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Afshar experiment questions - PROPONENTS only:P

  1. Jan 22, 2008 #1
    Why didn't the afshar experiment get more "luuv" and was dismissed and critized so hard?
    All he did was configure a experiment which showed Bohr had it wrong and helped us move on from a 80 year old "block".
    Ok, so we have to throw out MWI and CI (to me something I love), I can understand it's proponents are having a hard time accepting this, but it's phycists, not religion, when mother nature speak, you either listen or go to church.
    Why wasn't it recognized more mainstream?
    Also, which interpretations survive the ashar experiment besides Transactional Interpretation?
  2. jcsd
  3. Jan 22, 2008 #2


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    ALL interpretations survive the Afshar experiment. It complies entirely with a standard quantum calculation, which, in this case, even reduces to a classical optics calculation.

    It is only the "it's a wave" / "no it's a particle" mumbo jumbo which doesn't survive the Afshar experiment.
  4. Jan 22, 2008 #3
    Part of the debate is whether in fact he showed anything. Specifically, it's impossible to know whether the photons strike the same detector that they would have but for the grid. So it's impossible to know whether "which slit" information really was preserved. It's possible that the grid disturbed the photons in such a way as to re-randomize their path.

    I had an idea to try to use entangled particles to remove this ambiguity but thanks to Zelinger I'm struggling with how to make an interference pattern using entangled particles; it's quite difficult, if you believe Zelinger.

    That said, Afshar definitely showed that the photons physically avoid the reigons where the grid is located; the interference pattern is not merely a statistical curiosity. Whether this actually contradicts any current interpretation of QM, I am not sure.

    I do not understand why you think either MWI or CI must be thrown out because of this. I don't think Bohr ever denied the existence of space or that photons move through it...

    Not sure what this means either. If the "which-path" problem I referneced above were dealt with, I think it'd be pretty good evidence that geometry of the photon's path is governed by wave mechanics while the photon is unquestionably a particle. Sounds like it supports Bohm if you ask me.
    Last edited: Jan 22, 2008
  5. Jan 23, 2008 #4


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    This is because the "wave/particle" duality is a much too naive way of seeing things, and doesn't really correspond to the modern formalism of quantum theory. A photon is not a classical wave, and is not a classical particle, and doesn't "switch" pictures between them, although this was what was sometimes said in the early days of quantum theory, and these are relics which one still finds in some textbooks.

    What happens - no matter the interpretation - is that a quantum system starts out in a prepared state, which corresponds to a certain basis, which can be closer to "localised" (particle ?) or closer to "spread out" (wave ?) or whatever, depending on the exact preparation setup.
    Then, one has unitary evolution of that state (and if anybody now says that it is a "particle" or a "wave", he's fooling himself, although, again, one could analyse the "locality" of the evolving state vector, and whether or not it is more "lumped" or not, would make you say that it is "particle-like" or "wavelike", but most of the time, you have just a statevector evolving in hilbertspace which is NEITHER).
    Finally, you do an irreversible measurement in a specific measurement basis, which can be again, more "particle-like" (if you do a position-like measurement) or more "wavelike" if you do an interference experiment.

    Now, whether there is a genuine collapse, or whether this is only "in the eye of the beholder" is a matter of interpretation.

    But what is NOT in the eye of the beholder is that IN BETWEEN the set up basis (choice of the setup), and the measurement basis, there is simply unitary evolution of the state vector. No interpretation at all tells you what "aspect" this unitary evolution must have, and whether one must "choose between particles or waves" (it is NEITHER).

    But in this case of the Afshar experiment, things become in fact VERY SIMPLE, because it is a single-photon experiment, and the peculiarity of single-photon experiments is that one can use classical optics as the "unitary evolution". That is, you can give, to a single photon, the "statevector" of a classical EM field, evolve it as in classical electrodynamics, and then use the relative intensities of the fields, integrated over the detector surfaces, as the probabilities of detection. Mind you, it is just a mathematical trick to simplify the quantum-mechanical state vector evolution. But, keeping this in mind, as we know much more and have better intuition of classical EM, can you tell me if a light flash (seen as an EM wave) evolving through a screen with one slit open, is "particle-like" ? Yes, in a certain way: it is a LUMPED wavepacket after the screen. When it meets the grid, however, is scatters a bit, and propagates now as a more extended "wave" through the lens, to give you a more or less diffuse image on the detectors ---> probability of detection.
    If a light flash evolves through BOTH slits, it becomes TWO wavepackets (particle ? wave ? Two particles ? ), and the two wavepackets happen not to have any summed amplitude at the wires of the grid, so there is no scattering. The "wave" then propagates through the lens, and gives you two fine peaks of intensity in the image: the probabilities of detection.

    But who is to say that "the click on detector 1 was the wavepacket that went through hole 2" ? That doesn't mean anything ! We simply PREPARED a photon to go through the two slits, and we MEASURED a 50/50 chance for it to arrive on detector 1 or detector 2.
    As the two EM wavepackets came together, interfered at the level of the grid, and got separated again, who is to say what part of the EM field afterwards "came from the first packet" and what part of the EM field "came from the second packet" ? You simply can't do that, nor in classical EM, nor in quantum dynamics (which happen, in this case, to be mathematically identical).

    So the only thing the Afshar experiment does, is to point out an erroneous interpretation of assigning "wave or particle" natures to a quantum state IN BETWEEN preparation and measurement, which is IN ANY CASE an error.
    Last edited: Jan 23, 2008
  6. Jan 23, 2008 #5
    Well, the evolution of the state vector, correct me if I'm wrong, always evolves like a "wave" in the sense that if it starts out localized to a single point (i.e. single slit) it will spread out in a gaussian pattern, whereas if it starts out localized to two possible points (i.e. double slit) it will spread out into the familiar interference pattern. Hence the idea of the "pilot" wave. And I think that what the Afshar experiment shows, or at least was _intended_ to show, is that regardless whether you are ultimately able to determine the position of the photon, its state vector _always_ evolves like a wave.

    Exactly. So why is this not tantamount to saying there is a pilot wave governing the path of the photon?

    How can it be just a mathematical trick if the total number of photons detected in the Afshar experiment was unchanged by the presence of the grid? This mathematical trick describes real space that the photons really avoid. Leaving aside the issue of whether we can say which photon went through which hole, we cannot deny that virtually ALL the photons avoided the interference minima can we?

    I also do not understand this, because a minute ago you said that this was a single-photon experiment and that the classical EM wave was just a mathematical trick. We know that a single photon always goes through one slit and never both. And both detectors never go off at the same time. So how can you say that there is a "first packet" and "second packet" that contribute to the final detected photon unless you are now saying that the classical EM wave and the photon actually _are_ one and the same, which brings back the duality? In that case, I could see the argument that the EM wave that exists as an unfocused interference pattern at the grid becomes refocused afterwards to create a new photon travelling in one of two random directions.

    However, if you're going to treat the photon as a particle always, then you have to concede that one and only one photon travelled between the grid wires and emerged from the other side to hit the detector. Are you arguing that there's a 50/50 chance that the photon did a "magic bullet" and made a hard turn in mid-flight?

    It is only an error in CI.
  7. Jan 24, 2008 #6


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    Yes. That is, it follows unitary evolution between preparation and detection.
    But I thought that, no matter the interpretation, that is ALWAYS the case.

    Because the notion itself of "path of the photon" is not needed nowhere. Its quantum state evolves.

    But that is already ASSUMING that the photon "follows a path", and hence ASSUMING that it is classical-particle like.

    BTW, the "mathematical trick" I was talking about was the "trick" of using classical electrodynamics to "mimic" the (genuine) unitary state vector evolution, because in the case of single photon situations, both are mathematically equivalent.
    What I meant was that one should actually work out the unitary state vector evolution, but because people don't have much intuition for this, in this particular case, we can do with classical EM wave propagation, it will come out the same.


    No, on the contrary. If the photon is in a superposition of going through both slits, then you cannot say that it "only goes through one slit" ! That's exactly the kind of error that leads to all the fuzz around this experiment.

    The quantum state "photon at slit 1" is physically different from the quantum state "superposition of photon at slit 1 and slit 2".

    Of course, because *at detection* the wavefunction gives you a probability of detection. But at the slits, there's no detection, and hence the wavefunction doesn't represent any probability.

    No, the classical EM wave is not the photon, but the classical EM wave follows the same mathematics as the unitary evolution of the single-photon state.

    So with some stretch, you can consider the classical EM field to be the hilbert state vector of the photon.

    I don't see why you consider it to be a *new* photon. It is the evolution of the photon state vector. That said, if you consider photons to be the quanta of free field plane waves, yes. But in this kind of setups, it is better to consider photons to be the quanta of the modes of propagation of the setup.

    Consider an electron in a hydrogen atom. Now, consider that it is hit by a UV photon, and is set free (ionisation). Did the electron come from the "upper part" or the "lower part" of the s-orbital of the hydrogen atom ?

    I see the photon (or any other particle) in a quantum state compatible with the initial setup. That is, it is OR in the quantum state "top slit" (if only the top slit is open), it is OR in the quantum state "lower slit" (if only the lower slit is open), OR it is in the third distinct physical state which is |top slit> + |lower slit> (both slits open).
    Mind you that the superposition principle tells us that this is NOT a statistical mixture, but a distinct physical state.
    Next, this quantum state evolves unitarily throughout the setup, and this unitary evolution takes into account the presence of the grid. It is at this point that I used the "trick" of using classical optics (EM) to find it easier to visualise the unitary evolution (which is mathematically equivalent to it).

    Finally, the quantum state arrives at the detectors. At that point, the quantum state describes a probability distribution of detection.

    The state |top slit> gives rise to a blurry picture, with many hits on det1 and few on det2. The state |bottom slit> gives rise to a blurry picture, with many hits on det2 and few on det1. The state |top slit> + |bottom slit> gives rise to a sharp picture with 50% chance on det 1 and 50% chance on det 2.
    But it has no meaning to say that IF we hit det 1, then "it must have been that the initial state was actually |top slit>. No, the initial state was |top slit > + |bottom slit>. And 50 % chance to DETECT det 1 and 50% chance to DETECT det 2.

    The physical state |top slit > + |bottom slit> is NOT the same as the statistical mixture 50% top and 50% bottom, and this is the implicit reasoning error which is present when one "traces back" from det1 to "it must have been slit 1".
  8. Jan 24, 2008 #7
    Ok I understand your argument but I think your premise is that Bohr and CI are correct in that physical reality is _solely_ defined by what the human experimenter sees.

    Let me ask you a question then - what experimental evidence would, in your mind, disprove CI in favor of, say, a pilot wave theory or TI?

    And let me ask you another question. And I know this sounds stupid but I do not intend it to be facetious. The air molecules immediately beyond the slit "know" which slit the photon went through do they not? Let's put two clear pieces of glass beyond the slit. Those, too, "know" which slit the photon went through. Yet we'll still see the interference pattern right?

    Now let's put a miniature single photon source beyond each slit which, when it receives a photon, emits an _identical_ one and records the receipt. Suddenly we won't see an interference pattern. Why? Because human beings (as opposed to inanimate molecules) now "know" which slit the photon went through? So somehow the quantum state evolution depends on whether a human made the observaiton instead of a molecule? Does anybody really believe that?

    As someone (can't remember who) once put it: "We might not know whether Schroedinger's cat died, but the cat certainly knows."
  9. Jan 24, 2008 #8


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    Unless the different interpretations , CI, TI, MWI, deBB, Nelson etc make different predictions it is not possible to tell them apart by experiment.
  10. Jan 24, 2008 #9
    There is a difference between making no different predictions, and making no different predictions that you care about. CI only cares about where the particle is found because Bohr and Heisenberg told us that's all we should care about. I care about where the particle is before it's found and that's what the other interpretations are trying to address that's what people are _attempting_ to design experiments to test.
  11. Jan 24, 2008 #10


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    In all current interpretations, this is contradictory. You can only know where it is when it is found. This information doesn't tell you anything about where it was before your measurement. The Bohmians have been struggling for years to design an experiment to settle the issue, and have so far failed.

    And in my judgement the Afshar experiment also failed to do this.
  12. Jan 24, 2008 #11
    You might be right, but that doesn't mean it can't be done.

    In my view we're in a state not unlike physics before Michelson-Morely. Sooner or later someone will definitively show one way or another whether objective reality exists and, hopefully, someone else (or maybe the same person) will build a fundamental new theory from the result.

    Anyway no one answered my question. Why does the human observer collapse the wave function but the glass and/or air molecules that "see" the photon don't?
  13. Jan 25, 2008 #12


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    Well, the point of my earlier post is just that it can't be done.

    I believe reality is objective by definition. Let's hope for better theories, though.

    There is no wave function so it can't collapse or do anything. The 'collapse' is just a mathematical problem in CI.
  14. Jan 25, 2008 #13


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    No experiment can do this, as these interpretations are, as far as quantum theory is concerned, IDENTICAL in their experimental predictions. It is only when one would find / or not find a DEVIATION from the predictions of the quantum formalism (and we all agree that the Afshar experiment doesn't CONTRADICT the predictions of the quantum FORMALISM), and after one would have build a NEW theory, that some interpretations might lend themselves better to the change than others. But as long as standard quantum mechanics prevails, all interpretations are empirically indistinguishable - because they are ALL compatible with the quantum formalism.

    No ! The point is that nothing interacts in an irreversible way (that means, is in a different physical quantum state) after the experiment. Actually, if this IS the case, then the interference doesn't appear anymore. The quantum-mechanical explanation for this is that, if the state of another system (glass plate or anything), has interacted with the original photon system in such a way as to entangle with it, that we have the following:

    |glassplate0> (|slit 1> + |slit 2> ) before.

    after irreversible interaction:

    |glassplate +> |slit 1> + |glassplate -> |slit 2>

    Now, you can easily work out that concerning the photon system only (reduced density matrix), all observables to "slit 1 and slit 2" will behave as if the state changed from:

    "|slit 1> + |slit 2> " into a statistical mixture of 50% slit 1 and 50% slit 2.

    This is what one calls "decoherence". Another way of seeing this is that the sum |slit 1> + |slit 2> happens in the hilbert space of the photon only and gives you a new vector in that hilbert space which can be "longer or shorter", while the entangled state has "orthogonal components" of the other system added to it, so any norm will just add the amplitudes squared.

    So from the moment that there is interaction which leaves another system in a state which is "dependent on which way" (which means, in quantum speak, that this other system entangled with our system under study - the photon - during interaction). At that point, interference (the getting longer or shorter of the sum) disappears, because we now add the amplitudes squared.

    Nope, we WILL see interference. That's btw what happens in a laser amplifier.

    No, not at all. The state of the emitter will be identical whether or not it "absorbed and re-emitted" an identical photon or not, and hence will not entangle with the photon. So the photon system remains unentangled, and will hence show interference.

    However, if something entangles with the photon system (be it a single air molecule, or anything) then the interference will disappear.

    This phenomenon is called decoherence.

    EDIT: btw, this is the reason why "macroscopic quantum experiments" are so difficult to do - or impossible even. The "signature" of quantum mechanics is interference: that is: we have a system in a superposition of states which look classical, and we let this superposition evolve in such a way that it becomes a "classically observable state".

    We have initially a system in |A> + |B> (say, "slit 1" and "slit 2"), and we evolve it, by setting up an experiment, that this evolves into a new state which is observable (hit at peak of interference pattern, or at bottom, or focus through a lens at one detector, or another).
    So, |A> + |B> evolves into |C>.
    If we can now design or change the setup such that we have initially:

    |A> - |B> (that is the same states, but with a different phase relation), and we can have it evolve in a DIFFERENT observable state |D> (shifted interference pattern, other detector, ...) so that:

    |A> - |B> evolves into |D>

    then we have the typical setup which demonstrates quantum interference in the following way:

    If we apply only |A>, then we find 50% |C> and 50% |D>
    If we only apply |B>, then we find 50% |C> and 50% |D>

    If we now apply |A> + |B> then we only find |C>
    and if we apply |A> - |B> then we only find |D>

    Clearly, if we had a statistical mixture of 50% A and 50% B, we wouldn't find "only C" or "only D", so this demonstrates without ambiguity that there was a superposition of states, and not a mixture at the start.

    ALL quantum interference experiments are somehow modelled on this.

    But the difficulty with these experiments is that NOTHING may entangle with A or B during the evolution, not even a single molecule of air. If this happens, then this molecule will "entangle" and

    |A> + |B> will evolve into |mol1> |A> + |mol2> |B>

    and this will give rise to 50% |C> and 50% |D>.

    So the slightest entanglement will screw up your quantum interference experiment.

    The "bigger" your quantum systems, the more difficult it is to preserve it from interaction during evolution (even with a single molecule or photon or ....). For macroscopic systems, it is totally impossible to avoid this beyond extremely short timescales. Even the cosmic microwave background will end up interacting with the system. This is why quantum interference is hard or impossible to observe for bigger systems.
    Last edited: Jan 25, 2008
  15. Jan 25, 2008 #14
    There seem to be different opinions about what causes a collapse. A currently common opinion seems to be that any interaction which leaves a trace of information will cause a collapse.

    I'm wondering whether one could say it is an interaction of an (Heisenberg-) "uncertain" property with a "certain" property which will cause the uncertain one to become certain as well. Just speculating, but maybe the certain one will then perhaps be more uncertain?

    Whereas, again just speculating, when two uncertain properties interact, they cause interference. Sorry if this is nonsense... :smile:
  16. Jan 25, 2008 #15
    colorSpace - How dare you post speculation that is not mathematically sound! :)

    I'll try to cover you though.

    So what you're saying is

    Certainty + Certainty = Certainty (Classical)
    Certainty + Uncertainty = Certainty (QM wavefunction "collapse")
    Uncertainty + Uncertainty = Uncertainty (QM wavefunction "interference")

    Logically, the operation is A OR B, where "true/false" = "certain/uncertain"

    Interesting way of looking at it.
  17. Jan 25, 2008 #16
    vanesch, I won't requote your post because it contains a ton of (very helpful!) information. But let me just make sure I understand you correctly.

    In a quantum interference experiment involving interference patterns shown using single-photon sources, are you saying that the photons that are used to demonstrate the interference pattern have not interacted with a single molecule of air? How, then, do these experiments work when mirrors or lenses are employed? Surely these photons have "interacted" with those items as well? No lens is perfect... Isn't some physical property of the glass altered by the fact that the photon passed through it and isn't that detectable theoretically?

    Better yet, let's take the laser amplifier idea. So if the laser amplifier re-emits the photon _and_ sends a tiny electrical signal to a computer to indicate a hit, the interference pattern disappears, but if the laser amplifier re-emits the photon without sending such a signal, the interference pattern remains?

    Finally, again, let's talk about the Cat. So according to you, the Cat metaphor is wrong because, in fact, the quantum object (the quantum that triggers the cyanide capsule) actually has interacted with a ton of other things inside the box and caused an irreversible measurement to occur. Do I understand you right?
  18. Jan 25, 2008 #17


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    For those photons that make up the interference pattern, although there was interaction, the mirror or the glass should be in exactly the same quantum state before and after in most of the cases. That's what I'm saying. Now, you're right that no mirror or lens is perfect, and that means that there is *a small amplitude* to interact irreversibly:

    |mirror0> (|slit1> + |slit2>) becomes actually:

    0.99 |mirror0> (|bouncedslit1> + |bouncedslit2>) + 0.001 |mirror1> |bouncedslit1> + 0.001 |mirror2> |bouncedslit2>

    In other words, in MOST of the cases, after interaction, the mirror came back in its original state (or in a state which can't distinguish between slit1 and slit 2), and we have interference, but in a few cases, the mirror did alter itself (a dustspeck got displaced a bit or something) and then the state entangled itself with slit 1 or slit 2.

    What we will observe is a "diminished" interference pattern.

    You won't be able to build a laser that re-emits exactly one photon and that signals it. Because what you have is that a system (atom, crystal...) should get one state up and de-excite immediately to end up in exactly the same state (otherwise it won't be the same photon that you are emitting).

    Yes. There won't be a "superposition of cat states" by themselves, there will be a superposition of a "dead cat entangled with the air, the environment etc..." and a "living cat entangled with the air, the environment etc... ".
    To an observer (no matter the interpretation - I prefer MWI, but it doesn't matter), this will then appear as a statistical mixture 50% dead cat and 50% living cat.... exactly as we classically expect. No "cat -interference".
    Last edited: Jan 26, 2008
  19. Jan 26, 2008 #18
    Well it only matters in one respect. There's four possibilities:
    1) The cat is alive objectively.
    2) The cat is dead objectively.
    3) The cat is alive in one universe and dead in another universe, objectively. (MWI)
    4) The cat is neither alive nor dead until the observer sees it. (CI)

    The problem I have with #4 is that the cat knows whether he's alive or not. So unless MWI is true, naive CI is absurd.
  20. Jan 26, 2008 #19
    But how ? To reach this state of awareness the cat would need to have the capability of deriving the argument of the Cognito ergo sum of Descartes. Perhaps the cat can know it is confined within a box, as a form of perception, but I do not see how it can know that it is alive. Cat cannot transform perception into concepts of life and death. Show me a cat that knows it is alive, and I use it to start a new religion and get rich.
  21. Jan 27, 2008 #20
    Ok, you really have totally missed the point. And if you really think the laws of physics care about whether something "cogitet" or not, that's one of the most arrogant vitalist ideas I've ever heard.

    And by the way, just for the sake of argument, yes, the cat knows it's dying when the cyanide opens and it starts to feel like crap. Cats go off to be alone when they're about to die. Come to think of it, cats are more self-aware than most people I know.
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