Are circular polarizations a basis for any light polarization? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-21T02:50:13Z https://physics.stackexchange.com/feeds/question/43042 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/43042 2 Are circular polarizations a basis for any light polarization? Learning is a mess https://physics.stackexchange.com/users/14405 2012-10-30T13:21:41Z 2012-11-20T00:26:06Z <p>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.</p> <p>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.</p> https://physics.stackexchange.com/questions/43042/-/43045#43045 2 Answer by Kitchi for Are circular polarizations a basis for any light polarization? Kitchi https://physics.stackexchange.com/users/14473 2012-10-30T14:16:29Z 2012-11-20T00:26:06Z <p>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.</p> <p>First, the following is a general expression for two waves that are out of phase : $$E_{1} = E_{0x} \cos(kz - \omega t)$$ $$E_{2} = E_{0y} \cos(kz - \omega t + \phi)$$</p> <p>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, </p> <p>$$E_\text{circ} = E_{0x}(\cos(kz-\omega t) + \sin(kz-\omega t))$$</p> <p>Conversely, you can just as easily see how you can get linearly polarized light from two circular polarizations moving in opposite directions.</p> <p>$$E_\text{lin} = E_{0}(\cos(kz - \omega t) + \sin(kz - \omega t)) + E_{0}(\cos(kz - \omega t) - \sin(kz - \omega t))$$ $$\implies E_\text{lin} = 2E_{0} \cos(kz - \omega t)$$</p> <p>You could try this with two orthogonal elliptically polarized waves to convince yourself further.</p>