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Propagation of vectorial laser beams

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Abstract

The angular spectrum of a vectorial laser beam is expressed in terms of an intrinsic coordinate system instead of the usual Cartesian laboratory coordinates. This switch leads to simple, elegant, and new expressions, such as for the angular spectrum of the Hertz vectors corresponding to the electromagnetic fields. As an application of this approach, we consider axially symmetric vector beams, showing nondiffracting properties of these beams, without invoking the paraxial approximation.

©2012 Optical Society of America

1. INTRODUCTION

In many branches of physics, it is necessary to solve the vectorial wave equation in three dimensions. One of the best known and most general methods for the vectorial Helmholtz equation is the expansion in plane waves by representing the solution as a three-dimensional (3D) Fourier integral over the 3D reciprocal wavevector space. Especially in laser physics, the relevant electric and magnetic fields are in fact solutions having a directed-energy beam-character. That is, the fields propagate along a certain axis, called the optical axis and which we choose to be the z axis of the Cartesian coordinate system linked with the beam. A number of mathematical methods are known to treat this case efficiently [1,2]. In this paper, we consider three methods of solution, the angular spectrum representation of the beam, the TE/TM decomposition of the beam and finally the Hertz vectors of the beam. The last method is the least well known, despite the fact that it shows a major advantage by generating particularly simple expressions for the fields. We shall indicate the interrelations between the three methods. In a first step we show how to transform the Cartesian axes to an intrinsic coordinate system linked with the propagation constants of the beam.

Physically, the new base vectors are linked with the classically known p and s components of polarization optics, but their expression in a Cartesian or cylindrical coordinate system is not evident.

In a second stage, we determine the angular spectrum not of the electric and magnetic fields E and B, but of the Hertz potential vectors Pe and Pm corresponding to these vectors. Then we use the results of the first phase and switch to the new coordinates. New and simple formulas are obtained. This leads to an alternative angular spectrum representation of the solution of the vectorial Helmholtz equation.

As an application of the usefulness of this representation, we will briefly discuss diffraction-free vectorial beams. We will do this without the restriction to reduce the Maxwell equations to their paraxial form. The diffraction-free solution is hence a rigorous solution and not an approximate one, satisfying in particular the zero divergence Maxwell equation for the electric field, which a paraxial solution does not satisfy.

2. INTRINSIC COORDINATE SYSTEM

Let k be the wave vector of a plane wave

k=kxex+kyey+kzez=kter+kzez=ktcosφex+ktsinφey+kzez
and its magnitude is given by
k2=kx2+ky2+kz2=kt2+kz2.
In Eq. (1) we have denoted the split of the wave vector into its transverse (subscript t) and longitudinal part (subscript z). The k vector forms together with the z axis the meridional plane [3]. In optical terminology, we would call it the plane of incidence (on a mirror for example, where z is the normal to the mirror). The angle between the z axis and the k axis is the polar angle θ. The cylindrical coordinates are defined through their unit vectors er, ez, and eφ, φ being the azimuthal angle of the k vector in the transverse (x,y) plane.

The 3D locus vector R is defined by

R=xex+yey+zez=rer+zez=rcosαex+rsinαey+zez,
where we indicated the split of the locus vector in its transverse and its longitudinal part.

The key element in our approach is the point brought up in [2] and [4], which is that we should consider an intrinsic coordinate system. It contains three unit vectors: two of them are respectively perpendicular and parallel to the meridional plane, the third being the unit vector in the k direction:

s=eφp=kzkerktkezm=k/k.
These unit vectors, taken in the order s, p, m, form a right-hand sided coordinate system.

Note that s does not possess a z component. s is perpendicular to the meridional plane, whereas p is coplanar with it, just like m. s has only a transverse component, its azimuthal part, whereas p both has a transverse component, its radial part, and a longitudinal component. Although the definition of the s and p vector are classically known in the literature on optical polarization, their relation [Eq. (3)] with the cylindrical base vectors is not evident. An alternative but equivalent definition of the intrinsic coordinates is

s=ez×k|ez×k|=ez×kktp=s×kk=s×m.
The transformation of the Cartesian coordinates to the intrinsic coordinates is given by
s=kyktex+kxkteyp=kzkxkktex+kzkykkteyktkezm=kxkex+kykey+kzkez.
The inverse transformation from the intrinsic coordinate system to the Cartesian laboratory coordinate system is, after lengthy but straightforward calculations, given by
ex=kykts+kxkzkktp+kxkmey=kxkts+kykzkktp+kykmez=ktkp+kzkm,
which is a central result for the subsequent calculations. Note that s and p are in fact functions of kx and ky and should in fact be written as s(kx,ky) and p(kx,ky) when they will in a later paragraph appear behind the integration sign.

3. HERTZ POTENTIALS AND THEIR ANGULAR SPECTRUM REPRESENTATION

We will only consider free-space situations: the sources are infinitely far away from our observation region. The electric and magnetic fields can be defined in S.I. units in terms of their scalar and vector potential [5]:

E=gradϕAtB=rotA,
where the potentials are linked by the Lorentz gauge
divA+ε0μ0ϕt=0.
An equivalent and alternative set to these potentials are the Hertz electric and magnetic vector potentials Pe and Pm, which in free space are, just like E and B, also solutions of the homogeneous vectorial d’Alembert wave equation
2P1c22Pt2=0;=e,m.
The link between the regular potentials and the electric Hertz vector is given by [6]:
ϕ=divPeA=ε0μ0Pet
and for the magnetic Hertz vector by
ϕ=0A=rotPm.
Combining this with Eq. (6), hence eliminating the scalar and vector potential, results for the two cases resp. in the expression of the fields as a function of the Hertz vectors. For the electric Hertz vector:
E=graddivPeε0μ02Pet2=rotrotPeB=μ0H=ε0μ0t(rotPe).
And for the magnetic Hertz vector:
E=t(rotPm)B=μ0H=graddivPm1c22Pmt2=rotrotPm.
Because of the source-free space, the magnetic induction B reduces to the magnetic field strength H. By adding Eqs. (7) and (8), we find the general expression for the fields in terms of the Hertz vectors:
E=graddivPe1c22Pet2(rotPm)tB=μ0H=ε0μ0(rotPe)t+graddivPm1c22Pmt2.
Note that the dimensions of the electric Hertz vector are Vm and Vs for the magnetic Hertz vector.

These expressions do not look at first sight like simplifying life. The use of the Hertz potentials is advantageous, however, in case the current sources are all parallel to a common direction, say the z direction [57]. In this situation, and far from the sources, Eq. (7) for the electric Hertz potential reduces to the set

Ex=2PezxzEy=2PezyzEz=2Pezz2ε0μ02Pezt2Hx=ε02PezytHy=ε02PezxtHz=0,
which show that the electric Hertz vector Pe reduces to its scalar component Pez:
Pe=(0,0,Pez)
This fact enormously simplifies the mathematical manipulations and was exploited to analyze a number of vectorial diffraction effects in [3]. The attractive aspect of the “scalar” Hertz vector representation as compared to, e.g., the mathematically equivalent modal representation is that it arrives at the same rigorous results as the modal representation, but in a much shorter and more elegant way. To give a quick example of its usefulness, consider the electric field of an electric dipole oscillating along the z axis. In every direction starting from the origin, the electric dipole field has another orientation but the corresponding Hertz vector is nevertheless everywhere in space a scalar, oriented along the z axis, independent of the radiation direction [3].

For laser applications, we are exclusively interested in directed-energy solutions of the vectorial Maxwell equations, i.e., in “beams,” rather than in “fields.” Of course we then take the z direction as the propagation direction of the field energy. So, using the Hertz potentials, vectorial beams can be completely described by a scalar function. Specifically, whereas Pe(R,t) in general is a solution of the vectorial wave equation, its z component Pez(x,y,z,t) in Eq. (11) is now a scalar solution of the Helmholtz wave equation with the boundary condition Pez(x,y,0,t)=Pez(x,y)exp(jωt). The two-dimensional (2D) Fourier transform of the Hertz potential Pez(x,y) in the plane z=0, is defined as

pez(kx,ky)=k2Pez(x,y)exp[j(kxx+kyy)]dxdy
and is called its angular spectrum. The factor k2 before the integration signs is required to give the spectrum the same physical dimensions as the original. The inverse Fourier transformation is given by
Pez(x,y,z=0)=Pez(x,y)=14π2k2pez(kx,ky)exp[j(kxx+kyy)]dkxdky.
The factor of 1/4π2 appears for mathematical reasons, the factor k2 for physical reasons. Note that pez is not the angular spectrum of the electric field in the plane z=0, but rather of the electric Hertz vector. The general solution of the scalar wave equation for Pez(x,y,z,t) in the halfspace z>0 is now given by [8]
Pez(x,y,z,t)=exp(jωt)14π2k2pez(kx,ky)exp[j(kxx+kyy+kzz)]dkxdky,
which is called the angular spectrum representation of the electric Hertz potential Pez(x,y,z,t). Equation (14) shows that for beam-like solutions, we do not need a 3D Fourier transform as solution of the wave equation, the dimensionality of the problem is reduced by one unit.

A. Vectorial Angular Spectrum Representation

A scalar field E(R) satisfying the scalar Helmholtz equation, can be represented as in Eq. (14) by its angular spectrum representation, i.e., the beam is linearly polarized. What happens if the beam has a vectorial character? Just by extending the argument from one dimensional (1D) to 3D, we look for a representation like

E(R)=14π2k2e(kx,ky)exp[j(kxx+kyy+kzz)]dkxdky.
But once Ex(R) and Ey(R) are known, then Ez(R) is fixed, because the electric field has no divergence. We hence need only two scalar functions to determine the 3D electric field. This should be reflected in the form of the vectorial angular spectrum.

Assuming harmonic time variation, we now take the 2D Fourier transform of E and H at z=0 in Eq. (9) and call them the vectorial angular spectra e and h of the field vectors, or the spectral field vectors in short

e(kx,ky)=k2E(x,y,0)exp[j(kxx+kyy)]dxdy
h(kx,ky)=k2H(x,y,0)exp[j(kxx+kyy)]dxdy.
It should be noted that the vectorial character of the angular spectrum, i.e., its polarization state, is the same as that of the original field. Hence e and h inherit the polarization state of E and H (and also their physical dimensions) Since the divergence of E and H is zero in free-space, this means for their spectral vectors
e·k=h·k=0,
i.e., the spectral vectors are transverse to the wave vector, just like the original field vectors.

We now take the 3D Fourier transform of the field Eq. (9) and use the theorem that the Fourier transform of the derivative of a function is equal to jk times the Fourier transform of the original function. In working out the expressions for z=0, one should take into account relations such as

x[pe(x,y,0)]=pex|z=0,
which is easy to prove by writing out the derivatives according to their definition as a limit of the differential quotient for Δx0. Finally, we arrive at the angular field spectra as functions of the angular Hertz spectra
e(kx,ky)=(k.pe)k+k2pekck×pm
h(kx,ky)=ε0kck×pe(k.pm)k/μ0+k2/μ0pm,
where we also made use of the dispersion relation ω=kc to further simplify the expressions. So, in reciprocal k space, the relations between the field spectra and the Hertz spectra are no longer differential equations but become vector equations.

What we will do in this paragraph is work with the spectral field vectors e and h to finally obtain an angular spectrum representation of the fields E and H, through the intermediary of the Hertz vectors and their angular spectrum representation, according to the Fourier transforms indicated in the following diagram:

E,He,hPe,mpe,m.
Note: the spectral vectors pe and pm should not be confused with the intrinsic vector p.

B. Transverse Spectral Hertz Vectors

Next, we proceed to decompose each angular vectorial spectrum of the Hertz potentials pe and pm in a part that is parallel to k and a part that is perpendicular to k:

p(kx,ky)=Π+Π=e,mk.Π=0k×Π=0.
On substituting Eqs. (19) in Eq. (17,18), the parallel components of the spectral Hertz vectors all drop out and what rests is
e(kx,ky)=k2Πe(kx,ky)kck×Πm(kx,ky)h(kx,ky)=k2μ0Πm(kx,ky)+kcε0k×Πe(kx,ky).

C. TE/TM Decomposition

The considerations in Subsection 3.A were valid for fields in general. Because we consider beam propagation along the z axis, the TE/TM decomposition theorem for the electric and magnetic field can be invoked here [9], using the z axis as axis along which the decomposition is stated:

{E,H}={ETE,HTE}+{ETM,HTM}ETE·ez=0;HTM·ez=0.
The transverse character of the field vectors E and H now has to be conveyed to the spectral vectors e and h. We will proceed by taking a detour and first determine how e and h depend on the spectral Hertz vectors. The reason is, as we will see shortly, that the spectral Hertz vectors take on very simple expressions. And once the spectral Hertz vectors are known, it is again easy to deduce the spectral field vectors from them.

We now consider a source of oscillating dipoles, all oscillating in the same direction, which we choose to be the z direction. The Hertz vectors have the same orientation as the currents [6]; hence the Hertz vectors have only a z component. We assume that the dipole source is located at z= so that the halfspace z>0 is source free and the field differential equations become homogeneous and hence simpler. In this halfspace, the Hertz vectors keep their z orientation. From here on, we assume that the Hertz vectors have only a z-dependent component:

Pe(R)=Pez(R)ez;Pm(R)=Pmz(R)ez.
Hence, also their angular spectra pe and pm are only z dependent. Pez and Pmz are solutions of the homogeneous, scalar Helmholtz equation.

We now revert to the angular spectra of the Hertz vectors.

D. TM Case

For the moment, we only know that, by Fourier transforming the original electric Hertz vector, we obtain

pe=pezez.
Now, transforming Eq. (21), we see that hTM·ez=0. We use Eq. (20) to calculate this scalar product. This leads after some algebra to: pm=0. Next to the coordinate transformations Eq. (4,5) for the intrinsic vectors, this small equation is the second central result of this paper. It shows that the spectral magnetic Hertz vector is zero for a TM beam, leading to extremely simple equations, despite the vectorial character of the beam. This is shown by again using the intrinsic coordinate transformations, so that the angular spectra now become
eTM(kx,ky)=pez[kz(kxex+kyey)(kx2+ky2)ez]=kktpez(kx,ky)p(kx,ky)aTMp
hTM(kx,ky)=ε0kck×ezpez=ε0kc(kyexkxey)pez=ε0μ0kktpez(kx,ky)s(kx,ky)bTMs.
The angular field spectra are shown to be simple functions of the intrinsic vectors and the scalar angular electric Hertz spectrum. In Eq. (23), the magnetic vector only depends on s, which has no z component, according to Eq. (3); so therefore the fields given by Eq. (22,23) are indeed TM. This is in complete agreement with Eqs. (10) which show that the magnetic field has no z component. [Eq. (10) is not written in the Fourier domain, but for the polarizations this does not matter.]

E. TE Case

In a similar way as in the former case, this corresponds to:

pe=0pm=pmzez.
With this substitution, Eqs. (14) and (15) become, making use of the transformations (5):
eTE(kx,ky)=kck×ezpmz=kc(kyexkxey)pmz=ckktpmz(kx,ky)s(kx,ky)aTEs
hTE(kx,ky)=1μ0pmz[kz(kxex+kyey)(kx2+ky2)ez]=1μ0kktpmz(kx,ky)p(kx,ky)bTEp.
Hence we see again that the spectral vectors have become very simple functions of the intrinsic vectors and the scalar spectral magnetic Hertz potential.

Note that the magnitudes of the magnetic spectral vectors are just proportional copies of the magnitudes of the electric spectral vectors.

F. Spectra of the Transverse Hertz Vectors

The fact that the spectral vectors only have a z dependence further simplifies their general polarization expressions (18) and (19). Comparing Eq. (22) with Eq. (17) gives

eTM=k2Πe=kktpezp,
so we deduce that
Πe(kx,ky)=ktkpez(kx,ky)p.
In the same way, we find:
Πm(kx,ky)=ktkpmz(kx,ky)p.
So finally in the spectral domain, the relations between the fields and their Hertz vectors is
eTM=k2ΠehTM=kcε0k×ΠeeTE=kck×ΠmhTE=k2μ0Πm.

G. Angular Spectral Representation

The electric field of the beam propagating along the z axis has been decomposed in its TE and TM components:

E(R)=ETM(R)+ETE(R)
and its angular spectrum, given by Eq. (22) and (24) in
e(kx,ky)=eTM(kx,ky)+eTE(kx,ky).
This means that the angular spectral representations of the fields look like
ETM(R)=1(2π)2Πe(kx,ky)exp(jk.R)dkxdkyHTM(R)=cε0(2π)2kk×Πe(kx,ky)exp(jk.R)dkxdkyETE(R)=c(2π)2kk×Πm(kx,ky)exp(jk.R)dkxdkyHTE(R)=1(2π)2μ0Πm(kx,ky)exp(jk.R)dkxdky.
Or, alternatively,
ETM(R)=1(2π)2kktpez(kx,ky)pexp(jk.R)dkxdkyHTM(R)=1(2π)2ε0μ0ktpez(kx,ky)sexp(jk.R)dkxdkyETE(R)=c(2π)2ktpmz(kx,ky)sexp(jk.R)dkxdkyHTE(R)=1(2π)2kμ0ktpmz(kx,ky)pexp(jk.R)dkxdky.
For clarity and brevity, we have suppressed in the above integrals the dependency of the intrinsic vectors on kx and ky. Equation (30) show the electric and magnetic fields as functions of the angular spectra of the two Hertz vectors and of the intrinsic polarization vectors s and p. In the next paragraph, we will express the fields directly using their angular spectra. This will lead to a relation between the angular spectra of the field vectors and of the Hertz vectors.

4. ANGULAR SPECTRUM OF THE VECTOR FIELDS

The angular spectrum representation of the vector electric field has been given in [9], expressed in Cartesian coordinates as:

Ex(x,y,z)=14π2k2Ax(kx,ky)exp(jk.R)dkxdkyEy(x,y,z)=14π2k2Ay(kx,ky)exp(jk.R)dkxdkyEz(x,y,z)=14π2k2[kxkzAx(kx,ky)+kykzAy(kx,ky)]exp(jk.R)dkxdky.
Here again, we see that Az in fact is a function of Ax and Ay.

Following another strategy to solve the same vectorial wave equation, [1] started from the modal representation, as commonly used in guided wave problems, combining it with the angular spectrum representation. The advantage of this approach is that it clearly identifies and separates the contributions of the transverse electric fields (TE and TM) right from the start. In laboratory Cartesian coordinates, this finally leads to

E(R)=ETM(R)+ETE(R)=Ex(R)ex+Ey(R)ey+Ez(R)ezETM(R)=14π2k2[kzkkt(kxex+kyey)ktkez]ATM(kx,ky)exp(jk.R)dkxdkyETE(R)=14π2k2[kyktexkxktey]ATE(kx,ky)exp(jk.R)dkxdky.
The angular spectra ATM and ATE are in fact 2D Fourier transforms defined by the total fields Ex and Ey in z=0 [1]
ATM(kx,ky)=k2kktkz(kxEx+kyEy)exp[j(kxx+kyy)]dxdyATE(kx,ky)=k21kt(kyExkxEy)exp[j(kxx+kyy)]dxdy,
but due to the vector character of the fields, the mathematical expressions of the integrands are more complex than in the scalar case. Note again that we only need to know Ex and Ey. So [1] was able to separate the propagating field into two transverse contributions, which [2,4] did not indicate in this way. We will see, however, that in [4], essentially the same result was obtained. Expressions for the magnetic field are not given by [1]. It is nevertheless clear that ETE is lacking a z component, as it should for the electric field of a TE mode.

Again using the vector transformations (4), these expressions (32) now take on a very simple form:

ETM(R)=14π2k2pATM(kx,ky)exp(jk.R)dkxdkyETE(R)=14π2k2sATE(kx,ky)exp(jk.R)dkxdky
so that the vectorial angular spectrum of the total vectorial electric field E(R)=ETM(R)+ETE(R) is given by
A(kx,ky)=ATMp+ATEs,
which is exactly the form mentioned in [4], although in a slightly different format, denoted there as A=App+Ass without explicitly pointing to the transverse character of the contributions As and Ap.

The same reasoning can be repeated for the magnetic angular spectra:

B(kx,ky)=BTMs+BTEp.
The vectorial angular spectrum representation of the total fields now become compactly
E(R)=14π2k2A(kx,ky)exp[j(kxx+kyy+kzz)]dkxdkyH(R)=14π2k2B(kx,ky)exp[j(kxx+kyy+kzz)]dkxdky.
Now, going back to the spectral Hertz vectors, Eqs. (2225), and comparing these with Eq. (30,34,35), we see that
aTM=ATMbTM=BTMaTE=ATEbTE=BTE.
This means that the angular spectra of the field vectors and of the Hertz vectors are related as
ATM=kktpezBTM=ε0μ0kktpez=ε0μ0ATM
ATE=ckktpmz=μ0ε0BTEBTE=1μ0kktpmz.
Alternatively, in function of the spectra of the transverse Hertz potentials:
ATM=k2ΠeBTM=ε0μ0k2ΠeATE=μ0ε0k2ΠmBTE=k2Πm.

5. AXIALLY SYMMETRIC FIELDS

We now work out the general Eq. (34) for the special case of axial symmetry of the fields around the z axis.

E(R)=Er(r,z)er+Eα(r,z)eα+Ez(r,z)ez=ETM(r,z)+ETE(r,z).
Both [1] and [2] have considered this case. The angular spectrum is independent of the azimuthal angle φ in reciprocal k space and only depends on kt. Equation. (34) becomes [1]
ETM(r,z)=12πk30k[jkzJ1(ktr)erktJ0(ktr)ez]ktATM(kt)exp[jkzz]dktETE(r,z)=jeα2πk20kATE(kt)J1(ktr)ktexp[jkzz]dkt,
where the 2D Fourier transforms for the angular spectra now reduce to 1D Hankel transforms
ATM(kt)=2πk3jk2kt20Er(r,z=0)J1(ktr)rdrATE(kt)=2πjk20Eα(r,z=0)J1(ktr)rdr.
J0(z) and J1(z) are the well known Bessel functions. If we rewrite Eq. (39) by just keeping the vectorial part and the differentials in place, and absorbing the other contributions in the symbol (.), we can more clearly see how the unit vectors are transformed by changing from Cartesian to cylindrical coordinates. Note that eφ as integration variable in k space is transformed into eα in R space, which is independent of kx and ky and hence can be brought in front of the integration sign for the TE-field:
ETE=s(.)dkxdky=eφ(.)dkxdky=eα(.)dkt.
The electric TE-field is purely azimuthally polarized. For the TM-field, this simplification does not occur:
ETM=p(.)dkxdky=(kzkerktkez)(.)dkxdky=(J1kzkerJ0ktkez)(.)dkt.
The TM-vector keeps its transverse and longitudinal contributions in place, although now with other weighting coefficients, which are Bessel functions. Also here, the unit vectors can be brought in front of the integration symbol, since they are independent of kt.

6. NONDIFFRACTING VECTORIAL BEAMS

The best known nondiffracting scalar beam in the laser literature is the Bessel beam J0(ktr). This beam is linearly polarized. As a short application of Eqs. (39,40), we will consider the diffraction-free propagation of an azimuthally polarized vectorial beam and extend in this way the scope of [10], which analyzed scalar beams.

The general solution of the scalar wave equation for a diffraction-free beam is given [10] as

u(x,y,z)=exp(jkzz)02πa(φ)exp[jkt(xcosφ+ysinφ)]dφ
and is known under the name “Whittaker integral” in the mathematical literature. Physically, it represents a superposition of plane waves with amplitude coefficient a(φ) and with their wave vector situated on a cone with base radius kt and height kz. In other words, their angular spectrum in reciprocal k space is given by A=δ(ktkt). In the vectorial case, we use this same angular spectrum to describe nondiffracting vector beams and substitute it in the TE-field expression of Eq. (14) to arrive immediately at
ETE(r,z)=j2πJ1(ktr)exp(jkzz)eα.
The vectorial Bessel beam (41) is hence nondiffracting since the transverse part is not depending on the propagation coordinate z. This result was also obtained in [11], but by the method of separation of the variables of the paraxial wave equation. A mathematically equivalent statement is that the J1 Bessel function can be represented as a Whittaker integral. Its explicit form can be found in [3].

7. CONCLUSIONS

In summary, we have pointed out the relevance of using an intrinsic coordinate system to unify the existing angular spectrum representations. We have given the transformation formulas from intrinsic to Cartesian coordinates. The introduction of intrinsic coordinates much simplifies the angular spectral representation of vectorial beams. We have introduced the Hertz vector potential oriented along the z axis and have combined it with the intrinsic coordinates to represent TE and TM beams in a particularly simple and elegant form. This leads also to the mathematical link between the angular spectra of the fields and of their Hertz vector. Next, we derived expressions for a diffraction-free vectorial beam, without invoking the paraxial approximation.

REFERENCES

1. H. Guo, J. Chen, and S. Zhuang, “Vector plane wave spectrum of an arbitrary polarized electromagnetic wave,” Opt. Express 14, 2095–2100 (2006). [CrossRef]  

2. C.-F. Li, “Integral transformation solution of free-space cylindrical vector beams and prediction of modified Bessel-Gauss vector beams,” Opt. Lett. 32, 3543–3545 (2007). [CrossRef]  

3. A. Nesterov and V. Niziev, “Vector solution of the diffraction task using the Hertz vector,” Phys. Rev. E 71, 046608 (2005). [CrossRef]  

4. C.-F. Li, “Unified theory for Goos-Hänchen and Imbert-Fedorov effects,” Phys. Rev. A 76, 013811 (2007). [CrossRef]  

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6. J. Van Bladel, Electromagnetic Fields (McGraw-Hill, 1964).

7. P. Varga and P. Török, “Exact and approximate solutions of Maxwell’s equations for a confocal cavity,” Opt. Lett. 21, 1523–1525 (1996). [CrossRef]  

8. C. Someda, Electromagnetic Waves, 2nd ed. (CRC, 2006).

9. D. Rhodes, “On the stored energy of planar apertures,” IEEE Trans. Antennas Propag. 14, 676–683 (1966). [CrossRef]  

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Equations (64)

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k=kxex+kyey+kzez=kter+kzez=ktcosφex+ktsinφey+kzez
k2=kx2+ky2+kz2=kt2+kz2.
R=xex+yey+zez=rer+zez=rcosαex+rsinαey+zez,
s=eφp=kzkerktkezm=k/k.
s=ez×k|ez×k|=ez×kktp=s×kk=s×m.
s=kyktex+kxkteyp=kzkxkktex+kzkykkteyktkezm=kxkex+kykey+kzkez.
ex=kykts+kxkzkktp+kxkmey=kxkts+kykzkktp+kykmez=ktkp+kzkm,
E=gradϕAtB=rotA,
divA+ε0μ0ϕt=0.
2P1c22Pt2=0;=e,m.
ϕ=divPeA=ε0μ0Pet
ϕ=0A=rotPm.
E=graddivPeε0μ02Pet2=rotrotPeB=μ0H=ε0μ0t(rotPe).
E=t(rotPm)B=μ0H=graddivPm1c22Pmt2=rotrotPm.
E=graddivPe1c22Pet2(rotPm)tB=μ0H=ε0μ0(rotPe)t+graddivPm1c22Pmt2.
Ex=2PezxzEy=2PezyzEz=2Pezz2ε0μ02Pezt2Hx=ε02PezytHy=ε02PezxtHz=0,
Pe=(0,0,Pez)
pez(kx,ky)=k2Pez(x,y)exp[j(kxx+kyy)]dxdy
Pez(x,y,z=0)=Pez(x,y)=14π2k2pez(kx,ky)exp[j(kxx+kyy)]dkxdky.
Pez(x,y,z,t)=exp(jωt)14π2k2pez(kx,ky)exp[j(kxx+kyy+kzz)]dkxdky,
E(R)=14π2k2e(kx,ky)exp[j(kxx+kyy+kzz)]dkxdky.
e(kx,ky)=k2E(x,y,0)exp[j(kxx+kyy)]dxdy
h(kx,ky)=k2H(x,y,0)exp[j(kxx+kyy)]dxdy.
e·k=h·k=0,
x[pe(x,y,0)]=pex|z=0,
e(kx,ky)=(k.pe)k+k2pekck×pm
h(kx,ky)=ε0kck×pe(k.pm)k/μ0+k2/μ0pm,
E,He,hPe,mpe,m.
p(kx,ky)=Π+Π=e,mk.Π=0k×Π=0.
e(kx,ky)=k2Πe(kx,ky)kck×Πm(kx,ky)h(kx,ky)=k2μ0Πm(kx,ky)+kcε0k×Πe(kx,ky).
{E,H}={ETE,HTE}+{ETM,HTM}ETE·ez=0;HTM·ez=0.
Pe(R)=Pez(R)ez;Pm(R)=Pmz(R)ez.
pe=pezez.
eTM(kx,ky)=pez[kz(kxex+kyey)(kx2+ky2)ez]=kktpez(kx,ky)p(kx,ky)aTMp
hTM(kx,ky)=ε0kck×ezpez=ε0kc(kyexkxey)pez=ε0μ0kktpez(kx,ky)s(kx,ky)bTMs.
pe=0pm=pmzez.
eTE(kx,ky)=kck×ezpmz=kc(kyexkxey)pmz=ckktpmz(kx,ky)s(kx,ky)aTEs
hTE(kx,ky)=1μ0pmz[kz(kxex+kyey)(kx2+ky2)ez]=1μ0kktpmz(kx,ky)p(kx,ky)bTEp.
eTM=k2Πe=kktpezp,
Πe(kx,ky)=ktkpez(kx,ky)p.
Πm(kx,ky)=ktkpmz(kx,ky)p.
eTM=k2ΠehTM=kcε0k×ΠeeTE=kck×ΠmhTE=k2μ0Πm.
E(R)=ETM(R)+ETE(R)
e(kx,ky)=eTM(kx,ky)+eTE(kx,ky).
ETM(R)=1(2π)2Πe(kx,ky)exp(jk.R)dkxdkyHTM(R)=cε0(2π)2kk×Πe(kx,ky)exp(jk.R)dkxdkyETE(R)=c(2π)2kk×Πm(kx,ky)exp(jk.R)dkxdkyHTE(R)=1(2π)2μ0Πm(kx,ky)exp(jk.R)dkxdky.
ETM(R)=1(2π)2kktpez(kx,ky)pexp(jk.R)dkxdkyHTM(R)=1(2π)2ε0μ0ktpez(kx,ky)sexp(jk.R)dkxdkyETE(R)=c(2π)2ktpmz(kx,ky)sexp(jk.R)dkxdkyHTE(R)=1(2π)2kμ0ktpmz(kx,ky)pexp(jk.R)dkxdky.
Ex(x,y,z)=14π2k2Ax(kx,ky)exp(jk.R)dkxdkyEy(x,y,z)=14π2k2Ay(kx,ky)exp(jk.R)dkxdkyEz(x,y,z)=14π2k2[kxkzAx(kx,ky)+kykzAy(kx,ky)]exp(jk.R)dkxdky.
E(R)=ETM(R)+ETE(R)=Ex(R)ex+Ey(R)ey+Ez(R)ezETM(R)=14π2k2[kzkkt(kxex+kyey)ktkez]ATM(kx,ky)exp(jk.R)dkxdkyETE(R)=14π2k2[kyktexkxktey]ATE(kx,ky)exp(jk.R)dkxdky.
ATM(kx,ky)=k2kktkz(kxEx+kyEy)exp[j(kxx+kyy)]dxdyATE(kx,ky)=k21kt(kyExkxEy)exp[j(kxx+kyy)]dxdy,
ETM(R)=14π2k2pATM(kx,ky)exp(jk.R)dkxdkyETE(R)=14π2k2sATE(kx,ky)exp(jk.R)dkxdky
A(kx,ky)=ATMp+ATEs,
B(kx,ky)=BTMs+BTEp.
E(R)=14π2k2A(kx,ky)exp[j(kxx+kyy+kzz)]dkxdkyH(R)=14π2k2B(kx,ky)exp[j(kxx+kyy+kzz)]dkxdky.
aTM=ATMbTM=BTMaTE=ATEbTE=BTE.
ATM=kktpezBTM=ε0μ0kktpez=ε0μ0ATM
ATE=ckktpmz=μ0ε0BTEBTE=1μ0kktpmz.
ATM=k2ΠeBTM=ε0μ0k2ΠeATE=μ0ε0k2ΠmBTE=k2Πm.
E(R)=Er(r,z)er+Eα(r,z)eα+Ez(r,z)ez=ETM(r,z)+ETE(r,z).
ETM(r,z)=12πk30k[jkzJ1(ktr)erktJ0(ktr)ez]ktATM(kt)exp[jkzz]dktETE(r,z)=jeα2πk20kATE(kt)J1(ktr)ktexp[jkzz]dkt,
ATM(kt)=2πk3jk2kt20Er(r,z=0)J1(ktr)rdrATE(kt)=2πjk20Eα(r,z=0)J1(ktr)rdr.
ETE=s(.)dkxdky=eφ(.)dkxdky=eα(.)dkt.
ETM=p(.)dkxdky=(kzkerktkez)(.)dkxdky=(J1kzkerJ0ktkez)(.)dkt.
u(x,y,z)=exp(jkzz)02πa(φ)exp[jkt(xcosφ+ysinφ)]dφ
ETE(r,z)=j2πJ1(ktr)exp(jkzz)eα.
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