Charles, to prove the equivalence of QM and BM predictions you need, of course, some additional assumptions:  First of all, quantum equilibrium (BM outside the equilibrium is a well-defined theory which gives completely different predictions than QM) and, second, that you use measurement devices such that different results correspond to different positions of some part of the measurement device.  This condition is more or less formal, given that once a classical measurement result is obtained, one can always transform it into one of this form simply by writing down the results (so that ink atoms have different positions for different results).

And, by the way, sentences like ".. neither Bohm nor any of the Bohmians have any understanding of quantum mechanics whatsoever" have a straightforward mirror translation from crank to human language:  You don't have any understanding of quantum mechanics whatsoever.  

Dear Prof. Menon,

QFT cannot resolve any fundamental problem like the measurement problem of QM since it bases itself on QM and all the problems just carry over smoothly to the QFT domain. Unless we have some fundamentally new inputs from outside, QFT is not going to help resolve the problems of QM.

Regards,

Rajat

Dear Prof. Pradhan

However, QFT is a bit more expanded than QM. For example, particle creation and annihilation are not included in QM. Even the full hydrogen spectra (i mean, the Lamb shift, etc.) is not accounted with in QM. The complete theory of electron-electromagnetic  interaction is QED, is it not ?. that is why i was doubting the possibility of resolving the measurement problem with in QFT.

Thank you very much. Menon

Dear Prof. Menon,

I think that QFT is a necessary but not sufficient element of solving the quantum-classical transition puzzle. Since the problem has not yet  been solved one can have guesses only. QFT is an extension of quantum mechanics for processes with changing number of degrees of freedom, as you have mentioned. It is a necessary extension because the quantum-classical transition is a multi-particle process. The irreversible step, separating the quantum and classical domains is similar to a dynamical breakdown of a symmetry, a phase transition. The description of this feature needs QFT. But we are far from solving the puzzle: the irreversible step, decoherence, is but a necessary condition of the transition.

But I believe, and this is a personal impression only, that QFT has the capacity to solve the puzzle in a pragmatic manner. What I mean is that we can, in principle, describe the dynamical narrowing of the quantum uncertainty during the measuring process and thereby reducing the indeterministic nature, the amplitude of quantum fluctuations, to a level which is compatible with the macroscopic observation. This result, accessible within QFT, would suggest that quantum physics is always the correct way of looking on events which sometime, when the environment is sufficiently soft, can be approximated, cf. phase transitions in large but finite systems, by classical physics.

Best regards,

Janos

Dear Dr. Rajat Pradhan!

Equations of QFT are different from equations of QM of particles! In particular superposition principle is not valid in QFT.

Quantum field theory is standard quantum theory, for the case where the configuration is described by a field. The measurement problem, in those interpretations where it appears (in de Broglie-Bohm interpretation where is no such problem) it appears independent of the choice of the configuration space Q.  So, using fields as configurations only adds problems, but does not solve them. 

I would be skeptical of such claims: theories have a region of validity, where they are excellent. Ordinary Schroedinger quantum mechanics is valid when the actions of the motions are of order hbar and th speed low with respect to c. In thiese cases, ordinary quantum mechanics works perfectly at the quantitative level.

Measurement and its problems arise when we cross from the quantum domain to the classical: we magnify the effect of a quantum interaction until it becomes observable at the macroscopic level. So, in a sense, we leave quantum mechanics not through the door of high velocities, but through that of high actions. A prefectly non-relativistic measurement can nevertheless exhibit the several problems of measurement. The issue is rather to find a uniform treatment that includes both a semiclassical system, an environment (necessary, because an important component of measurement is decoherence by the environment) and an approximately classical measuring apparatus. Again, measurement can be entirely non-relativistic, and yet measure. So we should not need QFT. It is a many-particle process, to be sure, and we will need many-body theory. But not QFT in the sense of a relativistic theory, or a theory involving fields.

The formalism of QFT is in principle quite different from that of ordinary Quantum mechanics: the operators one deals with are not the wave function. If we were looking at comparable objects, QFT would in fact satisfy superposition.

Dear All,

It is interesting to hear from so many people on this tricky issue. I agree with those who say that QFT is no better in solving the measurement problem. The measurement problem cannot be solved unless we address the conscious observer, the measurer as part of the description. Superposition only adds more colour to the whole thing. The essential problem is how knowledge arises in the observer about what is observed---the  theory of perception or mind-brain connection. Even measurements in QFT require knowledge of the meausred observable to arise in the observer's mind. They are no different from ordinary classical or QM measurements from this standpoint. That it is difficult to tackle the issue of consciousness within physics is of course a different matter altogether. But that should not deter us from looking at the problem squarely.

The well-accepted classical scientific position which goes by the name "objective realism" is called into question by QM.  Is reality created by/through observation, or does it exist out there independent of observation by observers? This is the crux of the philosophical issue at hand.

If we are not to sort this out then we can continue to argue in favor or against any contrived theory or interpretation to our liking and taste.

Regards,

Rajat

The collapse of the wave function in Bohmian version of quantum mechanics is very clear. Please see: The quantum theory of motion, by : P.R.Holland , chapter 8

Charles, I have read your paper, and while I waited to come to the interesting part, which shows that dBB theory is somehow inconsistent, I ended up reading "conclusion".  The only argument around equation 4.2 is funny - of course, a hidden variable theory contains hidden variables, and, once the phase of psi is unobservable, ok, it is hidden too.  Thus, v is not uniquely defined by observables. Funny. Hidden variable theories cannot be consistent because they contain hidden variables. 

This is the same funny type of argument which rejects dBB theory because it needs a preferred frame, and a preferred frame is not allowed, because it would have to be hidden. 

Charles: I must say I have been sorely confused as well around your equation 4.2. Surely v does not depend on phase, since it depends on the quotient nabla psi/psi. Nor have I found any derivation of a contradiction. You just argue that someone's proof is wrong. But it would be nice to know, more or less precisely, whether the statement `` if, in the initial condition, particles have random distribution |ψ|^2, then (4.2) is such that at a later time they will still have random distribution |ψ|^2, where ψ evolves according to the Schrödinger equation'' is true or false, and if false, why.

Charles, ok, let's look at the difference between hidden and arbitrary, as far as this makes sense. A hidden variable theory contains variables which cannot be directly observed and measured. For a positivist this makes them arbitrary. But in the hiddden variable theory they follow laws - of course, hypothetical laws, but all laws, all physical theories, have hypothetical character. Therefore, their arbitraryness is usually restricted to their initial values.  Instead, for arbitrary variables there would be no law. 

For the phase, only the initial value is unspecified.  Once the initial value is fixed, the Schroedinger equation already defines its further evolution.  Thus, nothing arbitrary, only hidden. 

Then, of course, if a theory contains logical contradictions, one can derive everything. But every theory has the property that there may be people who do not understand the theory, and make errors in their own derivations, can "derive" everything and then claim to have derived a contradiction. 

The funny situation is, in this case, one cannot simply tell him "sorry, but X in your derivation is wrong, because the theory says "not X". The answer is "so what, once the theory is inconsistent, X holds in this theory too".  

In this particular situtation, it is even worse, because I have been even unable to identify the part of your text where a contradiction is derived. Others have, obviously, similar problems.

So, I have to give up. 

The Schroedinger equation defines the evolution of psi(q,t), the BM equation defines the evolution of q(t).  These are different objects in BM.  So, two objects - two evolution equations - no contradiction.

BM does not make any simultaneous assumption of a definite hidden value for p, so there is no problem at all for this.  Of course, p=mv is not a BM formula.  

(The results of momentum measurements are also predefined, but depend on the state of the devices used for the momentum measurement. Thus, the momentum is not a hidden variable of the state itself, but a contextual variable. )

First, no, q is supposed to determine the actual position, not a probability.  And psi(q) is supposed to determine the probability only in a very particular state known as quantum equilibrium. Outside quantum equilibrium there is no connection between psi and probability distributions for q. 

What is necessary to prove is the property that if the state is in quantum equilibrium, it remains in quantum equilibrium forever.  The proof is simple and well-known and has been done already by Bohm. 

And, no, the wave function only defines a probability distribution for results of momentum measurements, and also only in the particular case of quantum equilibrium.

If some texts use p=mv in some context, this is their problem. The dBB theory is innocent for errors of various authors. But, in this case it is probably not an error of the authors, but of one particular reader, SCNR.

Obviously already in the ground state of the harmonic oscillator (or, more general, every stable eigenstate in one-dimensional theory), where the wave function is real, thus, we have v=0 exactly.  But momentum measurements give an uncertainty for the measurement results, thus, not exactly p=0.  So, there is no p=mv, if p is the actual measured momentum and v the actual velocity, as defined by the guiding equation at measurement time. 

If it is a good idea to derive the guiding equation I doubt. Whatever, something additional has to be postulated, or the guiding equation itself, or something else which would allow to derive it.  To use here some p=mv in average or classical limit or some other context where it would make sense would be unreasonable, because it would restrict the dBB theory to theories where at least in the classical limit p=mv holds. But dBB theory is more general. 

Charles: no: you take nabla psi/psi which is *independent of phase as it stands*.. Then you take the imaginary part, and if you like the huyperbolic cosine of the cube root, and it is still independent of phase.

Charles: true, but a space dependent phase is not arbitrary. Indeed, e^{i p x} and 1 differ only by a space-dependent phase, but surely, they are observably different. A global phaase cancels between nabla psi and psi. So the velocity as defined by Bohm is a phase independent quantity, and it is not unphysical: to evry state psi(x) I can assign an unique velocity by (4.2) at every point. Now this velocity is not a ``global velocity'' of the wave packet. Nor does Bohm claim that it should be. The only claim is the (well-known) continuity equation: if you have a cloud of classical particles, the cloud having density |psi(q, t=0')|^2 and all particles move without interacting with a velocity determined by that rule, and if additionally psi(q, t) obeys the time-dependent Schroedinger equation with a Hamiltonian of the form kinetic + potential energy, then the cloud maintains a density |psi(q, t)|^2. That is easy to show. What it means is another issue. But if you do imagine a classical particle surfing on psi(q) via that rule---much as a charged particle is influenced by an electric field and has an acceleration proportional to the field---then if the initial probability of finding this classical particle at q is |psi(q, t=0')|^2 then, under appropriate hypotheses, the probability of finding the particle at q at time t will be given by |psi(q, t)|^2. This essentially follows from the similar result for the cloud.

Also, I believe it is misleading to say that the position of the particle obeying equation (4.2) is a ``hidden'' parameter. It is in fact so far from being hidden, that it is often said that this position is what we really measure.

None of this is to say that Bohmian mechanics has no problems. Locality is surely a big one. But it does have the merit of clarity in several respects. It does not require the peculiar collapse of the wave function. I was looking for a true contradiction, and was not able to see it in your arguments.

As for the Jauch-Piron proof, as well as the von Neumann proof, they claim to lead any theory with hidden parameters that has the same predictions as QM to a contradiction. It would be nice to see this program carried out in the case of Bohmian mechanics, since it does claim to lead to the same predictions as QM.

Charles: my apologies for contradicting. There is a very simple proof. Define rho(q,t) to be |psi(q,t)|^2, j(q,t) (the current, a vector if you wish), as rho(q, t)*v(q, t), where v(q, t) is defined by (4.2). It is then a matter of elementary calculus to show that the continuity equation is obeyed if psi(q, t) satisfies the time-dependent Schroedinger equation: partial_t rho(q, t) + div j=0. Now a cloud of particles obeying the dynamic described above will satisfy a continuity equation. Its particle current, since there are no interactions nor any diffusion, will be purely convective and given by

j(q, t) = rho(q, t)*v(q, t)

as above. The continuity equation we have just seen to hold for the wave function, thus shows that rho(q,t)=|psi(q, t)^2| is a possible solution for the cloud density, and since the process is deterministic, it is the only solution.

I do not believe I have conflated anything so far. We have a field psi(q, t), the field equation of which is the time-dependent Schroedinger equation, and an equation of motion for the particle subjected to the field, namely qdot=v(q, t), where the field is purely external, just as in many common electrodynamics problem: how does a test charge move in an EM field given in such and such a way. As far as interpretation goes, we say that the particle's position is the one we measure, its velocity is inaccessible to us. 

This point deserves to be  emphasized: the only thing we can measre is the position of the particle. The velocity is unobservable, unless we by some other method know the wave function. So the velocity is in fact the hidden parameter. Could we measure it, we would indeed have drastic contradictions with QM, not to speak, conceivably, of faster-than-light signalling. But if we stick to the assuunption that v, while existing and determined by (4.2) is not measurable, I see no problem. 

As to the other proofs: I agree that they are of interest, but, as all mathematical theorems, they have hypotheses as well conclusions. Does Bohmian mechanics satisfy the argunents' hypotheses, and is it thus inconsistent? Bell claims not. It would therefore be nice to see that. So far I have not been able to glean this result from your article.

Once you remove the walls, the wave function changes, and, in correspondence with this change, v changes too, following the guiding equation. 

And, given the choice between you and Bell, I think it is you who should not be taken seriously. Learn BM, find the errors in your papers, and then come back. 

Without being polemical, I am afraid Ilja is correct: you have a cloud of particless uniformly distributed inside the well. All the velocities are zero, because the function is real. Now take the walls away: for the first instant, the particles do not move, since all their veolocities are zero. But also during that first instant the wave function only changes quadratically. So for the first instant in time, things are consistent. That they remain consistent follows from the elementary computation I have sketched above. Indeed, if you follow the time evolution of psi(q, t) after getting rid of the walls, the function rapidly becomes complex near the borders and the particles leave. But I do agree with you fully ``it is a question of mathematics, which depends upon no individual.'' I am sure you can follow the elementary derivation I have sketched above. Saying: I will not look at this, because I am already convinced by other arguments, is not quite OK.

Again: what do you mean by ``particles inside the position of the well will start to move at a rate consistent with quantum theory''? If you mean, the distribution of velocities is not compatible with the momentum disribution, that is true, but, as I have tried to point out, besides the point. If you mean that the motion of a particle cloud will not keep the correct density |psi(q, t)|^2 as time goes on, I think you are on the losing side of a straightforward mathematical argument. 

You say: ``All one can say is that if measurement in qm is described by Hilbert space, then it is not possible to add variables which would describe an underlying determinism in Nature. Since that appears to be what BM claims to do, BM is not possible.''. I do not quite know what you mean by the hypothsesis: "if measurement in QM is described by Hilbert space". I do not believe Bohmian menchanics claims to do this. What it does claim to do is, within a certain set of prescriptions, is to reproduce the predictions of QM---the predictions of what will be observed at what frequency---at he quantitative level, all the while maintaining a picture involving particles with known positions and well-defined, though unmeasurable, velocities. 

In any case, I would like to see some kind of argument as to why Bohmian mechanics is contradictory, specifically applied to Bohmian mechanics. That is, say, if von Neumann's or Jauch-Pirons arguments can be transliterated to show inconsistency of Bohmian Mechanics, that would be of interest. Whether Bell got von Neumann's argument right or not is not the most important issue here.

As to the proofs in Section 3 and 5, I was simply rather confused. The von Neumann proof seems to presume a great amount of Hilbert space formalism. It is true that a theory with dispersion-free states must have a structure quite different from QM. This follows, in its simplest form, from the uncertainty relation: any state from the Hilbert space cannot have simultaneously well-defined position and momentum. But Bohmian mechanics uses a ``state'' altogether different. So I do not see how this proof can be applied to Bohmian mechanics. My insistence on the example of Bohmian mechanics being treated explicitly is precisely because its formulae and the concepts are not so open to metaphysical tergiversation. I am particularly left in the dark with your treatment of quntum logic. How can ordinary language have another structure than, essentially, that of a Boolean lattice? The Hilbert space projector lattice is surely a good description of what happens when you perform elementary measurements on a quantum system, but I do not understand your claim that ordinary language can be understood in this way. 

So I come back to BM, which is concrete. If your arguments are sound, apply them by computation to BM.

As to the technical problem of the cloud density maintaining itself, I really believe you should do the computation as I have outlined it:  work out the continuity equation for |psi(q, t)|^2 and show that it is identical to the hydrodynamic equation of motion of a cloud of particle obeying the guiding equation for the velocities of the particles. But it really *is* apurely technical question, there are no conceptual issues involved. And *this* is the reason why I would like to understand whether the other proofs you give can actually be adapted to this concrete case, where the system's behaviour is not a matter of opinion but of straightforward computation.

Chatles, "... since BM does indeed claim the uncertainty relation while at the same time allowing simultaneous definite, if unknown, values of position and momentum."

No, BM does not make claims about hidden momentum values. Again, forget about p=mv. Momentum measurements are possible, but their result is not defined by mv, but depends even on the position of the pointers of the momentum measurement device. So, whatever quantum momentum measurement measures, it is not mv of the configuration itself. 

Then, please tell us what, in your opinion, is the QM prediction for position measurement at time t if not |psi(x,t)|^2.  

Whatever somebody uses to "find" the guiding equation does not matter, BM contains, together with the Schrödinger equation, the guiding equation as an additional equation, which has to be postulated in one form or another.  Which particular axiom is added, the guiding equation itself or something else which allows to derive the guiding equation, is completely irrelevant, and, therefore, particulat consideration using some p=mv postulate are irrelevant too. 

Of course, names are names, one can name p=mv the "true momentum", big deal.  What remains, and what is essential, is something different:  If we postulate p=mv, then this p is unobservable, and is not measured by the quantum operator P = -ihbar nabla, which is what is usually named "momentum" in QM.  

And, if rho(x,0) = |psi(x,0)|^2 is the initial distribution, then it follows from the guiding equation for the whole ensemble and the Schrödinger equation for psi(x,t) that the distribution rho(x,t) = |psi(x,t)|^2 for all times.  This is a simple mathematical theorem.  I don't know which part of this simple theorem is not clear to you, so sorry if my first guess was wrong. 

A picture, words and some confuse feeling as proof that a mathematical theorem is false. 

And, of course, the v of the guiding equation has nothing to do with an operator equation.  

Charles: As a matter of fact there *is* a theorem, showing that a cloud of particles does indeed maintain the quantum density |psi|^2. And if you know a mistake in this theorem, it would be good for you to say so in formulae, or at the very least with clearer concepts than ``they use results for operator equations when they are working with c-number equations''. Such a simple minded error is surely not present.

A summary of the proof consists in saying that psi(x, t) satisfies a continuity equation of the type partial_t rho(x, t) = -div (rho*v(x, t)), where v(x ,t) is given by the Bohm formula. This follows from the fact that psi(x, t)  satisfies the time dependent Schroedinger equation, which is a basic assumption of Bohmian mechanics. The next step is to see what the particles do. If you define the particle density as D(x, t), then it also satisfies a continuity equation, partial_t D(x, t) = -div(D(x, t) * v_particle(x, t)) which is the same as the equation satisfied by rho(x, t), if the particles velocities v_particle(x, t) are linked to the velocity v(x, t) defined by the field. But this is precisely the Bohmian guiding condition. In all of this there is not the shadow of a confusion between c-numbers and operators.

As for your counterexample, it is simply contradicted by the very figure you publish: near the discontinuity we have immediately very large gradients, which immediately lead to large velocities for the particles led by the psi field. Note in particular that Bohm's equation is *first order*, so a large gradient immediately leads to a large velocity. There is no problem with inertia.

I have two equations

i dot psi = - del^2 psi + V psi

dot Q(t) = Im [ grad psi(Q(t), t)/psi(Q(t), t) ]

The particle velocity as such is therefore given by the gradient of psi divided by psi. It is not any kind of superposition. The crucial result is that, if particles start out with a density, or a probability, given by |psi(x, t=0)|^2, then this density remains equal to |psi(q, t)|^2 for all times, if psi(x, t) satisfies the first equation above, and the position Q(t) of the particle the second.

From the first equation one deduces a continuity equation for |psi(x ,t)|^2, from the facttha I have a cloud of classical particles moving at a preassigned volcity, one obtains another continuity equation, this time for the particle density. Due to the second equation, the two equations coincide, thus leading to the theorem.

Which equations here are ``alike, yet describing completely different mathematical procedures''? Rather, it seems to me that I have equations which are mathematically identical, and which thus have the same solutions. 

I do not believe a numerical experiment can show much as opposed to a proof. However, in your figure, I see something quite different: I do see large gradients in the cloud. As a test, I propose the following: do not show so much of the outside region (we can all agree that there is fairly quick, though not instantaneous, radiation). So show what happens inside the cloud, with perhaps just a bit of what is outside, and go to times as short as possible. I think you will see that even at very short times the velocities are amply large enough for the particles to leave the center of the cloud. In fact, I believe the velocities will oscillate wildly and have an amplitude that diverges as 1/sqrt(t) as t->0. Check it out.

You say ``Clearly the Schrödinger equation does not describe a pack of particles, each with different velocities, but it is a wave equation''. True, and I know what the Schroedinger equation is supposed to mean as well as you do. But you can also give another reading to the Schroedinger equation. You may think of it as some kind of field, which guides a particle, the initial position of which is random, with a probability density given by |psi(x, t=0)|^2. The point is, that however weird this may sound, it is actually consistent. If you start the particle in this manner, the probabilities of observing the particle at position x at time t will be given by QM. I have shown you a mathematical proof of this, so an answer should not be at the qualitative level, but rather in mathematical arguments.

We do not seem to see the same things in the same picture. I see fairly large gradients in the cloud (not comparing them to outside the cloud, I am just asking, are these gradients, and hence velocities, large enough to get some particles that originally were in the cloud, to outside the cloud. Once the particles get outside the cloud, they are subject, as you correctly point out, to huge gradients). Finally, I would argue that  the gradients inside the cloud at t_1 are larger  than at 2t_1, vindicating my assumption that the velocities inside the cloud actually diverge at t=0.

I do not understand your objection. The continuity equation I use in qm does not involve operators, nor superposition, nor anything of the kind. It is a conservation law for the norm squared of the psi function, and the current, which is what enters in the divergence, is not an operator, but simply psi^* grad psi - psi (grad psi)^*. This is a simple *local* calculus identity and does not involve operators, superposition or the like. it merely states how much ``probability density'' escapes from an infinitesimal volume in terms of a local current.

That the standard intepretation of qm gives to psi a meaning wholly different from that of a classical field is clear to me. I also believe this usual interpretation is rather more satisfactory than any other. But an interpretation like Bohm's may fail to satisfy my esthetic sense, as indeed it does, without being actually contradictory.

I shall try to give an explicit solution of your example. It will take some amount of work, though, and more formulae than I can possibly cram in a post.

To your other remark: what can possibly be ambiguous about saying that I have a quantity psi(x,t) satisfying a PDE (which happpens to be the time-dependent Schroedinger equation) and noting that the intensity of the wave |psi(x, t)|^2 satisfies a continuity equation with a  local current psi^*(x, t) grad psi(x, t) - psi(x, t) (grad psi(x, t))^*? This continuity equation follows trivially from the Schroedinger equation. Whether you want to call it quantum mechanical or not is a problem in nomenclature. I do not insist that it be called quantum mechanics. But we are talking of the equations which are solved when we solve problems in quantum mechanics. These equations can be solved and understood with a different interpretation than the one you have in mind, in terms of observables, Hilbert space etc... I am not in any way saying your interpretation has any problems. I am saying that viewing the TDSE as a field equation, with the field acting via the guiding equation on particles, while metaphysically profoundly different, is mathematically identical. Of course, attempting to cast Bohmian mechanics as part of Hilbert space quantum mechanics (``proper ket and operator notations'') is bound to fail, since you are imposing on a given theory an interpretative framework wholly inadequate to it. The particle position, and velocity, do not fall in these categories. The framework in which we should think of Bohmian mechanics is rather that of electrodynamics: in ED, there is a field, and it pushes particles around. Indeed, the approximation of an external field with no reaction from the charged particle is often made. So here we have a psi field, and particles pushed by it. If the particle initially has a random position with a distribution given by the psi field's intensity, then this state of affairs persists for ever. That much follows clearly from elementary computations.

I give up. But since you claim to have a mathematical difference, show it! In formulae, not in metaphysical words. A current is a current, it does not matter---to the mathematics---whether you choose to call it quantum or classical. The Schroedinger equation obeys a continuity equation in classical c-number terms. I hve shown this to you. Do you find an error in the proof or do you agree that there is a c-number continuity equation? I do claim that the answers of Bohmian mechanics are the same as QM, as far as the distribution of position of the particle (for all times) is concerned. In fact, I do not claim originality: this is just a rephrasing of things Bohm and many others have said. The continuity equation can be found at the beginning of Landau's Quantum mechanics. The proof is general, therefore your counterexammle must be wrong, though calculating it in detail will be hard

Dear All

Thanks for bringing many different points. 

regards, Menon

Is equation (1) i¡of the manuscript I sent you what you call a c-number equation? My derivation of (1) fits in 4 lines. Which of these lines is wrong? Saying I treat a quantum current as a c-number is a verbal statement. i have shown you formulae whereby I derive a mathematical result. Show me the error in the formulae and I shall stop arguing

So we are back to notations. However, I would insist that everything be considered as functions of q and t. A solution of the Schroedinger equatioon can always legitimately be so considered. Then there is no ambiguity: I am always talking about pointwise identities, not L^2. You may argue that this is not quantum mechanics. I agree, but we would then again be talking about semantics. It should not be a surprise that hidden variable theories are not quantum mechanics.

BM does not have the same mathematical structure as QM: this much I have never disputed. The result I have shown you proves that, in a specific sense involving distribution of positions, BM does reproduce the predictions of QM. The fact that the two theories have different mathematics does not mean that in a well defined sense their perdictions cannot be equivalent, as indeed they are. 

Charles, to prove the equivalence of QM and BM predictions you need, of course, some additional assumptions:  First of all, quantum equilibrium (BM outside the equilibrium is a well-defined theory which gives completely different predictions than QM) and, second, that you use measurement devices such that different results correspond to different positions of some part of the measurement device.  This condition is more or less formal, given that once a classical measurement result is obtained, one can always transform it into one of this form simply by writing down the results (so that ink atoms have different positions for different results).

And, by the way, sentences like ".. neither Bohm nor any of the Bohmians have any understanding of quantum mechanics whatsoever" have a straightforward mirror translation from crank to human language:  You don't have any understanding of quantum mechanics whatsoever.  

Your "proof", which simply restates von Neumann without showing the slightest understanding of the counterarguments, as evidenced by "Bell who denies that = +, although this is a trivial consequence of linearity", proves that indeed you have no understanding nor of quantum theory, nor of the counterarguments proposed against von Neumann's proof. 

Of course, Bell does not "deny  = +"  (which is nonsensical) but denies that  = + requires linearity.  The measurement described by the operator A+B may have no relation at all to the results of measurements described in the quantum formalism by the operators A and B.  So, for a particular hidden variable z (A+B) for z may be different from A for z + B for z.  

So, please learn about contextuality and come back.  

 ".. neither Bohm nor any of the Bohmians have any understanding of quantum mechanics whatsoever"

What about Yakir Ahranov? I think with these sentences we are overcoming the limits..

http://advances.sciencemag.org/content/2/2/e1501466.full

Dear colleagues!

By the way I revised my paper and now it is published in the journal:

Melkikh A.V. Nonlinearity of quantum mechanics and the solution of the problem of wave function collapse. Communications in Theoretical Physics. 2015. V.64, No. 1, 47-53.

You can find it on my page.

Francis, the thesis that the linearity of some average values requires linearity of all measurement results predicted by a hidden variable theory is simply wrong.  It would be, of course, the first try once one tries to develop such theories, but not more.

Then, the point is not the repetition of what von Neumann acknowledged, but that this destroys the only hope to prove that indeed there has to be a linear relation for all particular measurement results.  

And contextuality is the key for understanding your error. There is no such animal as "the hidden variable" in a contextual world.  In such a world, the particular measurement result will depend on hidden variables of the system as well as on hidden variables of the measurement device.  So, there is no single hidden variable which defines all measurement results of all measurements together.  

Charles, you seem to create some strawman named "classical determinism". Of course, dBB theory is a classical deterministic theory, and a counterexample to your impossibility claim.  It is well-known that it is not in conflict with von Neumann's theorem, given that it is a contextual theory.  

Your argument fails because a deterministic hidden variable theory is not obliged to predict results of observations which are not made.  If we have several different, incompatible measurements, say, A, B, C=A+B (the last in the quantum operator formalism) and only one of the measurements can be made, you can combine the hidden variable of the system with that of the measurement device actually used, but not with all of them, including those not actually used.  So your construction of a single hidden variable fails in a contextual theory like dBB. 

QFT is just Quantum Mechanics applied to fields, it doesn't incorporate more the corpuscle aspect than a single particle in a harmonic potential.  Though, in the case of QFT, the collapse to a well defined value of the field occurs locally.  But when for example a photon is absorbed, it is a mode that is spread in the whole space. So I think what is gained on one side is lost on the other side.  QFT doesn't solve the information anomaly and leads as well to Bell type correlations.

As for the consensus, it is quantum logics, and shut up and calculate.

Charles, nobody claims that the mathematics of von Neumann's theorem is wrong.  The point is that one of the assumptions of the theorem does not hold for dBB theory.  Which you should know if you have read Bell.  

Nobody has said that von Neumann was silly or foolish.  What some people may have said is that to claim that the assumption of linearity has to hold in any reasonable hidden variable theory may be foolish.  

Last but not least, I do not align. If you find some cranks and align them with me, this is your personal ad hominem attack, which discredits you not me. 

Claude, QFT simply prefers the Heisenberg formalism because it hides the global character of the collapse.  But it is present there as well, and is as global as in the Schrödinger picture, simply because the picture does not change the theory, only the presentation.  

Charles, again, every quantum measurement can be realized, equivalently, by a measurement where at least some parts of the measurement device (needles, ink of printers) have different positions for different measurement results. The quantum prediction does not depend on this choice of the basis for the measurement device, so we can use this particular choice to prove the equivalence with dBB theory.  Which representation we use, instead, for the measured system is also our free choice, the quantum prediction does not depend on this choice, thus, we can use configuration space representation as well for the system itself.  

Which operator is measured by the measurement process is a completely different question, it does not depend on the basis one uses to compute the Schrödinger evolution for the whole system, but depends on the interaction Hamiltonian between the system and the measurement device.  If the operator X should be measured, the simplest model is an interaction Hamiltonian \(H_{int}=\partial_{needle} X\) which, for eigenstates of X, shifts the needle with a velocity depending on the eigenvalue.  

Then, what defines the result is the pair \((X_{device}, X_{system})\).  This pair has, in any actual measurement, some initial values and the guiding equation defines how they involve.  But it does not tell us anything about other measurements, with other measurement devices, which may have completely different configuration spaces, and, moreover, have a completely different interaction Hamiltonian (which defines what is measured), thus, will even with equal starting values have completely different results.  And the initial values of measurement devices of measurements which do not happen do not have to exist, even in a deterministic theory.