How do entanglement experiments benefit from QFT (over QM)?

[vanhees71]

Situations like these are precisely what induced Heisenberg 1927 to talk about state reduction (aka reduction of the state vector, aka collapse). That you don"t like the commonly used words for it doesn't mean that you don't make use of the same concept.

The important difference of my view to Heisenberg's is Heisenberg's claim that this is something outside of quantum theory. I have not seen a single convincing argument that the understanding of a "beam dump" needs any other laws than the usual quantum theoretical laws about the interaction of particles with other particles forming the beam-dumping material.

 

[atyy]

QM mathematically violates classical probability, thus is not contained in it. It is contained in a classical theory with a certain kind of epistemic restriction such as Bohmian Mechanics at equilibrium. However note that in equilibrium we are violating Kolmogorov's axioms anyway due to how the epsitemic restriction functions. I'll say more about this in a while as it links into Spekkens model. Bohmian Mechanics and other hidden variable theories replicate much of QM by this restriction alone, you only need the nonlocality/retrocausality to violate CHSH or Bell inequalities.

Hmmm, does BM at equilibrium really break the Kolmogorov axioms? Why can't we just put it down to contextuality, which if I understand correctly, just means that if you set up an experiment to measure position, then you cannot also measure momentum.

So in no sense is QM contained in classical probability. Mathematically classical probability is a subset of quantum probability not the other way around. It's like saying curved spacetime is contained in flat spacetime because the latter might turn out to be the correct description of nature.

Hmmm, I do tend to think that curved spacetime is contained in flat spacetime.

I think more so one should say that a truly minimal view is neutral to there being a deeper theory where classical probability theory holds.

Yes, that is what I mean, though I guess I'm not sure what the distinction is between that and saying that classical probability contains QM.

However it would have to acknowledge that as far as we can tell now and operationally in labs preparations do not constitute ensembles. Regarding your previous statement:

Breaking the total law is not consistent with classical probability mathematically. It may be the case that there is a deeper theory which uses classical probability but that is a separate statement. Similarly a Newton-Cartan bundle is not consistent with a Lorentzian metric theory, but the deeper gravitational turned out to involve such.

Generally, my instinctive understanding of why QM does not prepare ensembles in the same sense as classical probability is that for classical probability, a mixed state is a unique combination of pure states, whereas in QM a mixed state is not a unique combination of pure states. Thus I would say that the ensemble in QM is underspecified in terms of what subensembles constitute it, not that the subensembles cannot exist until measurement.

Also note that from contextuality even still in such a deeper theory a preparation does not constitute an ensemble for most observables. Although classical probability is restored we cannot view our preparation as an ensemble for observables like angular momentum, but only the hidden $\lambda$.

Yes, I agree.

 

[DarMM]

Are you saying the protons in the LHC do not have the very well determined momentum? This claim contradicts the very functioning of the entire device!

No I'm not saying that. I'm just saying basic aspects of contextuality and quantum probability.

LHC beams have very well determined momenta for momentum measurements as shown by the tightness of the resulting momentum distribution.

However you cannot consider the beams as being an ensemble of different momenta independent of momentum measurements purely from the preparation.

 

[DarMM]

Hmmm, I do tend to think that curved spacetime is contained in flat spacetime

Well then it's just a different use of the word "contain".

I would have considered curved spaces to be mathematically more general than flat spaces thus not contained in flat spacetime. You're using it to mean "May ultimately be a physical limiting case in some sense of...".

Mathematically the theory of curved spaces is not contained in the the theory of flat spaces, but physically the flat space theory could be correct. It's a separate notion.

Hmmm, does BM at equilibrium really break the Kolmogorov axioms?

Yes, there are restrictions in the lattice of events that you don't have in Kolmogorov's axioms.

Yes, that is what I mean, though I guess I'm not sure what the distinction is between that and saying that classical probability contains QM

It's as I said above.

Mathematically quantum probability is more general. However you are discussing physically how any given mathematical structure may only arise in a specific physical limit of another theory. Our two notions of "contain" were different.

In the case you're talking about we don't find out that classical probability theory contains quantum probability theory, that's impossible as the latter is more general. Rather we find that the correct hidden variable theory contained an epistemic special case isomorphic to a quantum probability theory.

My only problem is that under this definition in some sense any theory is contained by almost anything as it could be wrong and be a limit of something else entirely different.

My statement more considered QM as it is now where we seem to not have a common sample space for our observables from their operational statistics and thus we currently have no grounds to accept a preparation as constituting an ensemble.

Generally, my instinctive understanding of why QM does not prepare ensembles in the same sense as classical probability is that for classical probability, a mixed state is a unique combination of pure states, whereas in QM a mixed state is not a unique combination of pure states. Thus I would say that the ensemble in QM is underspecified in terms of what subensembles constitute it, not that the subensembles cannot exist until measurement.

I wouldn't say this as even for a pure states we lack a common sample space which prevents one thinking of preparations as ensembles.

 

[A. Neumaier]

The important difference of my view to Heisenberg's is Heisenberg's claim that this is something outside of quantum theory. I have not seen a single convincing argument that the understanding of a "beam dump" needs any other laws than the usual quantum theoretical laws about the interaction of particles with other particles forming the beam-dumping material.

Heisenberg didn't think of state reduction as being outside of quantum theory but (like most physicists since him) as being an aspect of it.

Jede Ortsbestimmung reduziert also das Wellenpaket wieder auf seine ursprüngliche Grösse

The state of the system after the observation must be an eigenstate of [the operator corresponding to the observable] $\alpha$, since the result of a measurement of $\alpha$ for this state must be a certainty.

 

[A. Neumaier]

It means that something happens at a distance - namely that nature cooperates globally at long distance to ensure that the perfect nonclassical correlations predicted by quantum mechanics in certain experiments actually happen. But it cannot be controlled hence is a passive happening (a ''passion'') rather than an active one (an ''action''). In spite of (and consistent with) the locally induced interactions!

What do you mean by "nature cooperates globally".

Nature ensures in perfect correlation experiments with entangled photon pairs (the ''certain experiments'') that whenever Alice measures $Ak$ to get $a_k$ then Bob measures $Bk$ and also gets $a_k$, while for Bob it seems that his results are random. In spite of (and consistent with) the locally induced interactions!

 

[vanhees71]

No I'm not saying that. I'm just saying basic aspects of contextuality and quantum probability.

(a) LHC beams have very well determined momenta for momentum measurements as shown by the tightness of the resulting momentum distribution.

(b) However you cannot consider the beams as being an ensemble of different momenta independent of momentum measurements purely from the preparation.

For me (a) and (b) are contradicting each other since for me (a) is what I understand as an ensemble of protons with pretty well defined momenta. It's something I'd expect to be quite well described by a wave function sharply peaked in momentum space (or the appropriate formulation in QFT as the correspondingly smeared creation operator applied to the vacuum state).

How can the state concept of QT make physical sense, if (a) doesn't define an ensemble of protons with pretty well determined momentum?

 

[DarMM]

For me [a] and are contradicting each other since for me [a] is what I understand as an ensemble of protons with pretty well defined momenta. It's something I'd expect to be quite well described by a wave function sharply peaked in momentum space (or the appropriate formulation in QFT as the correspondingly smeared creation operator applied to the vacuum state).

That's just Kochen-Specker contextuality and quantum probability though. It's not my personal view or interpretation.

How do you view the Kochen-Specker theorem then? That might be an easier way to pin point the misunderstanding.

 

[DarMM]

Is the protean nature of ensembles in QM a weakness in the minimalist ensemble interpretation?

My understanding so far: The theory of a given system is the double $(H,\rho)$, the dynamics and the preparation. I.e. All physical content is contained in these terms. The triple $(H,\rho,\sigma)$ describes an ensemble in terms of possible outcomes of a measurement (or possible outcomes of a sequence of measurements), where $\sigma$ is the set of possibilities. The triple $(H,\rho,\sigma')$ describes an ensemble in terms of a different, incompatible set of measurement possibilities $\sigma'$.

Could we say the physical content of the triples $(H,\rho,\sigma)$ and $(H,\rho,\sigma')$ is the same, and the choice of one over the other is merely a choice of appropriate descriptive terms for a measurement context. I.e. A choice of measurement context does not change any physical content of the preparation. It merely constrains the physicist to use a description appropriate for that context.

[edit] - Added some clarification.

Forgot to respond to this. Yes indeed and it's essentially that constraint that prevents one viewing it as an ensemble. Only after such a "choice" does QM give a well defined statistical population.

 

[A. Neumaier]

for me [a] is what I understand as an ensemble of protons with pretty well defined momenta.

But before you had said,

In my example the LHC is a "preparation machine" for (unpolarized) proton beams with a quite well-defined momentum and energy. These beams are prepared such that they collide in specific points along the beam line. For me the preparation procedure delivers a well-defined ensemble of colliding proton beams.

An ensemble of proton beams (prepared moving blops consisting of many protons) is not an ensemble of protons. Protons are never seen in the LHC experiments, prepared are the blops
and measured are the traces of the collision products. In the spirit of the quote by Peres, this is what you have in the labs, not protons.

 

[vanhees71]

I don't see, why my notion of ensembles as the interpretation of quantum states should violate the KS theorem at all. My point is that the state for itself defines an ensemble, since I'm still free to measure whatever I can measure (restricting myself to precise PV measurements, I can always measure any set of compatible observables I like, independent of the state preparation).

E.g., if I prepare a proton beam polarized in $z$ direction, I'm still free to measure any spin component of each proton in this beam I like. Accordingly, given this state I know for any spin component the probabilities for measuring one of the two values $\pm \hbar$. So no matter, which (quantum theoretically sensible) measurement I perform on my "ensemble" I have well-defined probabilities. That's why I think it's too narrow to say the ensemble is not only given by the quantum state alone but only in context of the measurement to be performed on it. I think that's also demonstrated by the correct prediction for the (probabilistic) outcome of "delayed-choice measurements".

 

[vanhees71]

But before you had said,

An ensemble of proton beams (prepared moving blops consisting of many protons) is not an ensemble of protons. Protons are never seen in the LHC experiments, prepared are the blops
and measured are the traces of the collision products. In the spirit of the quote by Peres, this is what you have in the labs, not protons.

This is semantics. As any object also a proton is defined by its properties. In fact in QT objects are much more definitely defined than in classical physics since any proton is completely indistinguishable from any other proton. A beam of protons consists of many protons forming an ensemble. You can of course argue whether a specific bunch in the LHC is an ensemble of independently prepared single protons.

 

[A. Neumaier]

for me [a] is what I understand as an ensemble of protons with pretty well defined momenta.

But before you had said that it is an ensemble of proton beams with pretty well defined momenta,
not an ensemble of protons. Protons are never seen in the experiment, prepared are the bunches.

This is semantics.

Yes, and semantics (meaning) counts in arguments about the meaning of concepts.

A beam of protons consists of many protons forming an ensemble.

A single beam does not, according to your definition:

An ensemble is a collection of independent equally prepared systems.

The protons in a single beam are neither independent nor a collection. They are not even distinguishable, one can point to none of them, only to the time-dependent multiparticle state formed by the whole bunch.

 

[vanhees71]

Well, then how do you explain that the LHC measures outcomes precisely in accordance with the standard model assuming two protons in the initial state. Of course the answer is simply that the bunches are still dilute enough that for FAPP you can assume that only a single pp collision occurs in each interaction of the bunches.

 

[A. Neumaier]

Well, then how do you explain that the LHC measures outcomes precisely in accordance with the standard model assuming two protons in the initial state. Of course the answer is simply that the bunches are still dilute enough that for FAPP you can assume that only a single pp collision occurs in each interaction of the bunches.

I explain it by ''shut up and calculate''. In that mode all inquiries about the precise meaning of the concepts involved is meaningless, as the meaning is left to the discretion of everyone. On this level it is alright to equate ensemble with preparation, as such ''equations'' are only proxies for intuitive reasoning.

But if one starts to inquire into the meaning of the concepts used (as in many of these foundational threads), one finds them problematic and often inconsistent in how they are used.

 

[vanhees71]

Well, one can also overprobemetize the problems.

 

[DarMM]

My point is that the state for itself defines an ensemble

But it simply doesn't as a mathematical fact. An ensemble is an approximate realisation of a sample space. In quantum probability the state alone does not give one a well defined sample space due to Kochen-Specker contextuality. That's all there is to it.

What would help me is understanding what you think the Kochen-Specker theorem implies. Then we could more easily discuss this since currently what you are saying seems in direct contradiction to it.

 

[atyy]

Well then it's just a different use of the word "contain".

I would have considered curved spaces to be mathematically more general than flat spaces thus not contained in flat spacetime. You're using it to mean "May ultimately be a physical limiting case in some sense of...".

Mathematically the theory of curved spaces is not contained in the the theory of flat spaces, but physically the flat space theory could be correct. It's a separate notion.Yes, there are restrictions in the lattice of events that you don't have in Kolmogorov's axioms.It's as I said above.

Mathematically quantum probability is more general. However you are discussing physically how any given mathematical structure may only arise in a specific physical limit of another theory. Our two notions of "contain" were different.

In the case you're talking about we don't find out that classical probability theory contains quantum probability theory, that's impossible as the latter is more general. Rather we find that the correct hidden variable theory contained an epistemic special case isomorphic to a quantum probability theory.

My only problem is that under this definition in some sense any theory is contained by almost anything as it could be wrong and be a limit of something else entirely different.

My statement more considered QM as it is now where we seem to not have a common sample space for our observables from their operational statistics and thus we currently have no grounds to accept a preparation as constituting an ensemble.

I guess it is the difference between saying that determinism is a special case of randomness (which is mathematically true, since one can use the delta measure), or saying that randomness is a special case of determinism (which is not mathematically true, but is physically true in the sense that we can consider statistical mechanics as arising from Newton's laws for many particles and ignorance of the exact state).

I wouldn't say this as even for a pure states we lack a common sample space which prevents one thinking of preparations as ensembles.

I guess, reading your replies to @vanhees71, that this is due to the KS theorem. But I thought the point of the KS theorem is that QM is contextual? What has contextuality got to do with an inability to consider preparations as ensembles?

 

[Elias1960]

QM mathematically violates classical probability, thus is not contained in it. It is contained in a classical theory with a certain kind of epistemic restriction such as Bohmian Mechanics at equilibrium.

This makes no sense. BM is conceptually a completely classical, deterministic theory, and quantum equilibrium states are well-defined probabilistic states defined on the configuration space.

However note that in equilibrium we are violating Kolmogorov's axioms anyway due to how the ep[is]temic restriction functions. I'll say more about this in a while as it links into Spekkens model. Bohmian Mechanics and other hidden variable theories replicate much of QM by this restriction alone, you only need the nonlocality/retrocausality to violate CHSH or Bell inequalities.

All you need is, indeed, a preferred frame. Everything else (logic, probability theory, local configurations) are completely classical. Epistemic restrictions (as far as that means that we are unable to prepare some states but are restricted to prepare only states in quantum equilibrium) do not lead to any violations of Kolmogorovian probability.

By the way, the quantum theory fits into Kolmogorovian probability in a quite trivial way, which is described in Kochen, S., Specker, E.P. (1967). The Problem of Hidden Variables in Quantum Mechanics, J. Math. Mech. 17(1), 59-87 on page 63. They have combined this with some bad words about it, to motivate some additional restrictions (non-contextuality) for "good" hidden variables, which they prove are incompatible with quantum theory.

 

[DarMM]

This makes no sense. BM is conceptually a completely classical, deterministic theory, and quantum equilibrium states are well-defined probabilistic states defined on the configuration space.

All true. I don't know how it affects what I say though.

Quantum theory generalizes classical probability theory. That observation is decades old by now.

By the way, the quantum theory fits into Kolmogorovian probability in a quite trivial way, which is described in Kochen, S., Specker, E.P. (1967). The Problem of Hidden Variables in Quantum Mechanics, J. Math. Mech. 17(1), 59-87 on page 63. They have combined this with some bad words about it, to motivate some additional restrictions (non-contextuality) for "good" hidden variables, which they prove are incompatible with quantum theory

Can you show me where in that paper they show that quantum theory "trivially" fits into Kolmogorovian probability in a way that isn't essentially the sense @atyy and I have mentioned.

 

[DarMM]

Every classical probability theory is in essence the spectral theory of some set of commuting operators $\mathcal{A}$ and some normalized state on them $\rho$.

Since all elements of $\mathcal{A}$ commute and from the spectral theorem we know that any operator can be represented as multiplication on its spectrum, we then are able to represent all of these operators as functions $f$ on some common sample space $\mathcal{M}$ with the state $\rho$ then becoming a probability measure on $\mathcal{M}$. Gelfand's representation theorem thus tells us that all commuting C*-algebra give a probability model and that all probability models have random variables forming a C*-algebra. Thus we have a direct correspondence between Kolmogorov's theory and commutative C*-algebras.

An alternate more general probability theory can be developed then by having the C*-algebra be non-commutative. This is a decades old observation. In the same sense that non-commutative algebras are more general than commutative ones, thus quantum probability is more general than classical probability.

 

[DarMM]

I guess, reading your replies to @vanhees71, that this is due to the KS theorem. But I thought the point of the KS theorem is that QM is contextual? What has contextuality got to do with an inability to consider preparations as ensembles?

Quantum Theory doesn't have a single sample space, that immediately blocks preparations being ensembles. Unless one wants to move beyond the theory as it is that's the end of the story.

However even if you try to have one sample space, such as in a hidden variable theory, the KS theorem tells you it has to include the device as a label on the outcomes. Thus the preparation is for system-device pairs and so the device has to be mentioned.

 

[Mentz114]

But it simply doesn't as a mathematical fact. An ensemble is an approximate realisation of a sample space. In quantum probability the state alone does not give one a well defined sample space due to Kochen-Specker contextuality. That's all there is to it.

What would help me is understanding what you think the Kochen-Specker theorem implies. Then we could more easily discuss this since currently what you are saying seems in direct contradiction to it.

I thought that K&S only applies to unitary projections $|\psi\rangle\langle \psi|$ and thus cannot apply to momentum.

 

[DarMM]

I thought that K&S only applies to unitary projections $|\psi\rangle\langle \psi|$ and thus cannot apply to momentum.

An actual momentum measurement is either a projection or a POVM to which the Kochen-Specker theorem applies.

 

[Mentz114]

An actual momentum measurement is either a projection or a POVM to which the Kochen-Specker theorem applies.

I don't see that how that follows from the first paragraph in

Two simple proofs of the Kochen-Specker theorem
Asher Peres
Phys. A: Math. Gen. 24 (1991) L175-LI78

Am I missing something ?
Do you have another reference.

 

[Auto-Didact]

I guess it is the difference between saying that determinism is a special case of randomness (which is mathematically true, since one can use the delta measure), or saying that randomness is a special case of determinism (which is not mathematically true, but is physically true in the sense that we can consider statistical mechanics as arising from Newton's laws for many particles and ignorance of the exact state).

Scientifically - i.e. from a foundational perspective - only one of the two above is true; w.r.t. curved space in mathematics as an example: the theory of flat geometry is a subset of the more general theory of curved geometry, not the other way around.

The fact that curved space can be approximated by flat space is foundationally completely uninteresting, even if it is practically interesting precisely because you can use approximations. Mathematically this distinction is absolutely clear; in fact, almost every famous physicist of the generation who lived through the QT revolution understood these issues deeply and it is only modern physicists who don't understand it.

This misunderstanding among modern physicists happens because experimental QT went through a golden age, creating a hope or even belief that the theoretical problems would be resolved. Alas, the problems weren't resolved; instead the very context of the problem was altered. This lead modern physicists - after decades of having a more experimental edge versus a theoretical edge - to just tend to disregard and pretend such issues were 'not real, problems' exactly like their teachers told the theorists during the height of the golden age of experimental QT.

This cavalier attitude has become so dominant in the practice of physics that it even permeates almost all (undergraduate) textbooks and curricula, i.e. the bias has become institutionalized: this has lead to most physicists opting in favor of the culmination of experimental QT: the modern instrumentalist philosophy of theoretical science - which they erroneously also believe actually resolves these issues. When it happens that physicists find it normal to say that two opposite mathematical statements are actually saying the same thing then clearly there is a problem; this is hand waving at its worst!

The modern instrumentalist philosophy of theoretical science and its poster-child - i.e. SR-based QFT - does not and cannot resolve the original issues of QT; it automatically fails foundationally because SR already failed foundationally as everyone already knew back then. Instead of admitting this - as the theorists of old immediately did - they evade the question and pretend that there are no serious problems left to address; from a mathematical standpoint, this is clearly unacceptable which is exactly why it isn't accepted neither in mathematics nor in mathematical physics, hence the concerted effort of constructive QFT and so on.

 

[DarMM]

I don't see that how that follows from the first paragraph in

Two simple proofs of the Kochen-Specker theorem
Asher Peres
Phys. A: Math. Gen. 24 (1991) L175-LI78

Am I missing something ?

Why would you think that paper would tell you the type of operator that represents a momentum measurement? It's talking about assigning values to PVMs in general, it's not going to go through POVMs for momentum.

 

[atyy]

Quantum Theory doesn't have a single sample space, that immediately blocks preparations being ensembles. Unless one wants to move beyond the theory as it is that's the end of the story.

However even if you try to have one sample space, such as in a hidden variable theory, the KS theorem tells you it has to include the device as a label on the outcomes. Thus the preparation is for system-device pairs and so the device has to be mentioned.

So does this mean that in BM with equilibrium, since the preparation includes the device, that neither BM nor QM have a single sample space?

 

[Elias1960]

All true. I don't know how it affects what I say though.
Quantum theory generalizes classical probability theory. That observation is decades old by now.

There is no quantum "generalization of classical probability theory" as well as "quantum logic" is not a "generalization of logic", because both "generalizations" have nothing to do with the original, "generalized" thing. Logic, as well as probability theory (which is, following Cox and Jaynes only the "logic of plausible reasoning"), define how to handle propositions which have truth values "true" or "false" and nothing else. If you look at those "generalizations", they are about something different, the "and" and "not" operations are simply not the logical "and" and "not" operations of some statements with truth values in quantum theory, they correspond to such operations only in sloppy language. So, no wonder that they do not fulfill all axioms of classical logic. The appropriate (not misleading) name for "quantum logic" is lattice theory. If there is already an established name for "quantum probability theory" I don't know. But if this construction deserves a name at all (that means, if it is interesting enough for mathematicians to care about it), it would have to be named differently to avoid confusion among physicists.

That such misleading notions cause confusion is obvious. Classical logic (as well as classical probability theory understood as the "logic of plausible reasoning" are not physical laws, but laws of thinking. As laws of thinking, they are superior to physical laws, because they are (and have to be) applied if we evaluate various proposed physical theories or think about the consequences of the outcomes of particular experiments. So, any theory which is in conflict with classical logic has to be rejected based on this fact as logically inconsistent. And the same holds for any theory which is in conflict with the logic of plausible reasoning.

Can you show me where in that paper they show that quantum theory "trivially" fits into Kolmogorovian probability in a way that isn't essentially the sense @atyy and I have mentioned.

It explicitly constructs that "common sample space" the existence of which you have explicitly denied:

I wouldn't say this as even for a pure states we lack a common sample space which prevents one thinking of preparations as ensembles.

In fact, that "common sample space" always exists, based on Stone's representation theorem for Boolean algebras, and the Kochen Specker p. 63 construction is essentially the application of that theorem to quantum theory.

 

[Elias1960]

Quantum Theory doesn't have a single sample space, that immediately blocks preparations being ensembles. Unless one wants to move beyond the theory as it is that's the end of the story.

However even if you try to have one sample space, such as in a hidden variable theory, the KS theorem tells you it has to include the device as a label on the outcomes. Thus the preparation is for system-device pairs and so the device has to be mentioned.

Just to add, because this combination has possibly caused some misunderstanding between us:

I'm a mathematician by education and the first claim I interpret as that such a single sample case cannot exist. Not as what is claimed in the second part, namely you can have it, but in this case, you have to make the sample space a little bit larger.

 

[DarMM]

There is no quantum "generalization of classical probability theory"

There literally is. Debating this would be like us debating if differential geometry exists as a field. See Summer and Redei's classic introductory paper on the topic:
https://arxiv.org/abs/quant-ph/0601158
The rest of your post concerns quantum logic which I didn't mention.

It explicitly constructs that "common sample space"

Where do they do that for quantum theory?

 

[DarMM]

Just to add, because this combination has possibly caused some misunderstanding between us:

I'm a mathematician by education and the first claim I interpret as that such a single sample case cannot exist. Not as what is claimed in the second part, namely you can have it, but in this case, you have to make the sample space a little bit larger.

A bit larger? For no longer are the outcomes labelled by $E$ some POVM element, but by $E, M$ with $M$ any POVM containing $E$. Thus the sample space is infinitely larger.

To some degree what Hidden variable theories are doing is like generalizations of Nash's embedding theorem. There we find that any 4D Lorentzian manifold can be represented as a subspace of a Minkowski space of some dimension 231. This whole line of discussion would be like trying to deny pseudo-Riemannian geometry as a field or that General Relativity has curved spacetimes because Nash's embedding theorem holds.

In fact it is even worse since it would be as if we found the Minkowski space had to be infinite dimensional to embed a generic 4D curved spacetime. I think it is worth thinking about this analogy as it is exactly what hidden variable theories do mathematically. We restore classical physics at the cost of an infinite number of contextual degrees of freedom.

Again my original comments with @vanhees71 were about QM which does not have a single sample space. I think it should be possible to make this statement without having to talk about embeddings in unverified infinite dimensional hidden variable theories. Just like I could say something is not true in General Relativity and verified experimentally without discussing a possible embedding in a 231 dimensional Minkowski background.

 

[Elias1960]

There literally is. Debating this would be like us debating if differential geometry exists as a field. See Summer and Redei's classic introductory paper on the topic:
https://arxiv.org/abs/quant-ph/0601158

So, the abstract already contains appropriate (non-misleading) terms for this, "noncommutative measure theory" and "von Neumann algebras". What I criticize is not that mathematical structures which do not fulfill all axioms of probability theory are studied by those interested in such abstract mathematics, but that it is claimed that quantum theory somehow requires such a generalization of probability theory.

Where do they do that for quantum theory?

The reference, again,
Kochen, S., Specker, E.P. (1967). The Problem of Hidden Variables in Quantum Mechanics, J. Math. Mech. 17(1), 59-87. They do it on page 63, as already mentioned.

 

[vanhees71]

But it simply doesn't as a mathematical fact. An ensemble is an approximate realisation of a sample space. In quantum probability the state alone does not give one a well defined sample space due to Kochen-Specker contextuality. That's all there is to it.

What would help me is understanding what you think the Kochen-Specker theorem implies. Then we could more easily discuss this since currently what you are saying seems in direct contradiction to it.

The Kochen-Specker theorem formally demonstrates that QT is not consistent with the assumption that all observable are determined and makes this statement quantitatively testable by experiment, which confirm it.

I don't see, why a preparation procedure doesn't define an ensemble, because an ensemble does not depend on the assumption that all observables take determined value. It's just a collection of objects, where some properties are determined to a certain degree, enabling one to measure observables (no matter whether they take determined values before the measurement or not), leading to random results which can be statistically analyzed to test the probabilistic predictions of QT. That's why I don't understand the effect of the KS theorem on the notion of "ensemble".

 

[vanhees71]

Quantum Theory doesn't have a single sample space, that immediately blocks preparations being ensembles. Unless one wants to move beyond the theory as it is that's the end of the story.

However even if you try to have one sample space, such as in a hidden variable theory, the KS theorem tells you it has to include the device as a label on the outcomes. Thus the preparation is for system-device pairs and so the device has to be mentioned.

I don't need a single sample space to define an ensemble. Of course you can only prepare ensembles with properties that make sense within QT. If you could experimentally demsonstrate that you are able to prepare an ensemble violating the restrictions of QT (e.g., the Heisenberg uncertainty principle for position and momentum) that would be a disproof of QT, but no such case is known today.

Of course to define the "random experiment" in the sense of probability theory completely you need to completely specify the measurement setup, i.e., whether you measure the spin-x component or the spin-y component in a Stern-Gerlach experiment. Nevertheless, the silver atoms leaving the oven define an ensemble with more or less sufficiently determined properties (distribution of the momenta in beam direction determined by the temperature of the Ag vapor in the oven and more or less defined position in transverse direction determined by the aperture of the slits in front of the oven's opening) due to this specific preparation procedure. Independent from this one can position the magnetic field in any direction one likes, measuring the so defined component of the magnetic moment in this direction.