A Copenhagen: Restriction on knowledge or restriction on ontology?

[DarMM]

I believe that 1. and 2. are a consequence of 3., and my previous remarks apply to 3.

Yes that is correct, they're all ultimately the same thing in a sense.

 

[DarMM]

There is no need for multiple probability spaces; the only probability space you need is the probability space for those macroscopic configurations.

I've tried to work this out, but I'm not seeing it. Can you give an example with some mathematical details, even sketched not necessarily in full detail.

Like what exactly are the macroscopic degrees of freedom, what sample space do they use and how do you still end up with CHSH violations despite the single sample space?

 

[atyy]

I still think this is too strong. You are using the lack of a solution given to the measurement problem to dismiss any insight from the different probabilistic structure of the theory. The fact of different sample spaces has many implications in Quantum Information, it's not just nothing or a red herring because it doesn't provide a solution to the measurement problem.

Well, it seems to me that the various ways of saying that quantum mechanics is local is just a matter of shunting the issues that are of interest elsewhere---onto the measurement problem, or the single outcome problem.

Is this related?
Eric G. Cavalcanti
https://arxiv.org/abs/1602.07404

 

[zonde]

I've tried to work this out, but I'm not seeing it. Can you give an example with some mathematical details, even sketched not necessarily in full detail.

Like what exactly are the macroscopic degrees of freedom, what sample space do they use and how do you still end up with CHSH violations despite the single sample space?

Have you seen Eberhard's proof of Bell inequalities? It might be the thing you are asking for. It is contained in this paper: https://journals.aps.org/pra/abstract/10.1103/PhysRevA.47.R747. This paper is behind paywall but I have posted the part of Bell inequality proof here: https://www.physicsforums.com/threa...y-on-probability-concept.944672/#post-5977632

 

[Demystifier]

What's local are the interactions

Interactions between what? Between field operators, obviously. But according to the statistical ensemble interpretation, the field operator is a tool to analyze the ensembles of systems, not the individual systems. Hence the local interactions are interactions between the ensembles of systems, not between the individual systems.

So what then are the interactions between the individual systems? The statistical ensemble interpretation of QFT does not tell. But the Bell theorem tells us that, if interactions between individual systems exist at all, then those interactions are nonlocal.

 

[vanhees71]

This is also just personal opinion. My personal opinion is that the interactions described by QFT are local for each individual system. For the same reasons you give for your opinion that cannot be ruled within QFT.

The Bell theorem is about a fictitious deterministic local theory, fullfilling the validity of Bell's inequalities, which is ruled out by tremendous significance and rather verifies the predictions by (local!) QFT. It's also about the probabilistic statements of both the ficticious deterministic local theory and QFT. To Bell's dismay QFT is valid, while any deterministic local theory is ruled out. That's the great achievement by Bell: His work has brought philosophical gibberish a la EPR to a scientific statement that can be tested by experiments, and QFT delivers the correct description but not any deterministic local theory. Whether or not there is a deterministic non-local theory that describes this well-established facts, I cannot say, because there seems to be no such thing yet. It's understandable, because it's very hard to conceive a non-local description that is consistent with the causality structure of special (let alone general) relativity.

Note that above, I mean local/non-local in the sense of interactions!

 

[DarMM]

Have you seen Eberhard's proof of Bell inequalities? It might be the thing you are asking for. It is contained in this paper: https://journals.aps.org/pra/abstract/10.1103/PhysRevA.47.R747. This paper is behind paywall but I have posted the part of Bell inequality proof here: https://www.physicsforums.com/threa...y-on-probability-concept.944672/#post-5977632

Thanks for the link. It's not what I need, but it is very interesting.

 

[stevendaryl]

I've tried to work this out, but I'm not seeing it. Can you give an example with some mathematical details, even sketched not necessarily in full detail.

Okay, here's a sketch of the idea, which isn't all that profound. The minimal interpretation of quantum mechanics basically says that if you set up a system so that it is described by the state $|\psi\rangle$ and you measure some observable $A$ of the system, then you will get an eigenvalue $a$ of the corresponding operator with a probability given by: $P_a = \langle \psi|\Pi_{A,a}|\psi\rangle$, where $\Pi_{A,a}$ is the projection operator, which has eigenvalue 1 on any state in which $A$ has definite value $a$, and has eigenvalue 0 on any state in which $A$ has a definite value other than $a$.

Now, the above prescription has the phrase "you measure some observable $A$". What does that mean? Well, to measure an observable of a system means to set up an interaction between that system and a measuring device so that distinct values of that observable lead to macroscopically different states of the measuring device. In other words, for every possible value $a$ of the microscopic observable, there is a macroscopic configuration $C_a$ of the measuring device, so that the measuring device reliably ends up in configuration $C_a$ whenever the system of interest has value $a$. For now, a macroscopic configuration is basically whatever description of a macroscopic system that one could obtain by inspection: This red light is on. There is a black dot here rather than there. This display shows such and such value. The Geiger counter is clicking.

Now, although it's an enormously difficult to completely nail down the details, I think that most physicists are fairly confident that measuring devices themselves are described by the same quantum mechanics as the systems being studied. In principle, even if intractable in practice, one could do a full-fledged quantum mechanical analysis of the system + measuring device + relevant environment, and one would find something like this:

$|\psi_a\rangle \otimes |start\rangle \Rightarrow |\psi_a\rangle \otimes |C_a\rangle$

where $|\psi_a\rangle$ is the state of the system of interest when it is an eigenstate of $A$ with eigenvalue $a$, and $|start\rangle$ is the initial state of the measuring device + environment, and $|C_a\rangle$ is the state of the measuring device plus environment after the measurement takes place.

I say "something like this" rather than exactly this because the reality is much more complicated. There is no single state of measuring device + environment, there are enormously many microscopically distinguishable states corresponding to any macroscopic description, and a measurement process is an irreversible change, which is hard to describe using quantum mechanics. But all difficulties aside, I think most people are confident that there is nothing going on in a measurement process that isn't in principle describable by quantum mechanics.

So to the extent that my sketch can be accepted as approximately correct, with a large grain of salt, we can ask what happens if you use the same measuring setup to measure the microscopic system when it is not in an eigenstate of $A$. Well, since the evolution equations of quantum mechanics are linear, it follows that a superposition of initial states would lead to a superposition of final states:

$\sum_a \alpha_a |\psi_a\rangle \otimes |start\rangle$
$\Rightarrow \sum_a \alpha_a |\psi_a\rangle \otimes |C_a\rangle$

Then the Born rule saying that there is probability $|\alpha_a|^2$ that the microscopic system will be measured to have eigenvalue $a$ is essentially the same as saying that the measuring device will later be found to be in the configuration $C_a$ with probability $|\alpha_a|^2$. So the Born rule for the microscopic system presumably follows from the Born rule for the measuring device plus the definition of what it means to measure something.So my claim is that the empirical content of the minimal interpretation is equivalent (in principle) to the following recipe:

1. Describe the whole universe (or the part that's relevant) as a quantum system.
2. Let that system evolve according to the usual unitary rule.
3. Then decompose the final state into a superposition of macroscopically distinguishable states, each of which has definite values for all macroscopic properties.
4. Assume that the macroscopic system will be found in exactly one of those states, with a probability given by the square of the corresponding amplitude.

The Born rule for measurements would (I claim) follow from the above recipe, together with the definition of what it means to measure a microscopic quantity.

Mathematically, I think the above recipe could be formulated in terms of projection operators. Presumably any fact about the world that could be verified by observation such as "there is a black spot on the left photographic plate" corresponds to a claim of the form that some coarse-grained observable has a value in some range. Such claims can be formulated in terms of projection operators. So for every such macroscopic statement $c$, there is presumably a corresponding projection operator $\Pi_c$. If the observables are taken to be low-enough precision and coarse-grained enough, then all the corresponding projection operators are approximately commuting. Which means that we talk in terms of macroscopic configurations, which are just maximal collections of compatible macroscopic claims. So we can in principle come up with projection operators $\Pi_j$ on the state of the universe such that for $j \neq k$, the projection operators $\Pi_j$ and $\Pi_k$ correspond to macroscopically distinguishable states of the universe.

Then given an initial state of the universe $|\psi_0\rangle$, the probability that the universe will be in macroscopic configuration $j$ at a later time $t$ would be given by:

$P_j(t) = \langle \psi_0 | e^{+iHt} \Pi_j e^{-iHt} \rangle \equiv \langle \psi_0 |\Pi_j(t)| \psi_0 \rangle$

where $\Pi_j(t)$ is the time-dependent Heisenberg operator corresponding to $\Pi_j$:

$\Pi_j(t) = e^{+iHt}\Pi_j e^{-iHt}$

If we could actually calculate $P_j(t)$, that would give the entire empirical content of quantum mechanics.

 

[stevendaryl]

This is also just personal opinion. My personal opinion is that the interactions described by QFT are local for each individual system. For the same reasons you give for your opinion that cannot be ruled within QFT.

The empirical content of QFT is, like the empirical content of nonrelativistic quantum mechanics, composed of two distinct pieces: (1) You calculate amplitudes for processes using the Schrodinger equation (non-relativistically), or using the S-matrix (relativistically). (2) You square the amplitudes to get probabilities for measurement results. The issue for locality in quantum mechanics is not, and has never been, about (1). It's about (2).

 

[DarMM]

@stevendaryl I'll just need to think for a bit, but how does this differ from consistent histories?

 

[Demystifier]

This is also just personal opinion. My personal opinion is that the interactions described by QFT are local for each individual system. For the same reasons you give for your opinion that cannot be ruled within QFT.

My "personal" opinion is supported by many published papers and books, including the book by Balentine. What published work can be used to support your opinion? Namely the opinion that, in the statistical ensemble interpretation, the interactions are local not only on the ensemble level, but also at the individual one.

 

[stevendaryl]

@stevendaryl I'll just need to think for a bit, but how does this differ from consistent histories?

I think it's basically the same thing. But I don't see it as a different interpretation of quantum mechanics. It's basically the same as the minimal interpretation, but restricted to macroscopic observables. The nice thing about macroscopic observables is that they don't need a second system measuring them, so instead of saying "the observable will be measured to be an eigenvalue with such-and-such probability", you can just say "the observable will have this value with such-and-such probability". Measurement becomes irrelevant. At the cost of explicitly treating macroscopic variables as more special than microscopic variables.

Consistent histories doesn't explicitly make the microscopic/macroscopic distinction. It says you can take any collection of mutually commuting observables and compute the probabilities associated with their history. But I find that a little unsatisfying. What determines which collection is used? Does consistent histories imply a doubly-multiple Many Worlds, where not only are there different possible histories corresponding to different values for a fixed set of observables, but also different possible histories corresponding to different choices of the commuting observables?

 

[DarMM]

Okay that makes sense. Consistent histories still has multiple sample spaces though in the sense of history sets which cannot be combined or reasoned about together. Chapter 25 of Griffiths book has some good examples.

 

[A. Neumaier]

Okay that makes sense. Consistent histories still has multiple sample spaces though in the sense of history sets which cannot be combined or reasoned about together. Chapter 25 of Griffiths book has some good examples.

But why is this relevant? Only one history can be realized. The others don't matter.

If the history includes all repetitions of experiments then the probabilistic aspects must be deducible from the single realized history, just as we learn about empirical probabilities from looking at the history availsble to us.

In classical mechanics, only selective histories are permitted, which makes classical mechanics highly predictive and lack paradox. On the other hand, in quantum mechanics, all histories are permitted, which makes quantum mechanics strictly speaking totally nonpredictive unless you add an external selection criterion for which histories to permit.

 

[DarMM]

But why is this relevant? Only one history can be realized.

Of course, but it doesn't remove that mathematical structure from the theory, the counterfactual indefiniteness. Even Consistent History authors have it as a major feature of the theory.

The point is that even if only one history occurs it cannot be considered as a history where for spin measurements the whole vector $(S_x, S_y, S_z)$ occur, it will only contain a $S_z$ event say. Which is what would make QM different from a theory where the world was truly random but driven by a classical stochastic process.

 

[stevendaryl]

Of course, but it doesn't remove that mathematical structure from the theory, the counterfactual indefiniteness. Even Consistent History authors have it as a major feature of the theory.

The point is that even if only one history occurs it cannot be considered as a history where for spin measurements the whole vector $(S_x, S_y, S_z)$ occur, it will only contain a $S_z$ event say. Which is what would make QM different from a theory where the world was truly random but driven by a classical stochastic process.

It's a stochastic theory for a commuting subset of observables. As I sketched, if you have an enumeration of all possible macroscopic configurations, then quantum mechanics gives you a probability $P_j(t)$ of the universe being in state $j$ at time $t$. The macroscopic configuration doesn't say anything about microscopic observables such as the components of spin of individual electrons (except to the extent that those can be inferred from macrosopic information).

 

[DarMM]

It's a stochastic theory for a commuting subset of observables. As I sketched, if you have an enumeration of all possible macroscopic configurations, then quantum mechanics gives you a probability $P_j(t)$ of the universe being in state $j$ at time $t$. The macroscopic configuration doesn't say anything about microscopic observables such as the components of spin of individual electrons (except to the extent that those can be inferred from macrosopic information).

Yes, but you have multiple stochastic theories for each set of commuting macroobservables, unlike Classical Probability theory. You don't have a single space for all the macroscopic configurations, you can only form sample spaces for mutually commuting ones.

Would it be right to say that what you are getting at is that only $S_z$ is amplified up to the macroscopic level and thus only it constitutes a macroscopic outcome, i.e. if you take QM as a theory of stochastic macroscopic observables the lack of a common sample space arises purely from the fact that only $S_z$ "rises up".

 

[stevendaryl]

Yes, but you have multiple stochastic theories for each set of commuting macroobservables, unlike Classical Probability theory.

But macroscopic variables all commute (at least approximately). You can't know the position and momentum of an electron at the same time, but you know the approximate position and approximate momentum of a baseball at the same time.

Would it be right to say that what you are getting at is that only $S_z$ is amplified up to the macroscopic level and thus only it constitutes a macroscopic outcome, i.e. if you take QM as a theory of stochastic macroscopic observables the lack of a common sample space arises purely from the fact that only $S_z$ "rises up".

If $S_z$ is measured, then its value becomes part of the macroscopic configuration. If it isn't measured, then it isn't a part of the macroscopic configuration. Microscopic variables are involved in computing the macroscopic probabilities, $P_j(t)$, but they aren't assumed to have values (or associated probability distributions). Probability distributions only apply to values of macroscopic observables, and they are all approximately commuting.

 

[DarMM]

But macroscopic variables all commute (at least approximately).

Yes, but this is essentially answered by the next part of your post. The macroscopic observables in a history where $S_z$ becomes part of the macroscopic configuration don't commute with the macroscopic observables where $S_x$. This of course is really just a statement that there is no such thing as a history where $(S_x,S_y,S_z)$ becomes part of the macroscopic observables.

However this doesn't really invalidate what you are saying. Of which I will say more in a second, I just want to check that you agree with the paragraph above.

 

[stevendaryl]

Yes, but this is essentially answered by the next part of your post. The macroscopic observables in a history where $S_z$ becomes part of the macroscopic configuration don't commute with the macroscopic observables where $S_x$. This of course is really just a statement that there is no such thing as a history where $(S_x,S_y,S_z)$ becomes part of the macroscopic observables.

That's true.

There is no joint probability distribution describing $S_x$ and $S_y$. However, if you lift it to the macroscopic level, there can be a joint probability distribution describing the two situations: "I measured $S_x$ and found it to be $+\frac{1}{2}$" and "I measured $S_y$ and found it to be $+\frac{1}{2}$". Those two situations are exclusive: if one is true, the other is false. This is in contrast to the microscopic situation, where it is not possible to give simultaneous truth values to $S_x = + \frac{1}{2}$ and $S_y = + \frac{1}{2}$.

 

[bhobba]

Quantum mechanics does not unsettle me.

Me either. Nor does classical mechanics which we now know depends crucially on QM. Things progress and people always want to lift the veil as see what future research will reveal or participate in that research. The thing is this one has proven HARD and progress is slow.

Thanks
Bill

 

[DarMM]

That's true.

There is no joint probability distribution describing $S_x$ and $S_y$. However, if you lift it to the macroscopic level, there can be a joint probability distribution describing the two situations: "I measured $S_x$ and found it to be $+\frac{1}{2}$" and "I measured $S_y$ and found it to be $+\frac{1}{2}$". Those two situations are exclusive: if one is true, the other is false. This is in contrast to the microscopic situation, where it is not possible to give simultaneous truth values to $S_x = + \frac{1}{2}$ and $S_y = + \frac{1}{2}$.

There isn't a joint probability distribution for them as the two events belong to two incompatible histories. They are exclusive correct, but the rest of the properties of Boolean logic are not well defined for them and thus without a common Boolean lattice you cannot build a probability distribution for both.

 

[stevendaryl]

There isn't a joint probability distribution for them as the two events belong to two incompatible histories. They are exclusive correct, but the rest of the properties of Boolean logic are not well defined for them and thus without a common Boolean lattice you cannot build a probability distribution for both.

They belong to different histories, but not incompatible histories.

Look, if I flip a coin and if it's heads, I go left and if it's tails I go right, then there are two possible histories: one in which I flip the coin and go left and one in which I flip the coin and go right. Those are alternative, but they are not incompatible in the sense of probability distributions. I can give a probability of 50% to each.

This is in contrast to the two histories (1) an electron has spin-up in the x-direction, and (2) an electron has spin-up in the y-direction. Those two possibilities cannot be given consistent probabilities.

 

[DarMM]

To construct a probability distribution for both you need a Boolean lattice. Where are $A \land B$ and $A \lor B$ for those two events?

 

[Josh0768]

To my mind, the spirit of the Copenhagen interpretation refers to the ontology. As Heisenberg puts it:

"However, all the opponents of the Copenhagen interpretation do agree on one point. It would, in their view, be desirable to return to the reality concept of classical physics or, to use a more general philosophic term, to the ontology of materialism. They would prefer to come back to the idea of an objective real world whose smallest parts exist objectively in the same sense as stones or trees exist, independently of whether or not we observe them."

In case one gives up the concept of 'physical realism' - a viewpoint according to which an external reality exists independent of observation – and doesn’t insist on thinking about quantum phenomena with classical ideas, quantum mechanics doesn’t unsettle anymore.

Hey, could you tell me what book/paper that quote is from?

 

[stevendaryl]

To construct a probability distribution for both you need a Boolean lattice. Where are $A \land B$ and $A \lor B$ for those two events?

$A \wedge B$ is the impossible event. You can't measure both spin in the x-direction and spin in the y-direction.

Let's make a complete list of possible histories for such an experiment:

1. We measure the spin in the x-direction, and find it is $+ \frac{1}{2}$
2. We measure the spin in the x-direction, and find it is $- \frac{1}{2}$
3. We measure the spin in the y-direction, and find it is $+ \frac{1}{2}$
4. We measure the spin in the y-direction, and find it is $- \frac{1}{2}$

I don't understand what problem you see in giving a probability distribution to those 4 possibilities. For example, give all 4 the probability of 25%. They are exclusive events, in that the intersection of any two of them is the impossible event.

 

[DarMM]

No problem thus far. Just a quick question first. The probabilities you have there are you imagining something like having a system in a $|z = +\frac{1}{2}\rangle$ state and then flipping a coin to see if you measure $S_x$ and $S_y$?

 

[stevendaryl]

No problem thus far. Just a quick question first. The probabilities you have there are you imagining something like having a system in a $|z = +\frac{1}{2}\rangle$ state and then flipping a coin to see if you measure $S_x$ and $S_y$?

Yes.

I took a peak at the Wikipedia article on consistent histories (https://en.wikipedia.org/wiki/Consistent_histories), but I'm not sure that I understood it. The criterion given there was that

$Tr(C_{H_i} \rho C_{H_j}^\dagger) = 0$ whenever $i \neq j$, where $C_{H_i}$ is the product of the Heisenberg operators corresponding to the projection operators in history $i$, and $\rho$ is the initial density matrix.

In the simplest case, we have one-event histories, in which case $C_{H_i}$ is just the projection operator corresponding to the 4 possibilities I gave above.

So let $|\psi_j\rangle \langle \psi_j|$ be the projection operator corresponding to possibility $j$. Let $|\psi_0\rangle \langle \psi_0|$ be the initial density matrix (assumed to be a pure state for simplicity). Then

$C_{H_i} \rho C_{H_j}^\dagger = |\psi_i\rangle \langle \psi_i|U|\psi_0\rangle \langle \psi_0|U^{-1}|\psi_j\rangle \langle \psi_j|$

where $U$ is the evolution operator. Taking the trace gives:

$\langle \psi_j | \psi_i \rangle \langle \psi_i | U |\psi_0\rangle \langle \psi_0|U|\psi_j \rangle$

So if the matrix element $\langle \psi_j | \psi_i \rangle$ is zero, then the two histories are consistent, according to Wikipedia's criterion.

 

[DarMM]

Okay so then if you are measuring a spin-1 particle you will have nine possible macrohistories or if you are measuring two spin-$\frac{1}{2}$ particles (and keep to only measuring $S_x$ and $S_y$) you will have sixteen possible macrohistories.

Then basically there will be some states where the probabilities they give for those histories cannot result from a common measure you place over the whole set of nine/sixteen.

 

[A. Neumaier]

if you are measuring two spin-2 particles (and keep to only measuring $S_x$ and $S_y$) you will have sixteen possible macrohistories.

25, not 16?

 

[DarMM]

25, not 16?

Wrong spin! Don't want to force poor @stevendaryl to detect gravitons!

 

[stevendaryl]

Okay so then if you are measuring a spin-1 particle you will have nine possible macrohistories or if you are measuring two spin-$\frac{1}{2}$ particles (and keep to only measuring $S_x$ and $S_y$) you will have sixteen possible macrohistories.

I don't think it needs to be that complicated. I was thinking of a single particle initially spin-up in the z-direction. I flip a coin and either measure its spin in the x-direction, or in the y-direction. So there are 4 possible final states: $|X,up\rangle, |X,down\rangle, |Y,up\rangle, |Y, down\rangle$.

Then basically there will be some states where the probabilities they give for those histories cannot result from a common measure you place over the whole set of nine/sixteen.

I don't think that's true. If two possibilities are exclusive (they can't both happen), then they are consistent. It's only when two different possibilities have a nonzero overlap that there is an incompatibility. There is no overlap between "I measured spin in the x-direction and got spin-up" and "I measured spin in the y-direction and got spin-up".

 

[DarMM]

I don't think it needs to be that complicated.

It does. I've no issue with your original example for in that case one can construct an overall probability measure. However for measurements on any quantum system with Hilbert space dimension $d \geq 3$ what you are describing will not work. I simply listed $d = 3,4$ examples.

 

[DarMM]

Is this related?
Eric G. Cavalcanti
https://arxiv.org/abs/1602.07404

Just need time to absorb this, perhaps a day or two.

 

[stevendaryl]

It does. I've no issue with your original example for in that case one can construct an overall probability measure. However for measurements on any quantum system with Hilbert space dimension $d \geq 3$ what you are describing will not work. I simply listed $d = 3,4$ examples.

Give a concrete example of what you're talking about. As I understand it (which I admit I only do superficially), histories that are mutually exclusive are always compatible. Histories with different macroscopic details are mutually exclusive, so they are always compatible.

 

[DarMM]

Give a concrete example of what you're talking about. As I understand it (which I admit I only do superficially), histories that are mutually exclusive are always compatible. Histories with different macroscopic details are mutually exclusive, so they are always compatible.

For the $d = 4$ case we can take a pair of spin-$\frac{1}{2}$ particles and relate this back to entanglement.
The observables on one particle are $S_0, S_{\frac{\pi}{4}}$ and the observables on the other $S_{\frac{\pi}{2}}, S_{\frac{3\pi}{4}}$.

There are then four possible measurements and each measurement can have four possible outcomes.

So we would have histories like:
"I measured $S_0 = \frac{1}{2}$ on the first particle and $S_{\frac{\pi}{2}} = \frac{1}{2}$ on the second"

If you look at the probabilities the Bell state gives to these histories and you look at all 16 histories you'll see it isn't consistent with them being drawn from a single probability distribution.

This applies also outside of entanglement in the $d = 3$ case.

 

[DarMM]

Chapter 19 of Griffiths book demonstrates the difference between the macroscopic histories of a stochastic (in the Kolmogorov sense with a single sample space) world and a quantum mechanical world.

Particularly see 19.3 and 19.4 which directly discusses the counterfactual differences and the remarks on p.271

 

[stevendaryl]

For the $d = 4$ case we can take a pair of spin-$\frac{1}{2}$ particles and relate this back to entanglement.
The observables on one particle are $S_0, S_{\frac{\pi}{4}}$ and the observables on the other $S_{\frac{\pi}{2}}, S_{\frac{3\pi}{4}}$.

There are then four possible measurements and each measurement can have four possible outcomes.

So we would have histories like:
"I measured $S_0 = \frac{1}{2}$ on the first particle and $S_{\frac{\pi}{2}} = \frac{1}{2}$ on the second"

If you look at the probabilities the Bell state gives to these histories and you look at all 16 histories you'll see it isn't consistent with them being drawn from a single probability distribution.

This applies also outside of entanglement in the $d = 3$ case.

I would have to see this worked out. What you're saying doesn't seem correct to me.

We have 16 possibilities, which we can characterize by 4 numbers: $A_j, a_j, B_j, b_j$ where $A_j$ is the $j^{th}$ measurement on the first particle, $a_j$ is the $j^{th}$ result of that measurement (either "up" or "down"), $B_j$ is the $j^{th}$ measurement on the second particle, and $b_j$ is the $j^{th}$ result of that measurement. For maximally entangled anti-correlated spin-1/2 particles, if you measure one particle along axis $\overrightarrow{A}$ and the other along axis $\overrightarrow{B}$, then the probability that the results will be the same is $sin^2{\frac{\theta}{2}}$, where $\theta$ is the angle between the two axes. I'm not sure why you chose those particular orientations, but the probabilities work out to:

$$ \left[ \begin{array} \\ A & a & B & b & P(a,b|A,B)
\\ 0 & up & \frac{\pi}{2} & up & 0.25
\\ 0 & up & \frac{\pi}{2} & down & 0.25
\\ 0 & down & \frac{\pi}{2} & up & 0.25
\\ 0 & down & \frac{\pi}{2} & down & 0.25
\\ 0 & up & \frac{3\pi}{4} & up & 0.43
\\ 0 & up & \frac{3\pi}{4} & down & 0.07
\\ 0 & down & \frac{3\pi}{4} & up & 0.07
\\ 0 & down & \frac{3\pi}{4} & down & 0.43
\\ \frac{\pi}{4} & up & \frac{\pi}{2} & up & 0.07
\\ \frac{\pi}{4} & up & \frac{\pi}{2} & down & 0.43
\\ \frac{\pi}{4} & down & \frac{\pi}{2} & up & 0.43
\\ \frac{\pi}{4} & down & \frac{\pi}{2} & down & 0.07
\\ \frac{\pi}{4} & up & \frac{3\pi}{4} & up & 0.25
\\ \frac{\pi}{4} & up & \frac{3\pi}{4} & down & 0.25
\\ \frac{\pi}{4} & down & \frac{3\pi}{4} & up & 0.25
\\ \frac{\pi}{4} & down & \frac{3\pi}{4} & down & 0.25
\end{array} \right] $$

If we give a 25% probability for all 4 choices of which two measurements to perform, then you just multiply every conditional probability above by 0.25. What's wrong with considering that to be a probability space with 16 possibilities?

 

[DarMM]

Okay now take the observable $B_2$. Since this is a single sample space attempting to replicate the statistics, what value does $B_2$ take in each outcome such that its moments give you results replicating QM?

 

[stevendaryl]

I don't know what you mean. What are you saying is impossible?

The usual argument about Bell's inequality is not that it is impossible for a stochastic model to reproduce the statistics, but that it is impossible for a local model to reproduce the statistics. There is no claim that the stochastic model is local in the sense of Bell.

 

[DarMM]

I don't know what you mean. What are you saying is impossible?

The usual argument about Bell's inequality is not that it is impossible for a stochastic model to reproduce the statistics, but that it is impossible for a local model to reproduce the statistics. There is no claim that the stochastic model is local in the sense of Bell.

Much simpler than that. How do you define $B_2$ on each outcome so that it is a random variable on this sample space. Ignore Bell's theorem as such for now.

 

[stevendaryl]

Much simpler than that. How do you define $B_2$ on each outcome so that it is a random variable on this sample space. Ignore Bell's theorem as such for now.

I don't understand your question. There are 16 possible events. Each event has 4 components: $A, a, B, b$. You give a probability to each of the 16 combinations so that the probabilities add up to 1.

 

[DarMM]

I get that. How do you define $B_2$ as a random variable, i.e. a function on this space. So that you can compute things like $\langle B_2 \rangle$?

 

[stevendaryl]

I get that. How do you define $B_2$ as a random variable, i.e. a function on this space. So that you can compute things like $\langle B_2 \rangle$?

I didn't make any claims about $B_2$ as a random variable. I don't know what that means. There is one random variable, which is which event out of 16 possibilities. It can take values 1 through 16. If the value is 1, then in that history, the setting of $A$ is $0$, the setting of $B$ is $\frac{\pi}{2}$, the outcome $a$ is $up$, the outcome $b$ is $up$.

 

[stevendaryl]

I didn't make any claims about $B_2$ as a random variable. I don't know what that means. There is one random variable, which is which event out of 16 possibilities. It can take values 1 through 16. If the value is 1, then in that history, the setting of $A$ is $0$, the setting of $B$ is $\frac{\pi}{2}$, the outcome $a$ is $up$, the outcome $b$ is $up$.

You can certainly look at the subset of histories in which $B = \frac{3\pi}{4}$. For those histories, you can compute the expectation value of the result $b$.

 

[DarMM]

I didn't make any claims about $B_2$ as a random variable. I don't know what that means.

Really? $B_2$ is an observable quantitiy and will be encoded in a probability model as a random variable. Thus if you are using Kolmogorov probability $B_2$, $B_{\frac{3\pi}{4}}$ as we are calling it, has to be a random variable on that space. That's just how probability theory works.

You can certainly look at the subset of histories in which $B = \frac{3\pi}{4}$. For those histories, you can compute the expectation value of the result $b$.

Exactly. There is only a subset on which $B_{\frac{3\pi}{4}}$ is defined as a random variable. Thus in truth these are four separate Kolmogorov models with their own individual sample spaces.

 

[stevendaryl]

Really? $B_2$ is an observable quantitiy and will be encoded in a probability model as a random variable.

Okay, suppose that I take 16 cards. On each index card, I write down 4 quantities: $(A, a, B, b)$ where $A$ is either 0 or $\frac{\pi}{4}$ and $a$ is either $up$ or $down$ and $B$ is either $\frac{\pi}{2}$ or $\frac{3\pi}{4}$ and $b$ is again either $up$ or $down$. So I play lots and lots of rounds where I pick a card according to some probability distribution. Tell me in that situation what it means to say that $B_2$ is a random variable.

The random variable is which card I choose. There are 16 cards. The values of $a$ and $b$ and $A$ and $B$ are functions of this one random variable.

 

[DarMM]

Tell me in that situation what it means to say that $B_2$ is a random variable.

What you're missing here is that $B_2$ is a physical quantity. It's something you can observe. In a probability model it should therefore be a random variable. Note though what you are saying isn't entirely wrong. See the next the question below, this might serve the discussion better.

The usual argument about Bell's inequality is not that it is impossible for a stochastic model to reproduce the statistics, but that it is impossible for a local model to reproduce the statistics. There is no claim that the stochastic model is local in the sense of Bell.

Where is the nonlocality in your table?

 

[stevendaryl]

What you're missing here is that $B_2$ is a physical quantity.

In my card game, what corresponds to $B_2$?

 

[DarMM]

In my card game, what corresponds to $B_2$?

A value on the card I assume. Let's just focus on the nonlocality. In your table how does it manifest?