# Spin attributes of structured vector ﬁelds constructed by Hertz potentials

Z. Xiong1, 2 Zhen-Lai Wang2 Y. Liu1, 3 Corresponding author:
<EMAIL_ADDRESS>Bin Zhou1 1 Department of Physics, School of Physics,
Hubei University, Wuhan 430062, Hubei Province
2 School of Mathematics and Physics, Hubei Polytechnic University, Huangshi
435003, Hubei Province
3 Lanzhou Center for Theoretical Physics, Key Laboratory of Theoretical
Physics of Gansu Province, and Key Laboratory of Quantum Theory and
Applications of MoE, Lanzhou University, Lanzhou 730000, Gansu Province

###### Abstract

In this paper, we use the Hertz vector potential to define the electromagnetic
vector of different structured wavefields, and analyze the spin properties of
the wavefields. We show that for the single evanescent waves, the total spin
provides by the transverse spin and originates from the spatial inhomogeneity
of the momentum density of the field. However, for non-single evanescent wave,
there may be a part of the ”extraordinary” spin component $\textbf{s}_{E}$,
and the direction of $\textbf{s}_{E}$ is also perpendicular to the wave
propagation direction. In other words, it is transverse, but it does not
originate from the curl of the wave field momentum density. In addition, we
also calculate the spins of non-planar propagating waves, and analyze the spin
characteristics of these wave fields.

##  I Introduction

Light has spin angular momentum, which is related to the electromagnetic filed
distribution of the wavefields [1]. For a plane wave in free space or a wave
under the paraxial approximation, the spin angular momentum
$\textbf{s}\propto\sigma\hat{k}$. Where $\sigma=\pm 1$, corresponding to the
left-handed and right-handed circular polarization, respectively [1, 2, 3, 4,
5, 6]. In addition, the result also shows the two quantum spin states in
quantum optics [7, 8]. And $\hat{k}$ is the unit propagation direction of the
wave, it can be seen that the direction of the spin angular momentum s and the
wave propagation direction are parallel, in other words, the spin is
longitudinal.

In recent years, the emergence of transverse spin [9, 10, 11, 12]
$\textbf{s}_{t}$ perpendicular to the wave propagation direction has enriched
our understanding of spin angular momentum. This type of spin is ubiquitous in
structured wave fields, such as evanescent waves, focused beams, two-wave
interference, etc.[9, 10, 13, 14, 15, 16]. In addition, comparing with the
longitudinal spin $\textbf{s}_{l}$, the transverse spin $\textbf{s}_{t}$ has
extraordinary properties. For example, it is independent of the helicity of
the wave and can appear even for linearly polarized waves. This special
property lead to many important phenomena and applications possible [17, 18,
19]. So, there is a natural question, what causes this particular transcerse
spin?

A generally accepted explanation [9, 14, 15] is that the transverse
constraints $\nabla\cdot\textbf{E}=0$ results an imaginary longitudinal
electric field component, which parallel to the wave propagation direction.
From the transverse constraints, because of the imaginary longitudinal
electric component, the transverse and longitudinal field components have
$\pi$/2 phase difference, which causes the real-valued electric field vector
$\textbf{E}(\textbf{r},t)=\Re[\textbf{E}(\textbf{r})\exp(-i\omega t)]$ (where
symbol $\Re$ representes the real part) to rotate in the propagation plane,
resulting in a transverse spin perpendicular to the propagation direction.

In 2021, Shi _et al._ [20] proposed that the transverse spin of linearly
polarized evanescentcan be characterized by the wave field Poynting vector P
(or the kinetic momentum $\textbf{p}=\textbf{P}/c^{2}$ of the wavefields, c is
the speed of light in vacuum), as
$\textbf{s}_{t}\propto\nabla\times\textbf{P}$. which resembles the locked spin
momentum $\textbf{p}_{s}$ by spin angular momentum s i.e.
$\textbf{p}_{s}\propto\nabla\times\textbf{s}$. That is, the spin momentum of
light originates from the curl operation of the electromagnetic spin density.
Similarly, the transverse spin of the evanscent wave fields originates from
the spatial inhomogeneity of the momentum density of the field. Indeed, this
approach is beautiful and ingenious, and it can even be extended to other
propagation wave fields [21, 22]. However, it doesn’t always seem to be
complete.

In this paper, we use the Hertz vector potential [23, 24, 25, 26] to define
the electromagnetic field of different structured waves, and calculate the
momentum and spin angular momentum density distributions of these wave fields.
We find that for the non-single evanescent waves of H-type or E-type wave
[27], there are intrinsic attributes of transverse spin and longitudinal spin.
Only when the main polarization direction [28] of the evanescent wave is
aligning with the decaying direction, the total spin provides by the
transverse spin and originates from the spatial inhomogeneity of the momentum
density of the field, as
$\textbf{s}=\textbf{s}_{t}\propto\nabla\times\textbf{p}$. Otherwise, there
will be a part of the ”extraordinary” spin component $\textbf{s}_{E}$, and the
direction of $\textbf{s}_{E}$ is also perpendicular to the wave propagation
direction. In other words, it is transverse, but it does not originate from
the curl of the wave field momentum density. In addition, we also calculate
the spins of non-planar propagating waves, such as Bessel beam, Airy beam and
Gaussian beam, and analyze the spin characteristics of these wave fields. In
particular, for wavefields with evanescent tail wave characteristics, such as
Airy beams, can also exhibit evanescent wave-like spin-momentum locking. Our
results provide further insights into the understanding of spin angular
momentum of the optical fields.

##  II Main Content

###  II.1 Basic Equations

Hertz showed that in a source-free uniform linear isotropic medium,
electromagnetic field can be derived by a single vector potential $\bm{\Pi}$
[29]. The Hertz vector potential is an efficient mathematical formalism for
solving electromagnetic problems. Assuming that there is a time-dependent term
$e^{-i\omega t}$, the Hertz vector wave equation in the Cartesian coordinate
can be expressed as:

$\displaystyle\nabla^{2}\bm{\Pi}+k^{2}\bm{\Pi}=0$ (1)

where Hertz vector potential $\bm{\Pi}=\Pi(x,y,z)\hat{n}$,
$k=\omega^{2}\epsilon\mu$ is the wave number in the medium, $\epsilon$ and
$\mu$ are the permittivity and permeability of the medium. We emphasize that
$\hat{n}$ may take x-, y-, or z-directions, although the selection of
directions has no actual physical meaning. In isotropic media, the Hertz
vector wave equation Eq.(1) has two independent solutions $\bm{\Pi}_{e}$ and
$\bm{\Pi}_{m}$ called electric and magnetic Hertz potentials respectively
[30], which lead to E-type and H-type waves (In some references[21, 22, 26],
it is also called TM, TE wave) as [27]

$\displaystyle\rm E-type:\qquad\textbf{E}$ $\displaystyle=$
$\displaystyle\nabla(\nabla\cdot\bm{\Pi}_{e})+k^{2}\bm{\Pi}_{e}$ (2) H
$\displaystyle=$ $\displaystyle-i\omega\epsilon\nabla\times\bm{\Pi}_{e}$ (3)
$\displaystyle\rm H-type:\qquad\textbf{E}$ $\displaystyle=$ $\displaystyle
i\omega\mu\nabla\times\bm{\Pi}_{m}$ (4) H $\displaystyle=$
$\displaystyle\nabla(\nabla\cdot\bm{\Pi}_{m})+k^{2}\bm{\Pi}_{m}$ (5)

In a source-free homogeneous linear isotropic medium, the results obtained for
E-type and H-type waves are equivalent, and we only need to consider one of
them to define the electromagnetic field. Here, we choose $\bm{\Pi}_{m}$ to
define H-type waves.

For a vector wave field, the kinetic momentum density p can be divided into
orbital momentum $\textbf{p}_{o}$ and spin momentum $\textbf{p}_{s}$ [14, 16,
31, 32] i.e. $\textbf{p}=\textbf{p}_{o}+\textbf{p}_{s}$, where

$\displaystyle\textbf{p}_{o}$ $\displaystyle=$
$\displaystyle\frac{1}{4\omega}\Im[\epsilon\textbf{E}^{*}\cdot(\nabla)\textbf{E}+\mu\textbf{H}^{*}\cdot(\nabla)\textbf{H}]$
(6) $\displaystyle\textbf{p}_{s}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\nabla\times\textbf{s}$ (7)

with

s $\displaystyle=$
$\displaystyle\frac{1}{4\omega}\Im[\epsilon\textbf{E}^{*}\times\textbf{E}+\mu\textbf{H}^{*}\times\textbf{H}]$
(8)

where the symbol $\Im$ representes the imaginary part. s is the spin angular
momentum density. In addition, the kinetic momentum density p also be
represented by Poynting vector [1], namely

p $\displaystyle=$
$\displaystyle\frac{\epsilon\mu}{2}\Re[\textbf{E}^{*}\times\textbf{H}]$ (9)

According to Eqs.(1-9), we can obtain the spin and momentum density
distributions corresponding to different wave fields, and then analyze their
properties.

###  II.2 Calculation and Discussion

####  II.2.1 Spin properties of elliptically polarized plane waves

Firstly, we consider an elliptically polarized single evanescent wave, which
we know contains a spin component perpendicular to the direction of wave
propagation, called _transverse spin_ [9]. However, there is a different
understanding about the transverse spin in [21].

Assuming an elliptical polarization single evanescent wave propagating along
the z-direction and decaying in x-direction, the wave vector of the evanescent
can be written as $\textbf{k}=k_{z}\hat{z}+i\kappa\hat{x}$, where $k_{z}$ is
longitudinal wave number and $\kappa$ is decaying coefficient of the
evanescent wave. We can define the Hertz vector potential of the field as

$\displaystyle\bm{\Pi}_{m}$ $\displaystyle=$
$\displaystyle\frac{A}{\sqrt{1+|m|^{2}}}(\hat{x}+m\frac{k_{z}}{k}\hat{y})e^{ik_{z}z-\kappa
x}$ (10)

which is the solution of the Hertz vector wave equation 1. where A is the wave
amplitude, m is a complex number describing the elliptical polarization state
with $\sigma=\frac{2\Im m}{1+|m|^{2}}$ [9]. And $k\equiv|\textbf{k}|$ is the
wave number. Taking Eq.(10) into Eqs.(4-5), we can obtain the electromagnetic
field of this evanescent wave. And then, we can further get the momentum and
spin angular momentum by taking the electromagnetic vector into Eqs.(6-9), as

p $\displaystyle=$
$\displaystyle\frac{W}{\omega}(\sigma\frac{k\kappa}{k_{z}}\hat{y}+\frac{k^{2}}{k_{z}}\hat{z})$
(11) s $\displaystyle=$
$\displaystyle\frac{W}{\omega}(\frac{\kappa}{k_{z}}\hat{y}+\sigma\frac{k}{k_{z}}\hat{z})$
(12)

where the energy density W of the wave field is
$W=\frac{\mu}{2}k^{4}_{z}A^{2}e^{-2\kappa x}$. Eq.(12) shows that there is a
special spin component $s_{y}$ in this evanescent wave field, which called
_transverse spin_ in [9]. And the z-component called _longitudinal spin_.
However, there is no physical meaning in distinguishing longitudinal and
transverse directions solely based on the direction of the spin vector. So,
according the approach in [20, 21] to define the transver spin of this case
might be more reasonable, as

(a) (b)

Figure 1: The spin characteristics of single evanescent wave for Eq.(10) (a)
linear polarization (m=0, $\sigma=0$) (b) circular polarization (m=i,
$\sigma=1$)

$\displaystyle\textbf{s}_{t}$ $\displaystyle=$
$\displaystyle\frac{W}{\omega}[\frac{\kappa}{k_{z}}\hat{y}-\sigma\frac{\kappa^{2}}{kk_{z}}\hat{z}]$
(13)

Although there is a spin component parallel to the direction of wave
propagation in Eq.(13), it still can be seen as transverse spin [21]. In
addition, the longitudinal spin $\textbf{s}_{l}$ can be obtained by
subtracting the transverse spin $\textbf{s}_{t}$ from the total spin s, as

$\displaystyle\textbf{s}_{l}$ $\displaystyle=$
$\displaystyle\textbf{s}-\textbf{s}_{t}$ (14) $\displaystyle=$
$\displaystyle\hbar\sigma\hat{k}$

This is logical in physics because the elementary feature of longitudinal spin
$\textbf{s}_{l}$ for the same kind of wave-packet is constant. That is, the
transverse spin defined by [21] is reasonable. FIG. 1 shows the spin vector
distribution of the evanescent wave shown in Eq.(10) for linear and circular
polarization.

When the evanescent wave decaying parameter $\kappa=0$ i.e. $k=k_{z}$,
corresponding to the case of propagating wave, Eq.(10) will be transformed
into

$\displaystyle\bm{\Pi}_{m}$ $\displaystyle=$
$\displaystyle\frac{A}{\sqrt{1+|m|^{2}}}(\hat{x}+m\hat{y})e^{ikz}$ (15)

Similarly, we can get the spin characteristics of the plane wave field,

$\displaystyle\textbf{s}_{t}=0,\ \textbf{s}_{l}=\hbar\sigma\hat{k}$ (16)

As we konw, the momentum density of propagating plane wave is uniformly
distributed on the propagation plane. So the result obtained by calculating
its curl is of course 0. That is to say, its transverse spin
$\textbf{s}_{t}=0$, and the total spin s is provided by the longitudinal spin
$\textbf{s}_{l}$ i.e. $\textbf{s}=\textbf{s}_{l}\propto\hbar\sigma$.

From the above content, the definition of transverse spin $\textbf{s}_{t}$ in
[20, 21] are always valid for plane waves, whether they are evanescent waves
or propagating waves. However, for some non-single evanescent waves and some
structurally propagated waves, it seems that it is not enough to define the
transverse spin using the wave field momentum density inhomogeneity. As we
shall show soon, there may be an ”extraordinary” spin component
$\textbf{s}_{E}$, which direction is perpendicular to the local wave vector.
And it does not originate from the inhomogeneity of the wave field momentum
density.

####  II.2.2 H-type evanescent waves

In Sec. $\rm II$ A, we show that electromagnetic can be defined by a single
Hertz vector. Here, we use the magnetic Hertz vector potential to define the
H-type waves electromagnetic vector. We consider the evanescent wave case
firstly, for an evanescent wave decaying along the z direction, we can obtain
the magnetic Hertz potential by solving the Hertz’s vector wave equation in
Cartesian coordinate

$\displaystyle\nabla^{2}\bm{\Pi}_{m}+k^{2}\bm{\Pi}_{m}=0$ (17)

Hertz’s vector wave equation for source free homogeneously isotropic medium
can be reduced to a scalar wave equation [33], so Eq.(17) can be simplified to

$\displaystyle\nabla^{2}\Pi_{m}+k^{2}\Pi_{m}=0$ (18)

A feasible solution with

$\displaystyle\bm{\Pi}_{m}=F(x,y)e^{-\kappa z}\hat{n}$ (19)

where $F(x,y)$ fulfill $\nabla^{2}_{\bot}F+\beta^{2}F=0$ with
$\nabla^{2}_{\bot}=\frac{\partial^{2}}{\partial
x^{2}}+\frac{\partial^{2}}{\partial y^{2}},\beta^{2}=k^{2}+\kappa^{2}$,
$\kappa$ is the decaying coefficient of the evanescent wave, $\hat{n}$ can be
either in the x-, y- or z-direction.

Consider now $\hat{n}=\hat{z}$, in this case the Hertz vector potential can be
written as

$\displaystyle\bm{\Pi}_{m}=F(x,y)e^{-\kappa z}\hat{z}$ (20)

Taking Eq.(20) into Eqs.(4-5), we can get the electromagnetic field of the
evanescent wave in this case

E $\displaystyle=$ $\displaystyle i\omega\mu(\frac{\partial F}{\partial
y}\hat{x}-\frac{\partial F}{\partial x}\hat{y}+0\hat{z})e^{-\kappa z}$ (21) H
$\displaystyle=$ $\displaystyle(-\kappa\frac{\partial F}{\partial
x}\hat{x}-\kappa\frac{\partial F}{\partial
y}\hat{y}+\beta^{2}F\hat{z})e^{-\kappa z}$ (22)

The above formula Eqs.(22-22) is naturally satisfied Maxwell equations, which
is consistent with the electromagnetic field defined in [20]. And the final
result is naturally the same, as

$\displaystyle\textbf{s}_{t}$ $\displaystyle=$
$\displaystyle\frac{1}{2k^{2}}\nabla\times\textbf{p}$ (23) $\displaystyle=$
$\displaystyle\frac{\mu}{2\omega}\Im[\kappa\beta^{2}F^{*}\frac{\partial
F}{\partial y}\hat{x}+\kappa\beta^{2}F\frac{\partial F^{*}}{\partial
x}\hat{y}+\beta^{2}\frac{\partial F}{\partial y}\frac{\partial F^{*}}{\partial
x}\hat{z}]$ $\displaystyle=$ s

where symbol $*$ represents complex conjugate. The result of Eq.(23) show that
the wave field spin angular momentum s is only caused by the spatial
inhomogeneity of the wave field momentum density. And the total spin angular
momentum satisfies the spin-momentum locking [20, 34], which can be used as an
example to demonstrate the quantum spin Hall effect of light [35, 36].

As we emphasized earlier, the Hertz vector can take the x-, y-, and z-
directions. The selection of different directions has no actual physical
meaning, but will lead to different results. Here, we still assume that an
evanescent wave decays along the z-direction, but define the Hertz vector
along the x-direction, at this time the polarization of the wave field will be
changed.

For the case of $\bm{\Pi}_{m}=F(x,y)e^{-\kappa z}\hat{x}$, the electromagnetic
field of the evanescent wave can be expressed as

E $\displaystyle=$ $\displaystyle i\omega\mu(0\hat{x}-\kappa
F\hat{y}-\frac{\partial F}{\partial y}\hat{z})e^{-\kappa z}$ (24) H
$\displaystyle=$ $\displaystyle[(\frac{\partial^{2}F}{\partial
x^{2}}+k^{2}F)\hat{x}+\frac{\partial^{2}F}{\partial y\partial
x}\hat{y}-\kappa\frac{\partial F}{\partial x}\hat{z}]e^{-\kappa z}$ (25)

which still satisfies Maxwell equations. Taking Eqs.(24-25) into Eqs.(8-9),
one can obtain

$\displaystyle\textbf{s}=\frac{\mu}{2\omega}\Im\left(\begin{array}[]{cccc}k^{2}\kappa
F^{*}\frac{\partial F}{\partial y}+\kappa\frac{\partial^{2}F}{\partial
y\partial x}\frac{\partial F^{*}}{\partial x}\\\ \\\ \kappa\frac{\partial
F}{\partial x}(\frac{\partial^{2}F^{*}}{\partial x^{2}}+k^{2}F^{*})\\\ \\\
\frac{\partial^{2}F}{\partial y\partial x}(\frac{\partial^{2}F^{*}}{\partial
x^{2}}+k^{2}F^{*})\end{array}\right)e^{-2\kappa z},\
\textbf{p}=\frac{\mu}{2\omega}k^{2}\Im\left(\begin{array}[]{cccc}\kappa^{2}F^{*}\frac{\partial
F}{\partial x}+\frac{\partial^{2}F}{\partial y\partial x}\frac{\partial
F^{*}}{\partial y}\\\ \\\ -\frac{\partial F^{*}}{\partial
y}(\frac{\partial^{2}F}{\partial x^{2}}+k^{2}F)\\\ \\\ \kappa
F^{*}(\frac{\partial^{2}F}{\partial
x^{2}}+k^{2}F)\end{array}\right)e^{-2\kappa z}$ (36)

According the definition of transverse spin in [20, 21], one can obtain

$\displaystyle\textbf{s}_{t}$ $\displaystyle=$
$\displaystyle\frac{\mu}{4\omega}\Im\left(\begin{array}[]{cccc}\kappa
F^{*}\frac{\partial^{3}F}{\partial y\partial x^{2}}-\kappa\frac{\partial
F^{*}}{\partial y}\frac{\partial F^{2}}{\partial x^{2}}+2k^{2}\kappa
F^{*}\frac{\partial F}{\partial y}\\\ \\\ -(\kappa\frac{\partial
F^{*}}{\partial x}\frac{\partial^{2}F}{\partial x^{2}}+\kappa
F^{*}\frac{\partial^{3}F}{\partial x^{3}}+2\kappa^{3}F^{*}\frac{\partial
F}{\partial x}+2\kappa\frac{\partial^{2}F}{\partial y\partial x}\frac{\partial
F^{*}}{\partial y})\\\ \\\ -(\frac{\partial^{2}F^{*}}{\partial y\partial
x}\frac{\partial^{2}F}{\partial x^{2}}+\frac{\partial^{2}F}{\partial y\partial
x}\frac{\partial^{2}F^{*}}{\partial
x^{2}}+k^{2}F\frac{\partial^{2}F^{*}}{\partial y\partial
x}+\kappa^{2}F^{*}\frac{\partial^{2}F}{\partial y\partial x}+\frac{\partial
F^{*}}{\partial y}\frac{\partial^{3}F}{\partial x^{3}}+\frac{\partial
F^{*}}{\partial y}\frac{\partial^{3}F}{\partial y^{2}\partial
x}+\beta^{2}\frac{\partial F^{*}}{\partial y}\frac{\partial F}{\partial
x})\end{array}\right)e^{-2\kappa z}$ (42) $\displaystyle\neq$ s (43)

We can see that Eq.(43) and Eq.(23) show very different results. In the case
of Eq.(49), the total spin is not entirely provided by the transverse spin
caused by inhomogeneity momentum density. There should also be an
”extraordinary” transverse spin $\textbf{s}_{E}$, as

$\displaystyle\textbf{s}_{E}$ $\displaystyle=$
$\displaystyle\textbf{s}-\textbf{s}_{t}$ (49) $\displaystyle=$
$\displaystyle\frac{\mu}{4\omega}\Im\left(\begin{array}[]{cccc}\kappa\frac{\partial
F^{*}}{\partial y}\frac{\partial^{2}F}{\partial x^{2}}-\kappa
F^{*}\frac{\partial^{3}F}{\partial x^{2}\partial y}+2\kappa\frac{\partial
F^{*}}{\partial x}\frac{\partial^{2}F}{\partial y\partial x}\\\ \\\
\kappa\frac{\partial F}{\partial x}\frac{\partial^{2}F^{*}}{\partial
x^{2}}+\kappa F^{*}\frac{\partial^{3}F}{\partial x^{3}}+2\kappa\frac{\partial
F^{*}}{\partial y}\frac{\partial^{2}F}{\partial y\partial
x}+2\kappa\beta^{2}F^{*}\frac{\partial F}{\partial x}\\\ \\\
0\end{array}\right)e^{-2\kappa z}$

Certainly, in the case of single evanescent wave, the result of Eq.(49) is
zero i.e. $\textbf{s}_{E}=0$ and $\textbf{s}=\textbf{s}_{t}$. However, for
non-single evanescent wave, there may be non-zero spin $\textbf{s}_{E}$, it
does not originates from the inhomogeneity momentum density. And the
directions of $\textbf{s}_{E}$ may be perpendicular to the local wave vector.
That is, $\textbf{s}_{E}$ is also transverse. So, the definition of transverse
spin in [20, 21] doesn’t seem to be enough. In fact, for non-single evanescent
waves, only when the main polarization direction [28] of the evanescent wave
is aligning with the decaying direction, the total spin provides by the
transverse spin and originates from the spatial inhomogeneity of the momentum
density of the field, as shown in Eq.(23). Otherwise, there is an
”extraordinary” spin $\textbf{s}_{E}$, which is also transverse.

Here, we give an example, for a cosine wave [1, 20] propagating in the
y-direction and decaying in the z-direction,the wave vector is
$\textbf{k}=k_{x}\hat{x}+k_{y}\hat{y}+i\kappa\hat{z}$, its Hertz vector can be
expressed as

$\displaystyle\bm{\Pi}_{m}=\cos(k_{x}x)e^{ik_{y}y-\kappa z}\hat{x}$ (50)

Taking Eq.(50) into Eqs.(8-9), and Eq.(49), we can obatin

$\displaystyle\textbf{s}_{E}$ $\displaystyle=$
$\displaystyle\frac{\mu}{2\omega}k_{x}^{2}k_{y}\kappa e^{-2\kappa z}\hat{x}$
(51)

The momentum density and spin angular momentum density distribution of this
cosine field as shown in FIG. 2. Eq.(51) shows that there is an
”extraordinary” spin $\textbf{s}_{E}$, it is obvious that it is perpendicular
to the direction of wave propagation, so it should be transverse spin. But it
does not originate from the inhomogeneity momentum density. In other words, it
is not enough to use inhomogeneity momentum density to characterize the
transverse spin in this case.

Figure 2: Momentum density $\textbf{P}=P_{y}$, total spin angular momentum
density $\textbf{s}=s_{x}\hat{x}+s_{z}\hat{z}$, transverse spin
$\textbf{s}_{t}=s_{tx}\hat{x}+s_{tz}\hat{z}$ and longitudinal spin
$\textbf{s}_{l}$ angular momentum density distribution corresponding to cosine
evanescent wave of Eq.(50).

In addition, we also calculate the case of
$\bm{\Pi}_{m}=\cos(k_{x}x)e^{ik_{y}y-\kappa z}\hat{y}$, there is also a non-
zero $\textbf{s}_{E}$. When we change the direction of wave propagation, the
results are similar. That is, for single evanescent, the definition of
transverse spin in [20, 21] is absolutely valid. But for non-single
evanescent, only when the main polarization direction [28] of the evanescent
wave is aligning with the decaying direction, the definition is also valid.
Otherwise, it is not sufficient to rely solely on spatial inhomogeneities in
momentum density to define transverse spin.

####  II.2.3 H-type propagating waves

We have discussed the evanescent wave of H-type, and analyzed their spin
characteristics. In addition, we know that the propagating wave also has non-
trivial spin properties. For a wave propagates along a guided structure only
occurs in one direction [33], we assume that the wave propagates along the
z-direction. Therefore, the waves along a uniform guided structure have only
an $e^{ik_{z}z}$ dependence in the direction. This means that Hertz vector
potentials for two-dimensional uniform guided structures are of the form

$\displaystyle\bm{\Pi}_{m}=\psi(x,y)e^{ik_{z}z}\hat{n}$ (52)

where $\psi(x,y)$ fulfill $\nabla^{2}_{\bot}\psi+\zeta^{2}\psi=0$ with
$\zeta^{2}=k^{2}-k_{z}^{2}$, $k\equiv|\textbf{k}|$ is the wave number, and
$\hat{n}$ can be either in the x-, y- or z-direction.

Here, we still define $\hat{n}=\hat{z}$, so Eq.(52) can be written as

$\displaystyle\bm{\Pi}_{m}=\psi(x,y)e^{ik_{z}z}\hat{z}$ (53)

Figure 3: Momentum density $\textbf{p}=p_{z}$, total spin angular momentum
density $\textbf{s}_{\phi}$, transverse spin $\textbf{s}_{t\phi}$ and
extraordinary spin $\textbf{s}_{E\phi}$ angular momentum density distribution
of the Bessel beam in Eq.(62).

Taking Eq.(53) into Eqs.(4-5), we can get the electromagnetic field of the
propagating wave as

E $\displaystyle=$ $\displaystyle i\omega\mu(\frac{\partial\psi}{\partial
y}\hat{x}-\frac{\partial\psi}{\partial x}\hat{y}+0\hat{z})e^{ik_{z}z}$ (54) H
$\displaystyle=$ $\displaystyle(ik_{z}\frac{\partial\psi}{\partial
x}\hat{x}+ik_{z}\frac{\partial\psi}{\partial
y}\hat{y}+\zeta^{2}\psi\hat{z})e^{ik_{z}z}$ (55)

Taking Eqs.(54-55) into Eqs.(8-9), we can get the momentum and spin angular
momentum in this case. Here, we again use the definition of transverse spin in
[21], assuming it is valid, we can get

$\displaystyle\textbf{s}_{E}$ $\displaystyle=$ $\displaystyle\frac{\mu
k_{z}}{2\omega}\Re\left(\begin{array}[]{cccc}\frac{\partial^{2}\psi^{*}}{\partial
y^{2}}\frac{\partial\psi}{\partial y}-\frac{\partial^{2}\psi}{\partial
x\partial y}\frac{\partial\psi^{*}}{\partial
x}-\zeta^{2}\psi\frac{\partial\psi^{*}}{\partial y}\\\ \\\
\frac{\partial^{2}\psi^{*}}{\partial x\partial y}\frac{\partial\psi}{\partial
y}-\frac{\partial^{2}\psi}{\partial x^{2}}\frac{\partial\psi^{*}}{\partial
x}+\zeta^{2}\psi\frac{\partial\psi^{*}}{\partial x}\\\ \\\
0\end{array}\right)$ (61)

Similar to Sec.$\rm II$ $\textbf{B}.2$, for the non-planar propagating wave,
there also be non-zero spin E, and it is obviously transverse. So, the
definition of transverse spin in [21] is not complete. when we change the
direction of the Hertz vector along the x-, y- and z-direction, the results
are similar, there are non-zero $\textbf{s}_{E}$.

Here, we exemplify the zero-order Bessel beam [26, 37, 38, 39]. The reason for
considering the zero-order is to avoid the influence of spin-orbit coupling of
high-order Bessel beams [40, 41]. The Hertz vector potential of the Bessel
beam can be written as

$\displaystyle\bm{\Pi}_{m}$ $\displaystyle=$ $\displaystyle
J_{0}(rk_{r})e^{ik_{z}z}\hat{z}$ (62)

which is the solution of the Hertz vector wave equation in cylindrical
coordinates. The wave number fulfill $k^{2}=k_{r}^{2}+k_{z}^{2}$. Taking
Eq.(62) into Eqs.(4-5), Eqs.(8-9) and Eq.(61), where $\psi=J_{0}(rk_{r})$, we
can obtain the momentum density and spin density distributions of the zero-
order Bessel field as shown in FIG. 3.

p $\displaystyle=$
$\displaystyle\frac{\mu}{2\omega}k^{2}k^{2}_{r}k_{z}J_{1}(rk_{r})^{2}\hat{z}$
(63) s $\displaystyle=$
$\displaystyle-\frac{\mu}{2\omega}k^{3}_{r}k_{z}J_{0}(rk_{r})J_{1}(rk_{r})\hat{\phi}$
(64) $\displaystyle\textbf{s}_{t}$ $\displaystyle=$
$\displaystyle-\frac{\mu}{2\omega}k^{2}_{r}k_{z}[k_{r}J_{0}(rk_{r})J_{1}(rk_{r})-\frac{J_{1}(rk_{r})^{2}}{r}]\hat{\phi}$
(65) $\displaystyle\textbf{s}_{E}$ $\displaystyle=$
$\displaystyle-\frac{\mu}{2\omega}k^{2}_{r}k_{z}\frac{J_{1}(rk_{r})^{2}}{r}\hat{\phi}$
(66)

From Eq.(63) we can see that the Bessel field momentum only contains the
longitudinal component z, the total spin angular momentum only contains the
azimuth component, that is, the spin is azimuth polarization [42, 43], and
there is an ”extraordinary” transverse spin $\textbf{s}_{E}$ along the azimuth
direction. In addition, we also calculate the case of
$\bm{\Pi}_{m}=J_{0}(rk_{r})e^{ik_{z}z}\hat{x}$ and
$\bm{\Pi}_{m}=J_{0}(rk_{r})e^{ik_{z}z}\hat{y}$, and there are still non-zero
$\textbf{s}_{E}$.

(a) (b) (c)
(d) (e) (f)

Figure 4: The spin angular momentum (a) and momentum (b)-(c) density
distributions of the Airy beam in Eqs.(72-73), and (d)-(f) are total spin,
transverse spin and extraordinary spin angular momentum density respectively
of the Airy beam in the case of $\bm{\Pi}_{m}=\Pi_{m}\hat{x}$, here we choose
$k=x_{0}=1$. The orange arrow represent the electric field vector distribution
of Eq.(75)

####  II.2.4 The solution of Hertz vector wave equation under paraxial
approximation

Previously, we mainly discussed the exact solution of the Hertz vector wave
equation, so as to solve the Hertz vectors of different wave fields, and then
solve the electromagnetic vectors corresponding to different wave fields. In
this Sec., we will obtain the Hertz vector by solving the Hertz vector wave
equation under the paraxial approximation [1, 44, 45, 46], and analyze the
spin vector distribution of its wave field.

Under the paraxial approximation
$|\partial^{2}_{z}\psi|\ll|2k\partial_{z}\psi|$, we can obtain the paraxial
wave equation [47, 48],

$\displaystyle
i\frac{\partial\psi}{\partial\xi}+\frac{1}{2}\frac{\partial^{2}\psi}{\partial
s^{2}}=0$ (67)

where $\psi$ is the electric field envelope, $s=x/x_{0}$ represents a
dimensionless transverse coordinate, $x_{0}$ is an arbitrary transverse scale,
and $\xi=z/kx^{2}_{0}$ is a normalized propagation distance, k is the wave
number. We can also replace the electric field envelope $\psi$ in Eq.(67) with
Hertz vector potential $\bm{\Pi}_{m}$ to make it a Hertz potential wave
equation under the paraxial approximation, as

$\displaystyle
i\frac{\partial\bm{\Pi}_{m}}{\partial\xi}+\frac{1}{2}\frac{\partial^{2}\bm{\Pi}_{m}}{\partial
s^{2}}=0$ (68)

Obviously, one of the accelerating solutions of Eq.(68) is the well-known Airy
function with the characteristic infinite oscillatory tail, which can be
written as

$\displaystyle\bm{\Pi}_{m}=Ai(s-\frac{\xi^{2}}{4})e^{\frac{is\xi}{2}-\frac{i\xi^{3}}{12}}\hat{n}$
(69)

Similarly, the polarized direction $\hat{n}$ can be chosen as x-, y- and z-
directions. For the convenience of calculation, here we choose
$\hat{n}=\hat{y}$, so the Hertz vector of the propagating Airy beam is
$\bm{\Pi}_{m}=Ai(s-\frac{\xi^{2}}{4})e^{\frac{is\xi}{2}-\frac{i\xi^{3}}{12}}\hat{y}$,
taking it into Eqs.(4-5), one obatians the electromagnetic vector of the
H-type wave, as

E $\displaystyle=$
$\displaystyle\omega\mu[(\frac{x\mathcal{A}}{2kx^{3}_{0}}-\frac{z^{2}\mathcal{A}}{4k^{3}x^{6}_{0}}+\frac{iz\mathcal{A}^{{}^{\prime}}}{2k^{2}x^{4}_{0}})\hat{x}$
(70)
$\displaystyle+(\frac{-z\mathcal{A}}{2kx^{3}_{0}}+\frac{i\mathcal{A}^{{}^{\prime}}}{x_{0}})\hat{z}]e^{\frac{ixz}{2kx^{3}_{0}}-\frac{iz^{3}}{12k^{3}x^{6}_{0}}}$
H $\displaystyle=$ $\displaystyle
k^{2}\mathcal{A}\hat{y}e^{\frac{ixz}{2kx^{3}_{0}}-\frac{iz^{3}}{12k^{3}x^{6}_{0}}}$
(71)

where $\mathcal{A}=Ai(\frac{x}{x_{0}}-\frac{z^{2}}{4k^{2}x^{4}_{0}})$ and
$\mathcal{A}^{{}^{\prime}}=Ai^{{}^{\prime}}(\frac{x}{x_{0}}-\frac{z^{2}}{4k^{2}x^{4}_{0}})$
represents the first derivative of the Airy function. And then, we can get the
momentum and spin angular momentum properties of this Airy beam by taking the
electromagnetic field into Eqs.(8-9), the spin angular momentum density
distribution as shown in FIG. 4.

s $\displaystyle=$
$\displaystyle-\frac{\mu}{4\omega}\frac{xk}{x_{0}^{4}}Ai(\frac{x}{x_{0}}-\frac{z^{2}}{4k^{2}x^{4}_{0}})Ai^{{}^{\prime}}(\frac{x}{x_{0}}-\frac{z^{2}}{4k^{2}x^{4}_{0}})\hat{y}$
(72) p $\displaystyle=$
$\displaystyle\frac{\mu}{2\omega}k^{4}[\frac{z}{2kx^{3}_{0}}Ai^{2}(\frac{x}{x_{0}}-\frac{z^{2}}{4k^{2}x^{4}_{0}})\hat{x}$
(73) $\displaystyle-$
$\displaystyle(\frac{z^{2}}{4k^{3}x^{6}_{0}}-\frac{x}{2kx^{3}_{0}})Ai^{2}(\frac{x}{x_{0}}-\frac{z^{2}}{4k^{2}x^{4}_{0}})\hat{z}]$
$\displaystyle\textbf{s}_{t}$ $\displaystyle=$
$\displaystyle\frac{1}{2k^{2}}\nabla\times\textbf{p}=\textbf{s},\quad\textbf{s}_{E}=0$
(74)

where $Ai^{\prime}$ is the first derivative of the Airy function. Eq.(72)
shows a special result that an Airy beam propagating in the x-z plane has a
spin angular momentum s along the y direction, and the spin is completely
caused by the spatial inhomogeneity of the beam momentum density transverse
spin $\textbf{s}_{t}$ provided. That is, the definition of transverse spin in
[20, 21] also applies in this case.

Figure 5: The phase distributions of the two electric field components and its
phase difference of Eq.(70).

However, when we choose $\hat{n}=\hat{x}$, at this time, the form of the wave
field will change. So we may get different results. In the case of
$\bm{\Pi}_{m}=Ai(s-\frac{\xi^{2}}{4})e^{\frac{is\xi}{2}-\frac{i\xi^{3}}{12}}\hat{x}$,
we can get the electromagnetic vector as

E $\displaystyle=$
$\displaystyle\omega\mu(\frac{-x\mathcal{A}}{2kx^{3}_{0}}+\frac{z^{2}\mathcal{A}}{4k^{3}x^{6}_{0}}-\frac{iz\mathcal{A}^{{}^{\prime}}}{2k^{2}x^{4}_{0}})\hat{y}e^{\frac{ixz}{2kx^{3}_{0}}-\frac{iz^{3}}{12k^{3}x^{6}_{0}}}$
(75) H $\displaystyle=$
$\displaystyle[(\frac{x\mathcal{A}}{x^{3}_{0}}-\frac{z^{2}\mathcal{A}}{2k^{2}x^{6}_{0}}+\frac{iz\mathcal{A}^{{}^{\prime}}}{kx^{4}_{0}}+k^{2}\mathcal{A})\hat{x}$
(76)
$\displaystyle+(\frac{i\mathcal{A}}{2kx^{3}_{0}}-\frac{3xz\mathcal{A}}{4k^{2}x^{6}_{0}}+\frac{z^{3}\mathcal{A}}{4k^{4}x^{9}_{0}}-\frac{iz^{2}\mathcal{A}^{{}^{\prime}}}{2k^{3}x^{7}_{0}}$
$\displaystyle+\frac{ix\mathcal{A}^{{}^{\prime}}}{2kx^{4}_{0}})\hat{z}]e^{\frac{ixz}{2kx^{3}_{0}}-\frac{iz^{3}}{12k^{3}x^{6}_{0}}}$

we can also obtain the phase distributions of the electric field and the phase
difference between the two electric filed components as shown in FIG. 5. And
the we can get the spin and momentum density by taking the electromagnetic
vector into Eqs.(8-9), which shown in FIG. 4.

s $\displaystyle=$
$\displaystyle\frac{\mu}{2\omega}[\frac{-x\mathcal{A}^{2}}{2kx^{6}_{0}}+\frac{z^{2}\mathcal{A}^{2}}{4k^{3}x^{9}_{0}}-\frac{k\mathcal{A}^{2}}{2x^{3}_{0}}+\frac{z^{2}\mathcal{A}\mathcal{A}^{{}^{\prime}}}{2kx^{7}_{0}}-\frac{x^{2}\mathcal{A}\mathcal{A}^{{}^{\prime}}}{2kx^{7}_{0}}-\frac{kx\mathcal{A}\mathcal{A}^{{}^{\prime}}}{2x^{4}_{0}}]\hat{y}$
(77) p $\displaystyle=$
$\displaystyle\frac{\mu}{2\omega}\left(\begin{array}[]{cccc}\frac{3x^{2}z\mathcal{A}^{2}}{8kx^{9}_{0}}+\frac{z^{5}\mathcal{A}^{2}}{16k^{5}x^{15}_{0}}-\frac{5xz^{3}\mathcal{A}^{2}}{16k^{3}x^{12}_{0}}-\frac{z\mathcal{A}\mathcal{A}^{{}^{\prime}}}{4kx^{7}_{0}}+\frac{z^{3}\mathcal{A}^{{}^{\prime}2}}{4k^{3}x^{11}_{0}}-\frac{xz\mathcal{A}^{{}^{\prime}2}}{4kx^{8}_{0}}\\\
\\\ 0\\\ \\\
\frac{kx^{2}\mathcal{A}^{2}}{2x^{6}_{0}}-\frac{xz^{2}\mathcal{A}^{2}}{2kx^{9}_{0}}+\frac{z^{4}\mathcal{A}^{2}}{8k^{3}x^{12}_{0}}+\frac{xk^{3}\mathcal{A}^{2}}{2x^{3}_{0}}-\frac{kz^{2}\mathcal{A}^{2}}{4x^{6}_{0}}+\frac{z^{2}\mathcal{A}^{{}^{\prime}2}}{2kx^{8}_{0}}\end{array}\right)$
(83) $\displaystyle\textbf{s}_{t}$ $\displaystyle=$
$\displaystyle\frac{1}{2k^{2}}\nabla\times\textbf{p}=s_{t}\hat{y}\neq 0,\
\textbf{s}_{E}=\textbf{s}-\textbf{s}_{t}=s_{E}\hat{y}\neq 0$ (84)

Eq.(84) shows a very different result with Eq.(74), there is non-zero
extraordinary transverse spin in this case as showned in FIG. 4 (d).
Similarly, we also calculate the case of
$\bm{\Pi}_{m}=Ai(s-\frac{\xi^{2}}{4})e^{\frac{is\xi}{2}-\frac{i\xi^{3}}{12}}\hat{z}$,
and there is similar result.

In addition, another typical solution of the wave equation in the paraxial
approximation is the Gaussian beam [1, 14, 44, 45, 46]. The Hertz vector wave
equation Eq.(2) under paraxial approximation
$|\partial^{2}_{z}\Pi|\ll|2k\partial_{z}\Pi|$ and
$|\partial^{2}_{z}\Pi|\ll|\partial^{2}_{x}\Pi|,|\partial^{2}_{y}\Pi|$ can be
expressed as

$\displaystyle\nabla^{2}_{\bot}\bm{\Pi}+2ik\partial_{z}\bm{\Pi}=0$ (85)

A feasible solution in Eq.(85) is the Gaussian beam, as

$\displaystyle\bm{\Pi}_{m}=A_{0}\frac{z_{R}}{z-iz_{R}}e^{\frac{ik\rho^{2}}{2(z-iz_{R})}}\hat{n}$
(86)

where $A_{0}$ is a constant field amplitude, $z_{R}=\frac{kw^{2}_{0}}{2}$ is
the Rayleigh diffraction length, and $w_{0}$ is the beam waist [49].

When we choose $\hat{n}=\hat{x}$, so we can get the electromagnetic field by
taking Eq.(85) into Eqs.(4-5), as

E $\displaystyle=$ $\displaystyle\omega\mu
A_{0}[0\hat{x}+(\frac{k\rho^{2}z_{R}}{2(z-iz_{R})^{3}}-\frac{iz_{R}}{(z-iz_{R})^{2}})\hat{y}$
(87)
$\displaystyle+\frac{kyz_{R}}{(z-iz_{R})^{2}}\hat{z}]e^{\frac{ik\rho^{2}}{2(z-iz_{R})}}$
H $\displaystyle=$ $\displaystyle
A_{0}[(-\frac{k^{2}x^{2}z_{R}}{(z-iz_{R})^{3}}+\frac{ikz_{R}}{(z-iz_{R})^{2}}+\frac{k^{2}z_{R}}{z-iz_{R}})\hat{x}$
(88)
$\displaystyle-\frac{k^{2}xyz_{R}}{(z-iz_{R})^{3}}\hat{y}+(\frac{k^{2}\rho^{2}xz_{R}}{2(z-iz_{R})^{4}}-\frac{2ikxz_{R}}{(z-iz_{R})^{3}})\hat{z}]e^{\frac{ik\rho^{2}}{2(z-iz_{R})}}$

we can get the phase distributions of the electric field near the focal plane
(we choose z=0.001), as shown in FIG. 6 , and then we can further get the
momentum and spin density distribution of the Gaussian beam by Eqs.(8-9), as

(a) (b) (c)

Figure 6: The two components of electric field $E_{x}$, $E_{z}$ phase
distribution (a)-(b) and their phase difference (c), here we choose $z_{R}=1$
and $z=0.001$.

s $\displaystyle=$ $\displaystyle s_{x}\hat{x}+s_{y}\hat{y}+s_{z}\hat{z}$ (89)
$\displaystyle s_{x}$ $\displaystyle=$ $\displaystyle\frac{\mu
A^{2}_{0}}{2\omega}[\frac{k^{3}yz^{2}_{R}}{(z^{2}+z^{2}_{R})^{2}}-\frac{k^{4}\rho^{2}yz^{3}_{R}}{2(z^{2}+z^{2}_{R})^{3}}$
(90)
$\displaystyle+\frac{2k^{3}x^{2}yz^{2}_{R}}{(z^{2}+z^{2}_{R})^{3}}-\frac{k^{4}\rho^{2}x^{2}yz^{3}_{R}}{2(z^{2}+z^{2}_{R})^{4}}]e^{-\frac{k\rho^{2}z_{R}}{z^{2}+z^{2}_{R}}}$
$\displaystyle s_{y}$ $\displaystyle=$ $\displaystyle\frac{\mu
A^{2}_{0}}{2\omega}[\frac{k^{4}x^{3}\rho^{2}z^{3}_{R}}{2(z^{2}+z^{2}_{R})^{4}}+\frac{k^{3}x\rho^{2}z^{2}z^{2}_{R}}{2(z^{2}+z^{2}_{R})^{4}}-\frac{k^{3}x\rho^{2}z^{4}_{R}}{2(z^{2}+z^{2}_{R})^{4}}$
(91)
$\displaystyle-\frac{3k^{4}x\rho^{2}z^{2}z^{3}_{R}}{2(z^{2}+z^{2}_{R})^{4}}+\frac{k^{4}x\rho^{2}z^{5}_{R}}{2(z^{2}+z^{2}_{R})^{4}}-\frac{2k^{3}x^{3}z^{2}_{R}}{(z^{2}+z^{2}_{R})^{3}}$
$\displaystyle+\frac{2k^{2}xz^{3}_{R}}{(z^{2}+z^{2}_{R})^{3}}+\frac{2k^{3}xz^{2}z^{2}_{R}}{(z^{2}+z^{2}_{R})^{3}}-\frac{2k^{3}xz^{4}_{R}}{(z^{2}+z^{2}_{R})^{3}}]e^{-\frac{k\rho^{2}}{z^{2}+z^{2}_{R}}}$
$\displaystyle s_{z}$ $\displaystyle=$ $\displaystyle\frac{\mu
A^{2}_{0}}{2\omega}[\frac{k^{3}xyzz^{2}_{R}}{(z^{2}+z^{2}_{R})^{3}}-\frac{2k^{4}xyzz^{3}_{R}}{(z^{2}+z^{2}_{R})^{3}}]e^{-\frac{k\rho^{2}}{z^{2}+z^{2}_{R}}}$
(92) p $\displaystyle=$ $\displaystyle p_{x}\hat{x}+p_{y}\hat{y}+p_{z}\hat{z}$
(93) $\displaystyle p_{x}$ $\displaystyle=$ $\displaystyle\frac{\mu
k^{2}A^{2}_{0}}{2\omega}[-\frac{k^{2}x\rho^{2}zz^{3}_{R}}{(z^{2}+z^{2}_{R})^{4}}+\frac{2kxzz^{2}_{R}}{(z^{2}+z^{2}_{R})^{3}}$
(94)
$\displaystyle+\frac{k^{3}x\rho^{4}zz^{2}_{R}}{4(z^{2}+z^{2}_{R})^{4}}+\frac{k^{3}xy^{2}zz^{2}_{R}}{(z^{2}+z^{2}_{R})^{3}}]e^{-\frac{k\rho^{2}}{z^{2}+z^{2}_{R}}}$
$\displaystyle p_{y}$ $\displaystyle=$ $\displaystyle\frac{\mu
k^{2}A^{2}_{0}}{2\omega}[-\frac{k^{3}x^{2}yzz^{2}_{R}}{(z^{2}+z^{2}_{R})^{3}}+\frac{k^{3}yzz^{2}_{R}}{(z^{2}+z^{2}_{R})^{2}}]e^{-\frac{k\rho^{2}}{z^{2}+z^{2}_{R}}}$
(95) $\displaystyle p_{z}$ $\displaystyle=$ $\displaystyle\frac{\mu
k^{2}A^{2}_{0}}{2\omega}[-\frac{k^{2}x^{2}z^{3}_{R}}{(z^{2}+z^{2}_{R})^{3}}+\frac{kz^{2}_{R}}{(z^{2}+z^{2}_{R})^{2}}-\frac{k^{2}z^{3}_{R}}{(z^{2}+z^{2}_{R})^{2}}$
(96)
$\displaystyle+\frac{k^{3}x^{2}\rho^{2}z^{2}_{R}}{2(z^{2}+z^{2}_{R})^{3}}-\frac{k^{2}\rho^{2}z^{3}_{R}}{2(z^{2}+z^{2}_{R})^{3}}-\frac{k^{3}\rho^{2}z^{2}z^{2}_{R}}{2(z^{2}+z^{2}_{R})^{3}}$
$\displaystyle+\frac{k^{3}\rho^{2}z^{4}_{R}}{2(z^{2}+z^{2}_{R})^{3}}]e^{-\frac{k\rho^{2}}{z^{2}+z^{2}_{R}}}$

If we consider the simplest case of the focal plane $z=0$, according the
definition in [21], we can obtain the tansverse spin $\textbf{s}_{t}$ and non-
zero extraordinary spin $\textbf{s}_{E}$ of the Gaussian beam, as shown in
FIG. 7.

$\displaystyle\textbf{s}_{t}$ $\displaystyle=$
$\displaystyle\frac{1}{2k^{2}}\nabla\times\textbf{p}$ (97) $\displaystyle=$
$\displaystyle s_{tx}\hat{x}+s_{ty}\hat{y}$ $\displaystyle s_{tx}$
$\displaystyle=$ $\displaystyle\frac{\mu
A^{2}_{0}}{2\omega}[\frac{2k^{3}x^{2}y}{z^{4}_{R}}-\frac{k^{4}x^{2}y}{2z^{3}_{R}}-\frac{k^{4}x^{2}y^{3}}{2z^{5}_{R}}+\frac{k^{3}y^{3}}{2z^{4}_{R}}$
(98)
$\displaystyle-\frac{k^{4}y^{3}}{2z^{3}_{R}}-\frac{3k^{2}y}{2z^{3}_{R}}+\frac{3k^{3}y}{2z^{2}_{R}}-\frac{k^{4}x^{4}y}{2z^{5}_{R}}]e^{-\frac{k\rho^{2}}{z_{R}}}$
$\displaystyle s_{ty}$ $\displaystyle=$ $\displaystyle\frac{\mu
A^{2}_{0}}{2\omega}[\frac{k^{4}x^{3}y^{2}}{2z^{5}_{R}}-\frac{5k^{3}x^{3}}{2z^{4}_{R}}+\frac{k^{4}x^{3}}{2z^{3}_{R}}-\frac{k^{3}xy^{2}}{z^{4}_{R}}$
(99)
$\displaystyle+\frac{k^{4}xy^{2}}{2z^{3}_{R}}+\frac{5k^{2}x}{2z^{3}_{R}}-\frac{3k^{3}x}{2z^{2}_{R}}+\frac{k^{4}x^{5}}{2z^{5}_{R}}]e^{-\frac{k\rho^{2}}{z_{R}}}$
$\displaystyle\textbf{s}_{E}$ $\displaystyle=$
$\displaystyle\textbf{s}-\textbf{s}_{t}$ (100) $\displaystyle=$
$\displaystyle\frac{\mu
A^{2}_{0}}{2\omega}[(-\frac{k^{3}y}{2z^{2}_{R}}-\frac{k^{3}y^{3}}{2z^{4}_{R}}+\frac{3k^{2}y}{2z^{3}_{R}})\hat{x}$
$\displaystyle+(\frac{k^{3}xy^{2}}{2z^{4}_{R}}-\frac{k^{3}x}{2z^{2}_{R}}-\frac{k^{2}x}{2z^{3}_{R}})\hat{y}]e^{-\frac{k\rho^{2}}{z_{R}}}$

Similarly, we can also choose $\hat{n}=\hat{y}$, after a series of tedious
calculations, we can obtain the momentum and spin angular momentum density
distribution at this time, and there is still a non-zero $\textbf{s}_{E}$ as

$\displaystyle\textbf{s}_{E}$ $\displaystyle=$
$\displaystyle\textbf{s}-\textbf{s}_{t}$ (101) $\displaystyle=$
$\displaystyle\frac{\mu
A^{2}_{0}}{2\omega}[(\frac{k^{3}y}{2z^{2}_{R}}+\frac{k^{2}y}{2z^{3}_{R}}-\frac{k^{3}x^{2}y}{2z^{3}_{4}})\hat{x}$
$\displaystyle+(\frac{k^{3}x}{2z^{2}_{R}}-\frac{3k^{2}x}{2z^{3}_{R}}+\frac{k^{3}x^{3}}{2z^{4}_{R}})\hat{y}]e^{-\frac{k\rho^{2}}{z_{R}}}$

(a) (c) (e)
(b) (d) (f)

Figure 7: The total spin (a)-(b), transverse spin (c)-(d) and extraordinary
spin (e)-(f) angular momentum density distributions of the Gaussian beam of
Eq.(86) with $\bm{\Pi}_{m}=\Pi_{m}\hat{x}$, here we choose $\lambda=632.8nm$
and $w_{0}=1\mu m$.

##  III Conclusion

We have carried out an expansion on the basis of [21]. We know that the
transverse spin is an intrinsic property of all structural wave fields, and it
is necessary to explore its physical origin. Compared with the traditional
method of judging whether the transverse spin is perpendicular to the
direction of the wave propagation and the direction of the spin vector, there
is no physical meaning. However, it seems that it is reasonable to judge the
transverse spin by the spatial inhomogeneity of the momentum density of the
wave field, because for the structural wave field, its momentum density is
always non-uniform distributed, so there will be a transverse spin produce.
However, it is not enough to define transverse spin by momentum density alone.
In addition to the transverse spin caused by inhomogeneity momentum, there may
be another part of ”extraordinary” spin, it is also transverse. In particular,
for single evanescent waves and non-single evanescent wave whose attenuation
direction is the same as the main polarization direction, the ”extraordinary”
spin vanishes. And for non-planar propagating waves, the transverse spin
usually provides by inhomogeneities momentum and non-zero ”extraordinary”
spin.

## Appendix

In a general scenario from any waves either evanescent or propagating, we
derive the following curl of momentum from Maxwell equations in Gaussian
units,

$\displaystyle\nabla\times\mathbf{p}$ $\displaystyle=$
$\displaystyle\nabla\times\mathbf{p}_{\rm
O}+\frac{1}{2}\nabla\times\nabla\times\mathbf{s}=-\frac{1}{2}\nabla^{2}\mathbf{s}+\nabla\times\mathbf{p}_{\rm
O}$ (102) $\displaystyle=$ $\displaystyle
k^{2}\mathbf{s}-\frac{g}{\omega}\nabla\times\mathbf{A_{1}}+\nabla\times\mathbf{p}_{\rm
O}.$ (103)

Here

$\nabla\times\mathbf{A}_{1}=\Im\left(\begin{array}[]{ccc}\nabla
E_{y}^{*}\cdot\nabla E_{z}+\nabla H_{y}^{*}\cdot\nabla H_{z}\\\ \nabla
E_{z}^{*}\cdot\nabla E_{x}+\nabla H_{z}^{*}\cdot\nabla H_{x}\\\ \nabla
E_{x}^{*}\cdot\nabla E_{y}+\nabla H_{x}^{*}\cdot\nabla
H_{y}\end{array}\right).$ (104)

According to Eq. 102, we obtain two wave equations,

$\displaystyle\nabla^{2}\mathbf{s}+2k^{2}\mathbf{s}=\frac{2g}{\omega}\nabla\times\mathbf{A}_{1},$
(105) $\displaystyle\nabla^{2}\mathbf{p}_{\rm S}+2k^{2}\mathbf{p}_{\rm
S}=-\frac{g}{\omega}\nabla^{2}\mathbf{A}_{1}.$ (106)

We may choose $A_{1}$ as

$\mathbf{A}_{1}=\frac{\omega}{g}\left(\frac{1}{2}\nabla^{2}+k^{2}\right)\mathbf{A_{\mathbf{s}}},$
(107)

where $\mathbf{A_{\mathbf{s}}}$ serves as the potential of $\mathbf{s}$.

$\displaystyle\mathbf{s}=\nabla\times\mathbf{A_{\mathbf{s}}},$ (108)
$\displaystyle\mathbf{A_{\mathbf{s}}}=\frac{1}{2\pi}\int_{V}{\rm
d}\tau^{\prime}\frac{\mathbf{p}_{S}}{|\mathbf{r}-\mathbf{r}^{\prime}|}.$ (109)

###### Acknowledgements.

Z. X. and Y. L. are supported by Young Scientist Fund [NSFC11804087], Natural
National Science Foundation[NSFC12047501]; Science and Technology Department
of Hubei Province [2022CFB553, 2018CFB148]; and Educational Commission of
Hubei Province of China [Q20211008]. Y. L. and B. Z. are supported by National
Natural Science Foundation of China [NSFC12074107], Science and Technology
Department of Hubei Province [2022CFA012], and Educational Commission of Hubei
Province [T2020001].

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