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  What does John Conway and Simon Kochen's "Free Will" Theorem mean?

+ 6 like - 0 dislike
3171 views

The way it is sometimes stated is that

if we have a certain amount of "free will", then, subject to certain assumptions, so must some elementary particles."(Wikipedia)

That is confusing to me, but it seems to be an amazing theorem. It has been interpreted as ruling out hidden variable theories, but there is still some dissent. Lubos has a good discussion of it on his blog in the birthday blog for John Horton Conway. I assume it means that the outcomes of microscopic measurements are not deterministic.

What does the theorem assume and what does it prove?

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Gordon
asked Jan 29, 2011 in Theoretical Physics by Gordon (400 points) [ no revision ]
Most voted comments show all comments
@Gordon Wilson, @Bruce, @Marek: Ok, so the name is rather misleading, but it seems to be genuine! "Free will" is an inherently philosophical term, and physicists should not be using it. Still, it seems like they got an unusual bout of romanticism in this case!

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Noldorin
@Marek: It's Bell's Thereom++ and another go at clockwork universe ==> no freewill. An all you can eat buffet of both physics and philosophy. On-topic as far as I'm concerned, but I've said all I will about the philosophical aspects.

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user dmckee
@MBN-Of course the axioms are not proven. Axioms are never proven. One proposes axioms which have to be accepted in order to prove a theorem. If you accept the axioms as self-evident, or as demonstrably true, you use them as the starting point to derive and prove a theorem.

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Gordon
Oh, I see the confusion--in my response to Noldorin fourth down, I meant that the theorem has been proven, given the axioms, not that the axioms were proven...not expressed clearly....:)

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Gordon
Ok, then my remark is that they have proven their consistency, which is not always easy, and it means that there is no chance that the theorem is "stupid" because the axioms are inconsistent. I think it is a very good result. May be more logic than physics but very good. There is a talk by Conway available online but I am too lazy to search for it again. May be someone can give the link. He gives a very good exposition and makes some great comments.

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user MBN
Most recent comments show all comments
Is it related - Intelligent(?) Particles..?

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user voix
@Gordon: So what? I have to behave as if we do have freewill. ::following the inescapable logic of his own past history, the author's face breaks into a very slow grin::

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user dmckee

2 Answers

+ 4 like - 0 dislike

I shall attempt here to give an explanation of the meaning of the theorem with a limited background. Issues such as the validity of the proof I shall leave aside.

The Free Will Theorem (assuming SPIN, TWIN, and FIN)]. "If the choice of directions in which to perform spin 1 experiments is not a function of the information accessible to the experimenters, then the responses of the particles are equally not functions of the information accessible to them."

This theorem is a combination of ingredients which explain the hypotheses: The EPR setup of two particles (the SPIN and TWIN axioms); special relativity in a limited form (the FIN (later MIN) axiom); the Kochen-Specker "non-existence of a function" theorem/paradox.

So loosely we can imagine the typical EPR setup with two spacelike separated Observers making independent "choices" as to which of several axes to measure spin in two correlated particles: A and B, say. Let A's measurement be at time $t_A$. Then the conclusion is that no function of past properties (past light cone to $t_A$) predicts the outcome for A.

So the key phrase is "no function".

This theorem was primarily motivated to further exclude a fully deterministic interpretation of QM (obviously with hidden variables). Such hidden variables would give rise to a function of them - which here doesnt exist.

A secondary issue - where the "free will" comes from - is whether this just reconfirms an essential randomness in QM. Well the argument here, I believe, is that they interpret "random" to mean "a random function of" - but since no function exists even a random function doesnt exist (of any earlier properties). The responses of the particle A is determined at time $t_A$ based on no prior information (or information on B) - just the "free" choices of the experimenters as to what to measure.

A link to the Strong Free Will Paper: http://www.ams.org/notices/200902/rtx090200226p.pdf

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Roy Simpson
answered Jan 29, 2011 by Roy Simpson (165 points) [ no revision ]
It seems like a concise and fair presentation, @Roy Simpson! +1.

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Luboš Motl
+ 1 like - 0 dislike

The meaning is that either one shows that one or more of the axioms or some hidden premise is not acceptable for some reason, or one accepts the conclusion; or one can live in a state of "there's something wrong with that, I just can't find it" for as long as forever takes. If that means sleepless nights, so be it.

FWIW, a long time ago I chose to stop talking about particles. I haven't followed the Conway-Specker literature, since I argued some years ago that Bell inequalities cause almost no problem for random fields and no-one has so far contradicted that argument [go at it now, if you like, "Bell inequalities for random fields", J. Phys. A: Math. Gen. 39 (2006) 7441-7455, cond-mat/0403692, if you haven't previously decided it's wrong and decided not to publish a rebuttal]. Conway and Specker introduce assumptions that are fairly reasonable for classical particles but it's easy to exhibit that they do not hold for random fields.

The MIN axiom, in particular, is generally not true for quantum fields, because there are correlations at space-like separation in QFTs, without, because of microcausality, any need for there to be transmission of causal effects between space-like separated regions. The MIN axiom is also generally not true for random fields. There are correlations, but there is no causality, because there is microcausality at time-like and light-like separation as well as at space-like separation. The correlations, to your taste, can either be pre-existing (the Conspiracy Loophole, if you like, but the conspiracy is probabilistic, not deterministic) or, instrumentally, they're there just because they're there (that is, we observe them, so they're there). The FIN axiom is harder, because then one has to be convinced that negative frequency components of the random field do not cause trouble, but of course classical electromagnetism lives happily with negative frequencies without problems of causality outside the light-cone, negative frequencies occur in QFT loop diagrams without difficulty, and also frequency is not linearly related to energy for classical fields.

For random fields, since they're not well known, you can try my approach in "Equivalence of the Klein-Gordon random field and the complex Klein-Gordon quantum field", EPL 87 (2009) 31002, arXiv:0905.1263 quant-ph, where a few references to other people's work can also be found. It's a niche, of course.

Move on, is what this theorem means; in my case on to random fields, but it's best if we all make different choices. You might, alternatively, be happy to drop locality and/or classical states and observables and/or something else, or to lose sleep.

On the other hand, I think you could do much worse than to accept the answer on Luboš Motl's blog, which I found very interesting, so I voted up his comment on Roy Simpson's Answer as proxy.

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Peter Morgan
answered Jan 30, 2011 by Peter Morgan (1,230 points) [ no revision ]
Peter, Thanks (I think). These Random Field papers seem very interesting, and one has to watch the minefield of axioms and assumptions in this area, but I would like to challenge one aspect of the Bell claim. In fact the Bell paper admits, does it not, that it could not find an axiom allowing the derivation of the actual $2\sqrt2$ bound in the Bell Inequalities? I havent studied the other papers cited to see what they propose. Also MIN is about experimenter independence rather than field correlations per se.

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Roy Simpson
You need my approbation after Luboš's imprimatur? Yes, I can't explain the $2\sqrt{2}$. The nature of Experimenter independence depends on where the Heisenberg cut is. If they're inside the cut, they're correlated, qua being in a QFT state. Aren't we being Wigner, looking at our friends doing the experiment? If it's the vacuum state, the correlation is minimal, but if they've created an experiment that's designed to exhibit correlations at significant space-like separation (not easy), different story. Anyway, eg, Gregor Weihs's choices are made for him by two quantum mechanical mechanisms.

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Peter Morgan
Thanks for your comments and papers. I wish I could split the checkmark ;)

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Gordon
Thanks for the kind words, Gordon. Roy's answer is worthy. Being new here, my only wish is to have enough rep to be able to cut some of it off to put a premium on a question, and now I'm close! Like every mathematician on the site, I imagine, perhaps too cynically, I've been thinking of ways that rep might be gamed, and it looks as if there are so many ways that to succeed would be no achievement. For the time being, I've concluded that the coolest thing is for one's rep graph to be a sawtooth.

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Peter Morgan
Thanks again Peter. The main benefit of the early Rep points are the Privileges to use the site properly. I can envisage some Stack Q&A connected with these modern axiomatic QM/QFT theories. Then you will get lots more points!

This post imported from StackExchange Physics at 2014-05-14 20:02 (UCT), posted by SE-user Roy Simpson

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