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Bohm Interpretation and Bell Theorem

  1. Feb 16, 2009 #1

    NJV

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    I recently read about the pilot wave theory and the Bohm interpretation, and I must say I really like the idea. The only of my confusions it leaves unchanged are those concerning the EPR paradox. The Bohm interpretation does not seem to answer the question of the dreaded spokhafte Fernwirkung. Yet, I read that

    "The Bohm interpretation describes [the experiment described in the Bell Theorem] as follows: to understand the evolution of these particles, we need to set up a wave equation for both particles; the orientation of the apparatus affects the wave function. The particles in the experiment follow the guidance of the wave function; we don't know the initial conditions of these particles, but we can predict the statistical outcome of the experiment from the wave function. It is the wave function that carries the faster-than-light effect of changing the orientation of the apparatus."
    (source: Wikipedia; http://en.wikipedia.org/wiki/Bohm_interpretation#The_Einstein-Podolsky-Rosen_paradox )

    How does the orientation of the apparatus affect the wave function? Can someone explain just how, according to the Bohm Interpretation, the wave function of these particles could have such superluminal effect? Is there any theory about this at all? From the way Wikipedia put it, I got the impression there was more to be said about it.
     
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  3. Feb 17, 2009 #2

    Demystifier

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    The apparatus also also has its wave function. Different orientations mean different wave functions. However, the apparatus interacts with the system you want to measure, which means that the wave functions of the apparatus and the system influence each other. More precisely, these two wave functions become entangled, so that at the end you do not have separate wave functions of the apparatus and the system, but a single wave function that describes both.

    Just take a look at the equation that describes the motion of particles for a wave function of two or more particles. For example, take a look at Eq. (15) in
    http://xxx.lanl.gov/abs/quant-ph/0512065 [AIP Conf.Proc.844:272-280,2006]
    This equation clearly shows that a force on a particle depends on the instantaneous positions of all other particles.
     
    Last edited by a moderator: Apr 24, 2017
  4. Feb 17, 2009 #3

    NJV

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    Thanks for the quick reply.

    That's clarified, thanks. I'm not sure if I've understood the rest, though:

    In other words, the force that affects either of the entangled particles also affects the other because both depend on the instantaneous positions of the same particles?
     
  5. Feb 17, 2009 #4

    Demystifier

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    Yes.
     
  6. Feb 17, 2009 #5

    NJV

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    Thank you. I stand demystified. :)
     
  7. Feb 17, 2009 #6

    DrChinese

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    I have a few questions. An entangled pair is created at T0. You measure Alice at time T1 and Bob at time T2 (where T0<T1<T2), and the measurement apparati settings are changing at all points in time between T0 and T2.

    1. How is Bob able to be sensitive to the observation of Alice at T1? It seems Bob would need to ignore the settings at Alice at all points in time other than exactly at T1.

    2. In addition, the activities of all of the other particles in the universe which exert a non-local influence on Bob must exactly have an identical influence on Alice, and yet Bob is around longer and therefore should be influenced in ways that wouldn't affect Alice. So it seems this would provide an influence that would tend to remove the correlation between Alice and Bob.

    Thanks,
     
  8. Feb 18, 2009 #7

    Demystifier

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    No. The Bob's observations at T2 are determined by the Alice's state at T2. But why then it is also correlated with the Alice's state at T1? Because the evolution of Alice's state is deterministic, so the Alice's state at T2 is correlated with Alice's state at T1. Typically, the Alice's observable measured at T1 does not change at all during the further evolution from T1 to T2.

    If other particles have a significant influence on Bob, then these other particles cause the environment-induced decoherence. Yes, it will affect Alice too and the correlation between Alice and Bob will be effectively destroyed. There is no big difference between standard QM and Bohmian QM in that regard.
     
    Last edited: Feb 18, 2009
  9. Feb 18, 2009 #8

    DrChinese

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    Thanks as always for helping me to understand this.

    So even changing Alice's measurement apparatus setting from X at T1 to Y at T2 would not affect the result of Bob's measurement at T2? I was thinking that the measurement apparatus itself was inducing an effect on Alice. And I would think that it would be significant, in order to comply with Bell.

    For example: Bob is being measured at 0 degrees at T2; and Alice at one or the other of two settings, either +120 degrees or -120 degrees relative to Bob, at earlier time T1. Let's say the apparatus at Alice switches from the one setting at T1 to the other at T2. We know the coincidence rate will be 25%. But the results at Bob must reflect whether the setting at Alice is the +120 or -120 setting, which nearly reverses the outcome at Alice. Obviously, to comply with Bell, the observer settings of Alice and Bob must somehow be factored in to make the numbers work. But what you are saying about the time evolution is hard for me to grasp. It seems like you could do experiments in which you vary the settings at Alice at T1 and T2 to either be the same or 120 degrees apart, and the results would reveal the non-local effect either way. (And yet I know there wouldn't really be any difference in the outcome no matter what happens after T1 at Alice.) Does where I'm going with this example make sense? In this case, we know decoherence has not set in.
     
  10. Feb 19, 2009 #9

    Demystifier

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    If I understood you correctly, this changing of the Alice measurement apparatus is something that an experimentalist makes buy her "free will", right? If it would affect Bob, it would mean that she could send a GENUINE information to Bob faster than light, whereas we know that it is not possible. From this, it follows that SUCH changing of the apparatus (by "free will") cannot influence Bob. It is equally true in both the standard and the Bohmian interpretation.

    But perhaps you meant some automatic deterministic change of the apparatus? Something determined by a predefined time-dependent Hamiltonian or by the unitary evolution of the wave function? In this case, it can be said that this change will influence Bob. (But in a sense, it can also be said that such a change is not a change at all because everything was actually predetermined already at T0, so there is no GENUINE transfer of information.) In the Bohmian interpretation, Bob's observations at T2 are determined by the Alice and apparatus state at T2. Again, if you know how is Alice and apparatus state at T2 deterministically related to that at t<T2, then you can also interpret Bob's observations at T2 as being (indirectly) determined by the Alice and apparatus state at t<T2.

    Or perhaps you meant some genuinely random change of the apparatus? Well, there are no genuinely random events in Bohmian mechanics.
     
    Last edited: Feb 19, 2009
  11. Feb 19, 2009 #10

    Demystifier

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    In THIS SPECIFIC situation that you describe, you are right that there wouldn't really be any difference in the outcome no matter what happens after T1 at Alice. On the other hand, Bohmian mechanics claims that it should be determined by the state at T2, which is AFTER T1. Is this a contradiction? No, because in THIS SPECIFIC situation, all the relevant properties of Alice do not change after T1.

    But why? What forbids their changing? The point is that at T1 a genuine measurement has been performed. For example, the spin in z-direction has been measured to be equal to +1. The crucial point is that a genuine measurement allways contains interaction with a large number of the degrees of freedom of the measuring apparatus, which implies that decoherence takes place. The effect of decoherence is a sort of stabilization of the measured observable (in this case spin in z-direction), so the spin in z-direction does not change during the further evolution. Since it is equal to +1 at T1, it is equal to +1 at T2 as well. In fact, if there was no such a stabilization, then we would not be able to actually register the spin by a macroscopic apparatus at T1 in the first place.

    I think I know what will be your next question: But what if, after T1, we perform a measurement of the spin in some other direction? My answer is ready, but I will wait for your explicit question. :)
     
    Last edited: Feb 19, 2009
  12. Feb 19, 2009 #11

    DrChinese

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    I was imagining some system whereby Alice's measurement setting either changed - or didn't change - between times T1 and T2 according to some pseudo-random mechanism: something sufficiently macroscopic as to not be directly connected to Alice or Bob. Now I realize that everything around us is connected via a relativistic cone going to the past, but I am assuming that would not itself be capable of directly influencing the results of Alice and Bob. If it did, that would in effect be the superdeterminism argument redux. And I don't think you are advocating that.
     
  13. Feb 19, 2009 #12

    DrChinese

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    Yes, that is exactly my next question. :smile:

    So we have +1 in z direction at T1 for both Alice AND Bob. Then Bob measures at some other setting at time T2 and sees results consistent with that per standard prediction. Hmmm. I guess that actually makes good sense. And it wouldn't be inconsistent with Bell. Am I getting close? :blushing:
     
  14. Feb 20, 2009 #13

    Demystifier

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    You certainly are. But let me clarify the notion of spin in Bohmian mechanics.

    In Bohmian mechanics, particles do not have spins. They only have positions and velocities. Spin is a property of the wave function (not of the particle), but wave function is not something that can be measured. All that can be really measured according to Bohmian mechanics - are particle positions. But then, what does it mean to measure the spin of a particle?

    Consider the Stern-Gerlach apparatus, which is supposed to measure the spin. Due to the interaction with the magnetic field, one usually says that particles of positive spin go in one direction while particles of negative spin go in the other direction. But honestly, all that we really determine by the experiment is the direction in which the particle goes. We do not really measure spin, we only interpret the measured particle position as being caused by the spin.

    Now back to Bohmian mechanics. Bohmian mechanics provides a mechanism that causes particles to go in one or the other direction. The wave function splits into two channels (localized wave packets), one having one spin and localized in one direction, the other having the other spin and localized in the other direction. The particle enters one and only one of these channels. Once it enters one of the channels, it stays there because the two channels do not overlap.

    Now what happens if spin is measured again, now in another direction? The wave function splits again, so now we have 4 channels, namely 2 new channels for each of the channels created by the first measurement. But particle will again enter only one of the 4 channels. More precisely, it will enter one of the 2 channels that emanate from the channel that particle entered during the first measurement. That explains why Bob is correlated not only with the result of the second measurement, but also with the result of the first measurement.

    Perhaps all this is not a direct answer to your question, but I hope that it helps.
     
  15. Feb 20, 2009 #14
    I've been following http://www.tcm.phy.cam.ac.uk/~mdt26/pilot_waves.html" [Broken]'s lecture course on Bohmian mechanics (though he calls it pilot-wave theory) if it's any help. All his slides are on the web site with links to loads of papers.. If they're teaching it at Cambridge University now it must be getting respectable.. :grumpy:
     
    Last edited by a moderator: May 4, 2017
  16. Feb 20, 2009 #15

    Doc Al

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    So have I, at least the gist of it. I highly recommend it to anyone interested in these matters. And Towler's got a sense of humor, too. :wink:
     
    Last edited by a moderator: May 4, 2017
  17. Feb 20, 2009 #16

    Demystifier

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    Beautiful! :smile:
     
    Last edited by a moderator: May 4, 2017
  18. Feb 20, 2009 #17

    DrChinese

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    OK, I think I get some idea about Bohmian spin, but... it follows the HUP just like position, momentum, etc. So equally as much an observable. Hmmm.

    But I think I am lost on the channels. :eek: How does what happens at Alice (which clearly is related to the observational setting at Alice) get transmitted to Bob? You were saying that particles move deterministically, but yet we know the "hidden variables" cannot have been encoded at the time the particles (Alice, Bob) are created. We know there is a dependency on the measurement at T1, in this example. So I can only suppose that at the T1 time of the Alice measurement that Bob instantly adjusts to match. Or...?
     
  19. Feb 23, 2009 #18

    Demystifier

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    That is true. But it happens at each time t, not only at T1. Still, T1 may be more interesting because at that particular time more dramatic changes may occur.
     
  20. Feb 23, 2009 #19

    DrChinese

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    Thanks for sticking with me. A few more questions:

    1. I assume that Alice and Bob, being entangled, have a special relationship not shared by other particles. Is that correct? Or are Alice and Bob equally influenced by other nearby particles as each other?

    2. From the lecture notes (#3, slide 31), Towler says that in a double slit setup: "(1) Each particle passes through one slit or the other, (2) Wave function single-valuedness implies no trajectory can cross or even intersect apparatus axis of symmetry - thus know which slit from on-screen particle position." What are your thoughts on this?

    It seems as if this interpretation would be open to physical testing. I mean, you could play tricks on the path of the pilot wave, which clearly is fairly large if it goes through the slit opposite the particle. True, standard QM has this same thing. But the difference is that in QM, the particle goes through both slits and there is no preferred slit. Therefore, in Pilot Wave, the difference is that the particle is guided by the influence of the pilot wave crests/valleys but goes through a slit that is known. Seems like that would lead to a way to physically measure/observe this asymmetry by exploiting differences in the evolution of the pilot wave and the underlying particle.
     
  21. Feb 24, 2009 #20

    Demystifier

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    It depends. If Alice and Bob are not entangled with other particles, then their relationship is not shared by other particles. Likewise, if Alice and Bob are entangled with other particles, then their relationship is shared by other particles. These statements are interpretation independent.

    In the Bohmian interpretation, this is absolutely correct.

    It cannot be tested. If you do the experiment with only one particle, you will see the asymmetry (you will observe the particle at one side of the axis only), but standard QM also predicts this asymmetry. If you repeat the experiment many times, you will not see the asymmetry because in average you will see an equal number of particles at both sides, which is again predicted by both standard and Bohmian QM. In the Bohmian interpretation, this is because some particles (50%) will go through one slit, while others will go through the other.

    By the way, there is a great difference between "knowing" the slit through which the particle passed (by calculating it using Bohmian theory) and actually measuring it. The actual measurement allways causes decoherence, which ruins a lot.
     
    Last edited: Feb 24, 2009
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