The Pros and Cons of Bohmian Mechanics

[DrChinese]

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.

 

[ttn]

1. BM can reproduce most of the predictions of QM.

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?

 

[El Hombre Invisible]

There was this:

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

which suggests there are testable differences between Bohm and QM.

 

[ttn]

There was this:
http://arxiv.org/abs/quant-ph/0206196
which suggests there are testable differences between Bohm and QM.

The suggestion is totally wrong. See, for example, http://arxiv.org/abs/quant-ph/0101132, http://arxiv.org/abs/quant-ph/0007068, http://arxiv.org/abs/quant-ph/0108038, ...

 

[El Hombre Invisible]

The suggestion is totally wrong. See, for example, http://arxiv.org/abs/quant-ph/0101132, http://arxiv.org/abs/quant-ph/0007068, http://arxiv.org/abs/quant-ph/0108038, ...

Excellent! I will read through these tonight. But, for now, are these a total disproof of the earlier experiment, or merely a difference of opinion? (Sorry I don't have time to read these yet.)

 

[ttn]

Excellent! I will read through these tonight. But, for now, are these a total disproof of the earlier experiment, or merely a difference of opinion? (Sorry I don't have time to read these yet.)

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.

 

[El Hombre Invisible]

Thanks ttn. I look forward to reading up on it.

 

[DrChinese]

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!

 

[NateTG]

Your (brutal) comments are invited!

Please justify that a and c are space-like seperated.

 

[DrChinese]

Please justify that a and c are space-like seperated.

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.

 

[DrChinese]

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? ... But maybe you had something else in mind that you think is a problem for BM even in the non-relativistic context?

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.

 

[NateTG]

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.)

 

[slyboy]

P(B+, a) = P(B+, c) is definitely true in Bohmian mechanics for equilibrium ensembles. However, I think you need P(B+| a, \lambda) = P(B+| c,\lambda) 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.

 

[DrChinese]

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.)

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. )

 

[DrChinese]

P(B+, a) = P(B+, c) is definitely true in Bohmian mechanics for equilibrium ensembles. However, I think you need P(B+| a, \lambda) = P(B+| c,\lambda) 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.

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:

P(B+| a, b) = P(B+| a', b)

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.

 

[NateTG]

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

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.

 

[NateTG]

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.

Actually, a hidden variable model is equivalent to assuming that the process is stochastic: Let's say we have some bunch of measurements M, 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 \lambda.

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.

 

[DrChinese]

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.

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.

 

[DrChinese]

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.

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.

 

[ttn]

So my next question is: does BM predict a violation of Bell Locality?

"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.

Bell Locality is exactly the following:
[2] P(B+, a) = P(B+, c)

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

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.

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."

 

[ttn]

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.)

Let's see it! Every standard derivation *I* know of assumes hidden variables. Can you cite a textbook or paper that has a derivation which doesn't need hv's? Or can you explain how such a derivation might go?

We want the weakest form of Bell Locality possible in order to have Bell's Theorem apply to the maximum range of theories.

Bell Locality is a well-defined, perfectly precise criterion. There aren't strong and weak forms of it.

 

[ttn]

Again, to be specific about Bell Locality:
P(A+, b) = P(A+, c)
P(A-, b) = P(A-, c)

Sorry, that ain't Bell Locality.

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!

This just proves you still don't know what Bell Locality means. You're obviously identifying Bell Locality with signal locality, and they just aren't the same thing!

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.

Look, Bohmian Mechanics makes the same predictions as OQM, so they're both equally in agreement with the data. They both predict that Bell's Inequality should be violated by a certain amount, and that is what is found in the experiments.

The issue here isn't whether these theories agree with experiment. They both do. The issue is: are they local theories? And the answer is: if what you mean by "local" is "Bell Local" then, no, neither is local. But you don't need the experiments to tell you that. You just look at the theories.

However, it would demonstrate unambiguously that BM is not "realistic".

Huh!?

But that's OK with BM too, as I don't think BM asserts that reality is observer independent.

Sure it does, if what you mean is that measurement is just another process obeying the same laws as non-measurement. That's true for Bohmian Mechanics (but false for OQM).

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. )

Only laughs of annoyed frustration.

 

[ttn]

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.

You have this just backwards, perhaps because you are confusing signal locality and Bell Locality. No experiment has ruled out "that the outcome at Alice is affected by the setting at Bob." In fact, just the opposite -- that the outcome on one side is affected by the distant setting/result is *precisely* what is ruled *in* by the empirical violations of Bell's inequality. The violation shows that Bell Locality cannot be true. But this doesn't mean that signal locality is false.

Moreover, what are you talking about when you say "there is NO SIMULTANEOUS a, b and c to begin with." You seem to think that the "a, b and c" are hidden variables or something. This is not true. "a, b and c" are simply three possible directions in space along which Alice or Bob could measure the spin of her/his particle. The hidden variables are the *outcomes* of those possible experiments, the A(a), A(b), A(c), etc. The hidden variable assumption is that each particle carries some kind of "instruction set" (which amounts to a list of values for A(a), A(b), A(c)). Those are the hidden variables. And every derivation of a Bell type inequality assumes the existence of those variables.

The other piece you continue to miss is that the *existence* of these hidden variables *follows* from the assumption of Bell Locality. That is (essentially, though of course not using the precise condition of Bell Locality) what EPR proves.

But stepping back, why are we even talking about any of this here? What relevance do you think any of this has to Bohmian Mechanics? You seem to be trying to construct some kind of argument that Bohmian Mechanics either (a) doesn't agree with experiment (b) isn't signal local. I really can't tell exactly what you're after, but it seems the whole approach is wrong-headed. Bohmian Mechanics agrees with experiment (that is, any experiment that orthodox non-relativistic QM can account for, at any rate) and it is signal local. But it, like OQM, violates Bell Locality. What more is there to say about Bohm's theory vis a vis locality? Why all this talk about Bell and Bell's Theorem? Bell's Theorem doesn't even *apply* to Bohmian Mechanics (obviously, since BM predicts violations of Bell's inequality!), because BM isn't a Bell Local theory. Bell's Theorem is that Bell Local theories must have predictions that are constrained by the inequality. So what in the world are you hoping to learn about Bohmian Mechanics by going down this Bell road? You don't *need* anything like Bell's Theorem to look at Bohm's theory and ask questions like: how does it work? Does it agree with the QM predictions? How does it manage that exactly? Is it a local theory? How come the non-locality can't be harnessed by humans to send signals? How does Bohm deal with particles with spin? How does Bohm's theory solve the measurement problem? What's all this about "contextual" properties in Bohmian Mechanics? etc.

 

[DrChinese]

But stepping back, why are we even talking about any of this here? What relevance do you think any of this has to Bohmian Mechanics? ... How does Bohm deal with particles with spin? How does Bohm's theory solve the measurement problem? What's all this about "contextual" properties in Bohmian Mechanics? etc.

Well, sorry, I think I drifted a little further than I intended. I was hoping to stay more on Bohmian stuff than EPR stuff here. I take the posting related to NateTG's discussion elsewhere - I don't want us to go around in circles anyway.

Can you tell me a little about how spin fits in? I am interested to learn more about that.

 

[NateTG]

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.

First off, the assumption that there is some suitable hidden state \lambda is (more or less) equivalent to assuming that the process is realistic and stochasitc. There's some i-dotting and t-crossing missing from this, but:
If all of the probabilities exist and are well defined, then we can clearly describe the behavior of the system by having \lambda be a particular set of measurement results, and \lambda can be assigned with the appropriate probability distribution - so we can safely assume that there is indeed some hidden state \lambda, even if it only exists in our minds.
Conversely, let's say that the systems state can be completely described by some hidden state \lambda which is selected from a set \Lambda. Then the system is clearly stochastic.
In a sense, \lambda is just a list of measurement results, and \Lambda is the set of possible measurement results, so summing over the possible results is the same as summing over \lambda.

 

[ttn]

Can you tell me a little about how spin fits in? I am interested to learn more about that.

In the simplest possible version, you just do the obvious thing: make the wave function a spinor (satisfying, e.g., the Pauli equation rather than the Schroedinger equation). The guidance formula for spinless particles (grad psi / psi) stays basically the same too -- you need only throw a psi-dagger in the numerator and denominator to contract the spin indices.

So the ontology for particles with spin is really no different from the ontology for spinless particles. There's just a particle being guided around by a wave -- it just happens that the wave has two components (i.e., is a spinor). So when a particle riding a certain wave packet enters, say, a Stern-Gerlach device, the usual Pauli-equation-type-time-evolution causes the wave packet to split into two disjoint lumps, one of which comes out "up" along the field, the other coming out "down." And the particle just (deterministically) ends up in one or the other of those two packets, depending on its initial position. And it's a theorem that, if the initial position is Born-rule distributed (i.e., the Quantum Equilibrium Hypothesis) then the probability for an "up" or "down" "measurement outcome" is just exactly what is predicted by orthodox QM. But unlike the orthodox theory, Bohm needs no strange measurement axioms, collapse postulates, or other "magic" to make this come out -- the particle is detected near the "up" output port of the device because (whether the detector is present there or not) that's where the particle *is*.

Notice that this means "spin" isn't really a property carried by the particle at all. The particle itself isn't rotating about an axis like a tiny basketball, the way students are taught to visualize spin (but then warned not to take such visualizations too seriously). It's just a point particle -- the "spin properties" are carried exclusively by the wave function.

But there's an even stronger sense in which particles (according to Bohm's theory) don't carry spin. Spin is a "contextual property". This means that the outcome you get depends not just on the operator corresonding to the type of measurement you're doing, but the *specific* implementation of that measurement (e.g., which other commuting operators are being measured simultaneously). David Albert cooked up the simplest possible example to illustrate this. Imagine a Bohmian spin 1/2 particle coming up toward a SG device that will "measure its spin in the z direction". And suppose the initial position of the particle is such that it comes out the "spin up" output port of the SG device. Now do a thought experiment: let that same initial particle+wave be incident on the same exact SG device, but let the SG device be rotated 180 degrees so that the B-field is now pointing in the -z instead of the +z direction. Note, according to QM, this is still a device that will "measure its spin in the z direction." But according to BM (and making some other minor assumptions that aren't worth mentioning) the particle (in, mind you, the exact same initial state) will now come out of the "spin down" port of the SG device (i.e., will emerge from the device going the same direction as before).

So two situations with identical incoming particles (and they are identical at the level of the uncontrollable "hidden variable", position, not just in the sense of having been prepared the same way) and two apparatuses that "measure the particle's spin in the z direction" give different outcomes! Shocking, right? Well, not really. It is only shocking if you insist on believing that the particle really genuinely has a property called "spin" which is literally "measured" (in the sense of revealing the pre-existing value of the property) in these experiments. But according to BM, that just isn't the case. Spin is a contextual property -- the value you get depends not just on the state of the particle, but on *how* you perform the measurement -- in other words, spin isn't really "measured" at all, not literally. To say that spin is a contextual property is really just a cumbersome way of saying it isn't a property at all. See the splendid article by Goldstein et al on "Naive Realism about Operators" for more detail on this important point.

I stress it here because Dr. Chinese seems hell bent on defining a "realistic theory" to be one that attributes definite pre-measurement values to all possible observables. Hence, according to him, Bohm's theory isn't "realistic" because it doesn't attribute definite spin values to particles. In fact it says spin isn't even a property that particles possess. Yet I think it should be crystal clear that Bohm's theory is entirely "realistic", if that just means that it denies all of the idealistic nonsense (esse ist percipi) of the orthodox theory (in particular, the idea that nothing exists until it is measured, measurement is a fundamentally different kind of process than non-measurement, etc.). It's just that in BM everything always comes down to the *position* of things, so all this talk of "measuring spin" has to be carefully parsed out in terms of positions (of detector needles, if nothing else). And then everything makes perfect sense, and you reproduce all the QM predictions with a totally straightforward *realistic* theory that is completely immune to the measurement problem that plagues OQM.

 

[slyboy]

But there's an even stronger sense in which particles (according to Bohm's theory) don't carry spin. Spin is a "contextual property". This means that the outcome you get depends not just on the operator corresonding to the type of measurement you're doing, but the *specific* implementation of that measurement (e.g., which other commuting operators are being measured simultaneously).

It is worth noting that any hidden variable theory is required to be contextual in this sense due to the Bell-Kochen-Specker Theorem. In fact, violations of Bell's inequality are also a kind of contextuality, since the result of measuring an operator at one location has to depend on the operator measured at the other location, and these two operators commute by virtue of spacelike separation of the measurements. However, BM is even more contextual than required by BKS because it can predict different results even when the set of operators corresponding to two different measurements are exactly the same. The Albert example is of this type.

It is also worth noting that position is also a contextual property in some circumstances. Although Bohmian particles have definite positions, there are some ways of measuring position that do not simply reveal the Bohmian position, but give a different value. Whilst these have all the properties we would expect from a position measurement in orthodox QM, and it is worth emphasizing that BM still gets all the empirical predictions right, they are not position measurements in the strict sense according to BM.

 

[eljose]

In fact perhaps BM is not true... but it muste be modified to include Complex trajectories, i believe in Complex more than in real trajectories,,..in fact there are several proofs of Complex trajectories to exist..for example:

1)e^{iS/\hbar}=e^{-S/\hbar} under setting t`=it, the last is

only the factor that appears in Statistical Physics,so the QFT is only an Statistical physics with infinite particles, but Statistical physics deals with classical particles so the QM is a classical mechanics with complex trajectories at a temperature kT=\hbar

Another example..Schroedinguer equation:

i\hbar\frac{\partial{\Psi}}{\partial{t}}=\frac{-\hbar^{2}\nabla^{2}{\Psi}{2m}

Under the change t`=it it becomes the diffusion equation (but the diffusion equation is a classical equation):

i\hbar\frac{\partial{f}}{\partial{t}}=\frac{-\hbar^{2}\nabla^{2}{f}{2m}

where the quantities D,and so on depends on h and m of the particle..

The last example is related also to Schroedinguer equation if we set the solution to it in the form e^{iS/\hbar} with S the action then we have that:

(\nabla{S})^{2}=n^{2}=2m((dS/dt)+ai\hbar(\nabla^{2}S)

this is the eikonal equation of the classical optics in a medium with n (refraction index) complex so the trajectories of the particle would be the trajectories of the light in a complex medium...

so we have to deal with coordinates in the form q(it)=a(t)+ib(t) there are trajectories in QM of course, but they are complex..:)

 

[lalbatros]

Unfortunately, this discusssion started without a definition of what is meant by BM.

For me BM is just another point of view on the Schrodinger equation. In this pov, the SE is equivalent to a modified classical particle motion coupled to a wave equation. The particle motion differs from pure CM by the additional quantum potential (force). The first interrest of this pov is of course the link with classical mechanic, since the quantum potential vanishes in the classical limit.

With this in mind, I am rather perplexed by the 5 affirmations by DRChinese:

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.

1) BM can reproduce all predictions of the SE, but its scope does not go beyond the SE
2) It is nonloncal just like the SE
3) It is neither more neither less deterministic than the SE
4) The SE does not either give a special status to the act of measurement. This is the domain of the postulates of QM, more useful for teaching than for the physics.
5) Spin is unknown by SE too.

Finally, I also think that the difficulties with relativity is simply similar to the difficulties found by Schrodinger. But as long as there is some classical limit (no particle creation e.g.), I would be confident that a Bohm pov can be found: the difference between the quantum theory and the classical limit highlights the quantum behaviour. If it keeps the form of a quantum potential, I don't know, but does it matter?

 

[DrChinese]

Unfortunately, this discusssion started without a definition of what is meant by BM.

Well actually I did about all I could. So all I can say is if it isn't accurate, blame me. Although ttn didn't think it was so bad.

Regardless, I would be very interested in learning anything you care to add. I have already learned that the Schroedinger Equation is used as in QM. And to that, it is possible to add particle position information which fills out the picture.

How does the Heisenberg Uncertainty Principle fit in?

 

[lalbatros]

DrChinese,

... So all I can say is if it isn't accurate, blame me ...

I have no intention to blame you, on the contrary. Your question is like an opportunity to ask about what is the Bohmian theory, what is the foundation paper, and all derived questions.

Personally I only know for sure that Bohmian mechanics is based on the (phase, amplitude) transformation of the SE.

This transformation is quite fundamental since it gives a direct link to classical mechanics: the phase is linked to the action while the amplitude is linked to the trajectories in the classical limit.

However, I take as obvious that these transformed equations together with the transformed boundary/initial conditions are equivalent to the original SE. Therefore I don't see how a so-called "Bohmian theory" could be more than a matter of interpretation and/or phylosophy. In particular, I think there is no experimental way to distinguish it from the SE theory and no additional prediction could be made.

But I am not negative about the "Bohmian theory", since it is equivalent to the SE. Taking such another point of view could reveal useful sometimes. Alternative interpretations are not practically useful as long as they keep the equivalence to the original theory. However, alternative interpretations could also suggest new theories which are not equivalent and useful if validated by experiment.

Now, considering the large domain of validity of the SE, how far away should we explore to find a no-mans-land where new theories, possibly inspired by Bohm, could be competitive? Note in addition, that this exploration should be limited to a domain where the Bohmian pov has a meaning: this excludes quantum field theory, at first sight.

Therefore my point of interrest are:

- what are the most 'advanced' formulations of the Bomhian pov
(have fields been included for example)
- are there some 'derived' theories on the market and what are their salient features ?
- what are the directions pointed by the Bohmian pov ?

Finally, I would like to mention that the most striking aspect of the Bohmian pov is related to the role played by the action. And in this sense it goes back to classical mechanics: why is a particle "guided" by the action and why the "least action" principle? There is only one addition, I believe, it is the quantum potential. One could consider that QM is the explanation for the "least action" principle, as it is usually suggested. But one could also envisage further understanding, since QM has not help us very much in this respect. Sure, the Bohmian pov is interresting.

 

[Careful]

Bohmian mechanics makes *exactly* the same statistical predictions as standard QM does (where standard QM *can*.). The only difference IMO is that one can speak of position and momentum of a particle *at the same time* (the famous EPR notion of reality/``answered´´ by complementarity). Therefore BM clearly suggests how to construct a *full* dynamics between a quantum particle and a *classical* gauge field such as the Maxwell waves. In that sense, Bohmian mechanics is *capable* of making more predictions. An example (which I gave already): in the ground state of the H-atom, the particle is standing still (the quantum diffusion force exactly compensates the static electric attraction) which means each H atom would have a *permanent* (this is a prediction Copenhagen can *never* make) electric dipole moment (contrary to observation) which could be measured (assuming classical interactions).

 

[ttn]

How does the Heisenberg Uncertainty Principle fit in?

I'm not sure exactly what you are looking for, but the HUP emerges in Bohm's Theory as simply a constraint on our knowledge of quantum systems (as opposed to a constraint on their objective definiteness). It is closely related to the quantum equilibrium hypothesis (QEH), which says the particle positions are Born-rule distributed. It is a trivial consequence of this that the position uncertainty for a particle's location (square root of the expectation value of x^2 minus the squared expectation value of x) matches that of OQM.

Statements about the uncertainty of the momentum are a bit more subtle. In principle, there is nothing indefinite about the momentum -- if you knew the exact position of the particle and the wave function, you'd be able to infer the precise velocity from the guidance formula. But in general, the QEH prevents you from knowing the position, and hence prevents you from knowing the velocity. *Plus*, the momentum which the HUP says something about the uncertainty of, is not the mass times the actual velocity of the particle at some moment, but, rather, the *outcome of a momentum measurement*. These are not the same thing -- according to Bohm, a "momentum measurement" does *not* simply reveal the value of the pre-measurement velocity (times mass). So to really understand the HUP in Bohm's theory, we need to know something about how measurement works (in particular, measurement of momentum). But suffice it to say that, for some given wave function and assuming the QEH, the product of the uncertainty in position and the "RMS spread of possible measured momentum values" (what we might call the momentum uncertainty) is greater than hbar/2, and the mathematical derivation of this just follows the standard derivations (with appropriate re-interpretation).

Here's a simple example that might clarify how this works. Take a particle in the ground state of the 1-D particle-in-a-box (infinite square well). The wave function is a sine curve with one hump that vanishes at both ends. Assuming the QEH, the uncertainty in position is just what we'd calculate normally, about half the width of the box or whatever. But according to Bohm's guidance formula, if we know the wave is in that ground state, the particle's velocity is zero, for any of its possible positions. It's just sitting there not moving. But does this mean "delta p" vanishes, so that the HUP is violated? Well, sure, in a sense, if that's what you want to define as "delta p". But the quantity "delta p" in Bohm's theory that maps on the traditional "delta p" from OQM is the RMS spread in measurable momentum values. So how does a momentum measurement work here?

One way to measure the momentum (which is nice b/c it doesn't require intervening with a detector) is to just remove the walls and let the particle fly free. In BM, what happens when you do that is that the wave function forms two packets which fly off in the two directions, with the particle getting carried away by one or the other of those two packets (with probability 50/50, and with the initial position determining which packet it ends up in). So then you wait a while until the packets are really far apart and measure the position of the particle to see how far it's gotten in a certain amount of time -- from which you can calculate the momentum it must have been moving with. And it turns out you are going to get either p = +p_0 or p = -p_0, depending on whether the particle went left or right. And so the RMS spread in possible measured p values is just about delta p = |p_0|, which turns out to match exactly what you'd calculate for the momentum uncertainty in OQM. So with *that* as your definition of "delta p", the HUP is satisfied, though it is satisfied in a kind of interesting way -- it turns out to be telling you something about how the initial spread of particle positions (which is related to the wave function by the QEH) relates to the spread of momentum values that could be measured if a momentum measurement were performed (these values being related to the wave function's Fourier transform, roughly speaking).

Hopefully that clarifies a bit, or will at least stimulate some further questions.

 

[ttn]

An example (which I gave already): in the ground state of the H-atom, the particle is standing still (the quantum diffusion force exactly compensates the static electric attraction) which means each H atom would have a *permanent* (this is a prediction Copenhagen can *never* make) electric dipole moment (contrary to observation) which could be measured (assuming classical interactions).

Not so. It's true that according to Bohm's theory the H atom in its ground state involves a stationary electron (and hence an edm). And, according to Bohm's theory, if you measure the position of the electron (i.e., the atom's edm) you'll get some nonzero value -- because the measurement just reveals the pre-existing position of the particle.

But you really think any of this conflicts with Copenhagen? Copenhagen says that the electron is in a big spherically symmetric schmear around the nucleus -- but only until you make a measurement, at which point the collapse causes the electron to pop into existence (as a particle) at some particular spot, with a probability distribution that precisely matches the distribution that Bohm attributed to the electron initially. So, in regular old QM too, one measures some nonzero edm.

Of course, in Copenhagen QM, if one repeats this measurement many many times, one will find that the measured edm's *average* to zero (because of the spherical symmetry of the initial wf). But then this too is true in Bohm's theory. The particle is just sitting there, but for each of the members of some large ensemble it will be sitting in a different place, with the positions governed by the Born rule. So for the same reasons (spherical symmetry of the wf) the average edm (over many trials) will tend toward zero.

So there is certainly *not* any conflict between Bohm's theory and experiment (nor between the predictions of Bohm's theory and the predictions of Copenhagen) here.

 

[Careful]

And, according to Bohm's theory, if you measure the position of the electron (i.e., the atom's edm) you'll get some nonzero value -- because the measurement just reveals the pre-existing position of the particle.
But you really think any of this conflicts with Copenhagen? Copenhagen says that the electron is in a big spherically symmetric schmear around the nucleus -- but only until you make a measurement, at which point the collapse causes the electron to pop into existence (as a particle) at some particular spot, with a probability distribution that precisely matches the distribution that Bohm attributed to the electron initially. So, in regular old QM too, one measures some nonzero edm.
Of course, in Copenhagen QM, if one repeats this measurement many many times, one will find that the measured edm's *average* to zero (because of the spherical symmetry of the initial wf). But then this too is true in Bohm's theory. The particle is just sitting there, but for each of the members of some large ensemble it will be sitting in a different place, with the positions governed by the Born rule. So for the same reasons (spherical symmetry of the wf) the average edm (over many trials) will tend toward zero.
So there is certainly *not* any conflict between Bohm's theory and experiment (nor between the predictions of Bohm's theory and the predictions of Copenhagen) here.

You did not understand the subtle point I was trying to make. I clearly stated that the statistics concerning measurements of *quantum* observables in both formalisms are the same what is basically your argument. What I said moreover is that BM intrinsically allows for MORE than that (for example you can state that the particle *IS* standing still in BM, you cannot claim that in the Copenhagen formalism): we can couple dynamically the particle to a *classical* EM field and consider measurement of that field by detectors around the atom. So we are not measuring an observable of the H atom, neither do we have any trouble with collapse of the state and entanglement of the photon states to the electron state; we are simply measuring a classical field generated by a classical stationary charge cloud (think about Hartree here) which shifts of course a bit during the measurement. Of course you cannot do that in the Copenhagen approach, but the Bohmian approach allows for such possibility. You are of course free to say that this is a bad idea and that one should restrict to direct measurements of quantum observables, but I see no logical reason to do so.

Moreover, it seems rather silly to claim that you can make a (perfect) position measurement of the electron in the H-atom. I ask you: if I put ONE H-atom in the middle of a ring of detectors which measure the strength of the EM field, what is going to be the outcome of that measurement (is there going to be any asymmetry or not?)?

 

[ttn]

You really did not understand the subtle point I was trying to make. I clearly stated that the statistics concerning measurements of *quantum* observables in both formalisms are the same what is basically your argument. What I said moreover is that BM intrinsically allows for MORE than that (for example you can state that the particle *IS* standing still in BM, you cannot claim that in the Copenhagen formalism): we can couple dynamically the particle to a *classical* EM field and consider measurement of that field by detectors around the atom. So we are not measuring an observable of the H atom, neither do we have any trouble with collapse of the state and entanglement of the photon states to the electron state; we are simply measuring a classical field generated by a classical stationary charge cloud (think about Hartree here) which shifts of course a bit during the measurement. Of course you cannot do that in the Copenhagen approach, but the Bohmian approach allows for such possibility. You are of course free to say that this is a bad idea and that one should restrict to direct measurements of quantum observables, but I see no logical reason to do so.

Well then I guess I still don't understand the point. Are you saying that it is *correct* to do the sort of thing you're talking about (couple the Bohmian H atom to a classical EM field, and then "classically measure" the field in order to learn something about the state of the H atom)? Because I think it's clear that it isn't correct to do that. Basically, you are just hiding what amounts to the act of measurement "off stage", by assuming a bunch of "classical" fields which couple to the H-atom in some specified way and which can be measured without disturbance because, by assumption, they are "classical." In other words, you are smuggling in a bunch of things that are "illegal" according to Bohm's theory. And by the way it's exactly the same with orthodox QM. I could say something like: according to OQM, the electron in the H atom is a spherically symmetric smear of charge when it isn't being measured. So instead of measuring it, let me put in a "classical EM field" which couples to that smeary charge distribution, and then I'll learn something about the charge distribution by "classically measuring" the EM fields that are produced by it. But that's illegal, right? If you do things correctly, you'll find that all of this business about letting the atom couple with an EM field, etc., constitutes a *measurement* so you'll never get to directly observe the spherical symmetry of the ground state wave function.

Assuming you're talking about the real world. Of course, you can imagine a different universe in which these "classical EM fields" really exist. And maybe you're right that in that universe, Bohm and OQM make different predictions. I don't really know or care about that.

On the other hand, maybe your point was just that, because Bohm provides a different ontology than OQM, you can sort of imagine things happening according to Bohm's theory, and what you imagine is different from what OQM says. That's no doubt true. I think it's one of the (several) virtues of Bohm's theory that it permits this clear, unambiguous visualization of (even) unobserved quantum systems. That is, it's really a quantum theory without observers. But it's a *fact* that these different-looking ontologies give rise to all the same empirical predictions. And, despite denying that you're denying that, I still can't help thinking that you're denying it. Your last sentence suggests that it's somehow arbitrary (or unnecessarily restrictive) to require that we only talk about measuring quantum observables. But, if the theory is right, this is what we can do, period. Bohmian Mechanics is not classical mechanics. It's not at all arbitrary to say: if this theory is right, here's what we can do and what we can't do. In other words, if Bohm's theory is right (or for that matter if orthodox QM is right) there is an extremely "logical reason" why we can't do the sort of thing you're suggesting -- namely, it's *impossible*.

 

[Careful]

Well then I guess I still don't understand the point. Are you saying that it is *correct* to do the sort of thing you're talking about (couple the Bohmian H atom to a classical EM field, and then "classically measure" the field in order to learn something about the state of the H atom)? Because I think it's clear that it isn't correct to do that. Basically, you are just hiding what amounts to the act of measurement "off stage", by assuming a bunch of "classical" fields which couple to the H-atom in some specified way and which can be measured without disturbance because, by assumption, they are "classical." In other words, you are smuggling in a bunch of things that are "illegal" according to Bohm's theory. And by the way it's exactly the same with orthodox QM. I could say something like: according to OQM, the electron in the H atom is a spherically symmetric smear of charge when it isn't being measured. So instead of measuring it, let me put in a "classical EM field" which couples to that smeary charge distribution, and then I'll learn something about the charge distribution by "classically measuring" the EM fields that are produced by it. But that's illegal, right? If you do things correctly, you'll find that all of this business about letting the atom couple with an EM field, etc., constitutes a *measurement* so you'll never get to directly observe the spherical symmetry of the ground state wave function.

Hi tnt,

Calm down ... I want to argue that there is something fishy about the reality aspect of BM. Of course I am aware that if you apply it ``correctly´´ there is no contradiction with observation, but I want to argue that this ``correctness´´ conflicts with the very meaning of reality. I shall explain in full detail what I mean. In Bohmian mechanics the wave function gives rise to an extra force (the quantal guidance mechanism); of course this force is not classical in the sense that it depends on the possible motion of the particle would it have been somewhere else. Still, the particle exists and follows a well defined trajectory determined by a solution of the Hamilton Jacobi equation with some energy. So, consider the stationary ground state of the H-atom, we know the particle stands still (which is impossible to tell in OQM). We can ask ourselves now, what is the external EM field produced by this configuration? We can take two points of view: (a) the H-atom is quantum mechanical and whatever influence it has is considered to be classical (b) the EM field itself is quantum mechanical. Now, there is *nothing* in QM which tells us that a classical massless field needs to be quantized (stricly speaking the Schroedinger equation only applies for massive particles.). First, consider (a), the reality criterion of BM tells us that the electron is standing still and therefore the EM field is *not* rotationally invariant (wrt. to rotations with the nucleus as centre), we do not know the axis of the dipole, but we do know that there is a dipole and a such a measurement shall reveal this. Standard quantum mechanics does *not* give us such information, and the best we can do is to compute the Coulomb potential corresponding to the nucleus and the charge density of the electron determined by the wave function which *is* for all practical considerations zero at a distance scales of 10^{-9} meters from the nucleus (this is the Hartree approach which is correct in this case) and the experimental outcome is *no* electric dipole. In case you do second quantization, both theories agree (perfect symmetry). I do not see why (a) is illegal; the classical EM field produces a kind of measurement in BM but is incapable of doing so in OQM.

You may not like this, but it is a logical reasoning. But again, you can just exclude this reasoning by hand, but for me it is a sign that there is something wrong with the reality criterion on BM.

Cheers,

Careful.

 

[ttn]

In Bohmian mechanics the wave function gives rise to an extra force (the quantal guidance mechanism); of course this force is not classical in the sense that it depends on the possible motion of the particle would it have been somewhere else.

I don't think that really captures what's non-classical about the force, but I don't think that's a main point.

Still, the particle exists and follows a well defined trajectory determined by a solution of the Hamilton Jacobi equation with some energy.

Yes, fine.

So, consider the stationary ground state of the H-atom, we know the particle stands still (which is impossible to tell in OQM). We can ask ourselves now, what is the external EM field produced by this configuration? We can take two points of view: (a) the H-atom is quantum mechanical and whatever influence it has is considered to be classical (b) the EM field itself is quantum mechanical. Now, there is *nothing* in QM which tells us that a classical massless field needs to be quantized (stricly speaking the Schroedinger equation only applies for massive particles.). First, consider (a), the reality criterion of BM tells us that the electron is standing still and therefore the EM field is *not* rotationally invariant (wrt. to rotations with the nucleus as centre), we do not know the axis of the monopole, but we do know that there is a monopole and a such a measurement shall reveal this. Standard quantum mechanics does *not* give us such information, and the best we can do is to compute the Coulomb potential corresponding to the nucleus and the charge density of the electron determined by the wave function which *is* for all practical considerations zero at a distance scales of 10^{-9} meters from the nucleus (this is the Hartree approach which is correct in this case) and the experimental outcome is *no* electric monopole.

OK, so then isn't it orthodox QM which is refuted by experiment? Because when you actually look for the edm (i.e., measure the EM field configuration near the atom) you *find* an edm.

Of course, OQM isn't really refuted, because what you calculated was just the expectation value for the edm operator. But my point is, if there's a problem here for anything, it's OQM -- which has to invoke a kind of "magic" (the collapse of the wf) to get from what it says is happening before we look (the charge is distributed symmetrically around the nucleus) to what we see when we do look (the electron is at some specific point, giving rise to a nonzero edm).

I just don't see what you think is the problem for Bohm here.

In case you do second quantization, both theories agree (perfect symmetry). I do not see why (a) is illegal; the classical EM field produces a kind of measurement in BM but is incapable of doing so in OQM.

I agree, I don't see any a priori reason why (a) is illegal. (Of course, it might turn out that EM fields aren't really classical, and that to get the right answer you need to quantize that too, presumably in a Bohmian kind of way if we're talking about extending the Bohmian picture of the H atom. But that's a different point, right?)

So the real question is: does an H atom produce an EM field in its vicinity that is the kind of field produced by an edm? If not, we need to go deeper into the question and maybe think about quantizing the field. But if so, Bohm has a ready explanation. OQM also has a ready explanation, though it is a bit fishier as it involves the collapse postulate.

So again, what's the point exactly? What's the problem with the "reality criterion on BM"?

 

[Careful]

**I don't think that really captures what's non-classical about the force, but I don't think that's a main point.**

I think it exactly captures it apart from some faster than speed of light motion due to a quasi local coupling between the phase and amplitude in the hyperbolic equations for the phase and amplitude.

The point is that, to my knowledge, the H atom in the ground state is measured *not* to have a permanent electric dipole moment (as I said the result from the Hartree approximation is correct here). To be entirely explicit; I you let this one H-atom stand for a long time between the detectors, then there will be a dipole detected *on average* (that is: just count the average field strengts measured by each detector and compare them) in scenario (a) in the Bohmian thought experiment, since a classical measurement on the EM *classical* field is not going to have any influence on the electron position in the H atom (remember: classical measurements on classical fields do not disturb).

 

[DrChinese]

If position does not commute with momentum in Bohmian Mechanics, how is that reconciled with the idea that position is fundamental?

 

[Careful]

If position does not commute with momentum in Bohmian Mechanics, how is that reconciled with the idea that position is fundamental?

Hi, this depends upon your notion of reality. Your position seems to be: the reality of a property of the particle IS the value of the corresponding observable obtained in a (future) experiment (the position Bohr takes). This is also the notion of reality used in the Kochen Specker theorems as far as I remember. My notion of reality is much more extended: for example a particle may have a well defined postion and momentum at the same time, but it may be that I cannot *measure* both at once since in a position measurement where I fire a ``photon´´ to the particle, the momentum is going to be disturbed. However, I can still build in a split between the quantum world and the classical one (such as happens in BM/they do not dispose of the Reduction). Of course, it would be preferred that the number which is produced (i.e. outcome of experiment) must result only from *physical* interactions and not through some magical collapse of the state. That is: I would like to precisely know how (and how much) the momentum is disturbed (which is not possible in BM) !

Cheers,

Careful

 

[ttn]

If position does not commute with momentum in Bohmian Mechanics, how is that reconciled with the idea that position is fundamental?

Following up on Careful's remarks... it really just isn't clear what you mean when you say that "position does not commute with momentum in Bohmian Mechanics." As I'm sure you know, BM isn't fundamentally framed in terms of operators. "Operators as observables" emerges from the basic structure of the theory as a theorem. So it is possible, I guess, to talk about position and momentum operators (which of course don't commute) but then we are just talking about one of the standard pieces of the quantum formalism which is in no way unique to (and is definitely not fundamental to) Bohm's theory.

But maybe the thrust of your question is simply this: how do we reconcile the uncertainty principle (which follows from the non-commutation of the x and p operators) with the fact that, in Bohm's theory, particles are postulated to always have definite positions -- and, for that matter, to be following definite trajectories and hence to have always some definite velocity (hence also mass times velocity!). Doesn't this violate the HUP? No, it doesn't, because in BM the HUP is a statement about the spread of *measured* (or possibly-measured) values, not a statement about the *actual* values. In the case of *position*, these aren't different, since measurement simply reveals the pre-existing position. But in the case of *momentum* they definitely are different (as Careful pointed out). If at some instant a Bohmian particle of mass m is moving with velocity v and then you make a "momentum measurement", you will (in general) *not* get the value mv as the outcome of the measurement.

The trapped particle-in-the-box that I mentioned before is the clearest example of this. If the wf is an energy eigenfunction, the particle will be stationary. But if you measure the momentum, you won't get zero (which one can visualize by imagining the measurement is made by removing the walls of the box and letting the particle fly free). So one gets, as the measurement outcome, (roughly) either plus or minus h-bar times k (where k is the wave number associated with the initial wf). So the "uncertainty in p" associated with the original state is just hbar*k. And this even though, in that initial state, the particle was *definitely* at rest!

Note by the way that the smallest possible value of k is pi/L, where L is the length of the box. And for that wf, the quantum equilibrium hypothesis yields that the "uncertainty in x" is (roughly) L/2. So the product of the "uncertainty in x" and the "uncertainty in p" is pi/2 -- just as the HUP says it should be.

Anyway, the point is: how this consistency with the HUP comes about in Bohm's theory is rather interesting and subtle, because the "uncertainties" in general refer not to any kind of objective indefiniteness in the initial state, but simply to our ignorance about the precise state -- in this case, our ignorance of the position of the particle in the wave and the *resulting* ignorance about what value will be obtained for momentum if a momentum measurement is performed.