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What does the probabilistic interpretation of QM claim?

 Quote by unusualname is deterministic chaotic dynamics the fundamental mathematical description of reality in your model?
The fundamental mathematical description of reality is standard quantum field theory, _not_ deterministic chaos. The latter is an emergent feature.

In my thermal interpretation of quantum physics, the directly observable (and hence obviously ''real'') features of a macroscopic system are the expectation values of the most important fields Phi(x,t) at position x and time t, as they are described by statistical thermodynamics. If it were not so, thermodynamics would not provide the good macroscopic description it does.

However, the expectation values have only a limited accuracy; as discovered by Heisenberg, quantum mechanics predicts its own uncertainty. This means that <Phi(x)> is objectively real only to an accuracy of order 1/sqrt(V) where V is the volume occupied by the mesoscopic cell containing x, assumed to be homogeneous and in local equilibrium. This is the standard assumption for deriving from first principles hydrodynamical equations and the like. It means that the interpretation of a field gets more fuzzy as one decreases the size of the coarse graining - until at some point the local equilibrium hypothesis is no longer valid.

This defines the surface ontology of the thermal interpretation. There is also a deeper ontology concerning the reality of inferred entities - the thermal interpretation declares as real but not directly observable any expectation <A(x,t)> of operators with a space-time dependence that satisfy Poincare invariance and causal commutation relations.
These are distributions that produce exact numbers when integrated over sufficiently smooth localized test functions.

Approximating a multiparticle system in a semiclassical way (mean field theory or a little beyond) gives an approximate deterministic system governing the dynamics of these expectations. This system is highly chaotic at high resolution. This chaoticity seems enough to enforce the probabilistic nature of the measurement apparatus. Neither an underlying exact deterministic dynamics nor an explicit dynamical collapse needs to be postulated.
 Sorry, but chaotic dynamics is an exact mathematical model, that's the whole point of it, you can't say it's "emergent". Sensitive dependence at infinitesimally small changes in the the dynamical parameters is part of the definition of chaotic dynamics. If you have a stochastic dynamics then you have stochastic dynamics, if you have deterministic dynamics then you have deterministic dynamics, there's no inbetween "emergent" type system.

Recognitions:
 Quote by unusualname Sorry, but chaotic dynamics is an exact mathematical model, that's the whole point of it, you can't say it's "emergent". Sensitive dependence at infinitesimally small changes in the the dynamical parameters is part of the definition of chaotic dynamics. If you have a stochastic dynamics then you have stochastic dynamics, if you have deterministic dynamics then you have deterministic dynamics, there's no inbetween "emergent" type system.
The world is not as black and white as you paint it!

The same system can be studied at different levels of resolution. When we model a dynamical system classically at high enough resolution, it must be modeled stochastically since the quantum uncertainties must be taken into account. But at a lower resolution, one can often neglect the stochastic part and the system becomes deterministic. If it were not so, we could not use any deterministic model at all in physics but we often do, with excellent success.

This also holds when the resulting deterministic system is chaotic. Indeed, all deterministic chaotic systems studied in practice are approximate only, because of quantum mechanics. If it were not so, we could not use any chaotic model at all in physics but we often do, with excellent success.

 Quote by A. Neumaier The world is not as black and white as you paint it! The same system can be studied at different levels of resolution. When we model a dynamical system classically at high enough resolution, it must be modeled stochastically since the quantum uncertainties must be taken into account. But at a lower resolution, one can often neglect the stochastic part and the system becomes deterministic. If it were not so, we could not use any deterministic model at all in physics but we often do, with excellent success. This also holds when the resulting deterministic system is chaotic. Indeed, all deterministic chaotic systems studied in practice are approximate only, because of quantum mechanics. If it were not so, we could not use any chaotic model at all in physics but we often do, with excellent success.
You either have deterministic laws at the fundamental level or you don't, why don't you just say you believe the universe is deterministic at the fundamental level, then I would understand you.

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 Quote by unusualname You either have deterministic laws at the fundamental level or you don't, why don't you just say you believe the universe is deterministic at the fundamental level, then I would understand you.
On the fundamental level, we have textbook quantum field theory. It doesn't matter for my interpretation whether or not there is an even deeper underlying deterministic level. So there is no need to commit myself.

 Quote by A. Neumaier On the fundamental level, we have textbook quantum field theory. It doesn't matter for my interpretation whether or not there is an even deeper underlying deterministic level. So there is no need to commit myself.
Ok, then if you don't mind I'll answer the thread title, the probabilistic interpretation of QM claims nature is fundamentally probabilistic, and this claim has stood the test of time since the late 1920s, ok?

 Quote by strangerep In that case, what is wrong with Mott's or Schiff's analyses (which apply for incident field carrying charge)? To me these seem adequate to account for the experimental observations.
I don't have access to Mott's and Schiff's writings. My only point was that it is unreasonable to represent 1 (one) electron by a continuous charge density wave. When we look at the electron experimentally, we often find it well-localized, i.e., within the space of one atom. And I find it rather difficult to imagine how a spread-out charge wave can condense to the atomic-size volume all by itself.

Eugene.

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 Quote by unusualname Ok, then if you don't mind I'll answer the thread title, the probabilistic interpretation of QM claims nature is fundamentally probabilistic, and this claim has stood the test of time since the late 1920s, ok?
If this were the only thing the probabilistic interpretation of QM claims, there were no point in doing QM, and there were no point for this thread.

By the way, the url in your profile is spelled incorrectly.

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 Quote by meopemuk I don't have access to Mott's and Schiff's writings.
 Quote by meopemuk My only point was that it is unreasonable to represent 1 (one) electron by a continuous charge density wave. When we look at the electron experimentally, we often find it well-localized, i.e., within the space of one atom. And I find it rather difficult to imagine how a spread-out charge wave can condense to the atomic-size volume all by itself.
You are confusing assumptions and knowledge.

We never ''look at an electron experimentally'' - we only infer its presence from a measured current or ionization track. Mott shows that this track is produced by a classical spherical wave impinging on the cloud chamber from a certain direction, which will determine the direction of the track produced at the atom that happens to fire. There is nothing counterintuitive about that. The uncertainty in the charge density inside the detector is much larger than the charge of one electron.

You _assume_ instead that this is caused by a single electron. And then you say that you find it because of the track. This is a simple instance of a self-fulfilling prophecy. http://en.wikipedia.org/wiki/Self-fulfilling_prophecy

 Quote by A. Neumaier By the way, the url in your profile is spelled incorrectly.
thanks! probably just as well since I haven't constructed M yet (need linear groups and propagation of states throughout the universe ;) )

 Quote by A. Neumaier You are confusing assumptions and knowledge. We never ''look at an electron experimentally'' - we only infer its presence from a measured current or ionization track. Mott shows that this track is produced by a classical spherical wave impinging on the cloud chamber from a certain direction, which will determine the direction of the track produced at the atom that happens to fire. There is nothing counterintuitive about that. The uncertainty in the charge density inside the detector is much larger than the charge of one electron. You _assume_ instead that this is caused by a single electron. And then you say that you find it because of the track. This is a simple instance of a self-fulfilling prophecy. http://en.wikipedia.org/wiki/Self-fulfilling_prophecy
Instead of a cloud chamber bombarded by a dense electron flow I would like to think about a cleaner setup in which definitely one and only one electron was emitted and then captured by an array of tiny detectors, such as CCD device. After the measurement we can be sure that only one detector in the array has "clicked" and the entire electron charge has been deposited inside this detector. If I apply your "charge density field" theory to this situation, I'll get a surprising conclusion that the extended charge distribution has collapsed to the volume of one micrometer-size detector all by itself and against the resistance of the Coulomb repulsion. I find this rather amazing.

Eugene.

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 Quote by meopemuk Instead of a cloud chamber bombarded by a dense electron flow I would like to think about a cleaner setup in which definitely one and only one electron was emitted and then captured by an array of tiny detectors, such as CCD device. [...]
You still haven't given a reference to an actual experimental setup that does this.
I.e., emits definitely one and only one electron, presumably with a momentum uncertainty
corresponding to a small solid angle that exactly encompasses the CCD device.

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 Quote by A. Neumaier [...] there are no no-go theorems against deterministic field theories underlying quantum mechanics. Indeed, local field theories have no difficulties violating Bell-type inequalities. See http://www.mat.univie.ac.at/~neum/ms/lightslides.pdf , starting with slide 46.
On slide 49, you begin a hidden-variable analysis of a particular
experiment with the following assumptions:

 (i) The source of beam 1 produces an ensemble of photons which is in the classical (but submicroscopic) state $$\lambda$$ with probability density $$p(\lambda)$$. (ii) Whether a photon created at the source in state $$\lambda$$ reaches the detector after passing the kth filter depends only on $$B_k$$ and $$\lambda$$. (This is reasonable since one can make a beam completely dark, in which case it carries no photons.) (iii) The conditional probability of detecting a photon which is in state $$\lambda$$ and passes through filter k when $$B_k = B$$ and $$B_{3 − k} = 0$$ is given by a functional expression $$p_k(B,\lambda)$$.
Later (after a QM analysis, etc), you say on slide 57:

 The experiment can be explained by the classical Maxwell equations, upon interpreting the photon number detection rate as proportional to the beam intensity. This is a classical description, not by classical particles (photons) but by classical waves.
I didn't see where you "explained the experiment by the classical
Maxwell equations" in these slides. (Or are you implicitly referring
to the arguments given in Mandel & Wolf?)

You go on to say:

 Thus a classical wave model for quantum mechanics is not ruled out by experiments demonstrating the violation of the traditional hidden variable assumptions. Therefore the traditional hidden variable assumption only amount to a hidden point particle assumption. And the experiments demonstrating [Bell inequality] violation only disproves classical models with point particle structure.
It's not clear to me where, in the hidden variable assumptions you
listed, one has assumed point particle structure.

 Quote by strangerep emits definitely one and only one electron,
I am not an experimentalist, so I can be mistaken. But I think that it should be possible to arrange emission of exactly one electron in a controlled fashion. For example, one can use a single radioactive nucleus, which experiences beta-decay.

You can say that the emitted electron flies in a random direction, so, most likely, it will not be found in our measuring device. But if we are persistent and prepare another radioactive nucleus, then another one... Eventually, we will be able to catch the electron and perfrom the experiment.

 Quote by strangerep presumably with a momentum uncertainty corresponding to a small solid angle that exactly encompasses the CCD device.

If this electron passes through a crystal, then the effect of "electron diffraction" occurs, which is basically similar to the occurence of interference picture in the famous double slit
experiment. There are probability peaks in certain directions of electron propagation. The preferential angles depend on the (1) initial electron's momentum, (2) type of the crystal lattice, (3) orientation of the crystal. So, it should not be difficult to arrange all components in such a way that the diffraction (or interference) picture covers the entire surface of the CCD device.

Eugene.

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 Quote by meopemuk [...] I think that it should be possible to arrange emission of exactly one electron in a controlled fashion. For example, one can use a single radioactive nucleus, which experiences beta-decay. You can say that the emitted electron flies in a random direction, so, most likely, it will not be found in our measuring device. But if we are persistent and prepare another radioactive nucleus, then another one... Eventually, we will be able to catch the electron and perform the experiment.
This is almost the "preparation" arrangement assumed in Mott's analysis (except
he uses alpha particles instead of electrons).

But you have not specified a way to observe the electron on its way from nucleus
to target (and I'm not sure what you meant by "catch the electron").

Instead, you're relying on random emission by nuclear decay. I don't see how this
is practical except by having a sample of the radioactive material with many nuclei,
and this leads to an ensemble of emitted electrons, with nonzero probability of
more than 1 electron in any given time interval.

 Quote by strangerep But you have not specified a way to observe the electron on its way from nucleus to target (and I'm not sure what you meant by "catch the electron").
There is no need to observe the electron on its way from nucleus to target. "Catching the electron" means registration of the hit by one detector in the CCD array.

 Quote by strangerep Instead, you're relying on random emission by nuclear decay. I don't see how this is practical except by having a sample of the radioactive material with many nuclei, and this leads to an ensemble of emitted electrons, with nonzero probability of more than 1 electron in any given time interval.
This is not easy, but in principle possible. One can make sure that the radioactive sample contains one and only one nucleus of the desired unstable type. As an exotic possiblity I can suggest using C60 buckyballs. Currently there are techniques allowing to place one foreign (e.g., radioactive) atom inside the buckyball sphere. Then, I guess, it might be possible to deposit just one such "stuffed" buckyball on the surface of the sample. This would arrange a "single electron" emitter.

Eugene.

EDIT: I've googled for "single electron source" and found a number of interesting references. So, I guess that preparation of one-electron states is a solved technical problem. See, for example,

J.-Y. Chesnel, A. Hajaji, R. O. Barrachina, and F. Frémont, Young-Type Experiment Using a Single-Electron Source and an Independent Atomic-Size Two-Center Interferometer. Phys. Rev. Lett. 98, 100403 (2007). http://prl.aps.org/abstract/PRL/v98/i10/e100403

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 Quote by meopemuk I've googled for "single electron source" and found a number of interesting references. So, I guess that preparation of one-electron states is a solved technical problem.
Although such papers are indeed interesting I came to the opposite
conclusion about it being a "solved technical problem" in the way you seem
to mean. I think such a description is an over-claim indeed.

The various setups I saw appear quite elaborate, specific to particular
applications that don't correspond easily to what you wanted (imho).

 See, for example, J.-Y. Chesnel, A. Hajaji, R. O. Barrachina, and F. Frémont, "Young-Type Experiment Using a Single-Electron Source and an Independent Atomic-Size Two-Center Interferometer." Phys. Rev. Lett. 98, 100403 (2007). http://prl.aps.org/abstract/PRL/v98/i10/e100403
Umm,... did you actually read this paper?
As usual, the devil is in the detail...

The experiment consists of an incident beam of alpha particles,
striking a gas of $$H_2$$ molecules. There's a particle reaction
chain in which the alpha particle becomes a doubly-excited helium atom
by capturing two electrons from a hydrogen molecule. The two resultant
protons move apart a little, and the doubly-excited helium atom decays,
re-emitting the electrons. Sometimes, one of the electrons is
re-emitted back towards the 2-proton target which acts like a 2-centre
scatterer. The resultant scattering pattern of such back-emitted
electrons is recorded. The "result" of the experiment is thus a
scattering cross section.

The experiment is called "single-electron" only because the probability
is extremely low that more than one electron is scattered by a given
2-proton scatterer. I.e., it's "single-electron" within the lifetime of
the 2-proton scatterer.

On the 2nd page, the authors clarify further that:

 Quote by Chesnel et al Since these individual scattering processes are repeated with similar initial conditions many times, what is actually measured here is the ensemble probability of the diffraction of just one single electron by one single two-center scatterer.
The results seem adequately accounted for by statistical field-theoretic analysis.