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Many experiments with entangled photons are sending them through different glass fiber cables (e.g. in opposite directions for spatial separation). The photons will inevitably be reflected many times in the cable before reaching the end of the fiber, correct? Why is this interaction with the optical barrier layer not a measurment, which to my understanding would destroy the entanglement?

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Position and spin are independent degrees of freedom. So some process equivalent to a measurement about position does not destroy the spin entanglement. –  Trimok Jun 18 at 8:18

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The interaction is not a measurement because the probability that it will produce a measurable change in the momentum of the reflecting object is extremely small. The reflector is in a mixed state in which its momentum has a range of values that is large compared to the momentum of the photon. So the shift in the mirror's momentum as a result of the reflection will be so small that it will not be detectable with very high probability. I don't know the exact numbers but let's say it's -1,000,000 to +1,000,000 in units of the momentum the photon will impart upon reflection. The photon is incident on the mirror and changes its momentum by +1 so that the range is now -999,999 to +1,000,001. So the probability is large that if you measure its state you won't detect any difference.

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I was under the impression that "measuring" is the pure act of interaction with other particles, no matter if there is some mind or apparature behind this which would assesses the quality of interaction. –  Vroomfondel Jun 20 at 14:08
    
The issue is not whether there is a mind it is just whether there is a record. You can have a record of something without anybody seeing it. Fossils were records of animals long before anybody dug them up. In the case of the mirror the issue is just whether the photon reflecting from the mirror generates a record and it doesn't with high probability. –  alanf Jun 20 at 14:33

It is a physical characteristic of reflection that there is no interaction with the photon which would destroy the photon and inhibit its further propagation.

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"It doesn't interact because it doesn't interact." That's not much of an explanation. –  Zack Jun 18 at 13:34
    
... it's as much of an explanation as you usually get in QPh. –  vaxquis Jun 18 at 13:37

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