I Is quantum weirdness really weird?

[rubi]

Say measurement of relative phase can theoretically give perfectly predictable result. Maybe that is what you are arguing about?

What I'm arguing about is the following. In your post #133 you claimed that the existence of the functions $A,B:\Lambda\rightarrow\mathbb R$ that Bell is evidently using, is not an additional assumption of Bell's theorem, but rather can be proved. So you made a mathematical claim (the existence of a proof). When someone makes a mathematical claim, the burden of proof lies with the person who makes the claim. Up to now however, you just made the claim and I was arguing that the claim is in conflict with the experimentally verified predictions of the GHZ state. However, I seem not to be getting through to you and it becomes quite cumbersome, so now I insist that the burden of proof lies on your side and I demand a mathematical proof for the claim of your post #133. I hope we agree that if something is not an assumption of a theorem, then there must be some assumption from which it follows. Please state this assumption mathematically and prove that it implies the existence of the functions $A$ and $B$.

 

[zonde]

If one assumes locality then CD is incompatible with QM. That is all that the Bell results show, i.e. local realism (= CD) is ruled out by QM.

If you assume locality, how do you see a way out in secur's example?

 

[zonde]

Please state this assumption mathematically and prove that it implies the existence of the functions $A$ and $B$.

No, better you state Bell's assumptions mathematically as you insist on mathematical treatment.
Two Bell's assumptions are:
1. Two measurements of entangled particles are independent.
2. When measurement angles are the same measurements give perfectly anticorrelated results.

 

[Zafa Pi]

Why don't you first understand the GHZ argument before making such claims?

Not only do I fully understand the GHZ argument, I can present it clearer and in less than 1/2 the space it took Mermin.

No, the GHZ experiment doesn't need any locality assumption.

If you are talking about the the QM measurements on the three entangled entities, say photons, then you are correct, and the same goes for measurements on the CHSH state which equals the state for secure's example = √½(|00⟩ + |11⟩). There is nothing special about the GHZ state in this matter. Now these measurement are used to show the classical conclusions (inequalities) are violated. However it is the classical case that needs locality in order to get the usual Bell inequalities. Without locality it is easy to violate the usual inequalities in a strictly classical fashion, no need for QM. If you wish I will demonstrate this in a clear and precise way.

 

[Zafa Pi]

If you assume locality, how do you see a way out in secur's example?

You assume realism = CD is false. I think i am going to need to present a more complete and clearer version of what secur did. Maybe tomorrow.

 

[zonde]

You assume realism = CD is false.

So this has to be part of secur example that is assumed to be false, right?

Assuming CD we can assign values for all, even though we can only do one of them.

Zafa Pi,
You still say that measurements are independent. How do you state this? My way would be that Alice can set her measurement angle to either 0 or 30 and Bob's result shouldn't change from that (Bob does not change his measurement angle). Do you see how counterfactual determinism is used in this explanation of "independent"? Can you explain "independent" differently without using counterfactual determinism?

 

[rubi]

No

So you refuse to provide a proof for the statement you made in post #133. Then I see no point discussing with you. Evidently you don't accept that mathematical claims must be proven.

better you state Bell's assumptions mathematically as you insist on mathematical treatment.

Bell's assumptions are
1. The existence of the functions $A(\alpha,\beta,\lambda)$ and $B(\alpha,\beta,\lambda)$.
2. Locality: $A(\alpha,\beta,\lambda)=A(\alpha,\lambda)$ and $B(\alpha,\beta,\lambda)=B(\beta,\lambda)$.
If you claim that Bell needn't explicitely assume 1., then provide a proof. I'm waiting.

Two Bell's assumptions are:
1. Two measurements of entangled particles are independent.
2. When measurement angles are the same measurements give perfectly anticorrelated results.

State these assumptions mathematically and prove Bell's inequality with them. If you can't do that, then your claim is unjustified. I've had enough of this vague language. Perform the mathematics you claim you can perform.

If you are talking about the the QM measurements on the three entangled entities, say photons, then you are correct, and the same goes for measurements on the CHSH state which equals the state for secure's example = √½(|00⟩ + |11⟩). There is nothing special about the GHZ state in this matter.

What is special about the GHZ state is that one doesn't need statistics to disprove the existence of elements of reality.

Now these measurement are used to show the classical conclusions (inequalities) are violated. However it is the classical case that needs locality in order to get the usual Bell inequalities. Without locality it is easy to violate the usual inequalities in a strictly classical fashion, no need for QM. If you wish I will demonstrate this in a clear and precise way.

I'm not talking about inequalities. What I'm arguing comes before you even attempt to prove an inequality from the GHZ state. The non-existence of elements of reality is nothing you can reproduce classically, not even without locality.

 

[zonde]

State these assumptions mathematically

Two Bell's assumptions are:
1. Two measurements of entangled particles are independent.
2. When measurement angles are the same measurements give perfectly anticorrelated results.
Two measurements are described by probabilities $P(A|\alpha,S)$ and $P(B|\beta,S)$ where A and B stands for measurement result of Alice and Bob respectively and S stands for entangled particles and any other shared information (source).
Independence assumption is stated as $P(A|\alpha,S)=P(A|\alpha,S,B,\beta)$ and $P(B|\beta,S)=P(B|\beta,S,A,\alpha)$
Perfect anticorrelations assumption is stated as $P(A,B|\alpha,\beta,S)=1$ when $A=not B$ and $\alpha=\beta$ and $P(A,B|\alpha,\beta,S)=0$ when $A= B$ and $\alpha=\beta$.

Is it ok so far?

 

[secur]

I find this very unfortunate, because it feeds into @zonde's objections. Your statement is false, it has everything to do with locality. If we have non-locality then Alice and Bob can conspire to make the measurements come out any way they want. S0 and A30 don't need to have only 25% mismatches they can have 100% mismatches.

Of course I could be wrong but I disagree. My example just shows what @rubi showed in previous post:

Yes, you can prove Bell-type inequalities for general random variables in a probability space. For example let $A,B,C,D:\Lambda\rightarrow\{-1,1\}$. Then it is easy to show that $\left|A(\lambda)B(\lambda)+A(\lambda)C(\lambda)+B(\lambda)D(\lambda)-C(\lambda)D(\lambda)\right|\leq 2$. Thus $\left|\left<AB\right>+\left<AC\right>+\left<BD\right>-\left<CD\right>\right|\leq 2$. It doesn't matter whether $A$, $B$, $C$ and $D$ represent locally separated observables of a physical theory or not.

The point was to make a simple proof dealing directly with the physics situation. No abstract math, just trigonometry and algebra. (Not that a little probability theory should confuse anyone). Most important: all the details contained in one post. The math shows explicitly what I mean by CD. If it means something different to you, just ignore that word.

Alice and Bob can conspire all they want. They can sit down together in the faculty club with coffee and donuts, or spend a year on sabbatical, to come up with those four sequences. The lists can be as artificial as they like; for instance they can make one all 1's, another alternating 1's and 0's, whatever. (That would violate QM also but I'll allow it anyway). It's impossible to satisfy the particular demands of QM that I mentioned, no matter how you do it.

Another way to see this is that Bohmian Mechanics is a consistent interpretation of QM, and it allows for the validity of CD (or determinism). However BM is nonlocal.

You're right, except that would be a different definition of CD. Bohm's pilot wave communicates non-local details of Bob's detector setting to Alice's electron. She gets different results depending whether he sets 0 or 30 degrees. With Bohmian mechanics you cannot write down my "four sequences", which apply regardless of the other station's detector settings.

If one assumes locality then CD is incompatible with QM. That is all that the Bell results show, i.e. local realism (= CD) is ruled out by QM.

You're right - except for that term "CD"! To me it's not the same as "determinism". Alice can deterministically (not probabilistically) get a predictable, definite sequence when Bob sets 0 degrees; and a different, determined, sequence when he uses 30. But (to me) CD doesn't allow that. My example shows that the property I'm calling CD is incompatible with QM, regardless of locality.

For some reason I can't get zonde to agree to my last sentence.

I wouldn't be surprised if @zonde uses the definition I gave for "CD". In that case, simply say "deterministic" instead and he'll start to agree.

Of course that demonstrates why we like to use math not words. Words can cause endless debate, via simple miscommunication, but not math. If the only accomplishment of my post is to remove the CD terminology confusion, it was worth the trouble.

What's proven with the violation of Bell's inequality is that nature cannot be described by a local deterministic model. ... that has been the great achievement of Bell's work: He provided a physically sensible criterion for what's called in a somewhat unsharp way by philosophers of science (including EPR themselves!) "local realism".

@vanhees71 is saying what you're saying, except correctly using the word "deterministic" not CD. AFAIK, IMHO.

[EDIT] It occurs to me, maybe your (@Zafa Pi) definition of CD is correct. I was assuming not, but don't really know (or care very much). So just consider the physics and math of my post. Forget I ever used that term "CD".

 

[rubi]

Two Bell's assumptions are:
1. Two measurements of entangled particles are independent.
2. When measurement angles are the same measurements give perfectly anticorrelated results.
Two measurements are described by probabilities $P(A|\alpha,S)$ and $P(B|\beta,S)$ where A and B stands for measurement result of Alice and Bob respectively and S stands for entangled particles and any other shared information (source).
Independence assumption is stated as $P(A|\alpha,S)=P(A|\alpha,S,B,\beta)$ and $P(B|\beta,S)=P(B|\beta,S,A,\alpha)$
Perfect anticorrelations assumption is stated as $P(A,B|\alpha,\beta,S)=1$ when $A=not B$ and $\alpha=\beta$ and $P(A,B|\alpha,\beta,S)=0$ when $A= B$ and $\alpha=\beta$.

Is it ok so far?

No, you have already made a hidden variables assumption here, by introducing the dependence on $S$. You have just stated it in terms of probabilities instead of functions. It's a non-trivial restriction on the set of allowed models.

 

[Zafa Pi]

So this has to be part of secur example that is assumed to be false, right?

Zafa Pi,
You still say that measurements are independent. How do you state this? My way would be that Alice can set her measurement angle to either 0 or 30 and Bob's result shouldn't change from that (Bob does not change his measurement angle). Do you see how counterfactual determinism is used in this explanation of "independent"? Can you explain "independent" differently without using counterfactual determinism?

Before I address your questions let me say:
1. If you google Bell's theorem you'll find that "all" say in some fashion or other that QM correlations cannot be replicated by local deterministic procedures. Or that local realism is ruled out by QM. Or post #141 by vanhees71. Local meaning no faster than light influence or communication.
2. I say that those QM correlations can be replicated by deterministic procedures. (notice the missing local)
3. That QM postulates entangled states and that QM predicts certain (probabilistic) outcomes/correlations on measurements of subsystems of of such states no more depends on locality than the prime number theorem.
4. Attempts to replicate the above QM predictions in the real world may require nonlocal phenomena. No one knows.
If anyone has a problem with any of the above four statements let me know.

Now the usual physical set up for the Bell experiment goes something like:
Alice and Bob are 2 light minutes apart and Eve is half way between and simultaneously sends a light signal to each. When Alice receives her signal she flips a fair coin. If it comes up heads selects either +1 or -1 by some objective procedure (we can duplicate the procedure) and we call that Ah. If she flips a tail she may do the same thing or something else to get At which also = 1 or -1. This takes less than 30 seconds. Bob goes through the same ritual to get Bh and Bt. E.g., it could happen that Bob rolls a die to determine Bt.

Bell's Theorem: Let Ah, At, Bh, and Bt be four numbers that are either 1 or -1. Assume that Ah = Bh (Ah•Bh = 1),
then we have Bell's Inequality: P(At•Bt = -1) ≤ P(At•Bh = -1) + P(Ah•Bt = -1). (Where P is probability)

Proof: P(At•Bt = -1) = P(At•Bt•Ah•Bh = -1) = P(At•Bh•Bt•Ah = -1) =
P({At•Bh = -1 and Bt•Ah = 1} or {At•Bh = 1 and Bt•Ah = -1}) =
P(At•Bh = -1 and Bt•Ah =1) + P(At•Bh = 1 and Bt•Ah = -1) ≤
P(At•Bh = -1) + P(Ah•Bt = -1) QED

Suppose that Alice selects 1 for both Ah and At, then she gets on the quikfone and tells Bob what she flipped (takes 10 seconds). Bob the Let's Bh = 1 and let's Bt = 1 if she said heads and let's Bt = -1 if she said tails. Bob is done before 30 seconds is up. Pr(At•Bt = -1) = 1, P(At•Bh = -1) = P(Ah•Bt = -1) = 0 so Bell's Inequality is violated.
Now no mathematical theorem can be violated by the goings on in the real world so some part of the hypothesis must also be violated. We don't have the four numbers, Bt is not a fixed value. This scenario requires non-locality.

If we assume locality Bob can't hear from Alice, in fact for all he knows Alice is dead, and vise versa, there is no communication. So if Bob flips heads his pick of Bh doesn't have anything to do with Alice's doings. If Bob rolls a die to determine Bt he still ends up with a fixed Bt. Same goes for Alice and we get our four numbers, so as long as Ah = Bh (e.g. they agreed before hand to make both = -1) Bell's Inequality is satisfied for their result.

But what if Eve sends each one photon from the state √½(|00⟩ + |11⟩) and Ah is the result of measuring her photon at 0º (Pauli Z) At is the result of measuring at 30º, While Bh is obtain by measuring at 0º and Bt by measuring at -30º (as "secure" suggested). Then P(At•Bt = -1) = 3/4, and P(At•Bh = -1) = P(Ah•Bt = -1) = 1/4. So Bell's inequality is violated. Now come to the crux, what part of the hypothesis is violated. Well either the QM business is nonlocal or not. If it is nonlocal we've already covered that, if not then what. We are assuming the QM business is local, it is not a given.

We know that Ah = Bh is satisfied, what's left from the hypothesis to violate? Here is what's said:
When the experiment is performed only two of the four numbers are found, say At and Bh, where do the other two come from? They come from the assumption of realism.
Wuz dat?
In essence it says we would have gotten something definite for the other two values call them Ah and Bt. The reality and properties of the photon would not have changed if Alice used 0º instead of 30º. In fact we know that Ah would = Bh be according to QM.
Counterfactual definiteness (CFD) is the ability to speak meaningfully of the definiteness of the results of measurements that have not been performed. (Wiki)
No, those unmeasured guys have no value, they can't be known, and cannot be used together with At and Bh in some formula otherwise we couldn't violated Bell's Inequality.
Under the assumption of locality the violation of Bell's inequality negates realism.
I'm uncomfortable. Suck it up and move on.

zonde your definition of independent is untestable. Would you have Bob measure again to see if his value didn't change with a new photon? You know he will get a different answer with P = ½. The old is now in the eigenstate of his measurement operator.
The only way I see to capture the idea of "independence" is to assume we have the four definite fixed (two of which are unknown) values and that assumption is called realism or CFD or determinism or hidden variables. The QM results are definitely not independent as r.v.s. What would you say if both Alice and Bob flipped fair coins to determine the four values?

 

[zonde]

Two Bell's assumptions are:
1. Two measurements of entangled particles are independent.
2. When measurement angles are the same measurements give perfectly anticorrelated results.
Two measurements are described by probabilities $P(A|\alpha,S)$ and $P(B|\beta,S)$ where A and B stands for measurement result of Alice and Bob respectively and S stands for entangled particles and any other shared information (source).
Independence assumption is stated as $P(A|\alpha,S)=P(A|\alpha,S,B,\beta)$ and $P(B|\beta,S)=P(B|\beta,S,A,\alpha)$
Perfect anticorrelations assumption is stated as $P(A,B|\alpha,\beta,S)=1$ when $A=not B$ and $\alpha=\beta$ and $P(A,B|\alpha,\beta,S)=0$ when $A= B$ and $\alpha=\beta$.

Is it ok so far?

No, you have already made a hidden variables assumption here, by introducing the dependence on $S$. You have just stated it in terms of probabilities instead of functions. It's a non-trivial restriction on the set of allowed models.

Sorry, but I don't get your objections. I am not introducing dependence between A,B and S. By including S into conditional probability I just state that my treatment will take into account possible dependence on S. This is obviously more general case than without included S. And there are visible variables included into S and these certainly have to be included into probability otherwise we can't analyze pairwise correlations.

You aggressively insisted on mathematical treatment as it was supposed to give you more clarity. But now it seems I have to explain to you basic things about probabilities. Then what's the point of switching over to math?

 

[zonde]

We know that Ah = Bh is satisfied, what's left from the hypothesis to violate? Here is what's said:
When the experiment is performed only two of the four numbers are found, say At and Bh, where do the other two come from? They come from the assumption of realism.

Do not mix realism here, it has nothing to do with that. Other two values come from counterfactual definiteness, let's stick to it.

In essence it says we would have gotten something definite for the other two values call them Ah and Bt. The reality and properties of the photon would not have changed if Alice used 0º instead of 30º. In fact we know that Ah would = Bh be according to QM.
Counterfactual definiteness (CFD) is the ability to speak meaningfully of the definiteness of the results of measurements that have not been performed. (Wiki)
No, those unmeasured guys have no value, they can't be known, and cannot be used together with At and Bh in some formula otherwise we couldn't violated Bell's Inequality.
Under the assumption of locality the violation of Bell's inequality negates realism.

I agree with what you say (if you replace "realism" with CFD).
But ... Bell's theorem does not speak about reality. Bell says that no local model that says something about individual detections can reproduce QM predictions. And any local (scientific) model has a property of CFD. So we can't say that reality is non-local but we can say that any scientific model of reality (that says something about individual pairs of entangled particles) has to be non-local.

 

[zonde]

Or post #141 by vanhees71.

@vanhees71 is using slightly different definition of "locality". His "locality" is no FTL signaling or more technically stated statistics of measurements commute (statistics are the same whether Alice has measured first or second).

 

[secur]

Counterfactual definiteness (CFD) is the ability to speak meaningfully of the definiteness of the results of measurements that have not been performed. (Wiki)

Since earlier I said the definition of CFD is unclear I'd like to address this. Of course I'd read it before.

1. CFD is not an ability. It's an assumption (statement, assertion) that confers the ability to ..., etc.
2. "Speak meaningfully" - So if the experimenters happen to be mute, they can't use CFD?
3. Here's a meaningful statement about the definiteness of the results of unperformed measurements: "unperformed measurements have no definite result". But that's not CFD, it's the exact opposite.
4. What's a measurement? This "definition" assumes it's specified by only Alice's setting. But we could insist the complete specification includes Bob's setting also. In that case we can, in fact, assign definite values to her unperformed measurements - different ones for different Bob-settings. But this is precisely what CFD is supposed to rule out.

I could point out more problems ... This definition is incoherent and worthless. We couldn't begin to understand it, except we already know what it's trying to say. The lesson: don't count on Wikipedia.

4. Attempts to replicate the above QM predictions in the real world may require nonlocal phenomena. No one knows.

That's right. So why do so many, such as Gell-Mann, say "there is definitely no nonlocal influence"? And why don't other physicists condemn such misinformation?

Bell's "theorem" is trivial math, and the setup is unnecessarily complicated. He's honored nevertheless for the concept: that it's possible to distinguish between QM and "local realistic" (whatever that means) models. The concept has been illustrated more directly in various ways. My example is as simple and clear as it gets. Your statement of the "theorem" would be excellent if you were talking to people who didn't already understand it, but you're not. You'd be better off sticking with my clarified example.

The only way I see to capture the idea of "independence" is to assume we have the four definite fixed (two of which are unknown) values and that assumption is called realism or CFD or determinism or hidden variables.

That's exactly what I said, with a few necessary details such as: we need to consider sequences of the four values.

So we can't say that reality is non-local but we can say that any scientific model of reality (that says something about individual pairs of entangled particles) has to be non-local.

As @Ken G (haven't seen him in a while) used to say - correctly - science allows us to say nothing at all about reality per se. We can only talk about models. Your statement is true for a restricted class of models, "called realism or CFD or determinism or hidden variables" (as @Zafa Pi sensibly puts it). But if the model is MWI, or many others, they can get the right results without non-locality - at the expense of introducing extremely far-fetched assumptions. Therefore I say non-locality is the simplest way to interpret QM results. I think you agree with that? But we can't go so far as to say that "all" possibly-valid models are non-local.

@vanhees71 is using slightly different definition of "locality". His "locality" is no FTL signaling or more technically stated statistics of measurements commute (statistics are the same whether Alice has measured first or second).

I think (not sure) you're right about vanhees71's view. The problem is, it doesn't admit the (valid) concept of non-local "influence". Since that's the whole point under debate, I wouldn't call his definition "slightly" different.

Off-topic: there's a famous John Cleese sketch in "Fawlty Towers" where he's talking to a couple of Germans who don't speak English. He says something like "Your room will be ready in half an hour, please wait in the lounge." They don't understand. So he says it again, this time with a fake German accent. They still don't get it. So he says it again louder and slower: "Your! ... room vill be ... Ready! In! ... Haff un hour! Please! To! ... vait in der Lounge!" And again, and again: same statement, but louder and louder.

A perfect analogy for this thread.

 

[PeterDonis]

why do so many, such as Gell-Mann, say "there is definitely no nonlocal influence"?

I think it's because they are using a different definition of "nonlocal", something like "violates relativistic causality". Bear in mind that Gell-Mann, and the others who I'm aware of that say things like this, are particle physicists who are used to working in the framework of quantum field theory. In QFT, "locality", or better, "relativistic causality", appears as the requirement that measurements at spacelike separated events must commute, i.e., their results must be independent of the order in which they are performed. All known experiments satisfy this criterion (including the EPR experiments that violate Bell's inequalities), so relativistic causality appears to be true. And since, intuitively, relativistic causality means that causal influences can't propagate instantaneously, they can only propagate at the speed of light, it seems to meet the requirements for a reasonable version of "locality:".

 

[A. Neumaier]

So why do so many, such as Gell-Mann, say "there is definitely no nonlocal influence"?

the requirement that measurements at spacelike separated events must commute, i.e., their results must be independent of the order in which they are performed. All known experiments satisfy this criterion (including the EPR experiments that violate Bell's inequalities), so relativistic causality appears to be true.

See my post on extended causality earlier this year.

 

[rubi]

Sorry, but I don't get your objections. I am not introducing dependence between A,B and S. By including S into conditional probability I just state that my treatment will take into account possible dependence on S. This is obviously more general case than without included S. And there are visible variables included into S and these certainly have to be included into probability otherwise we can't analyze pairwise correlations.

Wrong. By including $S$, you restrict the set of models that you consider to that particular subset which can be written in this particular form. Since not all possible models are hidden variable models (as I have proved in post #120), you have included an additional assumption, namely that your model is a hidden variable model.

You aggressively insisted on mathematical treatment as it was supposed to give you more clarity. But now it seems I have to explain to you basic things about probabilities. Then what's the point of switching over to math?

I'm certain that you can't explain me anything about probability theory, because I happen to be a professional scientist, while I very much doubt that you know the basics of probability theory yourself. Just a few weeks ago, you didn't even know how to multiply by the identity matrix, so you should be a bit more humble in your position. Of course, I insist on a mathematical treatment, because it exposes the assumptions of the theorem and it is absolutely apparent to anyone who knows the basics of probability theory (apparently you aren't one of them) that the hidden variable assumption is a non-trivial assumption.

 

[OCR]

quikfone

...

 

[secur]

Two Bell's assumptions are:

1. Two measurements of entangled particles are independent...

... Two measurements are described by probabilities $P(A|\alpha,S)$ and $P(B|\beta,S)$ where A and B stands for measurement result of Alice and Bob respectively and S stands for entangled particles and any other shared information (source).

@zonde is talking specifically about Bell's assumptions.

Wrong. By including $S$, you restrict the set of models that you consider to that particular subset which can be written in this particular form. Since not all possible models are hidden variable models (as I have proved in post #120), you have included an additional assumption, namely that your model is a hidden variable model.

That's true. But "hidden variables" was one of Bell's assumptions.

From "ON THE EINSTEIN PODOLSKY ROSEN PARADOX"

... Let this more complete specification be effected by means of parameters $\lambda$. ... Some might prefer a formulation in which the hidden variables fall into two sets, with A dependent on one and B one the other; this possibility is contained in the above, since $\lambda$ stands for any number of variables ...

 

[rubi]

That's true. But "hidden variables" was one of Bell's assumptions.

I agree and so does the rest of the physics community. However, zonde stubbornly denies this. He believes that he has summarized the assumptions correctly in his post #153.

 

[zonde]

Wrong. By including S, you restrict the set of models that you consider to that particular subset which can be written in this particular form.

Well, I was not saying anything about possible models yet. I just said what parameters I will use to describe measurements. Including more parameters than necessary should not restrict anyone.

Since not all possible models are hidden variable models (as I have proved in post #120), you have included an additional assumption, namely that your model is a hidden variable model.

Okay I am trying to decipher from your post #120 what would be example of some other possible model. I am considering this sentence:

In quantum mechanics, observables aren't functions $O_\xi : \Lambda\rightarrow\mathbb R$, but rather operators $\hat O_\xi$ that are defined on a Hilbert space.

Is it even meaningful?
It is rather in quantum mechanic where observables are functions (in particular operators). Classically they are just variables.
@Mentz114 seems to understand what you said so maybe he can provide some clue.

I'm certain that you can't explain me anything about probability theory, because I happen to be a professional scientist, while I very much doubt that you know the basics of probability theory yourself. Just a few weeks ago, you didn't even know how to multiply by the identity matrix, so you should be a bit more humble in your position.

Oh, but now I know what is "identity matrix" and what is "matrix ring". So I have learned something.

 

[zonde]

That's true. But "hidden variables" was one of Bell's assumptions.

I agree and so does the rest of the physics community. However, zonde stubbornly denies this. He believes that he has summarized the assumptions correctly in his post #153.

Well, at least I am on one side with Bell himself:

It is remarkably difficult to get this point across, that determinism is not a presupposition of the analysis.

 

[rubi]

Well, I was not saying anything about possible models yet. I just said what parameters I will use to describe measurements. Including more parameters than necessary should not restrict anyone.

Well, it does restrict the model to the set of models that admits a description in that form. That's a tautology and should be a completely trivial insight. If you start with a formula that doesn't include all possible models (and it doesn't), then you have made a non-trivial assumption. We can prove that the quantum model does't have this form (see post #120), so it is not included among the set of models that admit such a description.

Okay I am trying to decipher from your post #120 what would be example of some other possible model. I am considering this sentence:

Is it even meaningful?

Yes, of course it is meaningful. I just explained the Kochen-Specker theorem. You can read about it in most quantum mechanics textbooks, if you want to know more about it.

It is rather in quantum mechanic where observables are functions (in particular operators). Classically they are just variables.

Yes, and the theorem proves that quantum mechanics cannot be embedded into a classical theory by mapping quantum observables to classical observables. And if that were possible, you could also calculate their probability distributions and they would take the form you suggest, so it is also excluded that quantum mechanics can be embedded in the particular probability model you assumed. Your assumption is thus a non-trivial extra assumption.

Oh, but now I know what is "identity matrix" and what is "matrix ring". So I have learned something.

So you went from knowing not even the most basic thing in mathematics (the identity matrix) to understanding some high level probability theory in just two weeks? "Sounds reasonable."

Well, at least I am on one side with Bell himself:

No, you aren't. Assuming hidden variables and assuming determinism is not the same thing. What you need to assume is a hidden variables model. Those hidden variables may be deterministic or they may be stochastic. That doesn't matter and that's what Bell said in that quote.This discussion is becoming completely pointless. I have precisely pointed out where you introduce an extra assumption and proved that it is actually an extra assumption. It is kind of delusional to keep defending your position, especially since you are in disagreement with the whole physics community.

 

[Zafa Pi]

3. Here's a meaningful statement about the definiteness of the results of unperformed measurements: "unperformed measurements have no definite result". But that's not CFD, it's the exact opposite.

OK, "CFD is the exact opposite of unperformed measurements have no definite result". Wouldn't Occam prefer getting rid of opposite and no?

That's right. So why do so many, such as Gell-Mann, say "there is definitely no nonlocal influence"? And why don't other physicists condemn such misinformation?

If you think that's bad, what about all the luminaries that say, "QM is weird because you can send two particles far away from each other and one will have the opposite property from the other. Like if a pair of shoes is made and one of them is sent far away and you open the box containing one of them to find a left shoe then you will know the distant one is a right shoe."
I told Schlosshauer that it was in his decoherence book and he said he was embarrassed. After a talk by by Zeilinger in Portland I asked him why he lied to the audience. There was a hush, he cocked his head and asked where did I lie; I told him, he laughed and said it was a mixed audience I couldn't give them the Bell's Inequality. I said OK so your like a politician. When I was in Vienna at the 50th anniversary of Bell's theorem symposium he introduced me as "the guy that said I lie like a politician". Weinberg has done it, so has Griffiths, and Greene. Very naughty.

Your statement of the "theorem" would be excellent if you were talking to people who didn't already understand it, but you're not.

It seems to me there is unnecessary confusion in many threads due to different notation and definitions, so I wanted everyone on the same page (mine ).

The lesson: don't count on Wikipedia.

OMG, next you'll be telling me not to rely on the Old Testament a QM source. Where is the truth?

 

[Zafa Pi]

I agree with what you say (if you replace "realism" with CFD).

Realism, CFD, hidden variables, determinism, I can't tell the difference when it comes to Bell stuff. But perhaps you have different notions of what they mean than I do. I'll try to stick to CFD for you, it's even easier to write.

 

[vanhees71]

@vanhees71 is using slightly different definition of "locality". His "locality" is no FTL signaling or more technically stated statistics of measurements commute (statistics are the same whether Alice has measured first or second).

I'm more on the HEP/nuclear physics side. There "locality" means that in relativistic QFT you have a Lagrangian which is a polynomial in the fields and their derivatives at one space-time point. Together with microcausality, i.e., that you demand that local observables commute at space-like distances of their arguments, this implies that there is no causal information transformation over space-like distances, and that the linked-cluster principle holds.

This has to be distinguished from long-distance correlations describes by entanglement of observables concerning parts of a system detected at far-distant places. This of course holds in any QT, and entanglement is the really distinguishing feature between classical (deterministic) physics and quantum physics. Bell's work shows that the question, whether you can have a local determinstic hidden-variable theory which agrees with all (probabilistic!) predictions of QT is answered negatively: Bell's inequality can be violated by QT, but this doesn't mean "non-locality" but stronger "long-range correlations" than possible within any local deterministic model. Experiments show with overwhelming significance that QT is correct and thus rules out local deterministic models.

Also the EPR paper is not making Einstein's criticism very clear. Einstein didn't like the paper too much, and he wrote another paper on the subject where he makes clear that indeed his concern was more about the "non-separability" of properties, i.e., that you can have these very strong far-distant correlations described by entanglement. It's unfortunately a paper in German:

A. Einstein, Quantenmechanik und Realität, Dialectica 2, 320 (1948)

 

[A. Neumaier]

Einstein [...] wrote another paper on the subject where he makes clear that indeed his concern was more about the "non-separability" of properties, i.e., that you can have these very strong far-distant correlations described by entanglement. It's unfortunately a paper in German:

A. Einstein, Quantenmechanik und Realität, Dialectica 2, 320 (1948)

Here is an English translation of Einstein's Dialectica paper! In addition, this page also includes at the end an extended discussion.

 

[OCR]

But there remain always these qualitative and somewhat ambiguous statements [...] to express fuzzy concepts.

It seems to me there is unnecessary confusion in many threads due to different notation and definitions, so I wanted everyone on the same page (mine ).

[/PLAIN]
Quantum mechanics the way I see it.
(In the quantum mechanical forest few paths do not lead into nowhere land.)


The above link came ... from here .
Due to discussions ... occurring here.
For what it's worth... ? ... lol ... ( a joke... kinda )

OMG, next you'll be telling me not to rely on the Old Testament a QM source.

...

 

[secur]

Ok, now I see why everybody kept repeating Bell-type experiments, like John Cleese yelling at the unfortunate German couple. @zonde is the "German".

@zonde, please look at it this way. We've got entangled particles with state |11> + |00>. (Ignoring normalization). Suppose Alice uses 0 degrees setting on her detector and Bob uses 30. Suppose Alice gets 1, spin up. We can safely make the following "CFD 1" assertion:

CFD 1: "If Bob had used 0 setting also, A & B would have read the same result."

That's a counterfactual QM statement, nevertheless it's as reliable as any in classical mechanics. However we can't say this:

CFD 2: "If Bob had used 0 setting also, he would have measured 1."

Instead, if "we could run it again", (which of course we can't), it's a new QM calculation. Yes, Alice and Bob would definitely detect the same spin. But it might be up or down, 50 / 50 chance. The fact that A got 1 before is irrelevant.

Now, putting is simply but more or less accurately: Bell-type experiments and inequalities (including CHSH, GHZ, etc) prove you can't assert the CFD 2 assertion. If you do, you can't get the right experimental results. With or without non-locality.

Personally I don't find any of this excessively weird. It's good physics.

So, what part of it don't you accept?

I am saying that calculation is exactly the same whether you are calculating "what will happen if I will measure spin along x axis?" or "what would have happened if I would have measured spin along x axis?" as long as starting point (preparation of input state) is described exactly the same way.

Good question. If we can predict experimental results when the experiment "will be" run, the same should apply to what "would have" happened. That is, in fact, the case. Exactly like counterfactual past results, we can assert only CFD 1 about future results, not CFD 2.

A must equal B, but we can't know if it will be, or would have been, spin up or down. That's what Bell-type experiments prove.

 

[Zafa Pi]

Quantum mechanics the way I see it.
(In the quantum mechanical forest few paths do not lead into nowhere land.)The above link came ... from here .
Due to discussions ... occurring here.
For what it's worth... ? ... lol ... ( a joke... kinda )

Great references. I'm sure somewhere in the wiki article there must be a dead horse.

 

[Zafa Pi]

What is special about the GHZ state is that one doesn't need statistics to disprove the existence of elements of reality.

This was in response to my question that regarding the discussion of this thread there wasn't anything special about GHZ as opposed to any other Bell doings.
I agree that it's mighty cute that you only need one trial to contradict the equality in the GHZ theorem, as opposed to the needed statistics in secur's example at post #147, or CHSH et. al.

However, I wouldn't be to fast in asserting that the GHZ result doesn't need statistics. Here is GHZ:
The set up and notation are like mine at post #161 except we introduce a third player Carol that also receives a photon from Eve who uses the state |GHZ⟩ = √½(|000⟩ + |111⟩) to send three entangled photons.
GHZ Theorem: Assume that if only one of Alice, Bob, or Carol flips heads then the product of their values is -1. Then Ah•Bh•Ch = -1.
Proof: -1 = Ah•Bt•Ct = At•Bh•Ct = At•Bt•Ch, then the product of all three products = -1 = (At•Bt•Ct)²Ah•Bh•Ch ⇒ Ah•Bh•Ch = -1 (GHZ Equality). QED

Now if Ah, Bh, and Ch are the result of measuring with (Pauli)X, while the others employ (Pauli)Y the the hypothesis of the theorem is satisfied, but Ah•Bh•Ch = 1. The GHZ Equality is violated!

Just one trial of Ah•Bh•Ch does the trick. But how do we know the hypothesis is satisfied, e.g. that Ah•Bt•Ct = -1 for sure? It's akin to my saying I have a two headed coin and all you get to see is the results of flips. I would have to make a bunch before you were really convinced. Statistics.

 

[OCR]

... there must be a dead horse.

Lol, I get it... and she's still getting beat on ...

 

[zonde]

@zonde, please look at it this way. We've got entangled particles with state |11> + |00>. (Ignoring normalization). Suppose Alice uses 0 degrees setting on her detector and Bob uses 30. Suppose Alice gets 1, spin up. We can safely make the following "CFD 1" assertion:

CFD 1: "If Bob had used 0 setting also, A & B would have read the same result."

That's a counterfactual QM statement, nevertheless it's as reliable as any in classical mechanics. However we can't say this:

CFD 2: "If Bob had used 0 setting also, he would have measured 1."

Instead, if "we could run it again", (which of course we can't), it's a new QM calculation. Yes, Alice and Bob would definitely detect the same spin. But it might be up or down, 50 / 50 chance. The fact that A got 1 before is irrelevant.

Now, putting is simply but more or less accurately: Bell-type experiments and inequalities (including CHSH, GHZ, etc) prove you can't assert the CFD 2 assertion. If you do, you can't get the right experimental results. With or without non-locality.

Personally I don't find any of this excessively weird. It's good physics.

So, what part of it don't you accept?

Everything is fine. Well, CFD1 is of course non-local statement, while CFD2 is local statement. So if a theory can't make symmetrical CFD2 type claims about A & B we can't say it's local (meaning, A and B measurements are independent).

 

[secur]

I would call it non-local also. But others say it's not, because Alice and Bob commute (they live in the suburbs). It doesn't matter, as long as we agree what happens in the actual experiments.

 

[ljagerman]

Please explain or give simple example.
My position is that we use the model exactly the same way whether we ask "what will happen?" or we ask "what would have happened?". It's exactly the same input for the model and therefore it has to produce exactly the same output.

I also find that this way of thinking about QM makes sense: "Anything that can happen does happen," but each possible happening has a probability associated with it. The best that QM math can do is provide that probability. I can then say that a manifestation of quantum "weirdness" is something very unlikely indeed happening. Thus a plane bound to fly from New York to London can get there via Jupiter - a very weird but possible happening. A more familiar weird theoretically possible event is that for a few minutes all molecules of oxygen will be located in one remote corner of your room, and you will die from anoxia.

To make this more concrete I invoke Feynman's path integral approach to QM: Each molecule can traverse any possible path, each path has a probability, these probabilities can interfere with each other, and in our ordinary environment one path is usually by far the most likely; it enjoys the path of highest constructive interference and least destructive interference. Thus fortunately the probability that all molecules of oxygen moving to one corner of the room is extremely small in our lifetimes, but such weirdness is possible.

BTW, Feynman's system meshes neatly with what we see in two-slit experiments, the kind that are said to reveal wave-particle duality: The bands of an interference pattern that we see on a screen or photo arise from path interference.

 

[A. Neumaier]

this way of thinking about QM makes sense: "Anything that can happen does happen," but each possible happening has a probability associated with it.

This way of thinking also makes sense classically, but it explains nothing at all. Moreover, both classically and quantum mechanically, it is a gross violation of the basic philosophic principle called Ockham's razor.

 

[secur]

A more familiar weird theoretically possible event is that for a few minutes all molecules of oxygen will be located in one remote corner of your room, and you will die from anoxia.

Has nothing to do with the topic, but: no, you won't die from anoxia, in a minute or two. Instead you'll die in milliseconds from major trauma as all those air molecules immediately rush back into the rest of the room in a huge explosion. As pointed out by some PF poster a while ago; iirc, @Demystifier.

"Anything that can happen does happen", but each possible happening has a probability associated with it.

This way of thinking also makes sense classically, but it explains nothing at all. Moreover, both classically and quantum mechanically, it is a gross violation of the basic philosophic principle called Ockham's razor.

Is that true? "Anything that can happen does happen" is, in a sense, the simplest scenario, because if rejected you have to explain why particular things happen and not others. This is exactly parallel to the philosophically uncomfortable "collapse of the wave function". Why does it collapse to this branch of the wave function, and not all those others, some of which might indeed have higher probabilities? Applying Ockham to this question results directly in MWI.

 

[Zafa Pi]

This way of thinking also makes sense classically, but it explains nothing at all. Moreover, both classically and quantum mechanically, it is a gross violation of the basic philosophic principle called Ockham's razor.

William of Ockham was also known as William of Occam. Adeptly applying his razor you should use Occam.

 

[OCR]

When it's said...

Anything that can happen does happen...

And it's said...

...each possible happening has a probability associated with it.

Don't those statements become some what contradictory ?

Wouldn't it make more sense to say... "Anything that can happen... could, might, may, or even, can, etc. ... happen ?
Such ... "Anything that can happen, does can happen."

Don't you actual say that when you say... ?

The best that QM math can ...
can get there via Jupiter...
Each molecule can traverse any possible path...
these probabilities can interfere with each other...

I don't see the word does, or, do ... anywhere.

Instead you'll die in milliseconds from major trauma as all those air molecules immediately rush back into the rest of the room in a huge explosion.

I don't remember reading that... was it posted in this thread ?

 

[secur]

@OCR, yes it's contradictory to say "everything happens, but some things have less probability of happening" when talking about a single event. Indeed this is a big difficulty facing MWI proponents. But in normal QM situations (like calculating particle decays) we're talking about expectations, over many instances. Still I'm not sure that completely resolves the philosophical difficulty.

Someone - it may have been @Demystifier - corrected me in some other thread months ago, when I made the same comment about anoxia. Actually, depending how the air got into that room corner, you might have died then, if it all rushed over there very quickly.

 

[OCR]

Thank you for replying, secur ...

 

[OCR]

Adeptly applying his razor you should use Occam.

Correct !

 

[Zafa Pi]

Does anyone have a guess how 19th century physicists would have responded to:
Does "If Alice and Bob both perform experiment X they will get the same result." imply "If Alice performs X and gets value 1 and Bob performs Y and gets 2; he would have gotten 1 if he had performed X instead."?

I think they all would have said yes. Yet wouldn't those on this thread all say no?

 

[Zafa Pi]

In responding to zonde:

I'm certain that you can't explain me anything about probability theory, because I happen to be a professional scientist,

Wow!

 

[secur]

Does anyone have a guess how 19th century physicists would have responded to:
Does "If Alice and Bob both perform experiment X they will get the same result." imply "If Alice performs X and gets value 1 and Bob performs Y and gets 2; he would have gotten 1 if he had performed X instead."?

I think they all would have said yes. Yet wouldn't those on this thread all say no?

I'm pretty sure they would have responded "What the deuce does Alice have to do with physics?"

 

[Zafa Pi]

I'm pretty sure they would have responded "What the deuce does Alice have to do with physics?"

Ok, then how about Marie and Pierre (Curies). Anyway what's your answer to my question?

 

[Demystifier]

Someone - it may have been [USER=61953]@Demystifier - corrected me in some other thread months ago, when I made the same comment about anoxia. Actually, depending how the air got into that room corner, you might have died then, if it all rushed over there very quickly.[/USER]

It hasn't been me.

 

[A. Neumaier]

Anything that can happen does happen" is, in a sense, the simplest scenario

But a completely nonpredictive one, hence it is no science.

William of Ockham was also known as William of Occam. Adeptly applying his razor you should use Occam.

His razor was made for shaving scientific statements, not names.

 

[Zafa Pi]

His razor was made for shaving scientific statements, not names.

My bad.