Propagation of vectorial laser beams

Peter Muys

Laser Power Optics, Hooiland 3, B-9030 Gent, Belgium (peter.muys@gmail.com)

Received December 19, 2011; accepted January 23, 2012;posted January 27, 2012 (Doc. ID 159962); published April 18, 2012

The angular spectrum of a vectorial laser beam is expressed in terms of an intrinsic coordinate system instead ofthe usual Cartesian laboratory coordinates. This switch leads to simple, elegant, and new expressions, such as forthe angular spectrum of the Hertz vectors corresponding to the electromagnetic fields. As an application of thisapproach, we consider axially symmetric vector beams, showing nondiffracting properties of these beams, withoutinvoking the paraxial approximation. © 2012 Optical Society of America

OCIS codes: 260.1960, 260.5430, 260.2110.

1. INTRODUCTIONIn many branches of physics, it is necessary to solve thevectorial wave equation in three dimensions. One of the bestknown and most general methods for the vectorial Helmholtzequation is the expansion in plane waves by representing thesolution as a three-dimensional (3D) Fourier integral over the3D reciprocal wavevector space. Especially in laser physics,the relevant electric and magnetic fields are in fact solutionshaving a directed-energy beam-character. That is, the fieldspropagate along a certain axis, called the optical axis andwhich we choose to be the z axis of the Cartesian coordinatesystem linked with the beam. A number of mathematicalmethods are known to treat this case efficiently [1,2]. In thispaper, we consider three methods of solution, the angularspectrum representation of the beam, the TE/TM decom-position of the beam and finally the Hertz vectors of the beam.The last method is the least well known, despite the factthat it shows a major advantage by generating particularlysimple expressions for the fields. We shall indicate theinterrelations between the three methods. In a first step weshow how to transform the Cartesian axes to an intrinsiccoordinate system linked with the propagation constants ofthe beam.

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

In a second stage, we determine the angular spectrum notof the electric and magnetic fields E and B, but of the Hertzpotential vectors Pe and Pm corresponding to these vectors.Then we use the results of the first phase and switch tothe new coordinates. New and simple formulas are obtained.This leads to an alternative angular spectrum representationof the solution of the vectorial Helmholtz equation.

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

2. INTRINSIC COORDINATE SYSTEMLet k be the wave vector of a plane wave

k � kxex � kyey � kzez

� kter � kzez

� kt cos φex � kt sin φey � kzez (1)

and its magnitude is given by

k2 � k2x � k2y � k2z � k2t � k2z.

In Eq. (1) we have denoted the split of the wave vector into itstransverse (subscript t) and longitudinal part (subscript z).The k vector forms together with the z axis the meridionalplane [3]. In optical terminology, we would call it the planeof incidence (on a mirror for example, where z is the normalto the mirror). The angle between the z axis and the k axis isthe polar angle θ. The cylindrical coordinates are definedthrough their unit vectors er , ez, and eφ, φ being the azimuthalangle of the k vector in the transverse �x; y� plane.

The 3D locus vector R is defined by

R � xex � yey � zez

� rer � zez

� r cos αex � r sin αey � zez; (2)

where we indicated the split of the locus vector in its trans-verse 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 co-ordinate system. It contains three unit vectors: two of themare respectively perpendicular and parallel to the meridionalplane, the third being the unit vector in the k direction:

s � eφ

p � kzker −

ktkez

m � k∕k. (3)

990 J. Opt. Soc. Am. B / Vol. 29, No. 5 / May 2012 P. Muys

0740-3224/12/050990-07$15.00/0 © 2012 Optical Society of America

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 perpendi-cular to the meridional plane, whereas p is coplanar with it,just like m. s has only a transverse component, its azimuthalpart, whereas p both has a transverse component, its radialpart, and a longitudinal component. Although the definitionof the s and p vector are classically known in the literatureon optical polarization, their relation [Eq. (3)] with the cylind-rical base vectors is not evident. An alternative but equivalentdefinition of the intrinsic coordinates is

s � ez × kjez × kj �

ez × kkt

p � s × kk

� s ×m.

The transformation of the Cartesian coordinates to the intrin-sic coordinates is given by

s � −kykt

ex �kxkt

ey

p � kzkxkkt

ex �kzkykkt

ey −ktkez

m � kxkex �

kykey �

kzkez. (4)

The inverse transformation from the intrinsic coordinatesystem to the Cartesian laboratory coordinate system is, afterlengthy but straightforward calculations, given by

ex � −kykt

s� kxkzkkt

p� kxkm

ey � kxkt

s� kykzkkt

p� kykm

ez � −ktkp� kz

km; (5)

which is a central result for the subsequent calculations. Notethat s and p are in fact functions of kx and ky and should in factbe written as s�kx; ky� and p�kx; ky� when they will in a laterparagraph appear behind the integration sign.

3. HERTZ POTENTIALS AND THEIRANGULAR SPECTRUM REPRESENTATIONWe will only consider free-space situations: the sources areinfinitely far away from our observation region. The electricand magnetic fields can be defined in S.I. units in terms oftheir scalar and vector potential [5]:

E � −gradϕ −∂A∂t

B � rotA; (6)

where the potentials are linked by the Lorentz gauge

divA� ε0μ0∂ϕ∂t

� 0.

An equivalent and alternative set to these potentials are theHertz electric and magnetic vector potentials Pe and Pm,

which in free space are, just like E and B, also solutions ofthe homogeneous vectorial d’Alembert wave equation

∇2P• −1

c2∂2P•

∂t2� 0; • � e;m.

The link between the regular potentials and the electric Hertzvector is given by [6]:

ϕ � −div Pe

A � ε0μ0∂Pe

∂t

and for the magnetic Hertz vector by

ϕ � 0

A � rotPm.

Combining this with Eq. (6), hence eliminating the scalar andvector potential, results for the two cases resp. in the expres-sion of the fields as a function of the Hertz vectors. For theelectric Hertz vector:

E � grad div Pe − ε0μ0∂2Pe

∂t2� rot rotPe

B � μ0H � ε0μ0∂

∂t�rotPe�. (7)

And for the magnetic Hertz vector:

E � −∂

∂t�rotPm�

B � μ0H � grad div Pm −1c2

∂2Pm

∂t2� rot rotPm. (8)

Because of the source-free space, the magnetic induction Breduces to the magnetic field strength H. By adding Eqs. (7)and (8), we find the general expression for the fields in termsof the Hertz vectors:

E � grad div Pe −1c2

∂2Pe

∂t2−∂�rotPm�

∂t

B � μ0H � ε0μ0∂�rotPe�

∂t� grad div Pm −

1c2

∂2Pm

∂t2. (9)

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 direc-tion, say the z direction [5–7]. In this situation, and far fromthe sources, Eq. (7) for the electric Hertz potential reduces tothe set

P. Muys Vol. 29, No. 5 / May 2012 / J. Opt. Soc. Am. B 991

Ex � ∂2Pez

∂x∂zEy � ∂2Pez

∂y∂z

Ez �∂2Pez

∂z2− ε0μ0

∂2Pez

∂t2

Hx � ε0∂2Pez

∂y∂t

Hy � −ε0∂2Pez

∂x∂tHz � 0; (10)

which show that the electric Hertz vector Pe reduces to itsscalar component Pez:

Pe � �0; 0; Pez� (11)

This fact enormously simplifies the mathematical manipula-tions and was exploited to analyze a number of vectorialdiffraction effects in [3]. The attractive aspect of the “scalar”Hertz vector representation as compared to, e.g., the mathe-matically equivalent modal representation is that it arrives atthe same rigorous results as the modal representation, but in amuch shorter and more elegant way. To give a quick exampleof its usefulness, consider the electric field of an electric di-pole oscillating along the z axis. In every direction startingfrom the origin, the electric dipole field has another orienta-tion but the corresponding Hertz vector is nevertheless every-where in space a scalar, oriented along the z axis, independentof the radiation direction [3].

For laser applications, we are exclusively interested in di-rected-energy solutions of the vectorial Maxwell equations,i.e., in “beams,” rather than in “fields.” Of course we then takethe z direction as the propagation direction of the field energy.So, using the Hertz potentials, vectorial beams can be comple-tely described by a scalar function. Specifically, whereasPe�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 scalarsolution of the Helmholtz wave equation with the boundarycondition Pez�x; y; 0; t� � Pez�x; y� exp�−jωt�. The two-dimensional (2D) Fourier transform of the Hertz potentialPez�x; y� in the plane z � 0, is defined as

pez�kx; ky� � k2Z

∞

−∞

Z∞

−∞Pez�x; y� exp�−j�kxx� kyy��dxdy

(12)

and is called its angular spectrum. The factor k2 before theintegration signs is required to give the spectrum the samephysical dimensions as the original. The inverse Fourier trans-formation is given by

Pez�x; y; z � 0� � Pez�x; y�

� 14π2k2

Z∞

−∞

Z∞

−∞pez�kx; ky� exp�j�kxx

� kyy��dkxdky. (13)

The factor of 1∕4π2 appears for mathematical reasons, the fac-tor k2 for physical reasons. Note that pez is not the angularspectrum of the electric field in the plane z � 0, but ratherof the electric Hertz vector. The general solution of the scalarwave equation for Pez�x; y; z; t� in the halfspace z > 0 is nowgiven by [8]

Pez�x; y; z; t� � exp�−jωt� 1

4π2k2Z

∞

−∞

Z∞

−∞pez�kx;ky� exp�j�kxx

� kyy� kzz��dkxdky; (14)

which is called the angular spectrum representation of theelectric Hertz potential Pez�x; y; z; t�. Equation (14) shows thatfor beam-like solutions, we do not need a 3D Fourier trans-form as solution of the wave equation, the dimensionalityof the problem is reduced by one unit.

A. Vectorial Angular Spectrum RepresentationA scalar field E�R� satisfying the scalar Helmholtz equation,can be represented as in Eq. (14) by its angular spectrumrepresentation, i.e., the beam is linearly polarized. What hap-pens if the beam has a vectorial character? Just by extendingthe argument from one dimensional (1D) to 3D, we look for arepresentation like

E�R� � 1

4π2k2Z

∞

−∞

Z∞

−∞e�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 needonly two scalar functions to determine the 3D electric field.This should be reflected in the form of the vectorial angularspectrum.

Assuming harmonic time variation, we now take the 2DFourier transform of E and H at z � 0 in Eq. (9) and call themthe vectorial angular spectra e and h of the field vectors, or thespectral field vectors in short

e�kx; ky� � k2Z

∞

−∞

Z∞

−∞E�x; y; 0� exp�−j�kxx� kyy��dxdy (15)

h�kx;ky�� k2Z

∞

−∞

Z∞

−∞H�x;y;0�exp�−j�kxx�kyy��dxdy. (16)

It should be noted that the vectorial character of the angularspectrum, i.e., its polarization state, is the same as that of theoriginal field. Hence e and h inherit the polarization state of Eand H (and also their physical dimensions) Since the diver-gence of E and H is zero in free-space, this means for theirspectral vectors

e · k � h · k � 0;

i.e., the spectral vectors are transverse to the wave vector, justlike 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 deriva-tive of a function is equal to −jk times the Fourier transform ofthe original function. In working out the expressions forz � 0, one should take into account relations such as

∂

∂x�pe�x; y; 0�� �

∂pe∂x

����z�0

;

which is easy to prove by writing out the derivatives accordingto their definition as a limit of the differential quotient forΔx → 0. Finally, we arrive at the angular field spectra as func-tions of the angular Hertz spectra

992 J. Opt. Soc. Am. B / Vol. 29, No. 5 / May 2012 P. Muys

e�kx; ky� � −�k:pe�k� k2pe − kck × pm (17)

h�kx; ky� � ε0kck × pe − �k:pm�k∕μ0 � k2∕μ0pm; (18)

where we also made use of the dispersion relation ω � kc tofurther simplify the expressions. So, in reciprocal k space, therelations between the field spectra and the Hertz spectra areno longer differential equations but become vector equations.

What we will do in this paragraph is work with the spectralfield vectors e and h to finally obtain an angular spectrum re-presentation of the fields E andH, through the intermediary ofthe Hertz vectors and their angular spectrum representation,according to the Fourier transforms indicated in the followingdiagram:

E;H ⇄ e; h↓ ↑

Pe;m → pe;m.

Note: the spectral vectors pe and pm should not be confusedwith the intrinsic vector p.

B. Transverse Spectral Hertz VectorsNext, we proceed to decompose each angular vectorial spec-trum of the Hertz potentials pe and pm in a part that is parallelto k and a part that is perpendicular to k:

p•�kx; ky� � Π•

‖�Π•

⊥• � e;m

k:Π•⊥� 0

k ×Π•

‖� 0. (19)

On substituting Eqs. (19) in Eq. (17,18), the parallel compo-nents of the spectral Hertz vectors all drop out and what restsis

e�kx; ky� � k2Πe⊥�kx; ky� − kck ×Πm

⊥�kx; ky�

h�kx; ky� �k2

μ0Πm

⊥�kx; ky� � kcε0k ×Πe

⊥�kx; ky�. (20)

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

fE;Hg � fETE;HTEg � fETM;HTMgETE · ez � 0; HTM · ez � 0. (21)

The transverse character of the field vectors E and H now hasto be conveyed to the spectral vectors e and h. We will pro-ceed by taking a detour and first determine how e and h de-pend on the spectral Hertz vectors. The reason is, as we willsee shortly, that the spectral Hertz vectors take on very simpleexpressions. 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 oscil-lating in the same direction, which we choose to be the z

direction. The Hertz vectors have the same orientation asthe currents [6]; hence the Hertz vectors have only a z com-ponent. We assume that the dipole source is located at z � −∞

so that the halfspace z > 0 is source free and the field differ-ential equations become homogeneous and hence simpler. Inthis halfspace, the Hertz vectors keep their z orientation.From here on, we assume that the Hertz vectors have onlya z-dependent component:

Pe�R� � Pez�R�ez; Pm�R� � Pmz�R�ez.

Hence, also their angular spectra pe and pm are only z depen-dent. Pez and Pmz are solutions of the homogeneous, scalarHelmholtz equation.

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

D. TM CaseFor the moment, we only know that, by Fourier transformingthe original electric Hertz vector, we obtain

pe � pezez.

Now, transforming Eq. (21), we see that hTM · ez � 0. We useEq. (20) to calculate this scalar product. This leads after somealgebra to: pm � 0. Next to the coordinate transformationsEq. (4,5) for the intrinsic vectors, this small equation is thesecond central result of this paper. It shows that the spectralmagnetic Hertz vector is zero for a TM beam, leading to ex-tremely simple equations, despite the vectorial character ofthe beam. This is shown by again using the intrinsic coordi-nate transformations, so that the angular spectra now become

eTM�kx; ky� � −pez�kz�kxex � kyey� − �k2x � k2y�ez�� −kktpez�kx; ky�p�kx; ky�≜aTMp (22)

hTM�kx; ky� � ε0kck × ezpez � ε0kc�kyex − kxey�pez

������ε0μ0

rkktpez�kx; ky�s�kx; ky�≜bTMs. (23)

The angular field spectra are shown to be simple functions ofthe intrinsic vectors and the scalar angular electric Hertz spec-trum. In Eq. (23), the magnetic vector only depends on s,which has no z component, according to Eq. (3); so thereforethe fields given by Eq. (22,23) are indeed TM. This is in com-plete agreement with Eqs. (10) which show that the magneticfield has no z component. [Eq. (10) is not written in the Four-ier domain, but for the polarizations this does not matter.]

E. TE CaseIn a similar way as in the former case, this corresponds to:

pe � 0

pm � pmzez.

With this substitution, Eqs. (14) and (15) become, making useof the transformations (5):

eTE�kx; ky� � −kck × ezpmz � −kc�kyex − kxey�pmz

� ckktpmz�kx; ky�s�kx; ky�≜aTEs (24)

P. Muys Vol. 29, No. 5 / May 2012 / J. Opt. Soc. Am. B 993

hTE�kx; ky� � −1μ0

pmz�kz�kxex � kyey� − �k2x � k2y�ez�

� −1μ0

kktpmz�kx; ky�p�kx; ky�≜bTEp. (25)

Hence we see again that the spectral vectors have becomevery simple functions of the intrinsic vectors and the scalarspectral magnetic Hertz potential.

Note that the magnitudes of the magnetic spectral vectorsare just proportional copies of the magnitudes of the electricspectral vectors.

F. Spectra of the Transverse Hertz VectorsThe fact that the spectral vectors only have a z dependencefurther 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. (26)

In the same way, we find:

Πm⊥�kx; ky� � −

ktkpmz�kx; ky�p. (27)

So finally in the spectral domain, the relations between thefields and their Hertz vectors is

eTM � k2Πe⊥

hTM � kcε0k ×Πe⊥

eTE � −kck ×Πm⊥

hTE � k2

μ0Πm

⊥. (28)

G. Angular Spectral RepresentationThe electric field of the beam propagating along the z axis hasbeen 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 thefields look like

ETM�R� �1

�2π�2Z

∞

−∞

Z∞

−∞Πe

⊥�kx; ky� exp�jk:R�dkxdky

HTM�R� �cε0

�2π�2k

Z∞

−∞

Z∞

−∞k ×Πe

⊥�kx; ky� exp�jk:R�dkxdky

ETE�R� �−c

�2π�2k

Z∞

−∞

Z∞

−∞k ×Πm

⊥�kx; ky� exp�jk:R�dkxdky

HTE�R� �1

�2π�2μ0

Z∞

−∞

Z∞

−∞Πm

⊥�kx; ky� exp�jk:R�dkxdky. (29)

Or, alternatively,

ETM�R��−1

�2π�2k

Z∞

−∞

Z∞

−∞ktpez�kx;ky�pexp�jk:R�dkxdky

HTM�R��1

�2π�2�����ε0μ0

r Z∞

−∞

Z∞

−∞ktpez�kx;ky�sexp�jk:R�dkxdky

ETE�R��c

�2π�2Z

∞

−∞

Z∞

−∞ktpmz�kx;ky�sexp�jk:R�dkxdky

HTE�R��−1

�2π�2kμ0Z

∞

−∞

Z∞

−∞ktpmz�kx;ky�pexp�jk:R�dkxdky. (30)

For clarity and brevity, we have suppressed in the above in-tegrals the dependency of the intrinsic vectors on kx and ky.Equation (30) show the electric and magnetic fields as func-tions of the angular spectra of the two Hertz vectors and of theintrinsic 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 thefield vectors and of the Hertz vectors.

4. ANGULAR SPECTRUM OF THE VECTORFIELDSThe angular spectrum representation of the vector electricfield has been given in [9], expressed in Cartesian coordinatesas:

Ex�x; y; z� �1

4π2k2Z

∞

−∞

Z∞

−∞Ax�kx; ky� exp�jk:R�dkxdky

Ey�x; y; z� �1

4π2k2Z

∞

−∞

Z∞

−∞Ay�kx; ky� exp�jk:R�dkxdky

Ez�x; y; z� �−1

4π2k2Z

∞

−∞

Z∞

−∞

�kxkz

Ax�kx; ky�

� kykz

Ay�kx; ky��exp�jk:R�dkxdky. (31)

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 itwith the angular spectrum representation. The advantage ofthis approach is that it clearly identifies and separates the con-tributions of the transverse electric fields (TE and TM) rightfrom the start. In laboratory Cartesian coordinates, this finallyleads to

E�R� � ETM �R� � ETE�R�� Ex�R�ex � Ey�R�ey � Ez�R�ez

ETM�R� �−1

4π2k2Z

∞

−∞

Z∞

−∞

�kzkkt

�kxex

� kyey� −ktkez

�ATM�kx; ky� exp�jk:R�dkxdky

ETE�R� �−1

4π2k2Z

∞

−∞

Z∞

−∞

�kykt

ex

−kxkt

ey

�ATE�kx; ky� exp�jk:R�dkxdky. (32)

The angular spectra ATM and ATE are in fact 2D Fourier trans-forms defined by the total fields Ex and Ey in z � 0 [1]

994 J. Opt. Soc. Am. B / Vol. 29, No. 5 / May 2012 P. Muys

ATM�kx; ky� � k2Z

∞

−∞

Z∞

−∞

kktkz

�kxEx

� kyEy� exp�−j�kxx� kyy��dxdy

ATE�kx; ky� � k2Z

∞

−∞

Z∞

−∞

1kt�kyEx

− kxEy� exp�−j�kxx� kyy��dxdy; (33)

but due to the vector character of the fields, the mathematicalexpressions of the integrands are more complex than in thescalar case. Note again that we only need to know Ex andEy. So [1] was able to separate the propagating field intotwo transverse contributions, which [2,4] did not indicate inthis way. We will see, however, that in [4], essentially the sameresult was obtained. Expressions for the magnetic field arenot given by [1]. It is nevertheless clear that ETE is lackinga z component, as it should for the electric field of a TE mode.

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

ETM�R� �−1

4π2k2Z

∞

−∞

Z∞

−∞pATM�kx; ky� exp�jk:R�dkxdky

ETE�R� �−1

4π2k2Z

∞

−∞

Z∞

−∞sATE�kx; ky� exp�jk:R�dkxdky (34)

so that the vectorial angular spectrum of the total vectorialelectric field E�R� � ETM�R� � ETE�R� is given by

A�kx; ky� � ATMp� ATEs; (35)

which is exactly the form mentioned in [4], although in aslightly different format, denoted there as A � App� Asswithout explicitly pointing to the transverse character ofthe contributions As and Ap.

The same reasoning can be repeated for the magneticangular spectra:

B�kx; ky� � BTMs� BTEp. (36)

The vectorial angular spectrum representation of the totalfields now become compactly

E�R� � 14π2k2

Z∞

−∞

Z∞

−∞A�kx; ky� exp�j�kxx� kyy� kzz��dkxdky

H�R� � 1

4π2k2Z

∞

−∞

Z∞

−∞B�kx; ky� exp�j�kxx� kyy� kzz��dkxdky.

Now, going back to the spectral Hertz vectors, Eqs. (22–25),and comparing these with Eq. (30,34,35), we see that

aTM � ATM bTM � BTM

aTE � ATE bTE � BTE.

This means that the angular spectra of the field vectors and ofthe Hertz vectors are related as

ATM � kktpez

BTM � −

�����ε0μ0

rkktpez � −

�����ε0μ0

rATM (37)

ATE � −ckktpmz ������μ0ε0

rBTEBTE � 1

μ0kktpmz.(38)

Alternatively, in function of the spectra of the transverse Hertzpotentials:

ATM � k2Πe⊥

BTM ������ε0μ0

rk2Πe

⊥

ATE ������μ0ε0

rk2Πm

⊥

BTE � k2Πm⊥.

5. AXIALLY SYMMETRIC FIELDSWe now work out the general Eq. (34) for the special case ofaxial 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 spec-trum is independent of the azimuthal angle φ in reciprocal kspace and only depends on kt. Equation. (34) becomes [1]

ETM�r; z� �1

2πk3Z

k

0�jkzJ1�ktr�er

− ktJ0�ktr�ez�ktATM�kt� exp�jkzz�dktETE�r; z� �

−jeα2πk2

Zk

0ATE�kt�J1�ktr�kt exp�jkzz�dkt; (39)

where the 2D Fourier transforms for the angular spectra nowreduce to 1D Hankel transforms

ATM�kt� �2πk3j���������������k2 − k2t

pZ

∞

0Er�r; z � 0�J1�ktr�rdr

ATE�kt� � −2πjk2Z

∞

0Eα�r; z � 0�J1�ktr�rdr. (40)

J0�z� and J1�z� are the well known Bessel functions. If we re-write Eq. (39) by just keeping the vectorial part and the dif-ferentials in place, and absorbing the other contributions inthe symbol (.), we can more clearly see how the unit vectorsare transformed by changing from Cartesian to cylindrical co-ordinates. Note that eφ as integration variable in k space istransformed into eα in R space, which is independent of kxand ky and hence can be brought in front of the integrationsign for the TE-field:

ETE �ZZ

s�:�dkxdky �ZZ

eφ�:�dkxdky � eα

Z�:�dkt.

The electric TE-field is purely azimuthally polarized. For theTM-field, this simplification does not occur:

P. Muys Vol. 29, No. 5 / May 2012 / J. Opt. Soc. Am. B 995

ETM �ZZ

p�:�dkxdky �ZZ �

kzker −

ktkez

��:�dkxdky

�Z �

J1kzker − J0

ktkez

��:�dkt.

The TM-vector keeps its transverse and longitudinal contribu-tions in place, although now with other weighting coefficients,which are Bessel functions. Also here, the unit vectors can bebrought in front of the integration symbol, since they are in-dependent of kt.

6. NONDIFFRACTING VECTORIAL BEAMSThe best known nondiffracting scalar beam in the laser litera-ture is the Bessel beam J0�ktr�. This beam is linearly polar-ized. As a short application of Eqs. (39,40), we willconsider the diffraction-free propagation of an azimuthallypolarized vectorial beam and extend in this way the scopeof [10], which analyzed scalar beams.

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

u�x; y; z� � exp�jkzz�Z

2π

0a�φ� exp�jkt�x cos φ� y sin φ��dφ

and is known under the name “Whittaker integral” in the math-ematical literature. Physically, it represents a superposition ofplane waves with amplitude coefficient a�φ� and with theirwave vector situated on a cone with base radius kt and heightkz. In other words, their angular spectrum in reciprocal kspace is given by A � δ�kt − k0t�. In the vectorial case, weuse this same angular spectrum to describe nondiffractingvector beams and substitute it in the TE-field expression ofEq. (14) to arrive immediately at

ETE�r; z� �−j2π J1�ktr� exp�jkzz�eα. (41)

The vectorial Bessel beam (41) is hence nondiffracting sincethe transverse part is not depending on the propagationcoordinate z. This result was also obtained in [11], but by

the method of separation of the variables of the paraxialwaveequation. A mathematically equivalent statement is that the J1

Bessel function can be represented as a Whittaker integral. Itsexplicit form can be found in [3].

7. CONCLUSIONSIn summary, we have pointed out the relevance of using anintrinsic coordinate system to unify the existing angularspectrum representations. We have given the transformationformulas from intrinsic to Cartesian coordinates. The intro-duction of intrinsic coordinates much simplifies the angularspectral representation of vectorial beams. We have intro-duced the Hertz vector potential oriented along the z axisand have combined it with the intrinsic coordinates to repre-sent TE and TM beams in a particularly simple and elegantform. This leads also to the mathematical link between theangular spectra of the fields and of their Hertz vector. Next,we derived expressions for a diffraction-free vectorial beam,without invoking the paraxial approximation.

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