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Delayed Choice Quantum Eraser

  1. Jan 12, 2005 #1
    I have been contemplating this experiment for quite some time now, and I have some questions. Here is a preliminary copy of the paper describing the experiment, for reference.

    The first and most obvious question would be whether it is possible to actually predict the future, by replacing the beam splitters in the idler photon paths with movable (actually, my thought was LCD electrically switchable) mirrors, and making the idler path longer than the signal path; I believe that the answer to this must almost certainly be "no," because the interference shown by the signal photons cannot be separated left from right without the correlation of the idler photon impacts, but there is a vanishingly small chance based on the data in the paper (and I have seen a more complete version of the paper, and the results were the same) that there might be a way to see the difference between the interference and non-interference cases without reference to the idlers. As I say, I don't believe it will pan out, but the data presented in the paper do not exclude the possibility.

    But the first part that I really wanted some feedback on is:
    The positions of the mirrors that would replace the beam splitters in the left and right idler paths obviously determine whether the idler photons will go to the which-path detectors, or through the quantum eraser to the eraser detectors.
    Is the probability distribution of the signal photons that creates the interference interpreted as the effect, and the probability distribution of the idler photons either to the which-path detectors or to the eraser and eraser detectors interpreted as the cause of that effect?

    I have some reasons for interpreting it that way. First, there is obviously a correlation between the probability distribution of the signal photons and the probability distribution of the idler photons. Second, there is no other variable correlated to the probability distribution of the signal photons. Third, a change in the probability distribution of the idler photons such that they are all sent to the eraser and eraser detectors always results in a probability distribution of the signal photons that shows interference, and a change in the probability distribution of the idlers such that they all go to the which path detectors always results in a probability distribution of the signal photons that does not show interference.

    While I do know that the correlations between the probability distributions of the signal and idler photons prove a causal link between them, I do not know whether these three facts together constitute proof that the idler distribution is the cause and the signal distribution is the effect, and that is my question.

    There is more, but this post is long enough for now.
  2. jcsd
  3. Jan 12, 2005 #2
    Geez, I hate to bump my own post, but is there anyone here with the knowledge to give me any direction on this?
  4. Jan 13, 2005 #3


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    I discussed this paper already on this forum a while ago but I cannot find back the thread.
    Although it is a nice experiment and so on, I think it is a bit "oversold". You can give it the interpretation given by the authors if you want to, but in fact, something much simpler is going on. When you look at figure 3, you get an interference pattern, because you SELECT A SUBSAMPLE from all impacts at D0, which are coincident with a hit at D1. When you look at ANOTHER SUBSAMPLE, namely the impacts at D0 which are coincident with D2 (figure 4) you get a shifted interference pattern. This shift comes about, if you trace it back, to a difference in optical pathlengths in the polarizing beamsplitter, and this shift is utterly important.
    Finally, the third subsample of the D0 events gives you figure 5, the coincidence with D3. It doesn't show any interference, and is infact equal to the sum of the shifted interference patterns of figures 3 and 4.

    If the system is 100% efficient, then every event is classified in one of the 3 subsamples, and so at D0, without coincidence, you don't have any interference pattern. It is only when the coincidence information is used to do subsample selection that you see subsamples with or without fringes.

    The entire experiment is in fact based upon what I'd call a misunderstanding of quantum mechanics. It is assumed from the start that 'atom A OR atom B' emits a photon.
    But that's denying superposition !!
    The incoming pump photon is put in an entangled state:
    half goes to A, half goes to B.
    THIS ENTANGLED STATE IS TRANSMITTED TO THE ATOMS A AND B, so you cannot say that atom A is emitting OR atom B is emitting.
    This entanglement is further transmitted to the outgoing entangled photon pairs: we have in fact a superposition of 2 2-photon states, and the experiment analyses different correlations between these 2-photon states.

    The "quantum erasure" idea is based upon a misunderstanding and a TOO EARLY APPLICATION OF THE PROJECTION POSTULATE. You can only apply the projection postulate at the very end of your calculation.

    Mind you, I'm not critisizing the very nice experiment, and all that. I'm critisizing the hype around "quantum erasure" which makes people believe that according to whether we "look at remote data or not" we influence a physical phenomenon, and then leave them wonder why you cannot turn that into a faster than light telephone.

    Quantum theory, as it stands, doesn't allow for FTL telephones. I explained why in another post here ; it is due to the fact that all local observables derive their expectation values from the local density matrix, which traces out the remote state space from the overall density matrix, and hence is not influenced at all anymore by what happens to that remote state.

  5. Jan 13, 2005 #4
    Using the consistent histories (also "decoherent histories") approach to quantum theory of Gell-Mann, Griffiths, Hartle and Omnes to analyse the delayed choice paradox, Griffiths shows in his book Consistent Quantum Theory that there are actually many possible descriptions quantum theory allows of the events of the paradox but they can only be combined in certain ways in order to be consistent with quantum theory.

    The delayed choice paradox arises from combining certain incompatible descriptions of events to create the paradox and not being aware other compatible descriptions that can be combined that don't create a paradox.

    However, these compatible descriptions can have macroscopic quantum superposition states, as in Schrodinger's Cat, so it's perhaps no surprise that physicists are unconsciously avoiding them in this paradox as they aren't intuitive theoretically and decohere experimentally.

    The paradox is roughly like combining wave and particle descriptions at the same time and then declaring there to be a paradox, the consistent histories approach being very much the modern version of the principle of complementarity with its rules to avoid this.

    So according to this approach, the delayed choice paradox is something which arises from not being aware of macroscopic superposition states predicted theoretically and experimentally and combining descriptions in a way that makes sense to our minds used to classical physics and not following where quantum theory leads us.

    Since this is a delayed choice paradox thread, I just thought I'd mention this for anyone who might be interested. :smile:
  6. Jan 13, 2005 #5


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    That's exactly what I tried to point out. In fact, EPR and "delayed choice quantum erasure" experiments are more experimental indications of the physical reality of macroscopic superpositions than that they are indications of non-locality or delayed choice paradoxes. In fact, relativity tells us that non-locality and delayed choice paradoxes are of course one and the same.

    It all holds in one single rule: you apply the Born rule AT THE END of the calculation, for the final quantities, and not in the middle of your experiment.
    (and yes, this implies macroscopic superposition).
  7. Jan 13, 2005 #6
    Examining the diagram in Fig. 2, the exact position of the Glan-Thompson prism is not clear; not to mention Glan-Thompson is misspelled in the paper. However, I see no reason to conclude that the optical path lengths are different between left and right (or, if you prefer, A and B); while the path lengths for the idler and signal photons might be (and in the realization are, by 8ns) different, the path lengths for left and right for either the idler or signal photons need not be so. The fact that the phase shift is precisely π, placing the peaks of one interference pattern precisely overlapping the valleys of the other, argues that in fact that is not the reason for the phase shift.

    Of course it is; they are shifted by π! Since the graph in space of an interference pattern is a cos function, and we all know that a cos function added to a copy of itself with a phase shift of π yields a smooth function, precisely what is seen if there is no interference. So the question is, why should the phase shift be precisely π?

    It is not and cannot ever be. D0 is driven back and forth across the field of the interference with a stepper motor. Think about it.

    In the non-interference case, that is exactly true. How is it a misunderstanding of quantum mechanics to state that if you can recover the which-path information, you see no interference?

    On the other hand, note also that there is no interference in the combined patterns from the eraser; this is because it is not yet clear whether that information can be recovered. The operation of creating the idler photons, which makes which-path information theoretically available, has phase shifted the two interference patterns exactly enough to cause their superposition to yield what appears to be non-interference; only by scrambling the two possible states (co-incident with a signal photon from A/left, or B/right) can the interference be recovered. This is conceptually equivalent to the notion in Feynman's gedankenexperiment based on the Young dual-slit experiment, done with electrons, in which the slit through which the electron passed is identified by placing a very sensitive coil transducer around each slit. The paths of the electrons are "smeared" by their interaction with the transducer just enough to eliminate the interference. I ask again, how is this a misunderstanding of quantum mechanics?

    Now, this is a misunderstanding of quantum mechanics. First, a single photon cannot be entangled with itself. Second, this is an all-or-nothing interaction here, and there are only three alternatives: the whole photon goes to A and generates an entangled pair, the whole photon goes to B and generates an entangled pair, or the whole photon cannot be shown to have gone to either A or B, and generates an entangled pair that cannot be shown to have come from either A or B.

    We interpret this third case not as half of one and half of the other, but as both. Repeated analysis and experimentation has left us no choice but to assume in the interference case that the "particle" has not traversed one slit or the other, but both slits. And not half of it went through one and half through the other; both at the same time. The period of the interference (in the Young dual-slit experiment of course; the β-BBO crystal in the DCQE means that the wavelength is doubled there, in order to produce the idler photons) shows that the wavelength of the interfering waves is not twice the incoming wavelength; it is the incoming wavelength. Thus, there can be no question of the individual photons having divided in half and each half traversed one of the two slits.

    There is no entangled state at that point; a single photon cannot be entangled with itself. It requires two photons minimum to be entangled, and there aren't two photons until AFTER the β-BBO has completed SPDC and created them. Any other assumption leads to a violation of conservation laws.

    While I repeat that there is no entangled state until the pairs are produced, you are correct that there is a superposition of two eigenvalues into an eigenstate. The two eigenvalues are "went through A" and "went through B." Each has a probability of 0.5. It is only if this eigenstate is never collapsed into a state that we ever see interference. And in this experiment, that happens if the idler photons go to the eraser, that is, the beam splitter labeled "BS" in the diagram.

    As I have said many times elsewhere, the remarkable thing about a dancing bear is not how well the bear dances. And the remarkable thing about this experiment is not that we cannot see interference directly; it is that we can recover interference at all by any means whatsoever.

    The projection postulate is properly applied at the time when information becomes available about which path the photon took; and that time is the generation of the idler and signal photons, because the idler photons can show the which-path information and therefore constitute a measurement. What is remarkable is that we can retroactively cancel the measurement (quantum erasure) and recover the interference.

    I agree it's unlikely, but you have misinterpreted the potential results; they are not an FTL telephone, but a device that predicts the future. And I expect that you cannot do that either; but the data do not rule it out, so far.

    I have extracted the data points from the two graphs in Fig. 3 and Fig. 4, and there remains variation from the mean curve that is greater than the variation in Fig. 5. However, I suspect that this is merely a measurement artifact that would disappear if more careful measurement were done; for instance, the data points in Fig. 5 are at half the frequency of those in Fig. 3 and Fig. 4. But it is worth measuring to see if it can be seen. It would be a beautiful demonstration of QM to show that interference cannot be extracted from the signal photons by any means that does not use the idler photons.

    I agree that it should not be possible if QM is correct. However, the reasons you state do not necessarily make it so.

    -Da Schneib
  8. Jan 13, 2005 #7
    Not only that, but interpretation aside, is anyone going to answer my question? The results of the experiment are clear; there is no interpretation necessary. Is there a cause-to-effect relationship (as opposed to a merely "causal" without presumption of one or the other being cause or effect) relationship between the determination of which-path information and the appearance of interference?
  9. Jan 13, 2005 #8


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    I'll try to show what I meant - which should address a lot of your criticisms - by a half-symbolic calculation.

    First, there is the pump-photon, which is written as a superposition of its position through slits A and B:

    |pump> = 1/sqrt(2) (|pA> + f |pB>)

    I introduced a phase factor f, this will prove to be usefull later on.

    The interaction with the crystal (or, in the original proposal, with the atoms, I will take that example, it is symbolically the same) leads then to an entangled state of the excited states:
    |excited atoms state> = 1/sqrt(2) (|A> + f |B>)
    and then leads to the photon pair state:
    |photon pair> = 1/sqrt(2) (|A0> |Ax> + f |B0> |Bx>)

    The 0 photon is the one heading for the D0 detector, and the x-photon is the one heading for the complicated mixer setup.
    Now, we should rewrite the |Ax> and the |Bx> states of the mixer-photon as a function of the detection eigenstates, |D1>, |D2>, |D3> and |D4>.
    We do this by tracing them through the apparatus.
    |Ax> can be written as 1/sqrt(2) ( |D3> + 1/sqrt(2) ( |D1> + |D2> ) )
    and |Bx> can be written as 1/sqrt(2) (|D4> + 1/sqrt(2) ( |D1> - |D2> ) )

    The second part can be checked by noticing that the beam splitter BS gives pure D1 counts for a certain phase combination of Ax and Bx, and that it will give pure D2 counts for a 180 degree phase shift. This is what I called the different optical path lengths. The physical reason is that in one case you go through the beamsplitter, and in the other case, you reflect off it, which always gives you a 180 degree phase shift (otherwise it is not a 50-50 BS).

    Substituting, we find our photonstate now in the D basis:

    |photon pair> = 1/2 { |A0> (|D3> + 1/sqrt(2) (|D1> + |D2>))
    + f |B0> (|D4> + 1/sqrt(2) (|D1> - |D2>) ) }

    = 1/2 {|A0> |D3> + f|B0> |D4> + 1/sqrt(2) (|A0>+f|B0>) |D1>
    + 1/sqrt(2) (|A0> - f|B0> ) |D2> }

    And you read off easily the results:
    In coincidence with |D3> you get |A0> in the D0 pipe: no interference.
    Same with |D4> ; you get |B0> in the D0 pipe.

    In coincidence with |D1>, you find |A0>+f|B0> : interference between an "A beam" and a "B-beam".
    In coincidence with |D2>, you find |A0> - f|B0>: a similar interference pattern, but with a 180 degree phase shift.

  10. Jan 13, 2005 #9


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    I will now address your criticisms verbally, but have a look at my small calculation in the other post, please ; it is the essence of what I had in mind.

    I probably expressed myself badly: I was pointing to a half-wavelength pathlength difference due to the beamsplitter (sign difference between transmitted and reflected beam).

    That's an experimental detail: think of a position-sensitive detector which can measure the impact position of the photon at D0.
    What I meant was: D1, D2, D3 (and D4) cover all cases, the second photon has to end up in one of those 4 detectors.

    That's a correct statement. The less "correct" statement is that you have "information which you erase"... You only measure correlations, which themselves are the result of (superpositions of) product states.

    It is what I call "overselling". The only true information that is available is the correlations at the end of the experiment. I don't know what it means to have potentially available information which is then erased.

    But that's denying the essence of quantum theory ! The exciting photon (pump) was in a superposition of positions at A and B. This superposition is maintained, and the excitations at A and B are then in an identical superposition. The whole story of "which way" information, which denies the superposition of these states, is what gives rise to all these "erasure paradoxes". Have a look at my very simple calculation.
    Your third case is nothing else but the interference terms due to this superposition which you deny.

    You missed what I said: a single photon can of course be in a superposition of 2 position states, A and B. That doesn't mean that I think that it is sliced in 2 half photons of course. It is this position superposition which is then propagated and gives rise to all the results.

    That is only remarkable if you think that the pump photon cannot be in a superposed state, half at A and half at B. But if you do accept that, by simply applying quantum mechanics (and no projections) nicely all the way, a lot of fuzz falls away.

    No, the projection postulate must be applied when you calculate your final quantities, which are here correlations between D0 and D1 or D2 or D3...
    Sometimes, you can get away with intermediate use of the projection postulate, but not in these entangled cases.

  11. Jan 13, 2005 #10
    I guess I'm more comfortable discussing it than dissecting the math.

    I guess I'm not quite clear on what component you were talking about with the "polarizing beamsplitter." I thought you meant the combination of the β-BBO crystal and the Glan-Thompson prism. I don't see any other component that could be described as a "polarizing beam splitter" in the optical path between the slits and the D0 signal photon detection area, though, so I'm still not clear on exactly what "half-wavelength pathlength difference" you are referring to.

    I can see a half-wavelength pathlength difference between either signal path and its corresponding idler path; that is introduced by the β-BBO crystal, and accounts for the phase orthogonality between the signal photons and the idler photons (and thus the ability to sort them in a Glan-Thompson prism). But I cannot see any way that phase difference could make a difference in the positions of the interference patterns; because I also cannot see any difference in the idler photon paths between left and right (or, if you prefer, A and B).

    I guess I want you to explain in precise terms what exact component you believe is introducing this difference, and how the introduced phase difference accounts for the "phase shift" between the two interference patterns.

    But because not every photon that is measured at D0 can be correlated with one measured at one of D1-D4, there is no reason to believe that there would be any visible interference anyway.

    I guess I'm subconsciously making reference to a different state of the experiment (and I guess it's fair for me to do so, because it was explicitly mentioned in my OP): what if the BSA and BSB (which send idlers either to D1/D2 or D3/D4, randomly) were replaced with mirrors that always sent the idlers to the eraser D1/D2? Even in this case, the device cannot be anything like efficient, simply because we cannot detect corresponding signal photons for every idler; and conversely, even if we could, by using a CCD sensor with a maximum efficiency at 500nm of about 96%, unless D1-D4 have 100% quantum efficiency, we cannot guarantee to detect every corresponding idler photon for every signal photon that is detected. I'd estimate about 90% maximum efficiency, if we used state-of-the-art CCDs everywhere.

    Your conclusion was, "It is only when the coincidence information is used to do subsample selection that you see subsamples with or without fringes." I have to agree with that; but I'm not sure why you felt you needed to precede it with, "If the system is 100% efficient, then every event is classified in one of the 3 subsamples, and so at D0, without coincidence, you don't have any interference pattern." I agree with it without agreeing with (what appeared to be) the premise; however, I also don't think it really was a proper premise, and I still don't see why it is important that the interference is a subset. And last but not least, after the experimental modification I suggested with mirrors, there are only two possible alternatives, not three: no idler photons go to D3 or D4, so there are no non-interfering signal photons.

    OK, then explain the lack of visible interference in the D0 signal detection area when the mirrors are turned to send all the idler photons to D1 and D2. You can't see interference even then- without the correlation. But under what circumstances do you not see interference of two waves? Only when you have which-path information. So I have to dispute your claim that there is no "information which you erase." It appears to me that in this experiment, we have disrupted the photon paths through obtaining which-path information and destroyed the interference- and that we get the interference back when we destroy the which-path information. But in order to protect causality, we cannot see that interference without looking at the correlation between the idler photon detection events and the signal photon detection events. Thus, we preserve both of:
    1. Quantum mechanical waves always interfere, unless there is which-path information.
    2. Obtaining which-path information always destroys quantum mechanical wave interference.
    Both must be true; if there were no interference recoverable, then the first would be rendered untrue, and it would call everything all the way back to Young's original dual-slit experiment into question. If there were any way to see interference when which-path information had been obtained, then causality would be macroscopically violated.

    Then you haven't thought about the experiment when all the idler photons are sent to the eraser. Think about it again that way. Then you'll begin to see the very narrow tightrope that nature has to walk for this experiment to work exactly the way it does; causality must not be violated, but the wave nature must be visible whenever it should be seen.

    First, what you specifically said is shown above. Second, I still don't see any entangled state. Nothing you said following this has convinced me that this has anything to do with entanglement. Entanglement and superposition are not the same; they are different. Had you said that the incoming pump photon is superposed into an eigenstate whose eigenvalues are it goes to A or it goes to B, with 50% probability for each alternative, then I might have agreed with you. But I would also have pointed out that just because it is in such an eigenstate does not mean that that eigenstate ever has to collapse into a state, and realize one of the eigenvalues; and in fact, if it does not so collapse, if neither eigenvalue is realized, then we see interference! And that, my friend, is the essence of quantum mechanics!

    Feynman once said that you can see everything there is to know about quantum mechanics in the dual-slit experiment with single electrons. I tend to agree with him strongly. He of course didn't mean that all the math is there; what he meant is that all the most important concepts are there.

    You apparently did not understand the implications of what I said. I do not deny that such a superposition exists; I was referring to it. It is patently obvious that it does. What I deny is that a superposition represents an entangled state in this case. The entanglement does not occur until after the initial superposition.

    What I was trying to indicate was exactly that superposition; my first case is the first eigenvalue of the eigenstate, my second case is the second eigenvalue of the eigenstate, and my third case is the uncollapsed eigenstate, which may or may not (depending on whether you erase or read the which-path information) collapse into a state of one of the two eigenvalues. If it does collapse, then the corresponding signal photon will not be part of an interference pattern; in other words, its probability distribution will be a smooth function of the distance from the center of the D0 detection area. On the other hand, if the eigenstate is never collapsed into a state, then the corresponding signal photon will be a member of one of two classes of photons that form interference patterns in the D0 detection area.

    Now, represent me the possibility that that eigenstate will not collapse into a state of one of the eigenvalues. That is what is missing from your mathematics above. And that's why I don't critique them directly. I don't think you're applying them right.

    No, actually, I have quoted exactly what you said above: "The incoming pump photon is put in an entangled state: half goes to A, half goes to B." And you didn't say "superposition;" you said "entangled." The two are different. Sometimes simultaneous, yes; but different. The "half" part I'll leave to anyone who cares to read it.

    And again we have an apparent misunderstanding of the meaning of "superposition." It is not "half at A and half at B;" it is at both, and neither. It has a 50% probability of having been at A, and it has a 50% probability of having been at B. But there is a third alternative: it was at both, or neither, and we see interference. And you have not addressed this third alternative. When your eigenstate collapses due to a measurement, then there will be no interference, whichever way the collapse occurs, because the eivenvalue of the state, whichever it is, will preclude seeing interference. If and only if that eigenstate never collapses into a state will we see interference.

    Personally, I think that seeing interference is not "fuzz;" but you are welcome to your opinion.

    I don't think that you can prove that, and I assert that the projection postulate can be used any time there is a measurement. And producing the idler photons is a measurement; it measures which-path information. That is the reason that the interference cannot be seen. As I have said repeatedly, what I find remarkable, and a beautiful illustration of quantum logic, is that there is any way at all to recover the interference.

    -Da Schneib
  12. Jan 13, 2005 #11
    This right here is what I don't understand. I thought this was what you were talking about with the "half-wavelength pathlength difference due to the beamsplitter." Tell me the exact physical origin of this "phase factor."
  13. Jan 13, 2005 #12
    ...and I still haven't received an answer to my question!
  14. Jan 13, 2005 #13


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    You don't really need it, strictly speaking, because you can absorb it into the state it goes with. I wanted to do something nasty with it. It is NOT the phasefactor that is responsible for the interference pattern, but can be used to do so. But I realised that that would make the issue confused.
    So put it equal to 1.

    It has nothing to do with the lambda/2 shift ; this one comes from the |D1> + |D2> and the |D1> - |D2> sign difference when I rewrite the Ax and Bx states as a function of the detector eigenstates.

    Last edited: Jan 13, 2005
  15. Jan 13, 2005 #14


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    Mine is the opposite: I'm probably much clearer on the math than the words.

    Sorry, that' s an error on my part. The BS is not a polarizing one. I have to say that when I first commented the article, I confused it with one, very similar, that I already commented here (and in fact, the issues are very very close). There the "which path" information resided in the polarization of the photon, and they used a setup with polarizing beam splitters and quarter wavelength plates to "allow or deny" the extraction of the polarization information and hence the path information.
    Just read "beam splitter". I'm talking about BS in the figure.

    If you have a setup like the one with the beam splitter BS, and you have, simultaneously, two incoming beams interfering in it, then, according to the phase relationship of these two incoming beams, you can send all the outgoing light to D1 or to D2. The difference in phase relationship is exactly 180 degrees. So you can re-express this in the following way: the difference between going through, or reflect from the BS is given by a lambda/2 extra optical pathlength. But I now realise that it can be confusing to say it this way.
    The point is simply that when we have a photon in a state
    |Ax> + |Bx> it will go for sure to detector D1, and if we have a photon in a state |Ax> - |Bx> it will go for sure to detector D2.
    We can then invert this base transformation, and rewrite Ax and Bx as a function of the D1 and D2 states. And the minus sign of Bx between D1 and D2 is that famous 180 degree phase difference.

    In a 100% efficient setup, I would think that for every D0, there is one D1-D4 click, no ? Where does the other photon end up then ??

    I'm of course reasoning in a simplified way, not taking into account efficiencies, losses etc... just on a gedankenexperiment with a comparable setup. The true experimental analysis of the setup is more complicated, but we're talking about principles here.

    I wasn't actually answering your post, but commenting on the paper (which initially I even took for another, very similar one - so I start to see why my comments were confusing :-)
    I stressed the subsample thing because the way these papers are formulated seem to give the impression that a poor soul looking at the interference picture are D0 will see fringes or not, according to whether someone else, far away, has put in a mirror in the other branch or not. THIS IS NOT TRUE.
    Locally, the D0 picture is ALWAYS the same (in this case, a blob without interference).
    It is only when you get information from measurements done by that other person, and you do that to select a subsample of the recorded clicks at D0, that you see an interference pattern arrise or not. So you cannot use it to transmit a message to D0 by having him, on his own, see appear, or not, interference fringes. Maybe this was clear to you. It is often the question that comes back.

    I'll comment on other issues later...

  16. Jan 14, 2005 #15


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    Several points have to be made here.
    The first one is "why don't we have interference at D0".
    The answer is that, because of the entanglement of the "D0" photon with the "mixer" photon, the local state at D0 is not a pure state anymore but an (improper) mixture obtained by tracing out the "mixer" states out of the overall density matrix. You then get the local density matrix

    ( 1/2 0 )
    ( 0 1/2)

    which describes a non-interfering statistical mixture of two incoherent sources at A and at B.
    I can do the calculation explicitly if you want to.

    I realise now that this is in fact a beautiful experimental illustration of decoherence theory ! When considered locally at D0, the beams coming from points A and B have "decohered" with "the environment" which is now still limited to the mixer photon. So if we can distinguish the states of the mixer photon, we can still potentially find correlations which show interference.

    The second point is your "eraser" which doesn't in fact erase anything.
    Do you realise that the socalled eraser, where the two paths are mixed up in BS, is just a linear transformation of basis (my transformation with D1 and D2).
    You can UNDO that with another BS, at least in principle: if, instead of a detector D1, you put mirrors so as to redirect the "D1" beam onto a new beam splitter, and you do the same with D2, recombining the D1 and the D2 beams, if you choose the pathlengths right, what will come out of this new beam splitter will be your A or B path information. So, no, the mixer didn't ERASE any information, it just transformed two states |path A> and |path B> into two other states |D1> = |path A> + |path B> and |D2> = |path A> - |path B>
    Mixing |D1> and |D2> AGAIN in the same way would give us
    a |D1-bis> = |path A> and a |D2-bis> = |path B>.
    So if you do not put your detectors D1 and D2 there, but you send your beams off in space, YOU HAVEN'T erased your information. You could still capture them by mirrors on Jupiter, and reconstruct path A and path B. And still, you wouldn't see any interference in D0, because that's only determined by the LOCAL DENSITY MATRIX, which gives you an incoherent mixture.

  17. Jan 14, 2005 #16


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    Let's do some further calculations, a few lines of algebra are probably clearer and more precise than hundreds of lines of proza.

    I have my state as follows:

    I will work in the product basis:

    (|A0>|D1> ; |A0>|D2> ....|B0>|D1> ...)

    In that basis, the state at hand is written by the 8-tuple:

    (1/sqrt(8), 1/sqrt(8), 1/2, 0, f/sqrt(8), - f/sqrt(8), 0 , f/2)

    You can put f = 1 if you want to. It could also be used to indicate whether different phaseshifts from A and B to D0, along the D0 axis, will give rise to an interference pattern or not, but although correct, it is a bit messy.

    The density matrix is then given by (calculated with mathematica) the matrix in the attachment picture.

    How do you add an attachment ? It says "upload", but nothing seems to be attached ???
    In ugly numerical mathematica format, this gives:
    {{0.125, 0.125, 0.176777, 0., 0.125, -0.125, 0., 0.176777}, {0.125, 0.125,
    0.176777, 0., 0.125, -0.125, 0., 0.176777}, {0.176777, 0.176777, 0.25, 0.,
    0.176777, -0.176777, 0., 0.25}, {0., 0., 0., 0., 0., 0., 0., 0.}, {0.125,
    0.125, 0.176777, 0., 0.125, -0.125, 0.,
    0.176777}, {-0.125, -0.125, -0.176777, 0., -0.125, 0.125,
    0., -0.176777}, {0., 0., 0., 0., 0., 0., 0., 0.}, {0.176777, 0.176777,
    0.25, 0., 0.176777, -0.176777, 0., 0.25}}

    If you trace out the Dx degrees of freedom (by cutting the matrix in 4 (4x4) matrices), and you take the trace in each of these submatrices, you get the
    local (A0,B0) density matrix which is

    1/2 0

    0 1/2

    and that proves that the state at the D0 detector is an incoherent mixture of photons that come from A and photons that come from B, without any interference.

    Last edited: Jan 14, 2005
  18. Jan 14, 2005 #17


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    Up to here we are saying exactly the same (except that - I grant you that - I used "entanglement" too early: the pump photon is indeed in a superposition of states, but this turns immediately into an entanglement of the regions A and B: |A-emitting a pair>|B-not emitting a pair> + |A-notemitting>|B-emitting>). Yes, I was sloppy with the words.

    Granted, I used superposition and entanglement sometimes interchangedly, because the initial superposition turns quickly into an entanglement. But it was sloppy.

    It is this collapsing business which leads to all these "paradoxes" and so on. But if you do not insist on a "collapse" when a remote detector gives a click, but just apply linear quantum theory until you arrive at your final observations, all these paradoxes and mysteries disappear.

    I'm applying them correctly. The best proof is that I find back all the results that are observed. I'm simply not using "intermediate collapse". I work straight through everything with linear quantum mechanics, and use Born's rule to find the final probabilities of the quantities we will discuss, such as correlations.

    All right, all right :smile: Ok, I could be nasty and ask you mathematically what is the difference between a superposition and an entanglement, and you'd have a hard time if you are not allowed to define about which subspaces we're talking: I can turn any superposition mathematically in an entanglement and vice versa. But I won't be so nasty :-)

    When I say that, I am talking about the superposition of AMPLITUDES
    1/sqrt(2) (|A> + |B>).
    I'm not talking about a statistical mixture.

    But I did ! It is in the superposition of amplitudes.

    Nope, nothing ever collapses before the end of your experiment. Sometimes you CAN get away with it, because you will only consider one branch in the superposition in all what follows (like in a state preparation). But when you use collapse indifferently, in this kind of experiments, then you shouldn't be surprised to have "paradoxes" like "predicting the future" and so on. And be even more surprised that nature does it in such a way that you cannot use it really.

    No, it is not. Producing idler photons is just a physical process which can be described by linear quantum theory.
  19. Jan 14, 2005 #18
    I'm afraid the experiment is too complicated for any discussion here to be of much use. I'll have a little try, but everyone is bound to get confused!

    The interference patterns always depend on phase factors, of course -- ones introduced by means of path length changes -- but the way in which scanning the (invisible) pattern on one arm enables its existence to be revealed in the coincidences with the other is too subtle to be understandable without writing a complete paper.

    The general principle probably (though I still haven't looked at the actual paper you're talking about!) involves the existence of identical phase differences in the two arms, combined with the fact that these phase differences must always be initially (if frequencies are constant) one of just two values, 0 and 180 deg.. The reason the interference patterns are not immediately visible is that the two patterns, one for each phase value, are superposed and wash each other out. The use of coincidences enables the selection of just one of them.

  20. Jan 14, 2005 #19


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    ??? If you're scanning an image with a "point" detector, or you use a position-sensitive detector which gives you each time the impact position of a click (I'll avoid the word photon in order not to make you nervous :-), I don't see what's the difference ?

    Of course, without the use of photons, it is a bit hard to think of WHY these independent intensity measurements of classical beams by photomultipliers could ever disentangle interference patterns, but I leave it up to you to explain that :smile:

    As my back of the envellope quantum calculation shows, with the concept of photons and a bit of basic quantum theory, I haven't got the slightest difficulty in deriving these results.
    By results, I mean results of principle, like "I will clearly see fringes" or "I won't see fringes at all". A thorough analysis needs of course to take into account all effects of all optical elements, but nevertheless, we clearly see fringes (raw data :-) when we trigger in coincidence with D1 or D2, and we don't see when we don't take these coincidences, or with D3 or with D4, exactly as predicted by a simple calculation in QM, 5 lines long.
    So try to find a classical wave description that fits this (it may be more than 5 lines long)... I am curious :-)

  21. Jan 14, 2005 #20
    On edit: please be sure that you have read all of this post and the following two before you start responding. I don't believe that you have understood the thrust of my assertions; you are apparently assuming that I am arguing that it is possible to see into the future, and that is not what I am arguing at all. You will understand better if you read all three. End of edit.

    My problem with this is that, while I'm certain that the math properly describes the beginning and end of the experiment, I'm not certain it describes the middle.

    See, the way I'm used to relating to things, and using math, is you figure out a story about what happened, then you apply mathematics and observations to show whether that story could have happened that way or not; if not, then your story is wrong. You keep at this until you have a story that fits the facts. When you deal with quantum mechanics, that story is going to incorporate some elements that are impossible from the point of view of classical mechanics, and you have to be aware of where those elements are so that you don't wind up describing something that can't happen.

    Telling that story is what interpretations of QM are all about. That's what the CI originally was about, that's what Cramer's TI is about, that's what MWI is about, that's what Gell-Mann and those guys made Consistent Histories for, that's why Bohm made his non-local interpretation, and that's what Feynman's Many Paths is about. All of them allow you to tell a story about what took place. And if you can keep the peculiarities of QM and of your chosen interpretation in mind, then you can tell a consistent story (although, as I say, it will- not may, but will- incorporate some elements that are incompatible with classical mechanics, and therefore with what we think we see in the world around us).

    It's my belief that if you can't tell that story, then you don't understand what is taking place. I think that a great many physicists have missed that point; they (and you) seem to think that the entire story is in the math. I'm in the process of learning the math- and that's not easy for someone who's been out of college as long as I have (twenty years). But it's my experience in dealing with these types of things that once I have learned it, I'm going to find exactly what I'm saying- that it works to describe the beginning and end, but it doesn't describe the middle. Furthermore, I'd be willing to bet that once I know the math, I'll be able to show you exactly what's been left out, and what that implies about the story. The reason I believe this is because of all the trouble everyone's been to to create these interpretations in the first place. I believe that the physicists who have made the various interpretations have come to the same conclusion I have- that the math doesn't tell the whole story; it just describes the beginning state, and the end state. And if you had the right math, and there might be more than one (has to be more than one, if both MWI and Bohm's interpretation can both give correct results- and they can), then you can describe all of the middle stuff too. There are many ways to slice the cake and still get the right number of pieces.

    Let me give an example.

    Let's take a game of pool. We know that we will begin with the cue ball at one end of the table, and all the other balls racked up at the other end. We know that we will end up with all of either the stripes or the solids in pockets, and the eight-ball in a pocket. What you are telling me is that that is the entire story of the game. And I'm asking you, "what about the killer shot, a kick off the head rail and a combination of the twelve-ball into the nine, in the corner? Are you trying to tell me that shot never happened?" And you're saying, ""you can't prove it happened- the twelve and the nine are in the corner pocket, and by the way, that's all that's important!"

    Here's another example.

    There's a meteor falling out of the sky. On it's way to the ground to become a meteorite, it hits a plane, and the plane crashes, but the pilot was skillful enough to make it a ditch instead of just an impact, and all the people got out before the plane sank, so they were all saved. It also went through an office building, and five people's desks and the entire filing department of a major insurance company were destroyed, and three people died. Then it hit in the middle of the street and made a crater. And you're telling me that the only thing that's important is the crater, and that's all we can talk about. And I'm saying, "What about the plane? What about all the people that pilot saved? What about the filing department of that insurance company, and all the trouble that people who are trying to collect insurance will have? What about those people who died?" And your response is, "the only important thing is the crater and the meteorite at the bottom of it; you can't prove any of the rest of it ever happened. The math only talks about the crater and the meteorite."

    Do you see now my problem?

    I think I know the experiment you are talking about. I tried to find a really cool site that went into it in detail, but couldn't find a link or the site on google. I think this will make our conversation more profitable- we'll both be talking about the same experiment! :tongue2:

    OK, yes, I follow that now. It's actually a really good explanation of why the interference is recoverable- the phase of the idler photons controls which of D1 or D2 they go to, and their phase is determined by the original phase of the pump photon, and that phase also determines whether the signal photon is a member of the first or second interference pattern.

    To put it in terms of a story, the pump photon's phase results in a particular phase for the resulting signal and idler photons that are created in the β-BBO crystal; and the phases of those two photons are orthogonal. The phase of the signal photon determines whether it is in the first or the second interference pattern, or in the non-interference pattern. The phase of its associated idler then determines whether it goes to D3/D4 at BSA/BSB, or to the eraser; and if it goes to the eraser, then the phase determines whether it goes to D1 or D2. Do I have that right?

    Hrrrmmm, we're talking at cross-purposes. I wasn't talking about the 100% efficiency case, because it's impossible; that was clear from the two paragraphs that follow this one, which you have quoted below; your response there is clear enough. I was talking about the real world case; and real-world detectors might not click for a particular photon; so one might just be lost. I'm not sure it makes any sense to talk about the 100% efficiency case, because you can't ever see that in the real world. To answer your question, the other photon ends up not being detected, because the detectors aren't 100% efficient.

    Problem is, I'm pretty convinced that those principles can't be correctly applied without taking the real world into account. I guess you see now why I quibbled about 100% efficiency in the first place.

    As a matter of fact, I believe that you are correct- there cannot be a way to detect whether there is interference without the correlation to the events at D1 and D2- but it is only an opinion. So far, I can't prove you are correct, based on the experimental data. I recovered the data (photon counts for various positions of D0, variously correlated with D1, D2, and D3) from the charts in Figs. 3, 4, and 5, and analyzed it to see if I could prove that you can't see the difference between the non-interference case and the combination of the two interference patterns. The data were inconclusive; this is at least partly because they only took non-interference data at half the spatial resolution of the interference data. In fact, I saw what could be interpreted as a greater mean deviation from the "best curve" of the non-interference case in the combined interference data than in the non-interference data, leaving open the possibility that you could tell the difference; but it's inconclusive, because of the lower frequency of the non-interference data. I wish they'd taken more care to relate the non-interference and interference cases to one another rather than concentrating on the interference cases; I can understand it based on what they were trying to show, but it still is a little hinky. It leaves a loophole that you can drive a truck through.
    Last edited: Jan 14, 2005
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