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Delayed Quantum Erasure Double Slit with Observer Choice Available

  1. Sep 28, 2010 #1
    In the paper "Double-Slit Quantum Eraser" by Walborn, Cunha, Padua, and Monken (see http://grad.physics.sunysb.edu/~amarch/Walborn.pdf [Broken] and discussion of the paper at http://grad.physics.sunysb.edu/~amarch/ [Broken]), a double-slit quantum erasure experiment is performed with a delayed eraser. When the polarizer POL1 and Detector Dp are moved away from the beta-barium borate (BBO) crystal such that s-photon hits the detector Ds before the entangled p-photon hits the POL1 and Dp, an interference pattern is observed due to delayed quantum erasure. As such, one would surmise that the POL1 has some faster-than-light influence on the p-photon, which ultimately affects the entangled s-photon. The authors discuss that there was no choice available to an observer to influence the result in the time period after the detection of the s-photon and before the detection of p-photon. What would happen if there was a choice to affect the outcome after the s-photon hits Ds, but before the p-photon hits the POL1?

    That is, assume POL1 and Dp are moved even further away from the BBO. Assume also that (1) after the s-photon hits the Ds and the s-photon strikes the Ds consistent with interference, the POL1 is moved away from the path of the p-photon before the p-photon hits the POL1, and (2) after the s-photon hits the Ds and the s-photon strikes the Ds consistent with no interference, the POL1 is maintained in the path of the p-photon such that the p-photon hits the POL1. What will happen?

    In (1), we have a situation in which the s-photon hits the Ds consistent with interference, but the POL1 is moved before the p-photon hits the POL1 so that the polarity of the p-photon can be determined by the detector Dp. As such, the polarity of the s-photon can be determined and therefore the slit the s-photon went through can be determined. Such a result would imply that the s-photon would strike Ds consistent with no interference. In (2), we have a situation in which the s-photon hits the Ds consistent with no interference even through there is a delayed quantum erasure. Such a result would imply that delayed erasure does not work in this case.

    Is it possible that (1) cannot happen and that there are limits to the delayed erasure in (2)? That is, if we have the ability to move the POL1, will the s-photon always strike Ds consistent with no interference (because if the polarity of the p-photon were to be detected, the s-photon cannot possibly strike the Ds consistent with interference)? If there is faster-than-light influence on the p-photon, then one would expect that the s-photon would always strike the Ds consistent with interference, as one could not move the POL1 out of the way fast enough. But then, the Dp would tell us the polarity of the p-photon, and hence the polarity of the s-photon could be observed, which would allow one to know the slit the s-photon passed through, which would prevent the s-photon from striking the Ds consistent with interference.

    Or, is it possible that (2) cannot happen and that entanglement is destroyed by the movement of the POL1? That is, will the s-photon always strike the Ds consistent with interference? If that is the result, then (2) would not occur and therefore no limits on delayed quantum erasure would be implied. However, such a result would imply that moving the POL1 would destroy the entanglement of the photons, as determining the slit that the s-photon passed through should not then be possible given that the s-photon struck Ds consistent with interference. What is strange about this result is that the entanglement is destroyed after the s-photon strikes the detector, but before the polarity of the p-photon can be detected.
     
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  2. jcsd
  3. Sep 28, 2010 #2

    zonde

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    You have to take in account what is observed pattern of Ds detections if we completely ignore detections at Dp. Observed pattern is no interference pattern. So interference pattern can be observed only by means of postselection.

    For more realistic picture you can implement POL1 as fast switching polarizer that switches between 0° and 45° orientations. With 0° orientation with quarter wave plates before double slit we will observe no interference and with 45° orientation we will observe interference.

    Now because without postselection Ds is observed as having no interference it means that for the case of interference (after postselection) more detections of Ds at minimum are discarded as a result of postselection.

    Now we can easily conclude that you will have reduced detection rate at Dp for your case (2) if you pick detections at interference minimum as implementation of case (2) (that's needed because average detections at Ds are consistent with no interference).
     
    Last edited: Sep 28, 2010
  4. Sep 29, 2010 #3
    Thank you Zonde for responding. I'm not really sure whether you answered my question. I didn't fully understand what you meant by "discarded" with respect to "more detections of Ds at minimum (interference) are discarded as a result of postselection." For case (2), I agree there should be reduced detections or actually, no linearly y polarized or linearly x polarized detections, as the POL1 is maintained in position.

    I suppose that given the experimental setup of Walborn et al., an observer could not possibly know whether any given s-photon strikes Ds with interference or no interference. So, perhaps I can state my question differently. Assume the same experimental setup by Walborn et al. For the delayed erasure with QWP1, QWP2, and POL1 in place, the distribution of s-photon is an interference pattern as shown in FIG. 8 and FIG. 9. Now, if we were to run the experiment again, we would see that same interference pattern. Now, assume that immediately after the s-photon strikes Ds (which after tabulated will show an interference pattern), we change the fast polarizer orientation from 45° to 0° (and therefore allow the observer to choose to observe the which-path information -- see discussion in section B of paper). In that case, the POL1 would not change the linearly (x or y) polarized p-photon to a polarization with both x and y components. What would we see at the detector Dp? That is, would we see a photon that is linearly x polarized or linearly y polarized? If so, what would that tell us? Or, would we see a photon that is polarized with both x and y components?
     
  5. Sep 29, 2010 #4

    JesseM

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    In the delayed choice quantum eraser, no interference pattern will be seen in the total pattern of "signal" photons on the screen, regardless of whether you record or erase the "which-path information" from the "idler" photons. If you do choose to erase the information, you do it by forcing the idlers to go to a certain pair of detectors, and only by looking at the subset of signal photons whose idlers went to one of the two detectors (a 'coincident count' between signal photons and idlers at a particular detector) can you recover an interference pattern. There have been a few threads on the delayed choice quantum eraser (DCQE), see here for example. Here was my more detailed explanation from an earlier thread:

    Even in the case of the normal delayed choice quantum eraser setup where the which-path information is erased, the total pattern of photons on the screen does not show any interference, it's only when you look at the subset of signal photons matched with idler photons that ended up in a particular detector that you see an interference pattern. For reference, look at the diagram of the setup in fig. 1 of this paper:

    http://arxiv.org/abs/quant-ph/9903047

    In this figure, pairs of entangled photons are emitted by one of two atoms at different positions, A and B. The signal photons move to the right on the diagram, and are detected at D0--you can think of the two atoms as corresponding to the two slits in the double-slit experiment, while D0 corresponds to the screen. Meanwhile, the idler photons move to the left on the diagram. If the idler is detected at D3, then you know that it came from atom A, and thus that the signal photon came from there also; so when you look at the subset of trials where the idler was detected at D3, you will not see any interference in the distribution of positions where the signal photon was detected at D0, just as you see no interference on the screen in the double-slit experiment when you measure which slit the particle went through. Likewise, if the idler is detected at D4, then you know both it and the signal photon came from atom B, and you won't see any interference in the signal photon's distribution. But if the idler is detected at either D1 or D2, then this is equally consistent with a path where it came from atom A and was reflected by the beam-splitter BSA or a path where it came from atom B and was reflected from beam-splitter BSB, thus you have no information about which atom the signal photon came from and will get interference in the signal photon's distribution, just like in the double-slit experiment when you don't measure which slit the particle came through. Note that if you removed the beam-splitters BSA and BSB you could guarantee that the idler would be detected at D3 or D4 and thus that the path of the signal photon would be known; likewise, if you replaced the beam-splitters BSA and BSB with mirrors, then you could guarantee that the idler would be detected at D1 or D2 and thus that the path of the signal photon would be unknown. By making the distances large enough you could even choose whether to make sure the idlers go to D3&D4 or to go to D1&D2 after you have already observed the position that the signal photon was detected, so in this sense you have the choice whether or not to retroactively "erase" your opportunity to know which atom the signal photon came from, after the signal photon's position has already been detected.

    This confused me for a while since it seemed like this would imply your later choice determines whether or not you observe interference in the signal photons earlier, until I got into a discussion about it online and someone showed me the "trick". In the same paper, look at the graphs in Fig. 3 and Fig. 4, Fig. 3 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D1, and Fig. 4 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D2 (the two cases where the idler's 'which-path' information is lost). They do both show interference, but if you line the graphs up you see that the peaks of one interference pattern line up with the troughs of the other--so the "trick" here is that if you add the two patterns together, you get a non-interference pattern just like if the idlers had ended up at D3 or D4. This means that even if you did replace the beam-splitters BSA and BSB with mirrors, guaranteeing that the idlers would always be detected at D1 or D2 and that their which-path information would always be erased, you still wouldn't see any interference in the total pattern of the signal photons; only after the idlers have been detected at D1 or D2, and you look at the subset of signal photons whose corresponding idlers were detected at one or the other, do you see any kind of interference.
     
  6. Sep 30, 2010 #5

    zonde

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    No, distribution of s-photon is not interference pattern. Only coincidences from Ds and Dp make interference pattern. Single detections at Ds do not make interference pattern in any setup of this experiment.

    JesseM is talking about a bit different experiment but the "trick" is just the same. Interference pattern is seen only for coincidences of certain detector pairs.
     
  7. Oct 1, 2010 #6
    Zonde, thanks for answering my questions. Also, JesseM, thanks for getting at the root of my misunderstanding. The experiment in that paper now seems intuitive and quite clear that, at least with this experiment, one cannot affect the past with a future action, as the interference pattern is simply recovered through coincidences between subsets of the p and s photons.
     
  8. Nov 8, 2010 #7
    OK, I now understand that experiment (I think), but something is bothering me. If you replaced D1 and D2 with a single detector (located at "BS"), you apparently would not see any interference pattern for those coincidences at all (the superposition of R01 and R02). I don't understand why that would be. Why do you get a pattern when looking at erased information that has a 50% probability of going to D1 or D2, while you don't get any pattern if they all go to a detector located at BS? In the latter case, you still don't have any information about the path, so I would expect an interference pattern. And certainly if you then replaced BSA and BSB with mirrors as well, so everything ends up in the same detector. Actually, you might as well remove all of the detectors and mirrors, and there would still be no interference. Right?

    In the original double slit experiment, you got an interference pattern when you did not add any detectors, and the pattern disappeared when you merely added a detector to figure out which photons went through which slit. This seems to be in complete contradiction with what I understood from the erased information experiment.

    In fact, the only explanation seems to be that, if you set up an experiment that allows you to delay the decision whether or not to measure stuff (or if you try something like my Alice and Bob FTL experiment), nature will go "nice try, kiddo" and remove all interference patterns. Is that what would happen?
     
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