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In Kim experiment https://en.wikipedia.org/wiki/Delayed-choice_quantum_eraser I take out all the beam splitters, mirrors, and detectors for idler photons. But I leave idler photons travel freely in the universe. I look in D0 after I detect the last signal photon and I detect formless blob. I do not detect interference pattern even though I have no "which path" information just because In Principle "which path" information is incorporated in idler photons that travel and can be obtained any time in the future if one decides to. So, I look at D0 and definitely I SEE formless blob. Now, a year later I decide to put in only mirrors and detectors for IDLER photons(no beam splitters) that will provide me with "which path" information for every idler photon. After I detect the last idler photon I decide to look at D0 again. Will I see only two fringes on D0? If yes, does it mean that I have changed the past( the formless blob)? If No, does it mean that just because I looked at D0 a year ago and saw formless blob I will see a formless blob again(I cannot change the past)? What if I have not looked at D0 a year ago...will I see two fringes now?

Qmechanic
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1 Answers1

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There are a few misconceptions. When you look at D0 you see a formless blob.

The reason for this is two things:

  1. you created idler photons (this is where I disagree, but that is the accepted theory). In my reasoning, you need to create idler photons with SPDC, and interact with them. But the accepted theory is that you just need to create them (idler photons).

  2. you need to analyse and select (post selection) photons out of the D0 detections

During an experiment, detector D0 is scanned along its x axis, its motions controlled by a step motor. A plot of "signal" photon counts detected by D0 versus x can be examined to discover whether the cumulative signal forms an interference pattern.

https://en.wikipedia.org/wiki/Delayed-choice_quantum_eraser

If you want to see the pattern, you need to delete the idler photons, and do selection from D0 detections.