# Bohmian Interpretation: Equivalent or not to Standard QM

1. Apr 15, 2009

### JustinLevy

(My previous thread on Bohm Interpretation quickly became 4 pages of mostly unrelated discussion. So I'm restarting here with more precise questions.)

Okay, I've read some references on the Bohm Interpretation and I am still confused. As one paper mentioned, some extensions have stochastic elements, so the main purpose of these lines of interpretation seem to be to remove the "classical measurement" from the standard interpretation. Yet even that aspect seems unclear to me.

The fact that it hasn't been soundly refuted in 50 years suggests that I am still misunderstanding it. Continuing with that assumption, please, help me understand why it hasn't been refuted. For currently, the theories do not appear equivalent to me. I will try to be as clear as possible with my questions, so please try to be as clear as possible with your answers. In particular, please clearly define any terms you use that I appear to be misunderstanding, as I am probably unaware that I am misunderstanding them.

Question 1:
All measurements are supposed to boil down to a measurement of position in the Bohmian interpretation. What IS a position measurement in the Bohmian interpretation?

discussion:
comment A - In BM, the configuration and evolution of a system is specified in totality by the wavefunction, Hamiltonian, and particle positions.

comment B - In BM, the wavefunction is not a function of the particle position, but the possible particle positions (the position configuration space). Given N particles in D dimensions, the wavefunction maps points in the space Reals^(ND) to a complex number. The actual positions of the particles evolves according to a functional of the wavefunction. Note: the particle position evolution depends on the wavefunction but not the other way around.

comment C - Since the wavefunction evolution does not depend on the particle positions, the portion of the wavefunction corresponding to the measurement device cannot depend on the particle positions.

comment D - this leaves us in the same situation as standard quantum mechanics. A measurement just entangles you with what you are measuring. Nothing explains how you get a discrete answer instead of a superposition ... since the result up to this point should be a superposition. The very thing BM was supposed to supply, it appears to fail to accomplish. You need to rely on decoherence, or many minds, etc. other interpretations to get anything out despite adding additional information in.

Question 2:
The particle position is updated according to only the phase of the wavefunction, so BM uses a different definition of momentum than the standard quantum mechanics. This appears to make BM quantifiably falsifiable (ie. it is a different theory, not a different interpretation of the underlying math). What IS a momentum or momentum^2 measurement in the Bohmian interpretation?

discussion:
comment E - Since the ground state wavefunction of a bound system can often (always?) be written as a real valued function * exp(-i E hbar / t), the velocity of any particle in the system is exactly zero. The average velocity squared of a particle is therefore zero.

comment F - This alone appears to refute the BM theory. But also consider momentum conservation. Since the two theories differ, it is unclear to me how momentum can possibly be conserved in BM (especially when considering collisions of particles that can result in bound states).

comment G - Similarly, it is unclear how energy is conserved with this change in p^2 (proportional to kinetic energy). Regardless on whether they use the correct definition of KE, it is also unclear to me on general grounds if a conserved energy can be defined at all. While possibly a solved question, I'm unclear on whether the hamiltonian can be converted to a lagrangian which can be used with Noether's theorem if the number of variables are infinite (fields) AND the configuration has non-local interactions.

Question 3:
In standard quantum mechanics, spin is an internal degree of freedom of a particle. In BM, this degree of freedom is included in the wavefunction by borrowing from derivations in standard quantum mechanics. BM however does not give it a definitive value like it does position. Notice however, that since the wavefunction has two components for each spin (1/2) particle now, to consider evolution of the position, we MUST consider the position to have two components as well (which need not evolve similarly, so even the "real" particle positions are superpositions now!). And you cannot consider these as two individual particles, since they are not conserved (only by considering them together is particle number conserved). So in BM, while <Sx,Sy,Sz> can have any value, a system really truely does not have a known spin direction which seems contrary to a hidden variable theory and also which is distinct to orbital angular momentum due to real trajectories of particles. So what IS angular momentum and spin in the Bohmian interpretation?

discussion:
comment H - Since momentum is defined differently, again we have the same issue as before that it is unclear how angular momentum is conserved. However it now has the additional hurdle due to the true "unknowingness" of spin while the orbital trajectory is definite.

comment I - Because the L=1,m=0 state of the hydrogen atom can be written as a real valued wavefunction, again we have the issue that the particles are stationary. But now, the particles are stationary even though there is orbital angular momentum. So not only does it seem to contradict experiment, but it seems internally inconsistent.

comment J - A change in basis can change the wavefunction from real valued to complex (consider using the Lz basis, or the Lx basis). So the evolution of the position states seems to depend on the choice of basis. BM can't even seem to say definitively whether their hidden particles are moving or not.

comment K - Every derivation I've seen of the spin statistics theorem relied on locality. Is there a way to get this in Bohmian mechanics as well, or does it just borrow this verbatum from standard QM as well.

Any help in answering (please, not speculating, nor hand-waving, as the theory seems vague enough to me currently as it is) these questions would be greatly appreciated.

Last edited: Apr 15, 2009
2. Apr 16, 2009

### Demystifier

JustinLevy, since you ask too many questions, most of them being answered elsewhere, I have a deal for you. I will give you a paper where many of these questions are answered in a concise but pedagogic way. You will read it (only a few pages) carefully, and then you will tell me what is still unclear to you. After that, I will answer your remaining questions.
Deal?

3. Apr 16, 2009

### JustinLevy

Of course that is a deal, as that sounds perfect!

4. Apr 16, 2009

### Demystifier

OK, all you have to read for a start is the Appendix of
http://xxx.lanl.gov/abs/quant-ph/0208185
It is only two pages long, but read it carefully. Pay particular attention to the fact that it is a GENERAL theory of measurement, so it explains how measurement of ANY quantity is explained by BM in a manner consistent with the predictions of standard QM, including position, energy, momentum, angular momentum, spin, whatever.

5. Apr 17, 2009

### JustinLevy

That does not appear to answer my questions at all. To give extra benefit of the doubt, I read it again, thought about it that night, took a sleep on it, thought about it some more, and read it again today. It doesn't seem to answer any questions... to me it just rephrases stuff as if it is an answer.

Let me try to explain as succintly as possible. You say in your paper:
Thus, if there is such an initial distribution, then if you describe the entire system (measuring device and thing you are measuring) quantum mechanically you can never make a measurement of the position as the system will remain distributed as such. If the initial distribution is not as such, then Bohmian mechanics is not compatible with quantum mechanics. Either way, there is a problem.

When trying to explain measurement, invariably all the Bohmian interpretation sources I've read refer to the measurement of something classical. Essentially, they say: To measure a particle position couple it with a 'macroscopic' system and measure the position of the 'macroscopic' system. All you are doing is transferring the problem! You never actually answered how to measure anything.

6. Apr 17, 2009

### Demystifier

JustinLevy, I must admit that it is not clear to me what exactly is not clear to you.
But you could help me if you could explain what is YOUR favoured interpretation of QM.
In particular, what do you think about the von Neumann measurement scheme and what do you think about decoherence.

7. Apr 17, 2009

### JustinLevy

I prefer a mathematical model with local evolution and non-unitary "imposed" measurements, essentially the standard, as many call it here, "shut up and calculate" 'interpretation'. Basically, if there is something we don't quite understand yet (foundational issues) to me it seems more appropriate to limit the theory's range of applicability until we learn more experimentally, instead of ad-hoc removing symmetries that we see experimentally. If really pressed, I choose the "many worlds" interpretation since it maintains all previous features, and just essentially says "measurement" is our state entangling with the world ... it seems closest to what taking quantum mechanics 'literally' would result in.

However, it is useful to be aware of such discussion on foundational issues. If the standard truly was "shut up an calculate" we wouldn't discuss and marvel at Bell's inequalities. So I want to understand what is and isn't possible as far as interpretation. In this case, I really truly do not understand how Bohmian mechanics provides the same predictions in all situations. So there must be a flaw in my understanding of these foundational issues, and it can't hurt to be a little more well rounded in the subject.

What is not clear? I want to know how measurement works in your theory. Your answer boils down to we make a measurement by having a particle affect something else which we then measure. See the problem there? Your answer is circular.

Furthermore, your theory doesn't seem to even allow any measurement. Let's say we somehow knew the wavefunction and Hamiltonian of an entire experiment (in principle, this is possible in standard quantum mechanics). Can the experiment, in principle, determine the bohmian position of a particle? Because you say the wavefunction is always the statistical distribution of the particle position, the answer seems to have to be NO.

----

If you are having trouble understanding my confusion, let's focus on an example. Let's work with non-relativistic quantum mechanics. Consider the Stern-Gerlach experiment: a wavefunction of an electron and a measuring device start out separable. The electron starts as a small wave-packet with the particle having spin Sz=+1/2. Now it goes through a region of a magnetic field gradient in the y direction, which splits the wave-packet into two wave-packets. The electron now slams into a CCD imager. The wavefunction of the system is no longer separable. So, what do we measure? Let's say we knew the bohmian particle position ahead of time, can you show the math that explains how only one result comes out? I'm not sure how you'd even due this in Bohmian mechanics since the electron wavefunction has two components which seems to require the electron position to have two components ... so since Sz and Sy do not commute, it seems to me you still get a superposition of the position in Bohmian mechanics (and not just statistically, but an actual superposition of the Bohmian positions themselves).

Working out that example would be very helpful.

Last edited: Apr 17, 2009
8. Apr 17, 2009

### Demystifier

I think I understand now what you are saying.
The answer is the following. According to BM, everything consists of particles, just as in classical mechanics. In a measurement, particles affect other particles. The result of a measurement may be a position of a needle of the measuring apparatus, and this needle also consists of many particles. The role of wave function is analogous to a potential in classical mechanics, i.e., it plays a secondary role, because it only tells particles how to move for given initial conditions. Do you still find it circular?

See again the Appendix that I suggested to you. What you actually observe is the position of needle (or some other macroscopic configuration of particles of the measuring apparatus), represented by y. If the value of y turns out to be in the support of chi_a, then the measured particle behaves as if it was described only by psi_a. Furthermore, if your measurement procedure is aimed to be a measurement of the position of x, then psi_a is a localized wave function, in which case you know with certainty that the particle is localized within psi_a.
Is it clearer now?

We measure y. But y is strongly correlated with x due to (42). Therefore, we can say that we indirectly measure x as well.

Of course. y(t) is a single-valued function, so only one result can come out.

No, the electron position x(t) is a single-valued function as well, even though the wave function has two components.

See also Ref. [2] which also contains a good pedagogic explanation of the theory of quantum measurements.

9. Apr 17, 2009

### JustinLevy

No, the wavefunction is more than that. Because here you are also requiring the wavefunction to always give you the particle distributions. This is quite unlike a potential.

If the evolution makes the wavefunction of the needle become a superposition of two outcomes, but you see the needle in only one position, then you have violated the postulate I quoted from your paper.

No that is not clearer. You are presenting a dichotomy between the microscopic and macroscopic and saying essentially that we can obviously measure something macroscopic, so by entangling with something microscopic, that lets us measure the microscopic. But your theory claims to not make such a dichotomy and so it is not obvious how we measure anything let alone something macroscopic.

Furthermore, the wavefunction evolution doesn't depend on the bohmian particle positions. Therefore, for the wavefunction to collapse to a localised one due to the experiment, cannot depend on the bohmian particle positions, and thus can in no way be considered a "measurement" of the bohmian particle position.

Arrgh. I'm trying to figure out how to make a measurement in your theory. Your answer: "We measure y."
This is just transfering the measurement problem to a different entity.

The wavefunction will be in a super-position of two outcomes. If you actually measured y, then by your own explanations, the wavefunction should only be of that one value.

Then can you please show the math for that example. Seeing some actual math for an example could be very helpful. Math can speak clearer than prose.

That makes no sense. If the wavefunction has two components, then so does the magnitude and phase of the wavefunction. It is possible to apply your equations and get something like
$$v(t) = \frac{1}{m}\partial_x S(x,t) = \frac{1}{m} \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ -1 \end{array} \right)$$
The velocity is not single valued. How can you claim the position is?

Last edited: Apr 17, 2009
10. Apr 17, 2009

### Demystifier

It is true that the wave function plays this additional role, but this role can in fact be DERIVED from the first one, as a property that is valid in the equilibrium. (I will not present here this derivation, but I will give you the references if you wish.)

You mean
"This postulate provides that the statistical distribution of particle positions is given by |psi (x, t)|^2 for any time t, provided that this distribution is given by |psi (x, t_0) |^2 for some initial time t0."?
You misunderstood that. The particle allways has only one position, but the distribution says what is the probability of a particular position. Of course, this is an a priori probability, which makes sense only before you determine this position.

Well, in BM it is considered obvious that you can observe macroscopic things, simply because you are macroscopic as well. If you have problems with this, then you probably even have problems with measurements in classical physics. Do you? Anyway, I do not find any dichotomy with this, or at least not more than in classical physics.

The total wave function does not depend on particle positions. Yet, the conditional wave function, i.e., the wave function that describes your updated knowledge after you measured some particle positions, does depend on these positions.

Again, you should distinguish the total wave function form the conditional (effective) wave function.

Eq. (38) is valid without spin. With spin, (38) is slightly modified, such that velocity is single valued. If you want the exact equation in the case of spin, I can write this as well for you.

Do we make any progress?

Last edited: Apr 17, 2009
11. Apr 17, 2009

### JustinLevy

If the wavefunction initially gives the probability distribution, then it always does according to your theory. Therefore the total wavefunction (not some "conditional" wavefunction that you are introducing below) must represent the probability distribution at all times. If you have measured the position, and the total wavefunction does not collapse, you have violated one of your postulates.

If at any point the total wavefunction no longer holds to the "equilibrium hypothesis", then your theory differs in predictions from standard quantum mechanics.

How can you possibly not consider that a dichotomy!?
You are treating macroscopic objects differently from microscopic objects, despite claiming your theory doesn't.

Following Schrodinger's equation the wavefunction for your classical object will evolve a superposition of answers. What collapses the wavefunction to only one result? Based on your "conditional wavefunction" comments you seem to be devolving into a 'consistent histories'/'many worlds' interpretation.

Good, we have at least one definitive thing to start building a common ground from.

But the evolution only depends on the total wavefunction, correct?
Your equilibrium hypothesis only stays true if you consider the total wavefunction. If you now consider the probability to be expressed by the conditional wave function, then you have lost some probability (you have done a projection on the states, the very thing you seem to dislike that copenhagen does).

Yes, I would like to see that.
Again, a simple example presented with math would speak volumes here. Can you please work out the Stern-Gerlach experiment?

12. Apr 17, 2009

### Demystifier

Come on, you should first learn some basics of probability theory. It is pointless to discuss foundations of QM if, for example, you do not understand the concept of conditional probability. After that, you should learn some basics of classical statistical physics as well. Only then we can discuss foundations of QM.

I have no intention to teach you these non-quantum stuff, as I am sure that there are many people here that can do that as well.

13. Apr 17, 2009

### akhmeteli

Demystifier, I am afraid I am not sure how to answer the following question, so I wonder if you could explain:

What is the status of the projection postulate in dBB? Is it an approximation, or does it hold precisely, or maybe it does not hold at all?

There is the following phrase in the Appendix that you cited (and there are similar phrases in other texts on dBB measurement theory):

"To see how this problem resolves, it is essential to realize
that, when the system consisting of the variables $x$ and $y$ can be considered as a configuration suitable for measurement of the $x$ subsystem, the total wave function is {\em not} of the form $\psi(x,t)\chi(y,t)$. Instead, the interaction between the $x$ subsystem and the $y$ subsystem should be such that the total wave function takes the form
$$\label{app4} \Psi(x,y,t)=\sum_a c_a(t)\psi_a(x)\chi_a(y),$$
where the normalized wave functions $\chi_a(y)$ with different labels $a$ do not overlap in the $y$ space."

However, if some wave functions do not overlap at some time point, they'll overlap later, as wave functions disperse in space with time. Of course, you can say that the measurement device is macroscopic, so its wave function does not disperse, to all practical intents and purposes. However, strictly speaking, this is not a precise statement, but an approximation, so my conclusion is that the projection postulate is an approximation as well. Is this correct or not?

14. Apr 17, 2009

### JustinLevy

I do not appreciate you trying to sweep this away as me not understanding basic math.
I understand basic math just fine thank you. I have taken plenty of physics classes, taught physics classes, and have physics degrees.

I am trying very hard to understand what you mean here, and instead you just keep giving circular answers or resorting to what appears to be a micro/macro dichotomy or many worlds on a whim. I understand that you feel you are describing your theory clearly, but it is coming off very vague. I am not asking these questions in a vaccuum either ... I am trying to read up on it, but it appears there are physicists who believe BM is just many-worlds with additional hidden structure, which (to me) your views on measurement seem to support. So I am not alone in thinking these things. But I am trying, and asking earnest questions.

So let's please try this again.

So far we have agreed on:
A] The total wave function does not depend on particle positions.

Do you agree:
1] If at any time the total wavefunction no longer holds to the "equilibrium hypothesis", then your theory differs in predictions from standard quantum mechanics.

2] If you gain some information about the particle, so that you can neglect some of the total wavefunction and consider only the 'conditional wavefunction' then the total wavefunction no longer holds to the "equilibrium hypothesis".

3] The evolution of the particle positions depends on the total wavefunction, not the 'conditional wavefunction'.

Questions (hopefully just yes/no questions):
4] My understanding of your answer to "how do you make a measurement" was that you feel that regardless of whether the wavefunction of a macroscopic object evolves to contain a superposition of answers, the position of the object can be observed directly because it is "obvious" what the positions of "macroscopic things" are. Is this understanding correct?

5] Previously I asked if in a closed system in which the hamiltonian and initial wavefunction was known, if additional information as a bohmian particle position could be determined. My understanding is that your answer is yes. Is this understanding correct?

6] Consider an external observer also knew the hamiltonian and initial wavefunction of this box, and calculated the probabilities of some results when he looks in later. Would the experimenter in the box, using his additional knowledge of the system, be able to predict the final measurements to better accuracy than the experimenter outside the box?

I'm still interested in seeing the details of the evolution of an non-relativistic electron with a spin in an external magnetic field if you can provide a reference. Trying to find info, I found a paper claiming that there is experimental difference between the theories http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.3878v1.pdf

15. Apr 17, 2009

### conway

Justin, the last thing I want to do is derail your attempted engagement with Demystifier. It's obvious to me you're putting quite a lot of effort into trying to communicate here and I know how frustrating these discussions can be. I also somewhat share your frustration with what seems to be his glib answer as to what constitutes a measurement in the Stern-Gerlach experiment. So I don't really want to interfere in the discussion...BUT: I'm going to ask the question:

Is Bohmian Mechanics simply the identification of two real-valued functions, a position density and a velocity density, as derived from the ordinary wave function? And as such is the bottom line for any Bohmian calculation simply the idea that you transform back to the Schroedinger function, do the time evolution, and then re-convert back to the Bohmian position and velocities? If this is all there is to it, what insight can it possibly give to the problem of the collapse of the wave function?

Sorry if this if way off topic.

16. Apr 17, 2009

### zenith8

Look guys, can we not just give Demystifier a break here? He is genuinely trying to help and all your stuff about 'glib answers' and 'your theory' seems to me unnecessarily rude. Your (far too many) questions are all based on standard misunderstandings - they are not some deal-killing problems with an 80 year-old theory which you are the first to notice. Why don't you just read the standard references that you were actually given in the previous thread which you claim is 'mostly unrelated', instead of effectively asking him to copy them out for you?

In particular, the http://www.tcm.phy.cam.ac.uk/~mdt26/pilot_waves.html" [Broken] was suggested (the Lecture 4 on measurement theory answers most of your current questions).

Dr. Chinese also suggested:

Passon (2006): What you always wanted to know about BM but were afraid to ask...

Nikolic (2006): QM: Myths and facts

Passon (2004): Why isn't every physicist a Bohmian?

Tumulka (2007) Understanding Bohmian mechanics

Reading the above it seems to me you cannot continue to imply that the theory is incompatible with standard QM.

Just to state the answer to the question implied in your thread title, the Bohm interpretation/pilot-wave theory is certainly equivalent to standard QM. How could it be otherwise? As stated in the previous thread, the entire thing flows from standard QM by assuming the psi squared means the probability of the N particles being in those positions, rather than the probability of being found there in a suitable measurement. This assumption - hardly very revolutionary when you think about it - implies that the particles continue to exist between measurements.

In a stationary state, the probability distribution doesn't move. So the now continuously-existing particles must either be actually stationary or moving in some constant repeating pattern (distributed as psi squared over an ensemble). In a non-stationary state, the probability distribution changes in time, so the particles must move - the 'guidance equation' follows from the expression of the probability current of the moving distribution which is already in standard QM (it is often mistakenly presented as some extra maths grafted onto standard QM, which may account for your confusion).

It is the same mathematical theory, so it must make the same predictions. And for free, you get mathematical derivations of the Born postulate, the classical limit, measurement theory (with no collapse, splitting universes, observer-induced magic, or other weird stuff) etc. etc. It is a mathematical formulation of QM equivalent in status to (say) Feynman's path-integral formulation (rather than just an interpretation) because one can use it to do predictive calculations (e.g. calculate the propagator by integrating the quantum Lagrangian along the one actual path followed by the particle, rather than the classical Lagrangian along the infinite number of all possible paths as in Feynman).

A final interesting point is that while it is equivalent to standard QM, it is actually of wider scope in that it also describes the 'non-equilibrium' case where the particles are not distributed as the square of the wave function. In this respect one finds (theoretically or through numerical simulation) that particles with any initial distribution which are 'stirred' by the 'pilot-wave' psi in the way suggested become distributed as psi squared in the course of time, and once so distributed, remain so. It can be shown that psi squared is the only distribution with this property.

This is interesting since it seems to imply possible experimental tests of pilot-wave theory (admittedly in fairly extreme conditions such as the beginning of the universe/black holes etc..). See relevant work by Valentini, and the second half of Cambridge Lecture 5.

Last edited by a moderator: May 4, 2017
17. Apr 18, 2009

### Ilja

In the paper I'm currently writing I question this claim. My position is that measurements have to be (and are, in the standard equivalence proof) reduced to actual states of the beables (positions), but not to their measurements.

No. You have the actual values of the configuration. They allow to reduce the wave function of the universe to an effective wave function of the part we consider.

The definition is almost tautological

$$\psi(q_S) = \Psi(q_S, q_{env})$$

We do not have to measure the $$q_{env}$$, but have to use their actual values. Else, the explanation would be circular: A measurement explained by measurement results.

In some sense, you need some elements of decoherence to prove that $$\psi(q_S)$$ follows an effective Schroedinger equation. But only as far as decoherence is part of standard QM anyway, not on the fundamental (and wrong) applications of decoherence.

It does not make BM different from QM (falsifiable is something different - BM is as falsifiable as QM). Simply there are two notions of momentum - the result of the quantum measurement, and the expression mv for the velocity along the trajectory. Only the first is an observable in BM. The second isn't.

But there are situations where above have the same values: This is the classical case.

There is no such necessity. The evolution equation for the single position is in this case
$$mv = \hbar\Im (\psi^* \nabla \psi)/\psi^*\psi$$

No. Angular momentum, as an observable, has the known QM relation to momentum, as a QM observable, and nothing to do with some rotations of the particle position.

For every particular specification of the theory, there is a well-defined answer.

That for hidden variable theories there are different possible answers is quite straightforward, at least something we should expect. Else, they would not really be hidden.

Why not borrow? Once SQM is true (derived) in BM, we have any right to do this.

18. Apr 19, 2009

### JustinLevy

Ilja,
Thanks for the responses. When you finish your paper on measurement, please post a link here as I'd be interested in reading it.

-------
The original questions were not addressed specifically to Demystifier. I felt it appropriate to refer to some of his responses as "his" theory because 1) he started by giving me a reference to his paper and 2) from what I have read, his responses in this thread don't match what other people have presented for bohmian mechanics. I meant no offense by calling it "his" theory. I was merely trying to be specific in my responses.

In particular, he is claiming that a closed system can obtain more information than is obtained in the wavefunction. This lead to difficulty in understanding his positions, since it seemed to contradict what I've read of other presentations of BM. (The cambridge one says "In pilot-wave theory, [the wavefunction] is a real physical field which influences particle trajectories. At the same time it represents the greatest possible information about the system.")

If my frustration shows it was because he kept answering that to make a measurement on one thing we measure something else... which doesn't answer the question. I apologize if this frustration came off as rudeness.

That thread IS mostly unrelated. I only made one post in that thread and it became 4 pages of mostly other discussion. That doesn't mean some responses weren't useful, nor does it mean those were ignored.

As I mentioned, I have read several papers and presentations of the bohmian theory.
I've also read papers that seem to refute it. I'm not claiming I'm going to come up with anything new. I'm asking about what I do not currently understand, or feel is contradictory to claims, in order to better understand discusions of the theory.

The more I read, the more I am coming to the opinion that this theory is not equivalent to standard QM. But I don't feel I can really come to a conclusion without better understanding the bohmian theory ... I want to make sure that if I decide to dismiss it, that it is for a valid reason. If I learn why I can't dismiss it, then I learn something useful as well.

I am not a proponent of pilot wave theory, and even I find that webpage to be a disservice to the theory. It comes off like an almost crackpot page. Look how much time he wastes on non-scientific stuff and phrasing of 'persecution'. A scientific discussion would be to present the theory as succintly and non-ambiguously as possible, then demonstrate its usefulness by making contact with current theory and make some example calculations. Everything else only does disservice to the theory.

Regardless, his webpage does present a lot of material. It doesn't answer how a measurement is made in BM... he assumes we can measure some of the hidden variables for a macroscopic object, and with this assumption we can then map any measurement to a measurement of the macroscopic object hidden variables. Basically, he too tries to explain measurement by saying we entangle it with another object and then measure the position of that second object. This is circular... or even if accepted that some hidden variables can be obtained, would allow for distinctions with standard quantum mechanics (since a closed system can learn more than what the wavefunction contains).

This too assumes we can measure the position of macroscopic objects to explain measurement. This is circular, unless you want to introduce an artificial micro/macro dichotomy.

How could it not be!? Because showing that an ensemble of particles has the same distribution as the wavefunction^2 is not equivalent. It must reproduce all correlations of the distributions as well ... which as has been argued in the past, cannot be done.

No, the assumption is more than that. It is that the particle exists in one place at any specific time between measurements. This is indeed revolutionary, and from what I have read, incompatible with quantum mechanics.

Are you saying that despite the knowledge of position "effectively" collapsing the total wavefunction to the "conditional" wavefunction as Demystifier puts it, that over time the distribution must redistribute to the total wavefunction again? That seems to be what you are claiming here.

Besides, the fact that the distribution is ever less than that implied by the total wavefunction means the equilibrium hypothesis is wrong at that point and can be used to distinguish with regular quantum mechanics. I fail to see how the theories are equivalent.

Last edited by a moderator: May 4, 2017
19. Apr 20, 2009

### Demystifier

Yes, it is an approximation. However, due to decoherence, this is an extremely good approximation. Essentially, this approximation is as good as the second law of thermodynamics is a good approximation.

Last edited: Apr 20, 2009
20. Apr 20, 2009

### Demystifier

Agree.

Agree.

Agree and disagree. The evolution is described by both, because both predict the same evolution if the initial position is in the support of both. This is the crucial point.

Yes.

I don't understand the question. Can you rephrase it?

Yes.

The details are presented in the book
P. R. Holland, The Quantum Theory of Motion, Sec. 9.5
Unfortunately, it is not available online.