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Delayed choice and information erasure

  1. Jan 7, 2006 #1
    In double-slit experiment one can observe either an interference pattern or which-path information, but never both.
    Spontaneous parametric down conversion (SPDC) is used in the recent experiments to create entangled pairs of photons. One of them is used to observe the interference pattern existence and other is used to observe the which-path information. The experiment installation could be changed to allow the which-path information to be observed or to be erased. The experiments confirmed QM predictions. When the which-path information was available the interference pattern was not available and vice versa.

    The experimenters tried to “outsmart” the QM and changed the installation to make the interference pattern observation before the second photon is even reached the part of the apparatus where the which-path information is possibly erased. Nevertheless, the experiment behaved the same way. If you have an interference pattern you do not have which-path information and vice versa. It looks like the information travels back in time, or the photon anticipates what will happen.

    Now follows my question:

    Did anyone make the following experiment? The information eraser is turned off, so the which-path information is available. After the first photon is detected, but before the second photon enters the information eraser, the eraser is turned on and the which-path information is no longer available. What will be the result for interference pattern observation? What predicts QM? Do the prediction and the result agree?

    Regards,
    Upisoft
     
  2. jcsd
  3. Jan 7, 2006 #2

    JesseM

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    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://xxx.lanl.gov/PS_cache/quant-ph/pdf/9903/9903047.pdf

    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.
     
  4. Jan 7, 2006 #3
    Thank you. Your reply was very usefull. It answered the questions that puzzled me, but has opened new one. :)

    In this particular experiment one has to use both photons to reconstruct the interference pattern. I've seen documents describing another experiment. It was little different, and in this case one had to use the information from both entangled photons to reconstruct which-path information. My question now is if it's possible to use one of the photons for creating interference pattern only and the second one to carry which-path information only.
     
  5. Jan 8, 2006 #4
    JesseM thanks for your wonderful explanation. But as Upisoft says, this appears to raise more questions. Could you pls. provide a link (I assume it was on this forum) to the explanation that was given to you to demystify Scully.

    I keep on lamenting Scully, Zeilinger, Walborn and others who do such interesting research and come up with such fantastic results, put when they write on it, I think, they try to mystify it, causing humongous problems for us lays, who are trying to make sense out of these fantastic results. If only Scully had said what you have said above, and it would have saved me (and problem tens of thousands others) days of frustration and fretting.
     
  6. Jan 8, 2006 #5
    JesseM - a couple of questions - you imply that the TOTAL pattern detected by D0 is a Guassian, with the interference (and its dual) both embedded in this pattern (in addition to the Gaussians due to D3 and D4). Namely IP1 + IP2 == Gaussian. (where IP2 is the dual phase shifted interference pattern of D2). Also, I assume (correct me if I am wrong) that if BSA and BSB are replaced with mirrors, we will lose G3 and G4 (Gaussians) and we will get TOTAL = IP1 + IP2 = Gaussian.

    1- What if we remove BS, and bring the beams together to interfere on a screen. Thus we have lost which-way-info (WWI) and IP should appear. But would the dual IP appear too? It should become TOTAL = IP (namely we see an IP on D0). Simply because there is no dualism introduced by BS anymore.

    2- If we remove D1, D2, D3, D4, and BS, BSA, BSB altogether, what kind of TOTAL pattern will be registered at D0?

    Thanks in advance for your reply.

    zekise
     
  7. Feb 13, 2006 #6
    I just read about the delayed choice experiment paper by Scully et al. as well and I am glad to find a fresh discussion about it. I am also confused why we can't send information to the past by replacing the D3-D4 setup with the D1-D2 setup before the idler photons arrive and after we observe the distribution on D0.
    D3-D4 setup would cause no interference pattern (IP) at D0 since we have path info; while D1-D2 setup would cause IP at D0 since we have no path information.
    I don't understand why we need to separate photons detected by D1 and D2 to get the interference patterns. If we combine the photons received by both D1-D2, we should still get the IP since we still don't have the information which slit the photons went through.
    Thank you very much for your help.
     
    Last edited: Feb 13, 2006
  8. Feb 13, 2006 #7
    Herein lies the problem:

    You will NOT get IP on D0, no matter what you do at the idler side. You can erase, you can get path, you can delay choice, etc. and you will NOT get IP at D0. Instead, you will always see the same Gaussian at D0, no matter what.

    If you have path info at D3, D4, then the Gaussian at D0 is the sum of two Gaussians one from each slot.

    If you have erased the path info at D1, D2, the Gaussian at D0 is exactly the sum of a fringe and an anti-fringe.

    If you force an erase and record which idler photons go to D1 and which go to D2, and send this information classically to D0, then you can take the Guassian at D0, and separate it into the fringe and anti-fringe, by correlating the D0 registrations with that of D1 (fringe) and D2 (anti-fringe).

    Why do you get anti-fringe? Because of entanglement of the idler photon, which happens to be orthogonal with the signal photon, resulting in a phase shifted wavefunction. The two photons are really one elementary object. When one collapses, the registration is equally influenced by the wavefunction of the other. I understand that stochastically half the D0 registrations fall on the fringe, and half fall on the anti-fringe. At D0 you cannot tell which is which, unless you get access to the data in D1 or D2.
     
  9. Aug 4, 2007 #8

    nrqed

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    I know this is an old post but it is a fascinating one and it brings up some obvious questions. This seems to point out to a deeper level of subtlety in the quantum world (beyond the usual interference/non interference pattern story)!


    My question is about the situation in which all which-path information has been erased.

    To review the setup: the signal photons are sent to the screen. The idler photons are sent through mirrors which redirect them to a hlaf-silvered mirror beyong which there are two detectors. Once the idler is detected, there is no way path information. So one has interference patterns.

    Except that there is a twist. If I understood correctly JesseM, one will actually NOT see an interference pattern on the screen. It's only if one isolates the signal photons corresponding to the idler photons observed in one of the two detectors do we see an interference pattern.

    My question is why is theory predicting this? I mean, how does one show that this arises from theory? A naive analysis would say that since there are no which path information, there is an interference pattern and, morover, since the setup is completely symmetric, there is no difference between the patterns related to idlers detected at one detector with respect to the other detector.

    So my question is : why are the two interference patterns shifted relative to one another??


    Patrick
     
  10. Aug 4, 2007 #9
    Hi Patrick,

    where you say "there are no which path information" you overlook that you have instead measured phase information. (Just as the which path measurement had two possible results, the "eraser" is also a measurement with two possible results, in fact you're just measuring the same thing in a different basis. This measurement result is the token that's necessary to mitigate a relativity paradox.)

    The idler tells you something similar to the path length difference between the two paths (eg. were the parts in, or out, of phase with one another as they passed the slits), from which you can trivially deduce whether the corresponding signal photon will contribute to a fringe or anti-fringe pattern.

    I think about this and wonder, couldn't we prepare the beam so that the two paths are always in phase?...
     
  11. Aug 5, 2007 #10

    nrqed

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    Thanks a lot for your input. I still am a bit confused so I hope you can clarify things for me.

    I am assuming that the setup is completely symmetric, so I don't quite see how the path length difference enters. I mean, let's say that an idler photon is detected in D1. Then we have no which path information at all. The same for an idler detected in D2. But you are saying that there is a different information about the phase whether the idler is detected in D1 or in D2? Can you tell me more about why this is the case?

    Thank you for your input!


    Patrick
     
  12. Aug 7, 2007 #11

    nrqed

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    I am still confused by the fringe vs antifringe interference patterns in the
    experiment described in

    http://xxx.lanl.gov/PS_cache/quant-p...03/9903047.pdf


    I hope someone can help clarify (maybe cesiumfrog who started explaining to me).

    Let me describe the setup briefly.

    Two atoms are side by side. Each emits, one at a time, an entangled pair of photons. The two atoms essentially play the role of the two slits in a double slit experiment.

    let's say one of the two atoms emit a pair of photons. One photon goes right to a screen. That's the signal photon. An idler photon goes left.
    It hits a mirror and is redirected to a beam splitter. Upon hitting the beam splitter, it either goes through and is detected by a detector D1 or is reflected and is detected by a detector D2.

    If the second atom emits an entangled pair, exactly the same situation occurs through a different path completely symmetric to the first path.


    therefore the situation is clear. Detecting a photon at D1 or D2 does not tell us which atom the photon was emitted by and there is therefore no which path information. Therefore there is an interference pattern for the signal photons.

    However, it turns out that there is no interference pattern observed on the screen because the signal photons in coincidence with the idler photons detected at D1 produce a pattern out of phase by pi with the signal photons in coincidence with the idler photons detected at D2.

    Can anyone explain to me why this is so? I don't follow the explanation of the paper.

    Not only I don't see the origin of the phase difference but I don't even see why there could be an assymetry between the two results. If an idler is detected at D1, we can tell which atom had emitted the pair. Same for an idler detected at D2. But the setup is completely symmetric (unless I am missing something) so I can not see why the pattern of signal photons associated to idlers detected at D1 could be any different than the pattern of signal photons associated to idlers detected at D2.


    I hope someone can clarify this for me!

    Thanks


    Patrick
     
  13. Aug 7, 2007 #12
    I can really only give a hand wavy explanation (I'm hoping someone else knows more)..

    From a theoretical point of view, if in one basis the state can be measured as A or B, then in another basis it can be measured as A+B or A-B.

    From a more experimental point of view, you argue about symmetry, but a fringe pattern *is* symmetric, and an anti-fringe pattern is also symmetric. I guess you might consider that the entangled photons came from one slit. Or came from the other slit. Or came from both slits at the same time (equal phase) in a superposition. Or came at a different time from each slit (different phase) in a superposition. I have wondered if it has to do with the frequency down-conversion process that produces the photon pairs, noting that the pump intensity has maxima twice as often as the entangled photon's frequency (which suggests a good probability of quite likely paths, for a given photon, naturally starting out 180 degrees out of phase), but I don't believe you can "trick" QM with gimmicks like very short (single maxima) pump pulses.
     
    Last edited: Aug 7, 2007
  14. Aug 7, 2007 #13
    The way I see it is this: you don't know which slit the photon went through (and, in a certain sense, it went through both slits). So, there are two possible paths for the idler photon: one starting from the right slit, and one starting from the left slit.

    If you understand how a mach-zehnder interferometer works, you know that the phase difference between the two possible paths determines the probability that the photon will be detected on the left or the right after it reaches the final half-reflective mirror. (Short synopsis: if the left path is 1/4 wavelength delayed from the right path, then a reflection at the final mirror on the left will add another 1/4 wavelength of delay to the left reflected path, and this will negatively interfere so that the photon will always come out of the right side).

    What happens in the delayed choice is that the two paths for the idler are out of phase, and this phase determines the probability that the idler is detected at either the left or right. This phase is also correlated with the x position of the signal photon, so basically the idler photons tend to be detected at the left or right depending on where the signal photon landed. This is what produces the interference.
     
  15. Aug 8, 2007 #14

    nrqed

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    Thanks bruce2g and cesiumfrog for your replies.


    I kind of see what you are saying but I am still puzzled. Let me try to explain what puzzles me.

    There must be some asymmetry in the setup that I am missing.

    My question is: what determines if the fringe goes with the idler photons detected at D1 is a fringe or an antifringe? My whole confusion boild down to answering that question.

    My problem with the symmetry of the setup is that this. If I understand the setup correctly, flipping it over (in such a way that the detector D1 takes the place of D2 and vice versa), the whole setup is idnetical to what we started with. That is a problem because then there is no possibility for any argument to associate a specific type of pattern (whether it's a fringe or an antifringe) with a specific detector.

    Do you see what I mean? If you had any argument favoring an antifringe for signal photons associated idlers detected at D1, I could make you close your eyes, switch over the setup and you would make exactly the same argument about D2.

    Now, it's true that both a fringe and an antifringe are symmetric, as cesiumfrog pointed out. But to me there is still a breaking of the symmetry in associating one one of the two to idlers detected at D1 and the other to the idlers detected at D2. This is the part I don't understand.



    Maybe I am missing something about the beam splitter just before the detectors D1 and D2. Is its action completely symmetric? I am assuming that a photon coming from one side has a 50/50 chances of going through or being reflected and a photon coming from the other side also has a 50/50 chances of being reflected/going through. I am not sure if this is possible practically to have both surfaces acting this way. If one side would be completely reflective and the other side would be 50/50, then I could see how an asymmetry would arise.


    Thanks for your input! I hope my question makes sense.


    Patrick
     
  16. Aug 8, 2007 #15

    Cthugha

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    This is not true. So the first point of asymmetry is the position of D0. The fringes and antifringes are a result of measuring the coincidence counts depending on the position of D0. If you move it, the distances A-D0 and B-D0 change. This is rather trivial, but nevertheless important.

    The second asymmetry is due to the paths the photons take. A photon going from A to D1 is reflected at both beam splitters. A photon going from A to D2 is reflected once and transmitted once. Reflection causes a phase shift of pi. Transmission causes a phase shift of pi/2 (considering an ideal 50/50 beamsplitter). So photons from A have differing phase information on the two detectors (shifts of 2 pi or 3/2 pi). The same is true for photons from B. It is just the other way around.

    So let's have a look at R01. If you mirror the setup concerning both asymmetries you should get this result back. That means looking at R02 instead and setting x=-x will give you R01 again. So the setup is symmetric, if you mirror it along the line x=0. But this means changing the detector you look at (D1,D2) and the position of D0.
     
    Last edited: Aug 8, 2007
  17. Aug 8, 2007 #16

    nrqed

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    yes, I agree. But this is unrelated to the pattern one observes on the screen (that detector does not do anything in terms of which-path information or phase shift between the paths)
    yes, I agree that the two paths have different phases. The idler from A to D1 (two reflections) has a different phase than the photon from B to D1 (one reflection and one transmission).

    My question is about the difference between observing D1 and observing D2. The two paths leading to D2 also have the same relative phase difference (either two reflections or one reflection and one transmission). So why is the pattern on the screen different when one looks at the signal photons in coincidence with idler in D1 or idlers in D2?

    But I am imagining scanning the screen completely with D0. So let's say you scan the entire screen with D0, recording only the signal photons in coincidence with D1. Now you flip the setup, scan again the screen and record all the screen with the photons correlated with D2. Why would one get a different pattern on the screen?

    Thanks for your input. I am sure I am missing something very simple so be patient :redface:
     
  18. Aug 8, 2007 #17

    Cthugha

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    Oh, no. This is very related to the pattern you observe. The path lengths from A and B to D0 differ as well. Therefore the relative phase difference between photons from A and B arriving at D0 changes as well, if you change the position of D0. This is the reason, why you get fringes and antigringes.
    Changing from D1 to D2 changes the relative phase difference between A and B photons on these detectors from -pi/2 to +pi/2, but the phase difference between the photons on the way to D0 stays the same, so the sign changes on one side, but not on the other, which is a phase shift of pi.
     
    Last edited: Aug 8, 2007
  19. Aug 8, 2007 #18

    nrqed

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    I do see that the relative phase difference between the signal photons changes as D0 is moved, yes. I don't quite see yet how this is tied in with observing the idler at D1 or D2 and hence how it's related to a fringe vs an antifringe.
    I am starting to see where my confusion comes from. I did not realize that the phase difference between the idler photons had anything to do with the interference pattern on the screen. The way I see it, the interference pattern on the screen is due to the signal photons. So it seems to me that the only issue is whether we know from which slit (which atom) the signal photon comes from or we do not know from which slit it comes from. The interference pattern on the screen is due to the superposition of the two paths the signal photon may have taken (from atom A to the screen or from atom B to the screen). I though that the role of the idler was simply to provide which path information or lack thereof.

    But it seems that I am wrong. You are saying that the phase difference of the idlers plays a role in the interference pattern on the screen.
    I am trying to see how this fit with what I know about QM.
    Why don't we simply consider the two different paths the signal photons may take? What is the rationale here to look at the phase difference between the idlers? Shouldn't one consider the possible paths from emission to observation of the signal photons? What is the general rule?
     
  20. Aug 8, 2007 #19
    In these experiments, it is very difficult to get everything aligned perfectly (half a wavelength is very small compared to the length of an optics bench, so in practice the apparatus is asymmetric). By slightly nudging one of the components (which usually alters the path length asymmetry in a gradual manner), I think you'll find the two interference patters smoothly shift across the screen (if you think of the pattern as a sinusoid with a gaussian "envelope", the envelope should stay the same whilst the sinusoid moves left or right). The fringes and anti-fringes are always complementary, but I think I was too hasty in assuming that they would individually have the same symmetry ideally as the apparatus. With a "perfectly" symmetric (down the centreline) apparatus, if the two patterns turn out aligned such that one is the mirror image (down the centreline) of the other (but neither is symmetric individually down the same centreline) then your paradox would be resolved.
     
  21. Aug 8, 2007 #20

    Cthugha

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    Ah, now I see your problem.
    You have to conside that the two emitted photons are entangled and that you still get some information from the detector (D1 or D2), which counts the idler photon: the phase difference.

    Imagine D0 is at a position, where the path length from A to D0 is just lambda/4 (pi/2) shorter than the path from B to D0.
    Imagine for simplicity, that there are two photons emitted from A and B. The photon from A is emitted a bit later, so that there is constructive interference on D0. So they arrive in phase and the phase of A is the phase of B+ pi/2.

    They are entangled with the idler photons, so the idler photons are emitted with the same phase difference.

    Let's have a look at D1. A photon from A will be shifted by 2 pi. A photon from B will be shifted by 3/2 pi. So the total phase difference is pi as there was a difference of pi/2 beforehand. This means destructive interference.

    Now have a look at D2. Here a photon from B will be shifted by 2 pi and a photon from A will be shifted by 3/2 pi. So there is no overall phase difference. This means constructive interference.

    So the joint count rate will show a peak for R02 and it will be low or even zero for R01. This is the principle of the origin of the fringes, although the model is a bit simplified.

    It is a matter of complementarity. You can either have which way information or information about the phase difference.
     
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