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

[Mentz114]

No. There is a single sample space for each pair of Bell alignments, but not all of them together. All can be carried out. @Morbert is referring to the fact that their refinement cannot be carried out.It is directly relevant, per Fine's theorem it explains the violations of the Bell inequalities by QM.

It does not explain anything physical - or there would not be dozens of threads arguing this.
When the data from an epr experiment are divided into the four categories each contingency table has expected marginals of 1/2 which is experimentally verified.

 

[DarMM]

So what? Try the same for the sample space
$\left\{(A_{0},C_{0}), (A_{0},C_{1}), (A_{1},C_{0}),(A_{1},C_{1}),(B_{0},C_{0}),\ldots (B_{1},D_{1})\right\}$
The construction of a sample space is a triviality.

You can't put a probability measure on that space that matches quantum theory.

 

[zonde]

In QM the experiments of Alice and Bob are independent as shown by their marginals being unaffected by the other's choice of experiment and yet the pairs do not have a common sample space. So this does not follow.

The devil is in details. Marginals are unaffected but in experiments you have actual detections. QM looks only at statistics but detections themselves are physical facts ignored by QM. And that's where this independence assumption fails.

It doesn't change the outcomes of the other experiment or their probabilities.

It does not change probabilities but it changes outcomes. QM does not look at outcomes so you can't argue that QM is counterexample.

 

[DarMM]

It does not explain anything physical - or there would not be dozens of threads arguing this.
When the data from an epr experiment are divided into the four categories each contingency table has expected marginals of 1/2 which is experimentally verified.

Well it's a property of quantum theory. If it doesn't explain things to you then tough, there isn't anything I can do. That's the theory.

It does have a physical meaning. Only those variables you measure obtain a value.

 

[DarMM]

The devil is in details. Marginals are unaffected but in experiments you have actual detections. QM looks only at statistics but detections themselves are physical facts ignored by QM. And that's where this independence assumption fails.

It does not change probabilities but it changes outcomes. QM does not look at outcomes so you can't argue that QM is counterexample.

I can't use QM in a discussion about the predictions of quantum theory? This has transcended the farcical.

It doesn't even change the outcomes. The outcomes of the local experiments at Alice's location are not affected by the experimental choice at Bob's location.

Have you gone through Fine's theorem, Streater's monograph or any texts on quantum probability?

 

[zonde]

I can't use QM in a discussion about the predictions of quantum theory? This has transcended the farcical.

Of course you can use QM. But you can't use QM to argue about things on which QM is silent.

It doesn't even change the outcomes. The outcomes of the local experiments at Alice's location are not affected by the experimental choice at Bob's location.

Outcomes = sample space. You yourself say that there is no single sample space in QM. And yet you say QM does not change outcomes. It's different words but the same meaning.

 

[DarMM]

And yet you say QM does not change outcomes

"QM changing outcomes" is a meaningless phrase. I am saying that Alice's outcomes only depend on her choice of measurement.

At this point it's just people axe grinding.

For everybody else, check out Chapter 6 of Streater's text or Summers paper, they're a really nice run down. There is also these lecture notes:
http://info.phys.unm.edu/~crosson/Phys572/Physics572Index.html
And these:
https://www.math.tamu.edu/~jml/trentosurvey.pdf
I'll leave the choice of whether to believe Streater, Summers and other experts or @zonde and @Elias1960 up to you.

 

[Jimster41]

If you can carry out the experiment, you can build an appropriate sample space. But if you try to combine two incompatible sample spaces into a single common sample space a priori, you'll fail, and you will not be able to carry out the implied experiment.

To me the key thing is “a-priori”. The role of location in time order gets lost here a lot I think.

Not saying there is a thing that is really Time. I’m just saying Time order is a critical component of the description of nature being debated here.

The cat is alive and dead a-priori. Well ontologically - that is one odd cat.

I But cripes it is subtle, so I’m still probably not getting it.

 

[DarMM]

So what? Try the same for the sample space
$\left\{(A_{0},C_{0}), (A_{0},C_{1}), (A_{1},C_{0}),(A_{1},C_{1}),(B_{0},C_{0}),\ldots (B_{1},D_{1})\right\}$
The construction of a sample space is a triviality.

Just quickly for others, for choices given by their angle with the $z$-axis:
$$
A = S_{0}\\
B = S_{\pi/4}\\
C = S_{\pi/2}\\
D = S_{3\pi/4}
$$
The probabilities QM gives for each outcome pair $\left\{00,01,10,11\right\}$ are given by:

	0	1
0	$\frac{1}{4}\left(1 - \cos(\theta)\right)$	$\frac{1}{4}\left(1 + \cos(\theta)\right)$
1	$\frac{1}{4}\left(1 + \cos(\theta)\right)$	$\frac{1}{4}\left(1 - \cos(\theta)\right)$

Where $\theta$ is the difference in the angles for the two observables.

You'll see that if applied to @Elias1960 's set this gives results going over unity. Thus it isn't a sample space.

 

[PeterDonis]

for choices given by their angle with the zz-axis

Should $B$ be just $\pi / 4$ (instead of $5 \pi / 4$)?

 

[DarMM]

Should $B$ be just $\pi / 4$ (instead of $5 \pi / 4$)?

Sorry corrected now. I've put in the conventional choices with $B = \frac{\pi}{4}$ and $C = \frac{\pi}{2}$

 

[Elias1960]

You can't put a probability measure on that space that matches quantum theory.

I can. However you construct your experiment, you will with some quite classical probability make a choice what to measure, AC, BC, AD or BD. Let's name these classical probabilities $P(AC), P(BC),P(AD),P(BD)$. They sum up to 1 as classical probabilities. Then, for each of the four choices, you have the quantum probabilities for that particular experiment. Let's name them $Q_{AC}(0,0), Q_{AC}(1,0), Q_{AC}(0,1), Q_{AC}(1,1)$ and so on. They sum up to 1 for each of the quantum experiments considered separately.

So, the obvious rule is $P(A_0,C_0) = P(AC)Q_{AC}(0,0), P(A_0,C_1) = P(AC)Q_{AC}(0,1)$ and so on in the straightforward way.

 

[DarMM]

Things like $P\left(AC\right)$ refer to probabilities for choosing the equipment set up, they're nothing to do with the electron.

Your sample space there is the Cartesian product of $\Sigma_{1} \times \Sigma_{2}$ where $\Sigma_{1}$ is the space of my choices and $\Sigma_{2}$ is the space of electron pair observable outcomes. It's not a single sample space for the actual observables of the electron and things like $P\left(AC\right)$ are not electron observable probabilities and not part of the predictions of quantum theory.

You're avoiding the fact that the probabilities for the electron's outcomes exceed unity by weighting them by my set up choices. That's like saying that if classical statistical mechanics predicts a mercury fluid has a 40% chance of being 300K, you say "No, it's only 20% because I'm only going to choose a thermometer 50% of the time"

Also done in full generality your "construction" is going to be infinitely larger than the sample spaces in QM.

I'm not sure what you're even arguing for.

 

[atyy]

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.

Is it correct to say that Fine's theorem says that if the CHSH equality is violated, there cannot be a single sample space in the measurement outcomes?

If so, then wouldn't a BM-like theory that reproduces the violation of the CHSH inequality also not have a single sample space in that sense?

 

[DarMM]

Is it correct to say that Fine's theorem says that if the CHSH equality is violated, there cannot be a single sample space in the measurement outcomes?

If locality and no retrocausality are retained then yes.

If so, then wouldn't a BM-like theory that reproduces the violation of the CHSH inequality also not have a single sample space in that sense?

BM rejects locality and thus retains a common sample space. The quantum formalism retains locality and rejects a common sample space.

 

[Morbert]

But the proposal was exactly a simplified version of this: The combination of two incompatible sample spaces {A,B} and {C,D} using an experiment which obviously can be done.

I don't think it was a simplified version of this. But let my try to address possible ambiguities.

If we have a Hilbert space of the X apparatus $\mathcal{H}_X$, Y apparatus $\mathcal{H}_Y$, the microscopic system $\mathcal{H}_s$, and a coin $\mathcal{H}_c$, and if we model the subexperimental outcomes as $I_{\mathcal{H}_X\otimes\mathcal{H}_s} = P_A+P_B$ and $I_{\mathcal{H}_Y\otimes\mathcal{H}_s} = P_C+P_D$, as well as a coin flip $I_{\mathcal{H}_c} = P_{\rm heads} + P_{\rm tails}$ then we can use the projective decomposition$$I_{\mathcal{H}_X\otimes\mathcal{H}_Y\otimes\mathcal{H}_s\otimes\mathcal{H}_c}=P_{\rm heads}P_AI_{\mathcal{H}_Y} + P_{\rm heads}P_BI_{\mathcal{H}_Y} + P_{\rm tails}P_CI_{\mathcal{H}_X}+P_{\rm tails}P_DI_{\mathcal{H}_X}$$and build a sample space from these four mutually exclusive outcomes.However, if the samples spaces $\{A,B\}$ and $\{C,D\}$ are instead alternative results of measuring the microscopic system such that $I_{\mathcal{H}_s} = P_A+P_B = P_C+P_D$, then neither $I_{\mathcal{H}_s} = P_A+P_B+P_C+P_D$ nor $P_iP_j = \delta_{ij}P_i$ hold, and $\Omega = \{A,B,C,D\}$ is not a valid sample space of mutually exclusive alternatives, resolved by any experiment.

 

[Tendex]

DarMM said:
"I'll leave the choice of whether to believe Streater, Summers and other experts or @zonde and @Elias1960 up to you."

Would Kochen and Specker in the article communicated by Gleason that Elias1960 has referenced several times in this thread count as experts to you? Maybe experts enough to be believed?

They say referring to formula (1) in that article that defines the probability measure of the single sample space of a classical system in terms of a measurable susbset of the reals: "We may always introduce, at least mathematically, a phase space $\Omega$ into a theory so that (1) is satisfied."

(1)$$P_{A\psi}(U)=\mu_\psi (f_A^{-1}(U))$$ with $P_{A\psi}$ the probability measure assigned to an observable A and mixed state $\psi$

They then describe exactly how to obtain such single sample space in QM, and go on to explain the reasons to introduce what they call "this somewhat trivial construction" [ I have in previous posts commented about this physical triviality but the existence of this mathematical construction follows from consistency of any mathematical physics theory based on standard mathematical logic.] and one of the reasons is that it "indicates the direction in which the condition (1) is inadequate[for constructing a classical hidden variables theory given it applies also to QM]. For each state $\psi$ as interpreted in the space $\Omega$, the functions $f_A$ are easily seen to be measurable functions with respect to the probability measure $\mu_\psi$. In the language of probability theory the observables are thus interpreted as random variables for each state $\psi$. It is not hard to show furthermore that in this representation the observables appear as independent random variables"

(My own aclaratory comments are bewteen brackets)

So I'm still pondering why would anyone deny these facts with such zeal with counterarguments that are totally orthogonal to this construction that is not related to the Gelfand algebra that restricts to each specific quantum experiment and the predictions on them and is included in the generalization by definition. Is it too much to ask to discern between the phenomenology and the math restricted to it from the global consistence of a mathematical theory?

 

[DarMM]

Would Kochen and Specker in the article communicated by Gleason that Elias1960 has referenced several times in this thread count as experts to you?

Nothing in that article contradicts what I'm saying. It's just the ontological models framework as most generally formulated by Spekkens in an earlier form. Which is essentially the most general framework for non-retrocausal hidden variable theories. Indeed there one has that the random variable for a POVM outcome $\Gamma_{E}\left(\lambda\right)$ is required to be generalised to $\Gamma_{E,M}\left(\lambda\right)$ as I have already stated, with $M$ any partition of the identity containing $E$. In fact I've already dealt with this.

My point is that QM itself does not do this. QM itself has multiple sample spaces, as it must because it has a non-commutative C*-algebra.

So I'm still pondering why would anyone deny these facts with such zeal

Because they're not facts.

 

[Tendex]

Nothing in that article contradicts what I'm saying. It's just the ontological models framework as most generally formulated by Spekkens in an earlier form. Which is essentially the most general framework for non-retrocausal hidden variable theories. Indeed there one has that the random variable for a POVM outcome $\Gamma_{E}\left(\lambda\right)$ is required to be generalised to $\Gamma_{E,M}\left(\lambda\right)$ as I have already stated, with $M$ any partition of the identity containing $E$. In fact I've already dealt with this.

My point is that QM itself does not do this. QM itself has multiple sample spaces, as it must because it has a non-commutative C*-algebra.Because they're not facts.

Of course QM phenomenology has multiple sample spaces depending on the experiment, nobody denies this, but the generalization of the quantum mathematical theory also has Kochen and Specker construction.
Also what are you calling "QM itself"? That is non-standard. Having QM phenomenology and making the same predictions? BM, an interpretation of QM, so with the same empirical physics, can have a single sample space(as you have finally admitted to atyy), how is a QM interpretation with QM predictions and experiments not QM itself?
Is your own interpretation only QM itself?

 

[DarMM]

Also what hat are you calling "QM itself"? That is non-standard

A non-commutative C*-algebra with normed states on it? That is utterly standard. That is QM. A non-commutative C*-algebra has multiple sample spaces. The proof being there is no Gelfand homomorphism that covers the whole algebra. The End.

Having QM phenomenology and making the same predictions? BM, an interpretation of QM, so with the same empirical physics, can have a single sample space(as you have finally admitted to atyy)

I've been saying Bohmian Mechanics has a single sample space from the beginning, before you got involved. I've never denied this.

Look just look at post #376, I think it is clear you are not actually understanding the papers and material being referenced here.

Is your own interpretation only QM itself?

If the C*-algebraic structure and the associated dual state space of QM is "my interpretation" I hope I get my Nobel Prize soon.

 

[Jimster41]

Does that mathematically consistent single sample space have to contain a cat that is both dead and alive.

Trying to understand if the argument here is about what is mathematically consistent vs. what is both mathematically consistent and observable... or something like that.

 

[Tendex]

A non-commutative C*-algebra with normed states on it? That is utterly standard. That is QM. A non-commutative C*-algebra has multiple sample spaces. The proof being there is no Gelfand homomorphism that covers the whole algebra. The End.I've been saying Bohmian Mechanics has a single sample space from the beginning, before you got involved. I've never denied this.

Look just look at post #376, I think it is clear you are not actually understanding the papers and material being referenced here.If the C*-algebraic structure and the associated dual state space of QM is "my interpretation" I hope I get my Nobel Prize soon.

Then the BM interpretation of QM is QM itself according to you and it also has a single sample space as constructed by Kochen and Specher not restricted to the noncommutative algebra(do you understand that in this generalization of the probability the noncommutative algebra is a special case?) besides multiple sample spaces for predictions involving just the noncommutative algebra.

I won't continue this exchange, as it seems you only keep repeating the same mantra to save face.

 

[DarMM]

Then the BM interpretation of QM is QM itself according to you

No, Bohmian Mechanics has a completely different mathematical structure. A structure which as of yet has not been shown to give results compatible with QFT, and possibly (there isn't complete proof in this regard) replicates non-relativistic QM when in equilibrium if one ignores iterated Wigner's friend scenarios. It is a different theory that isn't fully developed as of 2019.

QM itself, the fully developed framework of non-commutative C* algebras and states and unitaries upon them is the theory which matches all experimental predictions and it constitutes a multi-sample space probability theory.

it also has a single sample space

Yes, as I have said from the beginning Bohmian Mechanics has a single sample space.

do you understand that in this generalization of the probability the noncommutative algebra is a special case?

It isn't. I would study how QM's algebra is embedded in Bohmian observable algebra, it's quite subtle.

I won't continue this exchange, as it seems you only keep repeating the same mantra to save face.

I'm just glad the whole field uses my terminology and classifications, at least we all get to save face together.

 

[atyy]

If locality and no retrocausality are retained then yes.

BM rejects locality and thus retains a common sample space. The quantum formalism retains locality and rejects a common sample space.

I don't understand how the quantum formalism can retain locality by rejecting a single sample space. If the quantum state is not real, then the quantum formalism is not nonlocal, thus saving locality in a sense. However, saying that the quantum state is not real does not mean that the quantum formalism is local, because realism is a precondition for locality. If the quantum state is real, then the quantum formalism is nonlocal.

So I link the nonreality of the quantum state to saving locality in some sense, whereas you link not having a single sample space to saving locality. Is the nonreality of the quantum state related to not having a single sample space (I don't see how it is)?

 

[DarMM]

because realism is a precondition for locality

I'll answer this tomorrow, but what exactly do you mean here?

 

[DarMM]

This is based on Wiseman's well known complete exposition of Bell's theorem:
https://arxiv.org/abs/1503.06413
Which is itself an outcome of work explicating the theorem such as those by Fine and Jarrett.

So Bell's theory assumes:

1. A measurement has a single objective outcome
2. Light Cones, space-like separation and hypersurfaces, i.e. the machinery of Minkowski spacetime, make sense for laboratory experiments. Rejecting this is essentially the EPR = ER idea.
3. For an event $A$ there is a hypersurface that separates events in the PAST of $A$ from those which have $A$ as their past.
4. A cause of an event is in its PAST. Note axioms (3) and (4) don't yet say the PAST is the past lightcone. They just amount to no retrocausality.
   
   That concludes the most basic assumption that both classical-like hidden variable theories and QM agree on
5. Free Choice. It is possible to choose settings in a Bell experiment in a way that is not correlated with the particles, i.e. no superdeterminism
6. Relativistic Causality. The PAST is the contained in the past light cone
7. Common causes. If $A$ and $B$ are correlated and they are not the cause of each other, then they have a set of common causes responsible for the correlation $\mathcal{C}$. This is just the preparation in a Bell experiment, i.e. an event without which there would be no correlation
8. Decorrelating Explanation. There is an event, some subset of the common causes, that decorrelates the experiments, i.e. the probabilities for $A$ and $B$ factor

The problem with many expositions of Bell's theorem is that they collapse these assumptions into a smaller list. For example Relativistic Causality and Decorrelating Explanation together are equivalent to Bell's assumption in his second (stronger) theorem of 1976 called Local Causality. Which is where I think your statement of "Realism is a precondition for locality" comes from.

So Hidden Variable theories like Bohm's reject Relativistic Causality , where as in QM itself it is Decorrelating Explanations that are violated which is equivalent to rejecting a single space.

This is because for all variables to be conditionable upon a single set of events $\lambda$ they need to be Random Variables on a single common sample space. If they're in different sample spaces this is impossible.

 

[Elias1960]

• Common causes. If $A$ and $B$ are correlated and they are not the cause of each other, then they have a set of common causes responsible for the correlation $\mathcal{C}$. This is just the preparation in a Bell experiment, i.e. an event without which there would be no correlation
  
• Decorrelating Explanation. There is an event, some subset of the common causes, that decorrelates the experiments, i.e. the probabilities for $A$ and $B$ factor

So Hidden Variable theorems like Bohm's reject Relativistic Causality , where as in QM itself it is Decorrelating Explanations that are violated which is equivalent to a single space.

This is because for all variables to be conditionable upon a single set of events $\lambda$ they need to be Random Variables on a single common sample space. If they're in different sample spaces this is impossible.

And this is already close to the key point. Namely, beyond the basic laws of plausible reasoning, which are Kolmogorovian probability with a single sample space (which can be constructed for QT following Kochen Specker) there is also another important law which has to be rejected by non-realistic interpretations: Namely, causality in general. Because of causality without the requirement of common causes, which have to decorrelate observed correlations is not worth much. The tobacco industry would be happy if they could handle some correlations between smoking and lung cancer in a similar way as the violations of the Bell inequalities, namely by rejecting the necessity to find such common causes sufficient to decorrelate those correlations. All they probably have to do is to refer to quantum strangeness, more explanation is no longer required.

But who would accept this? And why this would be rejected? Because of not only the laws of plausible reasoning but also the basic concepts of causality like the common cause principle, are metatheoretical laws of reasoning, part of the scientific method itself, and therefore not open to experimental falsification.

 

[PeterDonis]

For an event $A$ there is a hypersurface that separates events in the PAST of $A$ from those which have $A$ as their past.

This is inconsistent with #2. There is no "past" in Minkowski spacetime; there are only the past and future light cones and the spacelike separated region. (Note that this as you state it is also inconsistent with #6, since the boundary of the past light cone does not separate the past of $A$ with events that have $A$ in their past; the latter events are the future light cone, not the complement of the past light cone.)

If you're going to assume #2, I don't see the point of #3 and #4 at all.

 

[DarMM]

If you're going to assume #2, I don't see the point of #3 and #4 at all.

It's the assumption of topological triviality. Wiseman has more detail. It's saying that you can even get the notion of a flat spacetime of the ground and operationally validate if there is spacelike communication. It's not as such an assumption that the event strucutre of Minkowski space is valid, but that it can be operationally checked.

#3 and #4 are then further specifications that aspects of that structure are valid, so there is a point to them.

This is inconsistent with #2.

It isn't. It isn't rejecting what you mention.

 

[DarMM]

Note that this as you state it is also inconsistent with #6, since the boundary of the past light cone does not separate the past of AAA with events that have AAA in their past; the latter events are the future light cone, not the complement of the past light cone.

Why do you think it refers to the boundary of the past light cone?

 

[PeterDonis]

It's the assumption of topological triviality. Wiseman has more detail.

I'll take a look at the paper then. It might be that I'm simply not familiar with this way of describing things. But I still see a contradiction; see below.

Why do you think it refers to the boundary of the past light cone?

Axiom #3 says that there is some hypersurface that bounds the PAST of $A$ from events that have $A$ in their past. That axiom says that the entire spacetime is divided into only two regions: the PAST of $A$ and events that have $A$ in their past. In other words: the region containing events that have $A$ in their past must be the complement of the PAST of $A$.

Then axiom #6 says the PAST of $A$ is its past light cone; that means the boundary of the past light cone must be the hypersurface referred to by axiom #3 and the region containing events that have $A$ in their past must be the complement of the past light cone. But this is obviously false for Minkowski spacetime.

 

[A. Neumaier]

Axiom #3 says that there is some hypersurface that bounds the PAST of $A$ from events that have $A$ in their past. That axiom says that the entire spacetime is divided into only two regions: the PAST of $A$ and events that have $A$ in their past. In other words: the region containing events that have $A$ in their past must be the complement of the PAST of $A$.

No. By definition, a separating hyperplane of two sets A and B is any hyperplane such that A and B are on different sides of the hyperplane. A and B can be small sets!

 

[DarMM]

That axiom says that the entire spacetime is divided into only two regions

Perhaps this is how I phrased it, but it is saying that there is a hypersurface with those two sets of events on either side. Not that those two sets constitute all events.

 

[PeterDonis]

it is saying that there is a hypersurface with those two sets of events on either side. Not that those two sets constitute all events.

Ah, ok. So in Minkowski spacetime any spacelike hypersurface that contains event $A$ would satisfy #3.

 

[DarMM]

Ah, ok. So in Minkowski spacetime any spacelike hypersurface that contains event $A$ would satisfy #3.

Precisely.

 

[atyy]

I'll answer this tomorrow, but what exactly do you mean here?

There are (at least) two types of locality
(1) no faster than light transmission of information
(2) classical relativistic causality

Classical relativistic causality requires realism as a precondition, since it does not make sense for a cause not to be real. Hence if the quantum state is assumed not to be real, it cannot be claimed that the quantum formalism has classical relativistic causality. If the quantum state is not real, then it evades Bell's theorem, so Bell's theorem cannot be used to say that QM is nonlocal; at the same time, the lack of reality also means that it cannot be said that QM is local. If the quantum state is real, then Bell's theorem applies, and the quantum formalism has manifest nonlocality (collapse of the wave function).

You can find these sentiments in these papers:

https://arxiv.org/abs/0706.2661 by Harrigan and Spekkens: (footnote 16)
"Note that no notion of ‘realism’ appears in our implication.This is because there is no sense in which there is an assumption of realism that could be abandoned while salvaging locality. There is a notion of realism at play when we grant that experimental procedures prepare and measure properties of systems, but it is a prerequisite to making sense of the notion of locality. Norsen has emphasized this point [41, 42]."

https://arxiv.org/abs/1208.4119 by Wood and Spekkens (p2)
"Nonetheless,this is an improvement over the standard characterization of Bell’s theorem as forcing a dilemma between abandoninglocality and abandoningrealism. It has always been rather unclear what precisely is meant by “realism”. Norsen hasconsidered various philosophical notions of realism and concluded that none seem to have the feature that one couldhope to save locality by abandoning them [8]. For instance, if realism is taken to be a commitment to the existenceof an external world, then the notion of locality – that every causal influence between physical systems propagatessub luminally – already presupposes realism "

 

[DarMM]

requires realism as a precondition

First I'd ask what you mean by "realism" in precise mathematical terms.

There are (at least) two types of locality
(1) no faster than light transmission of information
(2) classical relativistic causality

What you are calling Classic Relativistic Causality is I think a combination of Decorrelating Explanations and Relativistic Causality, i.e. what Bell in his 1976 paper called Local Causality. Indeed QM does not have Local Causality. However as I mentioned in #426 you have to split the assumption up to see what is really being rejected, namely Decorrelating Explanations.

If you include hidden variables in the definition of Locality, then certainly one cannot have locality in the Quantum formalism. However all one is really saying is that it has no hidden variables, as that is what is really being rejected.

Wiseman wrote his paper in order to separate out the assumptions fully due to confusion like this.

Classical relativistic causality requires realism as a precondition, since it does not make sense for a cause not to be real. Hence if the quantum state is assumed not to be real, it cannot be claimed that the quantum formalism has classical relativistic causality

This seems to be identifying $\Psi$-ontic views with realism. I would have said that psi-onticism is a separate notion from "realism". $\Psi$-epistemic retrocausal theories are "realist" in the sense of having a hidden variable account, but the wavefunction is not "real" in them.

If the quantum state is not real, then it evades Bell's theorem

That's not directly true. Bell's theorem doesn't really assume anything to do with $\Psi$ in its formulation. Again a $\Psi$-epistemic retrocausal theory avoids Bell's theorem via the retrocausal elements, not so much the status of the wavefunction.

How QM itself avoids Bell's theorem is via dropping Decorrelating Explanations (which means multiple sample spaces). Bohmian Mechanics does it via dropping Relativistic Causality.

 

[DarMM]

And this is already close to the key point. Namely, beyond the basic laws of plausible reasoning, which are Kolmogorovian probability with a single sample space (which can be constructed for QT following Kochen Specker)

I've already described how that sample space is not something predicted by QM. That's a fact, a non-commutative C*-algebra cannot have a single sample space.

But who would accept this? And why this would be rejected? Because of not only the laws of plausible reasoning but also the basic concepts of causality like the common cause principle, are metatheoretical laws of reasoning, part of the scientific method itself, and therefore not open to experimental falsification.

Well it's what QM rejects, I am only reporting that. It's not "my view" or something, it's what the machinery of QM itself does.

 

[Elias1960]

How QM itself avoids Bell's theorem is via dropping Decorrelating Explanations (which means multiple sample spaces). Bohmian Mechanics does it via dropping Relativistic Causality.

QM itself avoids nothing because this is not part of the minimal interpretation. It is your preferred interpretation of QM which avoids it in this way.

I've already described how that sample space is not something predicted by QM. That's a fact, a non-commutative C*-algebra cannot have a single sample space.

There is no need for such a prediction because the construction of such a space is a straightforward and trivial thing. Take all the propositions in your field of discourse. They define a Boolean algebra. Then use Stone's theorem and get a sample space. Applied to QM, this gives the Kochen Specker construction.

The notion of "single sample space of a non-commutative C*-algebra" is obviously something completely different and irrelevant. It is your personal metaphysical choice that this structure is of some fundamental importance.

Well it's what QM rejects, I am only reporting that. It's not "my view" or something, it's what the machinery of QM itself does.

No, these are the metaphysical choices you have made by choosing your preferred interpretation of QM. You are aware that there exist other interpretations of QM which have a different view. A piece of mathematical machinery does nothing without your interpretational choices. Here, in particular, you make the assumption that that $C^*$ structure is something fundamentally important, and not a nice accidental consequence of the fact that the stochastic equations for $\rho(q)$ and $S(q)$ become linear equations for $\psi$ for the particular choice of $\hbar$ in $\psi=\sqrt{\rho}\exp(\frac{i}{\hbar}S)$.

I do not question the idea that this linearity is something really fundamental, and not an otherwise meaningless accident. But it is a metaphysical choice you have to make, and without this choice, QM tells you nothing.

So, it is your own metaphysical decision to reject the necessity of decorrelating explanations. I have explained the consequences of this decision if one would apply this not as a special excuse to handle the violations of the Bell inequalities, but would be applied consistently everywhere: It would destroy the scientific method completely. It may be nonetheless a reasonable decision for you, the tobacco industry possibly needs better public relation specialists, and for a scientist which could explain that because of quantum strangeness there is no necessity at all to decorrelate smoking from lung cancer they could pay a lot. But it remains your choice. You have the freedom to accept, as well, realist interpretations of QM which preserve the scientific method.

 

[DarMM]

QM itself avoids nothing because this is not part of the minimal interpretation. It is your preferred interpretation of QM which avoids it in this way

The mathematical formalism of QM does not have decorrelating explanations because the observables are not defined on a common sample space to condition the events on.

That's just part of the QM formalism, not "my interpretation".

This thread has become bizarre where basic mathematical facts of QM are being called "my interpretation".

The notion of "single sample space of a non-commutative C*-algebra" is obviously something completely different and irrelevant. It is your personal metaphysical choice that this structure is of some fundamental importance.

What?! QM has non-commutative C*-algebras, how is that my metaphysical choice?

Utter crankish garbage I have to say. Another thread where basic aspects of QM have to be defended and explained for page upon page to people at least four decades out of date.

It would destroy the scientific method completely. It may be nonetheless a reasonable decision for you, the tobacco industry possibly needs better public relation specialists, and for a scientist which could explain that because of quantum strangeness there is no necessity at all to decorrelate smoking from lung cancer they could pay a lot

 

[DarMM]

No, these are the metaphysical choices you have made by choosing your preferred interpretation of QM. You are aware that there exist other interpretations of QM which have a different view. A piece of mathematical machinery does nothing without your interpretational choices. Here, in particular, you make the assumption that that $C^*$ structure is something fundamentally important, and not a nice accidental consequence of the fact that the stochastic equations for $\rho(q)$ and $S(q)$ become linear equations for $\psi$ for the particular choice of $\hbar$ in $\psi=\sqrt{\rho}\exp(\frac{i}{\hbar}S)$.

Show me such a construction for QFT.

Even so it is an alternate construction. The mathematics of QM does not involve such things, an alternate formalism incapable of replicating QFT and not known to replicate all of QM (as of 2019) does, but that's not QM.

 

[PeterDonis]

Take all the propositions in your field of discourse. They define a Boolean algebra.

Not for QM they don't. For example, consider the two propositions relating to an electron that has just passed through a Stern-Gerlach device oriented in the $z$ direction and has come out the "up" output:

(1) This electron has $z$ spin up.

(2) This electron has $x$ spin up.

#1 has a well-defined truth value, namely "true". #2 does not have a well-defined truth value at all. Therefore no set of propositions that contains both #1 and #2 can define a Boolean algebra. But both propositions are part of the "field of discourse" of QM. (I believe this is the kind of thing @DarMM is referring to when he says QM requires multiple sample spaces.)

 

[DarMM]

Not for QM they don't. For example, consider the two propositions relating to an electron that has just passed through a Stern-Gerlach device oriented in the $z$ direction and has come out the "up" output:

(1) This electron has $z$ spin up.

(2) This electron has $x$ spin up.

#1 has a well-defined truth value, namely "true". #2 does not have a well-defined truth value at all. Therefore no set of propositions that contains both #1 and #2 can define a Boolean algebra. But both propositions are part of the "field of discourse" of QM. (I believe this is the kind of thing @DarMM is referring to when he says QM requires multiple sample spaces.)

Yes that is equivalent to what I am saying.

Labelling your two events as $A$ and $B$, a Boolean algebra requires both $A \lor B$ and $A \land B$ to exist which isn't true for such events. These are basic requirements of both Boolean algebras and $\sigma$-algebras (the latter are basically just a special type of Boolean algebra) and are required to define a measure. Since we don't have this we can't define a measure and thus we don't have a sample space containing both events.

 

[atyy]

First I'd ask what you mean by "realism" in precise mathematical terms.

In the context of the Wiseman paper, my term "realism" is used in the sense that if something is designated as contributing to a cause, then it is real.

Thus if the wave function is real, then QM manifestly violates relativistic causality. If it is not real, then relativistic causality need not be violated. However, relativistic "causality" in the sense of having "causes" is then empty, unless one uses a different definition of cause than as conventionally used in science - for example, one could try using notions discussed in https://arxiv.org/abs/1602.07404.

That is my interpretation of Wiseman's comment (p22):
"That is, even the process whereby, when Alice and Bob share a singlet state, a measurement by Alice in a certain basis causes the quantum state of Bob’s system to collapse instantaneously into one of the basis states, does not violate LOCALITY. Yet the very wording of the preceding sentence implies that the described process does violate RELATIVISTIC CAUSALITY. By contrast, operational quantum mechanics does not violate RELATIVISTIC CAUSALITY, because it does not entail any causal narrative involving quantum states, but simply uses them as computational tools. A more precise formulation of this idea will be given elsewhere [10]."

How QM itself avoids Bell's theorem is via dropping Decorrelating Explanations (which means multiple sample spaces). Bohmian Mechanics does it via dropping Relativistic Causality.

So Wiseman agrees with me that if the wave function is real, QM violates relativistic causality. However, it would seem that there is no single sample space whether or not the wave function is real. So I don't see how not having a single sample space allows relativistic causality to be saved, unless it has something to do with the wave function contributing or not contributing to a causal narrative.

 

[DarMM]

In the context of the Wiseman paper, my term "realism" is used in the sense that if something is designated as contributing to a cause, then it is real.

Is this a common cause or a decorrelating explanation though?

However, relativistic "causality" in the sense of having "causes" is then empty

This is tied into the question above.

So Wiseman agrees with me that if the wave function is real, QM violates relativistic causality.

Yes, if you have a wavefunction as a propogating wave you violate relativistic causality. Such a case is Bohmian Mechanics which as I said violates Relativistic causality and has a single sample space.

However, it would seem that there is no single sample space whether or not the wave function is real.

This is the point of error.

If the wavefunction is real then it defines an outcome and thus your sample space is:
$$\mathcal{H} \times \Sigma$$
with $\Sigma$ being possible additional variables. This is basically what is going on in Bohmian Mechanics.

If the wavefunction is not real, then outcomes are given by POVM elements and probabilities are given by contraction against elements of the dual to the algebra of which wavefunctions are a special case. This is the standard Quantum Formalism. Due to the algebra being noncommutative it cannot be given a common sample space and thus one loses decorrelating explanations and then avoids Bell's theorem.

 

[DarMM]

A toy example from classical probability might help.

Imagine our sample space is some finite set $\Omega$ and probability distributions on it $\rho(\omega)$.
We know $\rho(\omega) \in \mathcal{L}^{1}\left(\Omega\right)$.

If I claim the outcomes/set of events are given by elements or subsets of $\Omega$ my state space is finite, but if I say the probability distributions are the actual outcomes/events then my events are subsets of $\mathcal{L}^{1}\left(\Omega\right)$ which is an infinite-dimensional space. So changing the physical status of $\rho(\omega)$ results in an infinitely larger outcome space with very different properties.

This is essentially an aspect of the relation between standard QM and Bohmian Mechanics. Bohmian Mechanics gains a single sample space which is infinitely larger than any of the multiple ones in QM by promoting $\psi$ from being a (generalized) probability distribution to being an actual outcome.

Nobody has managed to prove such a promotion actually replicates QM completely. And there are no indications such a promotion works at all for QFT.

 

[Jimster41]

Sorry, What is L^1 there?

 

[atyy]

Is this a common cause or a decorrelating explanation though?

This is tied into the question above.

I'm not sure. Instinctively, I would say both a common cause and a decorrelating explanation. Can one have decorrelating explanation without common cause?

This is the point of error.

If the wavefunction is real then it defines an outcome and thus your sample space is:
$$\mathcal{H} \times \Sigma$$
with $\Sigma$ being possible additional variables. This is basically what is going on in Bohmian Mechanics.

If the wavefunction is not real, then outcomes are given by POVM elements and probabilities are given by contraction against elements of the dual to the algebra of which wavefunctions are a special case. This is the standard Quantum Formalism. Due to the algebra being noncommutative it cannot be given a common sample space and thus one loses decorrelating explanations and then avoids Bell's theorem.

Well, perhaps a point of error or maybe just not understanding your terminology. I have not understood till now why you say BM has a single sample space, but QM does not - to me neither have a single sample space in only the variables defined in Fine's theorem. It wasn't clear to me that you were referring to different sample spaces in each theory.

Anyway, if you mean QM rejects decorrelating explanation, then I understand what you mean (and yes, I do agree with Wiseman). I would, however, disagree on terminology. I would say QM is neutral on the rejection of decorrelating explanation, and it is neutral on the general issue of having no single sample space in arbitrary variables. I would say rejecting decorrelating explanation is an ultra-Copenhagenist interpretation, whereas the orthodox-style Copenhagen interpretation is neutral.

 

[Elias1960]

Show me such a construction for QFT.
Even so it is an alternate construction. The mathematics of QM does not involve such things, an alternate formalism incapable of replicating QFT and not known to replicate all of QM (as of 2019) does, but that's not QM.

The construction itself is Nelsonian stochastics. All it needs is to work mathematically is that the energy depends quadratically on the momentum variables. So, the mathematics of Nelsonian stochastics can be taken over to bosonic field theories in the same way as this can be done for Bohmian bosonic field theories, with the formulas given in

Bohm.D., Hiley, B.J., Kaloyerou, P.N. (1987). An ontological basis for the quantum theory, Phys. Reports 144(6), 321-375.

Not for QM they don't. For example, consider the two propositions relating to an electron that has just passed through a Stern-Gerlach device oriented in the $z$ direction and has come out the "up" output:
(1) This electron has $z$ spin up.
(2) This electron has $x$ spin up.
#1 has a well-defined truth value, namely "true". #2 does not have a well-defined truth value at all.

Means, it is not an adequately defined proposition of QM.

Because the field of discourse consists of propositions, what you have to do is to find out the propositions of your theories of interest (the field may contain many, all those discussed in a particular discourse). If among these theories is some theory THV where the electron has a well-defined spin in every direction, when the discourse can contain the proposition "THV holds and this electron has $x$ spin up", and this has a well-defined truth value (namely in this case "false" because such a THV has to be false given the violation of the Bell inequalities.) Once in QM itself there are no such propositions about values not measured, the statement "This electron has $x$ spin up." is simply not part of the discourse of QM. The statements would have to refer to results of measurements. So, a more adequate formulation would be "if passed through a Stern-Gerlach device oriented in the $x$ direction the "up" output would come out".

If you use statements A = "if passed through a Stern-Gerlach device oriented in the $x$ direction the "up" output would come out", B = "if passed through a Stern-Gerlach device oriented in the $y$ direction the "up" output would come out", you already have well-defined truth values and can use standard Boolean algebra operations with them. You cannot test them both, thus, you cannot establish their truth-value by observation, but this is not required. You can gain incomplete knowledge about them and apply the rules of classical probability theory without any problem, you can derive internal contradictions of various sets of propositions and so on.

Therefore no set of propositions that contains both #1 and #2 can define a Boolean algebra. But both propositions are part of the "field of discourse" of QM.

No. This is simply the error of "quantum logic": Inaccurate choices of propositions, using statements which, sloppily, could count as propositions in some interpretations making additional hypotheses but are not propositions in QM, with "and" and "or" operations sloppily defining other such non-propositions, so no wonder that the rules of classical logic are inapplicable.

 

[PeterDonis]

Means, it is not an adequately defined proposition of QM.

Sure it is. I can measure the spin in the $x$ direction.

a more adequate formulation would be "if passed through a Stern-Gerlach device oriented in the $x$ direction the "up" output would come out".

Which doesn't change the fact that this proposition does not have a well-defined truth value for an electron that has just come out of the "up" output of a Stern-Gerlach device oriented in the $z$ direction.

If you use statements A = "if passed through a Stern-Gerlach device oriented in the $x$ direction the "up" output would come out", B = "if passed through a Stern-Gerlach device oriented in the $y$ direction the "up" output would come out", you already have well-defined truth values and can use standard Boolean algebra operations with them.

Not for an electron that has just come out of the "up" output of a Stern-Gerlach device oriented in the $z$ direction. See above.

You cannot test them both, thus, you cannot establish their truth-value by observation, but this is not required.

Not required for what? Not required for QM, sure; but QM does not claim that there is a single Boolean algebra that captures all of these propositions. You are claiming that, so you can't wave your hands and say it's "not required" to test them both. Not being able to test them both is precisely what prevents there from being a single sample space.