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I am thinking about the property of circular polarization of light. I have two circularly polarized beams, one is rounding in positive direction, other is negative. My task is to figure out:

  1. if those two polarized beams are circular (I can use circular-polarizer detector to do so).
  2. i need to verify if those two polarized beams are just opposite. I don't have any idea but I am thinking shall I add those two beams, if they are circular but opposite, then the amplitude of the superposed beam should real (since the imaginary part cancel out). However, at this step, I don't know how to use that information to determine if the beams are two opposite circular polarization?
  3. To add two beams, will it work by just shining two beams onto the same photo detector? or I have to use some adder to add the beams first?
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There are a couple of things to consider. In your detection measurements, you are looking at energy, so polarization states are degenerate in this measurement. Polarization degree of freedom carries (loose terminology) angular momentum and a direct measurement of the angular momentum can be easily seen in light-matter interactions (my experience). To distinguish between polarizations, the easiest way is by using a high quality crossed polarizer arrangement. What wavelength are you working with? –  Antillar Maximus Dec 23 '12 at 15:37
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2 Answers

Look up Stokes Vectors and Mueller Matrices to treat these problems. Edward Collett's "Polarized Light" is a great book with a very simple, intuitive treatment.

To look at the combination of the two beams, they'd have to be at exactly the same frequency (OK if they come from the same laser) and phase, which is very hard to do.

The easiest is to transform the beams with waveplates, which are likely already in your setup. A quarter waveplate will transform a circularly polarized beam into a linearly polarized one.

Just like your two circularly polarized beams are expected to be orthogonal in the base of circular polarization (cw vs ccw), the two resulting beams after the waveplate will be orthogonal in the common sense.

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This won't be a complete answer but I can explain an answer for 2): If the two beams are just opposite, and we take that to mean they are identical except for their direction of the circular polarization, then the resulting wave is linearly polarized and has twice the original amplitude. To see this, we parameterize the electric field vector created by each of the two waves.

Wave 1 LH polarization

$E_{x_1}(t)= Acos(\omega t) $

$E_{y1}(t)= Asin(\omega t) $

Wave 2 RH polarization

$E_{x_2}(t)= Acos(-\omega t) = Acos(\omega t) $

$E_{y_2}(t)= Asin(-\omega t) = -Asin(\omega t)$

By the superposition principle, we can find the total field by summing the components of the two waves

$ E_x=E_{x_1}+E_{x_2}=2Acos(\omega t) $

$E_y=E_{y_1}+E_{y_2}=0$

Note that we can select a reference frame in which the waves' electric field vectors coincide at y=0 and t=0. This link has a decent animation of this.

As for the other parts of your question I am not entirely sure. If you can detect the circular polarization, haven't you already answered 1) yourself? As for 3), theoretically there is no difference between a linearly polarized beam and a superposition of circularly polarized beams of opposite frequencies. You should get the same result using either method, though there may be complications I haven't thought of:)

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