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Is dBB reproducing quantum results ?

  1. Apr 13, 2013 #1
    If we consider at time t=0 a particle at x=0 we get a gaussian distribution for time t>0 via the Schroedinger equation.
    Hence QM predicts that the particle can have travelled some distance at t>0 with given probabilities. see http://en.wikipedia.org/wiki/Uncertainty_principle

    However in dBB the speed of the particle is given by the guiding equation, namely the derivative of the wavefunction, http://en.wikipedia.org/wiki/De_Broglie–Bohm_theory
    This derivative is 0 forall t at x=0,
    hence the dBB-particle does not move at all, and hence dBB cannot reproduce QM results.

    What is wrong with this reasoning, since apparently a theorem (which?) proves that if dBB was refuted then QM were refuted too https://www.physicsforums.com/showthread.php?t=459148
     
    Last edited: Apr 13, 2013
  2. jcsd
  3. Apr 15, 2013 #2

    Demystifier

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    A similar critique of dBB theory has already been proposed. Nevertheless, the critique has not been valid, as explained in
    http://arxiv.org/abs/quant-ph/0305131

    In short, if you assume that initial wave function is a delta function, then such a wave function is a mathematically pathological object which never occurs in nature. If, on the other hand, you replace the initial delta function with a very narrow Gaussian, then the point x=0 has a measure zero, i.e., there is zero probability that the initial position will be exactly x=0. For any other initial position x close to 0 but not exactly 0, your argument does not longer work.
     
  4. Apr 15, 2013 #3

    martinbn

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    Demystifier, so you are saying that they indeed have different predictions, but in practice we will never (probability zero) will see a disagreement.
     
  5. Apr 15, 2013 #4

    Demystifier

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    You can put it this way, but a more convenient way to express it is to say that they are different theories with identical PROBABILISTIC predictions.
     
  6. Apr 15, 2013 #5
    I did some calculation with a gaussian distribution instead of a maladive Dirac delta (given by Wikipedia http://en.wikipedia.org/wiki/Uncertainty_principle under Constant momentum, setting p0=0)

    Psi(x,t)=1/Sqrt(1+iw0t)*exp(-x^2/x0/(1+iw0t))*N

    I get for (dPsi/dx)/Psi=-2x/x0/(1+iw0t)

    Then the guiding equation of dBB (I took http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory under overview says we have to take the imaginary part : hence the speed given by dBB is -xw0t/(1+w0^2t^2)

    So we see that for t>0 the speed is of opposite sign to the position, hence the particle goes towards 0.
     
    Last edited: Apr 15, 2013
  7. Apr 16, 2013 #6

    Demystifier

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    I think you made a sign error in taking the imaginary part. The imaginary part of 1/(1+i) is negative, not positive. To see this, multiply both the numerator and the denominator with (1-i).
     
  8. Apr 16, 2013 #7

    DrDu

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    and the probability amplitude is infinite, so what is infinite times 0?
     
  9. Apr 16, 2013 #8
    thanks demystifier i lose my skills. If we now imagine the particle a bit off 0 then it goes towards infinity always in the same direction (but after a measurement what happens ?)
    in qm if you measure the particle at place dx and after a while again then it could have gone in the other direction.(?)
     
  10. Apr 17, 2013 #9

    Demystifier

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    How old are you? :wink:

    The measurement changes the wave function, in both standard QM and dBB. If you measure the position at time t, then the wide Gaussian at t splits into a large number of narrow non-overlaping Gaussians at t+delta t, where delta t is time during which the measurement-causing interaction takes place. During the time delta t, the particle in dBB ends up in one and only one of these narrow Gaussians. Once the particle ends up in one of these Gaussians, the other narrow Gaussians do not longer influence the motion of the particle. From the point of view of the particle, it is effectively the same as if the wave function collapsed to the narrow Gaussian. That's how dBB explans the illusion of wave function collapse, without the actual collapse.
     
  11. Apr 17, 2013 #10
    Im 37 and former student. But the problem is that i had to stop due to health (mind) problems. It is probably now hopeless for me to go on with physics (maybe i should put this in the medical part)
     
    Last edited by a moderator: May 5, 2013
  12. Apr 17, 2013 #11

    Demystifier

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    Even if it is too late to become a professional physicist, it is certainly not too late to be a good amateur. From the posts I've seen I can tell that your physical and mathematical reasoning is quite good. The little mistake you made can happen to anybody. :smile:
     
  13. May 5, 2013 #12
    I got another point I don't understand. In QM the wavefunction delocalizes in position space, whereas the momentum space localizes around 0 (I'm not sure about this point).

    However in dBB the speed in function of time can be computed by solving the differential equation above and turns out to increase proportionally with respect to time (if I haven't made a mistake) ?
     
    Last edited: May 5, 2013
  14. May 6, 2013 #13

    Demystifier

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    I think you did. Speed does not need to increase with time.
     
  15. May 6, 2013 #14
    how do you compute this speed to see that ?

    Ah i think i found my mistake the speed is propto (1+w0^2t^2)^(1/2w0^2-1) so that if 2w0^2>1 the speed tends towards 0.

    Was this right ?
     
    Last edited: May 6, 2013
  16. May 7, 2013 #15

    Demystifier

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    I was talking about the speed for a general wave function. Concerning this special case, I will leave the calculation details to you. :smile:
     
  17. May 18, 2013 #16
    So are there non-probabilistic predictions on which the two theories differ? If so, would it in principle be possible to disprove Bohmian mechanics while keeping standard quantum mechanics intact?
     
  18. May 19, 2013 #17

    bhobba

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    M

    A few years ago there was a bit of work done on trying to experimentally test Bohm - check out:
    http://arxiv.org/pdf/quant-ph/0206196v1.pdf

    At the time I was really excited and thought it had been disproved. I got caught up in the hoopla - but there were these guys saying - hang on - somethings not right here - Bohm was deliberately concocted to be the same as bog standard QM - this should not be possible. I thought - they were just trying to deny the truth. But guess what - it turned out they were correct - there was a mistake and if I remember correctly it was also an incorrect use of the Dirac Delta function. Bottom line here is because of the way Bohm is designed its not possible to actually tell the difference.

    That does not apply to some other interpretations however - one of my favorite of that type is Primary State Diffusion:
    http://arxiv.org/pdf/quant-ph/9508021.pdf

    Nice theory - and I rather like it (yea I have a background in Stochastic Calculus, White Noise Functional's etc - nice math area as well - it also uses Rigged Hilbert Spaces) - but I heard a while ago it had been experimentally refuted.

    Thanks
    Bill
     
    Last edited: May 19, 2013
  19. May 19, 2013 #18

    Demystifier

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    People are trying to find out such a thing, but not yet completely successfully. For example, in
    http://arxiv.org/abs/1209.5196
    I argue that there are non-probabilistic predictions which can be used to experimentally distinguish different versions of Bohmian mechanics itself, and that experiments prefer the standard version of Bohmian mechanics. I believe it is at least a step.
     
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