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The Pros and Cons of Bohmian Mechanics

  1. Nov 28, 2005 #1

    DrChinese

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    This is a discussion aimed towards gaining a better understanding of Bohmian Mechanics (BM) - at least on my part. I would like to see BM in the best light possible.

    My starting point in this journey is a reference to Sheldon Goldstein's summary in the Stanford Encyclopedia: Bohmian Mechanics

    A couple of points for those of you new to this subject so we have a point of reference:

    1. BM can reproduce most of the predictions of QM.
    2. BM is "grossly non-local" (Bell) and is explicitly incompatible with Lorentz invariance.
    3. BM is deterministic - a Hidden Variable variable theory. Knowledge of a particle's initial position and its wave function is sufficient to predict its future.
    4. BM does not need to give a special position to the act of measurement as does QM, and therefore "solves" the measurement problem.
    5. Spin does not exist as a fundamental property of particles within BM, but the measurable effects of spin can be explained in some if not all cases.
     
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  3. Nov 29, 2005 #2

    ttn

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    Your 5-point summary seems generally accurate and fair. But I'm curious why you say "most" here. What predictions did you mean to exclude?

    Of course, presumably we're talking about "Bohmian Mechanics" to mean Bohm's version of non-relativistic QM. So there are, e.g., predictions of relativistic quantum field theory that BM doesn't reproduce. But then maybe there's a "Bohmian" version of quantum field theory that will reproduce those predictions. So as long as we're comparing apples to apples (i.e., BM to non-relativistic QM) I think it's correct to say that BM can reproduce *all* of the predictions of QM.

    But maybe you had something else in mind that you think is a problem for BM even in the non-relativistic context?
     
  4. Nov 29, 2005 #3
    There was this:

    http://arxiv.org/abs/quant-ph/0206196

    which suggests there are testable differences between Bohm and QM.
     
  5. Nov 29, 2005 #4

    ttn

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  6. Nov 29, 2005 #5
  7. Nov 29, 2005 #6

    ttn

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    Well, that probably depends on who you ask, but I certainly think they're a complete and total disproof. Basically, what all 3 of these papers point out is that Ghose's "Bohmian" prediction for the outcomes of the experiments is based (in part) on an assumption about the initial positions of the particles, and this assumption flagrantly violates the "quantum equilibrium hypothesis" which is one of the crucial principles of Bohm's theory. In short, Bohm says that the initial positions of the particles should be psi-squared distributed (ala the Born rule), but Ghose makes some other arbitrary assumption about the initial conditions. So that means either he used the wrong wave function (so that the whole time evolution and comparison to QM is all wrong), or he's using the right wave function but assuming we know more about the initial positions than we're permitted to know -- which means that his conclusions about where (according to Bohm) the particles will land, is only right for some tiny fraction of the actually-possible initial positions. And if one were to include all those other possible initial conditions, one would get (not at all surprisingly) the same predictions as QM.

    Or, to make it *really* brief, the quantum equilibrium hypothesis (QEH) *guarantees* that Bohm will always make the same predictions as regular QM. Ghose et al only derive an apparent disagreement by rejecting the QEH. But if you reject that, it's just stupidly obvious that Bohm won't make the same predictions as QM -- you don't need anything so complicated as a 2 slit experiment involving pairs of bosons to know *that*.

    So the whole thing is really completely bogus. But don't take my word for it. :smile:
     
  8. Nov 29, 2005 #7
    Thanks ttn. I look forward to reading up on it.
     
  9. Dec 1, 2005 #8

    DrChinese

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    OK, I understand that BM is non-local in the sense that particles that are space-like separated act as hidden variables. This is to say that BM is a non-local hidden variable theory.

    So my next question is: does BM predict a violation of Bell Locality? I know the usual answer is YES. But I wish to point out the following in rebuttal, from Bell's original paper:

    [1] "The vital assumption is that the result B for particle 2 does not depend on the setting a, of the magnet of particle 1, nor A on b."

    This says it all, notwithstanding his (2) in which separability is introduced. The fact is, there does not need to be separability to define Bell Locality. Bell Locality is exactly the following:

    [2] P(B+, a) = P(B+, c)

    Which is that the probability of seeing a + result at B does not change when the setting of particle 1 is varied from any a to any c. We already know and have known for a long time that:

    [3] P(B+, a) = P(B+, c) = .500 (for all settings of a, b, c)

    Is there is doubt whatsoever that [3] is experimentally true? Is there any doubt whatsoever that [1] is equivalent to [2]? I don't think either of these should be in question.

    [4] Bohmian Mechanics predicts [3] and therefore [1] as well

    In fact, if [4] were not true then BM would be experimentally falsified. Any theory would. Therefore I assert that Bohmian Mechanics is Bell Local per Bell's definition [1]. Lastly, the question is: does Bell use his own definition in his proof? Here the record is definitely unclear. I have absolutely no difficulty deriving Bell's Theorem with [2] and no variation whatsoever from this. I believe that the separability requirement is stricter than [2] and is not needed for Bell's conclusion. (However, I would not say this is a generally accepted opinion but I doubt many have given it much thought.)

    What is without doubt is that if [2] were not true, this fact could be exploited to send superluminal signals. Therefore, if signal locality is true then [2] is also true. However, [2] does not prove or otherwise imply that there cannot be a violation of signal locality. This equates to:

    [5] Violation of [1] is not compatible with signal locality
    [6] Violation of signal locality is compatible with [1]

    So result [6] means that BM could be explicitly non-local and still be Bell Local per [1]. Q.E.D.

    Your (brutal) comments are invited!!
     
  10. Dec 1, 2005 #9

    NateTG

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    Please justify that a and c are space-like seperated.
     
  11. Dec 1, 2005 #10

    DrChinese

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    To answer the question: a and c are alternate settings for particle 1 and so they are co-located at particle 1. b is the setting for space-like separated particle 2.

    We don't need any high-speed switching for the setting changes because we don't care if they are far enough away from each other to affect the outcome via signalling or not. All I care about is the the *setting* for measurement of particle 1 does not affect the *outcome* for measurements of particle 2. If this is true, then any signalling between particle 1 and particle 2 is irrelevent.
     
  12. Dec 1, 2005 #11

    DrChinese

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    That was not intended to detract from BM. There are obviously open questions about whether BM is less than, equal to, or greater than QM is some ways. I don't really know enough to appreciate the nuances on this and am still learning.
     
  13. Dec 1, 2005 #12

    NateTG

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    Then, isn't what you're describing signal locality, rather than Bell locality? That is, what you're describing is the property that the measurement results at one location are unaffected by which measurements are selected at another. While Bell locality is result independance - that is, that the results of a particular measurement cannot be affected by the result of an experiment that is not in it's past. (Naturally, in non-local theories like Bohmian mechanics, 'past' may be a bit ambiguous.)
     
  14. Dec 1, 2005 #13
    P(B+, a) = P(B+, c) is definitely true in Bohmian mechanics for equilibrium ensembles. However, I think you need [itex] P(B+| a, \lambda) = P(B+| c,\lambda) [/itex] to derive the Bell inequalities, i.e. you have to condition on the hidden state as well as the measurement settings. This is definitely violated in Bohmian mechanics.
     
  15. Dec 1, 2005 #14

    DrChinese

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    No, signal locality is different than Bell Locality. (Yes, I know all the definitions about the common light cone and past etc. but none of this is actually relevant to Bell's Theorem or Bell's Inequality. Look at the theorem itself.) Bell Locality is simply the assumption Bell needed to make to get the Theorem to work, over and above his fundamental assumption of "Bell Reality". Bell Reality + Bell Locality = Bell's Inequality. Again, to be specific about Bell Locality:

    P(A+, b) = P(A+, c)
    P(A-, b) = P(A-, c)

    The trick here is to recall WHY we need Bell Locality in the first place. The reason is that we want to (simultaneously) learn the values of 2 observables of particle 1, but we cannot do this directly. After all, in both classical mechanics and quantum mechanics we know that the results of an observation change the system being tested. So instead we measure one directly and the other indirectly - by reference to another particle which is entangled with the first. For our purposes, particle 2 is a clone of particle 1. But our results will be skewed if the process of the indirect measurement on particle 1 (which is done by measuring particle 2) disrupts our direct measurement of particle 1.

    Bell Locality protects us on this point. It says: hey, you can do whatever you want on particle 2 and it won't skew our results on particle 1. Then you are free to combine the main results of our Bell test (i.e. to determine if there is a violation of a Bell Inequality) on a series of particles pairs.

    Bell Locality was assumed in Bell's Theorem. He *had* to assume it, there was no other way to derive his theorem. But that is no longer necessary. What changed, you ask? We now have experimental confirmation of Bell Locality! You don't have to assume what you can prove!! Nearly every Bell test involves rotation of one or both measurement settings. If there had been any change in P(A+) while varying b we would have heard about it from the Nobel committee already because we would be sending FTL signals. But alas, the results are always the same: measurement of P(A+) always gives a mean value of .500 regardless of any change in b.

    What we can prove experimentally we can incorporate without assumption. So drop that assumption! It is extra baggage!!

    If Bell Locality - as I describe it - is false, then there is no signal locality as we have seen. On the other hand, there *could* exist non-local effects - even non-local signal effects - and Bell Locality is true. Maybe the non-local effects manifest themselves elsewhere, for exampe. And that would (still) make the results compatible with Bohmian Mechanics. However, it would demonstrate unambiguously that BM is not "realistic". But that's OK with BM too, as I don't think BM asserts that reality is observer independent.

    So despite it being billed as a Non-local Hidden Variable theory, I wonder if by Bell's definitions if BM should also be considered a Local non-Realistic theory. (I am hoping for a few laughs on that one. :tongue2: )
     
  16. Dec 1, 2005 #15

    DrChinese

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    P(B+, a) = P(B+, c): Yes, it really is a requirement of any theory now although that was definitely NOT true when EPR was written. And it couldn't be assumed before Aspect did his tests. The fact is that it is true whether or not there exist superluminal effects. So I have:

    [itex] P(B+| a, b) = P(B+| a', b)[/itex]

    So what does adding the hidden variables give us in Bell's Theorem? Bell used it because he needed to cover the bases. However, strictly speaking, it is not a requirement to prove the Theorem. (You can run through a standard derivation and see you don't need it.) We want the weakest form of Bell Locality possible in order to have Bell's Theorem apply to the maximum range of theories.
     
    Last edited: Dec 1, 2005
  17. Dec 1, 2005 #16

    NateTG

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    I don't understand what you're saying.
    I read the above as, "The probability that the measurement A returns a result of + if measurement b occurs is equal to the probability that measurement A returns a result of + if measurement c occurs."
    Now, if A is a measurement on a different particle than b or c, then this is clearly part of signal locality.
    Conversely, if A, and b, or A and c are measurements on the same particle, then we have simultaneous non-commuting measurements on the same particle, so the value of the probabilities cannot be experimentally determined, and their equality is certainly not assumed in Bell's theorem.
    Moreover, my understanding is that Bell Locality should be a stronger condition than signal locality. However, you describe it as a weaker condition. Your post has "if Bell Locality - as [Dr.Chinese] describes it - is false, then there is no signal locality as we have seen", and, according to that, signal locality implies Bell Locality - which conflics with the common notion that QM is signal local, but not necessarily Bell Local.
     
  18. Dec 1, 2005 #17

    NateTG

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    Actually, a hidden variable model is equivalent to assuming that the process is stochastic: Let's say we have some bunch of measurements [itex]M[/itex], none of which is in the past of any of the others. Then, if the results of the measurements is stochastically determined, the system can be modeled by assigning a hidden state [itex]\lambda[/itex].

    In all of the proofs that I have seen, the assumption of the hidden state is necessary in order to insure that the probabilities in question are actually well-defined. Without the hidden state (or the equivalent assumption that the system is stochastic), the proof becomes similar to multiplying and dividing both sides of an equation by zero.
     
  19. Dec 1, 2005 #18

    DrChinese

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    The definition of a "realistic" theory is that particles have observable attributes independent of the act of observation. This is all EPR says; and that is why EPR says QM is incomplete. It is not an assertion of EPR that there are hidden variables that predetermine outcomes. It is that the outcome values themselves exist independent of a measurement.

    And Bell follows that thinking completely. So if there are 2 simultaneous values for a single particle (corresponding to 2 different measurement settings), then there are 3 as well.

    a, b and c are different settings to measure observables on a single particle. But such a simultaneous measurement is not possible without disturbing the system under view. So if you measure the particle and a "clone", then you might be able to get 2 values simultaneously.

    In this example, we are testing the hypothesis - of Bell - that a single particle has 3 simultaneous values. I think your characterization is OK but let me repeat the experimental questions.

    I. Experimental test of Bell Locality so it does NOT need to be assumed a priori:

    P(Alice+ (at polarizer setting=a), Bob (any setting b)) =
    P(Alice+ (at polarizer setting=a), Bob (any setting c))

    and similarly (I left this out in the earlier post I think)

    P(Alice- (at polarizer setting=a), Bob (any setting b)) =
    P(Alice- (at polarizer setting=a), Bob (any setting c))

    We are looking at the variations of the setting for Bob and how it affects things over at Alice, but are not concerned with Bob's outcome in this statement. Because this scenario exactly - word for word - maps to Bell's statement as to his locality assumption. That being that the result at Alice is independent of the setting at Bob.

    The interesting thing: It just doesn't matter whether there is signal locality or not; if the particles are space-like separated or not; or if there are slower than light influences. None of these can matter in our experiment II IF the experimental result above is first proven. Therefore, there is no need to assume Bell Locality or locality of any kind. In fact, you are free to assume the opposite: that there are such effects because they just won't matter.

    II. Experimental test of Bell's Inequality

    This would test correlations between Alice and Bob once we have ruled out - by experiment - that the outcome at Alice is affected by the setting at Bob. So now we can see that the correlations are too strong to obey Bell's Inequality - because there is NO SIMULTANEOUS a, b and c to begin with.
     
  20. Dec 1, 2005 #19

    DrChinese

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    That's what I thought too, until I tried reconstructing the proof myself. I never made any reference to the Lambda and couldn't figure out why. Yet the result was identical.
     
  21. Dec 1, 2005 #20

    ttn

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    "Bell Locality" isn't a statement about what a theory predicts; it's a statement about how a theory works internally. BM just violates Bell Locality because of the way it works. The theory includes a mechanism by which a given particle's trajectory can be affected by stuff going on at spacelike separation, so it violates Bell Locality.


    Um, no, that isn't what Bell Locality says.

    No, no, no! Bell Locality says: the probability of seeing a + result at B does not change when we specify also the setting or outcome of the result at A -- *given* that we're already conditionalizing on a *complete description of the state of the particles in the past*.

    Seriously, you have to read Bell's article "La Nouvelle Cuisine."
     
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