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Must position be continuous in Bohmian mechanics?

  1. Nov 20, 2013 #1

    atyy

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    Here I'm thinking of a single free particle obeying the Schroedinger equation. The ensemble refers to multiple experiments with a single particle in which the initial wave function is the same.

    If I naively imagine that there is such a thing as a wave function that is delta function, in Bohmian mechanics, since the distribution of initial positions in the ensemble is given by the wave function, all particles in the ensemble presumably have the same position. This means that all members of the ensemble have the same trajectory, and there will be no future uncertainty in position. However, the wave function of a free particle does spread out since it is not an eigenstate of the Hamiltonian, and so the Bohmian prediction would not agree with quantum mechanics.

    Is my error in imagining that (due to wave functions being continuous) there is such a thing is a wave function that is a delta function?

    If that is the reason for the error, does this mean that Bohmian mechanics wouldn't work if position were discrete?
     
  2. jcsd
  3. Nov 21, 2013 #2

    Demystifier

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    I would say so. See also
    http://lanl.arxiv.org/abs/quant-ph/0305131

    Yes.
     
  4. Nov 21, 2013 #3

    bhobba

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    That's your first error - in practice that's not actually achievable - there will always be uncertainly in position that will grow.

    Your second one is the the uncertainly principle. Sure you know its position to a high degree of accuracy, to the point a delta function may even be a good model, but guess what? Its momentum is unknown so could scoot off anywhere and it's position is unknowable.

    The exact detail I don't know because I am far from any kind of expert on BM, but since it has been deliberately cooked up to be indistinguishable from the usual formalism what I said must be true.

    But for the detail an expert in it like Demistifyer needs to chime in.

    Added Later:
    I see he chimed in while I was writing my post.

    Thanks
    Bill
     
  5. Nov 21, 2013 #4

    JK423

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    atyy it seems that this is relevant to the continuity/discreteness of space we were discussing the other time, heheh :wink:, for which i got an infraction...! So i cannot talk about it otherwise i'll get banned :tongue2:

    Seriously now, i disagree with bhobba's statement,
    It's not that it's "in practice" not achievable, it's in principle not achievable; the delta function is not a proper quantum state at all, so there is nothing to be achieved. I'm really surprised that these people that Demystifier mentions in his paper used such an argument without proper checking.


    Amazing, never thought of that before.
     
    Last edited: Nov 21, 2013
  6. Nov 21, 2013 #5

    bhobba

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    I think you are using the word proper in a manner not used by virtually any author I am aware of - Von-Neumann being the only exception - and in fairness he was scathing. Dirac, Ballentine, Griffith, all allow a quantum state to be a delta function.

    It has issues with rigor you need a Rigged Hilbert Space to rectify (and it addresses Von-Neumann's concerns), and used naively you end up with really hairy stuff like the delta function squared, so care is required, but its a theoretically allowable state 100% for sure. If you don't allow it then you are limited to Von-Neumann's formalism, and cant use Dirac's approach, which most physicists would find unacceptable.

    Still it's of zero practical import since such is not physically realizable - its simply there for theoretical convenience. But its required for that or things are mathematically more difficult in practice - which is exactly why physics textbooks follow Dirac and not Von-Neumann.

    Looking at it another way, the physically realizable states are the subset of the Hilbert space in the Gelfland triple - they are continuously differentiable, are zero at infinity etc etc - all the nice properties you expect physically realizable states to have, but we need to introduce the super-set in the Gelfland triple that includes things like the Dirac delta function otherwise equations will not have solutions etc (eg a plane wave is not an allowable state) and things will be a lot more difficult mathematically.

    Thanks
    Bill
     
    Last edited: Nov 21, 2013
  7. Nov 21, 2013 #6

    JK423

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    Bill,
    i think you basically agree with what i am saying but you don't like the words that i use.
    The fact that we use it for theoretical convenience doesn't mean anything more than that. The fact is that a position eigenstate is not a quantum state since it's normalization gives infinity instead of unity, no need to cite Dirac or any bigshot for that. Use it incorrectly and you will find yourself in many paradoxical situations just because of "theoretical convenience".
     
  8. Nov 21, 2013 #7

    bhobba

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    Do you consider a plane wave a quantum state?

    Thanks
    Bill
     
  9. Nov 21, 2013 #8

    JK423

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    No, it's only an approximation of a quantum state.
     
  10. Nov 21, 2013 #9

    bhobba

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    Then, in the sense you mean it, I agree with you - but I do feel I must point out most physicists, and even guys like me whose background is of a more mathematical bent, probably wouldn't say that. It would make a bit of a nonsense of a number of standard analysis like crystal diffraction.

    Thanks
    Bill
     
  11. Nov 21, 2013 #10

    atyy

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    I think most physicists would say a dirac delta and plane wave are not proper quantum states, but they use them for convenience. But if they are allowed in the Rigged Hilbert Space formalism of QM, then can Bohmian mechanics work in a Rigged Hilbert Space? It would seem weird if Bohmian mechanics were compatible with QM in Hilbert space, but not with QM in a Rigged Hilbert Space. (I understand Demystifier's answer in post #2 in a Hilbert space, where as JK423 says, a dirac delta is not a proper quantum state.)
     
  12. Nov 21, 2013 #11

    bhobba

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    The way you view the Rigged Hilbert space formalism is you have physically realizable states that have very nice mathematical properties, the type of properties we generally assume in applied math. These would be the states in BM. These are expanded for mathematical convenience to include states that are not physically realizable by the mathematical trick of the dual. However we can find a physically realizable state that is arbitrarily close to these non-realizable states, in a certain mathematical sense anyway. At an intuitive level I think its best to think of wave-functions like the Dirac delta function as a physically realizable state that for all practical purposes behaves like the 'fictitious' state ie for a Dirac delta function its a continuously differentiable spike of very small width. That way you can do otherwise very nasty things like actually square a delta function and the mathematical formalism is a lot easier. Its sort of like considering dx a very small increment of x. Mathematically its wrong at all sorts of levels but its often done in applied math.

    My point is most consider states that are not strictly physically realizable like plane waves to be actual states so the distinction is a bit blurry.

    Thanks
    Bill
     
  13. Nov 21, 2013 #12

    atyy

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    I guess rigor is not always "rigor mortis" :tongue2: Yes, this seems to have BTSM implications, or maybe more like BBTSM implications. We'll have to discuss this in that forum (not here!) some time.
     
  14. Nov 23, 2013 #13

    atyy

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    In http://mattleifer.info/2012/02/26/quantum-times-article-on-the-pbr-theorem/ Matt Leifer writes "A trivial consequence of the PBR result is that the cardinality of the ontic state space of any hidden variable theory, even for just a qubit, must be infinite, in fact continuously so. This is because there must be at least one ontic state for each quantum state, and there are a continuous infinity of the latter. The fact that there must be infinite ontic states was previously proved by Lucien Hardy under the name "Ontological Excess Baggage theorem", but we can now view it as a corollary of PBR."

    I can't find a free version of Hardy's article, but PBR do make the statement in their paper http://arxiv.org/abs/1111.3328 .

    Hardy http://arxiv.org/abs/1205.1439 refers to a result of Montina http://arxiv.org/abs/0711.4770 "We consider the general class of ontological theories describing a quantum system by a set of variables with Markovian (either deterministic or stochastic) evolution. We provide the first proof that the number of continuous variables can not be smaller than 2N-2, N being the Hilbert space dimension."

    Is the requirement that position be a continuous variable in Bohmian mechanics related to Hardy's Ontological Excess Baggage theorem, and to Montina's proof?
     
    Last edited: Nov 23, 2013
  15. Nov 25, 2013 #14

    Demystifier

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    Yes, I think there is a relation, but I don't know how to prove it.
     
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