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I read in a book the following assertion.

In a double slit experiment photons are passed through the slits and detected at the end plate.

Each of the two slits has a quarter wave plate which alters the polarization of the photons that pass through it in a way different than the other QWP.

Thus a polarizing detecting barrier at the end plate can determine which slit the photon went through.

In such an experiment, there will be no interference pattern at the end plate. i.e. the wave functions collapse.

But if one does either of two things, the interference pattern shows up. Thus if either:

  1. the quarter wave plates are removed but the polarizing detecting barrier is kept.

OR

  1. the polarizing detecting barrier is removed but the quarter wave plates are kept.

Then the interference pattern is back.

A. Is this assertion correct?

B. If yes, then where does the wave function collapse when both are in place? At the plates or the end detector?

here's a quote from that book (Biocentrism ch.8 scroll down to "The Most Amazing Experiment"):

We’re back to QT’s complementarity – that you can measure and learn just one of a pair of characteristics, but never both at the same time. If you fully learn about one, you will know nothing about the other. And just in case you’re suspicious of the quarter wave plates, let it be said when used in all other contexts, including double slit experiments but without information-providing polarization-detecting barriers at the end, the mere act of changing a photon’s polarization never has the slightest effect on the creation of an interference pattern.

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  • $\begingroup$ It is a bit difficult to think about circularly polarized waves and about the superposition of waves with opposite handedness. It is easier to deal with linearly polarized light: vertically through the one slit, horizontally through the other. Adding those gives no interference pattern, $\endgroup$
    – user137289
    Commented Aug 5, 2019 at 21:17
  • $\begingroup$ @Pieter the point is that we have "which way" information when the QWPs and polarization detectors are in place $\endgroup$
    – michael
    Commented Aug 5, 2019 at 21:19
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    $\begingroup$ There will always be at least a single slit pattern present. The collapsed wave function does not mean no interference. $\endgroup$
    – Lambda
    Commented Aug 5, 2019 at 23:14
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    $\begingroup$ i broadened the question here physics.stackexchange.com/questions/495500/… $\endgroup$
    – michael
    Commented Aug 6, 2019 at 10:38
  • $\begingroup$ Maybe this experiment with electrons will help how trying to detect which slit turns into a different quantum mechanical experiment: instead of " two slits etc" it bec0mes "two slits +detector material for which way" phys.org/news/… $\endgroup$
    – anna v
    Commented Aug 6, 2019 at 11:52

1 Answer 1

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No the claim is incorrect. Assertion 2) is wrong. If the two slots pass orthogonal polarisation, linear or circular, then there is no interference at all. A polarisation sensitive detector then sees only a single slit diffraction pattern, either the left or the right one. An unpolarised detector sees a superposition of both single slit diffraction patterns without interference. Classical electromagnetism is sufficient to show this.

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  • $\begingroup$ the two slots change the polarization differently. the point is that by detecting the polarization at the end plate, you can know which slit the photon went through. are you saying this is not possible? $\endgroup$
    – michael
    Commented Aug 6, 2019 at 5:25
  • $\begingroup$ I updated the question. please see last paragraph $\endgroup$
    – michael
    Commented Aug 6, 2019 at 6:09
  • $\begingroup$ It is possible but there is no interference. $\endgroup$
    – my2cts
    Commented Aug 6, 2019 at 6:56
  • $\begingroup$ so you are saying that even though there is no "which way" information obtainable, the interference pattern is gone? $\endgroup$
    – michael
    Commented Aug 6, 2019 at 12:51
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    $\begingroup$ Yes. The which way info is in the wave, so there can be no interference. Whether you use it or not. $\endgroup$
    – my2cts
    Commented Aug 6, 2019 at 16:16

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