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Can hidden variable theories assign values to observables?

  1. Sep 1, 2013 #1

    naima

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    hi PF

    I read the no-go Kochen theorem using 18 vectors projection (here).
    One attribute a value to an observable A: v(A)
    This function is supposed to be linear and verify v(AB) = v(A) v(B)
    so v(AB - BA) = v(A)v(B) - v(B)v(A) = 0 (v is a real number function)
    v must assign v(id) = 1 because v(A id) = v(A)= v(A)v(id)
    If A B - B A = h Id
    v(AB - BA) = h
    Why have we to go further and look for non contextuality argument?
     
  2. jcsd
  3. Sep 2, 2013 #2

    Demystifier

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    I don't understand the question. Can you rephrase it?
     
  4. Sep 2, 2013 #3

    naima

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    As [x,p] = h Id
    we would have v(xp - px) = h v(Id) viz 0 = 1.
    Why do we have to find a 18 vectors exemple to show that one cannot find such a v?
    the wiki's proof is interesting but i think that it hides a simpler explanation.
     
  5. Jul 25, 2015 #4
    I don't know what function v is but if we take the quantum axiom it should be one of the eigenvalues and then it is surely not linear, we have :

    $$eig(A-B)\neq eig(A)-eig(B)\\
    eig(AB)\neq eig(A)eig(B)\\
    eig(A\otimes B)=eig(A)eig(B)$$
     
  6. Jul 25, 2015 #5

    micromass

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    The value ##v(A)## is only defined for projection operators, which the position and momentum are not.
    In another approach, the equality ##v(\alpha A + \beta B) = \alpha v(A) +\beta v(B)## only holds for comeasurable operators ##A## and ##B##, which again, the position and momentum are not.
     
  7. Jul 26, 2015 #6

    naima

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    Yes but for every possible eigenvector of the position operator there is a projector on this vector that tells you if the particle is at this given place. Can you give me well known observable projectors that are not made like that?
    As the question is about the v function which gives values to operators, why should it have to be linear or to verify v(f(O)) = f(v(O))?
     
  8. Jul 26, 2015 #7
    It really depends on the measurement setup you consider. For example the hydrogen atom you measure the energy and not the sum of measured kinetic and potential energy of the electron.

    For bell it is another setup we sum measurement result as opposed to measuring the sum.

    Thus the isomorphism v(A+B)=v(A)+v(B) should hold in the Bell case. To make this true one has to consider a particular operator operation + that satisfies this relationship.
    It is left as an easy exercise to find the operation for the Bell case.
     
    Last edited: Jul 26, 2015
  9. Jul 26, 2015 #8

    naima

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    The aim is to discard a maximun of hidden variables theories. Not only the simplest ones with linear rules! No?
     
  10. Jul 26, 2015 #9

    aleazk

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  11. Jul 26, 2015 #10
    Maybe we shall disprove anything hidden but where does that aim comes from ?

    Anyhow the term 'hidden' means it is not findable else you get a contradiction.

    it is mainly thought that there are no more fundamental level under quantum. If we take Goedel's theorem and admit quantum is complete this means the system of axioms is not strong enough to do meta-quantum in some sens we could not use quantum axioms to 'speak' about quantum mechanics which could make the theory not interesting.
     
    Last edited: Jul 26, 2015
  12. Jul 26, 2015 #11

    bhobba

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    Don't get too caught up in the usual proof of Kochen–Specker. Its just a simple consequence of the more powerful Gleason's theorem which I think is the best way to approach it. The reason its not usually done that way is Gleason has the reputation of being notoriously difficult to prove but like anything if you persevere a bit its not too bad:
    http://kiko.fysik.su.se/en/thesis/helena-master.pdf

    Thanks
    Bill
     
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