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I CFD - Counterfactual Definiteness

  1. Aug 15, 2016 #1
    I am thinking about a single run in a typical Bell-type scenario:

    A joint-measurement of an (appropriately prepared) entangled property is performed in spacetime regions A and B at spacelike separation.

    Regarding the above, in conjunction with Quantum Theory, would you say the following statement is true, false, or inapplicable? Why?

    The 'state of affairs' relevant to the outcome at A is independent of the setting at B.

    I am particularly interested in replies which assume a single (non-branching) universe.
    Nonetheless, replies from all perspectives are valued.
     
  2. jcsd
  3. Aug 15, 2016 #2

    DrChinese

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    My opinion is that the statement is false. :smile:

    Both settings (A and B) are inputs to something (I don't know what or where or when). Therefore the outcomes at A and B reflect in some manner the mutual relationship of both settings. Therefore they are not independent. Again, this is simply my view.
     
  4. Aug 16, 2016 #3
    DrChinese, thank you for your reply.

    (I too favour a view whereby the statement is false.)
    ____

    What then would you (especially, DrChinese) say about the following statement?

    This dependency is established through the agency of an 'influence'.

    True or false?

    (In my favoured view, this statement is also false.)
     
  5. Aug 16, 2016 #4

    Simon Phoenix

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    I don't really understand what you mean here by 'state of affairs'.

    Let's consider the usual scenario with our usual protagonists Alice and Bob each having one particle from a maximally entangled singlet state. What we can say is the following : the results obtained by Alice (Bob) are independent of the settings chosen by Bob (Alice).

    Suppose this were not the case and there were some measureable dependence on the settings then Bob could simply change his settings and convey information to Alice.

    But this, I feel, is not what you're asking - it would seem that you're asking about some kind of 'change of state' caused by a change in Bob's setting prior to Alice's measurement?

    So let's take a singlet state |01> + |10> and assume Bob performs some unitary transformation on his particle (essentially this is equivalent to a change of measurement setting). Then the new state would be |01'> + |10'> where the prime indicates the transformed basis. The reduced density operator for Alice's particle is unaffected by this unitary transformation, as are the statistics of her measurement results. The global quantum state has, however, changed.

    So yes I would say that in the sense above the 'state of affairs' is changed by Bob changing the measurement setting but this change has no observable consequences for Alice's measurements alone (i.e. considered separately from Bob's measurements).
     
  6. Aug 17, 2016 #5

    N88

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    The statement is true, based on my understanding of locality.
     
  7. Aug 17, 2016 #6
    Loosely, by the 'state of affairs' in a given region of spacetime, I mean all of the "goings-on" in that region.

    More generally, but again loosely, by the 'state of affairs' (not necessarily within spacetime) I mean all of the "goings-on" with regard to those things that have ontological status (possibly outside of spacetime).

    Suppose Alice obtained the results 0100101101, with Bob's setting fixed at b1.

    If her results are truly independent of his setting, and if the theory for the phenomenon is truly faithful to this fact, then no contradiction should arise in conjunction with the theory by supposing that Alice obtains the self-same results, i.e. 0100101101, if Bob's setting had been fixed at b2 instead of b1.

    Thus, "independence" (with regard to the phenomenon) and "faithfulness" (of the theory) imply CFD.

    Do you agree?

    Alice's outcome has a dependence on both the setting and the outcome of Bob. Although Bob is able to control his setting, he has no control over his outcome. For that reason, Alice is unable to detect a change in Bob's setting.

    Only if quantum states have ontological status would they be considered, by my intended meaning, to be a part of the 'state of affairs'.
     
  8. Aug 17, 2016 #7
    Thank you N88 for your reply.

    Please have a look at my reply above to Simon Phoenix. Do you have any comments?
     
  9. Aug 17, 2016 #8

    Simon Phoenix

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    No :-)

    The results obtained by Alice are statistically independent of the setting of Bob, or in other words, the probability that Alice measures up (or down) given that Bob's setting is b is the same as the probability that Alice measures up (or down) for all choices of b. Or in symbols P(+|b) = P(+).

    Alice and Bob's results are not, of course, statistically independent if they each have one of an entangled pair.

    Sorry I can't follow your reasoning here.
     
  10. Aug 17, 2016 #9

    bhobba

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    What was it Meatloaf said - You Took The Words Right Out Of My Mouth


    I don't know why, but there seems to be a lot of confusion about this - its simply a correlation with different statistical properties different to the classical case.

    Thanks
    Bill
     
  11. Aug 17, 2016 #10

    bhobba

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    Neither.

    QM is silent on it. The answer requires an interpretation.

    Pick an interpretation and I can often (but not always) answer it. For example here is Consistent Histories take:
    http://quantum.phys.cmu.edu/CQT/chaps/cqt24.pdf

    Then we have the Ignorance Ensemble. Unfortunately it doesn't answer it either, nor does Copenhagen - not all do. In fact most minimalist interpretations don't - that's because they are minimalist. In BM the systems are linked by the super-luminal pilot wave and that is the influence.

    Thanks
    Bill
     
  12. Aug 17, 2016 #11

    zonde

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    What is alternative to "influence"?
     
  13. Aug 17, 2016 #12
    Simon, thank you for your reply.

    I will need some time to think about it more, and also to think more about the position I think I am trying to express.
     
  14. Aug 17, 2016 #13
    As I see it:

    The 'state of affairs' in spacetime region A and the 'state of affairs' in spacetime region B are together in a condition of nonseparability.

    (i.e. the joint-state of Alice and Bob's measuring devices is nonseparable)

    OR

    The notion of a 'state of affairs' in a spacetime region is altogether invalid.
     
  15. Aug 17, 2016 #14
    Yes, but in a view whereby 'spacetime' has ontological status the quantum correlations carry with them implications that do not arise in the case of classical correlations.

    See my reply to zonde above.
     
  16. Aug 17, 2016 #15

    N88

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    1. As the OP, please define what you mean by CFD.

    2. Already we know that you hold true what I (based on my understanding of locality) hold to be false.

    3. Your reply to Simon begins (with my emphasis), "Loosely1, by the 'state of affairs' in a given region of spacetime, I mean all of the "goings-on" in that region. More generally, but again loosely2, by the 'state of affairs' (not necessarily within spacetime) I mean all of the "goings-on" with regard to those things that have ontological status (possibly outside of spacetime)."

    I see no need for Loosely1. Given loosely2, I'd welcome an example from the phrase that it qualifies.
     
    Last edited: Aug 17, 2016
  17. Aug 17, 2016 #16

    N88

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    Consider your CFD, our quantum world and the following thought-experiment:-

    As in Bell (1964), a quantum source SQ produces highly-correlated pairs of spin-half particles. As is well-known, these quantum correlations deliver an expectation E(a,b|Q) = -a.b.

    These quantum correlations are next reduced to 'classical' correlations by sandwiching SQ between two aligned S-G magnets. The new source, SC, is a 'classical' one because it produces pairs of spin-half particles correlated by having definite but antiparallel spins in one direction. These 'classical' correlations deliver an expectation E(a,b|C) = -a.b/2.

    Question:- What implications are to be derived from this fact: The original unfettered quantum source SQ produces particle pairs that are more highly correlated than the 'classical' source SC (the throttled quantum source)?
     
  18. Aug 17, 2016 #17

    bhobba

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    Bell type correlations are part of QFT which assumes SR and Minkowski space-time.

    People constantly tie themselves into knots about EPR, Bell etc. Its nowhere near as hard as they make out.

    Its simply a correlation and as such is excluded from the principle of locality in QFT which goes by the name of the Cluster Decomposition Property:
    https://www.physicsforums.com/threads/cluster-decomposition-in-qft.547574/

    QM as a theory is silent about things having properties if not observed - they may or may not depending on interpretation. All Bell shows is if you want to have properties when not observed you need non local influences. Don't want that and there is no issue at all - in fact you can even exclude correlations from considerations of locality so the Cluster Decomposition property makes sense. If you include it then things become much more complicated. Of course it proves Jack Shite - but we have this thing called Occam's Razor that says you do not complicate things unnecessarily.

    Thanks
    Bill
     
  19. Aug 18, 2016 #18

    zonde

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    I don't think these alternatives are consistent with scientific approach.
     
  20. Aug 18, 2016 #19
    ... In other words, "independent of" means "separable from and uninfluenced by", in the statement below:

    The 'state of affairs' relevant to the outcome at A is independent of the setting at B, and vice versa.

    NOTE: The above statement is rendered inapplicable, and therefore neither true nor false, when one asserts that

    the notion of a 'state of affairs' in a spacetime region is altogether invalid.

    Therefore, to say that the statement is false leaves but one alternative to "influence":

    the joint-state of Alice and Bob's measuring devices is nonseparable.

    What about Relational Blockworld for the but one alternative?
     
  21. Aug 18, 2016 #20
    I do not think I understand the question. So, I will answer a question similar to it. Then maybe you can clarify the question for me.
    ______________

    The pairs from SQ violate Bell's inequality, whereas the pairs from SC do not.

    Regarding the SQ-phenomenon, I am forced to conclude:

    Bell's principle of "Local Causality" is either violated or inapplicable.

    Regarding the SC-phenomenon, I am NOT forced to accept such a conclusion.
     
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