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

[A. Neumaier]

Of course, from a fundamental-physics point of view at the end you can say that all measurements are a determination of some parameter in a Hamiltonian.

No. Measuments, unlike the Hamiltonian, depend on the state of the system measured.

 

[vanhees71]

Measurements are independent of the state. You can prepare a system in any state you like (and are technically able to of course) and independent from this state preparation you can measure any observable of this system you can properly define by a measurement procedure.

To measure an observable there must be some interaction between the measurement device and the measured system, which is described by a corresponding interaction Hamiltonian. In this sense you always measure some parameter in the Hamiltonian.

E.g., in the SGE you use a magnet. The interaction Hamiltonian
$$H_{\text{int}}=-\frac{g q}{2m} \vec{s} \cdot \vec{B}$$
leads to the entanglement of the silver atom's spin with the magnetic moment in direction of $\vec{B}$ (of course with $\vec{B}$ chosen appropriately). The relation between the position and this to be measured quantity is quantitatively given by the parameters in the Hamiltonian. If you consider it as measurement of $g$ it's immediately clear that you measure "a parameter in the Hamiltonian". The choice of the magnetic field and the corresponding setup in the lab is independent of the state you prepare the silver atoms into be measured. Of course, you also have to choose the preparation procedure such that the interaction with the measurement device allows you to finally read off the to-be measured observable (in this case the gyrofactor $g$).

 

[Tendex]

I'm not talking about infinite spatial dimensions. You don't even need to make an infinity of observations. I'm talking about a mathematical fact of the theory. Show me a single sample space that doesn't need to postulate an infinite number of additional degrees of freedom.

Indeed it is a space with infinite dof, so? It is a mathematical space, it is you who is claiming that mathematically cannot be constructed, but not every space in a physical theory is actually empirically realized or "seen" in the mathematical theories that support a physical theory and less so in QFT, there is something called mathematical abstraction used here and since you are the one making the strong statement about impossibility of its construction in mathematical quantum theory you must show why a sample space of all conceivable macroscopic random quantum outcomes not locally constrained by choosing a local observable is forbidden in quantum theory irrespective of the fact we cannot "see" them(we are just allowed to "see" by definition the subsets determined by specific local observables, one can't help measurements are local). So if not all of the elements of a mathematical physics theory must be empirically realized why do you require it for this single sample space? It seems to me you are equating this space with the one in classical theory where no generalization to encompass classical and quantum probabilities has been made but one important difference is in the former one allows uncountable infinite dimensions.

The resulting sample space has observables far more general than those in quantum theory. Not even a finite subset of these have been seen.
You do realize that such an infinite dimensional sample space contains several observables that don't correspond to anything we've ever seen right?

Observables but we are talking about outcomes not restricted by specific observables. Again, lots of sets used in the mathematical construction of physical theories cannot be realized in the way you demand, more so in this case in which measurement outcomes are local and have to be incorporated mathematically as local properties(when true in small open domains means they are true globally without having to make an infinity of trials or observe directly their infinite subsets), are you maybe conflating outcomes and observables?
It is certainly not enough to use a confusing far fetched analogy with differential geometry(since again spacetimes in GR are usually meant to be 4 dimensional, not infinite dimensional)to show mathematically the truth of a statement as strong as yours.
You may say such construction is useless or "unphysical" in your opinion but haven't demonstrated it is not possible in QT.

 

[Tendex]

Where are you getting this from?

Nevermind, it is from informal talks and don't have time to check if he has published it formally. Feel free to edit it out that as I no longer can and it doesn't add anything.

 

[DarMM]

you must show why a sample space of all conceivable macroscopic random quantum outcomes not locally constrained by choosing a local observable is forbidden in quantum theory irrespective of the fact we cannot "see" them(we are just allowed to "see" by definition the subsets determined by specific local observables, one can't help measurements are local)

but haven't demonstrated

I have shown this. There is no Gelfand homomorphism of the Quantum algebra. That's an old theorem you'll find in books like Connes Non-Commutative Geometry.

far fetched analogy with differential geometry

The analogy is actually less far fetched. The supposed object of embedding in the GR case is closer to the normal theory.

 

[Elias1960]

That's not a counter-example. You've shown that the objects in quantum theory can be embedded in an infinite dimensional object not in quantum theory.
A Gelfand homomorphism is a map that takes C*-algebra elements and maps them to functions over a manifold. This manifold is then the sample space.
Quantum theory's observable algebra lacks a Gelfand homomorphism that covers all of the algebra. Thus it does not have one sample space. The end.
What you are doing is finding an algebra with infinite degrees of freedom with the quantum algebra embedded as a subset. Note though it's not a subalgebra, the embedding destroys some algebraic properties.

That makes no sense. The algebraic properties you want to preserve are irrelevant for probability theory.

Then the fact that this much larger algebra, with observables never seen in a lab, has one sample space you are taking as implying QM has one sample space.

This is a triviality. Because it is simply the same sample space. In the worst case, the subset of quantum theory cannot distinguish some elements of the sample space. So what? If one does not like this very much, one can factorize the sample space

This simply doesn't make any sense. As I said it's like embedding every spacetime from General Relativity in 231-D Minkowski and declaring GR deals with flat spaces.

Vague analogies do not count. As well as irrelevant C^* algebra structures. I have asked for the relevant mathematics of probability theory.

In fact it is worse, since the generalized Nash's embedding theorem tells us all properties of those manifolds are preserved, e.g. the curvature is retained as extrinsic curvature in the surrounding space. Where as the embedding destroys the algebraic relations in the quantum algebra.

If the embedding preserves the metric, then the subspace has the same intrinsic (not extrinsic) curvature. It is a property of this subspace. You have yet not told which properties of sets of probability distributions will be lost if restricted to subsets. All that comes to mind is the subset may not contain all affine combinations. But this example would be irrelevant given that the closure for affine combinations can be always defined in a trivial way. And the affine combination of two states of the subset is the same, as in the big set as in the subset.

 

[DarMM]

The fact that you consider C*-algebras, the mathematical objects which quantum observables are, to be irrelevant means discussion of the probabilistic structure of the theory cannot be had in any sensible fashion with you.

I've given the relevant mathematics already, references and the opinions of three experts in the topic. This is just crank level denial of the subject.

 

[Elias1960]

There are sophisticated applied mathematical models of entanglement built upon such objects which trivially subsume probability theory and at the same time are capable of unifying wide swaths of branches in mathematics in the process; to just ignore all of this based purely on the ideological reasons you posit, is to halt the march of science.

No. I have no aim to hide or ignore any mathematics. I just see no reason to reject anything in the Cox axiomatization of plausible reasoning. Just because the point of these axioms is the internal consistency of the plausible reasoning.

That is a very specific philosophy and a very premature one at that: you again assume that these axiomatizations are the logic of plausible reasoning, instead of a logic of plausible reasoning. Moreover, you are seemingly implicitly delimiting plausible reasoning to human reasoning while it has already been demonstrated empirically that there exists artificial algorithms which can reason in a totally foreign manner and while doing so sometimes get better answers than humans can with regard to certain kinds of questions.

Whatever, I prefer consistency in reasoning. If there are other forms of "reasoning", I suspect they are simply misnamed. As the use of the names of the logical operations in so-called "quantum logic". That AI will be in some aspects better than humans is trivial, but also irrelevant, given that humans are known for often reasoning inconsistently.

It isn't that much of a stretch to think that this is because these algorithms are actually utilizing undiscovered forms of mathematics which of course already exist and are consistent with these generalized probability theories; that is in fact exactly what would be needed to legitimize and normalize such generalizations more within the contemporary practice of mathematics and the sciences.

There are in fact other forms of plausible reasoning which were empirically discovered and are even formally utilized in actual practice which aren't isomorphic to either Kolmogorovian or Coxian axiomatization: possibility theory, quantum probability and fuzzy logic, just to name a few.

Of course, the logic of plausible reasoning as being probability theory is a quite new insight, so one has to expect that various other attempts to formalize plausible reasoning exist and have not yet been thrown away. Feel free to develop them. BTW, that they aren't isomorphic does not mean that they have to be in contradiction.

Especially in our modern computational era - which will some day be seen as the golden age of neural networks - where such alternate models are actually being implemented and studied not just as abstractions but as applied constructions, your stance is scientifically simply completely unjustifiable.

Except that you misrepresent my stance by suggesting I would like to like to "halt the march of science", even if all I suggest is to refrain from using misleading names like "quantum logic" for lattice theory.

 

[Tendex]

I have shown this. There is no Gelfand homomorphism of the Quantum algebra.

That homomorphism is only relevant for subsets of the space where an observable with noncommutative algebra has been picked, not for the space of all possible outcomes and all its possible subsets which is what we are talking about.

This reminds me (going back to analogies with differential geometry and GR) of someone that argued that because the Lorentz group was a subgroup of GL(4) that local GL(4)-valued functions on the frame bundle acted like local Lorentz transformations, wwll they don't, certainly not every element of the bigger group is an element of the subset.

 

[DarMM]

That homomorphism is only relevant for subsets of the space where an observable with noncommutative algebra has been picked, not for the space of all possible outcomes and all its possible subsets which is what we are talking about.

The space of all observables in quantum theory is a non-commutative algebra. I'm not talking about a subalgebra, I'm talking about the entire algebra.

What exactly is "an observable with a noncommutative algebra"?

 

[Tendex]

The space of all observables in quantum theory is a non-commutative algebra. I'm not talking about a subalgebra, I'm not talking about the entire algebra.

What exactly is "an observable with a noncommutative algebra"?

I was referring to outcomes determined by the space of quantum observables, algebras or subalgebras are not relevant here since we agreed we are dealing with a purely mathematical construction that doesn't necessarily have a empirical correspondence or the physical relevance of the noncommutative algebra for QM.
Such outcomes can be in subsets of the space of all possible outcomes(random macroscopic measurable elements) not subject to the noncommutative algebra (since no observable is picked, no physical local measurement must be performed to consider this space) or not? If not why not?

 

[DarMM]

I was referring to outcomes determined by the space of quantum observables

Outcomes are modeled in quantum theory as, in general, projectors in POVMs. The algebra of such projectors is non-commutative thus there is no Gelfand homomorphism.

 

[Tendex]

Outcomes are modeled in quantum theory as, in general, projectors in POVMs. The algebra of such projectors is non-commutative thus there is no Gelfand homomorphism.

Sure, but how is that a mathematical objection to have them as subsets of the infinite dimensional sample space of all possible outcomes not constrained by a specific performance of a physical measurement?, in a generalized probability that includes both classical commutative and noncommutative algebras, so the homomorphism that restricts to observable outcomes is not a restriction for the generalized sample space .

I mean I can see how you can argue against the physical relevance of constructing such a space, (and physical relevance is what Streater et al have in mind when discussing quantum probability), but arguing it's mathematically impossible is a different thing, you have to show it is a contradiction within the theory.

QFT is full of such constructions that are not observable but are important for the coherence of the theory(cluster decomposition comes to mind) and even to justify certain physical principles. The construction we are discussing doesn't seem important physically but it's existence is important mathematically if we want to keep using mathematical spaces based on classical mathematical logic.

 

[DarMM]

Sure, but how is that a mathematical objection to have them as subsets of the infinite dimensional sample space of all possible outcomes not constrained by a specific performance of a physical measurement?,

Because there is no Gelfand homomorphism enabling you to give the space of all possible outcomes a probability measure.

 

[Tendex]

Because there is no Gelfand homomorphism enabling you to give the space of all possible outcomes a probability measure.

There is, but not for all subsets of the single sample space. Look at page 96 of "Lost causes in and beyond physics" by Streater, it explains in which sense a Kolmogorov's theory is still the general framework, and only when turning to observables one has the Gelfand restriction.

 

[DarMM]

There is, but not for all subsets of the single sample space. Look at page 96 of "Lost causes in and beyond physics" by Streater, it explains in which sense a Kolmogorov's theory is still the general framework, and only when turning to observables one has the Gelfand restriction.

The page that says this:

This shows that there are some predictions of quantum theory that cannot be obtained from any Kolmogorovian theory.

I don't know how one can read that page to say Kolmogorov's theory is still the general framework. He literally says it is not.

 

[Auto-Didact]

Whatever, I prefer consistency in reasoning. If there are other forms of "reasoning", I suspect they are simply misnamed. As the use of the names of the logical operations in so-called "quantum logic". That AI will be in some aspects better than humans is trivial, but also irrelevant, given that humans are known for often reasoning inconsistently.

The argument I am making is far more subtle: you are delimiting reasoning to be in a single domain, i.e. the rational domain. In contrast, I - as well as the sciences in general - am including any possible domain including the conjunction of separate domains, i.e. not purely rational but also empirical. This latter strategy often turns out to be a superior one with respect to answering certain types of question: indeed, it is why physics doesn't need to rely purely on mathematics because experiment can guide us without actually calculating anything.

Except that you misrepresent my stance by suggesting I would like to like to "halt the march of science", even if all I suggest is to refrain from using misleading names like "quantum logic" for lattice theory.

The choosing of such misleading names is a double-edged sword, because the name is usually historico-structurally enlightening e.g. as in the moniker 'Newtonian' referring to a time period and an associated philosophy, but it tends to be non-descriptive i.e. misleading when taken literally. I agree with you that the nomenclature could be better chosen, but this is just an overly optimistic ideal, one I have learned to let go in the face of conventionalism. The reality remains that deciding nomenclature is the 'right' of the discoverer and/or happens when some convention is massively adopted by everyone following some source article.

 

[atyy]

@DarMM, if the Kochen-Specker theorem says that any theory that reproduces QM cannot have a single sample space, then doesn't it mean that BM (with equilibrium) also does not have a single sample space?

 

[A. Neumaier]

Sure, but this is the determination of a parameter in the Hamiltonian, not the measurement of an observable. The thermal interpretation differs from tradition only in the latter. For parameter determination there is no significant difference to the tradition.

Thus your observation does not affect the validity of the thermal interpretation.

Of course, from a fundamental-physics point of view at the end you can say that all measurements are a determination of some parameter in a Hamiltonian.

No. Measuments, unlike the Hamiltonian, depend on the state of the system measured.

Measurements are independent of the state. You can prepare a system in any state you like (and are technically able to of course) and independent from this state preparation you can measure any observable of this system you can properly define by a measurement procedure.

But unlike the Hamiltonian, the measurement results depend on the state. That's the whole point of Born's rule and its generalization.

And this is what makes the determination of parameters in a Hamiltonian a measurement procedure quite different from the measurements talked about in Born's rule.

 

[A. Neumaier]

@DarMM, if the Kochen-Specker theorem says that any theory that reproduces QM cannot have a single sample space, then doesn't it mean that BM (with equilibrium) also does not have a single sample space?

The Kochen-Specker theorem only says that QM itself cannot have a single sample space.

 

[DarMM]

@DarMM, if the Kochen-Specker theorem says that any theory that reproduces QM cannot have a single sample space, then doesn't it mean that BM (with equilibrium) also does not have a single sample space?

QM itself does not have a single sample space.

The KS theorem tells you that if you wish to have a theory with a single sample space it will need to be contextual.

Hardy's theorem then tells you that such a sample space will also need to be infinite dimensional.

 

[atyy]

QM itself does not have a single sample space.

The KS theorem tells you that if you wish to have a theory with a single sample space it will need to be contextual.

Hardy's theorem then tells you that such a sample space will also need to be infinite dimensional.

I understand BM is contextual, but does BM (with equilibrium) have a single sample space?

Also, can you point to a presentation of the KS theorem that talks about a single sample space? I don't understand what the term means. For example, the version at https://plato.stanford.edu/entries/kochen-specker/ doesn't mention it. It mentions value definiteness (VD) and noncontextuality (NC), and says that BM rejects both VD and NC.

 

[DarMM]

I understand BM is contextual, but does BM (with equilibrium) have a single sample space?

Indeed it does.

I don't understand what the term means

It's just its meaning from classical probability theory.

Also, can you point to a presentation of the KS theorem that talks about a single sample space?

First of all the version of KS there is not as fully general as the presentation to be found in Matt Leifer's paper here:
https://arxiv.org/abs/1409.1570
Leifer shows how to avoid the KS-theorem the random variable corresponding to a POVM outcome $E$ , initially $\Gamma_{E}\left(\lambda\right)$, has to be generalized to $\Gamma_{E,M}\left(\lambda\right)$ with a random variable for every parition of the identity $M$ that has $E$ as an element.

I would also note (and you will see this in Leifer's paper) that Bell's theorem and the KS theorem are ultimately related. The generalized KS theorem actually implies Bell's theorem.

That Bell's theorem is related to the number of sample spaces is a result known as Fine's theorem. It shows that assuming a single space (and locality and no retrocauslity, etc) gives the Bell inequalities. Two proofs are here:
https://arxiv.org/pdf/1403.7136.pdf
This is important to know because the lack of a single sample space is how QM itself manages to violate Bell's theorem. Violating it via nonlocality, retrocausality, etc is the approach of alternate theories.

 

[Elias1960]

The argument I am making is far more subtle: you are delimiting reasoning to be in a single domain, i.e. the rational domain. In contrast, I - as well as the sciences in general - am including any possible domain including the conjunction of separate domains, i.e. not purely rational but also empirical. This latter strategy often turns out to be a superior one with respect to answering certain types of question: indeed, it is why physics doesn't need to rely purely on mathematics because experiment can guide us without actually calculating anything.

I also rely on a lot of other things than reasoning in everyday life. In fact, I do not calculate probabilities if I use plausible reasoning in every life. But I think it is a remarkable insight that the laws of plausible reasoning have been identified with such a well-known and simple thing as probability theory, and that it can be derived by axioms which seem unquestionable if one wants to avoid inconsistency.

The choosing of such misleading names is a double-edged sword, because the name is usually historico-structurally enlightening e.g. as in the moniker 'Newtonian' referring to a time period and an associated philosophy, but it tends to be non-descriptive i.e. misleading when taken literally. I agree with you that the nomenclature could be better chosen, but this is just an overly optimistic ideal, one I have learned to let go in the face of conventionalism. The reality remains that deciding nomenclature is the 'right' of the discoverer and/or happens when some convention is massively adopted by everyone following some source article.

Here I can more or less agree, one has to live with the facts. But if it appears necessary, because some people seem really misguided, one also has to point out that the names are misleading. And if it is claimed that classical probability is no longer valid in quantum theory, because what holds there are various "generalizations" of probability theory, we have a clear case of such a misleading effect. The laws of plausible reasoning are (whatever they are) metatheoretical, and if some theory would be incompatible with them, the theory should be rejected as inconsistent. Laws of reasoning cannot be empirically falsified, because the very process of empirical falsification requires their application.

 

[zonde]

I don't understand this claim that QM does not have a single sample space.
Sample space is just a definition for a set of outcomes with few restrictions (from wikipedia):
- the outcomes must be mutually exclusive;
- the outcomes must be collectively exhaustive;
- we must remove irrelevant information from the sample space.

I consider an experiment E. It consists of choosing subexperiment X or Y and performing one of them. Experiment X has outcomes A and B, but experiment Y - outcomes C and D.
Sample space consists of A, B, C and D. It satisfies all three restrictions.
Why set of A, B, C and D can't be considered single sample space?

 

[A. Neumaier]

I don't understand this claim that QM does not have a single sample space.
Sample space is just a definition for a set of outcomes with few restrictions (from wikipedia):
- the outcomes must be mutually exclusive;
- the outcomes must be collectively exhaustive;
- we must remove irrelevant information from the sample space.

I consider an experiment E. It consists of choosing subexperiment X or Y and performing one of them. Experiment X has outcomes A and B, but experiment Y - outcomes C and D.
Sample space consists of A, B, C and D. It satisfies all three restrictions.
Why set of A, B, C and D can't be considered single sample space?

Because this set is not mutually exclusive in the sense implied by (the precise sources of) wikipedia. The probabilities for the four cases sum to 2 rather than to 1.

 

[zonde]

Because this set is not mutually exclusive in the sense implied by (the precise sources of) wikipedia. The probabilities for the four cases sum to 2 rather than to 1.

By the description of experiment they don't. Probabilities of performing subexperiment X or subexperiment Y sum to 1. Then probability of performing subexperiment X is split between outcomes A and B and probability of performing subexperiment Y is split between outcomes C and D. Of course the sum is 1 not 2.

 

[Morbert]

By the description of experiment they don't. Probabilities of performing subexperiment X or subexperiment Y sum to 1. Then probability of performing subexperiment X is split between outcomes A and B and probability of performing subexperiment Y is split between outcomes C and D. Of course the sum is 1 not 2.

In classical physics, different samples spaces can be related to one another by coarsening or refining, and there is a single sample space that is a common refinement of all others. In quantum physics, there isn't a common refinement.

 

[DarMM]

I don't understand this claim that QM does not have a single sample space.
Sample space is just a definition for a set of outcomes with few restrictions (from wikipedia):
- the outcomes must be mutually exclusive;
- the outcomes must be collectively exhaustive;
- we must remove irrelevant information from the sample space.

I consider an experiment E. It consists of choosing subexperiment X or Y and performing one of them. Experiment X has outcomes A and B, but experiment Y - outcomes C and D.
Sample space consists of A, B, C and D. It satisfies all three restrictions.
Why set of A, B, C and D can't be considered single sample space?

Because the four pairwise probabilities for Bell alignments $A, B, C, D$:
$$P\left(A,B\right), P\left(B,C\right), P\left(C,D\right), P\left(A,D\right)$$
can be proven to not be marginals of a single distribution $P\left(A,B,C,D\right)$ per Fine's theorem, thus they are not defined over a common sample space.

 

[zonde]

In classical physics, different samples spaces can be related to one another by coarsening or refining, and there is a single sample space that is a common refinement of all others. In quantum physics, there isn't a common refinement.

Are you saying that for my described experiment there is single sample space but when considering other experiments there is no common sample space?