Journal of Modern Physics, 2019, 10, 459-465
http://www.scirp.org/journal/jmp
ISSN Online: 2153-120X
ISSN Print: 2153-1196
An Evanescent Light Wave Cannot Possess a
Transverse Spin
Chunfang Li, Yunlong Zhang
Department of Physics, Shanghai University, Shanghai, China
How to cite this paper: Li, C.F. and Zhang,
Y.L. (2019) An Evanescent Light Wave Can-
not Possess a Transverse Spin.
Journal of
Modern Physics, 10, 459-465.
https://dx.doi.org/10.4236/jmp.2019.104031
Received: February 28, 2019
Accepted: March 19, 2019
Published: March 22, 2019
Copyright c 2019 by author(s) and Scientific
Research Publishing Inc.
This work is licensed under the Creative Com-
mons Attribution International License (CC
BY 4.0).
http://creativecommons.org/licenses/by/4.0/
Abstract
It is pointed out that the evanescent light wave occurring at total
reflection does not possess a transverse spin angular momentum as
Bliokh, Bekshaev, and Nori claimed recently in (2014) Nature Com-
munications, 5, 3300. This is not only because of the nonlocality
of the photon spin but also because the evanescent wave is such a
state whose angular momentum cannot be separated into spin and
orbital parts.
Keywords
Nonexistence, Transverse Spin, Evanescent Light Wave
1. Introduction
It was once believed [1] [2] that separating the photon angular momen-
tum into its spin and orbital parts is physically meaningless. Howev-
er, since the seminal work of Allen et al. [3], theoretical identification
of spin and orbital parts of photon angular momentum have drawn
much attention [4–14]. In a recent publication, Bliokh, Bekshaev, and
Nori [15] claimed that the evanescent light wave occurring at total
reflection has a transverse spin, which is independent of the polariza-
tion. They arrived at that conclusion by resorting to the so-called
local densities for the spin and orbital angular momentum (OAM),
which were constructed in Ref. [11].
Bialynicki-Birula [12] showed that even though the total angular
momentum of a light wave is local, after the splitting into spin and
orbital parts, “the locality is lost”. Furthermore, one of the authors
[16] found that the spin of the photon can be derived from a set of
two relativistic quantum equations for those states with respect to
which the momentum operator is Hermitian. The nonlocality of the
photon spin originates in the nonlocality of the photon itself that is
expressed by the relativistic quantum constraint. On the basis of
these discussions, we will demonstrate in this paper that the notion of
transverse spin advanced in Ref. [15] is physically incorrect.
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DOI: 10.4236/jmp.2019.104031 Mar. 22, 2019
459
Journal of Modern Physics
C. F. Li, Y. L. Zhang
2. Description of Demonstration Method
For clarity, let us first take a brief look at how Bliokh et al. came
to their conclusion. Consider a monochromatic light wave of angular
frequency ω in the air. The local spin density they defined reads
s = ψ† ˆSψ =
ˆΣ 0
Im(E∗ × E + H∗ × H),
γ
2
(1)
is a vector matrix, ( ˆΣk)ij = −iijk with ijk
where ˆS =
the Levi-Civit´a pseudotensor, ψ = γ
ˆΣ
0
E
2
H
is the spinor wave-
function with E and H the complex electric and magnetic vectors,
respectively, the superscript “†” denotes the conjugate transpose, the
superscript “∗” denotes the complex conjugate, Gaussian units with
γ = (8πω)−1 are used, and = 1 is assumed. They claimed that the
spin density is generated by the so-called spin part pS of the momen-
tum density in the following way,
s = x × pS,
(2)
where x is the position vector relative to the origin. Meanwhile, they
claimed that pS is expressed in terms of s as
pS =
∇ × s.
1
2
(3)
The local OAM density about the origin they defined is as follows,
l = x × pO,
(4)
where
γ
2
pO = Re(ψ† ˆP ψ) =
Im[(∇E) · E∗ + (∇H) · H∗]
(5)
is the so-called canonical part of the momentum density and ˆP = −i∇
is the operator for the canonical momentum. They required that the
sum of pS and pO should be equal to the momentum density that is
proportional to the Poynting vector and is expressed in terms of the
complex electric and magnetic vectors as
p = γk0Re(E∗ × H),
(6)
where k0 = ω/c is the wavenumber in the air. When applying Equa-
tion (1) to the evanescent wave that occurs at a total reflection, they
found that s is perpendicular to the propagation direction.
wave about the origin is given by x × pd3x, where the integrand is
It is well known [17] [18] [19] that the angular momentum of a light
known as the angular-momentum density about the origin,
j = x × p,
(7)
and p is the momentum density (6). As a physically meaningful no-
tion, the angular-momentum density has to be unique. So if s in
Equation (1) and l in Equation (4) are true local spin and OAM den-
sities, respectively, one must have s + l = j. This can also be seen
from pS + pO = p if s is generated by pS via Equation (2). To demon-
strate the nonexistence of the transverse spin in an evanescent wave,
it is enough to show that there is no such relation in that case. This
is done below.
DOI: 10.4236/jmp.2019.104031
460
Journal of Modern Physics
C. F. Li, Y. L. Zhang
3. Demonstration of Nonexistence of the
Transverse Spin
Because it was claimed in Ref. [15] that the transverse spin in an e-
vanescent wave is independent of the polarization, we consider such
an evanescent wave that occurs when a TM-polarized plane wave is
totally reflected by an interface between a dielectric medium of refrac-
tive index ni > 1 and the air of refractive index n0 = 1 as is shown in
Figure 1.
The complex magnetic vector of the evanescent wave in the air as-
sumes the following form,
H = A exp[i(k0 · x − ωt)]¯y,
x ≥ 0,
(8)
z − k2
where A is a constant, k0 = iκ¯x + kz ¯z is the complex wavevector,
κ = (k2
0)1/2 is the decay coefficient in the air, kz = ki sin θ,
ki = nik0, θ is the incidence angle that is larger than the critical angle
for total reflection θc = sin−1(1/ni), ¯x, ¯y, and ¯z are unit vectors along
the corresponding axes. The corresponding complex electric vector is
given by
E =
kz ¯x − iκ¯z
k0
A exp[i(k0 · x − ωt)],
x ≥ 0.
(9)
It is needless to say that the electric vector (9) is “perpendicular”
to the associated complex wavevector, E · k0 = 0, as the Maxwell
equation ∇ · E = 0 requires. Substituting Equations (8) and (9) into
Equation (6), we find for the momentum density,
p = γkz|A|2 exp(−2κx)¯z,
(10)
which is in the propagation direction, the z-axis. Upon substituting it
into Equation (7), we have for the angular-momentum density about
the origin,
j = γkz|A|2 exp(−2κx)(y¯x − x¯y).
It is transverse, having both x and y components.
(11)
Substituting Equations (8) and (9) into Equation (1), one gets
Figure 1. The evanescent wave that occurs at the total reflection of a
TM-polarized plane wave by an interface at x = 0 between two dielectric
media of refractive indices ni and n0.
DOI: 10.4236/jmp.2019.104031
461
Journal of Modern Physics
C. F. Li, Y. L. Zhang
s =
γkzκ
k2
0
|A|2 exp(−2κx)¯y,
(12)
which is in the transverse y direction. It is on the basis of this result
that Bliokh et al. [15] claimed that the evanescent wave possesses a
transverse spin. However, substituting Equations (8) and (9) into
Equation (5), one finds
pO =
γk3
z
k2
0
|A|2 exp(−2κx)¯z,
which, when substituted into Equation (4), gives
l =
γk3
z
k2
0
|A|2 exp(−2κx)(y¯x − x¯y).
(13)
(14)
Obviously, the sum of s and l in Equations (12) and (14) is different
from the angular-momentum density (11). Instead, one has
s + l =
γkz
k2
0
|A|2 exp(−2κx)[k2
zy¯x + (κ − k2
zx)¯y].
From this result it is concluded that the s in Equation (12) cannot
be the local spin density and therefore the evanescent light wave does
not possess a transverse spin.
Furthermore, substituting Equation (12) into Equation (3), one has
pS = − γkzκ2
k2
0
|A|2 exp(−2κx)¯z.
(15)
The sum of (13) and (15) is indeed equal to the momentum density
(10). But upon substituting Equation (15) into Equation (2), one
cannot find Equation (12). Instead, one obtains
x × pS = − γkzκ2
k2
0
|A|2 exp(−2κx)(y¯x − x¯y),
(16)
which, in addition to a y-component, has a x-component. This is un-
derstandable. Resulting from Equation (1), the s in Equation (12) is
supposed to be independent of the choice of a reference point. Nev-
ertheless, coming from Equation (2), expression (16) must depend on
the reference point, the origin. This further demonstrates that the s
in Equation (12) is not the local spin density. After all, as pointed
out by Bialynicki-Birula [12], “when one splits the total angular mo-
mentum into its orbital part and the intrinsic part, locality cannot be
preserved”.
It is noted that according to the definition in quantum mechanics
[20], the expectation value of the canonical momentum taken with
respect to state ψ is given by
ψ† ˆP ψd3x
ψ†ψd3x
.
ˆP =
If the evanescent wave (8) and (9) can be regarded as a photon
state, the resultant expectation value of the canonical momentum is
complex, ˆP = k0 = iκ¯x + kz ¯z. This shows that the evanescent
wave is such a state with respect to which the canonical-momentum
operator is not Hermitian. In other words, the canonical momentum of
DOI: 10.4236/jmp.2019.104031
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C. F. Li, Y. L. Zhang
the photon in the evanescent wave is not an observable from the point
of view of quantum mechanics. Accordingly, the corresponding OAM
represented by the operator x × ˆP is not an observable, either. As
a consequence, dividing the angular momentum of the photon in the
evanescent wave into spin and orbital parts is physically impossible.
This also explains why the angular momentum density j given by
Equation (11) is different from the sum of s and l in Equations (12)
and (14).
4. Discussions and Conclusions
On a final note, the photon in the evanescent wave (8) and (9) does
have a transverse angular momentum about the origin. To see this
in more detail, let us calculate the energy density of the evanescent
wave,
u =
1
16π
(|E|2 + |H|2) =
k2
z
8πk2
0
|A|2 exp(−2κx).
Considering that the energy of each photon is ω, we adopt the tech-
nique used in Ref. [6] to find the momentum of a single photon per
unit length along the propagation direction,
P = ω
x≥0 pdxdy
x≥0 udxdy
=
k2
0
kz
¯z.
Because of k0
kz
< 1, its magnitude, P = k2
0
kz
, is less than that of
the momentum of free photon in the air, k0. Correspondingly, from
Equation (11) we have for the angular momentum of a single photon
per unit length along the propagation direction,
J = ω
x≥0 jdxdy
x≥0 udxdy
= − P
2κ
¯y,
which is in the transverse y direction. According to the definition of
the angular momentum of a point particle about the origin, J = x×P ,
it follows from preceding two equations that the x-component of the
photon’s position vector x at any instant is x = 1
2κ if we consider
the photon as a point particle. That is to say, the photon behaves
as a point particle of momentum P along the propagation direction
with a distance 1
2κ away from the interface. The effect of such a
transverse angular momentum was observed twenty years ago [21] in
an experiment on combined Mie particles and was then explained as
the result of the vertical gradient of the longitudinal radiation pressure
that is expressed by the momentum density (10).
In conclusion, the evanescent light wave does not possess a trans-
verse spin not only because of the nonlocality of the photon spin but
also because the angular momentum of the evanescent wave cannot be
separated into spin and orbital parts.
Acknowledgements
This work was supported in part by the program of Shanghai Munic-
ipal Science and Technology Commission (18ZR1415500).
Conflicts of Interest
The authors declare no competing financial interests.
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C. F. Li, Y. L. Zhang
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