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I "Counterfactual definiteness" vs. "free will"

  1. Sep 11, 2016 #1
    Hi.

    What's the exact relation between the assumptions "counterfactual definiteness" and "existence of free will"? I (think I) know their "definitions" in QM:
    counterfactual definiteness: ability to speak meaningfully of the definiteness of the results of measurements that have not been performed
    free will: ability to choose experimental parameters completely independently of anything (more precisely: any joint distribution of the choice of experimental parameters and any past or present event must be a product distribution)

    But I'm not sure if those assumptions are equivalent or if one is stronger or if they even talk about the same matter.

    My thoughts:
    I think CFD must at least partly encompass free will, because in order to speak about measurements that have not been performed (because the experimental parameters have been chosen differently), one needs to assume that they could have been performed, i.e. the probability of the choice of more than one set of experimental parameters must have been nonzero.
    But I think this doesn't need the full extent of the definition of free will as stated above. It needs that conditioning on any past or present event does not lead to the trivial 1-0-distribution, not that such events must not influence the choice of experimental parameters at all.

    On the other hand: If free will exists and we have a useful theory of nature, we should be able to make predictions about any choice of experimental parameters, regardless of which one is actually chosen. So free will implies CFD.

    So is the free will assumption stronger than CFD?
     
  2. jcsd
  3. Sep 11, 2016 #2

    Nugatory

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    Staff: Mentor

    Rejecting CFD doesn't mean that you can't speak of measurements that haven't been performed, it means that you cannot speak of the result of such a measurement. So it's consistent to say that I could have chosen to perform measurement A instead of the non-commuting measurement B that I actually did. It doesn't follow from that that A has a definite value. On the contrary, when I chose to measure B not A I also chose to make the wave function not be an eigenfunction of A, thereby ensuring that A does not have a definite value.
     
  4. Sep 11, 2016 #3

    bhobba

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    Science Advisor
    Gold Member

    A point need to be made here.

    In conventional QM an assumption is made that observations appear in a completely common-sense classical world in no way dependent on conscious observers being present or not. This is a point of confusion because observer in QM is much more general than in conventional use.

    See the following:
    http://www.johnboccio.com/research/quantum/notes/paper.pdf

    Thanks
    Bill
     
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