Is quantum weirdness really weird?

[vanhees71]

What's proven with the violation of Bell's inequality is that nature cannot be described by a local deterministic model. Quantum theory is an example for a local indeterministic model. By construction relativistic QFT is local (and particularly microcausal!). Nevertheless you can have non-local correlations as described by entangled states which describe systems with parts that can be detected at a far distance and which don't have predetermined properties but very strong (sometimes 100%!) correlations. These correlations are stronger than possible in any local deterministic model, and 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".

 

[zonde]

Even if that were the case, it would still be completely irrelevant to the EPR argument, since you can only predict the value of the remote spin with certainty after you performed the measurement and you wouldn't disturb it anymore afterwards. Therefore, if the EPR argument was valid, there would have to be a corresponding element of reality, which the GHZ experiment proves to not exist.

And what about pure states? We can predict measurement outcome for pure state when it is eigenvalue of an operator.
And if we model measurement of entangled particle as changing the other particle to pure state I don't see how your argument holds.

The claim that the EPR argument allows you to prove the existence of functions $A,B:\Lambda\rightarrow\mathbb R$.

You will have to refresh my memory. In which post I claimed this?

 

[rubi]

It seems you are referring to this post of mine, right?

No, I was referring to your post #133.

And what about pure states? We can predict measurement outcome for pure state when it is eigenvalue of an operator.
And if we model measurement of entangled particle as changing the other particle to pure state I don't see how your argument holds.

I am arguing that the EPR idea that we can infer the existence of "elements of reality" using the EPR argument must be rejected, since the EPR argument would apply to the GHZ experiment as well, suggesting the existence of certain "elements of reality". However, we can prove that for the GHZ state the assumption of such "elements of reality" is inconsistent with the predictions of QM, which have been checked experimentally. Therefore, the EPR argument cannot be sound. Being able to predict something with certainty (without disturbing the system) does not necessarily imply the existence of some "element of reality", contrary to what the EPR argument would suggest. Of course, given an eigenstate of some operator, we can predict the outcome of certain experiments with certainty, but as I argued, that doesn't mean that we can infer the existence of an "element of reality", because the EPR argument just isn't sound.

We could now analyze why the EPR argument isn't sound and that would probably end up in a long discussion. But given the GHZ experiment, we must at least acknowledge that it isn't sound.

You will have to refresh my memory. In which post I claimed this?

You claimed it in post #133. At least that's how I understand that post.

 

[Zafa Pi]

However, we can prove that for the GHZ state the assumption of such "elements of reality" is inconsistent with the predictions of QM, which have been checked experimentally.

This, of course, is under the assumption of locality.

 

[Zafa Pi]

It's locality assumption that does not hold.

This was your response to my post #104, which asked:
Bell's inequality is the result of Bell's theorem. Experiments show Bell's inequality is not valid. Thus there must be some part of the hypothesis of Bell's theorem that is not valid. What do you think it is?

I agreed this was a valid option, and you also agreed that non-locality can't be proved.
As a theoretical question, suppose we assume locality, then what part of the hypothesis of Bell's theorem do you think is not valid? Do I need to state what I mean by Bell's theorem for you?

 

[zonde]

I am arguing that the EPR idea that we can infer the existence of "elements of reality" using the EPR argument must be rejected, since the EPR argument would apply to the GHZ experiment as well, suggesting the existence of certain "elements of reality". However, we can prove that for the GHZ state the assumption of such "elements of reality" is inconsistent with the predictions of QM, which have been checked experimentally. Therefore, the EPR argument cannot be sound. Being able to predict something with certainty (without disturbing the system) does not necessarily imply the existence of some "element of reality", contrary to what the EPR argument would suggest. Of course, given an eigenstate of some operator, we can predict the outcome of certain experiments with certainty, but as I argued, that doesn't mean that we can infer the existence of an "element of reality", because the EPR argument just isn't sound.

Look, it does not work without locality assumption. That "without in any way disturbing a system" is important part in EPR definition, and it really means without any disturbance. You can't ignore that.
And while "pure" measurements of pure states do not prove "elements of reality" they are consistent with them. So any part of QM that would use only "pure"measurements of pure states can't possibly falsify "elements of reality".
And then of course you can consider Bohmian mechanics as a hint that there is something wrong with your argument.

 

[secur]

Here's a simple proof that QM cannot be "counterfactual definite (CD)". Meaning that one cannot, in general, assign values for different experimental settings run on the same system at the same time, without contradiction. (Note, this is always possible classically, in theory if not in practice). It has nothing to do with locality.

Suppose Alice and Bob can each set their detectors to 0 or 30 degrees. Let the entangled electrons be in positive-parity "twin state", |11> + |00>, where 1 means spin up and 0, spin down. Suppose Bob's setting is opposite to Alice, so when they both set to 30, the total is 60 degrees.

They're able to run four distinct experiments, with settings (0,0), (0,30), (30,0), (30,30). Assuming CD we can assign values for all, even though we can only do one of them. That is, if we could run each experiment individually at the same time, we would get numbers from these assigned values. Suppose one experimental run generates values for N distinct electron pairs' detections. For convenience, suppose N is as large as we like, so statistical variance is negligible. There will be four sequences of length N: A0, A30, B0, and B30. Then QM says the following must happen.

First, A0 = B0. That is, for each of the N pairs in the two sequences, the values must match. To emphasize this fact let's call the single sequence S0.

S0 and A30 must have 1/4 mismatches, let's call that function M. I.e.,

M(S0, A30) = 1 - {sum from i=1 to n of abs(S0(i) + A30(i) - 1) / N} = sin^2(pi/6) = 1/4.

And the same is true for M(S0, B30).

Finally, we know from QM that M(A30, B30) must be sin^2(pi/3) = 3/4.

Obviously this is impossible. M(S0, A30) + M(S0, B30) = 1/2, setting an upper bound on M(A30, B30), less than 3/4.

QED.

This basically comes from Nick Herbert's 1985 book "Quantum Reality".

 

[Zafa Pi]

Here's a simple proof that QM cannot be "counterfactual definite (CD)". Meaning that one cannot, in general, assign values for different experimental settings run on the same system at the same time, without contradiction. (Note, this is always possible classically, in theory if not in practice). It has nothing to do with locality.

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.

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.
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.

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

 

[rubi]

This, of course, is under the assumption of locality.

Look, it does not work without locality assumption.

No, the GHZ experiment doesn't need any locality assumption. There is the so called GHZ state and it is incompatible with the idea of "elements of reality", no matter whether you perform the experiment with spacelike distances or not. You will get the same results in both cases. Why don't you first understand the GHZ argument before making such claims? I have given a link to an easily understandable article earlier.

That "without in any way disturbing a system" is important part in EPR definition, and it really means without any disturbance. You can't ignore that.

I don't ignore that. It is however irrelevant for showing the inconsistency of "elements of reality" with the predictions of QM and experiment. Please read the argument again. We can predict with certainty and without disturbing the system the value of Bob's spin after we have measured Alice's spin, both in the EPRB state and in the GHZ state. The EPR argument would then suggest the existence of an "element of reality". The EPR argument must thus be invalid. It's dead simple.

And while "pure" measurements of pure states do not prove "elements of reality" they are consistent with them.

No. The GHZ experiment shows that they aren't.

And then of course you can consider Bohmian mechanics as a hint that there is something wrong with your argument.

Bohmian mechanics is the perfect hint that my argument is completely correct. There are no element of reality for the spin values of the particles in Bohmian mechanics and Bohmians will agree to this. It is impossible. By the way, it is not "my" argument. It's GHZ's argument and I'm only repeating it. So you are in disagreement with highly respectable scientists.--
At this point, the discussion is becoming cumbersome. It makes no sense to discuss this if you don't make sure to have understood the GHZ experiment first.

Anyway, I have asked you earlier for a mathematical proof of the existence of the functions $A,B:\Lambda\rightarrow\mathbb R$, based on a mathematical formulation of the EPR argument. You have ignored this. I'm not wasting any more time on this unless you can present such a proof. It's pointless to keep discussing this with vague language like you do. You claim a mathematical statement, so you have to prove it.

 

[zonde]

It makes no sense to discuss this if you don't make sure to have understood the GHZ experiment first.

It was some time ago but I have read and analyzed GHZ experiments.

As I remember they compared QM prediction with non-contextual local hidden variable prediction.

Before going anywhere further I would like to add that I don't like EPR definition because it seems to exclude contextual hidden variables which certainly are realistic.
Say measurement of relative phase can theoretically give perfectly predictable result. Maybe that is what you are arguing about?

 

[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

[COLOR=#black]...[/COLOR]

 

[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?