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I apologize if my question does not make sense.(I'm teaching myself microscopy.)

So reading Fundamentals of Light Microscopy and electronic imaging by Douglas&Murphy, at one point the author mentions that light can be described as electromagnetic waves and then gives a picture similar to this one

http://en.wikipedia.org/wiki/Electromagnetic_radiation#mediaviewer/File:Electromagneticwave3D.gif.

Later on, when describing the "quality of light"(I quote him) he writes:

The kinds of light most frequently referred to in this text include:

• Monochromatic. Waves having the same wavelength or vibrational frequency (the same color).

• Polarized. Waves whose E vectors vibrate in planes that are parallel to one another. The E vectors of rays of sunlight reflected off a sheet of glass are plane parallel and are said to be linearly polarized.

• Coherent. Waves of a given wavelength that maintain the same phase relation­ ship while traveling through space (laser light is coherent, monochromatic, and polarized).

• Collimated. Waves having coaxial paths of propagation through space—that is, without convergence or divergence, but not necessarily having the same wave- length, phase, or state of polarization. The surface wavefront at any point along a cross-section of a beam of collimated light is planar and perpendicular to the axis of propagation.

My confusion stems from the paragraph when the author describes polarized light. In the general form of the waves, given in the picture, the electric field vectors are parallel. So, all light should be polarized, right?

Also, regarding the image describing the waves, does the pink arrow(which is a straight line), which is parallel to the x-axis(Am I right?) describe the path of a photon?

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    $\begingroup$ I'm pretty sure that the path of the photon is the pink arrow labelled elelctromagnetic wave di[rection]. $\endgroup$ – rodrigo Nov 13 '14 at 21:06
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    $\begingroup$ Think of it this way: that diagram you see is the wave for a single photon. The light that you see is the combination of many waves, pretty much all at different angles. $\endgroup$ – Kyle Kanos Nov 13 '14 at 21:21
  • $\begingroup$ @KyleKanos But in a beam of light don't all photons travel in the same direction? $\endgroup$ – shooting-squirrel Nov 13 '14 at 21:23
  • $\begingroup$ @shooting-squirrel: Are all the photons emitted from the sun traveling the same direction? $\endgroup$ – Kyle Kanos Nov 13 '14 at 21:24
  • $\begingroup$ @KyleKanos Ok, so photons are emmited radiantly from the source(always)? $\endgroup$ – shooting-squirrel Nov 13 '14 at 21:28
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The only requirement for light is that the electric field must be perpendicular to the magnetic field at any given point in time or space. This assumption arises naturally from Maxwell's equations.

The most intuitive way of thinking about light is with the picture you included of the light wave. However, you have to imagine an infinite number of light waves like the one pictured happening everywhere simultaneously. For simplicity just multiply your picture by several times and imagine several light waves travelling parallel to one other. If the electric field for all of the light waves is oriented in the same direction, you get something that looks like linearly polarized light.

However, since light is a wave, you can add together the electric and magnetic fields of all of these waves. You can create circular or elliptical of light by phase shifting axis of the electric field.

When physicists talk about about "unpolarized" light, they mean that the light is too unorganized to constitute using the word polarization. Imagine the same picture of several light waves travelling in parallel, but with the direction of the electric field for each wave randomly oriented. The sum of all the electric fields for all these waves would have a direction, but it would change very quickly due to random changes in the emission of your source light.

Incoherent light is produced by most everyday things since light is produced by the oscillations of atoms doing things independently. Most natural light is mostly unpolarized or incoherent except for reflecting off a material at non-normal incidence (See: Fresnel equations). Polarized light is commonly created with lasers or lcd screens or by using oblique incidence on a medium (see Brewster's Angle).

Wikipedia describes it well:

"Radiation is produced independently by a large number of atoms or molecules whose emissions are uncorrelated and generally of random polarizations. In this case the light is said to be unpolarized. This term is somewhat inexact, since at any instant of time at one location there is a definite direction to the electric and magnetic fields, however it implies that the polarization changes so quickly in time that it will not be measured or relevant to the outcome of an experiment. "

http://en.wikipedia.org/wiki/Polarization_(waves)#Unpolarized_and_partially_polarized_light

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The book is imprecise because there are other types of polarized light. Now, consider a point in the path of the wave, like the green point in animation you mentioned. In that image, the light is linearly polarized so the electric field draws a line at the location of the point. However, there is also circularly polarized light where the electric field draws a circle as in this picture enter image description here

Then, light is said to be polarized if the electric field at a given point draws a line, a circle, or in general, an ellipse. Unpolarized light occurs when you have many of these waves superimposed, which, results in an electric field that oscillates randomly in all directions.

This occurs whenever the source is a set of electrons which themselves oscillate in random directions such as in a hot wire.

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Under normal conditions, each photon can be thought of as a purely coherent entity with a definite polarization at all points in momentum (wavenumber) space. I discuss this notion in more depth in several other answers, notably this one here but the essential idea is this: lone photons propagate following Maxwell's equations (which are pretty much the Dirac equation for a massless, spin 1 particle), so a pure quantum one-photon state is specified by the complex quantum superposition weights of the two cicularly polarization eigenstates at each value of the wavevector $\vec{k}$ in momentum space.

So, at this one-photon level, yes - complicated entanglements aside - each photon interferes only with itself (Dirac's almost correct famous statement) and it has definite polarization, albeit a very complicated one (different superposition weights across a broad spectrum of frequencies).

Most natural sources however output these fundamentally coherent entities in a wide spread of different pure quantum states. So most light is in a mixed state (see also the Wikipedia articles on the Density Matrix and the Wigner's Friend thought experiment) : a classically probabilistic mixture of pure one photon states. This is the easiest way to think of depolarized light, whose classical description gets very involved indeed (Born and Wolf give a whole chapter to it - or at least to the classical description of the highly related notion of the partially coherent light). The full quantum description involving entanglement is also very thorny, but the basic intuitive notion of the mixed state is a much easier way than the classical description to understand depolarization.


Some answers to user Phonon's comment:

"very interesting take! [This post (Physics SE Question "Difference Between Spin and Polarization of a Photon")] may also interest you. (it still lacks a straight answer I'd say.) –"

I think that question shows that "polarization" - like all natural English (or German, Mandarin - any natural language) words - can become a bit vague and hence the need to encode one's ideas in mathematical language. The best definition of polarization I can think of off the top of my head is:

  1. Polarization is a word that, in its strictest sense, applies to pure quantum states; it applies most readily to one-photon Fock states or quantum pure coherent states;

  2. One specifies a one photon Fock state's polarization by defining the complex superposition weights of the left and right angular momentum eigenstates at each and every point $\vec{k}$ in momentum space. So polarization, in general, is a function of the form $\mathbb{R}^3\to\mathbb{C}\times \mathbb{C}$, i.e. two complex valued functions of the momentum space $\mathbb{R}^3$. Note that in general we are not talking about a monochromatic field; monochromatic fields look like spherical shells (defined by $c|\vec{k}| = \omega$) in momentum space. Now;

  3. We can broaden our polarization definition to partially polarized light by thinking of classical mixtures of pure quantum states. So now our partially polarized state becomes a density matrix function $\mathbb{R}^3\to \mathcal{M}(2,\,\mathbb{C})$ assigning a $2\times2$ Hermitian density matrix holding the Stokes parameters for each $\vec{k}\in\mathbb{R}^3$ momentum space.

Now then, the word "spin" has two, related meanings. The first is the spin state of a photon is defined in terms of the superposition weights of the eigenfunctions of the spin observables. If you consider the spin observable for the component of the angular momentum in the direction of $\vec{k}$ at all points in momentum space and write the photon's state as superpositions of the eigenstates of this observable, then you are indeed computing precisely the polarization as I have defined it above. So in this sense, spin and polarization are pretty much synonymous - or at least that's my understanding of the English (natural language) usage.

The second meaning of the word "spin" is as Lionel Brit's answer to that question discusses: when we say the photon has "spin 1" we are simply saying its state (a solution of Maxwell's equations) transforms in a certain way under the action of the Lorentz group. "Spin 1" names a certain representation of the Lorentz group.

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    $\begingroup$ +1, very interesting take! This post may also interest you. (it still lacks a straight answer I'd say.) $\endgroup$ – Phonon Mar 22 '15 at 11:59
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    $\begingroup$ @Phonon See my additions to the end of my answer. $\endgroup$ – WetSavannaAnimal Mar 23 '15 at 1:07
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    $\begingroup$ @Phonon BTW I hadn't heard of Hélène Grimaud: her intonation is breathtaking. Are you from France? $\endgroup$ – WetSavannaAnimal Mar 23 '15 at 1:18
  • $\begingroup$ Thanks for taking the time to write this up, pretty much got my answer. I agree with you on Grimaud, she really puts a lot of emotion into her playing (probably because she plays Brahms so much :), make sure to check them out) and gives a unique performance each time (leaving you with a different experience every time). She has a remarkable sense of legato and accentuation. You may also enjoy these: 1, 2, 3. (BTW I'm not french, but do live in France). $\endgroup$ – Phonon Mar 23 '15 at 10:02
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The picture of the wave you are looking at is already polarized and that is plane polarized light. But this is not the general case, waves emitted by any one molecule may be linearly polarized but an ordinary light source contains large number of molecules with random orientations,so the emitted light is random mixture of waves linearly polarized in all possible transverse directions. so light from ordinary light source is unpolarized.

So all light is not polarized ,you need a filter to creat polarized light from the light from ordinary sources. Also in addition,as @Asaf said there are also other types of polarization not only linear polarization. http://en.m.wikipedia.org/wiki/Polarization_(waves)

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See Wikipedia "Unpolarized and partially polarized light". It's written there

"Most common sources of visible light, including thermal (black body) radiation and fluorescence (but not lasers), produce light described as "incoherent". Radiation is produced independently by a large number of atoms or molecules whose emissions are uncorrelated and generally of random polarizations. In this case the light is said to be unpolarized."

(About the pink line, it shows indeed the direction of propagation, but the text is truncated.)

Good luck,

Sofia

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