5 / 2} (R) - (| n | - n) k^{- 1} r^{- 2} ψ_{| n | + 3 / 2} (R)] . \end{array}

For the Hankel beam with anticlockwise polarization, we have (α_x = 1 and α_y = –i in (3)):

E_{n y}^{-} (r, φ, z) = - i E_{n x} (r, φ, z) = - i \frac{\partial P}{\partial z} = - i^{n} π λ^{1 / 2} z r^{| n |} e^{i n φ} ψ_{| n | + 3 / 2} (R),

\begin{array}{l} E_{n z}^{-} (r, φ, z) = - \frac{\partial P}{\partial x} + i \frac{\partial P}{\partial y} = - e^{- i φ} (\frac{\partial P}{\partial r} - \frac{i}{r} \frac{\partial P}{\partial φ}) = \\ = \frac{1}{2} i^{n - 1} λ^{3 / 2} r^{| n | - 1} e^{i (n - 1) φ} [(| n | + n) ψ_{| n | + 1 / 2} (R) - k r^{2} ψ_{| n | + 3 / 2} (R)], \end{array}

\begin{array}{l} H_{n x}^{-} (r, φ, z) = \frac{- i}{k} (\frac{\partial E_{n z}^{-}}{\partial y} + i \frac{\partial E_{n x}}{\partial z}) = \\ = \frac{i^{n - 1} λ^{3 / 2}}{2} r^{| n |} e^{i n φ} {(| n | + n) (n - 1) k^{- 1} r^{- 2} e^{- 2 i φ} ψ_{| n | + 1 / 2} (R) + \\ + [| n | + 2 - (| n | + n) e^{- 2 i φ}] ψ_{| n | + 3 / 2} (R) - k (z^{2} + i r^{2} e^{- i φ} \sin φ) ψ_{| n | + 5 / 2} (R)}, \end{array}

\begin{array}{l} H_{n y}^{-} (r, φ, z) = \frac{- i}{k} (\frac{\partial E_{n x}}{\partial z} - \frac{\partial E_{n z}^{-}}{\partial x}) = \\ = \frac{i^{n} λ^{3 / 2}}{2} r^{| n |} e^{i n φ} {(| n | + n) (n - 1) k^{- 1} r^{- 2} e^{- 2 i φ} ψ_{| n | + 1 / 2} (R) - \\ - [| n | + 2 + (| n | + n) e^{- 2 i φ}] ψ_{| n | + 3 / 2} (R) + k (z^{2} + r^{2} e^{- i φ} \cos φ) ψ_{| n | + 5 / 2} (R)}, \end{array}

\begin{array}{l} H_{n z}^{-} (r, φ, z) = \frac{- i}{k} (- i \frac{\partial}{\partial x} - \frac{\partial}{\partial y}) E_{n x} = \frac{i}{k} e^{- i φ} (i \frac{\partial}{\partial r} + \frac{1}{r} \frac{\partial}{\partial φ}) E_{n x} = \\ = i^{n + 1} π λ^{1 / 2} z r^{| n | + 1} e^{i (n - 1) φ} [(| n | + n) k^{- 1} r^{- 2} ψ_{| n | + 3 / 2} (R) - ψ_{| n | + 5 / 2} (R)] . \end{array}

All these expressions (9)-(18) were verified by substitution into all eight scalar identities of the Maxwell's Eqs. (1) and by using finite-difference calculation of the partial derivatives.

From the comparison between (9)-(13) and (14)-(18) it is seen that if a Hankel beam with the positive topological charge n is circularly polarized clockwise, the amplitudes of the longitudinal projections $E_{n z}^{+}$ and $H_{n z}^{+}$ of the E-and H-field vectors have the topological charge n + 1, and vice versa, in the anticlockwise circularly polarized Hankel beam with positive topological charge n the amplitudes of the longitudinal projections $E_{n z}^{-}$ and $H_{n z}^{-}$ of the E- and H-field vectors have the topological charge n – 1. Similarly, if a Hankel beam with the negative topological charge n is circularly polarized anticlockwise, the amplitudes of the longitudinal projections $E_{n z}^{-}$ and $H_{n z}^{-}$ of the E-and H-field vectors have the negative topological charge –|n| – 1, and vice versa, in the clockwise circularly polarized Hankel beam with negative topological charge n the amplitudes of the longitudinal projections $E_{n z}^{+}$ and $H_{n z}^{+}$ of the E- and H-field vectors have the negative topological charge –|n| + 1. Similar results have been obtained for other vortices with the circular polarization [15–17]. The topological charge of the rest projections remains unchanged and equal to n. As a result, in the absence of an optical vortex (n = 0), the longitudinal components of the Hankel beam with clockwise and anticlockwise circular polarization are optical vortices with the topological charge of 1:

E_{0 z}^{\pm} (r, φ, z) = i π λ^{1 / 2} r e^{\pm i φ} ψ_{3 / 2} (R) = - i λ R^{- 2} (1 + i k^{- 1} R^{- 1}) r e^{i k R \pm i φ},

H_{0 z}^{\pm} (r, φ, z) = \pm i π λ^{1 / 2} z r e^{\pm i φ} ψ_{5 / 2} (R)

while on the optical axis (r = 0) only transverse components of the electromagnetic field take non-zero values (

E_{0 z}^{\pm} (r = 0, φ, z) = H_{0 z}^{\pm} (r = 0, φ, z) = 0

):

E_{0 y}^{+} (r = 0, φ, z) = i E_{0 x} (r = 0, φ, z) = - E_{0 y}^{-} (r = 0, φ, z) = π λ^{1 / 2} z ψ_{3 / 2} (z),

H_{0 x}^{+} (r = 0, φ, z) = - H_{0 x}^{-} (r = 0, φ, z) = \frac{i λ^{3 / 2}}{2} [2 ψ_{3 / 2} (z) - k z^{2} ψ_{5 / 2} (z)],

H_{0 y}^{+} (r = 0, φ, z) = H_{0 y}^{-} (r = 0, φ, z) = - \frac{λ^{3 / 2}}{2} [2 ψ_{3 / 2} (z) - k z^{2} ψ_{5 / 2} (z)],

At n > 1, all on-axis (r = 0) projections of the Hankel beam field equal zero. At n = 1, only longitudinal on-axis (r = 0) Hankel beam components with anticlockwise polarization are non-zero

E_{1 z}^{-} (r = 0, φ, z) = λ^{3 / 2} ψ_{3 / 2} (z),

H_{1 z}^{-} (r = 0, φ, z) = λ^{3 / 2} z ψ_{5 / 2} (z) .

As a result, at n = 1, instead of generating a doughnut intensity at small distances z, the Hankel beam with anticlockwise circular polarization at first generates a focal spot, which is then transformed into a ring, although in the center of the ring the intensity is small, but not zero. This can be seen in Fig. 1, which depicts intensity patterns for a Hankel beam with n = 1, characterized by linear (E_y = 0) (Figs. 1(a) and 1(d)), clockwise circular (Figs. 1(b) and 1(e)), and anticlockwise circular (Figs. 1(c) and 1(f)) polarization at distances z = λ/4 (Figs. 1(a)-1(c)) and z = λ/2 (Figs. 1(d)-1(f)). Shown in Figs. 1(a)-1(c) is the region –λ/2 ≤ x, y ≤ λ/2, while Figs. 1(d)-1(f) depicts the region –3λ/2 ≤ x, y ≤ 3λ/2.

Fig. 1 Intensity patterns I = |E_x|² + |E_y|² + |E_z|² from a Hankel beam (n = 1) with linear (а,d), clockwise circular (b,e) and anticlockwise circular (c,f) polarization at distances z = λ/4 (а-c) and z = λ/2 (d-f).

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Figure 1 suggests that while linearly polarized Hankel beams are devoid of circular symmetry, experiencing rotation in the near field upon propagation, circularly polarized Hankel beams are circularly symmetric, having a doughnut or circular intensity cross-section. The latter is also seen from the relation of the intensity for clockwise polarization, which is independent of the azimuthal angle:

\begin{array}{l} I_{n}^{+} (r, z) = π^{2} λ r^{2 | n |} (2 z^{2} + r^{2}) {| ψ_{| n | + 3 / 2} (R) |}^{2} + \\ + \frac{| n | - n}{2} λ^{3} r^{2 | n | - 2} {\frac{| n | - n}{2} {| ψ_{| n | + 1 / 2} (R) |}^{2} - k r^{2} Re [ψ_{| n | + 1 / 2}^{*} (R) ψ_{| n | + 3 / 2} (R)]}, \end{array}

where Re(...) is the real part of the complex number. The intensity of a Hankel beam with anticlockwise circular polarization is also independent of the azimuthal angle:

\begin{array}{l} I_{n}^{-} (r, z) = π^{2} λ r^{2 | n |} (2 z^{2} + r^{2}) {| ψ_{| n | + 3 / 2} (R) |}^{2} + \\ + \frac{| n | + n}{2} λ^{3} r^{2 | n | - 2} {\frac{| n | + n}{2} {| ψ_{| n | + 1 / 2} (R) |}^{2} - k r^{2} Re [ψ_{| n | + 1 / 2}^{*} (R) ψ_{| n | + 3 / 2} (R)]} . \end{array}

From (27), at n = 1, the on-axis intensity (r = 0) is seen to decrease with increasing z:

I_{1}^{-} (r = 0, z) = λ^{3} {| ψ_{3 / 2} (z) |}^{2} = \frac{λ^{4}}{π^{2} z^{4}} [1 + \frac{1}{{(k z)}^{2}}] .

Time-averaged power flow in the longitudinal direction is the same for both clockwise and anticlockwise circular polarizations:

\begin{array}{l} S_{n z}^{\pm} (r, z) = \frac{1}{2} Re {E_{n x}^{*} H_{n y}^{\pm} - E_{n y}^{\pm *} H_{n x}^{\pm}} = \\ = - \frac{1}{2} π^{2} λ r^{2 | n |} z (2 z^{2} + r^{2}) Im {ψ_{| n | + 3 / 2}^{*} (R) ψ_{| n | + 5 / 2} (R)} = \\ = \frac{1}{2} λ^{2} r^{2 | n |} z \frac{2 z^{2} + r^{2}}{R^{2 | n | + 5}} . \end{array}

It seems interesting that according to Eq. (28) the on-axis intensity is always non-zero ( $I_{1}^{-} \neq 0$ ), while from Eq. (29) it follows that there is no power flow along the optical axis ( $S_{1 z}^{-} = 0$ ). It can be explained since on the optical axis the Hankel beam with n = 1 has only longitudinal non-zero component (24), (25) of the light field. These longitudinal components do not affect the longitudinal power flow (29).

Comparison of Eqs. (26) and (27) at n > 0 shows that intensity of the Hankel beams with the clockwise and anticlockwise circular polarization is different. As it is shown below, it is due to the higher far-field divergence of the Hankel beam with the anticlockwise circular polarization compared to that with the clockwise circular polarization (Fig. 2), although the power flow (29) along the optical axis is the same for the beams of both polarizations. The Hankel beam (27) is diverging faster since the direction of rotation of the polarization plane coincides with the direction of increasing phase of the optical vortex (at n > 0), while the axial power flow (29) is the same since the transverse field components are the same in modulus: $| E_{x}^{+} | = | E_{x}^{-} | = | E_{y}^{+} | = | E_{y}^{-} |$ .

Fig. 2 Higher divergence and greater longitudinal component of the electric vector of the Hankel beam with anticlockwise circular polarization compared to that with the clockwise circular polarization (k₊ and k_– are the wavevectors).

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It is seen in Eq. (29) that this power flow is always non-negative, i.e. the Hankel beams do not possess the properties of tractor beams [19, 20].

3. Vector vortex Hankel beams with circular polarization in the far-field

From Eqs. (26) and (27) follows that we can use for intensity distributions of Hankel beams with negative topological charges:

I_{- | n |}^{\pm} (r, z) = I_{| n |}^{\mp} (r, z) .

Therefore, from now on, without loss of generality we will assume that n > 0, since equations for complex amplitudes are cumbersome if they are written for both positive and negative topological charges.

Although we imply here that the Hankel beam propagates in the positive direction of the optical axis z, this beam is highly non-paraxial and under the far-field we mean the area where R >> λ (area B in Fig. 3) instead of the area, where z >> r, λ, as it is usually meant for paraxial beams (area A in Fig. 3).

Fig. 3 Far field of the Hankel beam (area B in the blue beam) and for arbitrary paraxial beam (area A in the green beam).

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At R >> λ and for n ≥ 0 the relations are simplified. Making use of an approximate Hankel relation (x >> 1) [21] (Eq. (9).2.3)

H_{v}^{(1)} (x) \approx \sqrt{\frac{2}{π x}} {(- i)}^{ν + 1 / 2} e^{i x},

Equation (6) is rearranged to

ψ_{ν} (R) = \frac{H_{ν}^{(1)} (k R)}{R^{ν}} \approx {(- i)}^{ν + 1 / 2} \frac{λ^{1 / 2} e^{i k R}}{π R^{ν + 1 / 2}} .

Substituting (31) into (9)-(13), we obtain for a Hankel beam with clockwise polarization:

E_{n y}^{+} (r, φ, z) = i E_{n x} (r, φ, z) = - λ \frac{r^{n} z}{R^{n + 2}} e^{i n φ + i k R},

E_{n z}^{+} (r, φ, z) = - i λ \frac{r^{n + 1}}{R^{n + 2}} e^{i (n + 1) φ + i k R},

H_{n x}^{+} (r, φ, z) = - λ (i \frac{n + 2}{k R^{n + 2}} + \frac{i r^{2} e^{i φ} \sin φ - z^{2}}{R^{n + 3}}) r^{n} e^{i n φ + i k R},

H_{n y}^{+} (r, φ, z) = i λ (\frac{z^{2} + r^{2} e^{i φ} \cos φ}{R^{n + 3}} - i \frac{n + 2}{k R^{n + 2}}) r^{n} e^{i n φ + i k R},

H_{n z}^{+} (r, φ, z) = - λ \frac{r^{n + 1} z}{R^{n + 3}} e^{i (n + 1) φ + i k R} .

In a similar way, relations (14)-(18) to describe a Hankel beam with anticlockwise polarization in the far-field are:

E_{n y}^{-} (r, φ, z) = - i E_{n x} (r, φ, z) = λ \frac{r^{n} z}{R^{n + 2}} e^{i n φ + i k R},

E_{n z}^{-} (r, φ, z) = - i λ \frac{r^{n + 1}}{R^{n + 2}} (1 - \frac{2 i n R}{k r^{2}}) e^{i (n - 1) φ + i k R},

\begin{array}{l} H_{n x}^{-} (r, φ, z) = - \frac{λ}{k} \frac{r^{n}}{R^{n + 3}} {\frac{2 n (n - 1) e^{- 2 i φ}}{k r^{2}} R^{2} + \\ + i (2 n e^{- 2 i φ} - n - 2) R + k (z^{2} + i r^{2} e^{- i φ} \sin φ)} e^{i n φ + i k R}, \end{array}

\begin{array}{l} H_{n y}^{-} (r, φ, z) = - \frac{λ}{k} \frac{r^{n}}{R^{n + 3}} {i \frac{2 n (n - 1) e^{- 2 i φ}}{k r^{2}} R^{2} - \\ - (2 n e^{- 2 i φ} + n + 2) R - i k (z^{2} + r^{2} e^{- i φ} \cos φ)} e^{i n φ + i k R}, \end{array}

H_{n z}^{-} (r, φ, z) = λ \frac{r^{n + 1} z}{R^{n + 3}} (1 - \frac{2 i n R}{k r^{2}}) e^{i (n - 1) φ + i k R} .

Thus, instead of Eqs. (26)-(28), the far-field intensity distributions for clockwise and anticlockwise polarizations are given by

\begin{array}{l} I_{n}^{+} (r, z) = {| E_{n x} (r, φ, z) |}^{2} + {| E_{n y}^{+} (r, φ, z) |}^{2} + {| E_{n z}^{+} (r, φ, z) |}^{2} = \\ = λ^{2} \frac{r^{2 n}}{R^{2 n + 4}} (r^{2} + 2 z^{2}), \end{array}

\begin{array}{l} I_{n}^{-} (r, z) = {| E_{n x} (r, φ, z) |}^{2} + {| E_{n y}^{-} (r, φ, z) |}^{2} + {| E_{n z}^{-} (r, φ, z) |}^{2} = \\ = λ^{2} \frac{r^{2 n}}{R^{2 n + 4}} (r^{2} + 2 z^{2} + \frac{4 n^{2} R^{2}}{k^{2} r^{2}}) . \end{array}

It is seen in Eqs. (42) and (43) that in any point (r, z) for the anticlockwise polarization the far-field intensity is greater than that for the clockwise polarization (excepting on-axis points). It does not violate the energy conservation law, since the power flow through any plane, orthogonal to the propagation axis z, is the same for both polarizations (29). Since the transverse components have the similar intensity, it follows that the far-field intensity of the longitudinal field component is greater for the anticlockwise polarization. If n = 0 then $I_{0}^{+} (r, z) = I_{0}^{-} (r, z)$ = λ²(r² + 2z²)R^–4.

For clockwise polarization, the intensity maximum is found on a ring of radius

r_{\max}^{+} = z \sqrt{\frac{n - 3 + \sqrt{n^{2} + 2 n + 9}}{2}} .

At n = 0, the intensity maximum is at $r_{\max}^{+} = 0$ (resulting in an on-axis intensity peak instead of a light ring). At large topological charges (n >> 1), with increasing z, the radius of the intensity maximum increases as $r_{\max}^{+} = z \sqrt{n}$ . For anticlockwise polarization, the intensity maximum is defined by a more complicated relation:

(n - 3) \frac{r^{4}}{z^{4}} + 2 n \frac{r^{2}}{z^{2}} - \frac{r^{6}}{z^{6}} + \frac{4 n^{2} (n - 3)}{{(k z)}^{2}} \frac{r^{2}}{z^{2}} + \frac{4 n^{2} (n - 1)}{{(k z)}^{2}} - \frac{8 n^{2}}{{(k z)}^{2}} \frac{r^{4}}{z^{4}} = 0,

but putting z >> λ, the denominators in the three last terms are much larger than the numerators, allowing us to retain just the first three terms. In this case, (45) is rearranged, resulting in the intensity maximum relation identical to that for clockwise circular polarization:

r_{\max}^{-} \approx z \sqrt{\frac{n - 3 + \sqrt{n^{2} + 2 n + 9}}{2}} .

For circularly polarized Hankel beams, the Poynting vector S = Re{E* × H}/2 could be deduced from the relations for projections of the E-and H-fields in (9)-(18). However, the resulting cumbersome relations would be difficult to analyze. Only the longitudinal projection of the Poynting vector (29) is described by the simple expression for any value of z. Meanwhile, far-field projections of the Poynting vector of a circularly polarized Hankel beam are described by less sophisticated relations. The exact expression for the longitudinal projection of vector S is written above (Eq. (29)) and below we only give relations for transverse projections of vector S, which contribute to the on-axis projection of the AM vector. Then, for clockwise circular polarization:

\begin{array}{l} S_{n x}^{+} (r, φ, z) = \frac{1}{2} Re {E_{n y}^{+ *} H_{n z}^{+} - E_{n z}^{+ *} H_{n y}^{+}} = \\ = \frac{1}{2} λ^{2} \frac{r^{2 n + 1}}{R^{2 n + 5}} [(r^{2} + 2 z^{2}) \cos φ - (n + 2) \frac{R}{k} \sin φ], \end{array}

\begin{array}{l} S_{n y}^{+} (r, φ, z) = \frac{1}{2} Re {E_{n z}^{+ *} H_{n x}^{+} - E_{n x}^{*} H_{n z}^{+}} = \\ = \frac{1}{2} λ^{2} \frac{r^{2 n + 1}}{R^{2 n + 5}} [(r^{2} + 2 z^{2}) \sin φ + (n + 2) \frac{R}{k} \cos φ], \end{array}

and for anticlockwise circular polarization:

\begin{array}{l} S_{n x}^{-} (r, φ, z) = \frac{1}{2} Re {E_{n y}^{- *} H_{n z}^{-} - E_{n z}^{- *} H_{n y}^{-}} = \\ = \frac{1}{2} λ^{2} \frac{r^{2 n + 1}}{R^{2 n + 5}} {(2 z^{2} + r^{2}) \cos φ - (4 n \frac{z^{2}}{r^{2}} + n - 2) (\frac{R}{k}) \sin φ + \\ + \frac{2 n}{r^{2}} (2 n + 3) {(\frac{R}{k})}^{2} \cos φ - \frac{4 n^{2} (n - 1)}{r^{4}} {(\frac{R}{k})}^{3} \sin φ}, \end{array}

\begin{array}{l} S_{n y}^{-} (r, φ, z) = E_{n z}^{- *} (r, φ, z) H_{n x}^{-} (r, φ, z) - E_{n x}^{*} (r, φ, z) H_{n z}^{-} (r, φ, z) = \\ = \frac{1}{2} λ^{2} \frac{r^{2 n + 1}}{R^{2 n + 5}} {(2 z^{2} + r^{2}) \sin φ + (4 n \frac{z^{2}}{r^{2}} + n - 2) (\frac{R}{k}) \cos φ + \\ + \frac{2 n}{r^{2}} (2 n + 3) {(\frac{R}{k})}^{2} \sin φ + \frac{4 n^{2} (n - 1)}{r^{4}} {(\frac{R}{k})}^{3} \cos φ} . \end{array}

These expressions are cumbersome, but they allow obtaining of radial and azimuthal projections of the Poynting vector which are more simple and have more clear physical meaning since they are related with the beam divergence and with the beam angular momentum. For the clockwise circular polarization, we obtain:

\begin{array}{l} S_{n r}^{+} (r, z) = S_{n x}^{+} \cos φ + S_{n y}^{+} \sin φ = \frac{1}{2} λ^{2} \frac{r^{2 n + 1}}{R^{2 n + 5}} (r^{2} + 2 z^{2}) \equiv \frac{r}{z} S_{n z}^{+}, \\ S_{n φ}^{+} (r, z) = S_{n y}^{+} \cos φ - S_{n x}^{+} \sin φ = \frac{1}{2} λ^{2} (n + 2) \frac{r^{2 n + 1}}{k R^{2 n + 4}}, \end{array}

while for the anticlockwise circular polarization similar expressions are longer:

\begin{array}{l} S_{n r}^{-} (r, z) = \frac{1}{2} λ^{2} \frac{r^{2 n + 1}}{R^{2 n + 5}} [r^{2} + 2 z^{2} + \frac{2 n}{r^{2}} (2 n + 3) {(\frac{R}{k})}^{2}], \\ S_{n φ}^{-} (r, z) = \frac{1}{2} λ^{2} \frac{r^{2 n + 1}}{k R^{2 n + 4}} [4 n \frac{z^{2}}{r^{2}} + n - 2 + \frac{4 n^{2} (n - 1)}{r^{4}} {(\frac{R}{k})}^{2}] . \end{array}

It is seen in these equations that radial divergence $S_{n r}^{+} / S_{n z}^{+}$ of the Hankel beam with clockwise circular polarization is independent on the topological charge n and is equal to z/r, i.e. it is similar to that of the spherical wave (Fig. 4). In addition, the Hankel beam with zero topological charge has non-zero azimuthal component of the Poynting vector: $S_{0 φ}^{+} = - S_{0 φ}^{-}$ = λ²r/(kR⁴). This means, that for the clockwise polarization $S_{0 φ}^{+} > 0$ , i.e. the power flow rotates anticlockwise, while for the anticlockwise polarization, vice versa, $S_{0 φ}^{-} < 0$ and the power flow rotates clockwise. The radial divergence $S_{n r}^{-} / S_{n z}^{-}$ of the Hankel beam with the anticlockwise circular polarization always exceeds the radial divergence $S_{n r}^{+} / S_{n z}^{+}$ of the Hankel beam with the clockwise circular polarization. This explains why the intensity of the longitudinal field component is greater for the anticlockwise circular polarization.

Fig. 4 Radial divergence of the Hankel beam with clockwise circular polarization.

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Along with the forward-propagating Hankel beams, which travel predominantly in the positive direction of the optical axis z, we can consider back-propagating Hankel beams by changing variable z → f – z (f is the focal distance). According to (51) and (52), such back-propagating Hankel vortices are converging as sharply as converging spherical wave. Therefore, Hankel vortices have the potential to generate tight focus with high-concentrated light energy and with the doughnut intensity distribution, typical to the optical vortices.

The far-field z-projections of the AM density vector j = r × S for Hankel beams with clockwise and anticlockwise circular polarizations are given by

j_{n z}^{+} (r, φ, z) = r S_{n φ}^{+} = \frac{1}{2} λ^{2} (n + 2) \frac{r^{2 n + 2}}{k R^{2 n + 4}},

j_{n z}^{-} (r, φ, z) = r S_{n φ}^{-} = \frac{1}{2} λ^{2} \frac{r^{2 n + 2}}{k R^{2 n + 4}} [4 n \frac{z^{2}}{r^{2}} + n - 2 + \frac{4 n^{2} (n - 1)}{r^{4}} {(\frac{R}{k})}^{2}]

It can be shown that at r = 0, rather than tending to infinity, the AM density in (53) and (54) equals zero irrespective of polarization state. From (53) the total AM of the Hankel beam is seen to be infinite. Note that at n = 0, the Hankel beam still carries the spin angular momentum:

j_{0 z}^{+} (r, φ, z) = - j_{0 z}^{-} (r, φ, z) = \frac{λ^{2}}{k} \frac{r^{2}}{R^{4}} .

The non-zero spin angular momentum explains why there is a non-zero azimuthal component of the Poynting vector at n = 0.

For any n > 0, Hankel beam with the clockwise polarization has a positive axial AM (53), i.e. the power flow in such beam propagates along a right-handed helix. Hankel beam with the anticlockwise polarization propagates as a left helix at n = 0 and as a right helix at n ≥_□2. At n = 1 instead of Eq. (54) we obtain

j_{1 z}^{-} (r, φ, z) = \frac{λ^{2} r^{4}}{2 k R^{6}} (\frac{4 z^{2}}{r^{2}} - 1) .

It is seen in Eq. (56) that $j_{1 z}^{-} > 0$ at z > r/2 and, vice versa, $j_{1 z}^{-} \leq 0$ at z ≤ r/2 (Fig. 5), i.e. the power flow in the Hankel beam with anticlockwise polarization looks like two helices – right-handed near-axis helix nested into left-handed peripheral helix. This means that a Rayleigh microscopic particle, whose size is less than the wavelength, placed into different areas of the Hankel beam, rotates in different directions.

Fig. 5 Direction of longitudinal AM for the Hankel beam with anticlockwise polarization with unitary topological charge n = 1.

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It is seen in Eqs. (53)-(55) that the longitudinal projection of the angular momentum density is always positive with exception of the anticlockwise circularly polarized Hankel beam with zero or unitary topological charge. For n = 0, angular momentum density coincides with the spin angular momentum density and it is negative in every point in space.

From Eqs. (53)–(55) it follows that for both circular polarizations maximal density of the angular momentum of the zero-order (n = 0) Hankel beam is on a circle r = z. When the topological charge is n = 1, the circle of maximal AM density becomes dependent on the polarization. For clockwise polarization, this circle widens and its radius is $r = \sqrt{2} z$ , while for the anticlockwise polarization this circle is, vice versa, shrinks and its radius is $r = \sqrt{5 - \sqrt{21}} z$ ≈0.65z. Figure 6 shows distributions of the longitudinal projection of the angular momentum density for Hankel beams at z = 50λ with clockwise and anticlockwise circular polarization with zero and unitary topological charge. Calculation area –200λ ≤ x, y ≤ 200λ. Widening and shrinking of the AM density distribution for n = 1 (compared to n = 0) is seen in Figs. 6(c, d). In addition, it is clearly seen in Fig. 6(d) that at r = 2z = 100λ the longitudinal AM projection changes, i.e. at this value of radius $j_{1 z}^{-} = 0$ and a Rayleigh particle placed here would not rotate at all.

Fig. 6 Longitudinal projection of the angular momentum density for Hankel beams at z = 50λ with clockwise (a, c) and anticlockwise (b, d) circular polarization with topological charge n = 0 (a, b) and n = 1 (c, d). Red color – positive values, blue color – negative values.

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4. Conclusion

Summing up, we have derived relations to define all six projections of the E-and H-field vectors for circularly polarized vector vortex Hankel beams. These relations fully satisfy Maxwell's equations. For the non-negative topological charge, Hankel beams with clockwise and anticlockwise circular polarization have been shown to propagate in free space in a different manner. Thus, for a Hankel beam with the topological charge n ≥ 0 the amplitude of the longitudinal E-and H-field components has the singularity order of n + 1 and n – 1, respectively, for clockwise and anticlockwise circular polarization. Expressions to describe the intensity of the E-field in the entire space have been derived. The said relations have been found to differ for Hankel beams with clockwise and anticlockwise circular polarization, also demonstrating that the cross-section of circularly polarized Hankel beams shows circular symmetry. In the meantime, linearly polarized Hankel beams are devoid of circular symmetry [9]. Relations to describe projections of the Poynting vector and the AM for the far-field diffraction have also been derived. It is shown that Hankel beam with clockwise circular polarization has radial divergence (ratio between the radial and longitudinal projections of the Poynting vector) similar to that of the spherical wave, while the beam with the anticlockwise circular polarization has greater radial divergence. At n = 0, the circularly polarized Hankel beam has non-zero spin angular momentum. At n = 1, power flow of the Hankel beam with anticlockwise polarization consists of two parts: right-handed helical flow near the optical axis and left-handed helical flow in periphery. At n ≥_□2, power flow is directed along the right-handed helix regardless of direction of the circular polarization. Power flow along the optical axis is the same for the Hankel beams of both circular polarizations, if they have the same topological charge.

Funding

The Federal Agency for Scientific Organizations; Ministry of Education and Science of the Russian Federation; Russian Federation Presidential Grant For Support of the Leading Scientific Schools (NSh-9498.2016.9) and Young Candidates of Sciences (MK-2390.2017.2); Russian Foundation for Basic Research grants (15-07-01174, 15-47-02492, 16-29-11698, 16-47-630483).

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Circularly polarized Hankel vortices

Abstract

1. Introduction

2. Projections of the electromagnetic field vectors for clockwise and anticlockwise circular polarization

3. Vector vortex Hankel beams with circular polarization in the far-field

4. Conclusion

Funding

References and links

Cited By

Figures (6)

Equations (57)

Optics Express