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I Comments on Alice and Bob

  1. Aug 12, 2016 #1

    Paul Colby

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    Theory often affords multiple viewpoints or interpretations of phenomena. Not all of these need provide one with a feeling that they are sensible. For example, we could choose to view the classical trajectory of say a baseball as being “caused” or due to some future position and velocity and by convention simply integrate the equation of motion in time reverse order obtaining what we normally refer to as the initial conditions. That we can do this doesn’t in itself recommend it.

    I would like to layout a view of the classic EPR style experiment in which the decay of an S=0 state particle pair is measure by our friends Bob and Alice. My intent in doing so is to address in part what I feel is a common miss use of terms like “cause” and “effect” in regard to Alice and Bob’s measurement choices. It is clear that choice of measurement apparatus in QM very much is effects measurements, it’s the nature of the beast. It is also clear that Bell’s arguments are an important aspect of the phenomena. But, I’m left wondering in what sense does Alice’s measurements effect Bob’s? Well, there is a view point where Alice’s measurements simply are unaffected by Bob’s.

    Because QM measurements involve random events beyond our control things are complicated experimentally. No observation in this type of work is comprised of a single event. In this case what is a single event? We define a single event as a measurement on a single entangled pair of particles. This event, say event ##i##, is Alice’s measurement angle, ##\alpha_i## with her up or down result, ##a_i## obtained. Each of these are paired with Bob’s chosen angle, ##\beta_i##, and his up or down result, ##b_i##. We suppose that Alice and Bob dance around wildly choosing measurement angles at random (or not, it matters not) and we let them do so for a very long time. In this time we collect each event outcome on a card. This gives us a set of measurements, ##(\alpha_i,a_i,\beta_i,b_i)## which we put in a big deck. Now, since each event is independent of every other even, we may shuffle the deck freely. No matter how the deck is shuffled we will always have a potential time sequence of event which is completely consistent with QM and whatever Bell has to say. Each time the deck is shuffled we get a physically reasonable potential outcome of an experiment. Now, we could also choose any subset of cards and we will still have a consistent potential time sequence or outcome for an experiment.

    Now, I’m going to choose two sequences of events of identical in length, say ##N##, from our much bigger deck. The first is simply the first ##N## cards of the big deck. The second I am going to choose card ##k## such that ##(\alpha_k,a_k)## is the same as Alice’s results for the ##k##-th card in the first deck. What this does is it gives us two experiments in which Alice’s event sequence is identical in both decks. Both of these experiments could happen even though arranging it real time is not possible (well, very very unlikely at best). Both experiment time sequences are completely consistent with QM. At this point I ask; can we really claim that Bob’s measurements have effected Alice’s? How is such a claim made since both Alice’s results are identical between the two time sequences while Bob's are completely different?

    Now, Bob certainly does see a dependence on his choice of relative measurement angle to Alice’s. But this is to be expected since all Bob’s measurements depend on Bob’s choices in QM.
     
    Last edited: Aug 13, 2016
  2. jcsd
  3. Aug 13, 2016 #2
    This is not that easy like card games. First you need to know how polarizers works.
    We have a Malus-law for polarizer.
    https://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties
    Second important detail the independency of events.
    "Two events A and B are independent (often written as {\displaystyle A\perp B}https://wikimedia.org/api/rest_v1/media/math/render/svg/6522d65c3a0cf823fff40501a93b74e877a96f1e or {\displaystyle A\perp \!\!\!\perp B}[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/92e1d46c72b6742b13c8a88a72090534bd60c06f) [Broken] if their joint probability equals the product of their probabilities:"
    https://en.wikipedia.org/wiki/Independence_(probability_theory)
    Now we can start make "science"
    Suppose that there are two independent random events on both side of measurement. We can write up the probalility of common detection of photons.
    P=cos2(p-a)*cos2(p-b)
    p is the polarization of photon and a,b are the orientation of polarizers.
    Let a=0 b=90 p=45 the probability of common detection will be nonzero.
    The problem starting here because the equation of entangled photon of type1 is
    P=0.5*cos2(a-b)
    When a=0 and b=90 the probability will be ZERO.
    Well the detections of entangled photon are not INDEPENDENT events.
     
    Last edited by a moderator: May 8, 2017
  4. Aug 13, 2016 #3
    Its works by classical Malus-law when p=a or p=b only.
    /This called "spooky action at the distance" because you can rotate both polarizers freely./
    But the result will be symmetrical only if half of sample will be p=a another half will be p=b.
    Its looks like case of one photon and two polarizers but the lightsource alternately bounce between the two sides.
    Surprizely the equation is same as entanglement.
    P=0.5*cos2(a-b) when initial light is unpolarized.
     
  5. Aug 13, 2016 #4

    Paul Colby

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    The cards contain a record of actual data clicks of detectors and angle settings that have happened. All the QM is done by the time one has the cards filled out. Shuffling the deck and selecting subsets is allowed because the events are independent of one another. One may construct[1] two plausible measurement histories that are consistent with QM in every way and in which Alice has identical results between the two while Bob's differ between the two. This argument doesn't resolve anything and says nothing new. It does, however, provide a view in which Alice's measurements are unaffected by Bob's. The reverse also holds if Bob and Alice roles are reversed.

    [1] one may also construct statistical outliers. For example flipping two coins and always choosing Bob's to be heads. One may also with the simple picking scheme yield two very plausible measurement histories that are consistent with the expected statistics in every way.
     
  6. Aug 13, 2016 #5

    Paul Colby

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    I notice that a large section of my original post does not appear, however, when I attempt to edit the post the missing bits jump into existence as if by magic. I hate these web based things. Okay, I've restored the missing section. Sorry.
     
    Last edited: Aug 13, 2016
  7. Aug 13, 2016 #6

    Paul Colby

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    Nor have I ever claimed such. Each of the cards contains the results that occurred for a single entangled pair. The results are obtained from an actual measurement.
     
  8. Aug 13, 2016 #7
    I get it. The big different between cards and entanglement is the randomness. Better you think that there are two dice.
    If you roll a six with one the other dice has six also. This randomness is very important.
    In this metaphore east to see that is impossible to send message with it.
     
  9. Aug 13, 2016 #8

    Strilanc

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    So you record two runs of experiments, then re-arrange the list of results from the second run so that it matches up with the first run? And this is somehow an analogy for what happens during Bell tests? I don't see how they're equivalent in any relevant way.

    Yes you can create correlations by cherry-picking and splinching data. But then the data isn't going to be representative of what you measured anymore. Why are you even doing the experiment, if you're just going to ruin the data like that? Even just looking at subsets of the data can create apparent superpowers (e.g. post-selecting away wrong guesses to create the appearance of communication).

    The thing that makes Bell tests interesting is that you get the surprisingly-strong correlations in the raw data. It's due to the physics, instead of due to cherry-picking and editing.
     
  10. Aug 13, 2016 #9

    Paul Colby

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    I often encounter the statement that Alice's measurements are "affected" by Bob's choice. The construct I used produces two measurement histories that could actually happen and are not, when taken by themselves, statistical outliers. In them Bob's measurements depend on his choices of measurement angle relative to Alice's just as QM predicts, yet Alice's measurement angle choices and measured results are identical between the two experiments. My construction is required because the chance of two successive measurements of this kind, while not physically disallowed, are very very unlikely to occur in a real unedited measurement.

    I have in no way changed the statistics in either measurement. One certainly can by choice of cards, but I'm not in this case. All correlations predicted by QM are evident in both data sets.
     
  11. Aug 13, 2016 #10

    Strilanc

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    Yes, but you made them more likely by discarding data where they didn't happen. You're cherry picking. It's not enough that "it could happen".

    Basically you're describing an experiment where Alice picks a list of random settings, then uses it twice. Post-selecting doesn't avoid the consequences of using random data twice, because knowing that post-selection is going to happen can be used to gain the same advantage as knowing the same list of settings would be used twice.
     
  12. Aug 13, 2016 #11

    Nugatory

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    You will often hear this, but it's not an accurate statement of the physics involved, it's one of those things that you'll find in popularizations that try to explain the physics without demanding excessive rigor. (The only place it shows up in serious discussions is in the context of collapse interpretations, and there the conclusion is that collapse doesn't work well in analyzing spacelike-separated measurements on a quantum system). Thus, it is something of a strawman and there's only so much to be gained by refuting it instead of replacing it with a more correct and complete statement of the problem.

    One cup of coffee into my morning, I do find the question @Strilanc raises to be interesting: Is your procedure in fact statistically valid? Any approach that filters the raw data should be presumed bogus until proven otherwise requires a fair amount of justification. Gimme another cup of coffee and I may find this question less interesting, but I'm pretty sure that it will take more than handwaving to justify the approach.
     
  13. Aug 13, 2016 #12

    Paul Colby

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    Totally true. One (meaning me) must show that the statistics of each experiment taken in isolation is consistent (likely) given QM. The first experiment is taken directly from the first N cards and has no cherry picking and therefore can't be questioned. The second "cherry picked" experiment is constructed from the remaining cards not used in the first set. It is completely obvious that this cherry picking can be done to make anything happen. That's not the intent. The intent is to get a consistent history that isn't a statistical outlier when taken in isolation. This is what is meant by "could happen". Have I shown this, no. Is it possible, yes (he asserts without proof). I'm with Nugatory in that the question is likely not sufficiently of interest to people to warrant the effort. The reason to bring this up at all is the hope that this provides yet another way to look at the problem that might help others think about it.
     
  14. Aug 14, 2016 #13
    The dice metaphore is not really good because there is no measurement. Now I try another one.
    Now we have two coin. The coin has two state: heads or tails . It is like up and down spin of electron.
    The angle of measurement will be the tilt axis between coin and top of the table.
    If you drop like zero degrees then we got same state as initial state was. It is clear.
    But when you let fail into edge of coin /90 degrees/ then you can not tell the final state. The chance is 50-50 percent.
    This is works perfectly like
    https://en.wikipedia.org/wiki/Stern–Gerlach_experiment
    with entangled electrons.
    When you have two entangled coin then and drop it in 90 degress both, the result will be suprising:
    If one coin become tail the other will be head always.
     
  15. Aug 14, 2016 #14

    Strilanc

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    Drop two coins on their side, and you could get any of HH, TH, HT, or TT. They don't always disagree. The coin system has two output states and a continuous measurement angle, but doesn't work as an analogy because the coin outcome doesn't behave the way the quantum outcome does.
     
  16. Aug 14, 2016 #15

    stevendaryl

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    It seems to me that whether Alice's measurement choice has an effect on Bob's particle depends on whether quantum probabilities are objective or subjective. According to Alice, Bob initially has a 50/50 chance of observing spin-up or spin-down along the [itex]z[/itex] axis. If Alice measures the spin of her particle along the [itex]z[/itex] axis, then Bob's chances change to either [itex]0/100[/itex] or [itex]100/0[/itex]. From Alice's point of view, Bob's probabilities changed drastically and instantaneously. If probabilities are viewed as a subjective property (it reflects Alice's knowledge about Bob, rather than reflecting something directly about Bob), then maybe nothing objectively has changed in Bob's neighborhood. If probabilities are viewed as something objective, then it seems as if Bob's situation was changed by Alice's choices.
     
  17. Aug 14, 2016 #16
    you do not say?lol
    try to read again
    "If one coin become tail the other will be head always."
    This is kind of magical coins. Entangled. The essense is not what you get.
    The essence how magical coins behave randomess but connected. This works perfectly like real entanglement.
     
  18. Aug 14, 2016 #17
    Okay change the situation.
    Suppose that the coins are a little magnets
    and the table has hidden uniform magnetic field.
    Then its works perfectly.
    Quantum entanglement is solved.
     
  19. Aug 14, 2016 #18

    Nugatory

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    That doesn't work - in fact, it is impossible to construct any mechanism for manipulating the two objects (coins, dice, subatomic particles) that will reproduce all the correlations predicted by quantum mechanics. This is Bell's theorem: http://www.drchinese.com/Bells_Theorem.htm
     
  20. Aug 14, 2016 #19

    bhobba

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    No.

    The difference is cards are the same whether observed or not, but QM is silent on if things are the same regardless of observation - its very interpretation dependent.

    This is the rock bottom cause of Bell inequalities.

    There is a view, one I agree with but to be clear its simply a view, that entanglement is the the rock bottom essence of QM:
    https://arxiv.org/abs/0911.0695

    You need to read the above in conjunction with Hardy's seminal paper (and again I believe it is part of the essence of QM - but it's just a view which like any interpretation is no better or worse than any other - it simply reveals what I find elegant and beautiful but in an objective sense means Jack shite):
    http://arxiv.org/pdf/quant-ph/0101012.pdf

    Thanks
    Bill
     
  21. Aug 15, 2016 #20

    zonde

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    There are carefully chosen angles ##\alpha_k## and ##\beta_k## for which you can't carry out that procedure for several steps (same subset for Alice -> different subsets for Bob -> same subset for Alice for both Bob's subsets but different as a first Alice's subset) even if you allow arbitrary cherry picking. That is basis for Eberhard's version of Bell inequalities.
     
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