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Understanding Quantum Eraser Experiments

  1. May 5, 2015 #1
    I would like to start a discussion about the most basic principles behind quantum eraser experiments.
    I understand this has been debated here many, many times and for some of you there is nothing to talk about, but I still find it interesting and would like to get some of your opinions.
    I know there are many different experiments of this type that have been proposed and also realized.
    Some of them try to show some new aspect of it and become quite complex, but I would like to focus on the most simple ones. It seems like the basic concept is that if we have two particles which are entangled with respect to some observable, an action on one of them that deletes information will have some non-local implication on the other particle. I understand that depending on your favorite interpretation you may deny non-locality, but at least that's what at first sight it looks like. If you have the two particles with some 2-state observable maximally entangled, which means this is a singlet and you take some action on the particle on the right that will make it impossible to distinguish between these tow states, then in principle it should be possible to observe interference (after sending a few particles of course) on the other side right?. Well, I know you will say "No", this only happens if coincidence counting is done. I wonder how we can model this concept if we send the singlets one at a time, don't loose any, and don't have any noise or non-entangled particles. In real-life we would have to choose a particular type of particle such as a pair of photons produced by down-conversion, some kind of fermion or even atoms, and choose the particular observable that we are entangling. But maybe we can think of a generalized example without getting into the details. My understanding is that in principle what we have is complementarity between "path" distinguishability and interference. But I think you might have different concepts and opinions. If you would prefer to talk about a particular simple model, I am willing to discuss that.
  2. jcsd
  3. May 5, 2015 #2
    I don't understand this part -- action on one particle? What do you mean by that?
  4. May 6, 2015 #3


    Staff: Mentor

    It doesn't seem like that to me at all.

    To me 'it seems' that in some simple cases decoherence can be undone - which of course is exactly what's going in.

  5. May 6, 2015 #4


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    Gold Member

    You can think about generalized example but if you want to check your idea against reality you have to think about it in context of real experiments.
  6. May 6, 2015 #5
    I didn't say what type of action because I wanted to keep it general, not specific to a particular experiment. But I'll try to explain in more detail what I meant. Interference between two (past) states can only be observed when there is no permanent recording that shows the particle having been being in one state or the other. This would apply to the double-slit experiment. During certain time there may be a way to make a measurement and obtain one of the two states as a result. But if instead of making the measurement we do one of two things 1) let it evolve until we can't distinguish anymore between the two states or 2) do something to erase any information that may be present and which would allow to distinguish in which of the two states the particle was before. Then interference between the two states can be seen.
    In the case of the two entangled particles, we can say that each of them carries information about the other. One of the two particles could be manipulated in such a way that there is no way to distinguish between the two states, and then the other particle should be able to display interference. What kind of manipulation I am talking about? The one they do in all of these quantum eraser experiments. In the first thought experiment by Scully, the wall between two cavities was removed, in the one by Dopfer, the detector was placed at a position that would not be able to focus on the slits, in the one by Walborn, a polarizer is inserted. Then there is the one by Yoon-Ho Kim that is more complex as it uses two separate down-conversion sources. And of course there is Wheeler's initial thought experiment which did not involve entangled particles and in which I doubt we could talk about "erasure". We could get into one of these experiments, but I thought that if the principles behind all of them are basically the same, we could make an attempt to discuss those principles without getting into all the details of each experiment. But if you prefer to talk about one in particular, that's OK with me. I hope I made a little more clear what I had in mind during my initial post. If I didn't I apologize.
  7. May 6, 2015 #6
    According to the picture presented by Zurek, decoherence would imply an uncontrollable leak of information towards the environment, which would make it hard (perhaps not impossible) to "undo".
    I would have to look into each of the experiments, but from what I remember, the only stage at which there was decoherence was when the photons were detected.
    What you are saying makes some sense to me, specially when we talk about "erasure", but I don't recall any experiment where this could apply. Do you have a particular one in mind?
    Last edited: May 6, 2015
  8. May 6, 2015 #7


    Staff: Mentor

    It has been discussed here before:

    If you want to understand the exact mechanism in a particular set-up I will have to leave that up to an experimental type - but the principle is well known - in the quantum eraser all that's happening is you have undone decoherence.

  9. May 7, 2015 #8
    I know this question is a little off-topic, but as you consider decoherence and it's undoing as the explanation for quantum eraser experiments, do you think decoherence solves the measurement problem?
  10. May 7, 2015 #9


    Staff: Mentor


    Here is what it solves and does not solve:

    The issue is the difference between an improper and a proper mixed state.

  11. May 8, 2015 #10
    I did take a look at the article. So going back to two entangled particles running in opposite directions, how do you apply the concepts of improper and proper mixed states to quantum erasure? According to the article when describing one particle before the other has been measured it would be in an improper mixed state and after the other particle is measured, the first particle is in a proper mixed state if you haven't received information about the result of measurement. But when the other particle, even if it is far away and you can't know what the result of measurement is, you could conclude that it is in a definite eigenstate right?
    Now if you describe this in the composite Hilbert space of both particles, it looks like measuring the other particle would "collapse" the wavefunction and not only reduce the density matrix but leave only one of the diagonal elements on it right?
    I also don't see how all these ideas conflict with the picture I presented before.
  12. May 8, 2015 #11


    Staff: Mentor

    You will need to go though an actual set-up with an experimental type - which I am not.

    But in general when what you observe interacts with what you are observing it with, it becomes entangled with it, and a superposition gets converted to a mixed state. Its turns out in simple cases that entanglement can be undone. So what would happen is if you observe which slit it goes through it becomes entangled with what you are observing it with, and the superposition is now a mixed state. But it is possible to disentangle it so it becomes a superposition again.

    Here is a complete analysis of the closely related delayed choice experiment:

  13. May 10, 2015 #12
    That chapter you sent me the link to is part of Griffiths consistent histories book. I had read it before, and I always found it interesting.
    I was just hoping to have a little debate on this topic. But that's OK. Thanks anyway Bill.
  14. May 13, 2015 #13
    Ok. So, here's a generalized example.

    Consider the state

    |ψ> = N ( |a>|b> + |a'>|b'> ) .

    The probability density ρ(x) for finding system-1 at the point x, when no measurement is performed on system-2, is given by

    ρ(x) = |N|2 { |<x|a>|2 + |<x|a'>|2 + 2Re[<a|x><x|a'><b|b'>] } .

    Observe that the interference term (the third term) vanishes when the two system-2 states |b> and |b'> are orthogonal; i.e. <b|b'>=0.

    In the words of Zeilinger:

    Formally speaking, the states |a>1 and |a'>1 ... cannot be coherently superposed because they are entangled with the two orthogonal states |b>2 and |b'>2.

    Alternatively, one can put it this way:

    With no measurement performed on system-2, the 'which-way' information for system-1 remains fully intact (via its entanglement with the two orthogonal states of system-2) and so there is no interference.

    Let us then set the apparatus up such that |b> and |b'> are orthogonal, in which case we can write

    |ψ> = (1/√2) ( |a>|b> + |a'>|b'> ) .

    ... So how does one completely erase the 'which-way' information for system-1 encoded in system-2?

    Perform a measurement of |e> on system-2, where |e> has the property |<e|b>|=|<e|b'>|; that is,

    <e|b'> = α<e|b> , where α is a phase-factor.

    In that case (whenever the result |e> is obtained), the state |ψ> → |φ> where

    |φ> = N ( |a> + α|a'> ) |e> ,

    for which the interference characteristics of system-1 are fully manifest.

    This is what Zeilinger means by:

    ... the interference pattern can be obtained if one applies a so-called quantum eraser which completely erases the path information carried by particle 2. That is, one has to measure particle 2 in such a way that it is not possible, even in principle, to know from the measurement which path it took, b or b'.

    Here are the correspondences for the Walborn setup:

    |a> ↔ slit-1

    |a'> ↔ slit-2

    |b> ↔ left-circular polarization

    |b'> ↔ right-circular polarization

    |e> ↔ horizontal polarization
  15. May 13, 2015 #14
    I see here this correspondence is for a 'simplified' Walborn setup. You can see that setup at the bottom of page 341 [p. 6 in the PDF-viewer] at this link:


    In this setup, though, one cannot argue for nonlocality.
  16. May 13, 2015 #15
    Hello Eye_in_the_Sky. Thanks for your interesting posts. This morning I sent you an email.
    After looking at your post, I grabbed my (hard) copy of Zeilinger's "Experiment and Quantum Physics" which I think is the article you were looking at.
    I'll have to read it again. (I have read it many times already but that was a few years ago).
    However, what I see is that Zeilinger in page S290 talks about one pair of photons. His interpretation of quantum erasure is very clear. Now, I just read that page and I haven't read yet sections IV and V. But what I get from section III is that his argument, being based on just one pair of photons, would not (at least in principle) require coincidence counting for interference to appear. Of course in the real experiment due to technical reasons, you only see interference when you consider the photons that are entangled and discard all the others. This has been discussed here a lot many years ago and the discussion eventually died down. The arguments that were presented in favor of the idea that coincidence counting is needed in order to observe interference were not convincing to me. Tonight I'll take another look at the other sections of Zeilinger's paper so that I can see if he expresses (at least in that paper) some argument in favor of the need for coincidence counting. I think I have read before that he does think it is necessary.
    Now, going to Walborn's experiment, I don't see that one as much different than a typical EPR-Bohm experiment. I think describing a hypothetical experiment of that nature could clear things up. Let's say we have this pair of entangled particles (fermions) where like always one goes to the left (A) and the other to the right (B) and we record the arrival time for each photon. Let's say we measure spin in the vertical direction on A and in the horizontal direction on B. When we look at our measurements on A, we find that half of the time the particle has spin up and half of the time spin down. If we look for interference by recombining the paths in the Stern-Gerlach we don't find any. Same thing happens for particle B being measured in the horizontal direction. But if after the experiment (which was ideal and did only contain entangled photons) we select only those pairs (which we can do because we recorded the time of detection) where the photon B was detected as having horizontal spin pointing to the right (not looking at the paper from the top but looking along the direction of motion of the particle) we do see interference on particle A. Of course , to see interference on A we have to set it up to measure interference and not the up-down value on that particle. Now, if we select the pairs where the spin of B was pointing left, then we also see interference but the interference fringes are displaced with respect to the previous ones. So here there is no demonstration of non-locality (as you pointed out). Now Walborn seems to be saying that the idea that quantum-eraser experiments could be used for faster-than-light (he actually says back-in-time) communication is false and that his experiment confirms it. I think his experiment shows a particular setup in which faster-than-light communication is clearly non-feasible. But this does not prove that in a different type of setup it could not be done. Of course there is Eberhard's theorem, but that's a different story.
    As I said before, I will read Zeilinger's paper in more detail and then I'll post again.
  17. May 14, 2015 #16
    Oh, but we do need coincidence counting.

    Let's see what's going on.

    At the A-arm of the apparatus, the dots begin to fill the screen one, by one, by one. We keep track of the sequence of dots: dot-1, dot-2, ... dot-N. Once the dots have filled-in sufficiently, we find a bell-shaped distribution which shows no interference at all. This will be the case no matter what actions are performed at the B-arm of the apparatus.

    However, if we are told which of the dots on the screen (at the A-arm) correspond to a measurement of |e> (at the B-arm) for which the outcome was "yes", then we do indeed find that those dots taken alone will show an interference pattern.

    ... Coincidence counting required.
  18. May 14, 2015 #17
    The picture you are presenting seems to be similar to the one I mentioned in my previous post which was a simplified version of Walborn's experiment. Well, I think it was simplified at least at the conceptual level. Experimentally it would be harder to implement as it uses spin 1/2 particles instead of photons. Using your example, if you select those dots where the outcome is "no" when measuring |e> at the B-arm, do you see interference at the A arm?
    So we could keep track of all particles, assign a number to each pair and put them in a database right?
    Are you the same Eye I sent the email yesterday? I am wondering because each eye is connected to a different hemisphere in the brain. Perhaps one of your hemispheres is into classical physics and the other into quantum physics. Oh, changing the subject, I just read Zeilinger's section IV. I didn't find new evidence of his denial of the possibility of interference without coincidence-counting (I know that's his position, but at least he is not saying it in that section). What I did see is a train of thought much in sync with the Copenhagen interpretation and against realism. (no mention of non-locality in that section but it may be implied).
  19. May 15, 2015 #18
    Yes, you do.

    In general, you will see the combined intensities of two patterns:

    1) The anti-interference pattern ,


    2) A bell-shaped distribution which shows no interference .

    The first aspect will always be present.

    However, in an experiment for which P("yes")=P("no") (and, hence, both are ½), the second aspect vanishes.

    Note that, for any erasing experiment which can be 'mapped' to our generalized example, we must necessarily have P("yes")≤½. This follows from the conditions:

    |b> and |b'> are orthogonal ,

    |<e|b>| = |<e|b'>| ,

    and the original state is

    (1/√2) ( |a>|b> + |a'>|b'> ) .
    In the idealized scenario, yes.
  20. May 15, 2015 #19
    Did you really mean P("Yes")≤1/2 or you meant P(Yes)≠1/2?
    It looks to me that this scenario is exactly the same as both the Walborn experiment and the one I mentioned with spin 1/2 particles. Basically there are fringes and anti-fringes and when you add them up you get the Bell-shaped curve. In this case it is very clear that coincidence-counting is necessary.
    Now, let me make it clear. I am not saying that I am in favor of the idea that a quantum-eraser experiment can be done without coincidence-counting. All that I am saying is that the explanations I have seen (in my opinion) do not prove this concept in a simple and intuitive way. If the need (in a setup such as Dopfer's) for coincidence-counting is true, there should be a simple way to explain it. It should be an explanation as simple as the one you showed or the one I mentioned. But as I said, I think these both refer to a different setup and don't necessarily apply to something like Dopfer's experiment.
    I'll try to think about a convincing argument about the fundamental difference between these two experiments and explain it in my next post.
    Last edited: May 15, 2015
  21. May 15, 2015 #20
    OK, here it is. The main difference between the experiments we discussed above and Dopfer's is that in the latter the number of photons that form the interference pattern is the same as the ones that form the Bell shape curve. So we can't recover after-the-fact an interference pattern on the lower arm by selecting a sub-group of the photons according to the results of the measurement done on the upper arm (the Heisenberg lens arm) We can make the detector in the upper arm large enough that no photons will be lost. Now, there is something that complicates the picture. Zeilinger mentions (referring to Klishko) that this can be seen as if the source were the double-slit and the down-conversion crystal acted as a mirror. Now, when the (entangled) photons reach the double-slit, some pass through and the ones that don't go through get absorbed by the material of the double-slit. The "partners" of those photons that didn't go though will still be detected at the upper branch. So when we place the detector at the imaging plane we don't "see" the slits (but I guess if the detector was a very efficient CCD camera we could for each photon know if it's partner went through the slits by seeing at what position in the imaging plane it landed, and even determining through which slit it went. So this should destroy interference at the lower arm (behind the slits). Now if we put the detector at a distance of 2f, at which the image gets totally blurred, we are destroying which-way information and the interference pattern should show behind the slits. Each photon that reaches the detector in this position will be detected at a random position and there can be no inference from this as to which slit the other one went through. Even with the additional non-entangled (I mean non-entangled with the ones that went through the slits) photons in the upper branch, this should not provide any additional information that would allow us to find which-way information. So interference at the lower branch should be seen.
    Another thing that distinguishes this experiment from Walborn's or the one you showed is that we don't have an |e> measurement that gives a particular result (in your example it was "yes" or "no". In this case it could be argued that the detector placed at the 2f (blurring) distance is the one that does the |e> measurement because it is at that moment that which-way information is erased. OK, but you don't get "yes/no" or anything like that. The only thing you can measure at that detector is position, and it happens to be random, without providing any additional information (of course the other thing you measure is the time of arrival). And you get the interference pattern at the lower arm without having to select a sub-group of the photons based on the result of the upper branch measurement. Now, you may say that coincidence-counting is what does that selection, but besides filtering-out any "noise" detection events, coincidence counting does not seem to select a particular sub-set of the entangled photons. Unless you say that the arrival time at the upper branch detector is what is in this case represented by your |e>. That could at first sight be a feasible argument, but I doubt it could be made to work.
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