The Refutation of Bohmian Mechanics

[dextercioby]

In a universe consisting only of a proton and an electron, there is no way to do such a preparation.

I sense a conflict here between the meanings of concept of <statistics> which eventually lead to the different interpretations of the (possibly the same) theory.

In the Copenhagen interpretation, one could describe a universe made up of only one electron and one proton and purely from theory compute the probability of measuring the electron's position within a finite spatial volume.

The statiscal/ensemble interpretation however claims that quantum mechanics is a theory of infinitely many indentically prepared systems (in particular H-atoms), so that the universe is never made up of an electron & a proton, but from an infinity of such particles.

By reading the thread I sense that the Bohmian Mechanics (seen as a different interpretation, or even a different formulation of the same theory) uses the same meaning of statistics as the ensemble one, or ?

 

[Demystifier]

By reading the thread I sense that the Bohmian Mechanics (seen as a different interpretation, or even a different formulation of the same theory) uses the same meaning of statistics as the ensemble one, or ?

Yes. In fact, the Bohmian interpretation can be viewed as a specific realization of the general idea of ensemble interpretation.

 

[A. Neumaier]

I sense a conflict here between the meanings of concept of <statistics> which eventually lead to the different interpretations of the (possibly the same) theory.

In the Copenhagen interpretation, one could describe a universe made up of only one electron and one proton and purely from theory compute the probability of measuring the electron's position within a finite spatial volume.

The statistical/ensemble interpretation however claims that quantum mechanics is a theory of infinitely many identically prepared systems (in particular H-atoms), so that the universe is never made up of an electron & a proton, but from an infinity of such particles.

By reading the thread I sense that the Bohmian Mechanics (seen as a different interpretation, or even a different formulation of the same theory) uses the same meaning of statistics as the ensemble one, or ?

Unlike QM and its ensemble interpretation (which is valid only for small subsystems of the universe that can be identically prepared many times), Bohmian mechanics is heralded by its champions as a deterministic theory.

Deterministic foundations make sense only if they can describe the whole universe as a big deterministic system. This universe is at any time in a well-determined, unique state, and cannot be replicated many times. Thus any statistics (collected by an observer inside this single universe) must be the result of this deterministic dynamics in this particular state. Statistics in a deterministic theory is therefore the result of chaotic motion, not of a replication of the dynamics under different initial conditions.

As my example shows, Bohmian mechanics models systems where the postulated statistics does not follow from the dynamics but must be put in by hand as an assumption on multiply prepared systems. Only if the initial distribution matches exactly the required statistics, the outcomes are correct.

Thus Bohmian mechanics is no deterministic theory in the usual sense, since for its interpretation it needs to make assumptions of a stochastic nature that cannot be derived from the dynamics.

 

[Demystifier]

And if there are two devices with which a spin casn be measured, one would presumably need two pointer variables, and a larger space of wave functions. Now which of these description has a true ontological status?

If both devices exist in the laboratory (or elsewhere in the real world), then both are "preferred" and ontological. If only one exists, then this one is the preferred and ontological one. Simple, isn't it?

 

[A. Neumaier]

If both devices exist in the laboratory (or elsewhere in the real world), then both are "preferred" and ontological. If only one exists, then this one is the preferred and ontological one. Simple, isn't it?

And if none is in the laboratory?

Wouldn't you have to treat all the possible devices anywhere in the world as pointer variables parameterizing the spin wave function?

 

[Demystifier]

Only if the initial distribution matches exactly the required statistics, the outcomes are correct.

It was thought so in the past, but today it is known that this is not so. Even if the system starts with a different distribution, after a while it settles down towards the equilibrium distribution which coincides with the required quantum distribution. This is not a postulate, but a derived consequence of the Bohmian equations of motion.

 

[Demystifier]

And if none is in the laboratory?

Wouldn't you have to treat all the possible devices anywhere in the world as pointer variables parameterizing the spin wave function?

I said "(or elsewhere in the real world)".

 

[A. Neumaier]

I said "(or elsewhere in the real world)".

I understood that. Now there are lots of devices capable of measuring spins. Why are we allowed to treat the spin wave function to be psi(y) with just one pointer variable y rather than with one for each possible device?

 

[camboy]

Thus Bohmian mechanics is no deterministic theory in the usual sense, since for its interpretation it needs to make assumptions of a stochastic nature that cannot be derived from the dynamics.

Arnold - everything you say here seems to be based on the one afternoon you spent studying deBB sometime in the mid 1990s - this just does not justify your supercilious tone. Things do move on.

Today it is well known that the assumptions you refer to are not assumptions (even though they were presented as such by Bohm and by the early textbooks) but arise as a natural consequence of the dynamics. If the particles are not distributed as the square of the wave function, then they will become so under de Broglie-Bohm dynamical evolution (they will approach 'quantum equilibrium').

http://arxiv.org/abs/1103.1589" is a recent paper which demonstrates this particularly nicely.

Demystifier - I'm glad to see that you now contributing here. Looking back at the misinformation in this shocking thread it is good to see some posts from someone who knows what he is talking about. I'm only an amateur in deBB, and I thought you and zenith had given up (I know it must be very boring constantly having to refute ignorant arguments).

 

[A. Neumaier]

Today it is well known that the assumptions you refer to are not assumptions (even though they were presented as such by Bohm and by the early textbooks) but arise as a natural consequence of the dynamics. If the particles are not distributed as the square of the wave function, then they will become so under de Broglie-Bohm dynamical evolution (they will approach 'quantum equilibrium').

It doesn't for the ground state of the harmonic oscillator.

http://arxiv.org/abs/1103.1589" is a recent paper which demonstrates this particularly nicely.

A numerical simulation of a particular, simple case (as given in that paper) is no proof of your claim that ''If the particles are not distributed as the square of the wave function, then they will become so under de Broglie-Bohm dynamical evolution (they will approach 'quantum equilibrium').''

So now we have one example of analytically demonstrated non-approach to quantum equilibrium and one numerical simulation of approach to quantum equilibrium.

The paper starts off saying:

According to a recently-published (2008) encyclopedia of quantum mechanics [2] ''the conclusion seems to be that no generally accepted derivation of the Born rule has been given to date, but this does not imply that such a derivation is impossible in principle".

The counterexample of the ground state of the harmonic oscillator shows that such a derivation is impossible in principle.

Remark (1) on p.5 is also illuminating in that it shows that there are many inequivalent ontologies producing the same quantum equilibrium. At most one of them can be the real ontology. But all variants are (according to the Bohmian dogma) observationally fully equivalent to standard quantum mechanics. Thus it would be impossible to distingush between them experimentally. In stark contrast to what the remark (1) says.

 

[camboy]

It doesn't for the ground state of the harmonic oscillator.

It doesn't for any stationary ground-state of any system, but that's precisely the point - the 'simple example' given in the paper is of a non-stationary state.

In the extremely unlikely event of your harmonic oscillator having existed since the beginning of the universe as a stationary ground state without ever being perturbed by anything ever, then of course it is possible that it will still be out of quantum equilibrium. It's always possible to do that for a fake model system that exists all by itself in your imaginary universe.

However, the real universe has had a long and violent astrophysical history, which is why essentially every real subsystem we see today is in quantum equilibrium.

A numerical simulation of a particular, simple case (as given in that paper) is no proof of your claim that ''If the particles are not distributed as the square of the wave function, then they will become so under de Broglie-Bohm dynamical evolution (they will approach 'quantum equilibrium').''

It doesn't claim to be. The numerical simulation is an illustrative example which they use to compute timescales for relaxation to quantum equilibrium. More complicated systems would be expected to relax more quickly.

Without bothering to repeat them in their entirety, the article refers to well-known arguments that justify the expectation that all systems will relax to quantum equilibrium, subjet to certain assumptions about the initial conditions of the universe having no-fine grained microstructure (see references 1, 16, 20, 21). In this way, it is entirely analagous to the normal Boltzmann argument regarding relaxation to thermal equilibrium in classical statistical mechanics, and is as well established. It is merely a consequence of the fact that the particle distribution being equal to psi-squared has the maximum entropy.

So now we have one example of analytically demonstrated non-approach to quantum equilibrium and one numerical simulation of approach to quantum equilibrium.

No we don't, you're tossing off authorative-sounding arguments about something you've spent about 30 seconds reading, in an attempt to sound like you know what you're talking about.

The counterexample of the ground state of the harmonic oscillator shows that such a derivation is impossible in principle.

No it doesn't. You've merely artificially eliminated the mechanism responsible for relaxation to the Born rule, and pretended that this is the general case.

Remark (1) on p.5 is also illuminating in that it shows that there are many inequivalent ontologies producing the same quantum equilibrium. At most one of them can be the real ontology. But all variants are (according to the Bohmian dogma) observationally fully equivalent to standard quantum mechanics. Thus it would be impossible to distingush between them experimentally. In stark contrast to what the remark (1) says.

No it isn't impossible. Remark (1) itself points out that it is possible away from quantum equilibrium.

So, all your points are wrong. Want to try again?

 

[A. Neumaier]

No it isn't impossible. Remark (1) itself points out that it is possible away from quantum equilibrium.

If quantum nonequilibrium is possible then Bohmian mechanics deviates from standard QM.

But Bohmian mechnics claims to be equivalent to standard QM. You cannot have the cake and eat it.

 

[Demystifier]

If quantum nonequilibrium is possible then Bohmian mechanics deviates from standard QM.

But Bohmian mechnics claims to be equivalent to standard QM. You cannot have the cake and eat it.

The precise statement is:
Measurable statistical predictions of Bohmian mechanics IN QUANTUM EQUILIBRIUM are equivalent to those of standard QM.

BM also allows a possibility of being OUT of quantum equilibrium, but it does not predict that an out-of-equilibrium situation must occur. In fact, an out-of-equilibrium situation is a situation which, according to BM, is very difficult to achieve in practice.

 

[Demystifier]

Why are we allowed to treat the spin wave function to be psi(y) with just one pointer variable y rather than with one for each possible device?

Because a device modeled by a single pointer variable y is a good approximation, in the sense that nothing substantially changes if more pointer variables are added. No matter how many variables you add, you always get that the statistical distribution of the relevant variables (i.e., those variables the wave functions of which interact strongly with the measured system) is equal to the statistical distribution predicted by standard QM.

 

[A. Neumaier]

The precise statement is:
Measurable statistical predictions of Bohmian mechanics IN QUANTUM EQUILIBRIUM are equivalent to those of standard QM.

I conclude from your salomonic phrasing that, in general, Bohmian mechanics is _not_ equivalent to standard QM. Thus deviations are possible and should be looked for.

Note that Boltzmann's derivation of the approach to equilibrium need very stringent assumptions (ergodicity) that are only satisfied for few model systems. The universe (or even a flower) is far from being in equilibrium.

On the other hand, QM is universally applicable, even to flowers. Thus to establish practical equivalence of Bohmian mechanics with QM, one would need to show that quantum equilibrium holds for the whole universe and all its parts to which QM is applied. This cannot be done by a simple maximum entropy argument.

 

[Demystifier]

It doesn't for the ground state of the harmonic oscillator.

But it does for the ground state of the harmonic oscillator INTERACTING WITH A SUFFICIENTLY COMPLEX MEASURING APPARATUS. But it turns out that realistic measuring apparatuses are ALWAYS sufficiently complex.

 

[A. Neumaier]

Because a device modeled by a single pointer variable y is a good approximation, in the sense that nothing substantially changes if more pointer variables are added. No matter how many variables you add, you always get that the statistical distribution of the relevant variables (i.e., those variables the wave functions of which interact strongly with the measured system) is equal to the statistical distribution predicted by standard QM.

Which of the many existing pointer variables is the correct one? In particular, what is the correct model if no measurment device is close to the quantum system? Or if one removes one device in favor for another one?

Moreover, by this interpretation, spin is not a property of the particles but of the pointer!
This is very counterintuitive.

Suppose you do a long quantum calculation on a quantum computer, and decide at the very end what you are going to measure and with which device. Then according to the spin ontology, the whole quantum computer becomes an ontological property of the incidentally used measuring device...

A very strange world view, indeed.

 

[A. Neumaier]

But it does for the ground state of the harmonic oscillator INTERACTING WITH A SUFFICIENTLY COMPLEX MEASURING APPARATUS.

Making claims is easy. Do you have a proof of that? (Not just a numerical simulation with a single pointer variable.)

 

[Demystifier]

I conclude from your salomonic phrasing that, in general, Bohmian mechanics is _not_ equivalent to standard QM. Thus deviations are possible and should be looked for.

Yes, I agree with that.

Note that Boltzmann's derivation of the approach to equilibrium need very stringent assumptions (ergodicity) that are only satisfied for few model systems. The universe (or even a flower) is far from being in equilibrium.

First, one should distinguish quantum equilibrium from the (more familiar) thermodynamic equilibrium.

Second, even though we apparently do not live in the thermodynamic equilibrium, the fact is that we don't know why. It is one of the unsolved questions in statistical physics (and cosmology). Purely statistical arguments lead to the conclusion that we should expect to find nature much closer to the thermodynamic equilibrium than we actually do.

On the other hand, QM is universally applicable, even to flowers. Thus to establish practical equivalence of Bohmian mechanics with QM, one would need to show that quantum equilibrium holds for the whole universe and all its parts to which QM is applied. This cannot be done by a simple maximum entropy argument.

Actually, the simple maximum-entropy argument is enough. The fact that we (seem to) live in a quantum equilibrium is not surprising at all. What is surprising is that we don't live in a thermodynamic equilibrium.

 

[Demystifier]

Do you have a proof of that? (Not just a numerical simulation with a single pointer variable.)

Yes, and it can be found in many textbooks and reviews of BM. In particular, Chapter. 8 of
P. R. Holland, The Quantum Theory of Motion
is especially well written. If you REALLY want to understand how the theory of quantum measurements works, I highly recommend to read this chapter.

 

[camboy]

Dr. Neumaier,

I wonder - do you ever admit that you are wrong about anything ever? Essentially every statement of fact that you have made in this article has been shown to be due to a misunderstanding on your part seemingly due to your never having read any of the literature. And yet, you never say "Oh, thank you, Demystifer/camboy etc. for pointing that out to me". You just carry on to the next spurious thing "And what about this, what about that?" That's wrong too "So what about the other?".

It is not our responsibility to read the literature for you. And yet, from the beginning of this thread, you have made pompous statements implying that everyone who studies de Broglie-Bohm theory is an idiot. If you were not a respected Professor of Quantum Physics, and a writer of FAQS which are promoted by Physics Forums, this wouldn't matter, but you are and it does. As far as I'm concerned, you seem to be abusing your position to try to destroy a subfield of QM which happens not to accord wih your personal views and research goals. Students and even ordinary physicists who happen not to have read anything about deBB will quite naturally believe every weighty pronouncement that you make. More fool them.

Your ridiculous statements that "it doesn't matter that my opinion of deBB has not been published in the peer-reviewed literature" because "my views are so obviously correct" is not only pompous, but arrogant and self-serving. You are so used to people grovelling at you, that you have forgotten the responsibility that such a position brings. Now if I was a moderator, I would ban you. But of course, they're going to now ban me for being rude to you. Well so be it - someone has to say it.

 

[Demystifier]

Which of the many existing pointer variables is the correct one?

As I said, any and each one which strongly interacts with the measured system.

In particular, what is the correct model if no measurment device is close to the quantum system?

If no macroscopic measurement device is close to the measured quantum system, then no macroscopic measurement device can measure this quantum system (because there is no strong interaction). So in this case none of them is "correct".

Or if one removes one device in favor for another one?

The one (or more) which interacts strongly with the measured system, is the correct one.

Moreover, by this interpretation, spin is not a property of the particles but of the pointer!

That's true.

This is very counterintuitive.

Maybe to you, but to me other interpretations of QM are even more counterintuitive.

Suppose you do a long quantum calculation on a quantum computer, and decide at the very end what you are going to measure and with which device. Then according to the spin ontology, the whole quantum computer becomes an ontological property of the incidentally used measuring device...

Not exactly. Other devices and objects in the universe are also ontological, both before and after the measurement. However, only this particular device is strongly correlated with the microscopic degrees of the quantum computer. That's because, as I said many times above, only this particular device strongly interacted with the microscopic quantum-computer degrees.

 

[Demystifier]

Demystifier - I'm glad to see that you now contributing here. Looking back at the misinformation in this shocking thread it is good to see some posts from someone who knows what he is talking about. I'm only an amateur in deBB, and I thought you and zenith had given up (I know it must be very boring constantly having to refute ignorant arguments).

Thanks, camboy! I see that your understanding of deBB is also very good.

 

[A. Neumaier]

First, one should distinguish quantum equilibrium from the (more familiar) thermodynamic equilibrium.

Of course they are different. But the thesis about BM+quantum computing was using the statistical mechanics analogy.

Second, even though we apparently do not live in the thermodynamic equilibrium, the fact is that we don't know why. It is one of the unsolved questions in statistical physics (and cosmology). Purely statistical arguments lead to the conclusion that we should expect to find nature much closer to the thermodynamic equilibrium than we actually do.

No. There is nothing surprising in that we don't have thermal equilibrium. In statistical mechanics, The proof that equilibrium must be obtained is restricted to the very stringent assumption of ergodicity. Very few real systems are ergodic. Equilibrium is reserved for special systems that are more or less homogeneous.

Actually, the simple maximum-entropy argument is enough.

No. There must also be proof that entropy always increases, and that it increases to its maximum.
This is highly nontrivial in statistical mechanics, and wrong for most complex systems. Nature is full of systems that never reach the maximum entropy state.

It would be very surprising if the situation were different for quantum equilibrium, and that it is achieved without such stringent conditions. Can you point to an online source for the proof?

 

[A. Neumaier]

do you ever admit that you are wrong about anything ever?

If I see that I was wrong, I correct myself immediately. Thus any state of being wrong is very short-lived. If I repeat assertions that seem wrong to you then only because your arguments did not convince me.

to your never having read any of the literature.

You have not the slightest idea about how much literature I have read and how much I am reading
while preparing my contributions to this forum.

And yet, you never say "Oh, thank you, Demystifer/camboy etc. for pointing that out to me".

Scientific dispute takes the free offering of information as a given that doesn't need special thanks.
I also do not expect being thanked for the information I provide on this forum.

from the beginning of this thread, you have made pompous statements implying that everyone who studies de Broglie-Bohm theory is an idiot.

Please take this back. I never said such a thing.

If you were not a respected Professor of Quantum Physics,

You seem to say a lot without first checking the facts. I am not a professor of quantum physics.

Your ridiculous statements that "it doesn't matter that my opinion of deBB has not been published in the peer-reviewed literature" because "my views are so obviously correct"

I didn't say such a thing.

 

[A. Neumaier]

If no macroscopic measurement device is close to the measured quantum system, then no macroscopic measurement device can measure this quantum system (because there is no strong interaction). So in this case none of them is "correct".
.

But the system still has a dynamics. Though apparently not one easily described by BM.

 

[Demystifier]

Of course they are different. But the thesis about BM+quantum computing was using the statistical mechanics analogy.

No. There is nothing surprising in that we don't have thermal equilibrium. In statistical mechanics, The proof that equilibrium must be obtained is restricted to the very stringent assumption of ergodicity. Very few real systems are ergodic. Equilibrium is reserved for special systems that are more or less homogeneous.

No. There must also be proof that entropy always increases, and that it increases to its maximum.
This is highly nontrivial in statistical mechanics, and wrong for most complex systems. Nature is full of systems that never reach the maximum entropy state.

It would be very surprising if the situation were different for quantum equilibrium, and that it is achieved without such stringent conditions. Can you point to an online source for the proof?

OK, now we are at the territory of statistical mechanics. In statistical mechanics there is a Boltzmann H-theorem that provides that entropy never decreases. Perhaps this theorem is not perfectly rigorous (it does not use ergodicity), but in Bohmian mechanics there is an analogous equally (non)rigorous theorem that entropy measuring closeness to the quantum equilibrium also never decreases. So what we have in BM are
1) A theorem, probably not perfectly rigorous, and
2) Many explicit numerical simulations which agree with the theorem
I can agree that more work is needed in order to establish a completely satisfying proof, but the above is certainly a strong evidence (if not the proof).

 

[Demystifier]

But the system still has a dynamics.

Of course.

Though apparently not one easily described by BM.

Why do you think so?

 

[A. Neumaier]

Another, more general, point is at the heart of BM: Any observation eventaually is an observation of a POSITION variable (at the macroscopic level). This is why the Bohmian trickery is applicable in any situation

When we hear something, it can count as an observation, though no position variable is observed.

When we see a star, which position variable do we observe?

 

[A. Neumaier]

Why do you think so?

Because, as you said, none of the pointer variables is ''correct''. How else would you then describe the dynamics of an unobserved system of spins?