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Entanglement Rotation using a Faraday Rotator

  1. Mar 31, 2013 #1
    I've tried E-Composing this so many times and wadded up so much E-Notebook Paper I could fill up an E-Room with E-Paper Balls. What follows is a simple version:

    1. Assume two entangled particles emerging from a common source.
    2. There are two detectors separated by a great distance. Detector A will have the usual detection capabilities and the particle that travels to Detector A has an apparent free path to Detector A.
    3. Detector B is similar to Detector A. In front of Detector B, however, is a "Faraday Rotator" (See http://en.wikipedia.org/wiki/Faraday_effect and SciAm, "Centaurus A", November 1983 - Nice picture with statement, "The degree of rotation is proportional to the wavelength..."). The Faraday Rotator changes the angle of polarization.
    4. We turn on the Source and Detectors, leaving the Faraday Rotator off. Since we are good programmers, we build things so that we may let the computer find an alignment for the Detectors such that a Screen, apparently correctly, reads, "Entangled Particles Detected! Detectors are aligned!".
    5. We turn on the Faraday Rotator and rotate the polarization of Particle B by some amount, θ.

    6. Where do we set Detector B? Does anything change? If the Detectors were "in alignment" before, are they "in alignment" now? If the results indicate "Out of Alignment", do we rotate Detector B the same angle θ and re-establish alignment? As we map the angles and results, what does QM tell we will find?
    7. Important question for me: Does the Faraday Rotator alter the Polarization without determining the State of the Polarization?

    8. When I ran this experiment, I added another screen that was displayed at the end of the run: "Don't forget to pick up some aspirin on the way home. You've probably got a headache".

    Thanx,

    CW
     
  2. jcsd
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