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Collimation is clearly in reference to ray($\vec{k}_{xy}$ vector) orientation unlike waterfront continuity( $\phi_{xy}$ phase shift) described by plane-wave. Not to say that one is not directly associated with the other. However the purpose for separation of these two terms, rather than equating them, appears to be important.

Wikipedia states: "A perfectly collimated beam, with no divergence, cannot be created due to diffraction"

Makes sense, but why add the "with no divergence" and "perfectly" qualifiers in this sentence? Why are many other documents of optical science incapable of simply saying "A collimated beam cannot be created due to diffraction"? They seem to be required to use the word "perfectly" to mean indefinitely collimated.

Take for instance a lowest order Gaussian wave passing through a beam waist the intensity of the beam over (x,y) is not consistent for any given (x,y) aperture. However, if the intensity differentials are accounted for than assuming a small plane wave or collimated vectors will produce an exact mathematically and experimentally verifiable result. Thus, multiple texts and sources will unexpectedly state that the momentary wave at such a position is collimated or a plane-wave front.

Ultimately what is the essential difference between "A perfectly collimated wave" and "A perfect plane wave"? Can either or both exist in the universe, and under what conditions? Are the implications of adding "perfectly" to the word collimated the same as the implications of adding perfect to the word plane-wave, why or why not?

Don't forget about the waveguide point of view to this problem as well. Single mode fiber anyone?

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  • $\begingroup$ Well... SMF cannot stop temporal dispersion effects, if that makes you feel any better (or worse :-) ). It's just better at maintaining the input mode structure than large-core fiber types. $\endgroup$ – Carl Witthoft Dec 9 '13 at 19:47
  • $\begingroup$ SMF temporal dispersion is due to Material Dispersion or Waveguide Dispersion not Modal Dispersion. I thought only Modal Dispersion is a result of $\vec{k}_{xy}$ vector orientation $\endgroup$ – Luke Burgess Dec 9 '13 at 20:10
  • $\begingroup$ I believe that's correct. $\endgroup$ – Carl Witthoft Dec 9 '13 at 20:45
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In one sense you are right: the only free space "perfectly collimated" optical field is the plane wave in the sense that these are the only eigenfields of Maxwell's equations, being fields which conserve their form under propagation and only undergo scaling by an eigenvalue in such propagation. Since Maxwell's equations conserve energy in free space, propagation is represented by a unitary operator and the eigenvalue is therefore always the simple phase delay $e^{i\,\vec{k}\cdot \vec{r}}$. A wavefront of finite breadth must always undergo some "scrambling" by diffraction; Fourier analysis shows that it is always a superposition of plane waves with a spread of directions, so we can actually think of diffraction through a homogeneous medium as a three step process: (1) decompose the field at one infinite plane surface into a superposition of plane wave components through a Fourier transform; (2) impart the direction-dependent phase delay each of these constituent plane ways undergoes in crossing the homogeneous medium from infinite plane boundary to a parallel infinite plane boundary and (3) reassembling electromagnetic field from its constituent plane wave components at the output boundary. It is the fact that the phase delay is direction dependent (being $\delta\vec{z} \cdot \vec{k}$ radians where $\delta\vec{z}$ is parallel displacement between the input and output infinite planes) that scrambles the wavefront, thus begetting the phenomenon we call "diffraction". So it is the propagation distance dependent interference between a field's constituent plane waves.

So in this sense the only truly "collimated" field is indeed the plane wave. However the word "collimated" is to me a kind of "laboratory" word: one "collimates" a field, i.e. makes its phase front maximally flat, through a laboratory collimation procedure such as making adjustments to field-processing lenses and suchlike whilst the wavefront is observed through a suitable "collimation detector" such as a shear plate interferometer, wavefront sensor, wavefront camera or point diffraction interferometer. So a "collimated" field in this sense is one that has been flattened by such a procedure. Optically, a "perfectly" collimated field of this kind (i.e. with perfectly flat phasefronts but finite breadth) is actually the focal plane field of an electromagnetic field with an extremely small numerical aperture, with $NA$ of the order of $\lambda / w$ where $\lambda$ is the field's wavelength and $w$ its transverse breadth. It diverges, its phasefronts take on curvature exactly as would any other focal plane field and the mathematics describing this process is precisely the same as for any focal plane field of course; it's just that, owing to the extremely small numerical aperture, the divergence is very very slow with increasing axial distance from the focal plane.

For a Gaussian collimated field, the phase front at the focal plane is flat (as for any other collimated field), the phasefronts themselves are a family of ellipsoids and the integral curves of the rays (i.e. unit normals to the phasefronts) are the family of orthogonal hyperboloids (see the Wikipedia page for Gaussian beam).

An optical fibre is different again; here the matter of the fibre interacts with the electromagnetic field and there are many non-plane eigenfields (indeed all, even the radiation fields are non-plane). Again these fields are the ones that undergo only a simple scaling in propagation through the translationally invariant waveguide, and again, if the waveguide is made of lossless material, the scaling is a simple phase delay. The full tale is rather long winded: begin at chapter 11 of Snyder and Love, "Optical Waveguide Theory" (Chapman 1983) and keep reading! However you can get an intuitive feel for what's going on by looking in more detail into the Beam Propagation Method. The finite difference finding of a solution of Maxwell's equations is equivalent to solving a Master-like equation in the Lie group $U(N)$ for the square matrix $T \in U(N)$ which maps the input field to the output field with $N$ being of the order of the number of transverse grid points in the simulation (that's an awfully big Lie group!) depending on whether a vector scaler simulation is done:

$${\rm d}_z T(z) = K(z) T(z)$$

where $K(z) \in \mathfrak{u}(N)$ wanders around in the Lie algebra $\mathfrak{u}(N)$ of $N\times N$ skew-Hermitian matrices to represent the local action of diffraction and either focusing or de-focusing by the waveguide. A short, translationally invariant, length of waveguide has a transfer matrix of the form $\exp((D + L) z)$, where $D$ represents the diffraction through the "mean homogeneous" medium and $L$ the diffraction-free lensing of the waveguide's local profile. The beam propagation method implements "operator splitting", which is another way of saying splitting apart the effect of diffraction and lensing through the Trotter product formula of Lie theory:

$$\lim\limits_{m\to\infty}\left(\exp\left(D\,\frac{z}{m}\right)\,\exp\left(L\,\frac{z}{m}\right)\right)^m = \exp((D + L) z)$$

So the beam propagation method amounts to nothing more than a succession of diffractions through thin slivvers of homogeneous medium followed by the diffraction free lensing imparted by the equivalent local refractive index profile of each slivver. You can, with a bit of a stretch, imagine a step index refractive index profile to be a badly pixellated version of the smooth GRIN lens: the wave near the centre is delayed more by the higher refractive index of the core than the wave near the waveguide edges, so it tends to cancel out the divergence begotten of the pure diffraction of the foregoing slivver. It's not too hard to imagine that for certain special field shapes this lensing, although arising from a "bad" lens, exactly cancels the diffraction and one has a waveguide mode whose phasefront stays perfectly flat and whose transverse profile perfectly conserved as it propagates through the waveguide.

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  • $\begingroup$ Wow, from orthogonal hyperbolas to canceled diffraction, you got this one more than I could have hoped thanks! :) $\endgroup$ – Luke Burgess Dec 11 '13 at 3:53
  • $\begingroup$ @LukeBurgess Thanks. The hyperboloids are neat in that they help to visualise the diffraction limit: in the farfield they asymptote to cones of converging rays: near the focus they deviate from these cones and show that the wavefronts have a maximum curvature at the Rayleigh distance from the focal plane and then the curvature drops off as one moves nearer to the focal plane so that the focal plane field is flat. From the cone of rays, one would expect that the phasefront curvature diverges to infinity at the focal point (as the ray model says). Collimated fields have huge Rayleigh distances. $\endgroup$ – WetSavannaAnimal Dec 11 '13 at 5:17
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I might add a few commas to that Wikipedia sentence, as "A perfectly collimated beam*,* with no divergence*,* cannot..." to show informative rather than additional parameters.
To answer your question about "collimated" vs. "plane wave" , consider two point sources at th plane of focus of a lens. Each point source gives off spherical waves; the lens (ideally) converts each of these to a plane wave. The lens produces collimated light, but the two plane waves are going off in different directions, so the total lens output is collimated but not planar.

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  • $\begingroup$ I like your answer, and it gives me ideas regarding modifications to the question. However you never addressed the implications of "perfect" qualifying collimated or planewave. And, I quoted directly from wiki. If you think it needs commas, go to the wiki page and add them. en.wikipedia.org/wiki/Collimated_light $\endgroup$ – Luke Burgess Dec 9 '13 at 19:26
  • $\begingroup$ @LukeBurgess Sorry; I didn't mean to imply that you'd fouled up the Wikipedia quote. $\endgroup$ – Carl Witthoft Dec 9 '13 at 19:45
  • $\begingroup$ I was getting carried away with my comment, sorry. Do you have a wiki account, or is it ok with you if I use mine to change the sentience. $\endgroup$ – Luke Burgess Dec 9 '13 at 19:51

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