# Which solution to the electromagnetic wave equation is the most accurate model of monochromatic light?

When a photon is modeled as a monochromatic electromagnetic wave its electric and magnetic components are usually taken to be sine waves (for example here http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwv.html). I believe the practical reason for this is that any solution of the electromagnetic wave equation can be expressed as a sum of sine waves. But physically, when a photon can be interpreted as a wave, how is it best modeled? Do we have empirical evidence to think that it is best modeled by a single sine wave for the E and B fields, or if not which solution of the electromagnetic wave equation would best model it? Are there experiments that could show that light waves resemble more say square waves than sine waves?

Update

To rephrase the core of my question more properly: If a photon / electromagnetic wave travels through a point in space, in vacuum, and we measure the electric and magnetic fields at this point with a very high temporal resolution, would we measure the electric and magnetic fields to fluctuate exactly as sine waves, or as something else? Has such an experiment ever been made?

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By definition monochromatic light is a single sine wave. – ZachMcDargh Apr 15 '14 at 2:14
Monochromatic light is a wave of a single frequency, why would that wave be a sine wave? – user44558 Apr 15 '14 at 2:21

Are there experiments that could show that light waves resemble more say square waves than sine waves? Are there experiments that could show that light waves resemble more say square waves than sine waves?

Temporarily looking at your example of a square wave, a square wave of spatial wavenumber $k$ can be represented in a Fourier expansion as a superposition of sine waves with spatial wavenumbers $k,3k,5k...$ etc. Assuming that we are doing our measurements in a medium which has linear response to the fields (which is satisfied in most materials for fields with power densities $\ll10^8\mathrm{V/m}$, which virtually all optics experiments satisfy), the superposition principle holds, and so without loss of generality we can examine the effect of an experimental apparatus on each individual basis function, and then add together the results.

It's known that individual sine waves will diffract from a diffraction grating at well-defined angles, which is how monochromators work. So, if we had a square-wave source, we would see multiple harmonics split by the grating. Alternately, we could use a prism. In either case, if (quasi)monochromatic light really was a square wave, you'd be able to see the harmonics, but you don't.

(From a comment): You say "It's known that individual sine waves will diffract from a diffraction grating at well-defined angles", but how do we know that? Individual waves will diffract from a diffraction grating at well-defined angles, but how do we know these are sine waves?

It's not quite correct to say that "Individual waves will diffract from a diffraction grating at well-defined angles" in the sense that the diffraction pattern will take the form of a "picket fence" of single peaks. Read any intro optics text and you'll see examples (such as multiple-slit diffraction and diffraction gratings) where they temporarily assume that the electric field takes the form of sinusoids, and then go on to show that sinusoid waves diffract at integer multiples of an angle. The reasoning does not necessarily apply to arbitrary waves (but due to linearity, arbitrary waves can be handled by Fourier decomposition).

(From a comment): Also couldn't we represent a sine wave as an infinite superposition of square waves as well? Or at least as a superposition of periodic functions which are not sine nor cosine? In which case by your reasoning we would be able to see the harmonics even with a sine wave

While it's tempting to use verbal reasoning to come to conclusions in physics, you really have to do the math to make sure your words are correct. You are correct in saying that a decomposition in terms of sines is not necessary; by linearity, we can actually compute the physics using any periodic basis we want, such as square waves, and because of the linearity, the end result will always be the same.

To illustrate this, let's see what exactly happens when we attempt to diffract a sine wave and a square wave on a multiple-slit setup with 61 slits. Let the $k$th slit be located at a height $kd$ where $d$ is the slit spacing, and let the screen be located at a distance $R$ from the slits. At a location of height $y$ on the screen, the distance between the $k$th slit and the location is $$d(k,y)=\sqrt{R^2+(y-kd)^2}\approx \sqrt{d^2 k^2+R^2}-\frac{d k y}{\sqrt{d^2 k^2+R^2}}$$ and for an incident sine wave the field amplitude hitting the point becomes $$E(k,y)\propto\exp\left(\frac{2\pi i d(k,y)}{\lambda}\right)$$ and so the light intensity at that point becomes $$I(y)=\left|\sum_{k=-30}^{30}E(k,y)\right|^2$$ which basically looks like this:

expr = Sum[
Exp[2 \[Pi] I (Sqrt[d^2 k^2 + R^2] - (d k y)/Sqrt[
d^2 k^2 + R^2])/\[Lambda]], {k, -30, 30}];
Plot[Abs[expr /. {R -> 1, \[Lambda] -> 0.0000025,
d -> 0.00002}], {y, -0.3, 0.3}, PlotRange -> All,
PlotPoints -> 60, ImageSize -> 900, AspectRatio -> 0.3]


Meanwhile, we can do the exact same thing with a square wave beam hitting the slits. We have $$E(k,y)\propto\mathrm{SquareWave}\left(\frac{d(k,y)}{\lambda}\right)$$ and again the light intensity at that point becomes $$I(y)=\left|\sum_{k=-30}^{30}E(k,y)\right|^2$$ which basically looks like this:

expr = Sum[
SquareWave[(Sqrt[d^2 k^2 + R^2] - (d k y)/Sqrt[
d^2 k^2 + R^2])/\[Lambda]], {k, -30, 30}];
ListLinePlot[
Table[Abs[
expr /. {R -> 1, \[Lambda] -> 0.0000025, d -> 0.00002}], {y, -0.3,
0.3, 0.00005}], PlotRange -> All, AspectRatio -> 0.3,
ImageSize -> 900]


Notice that in addition to the main peaks, there are smaller sideband peaks with spacings which have spacings which are factors of $1,3,5,7,...$ times smaller that of the fundamental sequence. These are the harmonics I mentioned earlier, which as I mentioned before can be derived by considering the decomposition of the square wave into a sinusoid basis.

However, note that nowhere in the code above used to generate the image did I use sine waves! It was entirely based on a direct square wave summation. This is a good illustration of the fact that the choice of basis you choose to represent linear physics in is irrelevant.

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Thanks for the answer that's more what I'm looking for. You say "It's known that individual sine waves will diffract from a diffraction grating at well-defined angles", but how do we know that? Individual waves will diffract from a diffraction grating at well-defined angles, but how do we know these are sine waves? Also couldn't we represent a sine wave as an infinite superposition of square waves as well? Or at least as a superposition of periodic functions which are not sine nor cosine? In which case by your reasoning we would be able to see the harmonics even with a sine wave – user44558 Apr 15 '14 at 3:16
Sine waves are preferred stable solution in most media, because most media are linear. It can be shown that if the relation of current density to electric field is linear, the only way how field can be decomposed into individual stable components is with sinusoidal waves - they are "characteristic functions" of the underlying wave equation. Each sine wave with diff. frequency has diff. index of refraction and this leads to spatial separation of those components. – Ján Lalinský Apr 15 '14 at 6:08
In practice material medium is not exactly linear and the light is not exactly sinusoidal. Mathematically this can be handled by introducing some amount of higher harmonics. (I like your question. Most people do not realize that usage of sine waves is not entirely trivial). – Ján Lalinský Apr 15 '14 at 6:09
@DumpsterDoofus Thank you for the follow up. I'll have to refresh myself on wave optics, it's been a while. I think I have found a way to phrase my initial question properly: If a photon / electromagnetic wave travels through a point in space, in vacuum, and we measure the electric and magnetic fields at this point with a very high temporal resolution, would we measure the electric and magnetic fields to fluctuate exactly as sine waves, or as something else? Or is it impossible to make such a measurement for some reason? – user44558 Apr 18 '14 at 1:09
@user44558: For a single photon you have to do quantum optics to really get a good picture of what's going on, which I'm not well-versed in. However, for classical (ie, many photons) monochromatic light, the answer is yes, they are sine waves. This can in fact be experimentally measured, see for example the Wikipedia page on optical correlators and how they can be used to time-resolve a light electric field. When I briefly worked in an ultrafast lab, these wave measurements were a fairly routine measurement. – DumpsterDoofus Apr 18 '14 at 1:52

So, my best understanding:

The basic solution to the wave equation is $$\Psi(x,t)=Ae^{ikx-i\omega t}$$ Where the signs are arbitrary. If you combine this with the good old Euler Formula this expands to $$\Psi(x,t)=A\cos(kx-\omega t)+B\sin(kx-\omega t)$$ Where the imaginary part is absorbed into that B

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This is one particular solution to the wave equation. Light can sometimes be modeled as a wave, what empirical evidence is there that this solution is the most accurate model of such a wave? Is there any evidence at all? How do we know whether a monochromatic electromagnetic wave resembles more a sine wave, a square wave, a sawtooth, a wavelet, ... . You didn't seem to understand my question – user44558 Apr 15 '14 at 2:29
Okay. So this is the most basic equation. If we assume there are no free charges and that we are moving in vacuum, this is the solution to the wave equation. So it seems to be you are asking why we model light as a sinusoidal wave instead of anything else. So the answer to that is because the simple solution is sinusoidal, as I mentioned. If you have a square wave or a sawtooth or whatever, as long as it satisfies the wave eq, it works. – yankeefan11 Apr 15 '14 at 2:54
However, remember that you can decompose all of these waves into sine and cosines (fourier). But seeing as you asked for monochromaic waves, we only have one component. Shine a laser into a spectrometer and you will get a spike. The fourier transform of that delta is what? A sine wave – yankeefan11 Apr 15 '14 at 2:54