A computational model of Bell correlations

[entropy1]

What is the "speed of information"?

If information is available at A, then, to make it non-trivially available at B, that can't be done faster than lightspeed. Information in this case being knowledge about a situation in A.

 

[PeterDonis]

If information is available at A, then, to make it non-trivially available at B, that can't be done faster than lightspeed. Information in this case being knowledge about a situation in A.

None of this helps any. You still haven't rigorously defined what you mean by "information" (defining it as "knowledge" doesn't help because that term is just as vague as "information"), or rigorously explained why it has to "travel" somewhere. I know it seems intuitively obvious to you, but this is one of those cases where "intuitively obvious" doesn't cut it. You need to actually try to rigorously formulate your requirement in a way that can be mathematically analyzed and used to make predictions. (That's what Bell was trying to do with his definition of "locality".)

 

[Zafa Pi]

None of this helps any. You still haven't rigorously defined what you mean by "information" (defining it as "knowledge" doesn't help because that term is just as vague as "information"), or rigorously explained why it has to "travel" somewhere. I know it seems intuitively obvious to you, but this is one of those cases where "intuitively obvious" doesn't cut it. You need to actually try to rigorously formulate your requirement in a way that can be mathematically analyzed and used to make predictions. (That's what Bell was trying to do with his definition of "locality".)

Why not use your definition?

That is why it is said that no information can be transmitted in this manner--information transmission requires the "source" to be able to affect the "receiver" in some controllable way.

All that is needed to transmit is a pair of bits from Alice to Bob at FTL in order to conspire and get any correlations you want. Too bad it can't be done.

 

[Zafa Pi]

Why are you assuming that correlations are something that must "happen"? We don't know that. All we know are the correlations we calculate after the fact.

I may not be reading you right, but the correlations have been taking place in extreme abundance in quantum optics labs since 1981. That is why they must "happen".
I see it as the same as why apples must fall.

 

[entropy1]

None of this helps any. You still haven't rigorously defined what you mean by "information"

Perhaps that what travels at speed c or slower, which is practically everything.

And "traveling": Information X "travels" from A to B if B produces a record with meaning P if and only if A has produced record X with meaning P.

 

[PeterDonis]

the correlations have been taking place

Why must correlations be things that "take place"? As I said, we don't know that they "take place". All we know is that we can calculate them after the fact.

What does "take place" in quantum optics labs is that various preparation procedures are done, various measurements are made, and various results are recorded. Where in that do "correlations take place"?

This is the problem with using ordinary language instead of math. Ordinary language is vague, and it often includes implicit assumptions that are based on our classical intuitions and which are not valid--or at least not valid as assumptions--when quantum phenomena are involved.

Perhaps that what travels at speed c or slower, which is practically everything.

If "information" is defined as "what travels at speed c or slower", then by definition, information cannot travel at FTL. Problem solved. Are you satisfied with that?

Information X "travels" from A to B if B produces a record with meaning P if and only if A has produced record X with meaning P.

If A and B are spacelike separated, how do you know which one happened first? Your "if and only if" statement requires an invariant ordering: A has to come before B. But if A and B are spacelike separated, there is no invariant ordering.

 

[Zafa Pi]

What does "take place" in quantum optics labs is that various preparation procedures are done, various measurements are made, and various results are recorded. Where in that do "correlations take place"?

Smack-dab in the recorded results.

You strike me as a smart and educated guy so I believe you are thinking of something I am not picking up on.
Consider this example:
Experiment 1) trial consists of A preforming X and getting a 1 or 0, and B preforming Y and getting 1 or 0. After many runs the results look like flips of independent fair coins.
Experiment 2) trial is like above, but the coins are not independent, and A and B's results are always the same.

In 1) the results are uncorrelated (have correlation coefficient 0).
In 2) we find that the results are highly correlated (have correlation coefficient 1).
In 2) we find correlation. In measuring entangled particles we find correlations (depending on the axes of measurement).
Does my criterion for "find" make sense?

 

[PeterDonis]

Smack-dab in the recorded results.

I'm sorry, you still aren't grasping my point. See below.

You strike me as a smart and educated guy so I believe you are thinking of something I am not picking up on.

Indeed. The correlations are in the recorded results, yes. But the correlations in the recorded results are things that we extract from those results by analyzing them. That does not in any way mean the correlations "take place". The correlations are just mathematical calculations we make. At least, that's all that follows from your statement that the correlations are in the recorded results. So if you want to convince me that the correlations are something that had to "take place", you need to make some other argument.

Does my criterion for "find" make sense?

Sure, but it just means what I said above: we "find" the correlations by calculating them. It does not mean the correlations had to "take place". Something took place, yes. But that doesn't mean that something is properly described as "the correlations taking place".

 

[Zafa Pi]

So if you want to convince me that the correlations are something that had to "take place", you need to make some other argument.

I had a friend look at our discussion and I wrote 6 sentences to see if she could tell which ones you would find acceptable. She, and I were not sure.
At any rate I hope you agree that coral invasions take place.

 

[zonde]

Zafa Pi, I think you are making category error. Correlation is not a physical mechanism, instead physical mechanisms manifest themselves as correlations in data.

 

[Boing3000]

At any rate I hope you agree that coral invasions take place.

That coral invasion is in your past light cone.

As you describe the setup, the correlation take place some time/where between the start of the experiment, and the collection and transport of "information" for correlation checking where it is finally "placed".

That's a bit on the vague side of things.

 

[Zafa Pi]

Correlation is not a physical mechanism, instead physical mechanisms manifest themselves as correlations in data.

I agree, but so what?
I find birds chirping in the forest. I find beauty in the forest.
Two perfectly reasonable sentences in spite of birds chirping and beauty being different categories.
"A disagreement has taken place between A and B. " is also a valid sentence. Halfway between?

 

[zonde]

I find birds chirping in the forest. I find beauty in the forest.
Two perfectly reasonable sentences in spite of birds chirping and beauty being different categories.

That's because "find in" in two sentences have different meanings.
Let me rewrite your sentences like that:
I find that birds chirping takes place in the forest. I find that beauty takes place in the forest.
Now you can see that second sentence has lost it's meaning.

 

[entropy1]

Sure, but it just means what I said above: we "find" the correlations by calculating them. It does not mean the correlations had to "take place". Something took place, yes. But that doesn't mean that something is properly described as "the correlations taking place".

instead physical mechanisms manifest themselves as correlations in data.

The correlation is the result of dependence of the stochastic variables representing the local measurement outcomes (ie A and B). Could we speak of this "dependence" "happening"? In the sense that it "manifests itself"? Because dependence suggests a physical cause?

 

[PeterDonis]

At any rate I hope you agree that coral invasions take place.

Sure. But if we observed coral invasions taking place in two widely separated locations and calculated correlations between them (comparing how much distance the coral invaded per unit time in each place, for example), would you say those calculated correlations "take place"?

 

[entropy1]

You need to actually try to rigorously formulate your requirement in a way that can be mathematically analyzed and used to make predictions. (That's what Bell was trying to do with his definition of "locality".)

If I am not mistaking, Bell's theorem deals with models of local hidden variables. But does it deal with non-locality?

 

[PeterDonis]

Bell's theorem deals with models of local hidden variables

Bell's Theorem defines "locality" as follows: the function that describes the correlation between the measurement results factorizes, i.e., it is the product of two functions, each of which depends on the measurement settings at only one of the two measurements. It includes hidden variables in the mathematical expressions it uses, but they are integrated over and do not play any essential part in the proof of the theorem. The key condition is factorizability.

does it deal with non-locality?

In the sense that "non-locality" means the locality condition above is violated (as it is in QM).

 

[entropy1]

Bell's Theorem defines "locality" as follows: the function that describes the correlation between the measurement results factorizes, i.e., it is the product of two functions, each of which depends on the measurement settings at only one of the two measurements.

So does that correspond with:

1. At A, we know if the particle is detected;
2. At A, we know the angle of the detector;
3. At B, we know if the particle is detected;
   
4. At B, we know the angle of the detector;
   
5. But the correlation is a function of the difference of both angles.

so to establish the correlation, if we adopt realism and take the outcome of the measurement at the instant it is measured, we need the difference between both angles, so it can't be a factorizable function?

 

[Zafa Pi]

That's because "find in" in two sentences have different meanings.
Let me rewrite your sentences like that:
I find that birds chirping takes place in the forest. I find that beauty takes place in the forest.
Now you can see that second sentence has lost it's meaning.

Sorry, I find it poetic and embued with meaning. Perhaps you've missed your calling.
Also the sentence "A disagreement has taken place between A and B. " is common place, unambiguous, with the same categories as the one above. Now perhaps you and @PeterDonis might want to ask where it takes place. Halfway between? Could the ratio of distances between be irrational? I have no good answers for such questions.

I hope you don't accuse me of being anti-Semantic, but I would prefer to check out of this topic and return to the happy place of QM.

 

[Jilang]

Bell's Theorem defines "locality" as follows: the function that describes the correlation between the measurement results factorizes, i.e., it is the product of two functions, each of which depends on the measurement settings at only one of the two measurements. It includes hidden variables in the mathematical expressions it uses, but they are integrated over and do not play any essential part in the proof of the theorem. The key condition is factorizability.

I am unsure why one would assume only one of the settings counts. Would not both play a part?

In the sense that "non-locality" means the locality condition above is violated (as it is in QM).

 

[PeterDonis]

does that correspond with

I think you're making this more difficult than it needs to be. Have you read Bell's paper?

Here is Bell's line of reasoning:

We have a measurement of particle A and a measurement of particle B. These measurements take place at spacelike separated events.

At each measurement, we have settings, call them $a$ and $b$.

If we do a large number of runs, we can construct a function $E(a, b)$ which gives the degree of correlation between the results as a function of the measurement settings.

Bell's locality condition is then that this function can be factorized: $E(a, b) = F(a) G(b)$. Notice that each factor is a function of only one of the two measurement settings. He calls this condition "locality" because it appears to capture our intuitive notion that the results at one measurement should not depend on the settings at the other measurement, since they are spacelike separated. However, in the end it's just a mathematical condition; whether or not you think "locality" is a proper name for it is a matter of words, not physics or math.

Bell then shows that if the function $E(a, b)$ factorizes in this way, it must obey certain inequalities.

We know, however, that the function $E(a, b)$ which is predicted by quantum mechanics violates these inequalities.

Therefore, the function $E(a, b)$ which is predicted by quantum mechanics cannot factorize in the way described above, which means it violates Bell's locality condition. Therefore, quantum mechanics is "nonlocal" in this sense.

 

[PeterDonis]

I am unsure why one would assume only one of the settings counts. Would not both play a part?

See my response to @entropy1 just now.

 

[entropy1]

@PeterDonis
I think we agree. Basicly we mean the same thing. Thanks.

 

[rubi]

Bell's locality condition is then that this function can be factorized: $E(a, b) = F(a) G(b)$.

Check the proof again. That's not Bell's locality condition. The locality condition is that $P$ factorizes, if conditioned on the hidden variable $\lambda$: $P(a,b|\lambda) = P(a|\lambda)P(b|\lambda)$. The assumption of a conditioning on $\lambda$ is a crucial part of the proof. In general, $E(a,b)$ doesn't even factorize in perfectly local classical scenarios such as Bertlmann's socks. QM violates the assumptions of Bell's theorem by not allowing for a conditioning on $\lambda$.

 

[PeterDonis]

The locality condition is that $P$ factorizes, if conditioned on the hidden variable $\lambda$: $P(a,b|\lambda) = P(a|\lambda)P(b|\lambda)$.

Yes, you're right; $\lambda$ is integrated over, but the factorization occurs inside the integral.

 

[rubi]

Yes, you're right; $\lambda$ is integrated over, but the factorization occurs inside the integral.

Right. I just wanted to point out that knowing the correlation $E(a,b)$ isn't enough to conclude anything about locality. One must also assume a specific class of underlying models that explain how $E(a,b)$ comes about. This class of models is given by the introduction of some arbitrary variables $\lambda$ and the integral you mentioned. Bell concludes that any such model must be non-local. However, he can't conclude something about models that don't belong to this class. In QM, the function $E(a,b)$ isn't given by an integral over some variables $\lambda$, but rather by the expectation value of a product of operators. One could conclude something about locality in general, if all models would belong to this class. However, QM is a counterexample that shows that Bell's class of models isn't completely general.

 

[Jilang]

I understand that the probabilities don’t factorise, but I am curious as to whether the probability amplitudes do. Is there a consensus on that?

 

[zonde]

Right. I just wanted to point out that knowing the correlation $E(a,b)$ isn't enough to conclude anything about locality. One must also assume a specific class of underlying models that explain how $E(a,b)$ comes about. This class of models is given by the introduction of some arbitrary variables $\lambda$ and the integral you mentioned. Bell concludes that any such model must be non-local. However, he can't conclude something about models that don't belong to this class. In QM, the function $E(a,b)$ isn't given by an integral over some variables $\lambda$, but rather by the expectation value of a product of operators. One could conclude something about locality in general, if all models would belong to this class. However, QM is a counterexample that shows that Bell's class of models isn't completely general.

Your argument that QM can be considered as a counterexample relies on subjective idea that QM actually explains how $E(a,b)$ comes about.
But certainly there are a lot of people who would consider only spacetime events based model as an explanation of $E(a,b)$.

 

[PeterDonis]

Bell concludes that any such model must be non-local.

More precisely, he concludes that any such model that can reproduce the predictions of QM must be non-local.

QM is a counterexample that shows that Bell's class of models isn't completely general

I think this depends on what interpretation of QM one adopts. According to the "shut up and calculate" interpretation, yes, QM does not belong to Bell's class of models. But in the de Broglie-Bohm interpretation, for example, it does--the hidden variables are the particle positions. But because the particle motions are affected by the quantum potential, they are nonlocal, so the function inside the integral over the hidden variables does not factorize, and the Bell inequalities are violated. In this interpretation, of course, the standard formalism of QM that uses operators and expectation values is not fundamental; it's emergent from the underlying model.

 

[rubi]

Your argument that QM can be considered as a counterexample relies on subjective idea that QM actually explains how $E(a,b)$ comes about.
But certainly there are a lot of people who would consider only spacetime events based model as an explanation of $E(a,b)$.

It has nothing to do with spacetime events. Bell's class of models doesn't mention spacetime events either and in QM, the detections are of course events in spacetime too. The question is only whether the detected values at each spacetime event arise as functions of a hidden variable or not.

More precisely, he concludes that any such model that can reproduce the predictions of QM must be non-local.

Yes, you're right of course.

I think this depends on what interpretation of QM one adopts. According to the "shut up and calculate" interpretation, yes, QM does not belong to Bell's class of models. But in the de Broglie-Bohm interpretation, for example, it does--the hidden variables are the particle positions. But because the particle motions are affected by the quantum potential, they are nonlocal, so the function inside the integral over the hidden variables does not factorize, and the Bell inequalities are violated. In this interpretation, of course, the standard formalism of QM that uses operators and expectation values is not fundamental; it's emergent from the underlying model.

That's also correct, but most people (even Einstein) don't consider dBB to be a serious solution. It's toy model and it serves it purpose as such, but it has way more conceptual problems than it solves and it doesn't generalize to QFT, so it is essentially falsified already as long as the Bohmians don't fix it. My point is that there exist models that explain the correlations without introducing hidden variables and since such models exist, we know that Bell's class of models isn't the most general class. Hence, the violation of the inequality doesn't imply anything about locality unless we accept extra assumptions.

 

[stevendaryl]

I understand that the probabilities don’t factorise, but I am curious as to whether the probability amplitudes do. Is there a consensus on that?

I posted a note about this a year or so ago. Here's the summary:

In a spin-1/2 anti-correlated EPR experiment, the joint probability that Alice will measure spin-up at angle $\alpha$ and Bob will measure spin-up at angle $\beta$ is given by:

$P(A \hat B | \alpha, \beta) = \frac{1}{2} sin^2(\frac{\beta - \alpha}{2})$

According to Bell, to explain this via local hidden variables would require a parameter $\lambda$ and three probability distributions:

• $P(\lambda)$
• $P(A | \alpha, \lambda)$
• $P(B | \beta, \lambda)$

such that:

$P(A \hat B | \alpha, \beta) = \sum_\lambda P(\lambda) P(A | \alpha, \lambda) P(B | \beta, \lambda)$

(where the sum might be an integral, if $\lambda$ takes on continuous values). If there were such probability distributions, then there would be a straight-forward local explanation of the EPR result:

1. Each pair of particles has a corresponding "hidden variable" $\lambda$ selected according to the probability distribution $P(\lambda)$
2. When Alice measures her particle, she gets spin-up with probability $P(A|\alpha, \lambda)$
3. When Bob measures his particle, he gets spin-up with probability $P(B|\beta, \lambda)$
4. These probabilities are independent.

Alas, Bell showed that there were no such probability distributions. Now, you can ask the analogous question about amplitudes:

Let $\psi(A \hat B | \alpha, \beta)$ be the amplitude (square-root of the probability, basically) that Alice and Bob will get spin-up at their respective angles. So $\psi(A \hat B | \alpha, \beta) = \frac{1}{\sqrt{2}} sin(\frac{\beta - \alpha}{2})$ (times a phase factor). Then the analogous question for amplitudes is:

Does there exist a parameter $\lambda$ and amplitudes $\psi(\lambda), \psi(A | \alpha, \lambda), \psi(B | \beta, \lambda)$ such that:

$\psi(A \hat B | \alpha, \beta) = \sum_\lambda \psi(\lambda) \psi(A | \alpha, \lambda) \psi(B | \beta, \lambda)$ ?

The answer is straightforwardly "yes".

• Let $\lambda$ have two possible values, $\lambda_u$ and $\lambda_d$.
• Let the amplitude for these values be: $\psi(\lambda_u) = \frac{1}{\sqrt{2}}$ and $\psi(\lambda_d) = - \frac{1}{\sqrt{2}}$
• Let the conditional amplitudes be:
  
  • $\psi(A | \alpha, \lambda_u) = cos(\frac{\alpha}{2})$
  • $\psi(A | \alpha, \lambda_d) = sin(\frac{\alpha}{2})$
  • $\psi(B | \beta, \lambda_u) = sin(\frac{\alpha}{2})$
  • $\psi(B | \beta, \lambda_d) = cos(\frac{\beta}{2})$

Then $\psi(A \hat B | \alpha, \beta) = \sum_\lambda \psi(\lambda) \psi(A |\alpha, \lambda) \psi(B |\beta, \lambda)$
$= \psi(\lambda_u) \psi(A |\alpha, \lambda_u) \psi(B|\beta, \lambda_u) + \psi(\lambda_d) \psi(A |\alpha, \lambda_d) \psi(B|\beta, \lambda_d)$
$= \frac{1}{\sqrt{2}} (cos(\frac{\alpha}{2}) sin(\frac{\beta}{2}) - sin(\frac{\alpha}{2}) cos(\frac{\beta}{2})$
$= \frac{1}{\sqrt{2}} sin(\frac{\beta - \alpha}{2})$

I don't know physically what it means that amplitudes, rather than probabilities factor, but it shows that quantum problems are often a lot simpler in terms of amplitudes.

 

[Zafa Pi]

In response to: @PeterDonis saying: More precisely, he [Bell] concludes that any such model that can reproduce the predictions of QM must be non-local.

Yes, you're right of course.

You guys are far from alone on this.
So after looking at Bell's 1964 paper I thought that he showed: {Locality & hidden variables} is incompatible with the predictions of QM.
I would naively conclude that that any model that can reproduce the predictions of QM must not have both locality & hidden variables. And also this seems to me not the same as being non-local. For example, have locality but not hidden variables.

Where have I gone wrong?

 

[rubi]

In response to: @PeterDonis saying: More precisely, he [Bell] concludes that any such model that can reproduce the predictions of QM must be non-local.

You guys are far from alone on this.
So after looking at Bell's 1964 paper I thought that he showed: {Locality & hidden variables} is incompatible with the predictions of QM.
I would naively conclude that that any model that can reproduce the predictions of QM must not have both locality & hidden variables. And also this seems to me not the same as being non-local. For example, have locality but not hidden variables.

Where have I gone wrong?

You're correct. "Such" in our sentences referred to hidden variables models, so we agree with you. No hidden variables model can be local, but other models can be.

 

[Nugatory]

Where have I gone wrong?

$\overline{A\cap{B}}$ implies $\bar{A}\cup\bar{B}$ but not $\bar{A}\cap\bar{B}$. However, the situation here is somewhat more complex because the assumptions Bell made in his original paper do not exactly correspond to "hidden variables" and "locality"; they're closer to "the properties that a theory must have to 'complete' (in the EPR sense) QM".

 

[OCR]

However, the situation here is somewhat more complex...

I rather think that Mr. Pi will not consider that fact appreciably more tolerable...

As a simpleton I appreciate simplicity.