# Are circular polarizations a basis for any light polarization?

I was trying to solve an exercise in classical optics, consisting in finding the type polarization of certain field profiles. And, by analyzing this one: $$E_x(z,t) = |E|sin(kz-\omega t)\\ E_y(z,t)=|E|sin(kz-\omega t +\frac{\pi}{4})$$ I first found that it was the superposition of a elliptical polarization and a linear one by splitting the $sin$ in two, then I plotted it and discovered that in fact it's an inclined ellipse. It brought me to the conclusion that superposing a defined longitudinal and an elliptical gives another elliptical one. Now, I recall hearing that the elementary polarization are the two circular ones, which should imply that any elliptical or longitudinal can be decomposed into a sum of circular polarization, am I right? Because formally it's not evident to me.

ps: Also, what do the terminology TM and TE labeling the polarization of single photons refer to? I encountered them a few times, but they were never defined.

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To answer your first question - Basically, you can use any two orthogonal polarizations as your 'basis' to describe any other polarizations. Like light polarized along $x$ abd $y$, or light that is circularly polarized in opposite directions.

First, the following is a general expression for two waves that are out of phase : $$E_{1} = E_{0x} \cos(kz - \omega t)$$ $$E_{2} = E_{0y} \cos(kz - \omega t + \phi)$$

where $\phi$ is the phase difference between the waves. These would be two linearly polarized waves. If they are $\pi / 2$ out of phase with each other, and you superpose them, the resultant will be circularly polarized light. If $E_{0x} = E_{0y}$ then,

$$E_\text{circ} = E_{0x}(\cos(kz-\omega t) + \sin(kz-\omega t))$$

Conversely, you can just as easily see how you can get linearly polarized light from two circular polarizations moving in opposite directions.

$$E_\text{lin} = E_{0}(\cos(kz - \omega t) + \sin(kz - \omega t)) + E_{0}(\cos(kz - \omega t) - \sin(kz - \omega t))$$ $$\implies E_\text{lin} = 2E_{0} \cos(kz - \omega t)$$

You could try this with two orthogonal elliptically polarized waves to convince yourself further.

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TM and TE: not quite. If light is incident on a surface, then no transverse polarization is ever perpendicular to that surface. – ptomato Nov 2 '12 at 13:06
@ptomato - I'm not really clear on how TM and TE are defined, I'll probably edit that out of my answer. – Kitchi Nov 2 '12 at 13:30
As far as I know, TM and TE are defined in terms of waveguides. For more info, see Wikipedia: en.wikipedia.org/wiki/Waveguide_%28electromagnetism%29 – ptomato Nov 2 '12 at 15:47
I think if you see people using TM and TE to mean s and p polarization, it's probably wrong (although I admit to having done this myself in the past.) – ptomato Nov 2 '12 at 15:48
S is for German 'senkrecht'(perpendicular) and P for German 'parallel' (parallel), both referring to the plane of incidence. TM and TE (transverse magnetic and transverse electric) are equivalents to that. en.wikipedia.org/wiki/Polarization_(waves)#S_and_P_Polarization – roadrunner66 Mar 12 '13 at 8:31