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Using multiple polarizers (some of them beam-splitting to minimize losses) and a spiral alignment (more like this), one can, I gather, set the polarization of the majority of passing photons to a desired orientation.

What does this do to such photons' polarization-entangled twins?
Do they remain entangled?
What becomes of their polarization? Randomized?

kaay
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1 Answers1

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There's multiple ways to affect the polarization of a photon.

  • The mechanism you in the answer you lined to is a projective measurement of the polarization of one photon, followed by further projective measurements on that same photon. In that case, only the first measurement has any bearing on subsequent measurements of the polarization of the entangled counterpart.

    Thus, if the photons are in the polarization singlet state $|HV\rangle - |VH\rangle$, and the initial polarizer put in front of photon 1 is vertical (to be followed by diagonal and horizontal ones) then (in coincidence with a measurement of photon 1) the other photon will always be detected with a horizontal polarization.

  • On the other hand, there are mechanisms that act as unitaries, the simplest of which is a half-wave plate or a stretch of optically-active medium. This will have no effect on the entangled counterpart, or on the entanglement between them.

It's also important to note that there is a wide variety of types of entanglement between photons; this can be in polarization (so e.g. $|HV\rangle - |VH\rangle$ is different from $|VV\rangle + |HH\rangle$) but they can also be entangled in frequency, timing, angular momentum, direction, and so on, all of which are independent of polarization.

It's also a bit of a misstatement to say that your 'spiral' arrangement of polarizers sets the polarization of the 'majority' of the photons: it will determine the polarizations of all the photons that pass, but it will have a transmittance of 12.5%.

Emilio Pisanty
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  • I'll have to read up on what you wrote - thanks for the direction. How'd you get 12.5%? If they were only perfect absorptive polarisers, I thought that would be 0.5 after the first; times... cos(45\deg)^2 if they were at 45 degrees, or cos(22.5\deg)^4 if half that. 0.25, 0.36 and higher the more steps there would be. If the first one were a beam splitter, we wouldn't even lose the initial 50% (I think?). – kaay May 12 '17 at 10:56
  • And thank you for leading me to notice my mistake in the understanding of entanglement and results of measurement. Had to be pointed out. Speaking of which, I'm still curious what I missed about your 12.5% – kaay May 12 '17 at 11:46
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    @kaay That 12.5%=1/8 is a factor of 1/2 for the first polarizer, and then one factor of $\cos^2(45°)=1/2$ each for the subsequent ones. – Emilio Pisanty May 12 '17 at 13:44
  • Right. Malus' law. I got misled by this, which didn't use the square. – kaay May 12 '17 at 16:18