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Unit Circle and Quantum Predictions [hidden variable model agrees with QM?]

  1. Oct 8, 2014 #1
    I would like to reopen a discussion on assertions made by David Mermin such as: “There is no conceivable way to assign such instruction sets to the particles from one run to the next that can account for the fact that in all runs taken together, without regard to how the switches are set, the same colors flash half the time.” Taken from his 1985 article: Is the moon there when nobody looks? Reality and the quantum theory.

    He describes a simple gedanken demonstration with a source, and two detectors each with three switches. He then proceeds to explain how the data can be explained by quantum theory, that is, cos2120=1/4. And, that the data violates the Bell’s inequality and there can be no instruction sets. But, and striking (at least to me), he never made the connection between the gedanken model’s 3 switches and the cosine function. Without any mention of angles prior to this; it seem to appear out of nowhere.

    I suggest a model that uses angles, the unit circle, where the trig functions are defined. Place three points on the circumference of the unit circle 120 degrees apart and label them in any order Bob, Chris, and Alice. This guarantees they choose different angles. Each are free to choose an angle and their choices are independent of each other. For simplicity, in each trial Bob’s and Chris’s choices are compared to Alice’s. Now rotate all three points simultaneously by 60 degree increments and record the sequence of plus and minus given by the cosine of the angle for each. Example: B-, C-, A+ = Bob’s sign is negative, Chris’s is negative, and Alice’s is positive. You will generate 8 triplets. The unit circle cannot assign + + + or - - - triplets for these angles. I know what you are thinking. How can you generate 8 triplets when there are only 6 distinct sequences remaining? Two of them repeat as follows: Bob chooses 240 and gets B-, Chris chooses 120 gets C-, and Alice is left with 0 and gets A+. But Bob’s and Chris’s choices of angles can be reversed. Two ways to get the same sequence: - - +, this only occurs whenever Bob and Chris get the same signs as shown below for 1) & 2) or 7) & 8).

    240, 120, 0...........Bob/Alice....Chris/Alice.......(D = different signs, S = same signs)

    1) B-, C-, A+...............D....................D

    2) C-, B-, A+...............D....................D

    0, 120, 240

    3) B+, C-, A-...............D....................S

    4) C+, B-, A-...............S....................D

    180, 300, 60

    5) B-, C+, A+..............D....................S

    6) C-, B+, A+..............S....................D

    60, 300, 180

    7) B+, C+, A-..............D....................D

    8) C+, B+, A-..............D....................D

    Reordering the 8 triplets as BCA yields: (- - +), (- - +), (+ - -), (- + -), (- + +), (+ - +), (+ + -), (+ + -)

    P(S) = 4/16 = ¼ same as quantum theory!

    It is unlikely that quantum theory would have contradicted the unit circle. The fact that an analysis of the unit circle demonstrates the quantum prediction is indeed satisfying, but what are the implications? The hidden variable is the unit circle and the properties exist at all times, yet the above data (according to Mermin) violates Bell’s inequality. Does this discovery advance the debate over classical reality vs quantum reality, local vs nonlocal interactions, or Bell Theorem? Or, does it muddle the debate even more? Was Einstein both right and wrong? No pun intended.
  2. jcsd
  3. Oct 8, 2014 #2


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    If I understand what you're doing, you're maintaining the same relative angles between A, B, and C as you rotate the circle. It is possible to construct local realistic models that match the QM prediction under that condition. What's not possible is to match the QM prediction when A, B, and C independently choose their angles on each trial.
  4. Oct 8, 2014 #3


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    Your explanation is really unclear.

    • Don't label points with names of people. Points aren't agents.
    • Similarly, wording like "the unit circle assigns" is very confusing. How is the assignment made? What process is determining it? A person? Randomness?
    • The choice of angles can't be both independent and guaranteed to differ. Which is it?
    • Why does having "two ways to get --+" not also apply to the other states, and double the total from 6 to 12 instead of increasing it from 6 to 8?
    • Be more concrete about what is actually *happening* in the experiment. Like... a photon gets sent through a splitter to make an entangled pair. The information in the left photon of the pair is _____. In the right photon it's _____. When the switches are ______, the outcomes are determined by ______. Go through all of the cases.
    • Don't jump ahead to "What are the amazing consequences of this?!" before you've gotten past "Is there a simple mistake in this that I've missed?". There's almost always a simple mistake. In this case I think it's that the angle choices have to be worked out ahead of time or at least post-selected if they're not going to differ. This violates the assertion that they're independent.

    It would also be a good idea to start by clearly explaining how quantum mechanics gets the right answer. You should know how the switches relate to the cosine function.
  5. Oct 8, 2014 #4
    I disagree. It will give the same result for A, B, and C independently choosing an angle. I stated this in the post. No what order A, B, and C choose an angle only 6 sequences with two that repeat are possible. Rotation was mention only to demonstrate keep the points 120 apart and not a necessary requirement and any initial ϴ will give the same result. The P(s) = 1/4 for the complete data set will not change. If you disagree could you demonstrate it an example from the unit circle to prove your point.
  6. Oct 8, 2014 #5

    My post is clear. I will respond to why only two sequences repeat and not the others. Take the triplet (+, - ,-), now switch the first two to get (-, +, -).
    You get a different sequence. Same with (-, +, +) you get (+, -, +). Changing Bob and Chris choices gives different sequences! However as stated in the post clearly for the triplet (+, +, -) when you switch the first two you get (+, + -) back. It is counted twice because Bob's and Chris's angles have changed.
    For sequence 7) Bob's choice of angle was 60 degrees, but for 8) was 300 and the sign for cos 60 = cos 300 = +.
  7. Oct 8, 2014 #6


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    If the angles A, B, and C are chosen independently from the set {0°, 120°, 240°}, with uniform chance for each angle in the set, then the outcome A=B=C=0° occurs with probability 1/33 = 1/27. So clearly it is not guaranteed that the angles are always 120° apart if they are chosen independently.

    My guess is that you have something different in mind when you say "independently". That word has a very specific technical meaning.
  8. Oct 8, 2014 #7
    The post is only concerned with A, B, C choosing a different angle and quantum prediction of P(s) = P(cos 2120) = 1/4
  9. Oct 8, 2014 #8


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    That doesn't answer my question. How exactly are the angles being chosen?

    Is it like a dice roll where you choose uniformly between
    • A=0°, B=120°, C=240°
    • A=120°, B=240°, C=0°
    • A=240°, B=0°, C=120°
    • A=240°, B=120°, C=0°
    • A=120°, B=0°, C=240°
    • A=0°, B=240°, C=120°

    Does each detector get an angle assigned randomly and independently, but only the experiments where they didn't accidentally get the same angle are kept?

    Suppose I wanted to simulate your system with pen, paper, and dice. Please list exactly the steps I should follow: when to roll the dice, what to write down, what decisions to make based on outcomes, etc.
  10. Oct 8, 2014 #9


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    Sorry rlduncan. It is possible to cherry pick results and observers and make it look like your idea works, but you should have expected that any serious analysis will demonstrate errors.

    So here a couple that I am passing on. In your example, you show 2/8 as S for Bob/Alice (cases 4 and 6). That is correct. You also show 2/8 as S for Chris/Alice (cases 3 and 5). That is also correct. But oops, you forgot to show Chris/Bob. And guess what! That works out to be 4/8 (cases 1, 2, 7 and 8). When you average those for all observers, you get... 1/3. Not 1/4 as you announce. Your example is what is called "observer dependent" while a true realistic example must be observer independent.

    In fact, you should be able to see the problem easily by looking at your result triplets (- - +), (- - +), (+ - -), (- + -), (- + +), (+ - +), (+ + -), (+ + -). There are 8x3=24 permutations there. 4 permutations are --, 4 are ++, and the remaining 16 are either -+ or +-. That makes 8 as the Same, 16 are Different, for the result predicted by Mermin of... 1/3. While QM predicts 1/4.

    Using your logic, and getting to choose the observer pairs I want, I can give you any value from 0% to 100%. Please note that the part of your example where you rotate by 60 degrees each time is an unnecessary distraction. That merely serves to generate a sample of resultsets that is pseudo-random.

    The correct rules for this game are: a) You give the resultset as a batch of triples (as many as you like, values as you like). b) I pick the observer pair (which 2 of the 3 to consider) and I do this for each triple, but independently of knowing the outcome values you selected for each triple. c) The angles don't really matter, but the choices you gave in your example are the best: 0, 120, 240.
  11. Oct 8, 2014 #10
    Rolling dice will work. The probability you select same angle is 1/3 and selecting a different angle is 2/3. The probability of getting (+, +) or (-, -) for all runs is P(S) = (1/3)(1) + (2/3)(1/4). I have demonstrated the unit circle can reproduce the quantum theory prediction of 1/4 from the cos2120 for selecting different angles 1200 apart. I did not think it was necessary to include the trials of selecting the same angle because you always get the same result. No issue or conflict with quantum theory.
  12. Oct 8, 2014 #11

    Hello DrChinese

    Why would you include Chris/Bob???? It is irrelevant to the experiment! If Alice chooses 0, then there are 3 choices 0,120, and 240 for the distant location. Alice is at a distant location from Bob and Chris. Agree? When the angles are different we are only interested in comparing Alice/Bob and the counter-factual Alice/Chris. Agree?

    If you prefer we only need Bob at a distant location. Then yes we need the angle labels to distinguish Bob's two choices that are different from Alice's.
    Why would we compare Bob/Bob? Please clarify your position.
    Last edited: Oct 8, 2014
  13. Oct 8, 2014 #12


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    1. There is also counter-factual Chris/Bob, so no I completely disagree. That you need to consider this is simple to explain. You analyzed Bob/Alice and Chris/Alice and were amazed that the result is 1/4 (by ignoring Chris/Bob). Ok, now this time ignore Chris/Alice and consider Chris/Bob instead (also considering Bob/Alice as before). Presto, the results will now be 37.5% instead of 25%. That is a feature when YOU cherry pick the observer pairs. The observers make their choices freely and you are refusing to allow that. It does not work either if you hold one observer setting constant (Alice) and then allow the other to change as YOU want it to.

    So sorry, that does not pass the test.

    2. I never suggested this.
  14. Oct 8, 2014 #13


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    rlduncan: Another way to see the problem of your example is this: Let me pick whether to pick 2 of Alice (fixed at 0), Bob (fixed at 120) and Chris (fixed at 240) for each pair. I will do that below, but you do it for a new set of triples that you generate.

    Alice, Bob
    Alice, Chris
    Bob, Chris
    Alice, Bob
    Bob, Chris
    Alice, Bob
    Alice, Chris
    Bob, Chris
    Bob, Chris
    Alice, Bob

    You will find that your results converge to 1/3. Unless, of course, you cheat by hand picking to get the results knowing which 2 is picked from each triple. Because a random distribution will fail for you.
  15. Oct 8, 2014 #14
    1) Sorry, but ignore Chris/Alice which are at distant locations from each other and replace with Chris/Bob which are at the same location. This is a serious misunderstanding of the experimental setup! In an actual experiment there are just Bob and Alice. Bob has two choices that differs from Alice in which one is a counter-factual (Chris) which is compared to Alice if Bob had chosen that angle setting.

    2)Yes you can hold Alice constant and get the overall results just with fewer trials.
  16. Oct 8, 2014 #15


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    Thread locked, pending moderation.
  17. Oct 8, 2014 #16


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    After some discussion among the moderators, we're going to leave this thread locked.

    Explaining Bell's Theorem and the experimental support for it is a reasonable topic for Physics Forums, but that's not what where this thread is going.
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