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Yet another attack of Motl against realist interpretations
Lubos Motl has again attacked realist interpretations in general and de Broglie-Bohm theory in particular.  

A large part is simply namecalling:
Quote:I find it impossible to consider these people intelligent. ... Anti-quantum zealots ... Bohms and similar stinky Bolsheviks ...
And all the people persistently (in 2018) trying to negate the basic rules of the game as articulated in Copenhagen are morons – sadly, in most cases, pompous morons.
Fortunately, there are also some arguments. Let's take a look at them.

The first one uses the uncertainty relations.
Quote:Just try to imagine that you want to explain the general uncertainty principle in a "realist" i.e. fundamentally classical theory. So an anti-quantum zealot will probably admit that the operators are "useful" in some way but they're just a caricature of some "deeper", classical theory. Now you must ask: Can such a hypothetical classical theory have a justification for the uncertainty principle? A reason that implies that if A is accurately measurable in the prepared state, B must be less accurate, and vice versa? And can it get the right bound for any choice of A,B?

In a "realist" theory, if the outcome of the measurement of A and B that you're going to get is knowable in advance, there's just no reason why there should be some unavoidable uncertainty.
First of all, Motl confuses "realist" with "deterministic". Most realist interpretations are stochastic theories, thus, have an inherent uncertainty too.

Then, even in a deterministic theory there can appear probabilities. I would suggest him to study thermodynamics. Quite classical thermodynamics, with deterministic classical Hamiltonian mechanics as the starting point. He could easily learn that even if the fundamental theory is deterministic, what we can construct and measure will be uncertain.

In fact, the situation in dBB theory is quite similar. In thermodynamics, Boltzmann has derived his H-theorem and proven in this way that the states in thermodynamics move toward equlibrium. For dBB theory, Valentini has proven, using essentially the same techniques, his own subquantum H-theorem: A state which is more localised than a state in quantum equilibrium, which evolves following the equations of dBB theory, in quite short time approximates quantum equilibrium. Valentini and Westman have made some numerical computations to see how fast this happens, and it happened sufficiently fast:
Valentini, A., Westman, H. (2005). Dynamical Origin of Quantum Probabilities, Proc.Roy.Soc.Lond. A 461, 253-272, arxiv:quant-ph/0403034

Quote:In that world, graduate students don't have any fundamental obstacle that prevents them from approaching omniscient God increasingly closely
Wrong. In the dBB world, they are able to prepare states only in quantum equilibrium, because whatever they create, it approaches quantum equilibrium fast enough. In fact, the preparation procedure available to those graduate students fix only the wave function, not the configuration of the quantum system.

The next argument is the requirement that realist interpretations should somehow explain the uncertainty relation:
Quote:To sensibly claim that you have an alternative, you would actually have to offer a quantitative scheme that predicts the right lower bound for any choice of A,B. Even if the uncertainty is just an artifact of the apparatuses' imperfection, these apparatuses are still governed by the laws of physics and the laws of physics must have some explanation why their minimum uncertainty always seems to be what the uncertainty principle claims, right?
This is an obvious yet huge task that none of the Bohms and similar stinky Bolsheviks has even attempted to solve.
Because the way how to do this is well-known and simple. Combine the equivalence proof of dBB theory in quantum equilibrium with quantum theory, and then use the derivation of the uncertainty relations from any quantum theory textbook. In fact, Motl knows this, and argues in quite a strange way that this is somehow not good or so:
Quote:I think that all of them know that only the proper apparatus of quantum mechanics – in which the observables really are linear operators, and the calculable predictions really are subjective probabilities of outcomes – can achieve this triumph. The goal of the Bohmian, many-world, and similar theories is just to fake quantum mechanics – to "embed" quantum mechanics in some "realist" framework and claim that it's the better one.
So, Motl as the defender of True quantum mechanics starts to fight against some fake quantum mechanics. The fake quantum mechanics has, surprisingly, the same equations and makes the same experimental predictions, but it is a horrible fake!
Quote:But there's a problem with "faking". The things you're proposing are still "fake". If the comrades try to fake the capitalist economy but they impose all the Bolshevik constraints such as egalitarianism, they still get just the communist economy which totally sucks. The magic of capitalism and its prosperity strictly contradicts the communist axioms such as egalitarianism. You simply can't fake the capitalist economy within communism – and you can't fake quantum mechanics within a "realist" theory.
Your "realist" theory doesn't fundamentally associate the observable quantities with linear operators.
Maybe Motl cannot, but Bohm has done it 1952:
Bohm, D. (1952). A suggested interpretation of the quantum theory in terms of "hidden" variables, Phys.Rev. 85(2), 166-193
Hm, maybe this is not "fundamentally"? He has proven the equivalence only mathematically, and this is not sufficient? I don't know. I'm quite satisfied with a mathematical proof of equivalence, and if I have it, I feel free to apply the whole mathematical apparatus of quantum theory as if it is part of dBB theory too. This is the nice point of mathematics - you don't have to care if something is also fundamental - once you prove that the equations of dBB theory give in quantum equilibrium the equations of quantum theory together with the Born interpretation, everything is fine.

Quote:There's one cute, almost equivalent, way to kill the "realist" theories. And it's the universality of \(\hbar\).
Ups. \(\hbar\) is a universal constant in the equations of dBB theory as well as quantum theory. What could be the problem here?
Quote:This universality of Planck's constant is also totally incompatible with any realist theory simply because realist theories don't have and can't have any universal constant whose units are those of \(\hbar\). There's just no room for such a constant in classical or "realist" physics! The classical Hamiltonian dynamics is fully given by the Hamiltonian H whose units are just joules, but H isn't a universal constant and the scaling of H doesn't affect the evolution equation at all, anyway. All other universal constants in classical physics are various coefficients defining various terms in H etc. and those apply differently to different degrees of freedom – they are not universal. For example, if some students try to determine the energy-to-frequency ratio, \(\hbar\) from \(E=\hbar\omega\), "realist" theories predict that they must get different values of \(\hbar\) from different particle species etc.
Sorry, I don't get the point. Ok, classical Hamiltonian mechanics does not have a constant \(\hbar\). So what? The equations of dBB theory are not equations of classical Hamiltonian mechanics, but of dBB theory. And they have a constant \(\hbar\), in the equations. \(\hbar\) is, as in quantum theory, a constant fixed from the start in the equations of the theory.
Quote:So if many groups of graduate students try to extract a constant with units of \(\hbar\) from their experiment, it's basically guaranteed that each group will have a different answer for \(\hbar\): "realist" theories predict that nonzero quantities with the units of \(\hbar\) simply cannot be universal constants of Nature! This is perfectly falsified by Nature where \(\hbar\) may be extracted from infinitely many different experiments (with particles or fields or strings or branes of any kinds, or any combinations of those) and it always has the same value, despite the high precision of the modern experiments.
Is there anybody who is able to explain me what is the difference, in this question, between dBB theory and quantum theory? In both theories, \(\hbar\) is the constant used in the Schrödinger equation, which is, BTW, used in both theories in the same way.

Quote:"Realist" failure to get quantized measured values

The uncertainty principle is just one famous, and almost defining, consequence of the basic rules of quantum mechanics. But there are many others. Such as the quantized spectrum of the energy. ...Can a "realist" theory actually predict the discrete energy spectrum of atoms?
Of course, as Bohm has shown in 1952. By proving that realist dBB theory gives in quantum equlibrium the same predictions as quantum theory. After this, it remains to apply the proofs used in quantum theory. Just to inform Motl: Physical theories are not bounded by copyright restrictions, if a proof has been used in quantum theory, and a proponent of dBB theory likes the result, he is free to apply the proof in dBB theory too.

Quote:You may embed the mathematics of the wave functions in your "realist" theory. But the interpretation of the wave function will be wrong – the wave function will be misinterpreted as a classical wave – and this misinterpretation has far-reaching consequences.
Wow. I have to admit that this sounds interesting. I'm used to hear that considering other interpretations than the established ones is evil, because interpretations are anyway only metaphysics, without any consequences. Now it appears that this is wrong, and using a bad interpretation has even far-reaching consequences.

Quote:... every observable that you can measure will have a continuous spectrum.

The reason is utterly simple. If you interpret the wave function as a classical wave, your phase space S is a connected, infinite-dimensional continuum. It's as continuous as you can get. If your apparatus ends up measuring the energy of a photon, Eγ, you know that a priori, all positive values of the energy of a general photon must be allowed. If the transformation mapping the initial state to the final state is continuous in any way, it's obvious that you may perturb the desired final value of Eγ, run the evolution backwards, and find an appropriately perturbed initial state that leads to this non-quantized value of the photon's energy.
I look at some source of light through a spectrometer. I see spectral lines. Of course, the spectrometer has a continuous spectrum of possible results. Somehow only a few lines are used in this particular source. There is some accuracy of the device - so, the spectral line I see has some thickness. But it is, for all practical purposes, nonetheless a discrete spectrum. There are large parts of the spectrum it shows nothing.

What does dBB theory predict for this? The measurement process is described in dBB theory as an interaction with a macroscopic measurement device. What is measured is the position of the measurement device. Once the measurement device is big enough, one can easily see that the quantum potential of this device is negligible, so that it follows essentially a classical trajectory, which is easily observable. Now, the wave function of this measurement device will be, nonetheless, nonzero almost everywhere. So Motl is, of course, free to use the quantum method to catch a lion: Put a cage into the desert. The probability that the lion, via quantum tunneling, appears in the cage is non-zero - and quantum and dBB predictions agree about this. Wait.

So, assume Motl's waiting was successful and he has observed a pointer position telling him that the energy was some value which is completely off, in conflict with the quantum prediction. So what do we reach if we follow the the Bohmian trajectory backwards? The first problem is that what has happened was an interaction between the measurement device and the quantum system. What we have observed is,instead, only the position of the measurement device. The final position of the quantum system is unknown. All what we know is the wave function. But let's ignore this, so assume that we know also the position of the quantum system. what does this give? It gives the two initial positions. And we will see that the initial wave function for the same measurement process (the thing which is the same as for quantum theory) also gives these initial positions a nonzero probability.

What was Motl's error here? The whole process leads to a non-quantized value of the position of the measurement device, but not of photon's energy. These are different things.
Quote:Bohm's theory can't be constructed for relativistic particles such as photons (Quantum Electrodynamics) but a Bohmist would surely say that they explain the measured quantized energy because the pilot wave gets reduced to several beams and the real Bohmian particle is in one of them. Great but this sleight-of-hand won't work if you measure other observables that aren't reduced to positions – such as the voltages in our brain which is how we actually perceive things at the end.

The claim that Bohm's theory is unable to handle QED is simply wrong, and the first proposal to handle QED has been given in Bohm's original paper. A relativistic field theory in itself is not a problem at all - for a scalar field, the basic standard field-theoretic formulas are already sufficient. Gauge fields and fermions are more problematic, but the problems have nothing to do with relativity.
The key difference between the only approach known by Motl - the particle positions as the configurations - and the approach of dBB field theory is that in field theory the configuration space is the field configuration.

And this answers also the other objection - the idea that not all measurements can be reduced to measurements of the configuration (which is only in non-relativistic many-particle theory defined by particle positions). The voltages in our brain are classical field configurations of the EM field, thus, described by EM fields \(F_{\mu\nu}(x)\). In a field theoretic version of dBB electrodynamics, the configuration may be defined either by the potentials \(A_{\mu}(x)\) or the fields \(F_{\mu\nu}(x)\), in dependence on how gauge freedom is handled. But in both cases, the fields \(F_{\mu\nu}(x)\) are uniquely defined by the trajectory of the configuration, which is the trajectory of the field configuration. Photons are irrelevant - they have the same status as phonons in condenses matter theory: They are pseudo-particles with no fundamental relevance, and no measurement has to be reduced to a measurement of phonon positions. So, photon positions become similarly irrelevant in dBB field theory.

So far for the arguments of the first part. There is yet another argument, which I will handle separately, because it is based on Bayesian reasoning, and I favor a Bayesian interpretation of the wave function too. So, at least in this part, I'm in some sense on Motl's side. But only in some weak, general sense. Unfortunately, the argument itself is wrong too. But about this later.

To end this posting, some fun provided for free by Motl:
Quote:Now, this is an extremely general empirical fact – i.e. a fact that you may experimentally check in millions of different situations involving thousands of different physical systems, particles of all types, with or without spins, fields, strings, branes, whatever you like.
Emphasis mine, Wink
Let's consider now the "final hit" against realist interpretations:

Quote:If you make a measurement of the observable \(L\) and the wave function collapses to an eigenstate \(\ket\psi\) of that observable, all the parts of the wave function that "existed" (outcomes that were possible) before the measurement completely disappear and they have exactly zero impact on anything after the measurement of \(L\).
The erased parts of the collapsed wave functions are totally eradicated, totally forgotten. You may do anything to your experiment, try ingenious methods to persuade your clever apparatus to "remember" or "recall" the erased parts of the wave function. But your clever apparatuses will not be able to say anything about the number a that defined the state before the collapse
If you think about it for a second, this trivial fact totally contradicts any natural (not fine-tuned) "realist" theory. Take Bohmian mechanics as an example. In that "realist" theory, there's the objective pilot wave, a classical field/wave whose numerical values are chosen to fake the quantum mechanical wave function, and then there are the objective "real" positions of the particles.

Now, the pilot wave guides the "real" particle somewhere, and you may measure the real particle and say something about the spin – Bohmian mechanics doesn't allow the spin directly so the spin measurement has to be reduced to some measurement of the position. So the Bohmian particle is known to be at the place corresponding to "up". However, the pilot wave still exists in the region that would correspond to "down", too.

The point is that this "wrong part of the pilot wave" hasn't been cleaned or forgotten. This pilot wave is coupled to other degrees of freedom in the physical system so in principle, it should be observable. However, experiments clearly say that whatever you do, you just can't observe this "wrong part of the wave function". To avoid the contradiction with the basic empirical facts, the Bohmian mechanics really needs some "janitors" that remove the zombie parts of the pilot wave at places where the particle wasn't seen.

Let's start with the clarification that the argument is not against realist interpretations in general, but only against a particular variant of realist interpretations, namely those who give the wave function an ontological status.  There may be realist interpretations of quantum theory which give the wave function an epistemic interpretation. What makes them realist interpretations is that the wave function described incomplete knowledge about some well-defined reality. So, \(|\psi(q)|^2\) describes the probability that the configuration of the system is \(q\). For realist interpretations which interpret the wave function epistemically, as knowledge about some reality, the objection does not work at all. 

I think there are good arguments in favor of such an epistemic interpretation of the wave function, which is what I use in my <a href="http://ilja-schmelzer.de/quantum/">"minimal realist interpretation"</a>.  But the argument proposed here by Motl is not of this type. Because in the dBB interpretation one has to distinguish two variants of the wave function:  On the one hand, the wave function of the universe - which, according to the dBB interpretation, somehow objectively exists, but which remains unknown to us - and the effective wave function of a quantum subsystem of the whole universe.  And there is a simple formula which connects them:

\[ \psi_{effective}(q_{system},t) = \psi_{universe}(q_{system}, q_{rest-of-the-universe}(t),t)\]

Or similarly for the relevant part of the universe, where the rest of the universe is the macroscopic measurement device, which has a known, visible trajectory \(q_{rest-of-the-universe}(t)\).  And this effective wave function collapses.  

Now, the dBB wave function of the universe does not collapse.  So, it is, indeed, the superposition of a dead cat and a living cat in Schrödinger's experiment. But so what?  What we see is the configuration \(q(t)\), not the wavefunction \(\psi(q,t).\) This trajectory \(q(t)\) describes a particular configuration of the cat, which is either alive or dead, but not in some superposition. 

Once the cat lives, does it matter that the wave function is yet non-zero in imaginable worlds where the cat is dead? No. Once the measurement result is visible in a macroscopic measurement device, decoherence tells us that the other parts of the wave function, which correspond to other, non-observed measurement results, can no longer influence our state.

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