# What does this example say about the applicability of Bell's inequalities?

#### billschnieder

It is commonly believed that Bell's inequalities are a theoretical derivation of a condition that must be satisfied by locally causal theories. Therefore, it is often concluded that violation of these inequalities by experiments provides very strong evidence (if not conclusive proof, the only doubt being due to imperfect experiments) that reality is non-local.

However, I present a macroscopic example using coins where, Bell's theorem is always violated in the experiments. I would like to stimulate discussion of

1) why this specific example exhibits such behaviour?
2) where is the non-locality, or non-reality, or any other spooky conclusion?
3) what does this mean for the popular belief that Bell's inequalities are applicable to the EPRB experimental results?

Here it goes:

We have three coins labelled "a", "b", "c", if we toss all three a very large number of times, it follows that the inequality |ab + ac| - bc <= 1 will never be violated for any individual case and therefore for averages |<ab> + <ac>| - <bc> <= 1 will also never be violated (where heads = +1 & tails = -1).

Proof: For three coins with outcomes ±1, there are 8 possibilities. the LHS for each possibility is always <=1 as illustrated below:
a,b,c = (+1,+1,+1): |(+1) + (+1)| - (+1) <= 1, obeyed
a,b,c = (+1,+1,-1): |(+1) + (-1)| - (-1) <= 1, obeyed
a,b,c = (+1,-1,+1): |(-1) + (+1)| - (-1) <= 1, obeyed
a,b,c = (+1,-1,-1): |(-1) + (-1)| - (+1) <= 1, obeyed
a,b,c = (-1,+1,+1): |(-1) + (-1)| - (+1) <= 1, obeyed
a,b,c = (-1,+1,-1): |(-1) + (+1)| - (-1) <= 1, obeyed
a,b,c = (-1,-1,+1): |(+1) + (-1)| - (-1) <= 1, obeyed
a,b,c = (-1,-1,-1): |(+1) + (+1)| - (+1) <= 1, obeyed

The Bell inequality Challenge:
Find a locally realistic situation which violates the above inequality. According to Bell and proponents, this is impossible to do.

Experimental Violation:
We have 3 coins labeled "a","b","c", inside a special box. A button on the box releases only two of the coins at random when pressed. The coins must be returned to the box before the next press, therefore only two of the three coins can ever be outside of the box at the same time.

Since we can not toss all three at once, and since the inequality only contains averages of pairs of outcomes, we assume the inability to measure all three simultaneously is inconsequential. We decide to perform the experiment by tossing just the pairs a very large number of times and group the results into 3 runs for the pairs (a,b), (a,c), (b,c) tosses. Using the pairs, we calculate <ab>, <ac> and <bc>. Even though the data for each pair appears random, we find that from our data <ab> = -1, <ac> = -1 and <bc> = -1, which violates the inequality when substituted into the LHS. (i.e 3 <= 1 according to the inequality). This was supposed to be impossible according to the challenge!!! Does this mean "local causality" or realism is false?

The Explanation:
Consider the following: Each coin has a programmable bias which can be changed by the box just before it is released but not after. The box has an internal clock which keeps track of the time (t) in seconds. The above scenario [<ab> = -1, <ac> = -1 and <bc> = -1 ] can then easily be realized if the special box operates as follows:

Every time a button is pressed, calculate calculate sin(t) where t is the time read off the internal clock. If sin(t) > 0, program coin "a" to be biased for heads (+1) and coin "c" to be biased for tails (-1). Then randomly pick two of the three coins. If coin "b" is one of the picks, program coin "b" to be biased for tails (-1) if the other pick is coin "a", otherwise program coin "b" to be biased for heads (-1). If sin(t) <= 0 reverse all the signs.

Conclusion:
a) The box will always produce <ab> = -1, <ac> = -1 and <bc> = -1, no matter how many times the coins are tossed, (1 or 50 billion).
b) The results will be random
c) The inequality will always be violated no matter how many times the coins are tossed (1 or 50 billion)
d) The box and coins operate in a completely locally causal manner.
e) The result of each toss is non-contextual and predetermined from the moment the coins are produced
f) There are no loopholes in the experiment
f) There is no spooky business happening
g) YET Bell's inequality is violated.

How come?

Related Quantum Physics News on Phys.org

#### Nugatory

Mentor
Oh, I am going to be SOOOOOO sorry that I stepped into this..... But, altogether against my better judgement.... Here goes.....

With every button press, the box is producing a pair of coins, one biased for heads and one biased for tails. How is that any different than me randomly leaving one glove out of a pair behind when I leave home, and then "discovering" that there is a perfect anti-correlation between the handedness of the glove I have with me and the one that I left at home?

#### StevieTNZ

I guess the difference between the thought experiment and your explanation is that the experiment states there are pre-determined results, but bell's inequality is still violated.

#### billschnieder

With every button press, the box is producing a pair of coins, one biased for heads and one biased for tails. How is that any different than me randomly leaving one glove out of a pair behind when I leave home, and then "discovering" that there is a perfect anti-correlation between the handedness of the glove I have with me and the one that I left at home?
Hi Nugatory,
I'm not sure what you are asking but the difference is that the perfect anti-correlation is between the two coins actually produced and the person doing the experiment does not know anything other than what they actually measure and therefore can not use such hidden knowledge to calculate correlations.

#### rlduncan

You basically have taken up David Mermin’s challenge (Is the moon there when nobody looks? Reality and the quantum theory) to find an instruction set to account for the experimental results. Everyone recognizes the obvious instruction set, that is, the source guarantees the correlation (or anti-correlation) when choosing the same coin, orientation angle, etc. You now offer a second instruction set in the form of sin(t) when choosing different coins to demonstrate that Bell’s inequality can be violated for a locally realistic experiment when selecting two of the three coins. Sin(t) is significant in that it makes the connection to the EPRB experiments.

Why wouldn’t the instructions sets come in pairs? This seems necessary to explain the random results.

#### Fightfish

Oh, I am going to be SOOOOOO sorry that I stepped into this..... But, altogether against my better judgement.... Here goes.....
Haha, same sentiments here.

So here's my take:
Essentially there is nothing much different from your instruction set and an entangled pair. The reason why your experiment violates Bell's inequalities, I believe is because it, like the case of EPR pairs, does not exist in a definite state until you press the button (ie making the measurement).

One could see the EPR pairs as such: (consider the (|01> + |10>)/sqrt(2) case).
So, I have a source producing them. When I make a measurement on the first qubit, the following "program", similar to yours, may run:
Pick a number t. If sin(t) > 0, program qubit 1 to be |0> and qubit 2 to be |1>. If sin(t)<0, program qubit 1 to be |1> and qubit 2 to be |0>. (sin(t) is positive 50% of the time)
Why this violates Bell's inequalities is because the qubits do not exist in a definite state until I perform my measurement.

So, because your coins don't have a definite state until you press the button, they violate Bell's inequalities. But there is nothing strange or spooky there, because you have a machine that performs the program. But for two qubits separated over large distances? What sort of "program" and "system" can connect them? - That is where the strangeness and nonlocality comes in.

#### billschnieder

Essentially there is nothing much different from your instruction set and an entangled pair. The reason why your experiment violates Bell's inequalities, I believe is because it, like the case of EPR pairs, does not exist in a definite state until you press the button (ie making the measurement).
Not sure what you mean by definite state in the above. But each outcome of the toss is definitely predetermined. Even though you do not know the outcome of the toss until you toss the coins. Comparing to EPRB, the pressing the button would be equivalent to a source releasing two anti-correlated particles, or is that considered a "measurement" in EPRB too? And the measurement would be the tossing of the coins.

Why this violates Bell's inequalities is because the qubits do not exist in a definite state until I perform my measurement.
The coins in the above example exist in a definite state from the "source" (box) prior to being measured even though the outcome is unknown by the experimenter until after the measurement (toss).

So, because your coins don't have a definite state until you press the button, they violate Bell's inequalities. But there is nothing strange or spooky there, because you have a machine that performs the program. But for two qubits separated over large distances? What sort of "program" and "system" can connect them? - That is where the strangeness and nonlocality comes in.
Replace the box and button with the particle source in EPRB. The measurement then is only the tossing, not the pressing of the button. The box which produces the coins, has a definite non-spooky behaviour, each coin has a definite non-spooky behaviour. The behaviours pre-exist any measurement. Everything is local and realistic. If you like, we can separate the coins and send them off at the speed of light to remote locations where they are tossed.

Just to understand what you mean by definite state, do you think dynamic systems have a definite state?

#### billschnieder

Sin(t) is significant in that it makes the connection to the EPRB experiments.
Sin(t) does two things:
- is responsible for the random results.
- produces P(+1) = P(-1) = 0.5

Since the experimenters do not know anything about the how fast the internal clock of the box is or even that there is an internal clock, they have no way of selecting sin(t)>0 or sin(t) <=0. Thus they get random results which are symmetric around zero.

You are right, both of these features are very reminiscent of the EPRB experiments (on purpose).

#### Fightfish

Comparing to EPRB, the pressing the button would be equivalent to a source releasing two anti-correlated particles, or is that considered a "measurement" in EPRB too? And the measurement would be the tossing of the coins.
No. The releasing of two anti-correlated particles is not equivalent to the button pressing in your experiment. In your experiment, even before pressing the button, the outcome is already known to be anti-correlated, just that coin A, for instance can be either +1 or -1. So the source releasing the particles is just "setting up". While particle 1 has 50% probability of being spin up or spin down, it is neither spin up nor spin down until you make a measurement!

Pressing the button in your case IS a measurement, because it assigns a value to coin A. So before you toss, you have already "disturbed the system".

Going back to the EPR pairs, when we measure (for example with the polarising axis oriented in between the two polarisation modes), the photon actually experiences two things:
First, the photon is forced into being either horizontally or vertically polarised. ("Button pressing")
Second, the photon passes through the polariser with a probability given by Malus' Law. ("Tossing")

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#### San K

We decide to perform the experiment by tossing just the pairs a very large number of times and group the results into 3 runs for the pairs (a,b), (a,c), (b,c) tosses. Using the pairs, we calculate <ab>, <ac> and <bc>. Even though the data for each pair appears random, we find that from our data <ab> = -1, <ac> = -1 and <bc> = -1,
how do we get ab (or ac, bc) = -1?

not sure if I understand your experiement or nomenclature. however from what I understand-
half of the time ab will be plus 1 and half of the time it will be -1. the average of ab would then be zero.

#### Delta Kilo

Bill, what is the main assumption of Bell's theorem? Let me remind you, and I quote:
Bell said:
The vital assumption [2] is that the result B does not depend on setting $\vec{a}$, of the magnet for particle 1, nor A on $\vec{b}$.
Now, what have we here:
If coin "b" is one of the picks, program coin "b" to be biased for tails (-1) if the other pick is coin "a", otherwise program coin "b" to be biased for heads (-1).
Can you spot the difference?

The whole point of Bell setup is to have two spatially separated detectors where settings can be changed independently and at will and settings of one detector do not influence the results of another. Your model does not have anything like that. You cannot change settings at all, let alone independently and spatially separated.

Yes, the implications of Bell's theorem and it's experimental violations are profound, but the formulation and the proof are straightforward and uncontroversial. I mean, I could understand if it was your first post on the subject, but just how many times have we been through that A(a,λ) B(b,λ) stuff already? If you would bother to try to apply it to your setup, it would be immediately obvious that it does not fit, that in your case A in fact depends on b. I'm sorry to say but all this example shows is billschnieder is either trolling hard or is descending into crackpottery.

#### billschnieder

No. The releasing of two anti-correlated particles is not equivalent to the button pressing in your experiment. In your experiment, even before pressing the button, the outcome is already known to be anti-correlated.
Just like for the EPR case in which we know due to conservation of momentum that the particles are anti-correlated. What is the point?

So the source releasing the particles is just "setting up". While particle 1 has 50% probability of being spin up or spin down, it is neither spin up nor spin down until you make a measurement!
But that is just an assumption you are making. The difference between the real EPR experiment and the coin case above is that I have explained to you the detailed mechanism, which experimenters in the EPR case have no clue about. All they know is that they toss and get either a +1 and a -1. You can't argue that pressing the button in one case is different from "setting-up" in another based on having full information in one case and zero information in another.

Pressing the button in your case IS a measurement, because it assigns a value to coin A. So before you toss, you have already "disturbed the system".
Had I known that you would get hung up on the "pressing". I would have removed it completely. It is inconsequential. I could have said the box decides by itself at what time it releases the pair of coins without any pressing. How then would you argue that the system is "disturbed"?

#### billschnieder

how do we get ab (or ac, bc) = -1?

not sure if I understand your experiement or nomenclature. however from what I understand-
half of the time ab will be plus 1 and half of the time it will be -1. the average of ab would then be zero.
That is not correct. You missed this part:

Every time a button is pressed, calculate calculate sin(t) where t is the time read off the internal clock. If sin(t) > 0, program coin "a" to be biased for heads (+1) and coin "c" to be biased for tails (-1). Then randomly pick two of the three coins. If coin "b" is one of the picks, program coin "b" to be biased for tails (-1) if the other pick is coin "a", otherwise program coin "b" to be biased for heads (+1). If sin(t) <= 0 reverse all the signs.
The predetermined outcome for "a" is always opposite to that of "b". when the "ab" pair is produces Same for "ac" and same for "bc". So the products "a*b", "a*c" and "b*c" are always -1, therefore their averages after many tosses are always -1, not zero.

EDIT: There was a typo in the original where I said "heads (-1)" instead of heads "(+1)". That may have confused the issue.

#### billschnieder

Bill, what is the main assumption of Bell's theorem? Let me remind you ...
Despite the typo in the OP which used the phrase "Bell's theorem" in place of "Bell's inequalities", this is not a discussion of Bell's theorem but Bell's inequalities. The point is not to claim that the EPR exeriment is the same as the coin-toss experiment. So don't tell me that the example fails because "coins are not photons" or any similar claims etc.

The coin toss example above is self contained and parallels can be drawn. The inequalities are demonstrated to be valid in the theoretical treatment, yet violated in the experiment just like with Bell. The theoretical proof assumes all three outcomes from three coins are present at the same time, just like Bell case. The experimenters are limitted to measuring only two of the three outcomes at a time, just like in the Bell case. The experiment violates the inequalities, just like in the Bell case. The inequalities in both are Bell's inequalities. The calculation of expectation values in the experiments are done in the same way. In both cases, the outcomes are +1 or -1.

So the question again is this. How come the inequalities which were supposed to be valid for the "coin-toss" experiment as demonstrated earlier in the OP, get violated by the experiment? It is clear that the experiment is local and realistic.

The whole point of Bell setup is to have two spatially separated detectors where settings can be changed independently and at will and settings of one detector do not influence the results of another. Your model does not have anything like that. You cannot change settings at all, let alone independently and spatially separated.
Are you suggesting that if Alice and Bob in the real EPR experiment were given a fixed value of "a", "b", "c" and not allowed to change them, the inequality will never be violated? Or are you suggesting that if Alice and Bob were just 1m apart and not spacially separated in the real EPR experiment, the inequality will not be violated? Hopefully you now see (by answering these questions) that these issues you raise are irrelevant for the main point here. Now rather than throw accusations around, it could be usefuly if you calmed down and explain why you think the proven inequalities which appear valid from the theoretical proof in the OP, does not apply to the experiment described in the OP.

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#### rlduncan

Bill, you asked when comparing the theoretical and the experimental,

Why the violation?

1) The machine selects two of the coins at any one time instead of three. This has been discussed before.

2) The fact that the results for <ab>,<bc>, and <ac> = -1. Not sure of the deeper meaning of this fact. Surely, you had a reason for choosing this value as the outcome.

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#### lugita15

Are you suggesting that if Alice and Bob in the real EPR experiment were given a fixed value of "a", "b", "c" and not allowed to change them, the inequality will never be violated? Or are you suggesting that if Alice and Bob were just 1m apart and not spacially separated in the real EPR experiment, the inequality will not be violated?
It's not that the Bell inequality will never be violated in such a case, it's that the inequality won't be derivable in the first place. It's only when you assume that the particle behavior is independent of the settings of distant polarizers that you are able to derive the inequality.

#### Delta Kilo

Despite the typo in the OP which used the phrase "Bell's theorem" in place of "Bell's inequalities", this is not a discussion of Bell's theorem but Bell's inequalities.
...<groan>... it's the same bloody thing! The inequality is the main result of the theorem! It is only valid as long as the assumptions of the theorem are satisfied. In your case the vital assumption is not satisfied, so the inequality is inapplicable.

1. Do you agree that your setup does not satisfy Bell's "vital assumption", as quoted earlier?
2. Do you agree that the derivation of Bell's inequality requires "viltal assumption" to be valid?
3. Can you now make a great leap of logic and put 1. and 2. together and agree that Bells inequalities are inapplicable to your setup?

#### DrChinese

Gold Member
Every time a button is pressed, calculate calculate sin(t) where t is the time read off the internal clock. If sin(t) > 0, program coin "a" to be biased for heads (+1) and coin "c" to be biased for tails (-1). Then randomly pick two of the three coins. If coin "b" is one of the picks, program coin "b" to be biased for tails (-1) if the other pick is coin "a", otherwise program coin "b" to be biased for heads (+1)[corrected]. If sin(t) <= 0 reverse all the signs.
Bill,

This algorithm, by your own admission, produces -1 (perfect correlation) for every trial. No reason to put the other stuff in. You may as well say that a perfectly correlated pair comes out every time with a random orientation of +1-1 or -1+1.

So then the question becomes: does a perfectly correlated outcome violate a Bell Inequality? That is after all the criteria for the EPR state (elements of reality). Clearly the answer should be NO if this is a good analogy, because this case was considered specifically by Bell. He considered this as being simple.

But here the answer is actually YES (as you say), because we have not specified any angle settings. You get the same results at all angle settings, including 0 and 90 (which should be anti-correlated relative to each other).

So we conclude the following:

a) This analogy does not mimic a Bell setup as it produces predictions which are counter to any experiment. Ergo, it predicts a different result than QM, which matches experiment. Therefore the main result of Bell, stated below stands:

No physical theory of local Hidden Variables can ever reproduce all of the predictions of Quantum Mechanics.

b) It is not truly local, as the results are observer dependent in most cases (whenever b is one of the 3 selected, which occurs about 2/3 of the time). In other words, the outcome b is predicated on whether the observer gets a or c for the other coin. This is a contextual setup, which essentially will always violate the locality/separability condition. Note that b appears to be undefined when a and c are selected, though you could fix that and would need to if you claim to have a realistic example.

This analogy doesn't pass the test, no matter what you say about your Bell Inequality being violated.

#### billschnieder

...<groan>... it's the same bloody thing!
No it is not. Bell's theorem relates the inequalities to QM. The inequalities themselves do not need QM to be derived. Get it? Just because the inequalities are relevant for the theorem does not mean the inequalities by themselves are not worthy of independent discussion without being side-tracked by reference to the theorem.

1. Do you agree that your setup does not satisfy Bell's "vital assumption", as quoted earlier?
I disagree that the "vital assumption" quoted earlier is relevant to this thread. Simply look at the OP and explain why the proof of the inequality stated in the first part of the OP should not apply to the second part of the OP. Everything is contained in the openning post. Not need to waste time looking elsewhere.
2. Do you agree that the derivation of Bell's inequality requires "viltal assumption" to be valid?
Again, look at the openning post where the inequality is proven by brute force calculation of all the possibilities. Is it your claim that the inequality is different from Bell's? Just because it was proven differently? So your so called "vital assumption" may be interesting for another thread not this one.

3. Can you now make a great leap of logic and put 1. and 2. together and agree that Bells inequalities are inapplicable to your setup?
Again it is you who is failing to make the great leap of logic: We have the proof of the inequalities for three coins labelled "a", "b", "c", not angles. You do not argue that the inequality which is proven in the openning post is invalid. Neither do you argue that the inequality is not Bell's inequality. Then we have an example using three coins which violates the inequality just proven. The issue for discussion is: How come? Everything required for answering the question is contained in the openning post. You do not need any reference to QM predictions because there is none for the specific example with coins mentioned in the OP. We can go off trail discussing how "angles" are missing from the example, or how distant separation is missing, but nobody expects coins to be tossed at angles etc.

So again here are the issues for this thread.
1) Are the inequalities proven in the openning post valid for the situation described?
2) How come it is violated by the experiment described?
3) What does this mean about the applicability of inequalities of the type proven in the OP, to experiments of the type described in the OP?

#### billschnieder

Hi DrC,
Thanks for correcting the typo in your quote.
This algorithm, by your own admission, produces -1 (perfect correlation) for every trial.
Actually, it is perfect anti-correlation since the product of the two coins is always -1. Perfect correlation would be a product of +1 every time.

So then the question becomes: does a perfectly correlated outcome violate a Bell Inequality? That is after all the criteria for the EPR state (elements of reality). Clearly the answer should be NO if this is a good analogy, because this case was considered specifically by Bell.
If the question was changed to "does a perfectly anti-correlated outcome violate a Bell Inequality?", the is YES. To verify replace each of the expectation values with a -1 and you get 3 on the LHS.

You get the same results at all angle settings, including 0 and 90 (which should be anti-correlated relative to each other).
Again as I explained to DK, focusing on angles just muddies the water and takes us off topic. There are no angles involved in the example. "a", "b", "c" are just coins. In case you are tempted to think angles are relevant for "Bell's inequalities" as opposed to "Bell's theorem", remember than "a", "b", "c" in the mathematical derivation performed by Bell are just symbols. The derivation will proceed in exactly the same way to obtain the inequalities, no matter what those symbols represented. The algebra used to derive the inequalities makes absolutely no use of the idea that values for "a", "b", "c" can be changed, which means the inequalities are valid even if "a", "b", "c" are fixed. In our case "a", "b", "c" are just labels for three coins. The fact that the inequalities are valid even for the case in which "a", "b", "c" represent three coins is proven in the openning post. Just in case this is not clear still. Imagine that in the actual EPR experiment, the angles were fixed to just those in which a violation occured, and Alice and Bob were not allowed to change them. Would you argue in that instance that the inequality is not valid because they were not allowed to change the angles? The simple fact that a violation occured whether the angles could be changed or not is significant. That is the scenario you should be comparing this example with, if any.

So we conclude the following:

a) This analogy does not mimic a Bell setup as it produces predictions which are counter to any experiment. Ergo, it predicts a different result than QM, which matches experiment. Therefore the main result of Bell, stated below stands:
And what may those predictions of QM for a coin-toss experiment be? You are comparing apples and oranges here. No claim is made that this example reproduces QM results. The issue is the inequalities, period.
b) It is not truly local
Now that is an interesting claim. You think the example as described in the openning post is non-local? Please elaborate.

In other words, the outcome b is predicated on whether the observer gets a or c for the other coin.
But the coins are produced that way by the box due to conservation of it's mechanism of functioning.

This is a contextual setup, which essentially will always violate the locality/separability condition.
Really? At any one moment in time, each of the three coins has a definite outcome value. Prior to the tossing of the coins so it is not contextual. Unless you have a different meaning for the word.

Note that b appears to be undefined when a and c are selected, though you could fix that and would need to if you claim to have a realistic example.
I did not tell you what b was in that case but of course you understand that it could be easily fixed and it will be realistic and still violate the inequality.

So let us see:

- Every outcome is predetermined
- Every outcome is known for certain before the toss
- Every coin has a definite state before the toss
- The box and coins operate in a local and realistic manner.
- The outcomes are non-contextual (not obeserver dependent) -- there is nothing the observers can do to change the results except refuse to toss.

Yet the inequality is violated. Why?

#### billschnieder

It's not that the Bell inequality will never be violated in such a case, it's that the inequality won't be derivable in the first place.
Really? Look again at the openning post. The very first part provides a proof of the inequality for the very example involving three coins labelled "a", "b", "c". So your claim that the inequalities are not derivable is just not true.

#### JK423

Gold Member
This example that the OP gives is very good in showing the problems you're driven into when considering the results of mutually exclusive experiments, as done in the quantum case as well.
On one hand, i can see that this classical situation isn't entirely similar to the quantum case, since we have no freedom on which coins we will measure. You give me two specific coins, you tell me 'measure them' and this way the inequality is violated 'by hand'. It's like putting the numbers in the inequality by hand. If i was able to choose however which of the 3 coins to measure, the inequality would not be violated..
On the other hand, it's not yet clear to me what happens in the quantum case since we use results of mutually exclusive experiments to violate the Bell inequality.. Ofcourse, in this case we have freedom of measurement choice, something that isn't present in the OP's example. But still, im not sure yet if the freedom of choice is enough to assure that including results of exclusive experiments is consistent and doesn't lead to absurd results.
One first thought on this problem is to try to consider a classical experiment (using hidden varibles), where given freedom of measurement choice we can still violate Bell's inequality. If it turns out that this isn't possible, then we can safely conclude that there's something strange about quantum mechanics :). But if it is possible then we have a serious problem... My personal opinion is that it's not possible, but i'll give it a little more thought these days.

#### Delta Kilo

Is it your claim that the inequality is different from Bell's?
Yes. In your 'proof' the inequality applies to averages from the same series of $\{a_i,b_i,c_i\}$, where the average is defined as arithmetic mean of all elements in the series, even though you did not explicitly state it. It does not allow you to pick different subsets to calculate <ab>, <ac>, and <bc>. But of course, since you did not articulate your proof, you have missed this. If, instead of handwaving, you actually try to write out the steps, this is as far as you can get. Bell's inequality, on the other hand, applies to expectation values. It is a high time you learn the difference between the two.

Just because it was proven differently?
Of course. Applicability of any formula depends on the underlying assumptions.
1) Are the inequalities proven in the openning post valid for the situation described?
No.
2) How come it is violated by the experiment described?
Because your so-called 'proof' does not hold water.
3) What does this mean about the applicability of inequalities of the type proven in the OP, to experiments of the type described in the OP?
That you have to learn the difference between average and expectation value. Didn't I tell you that about a month ago?

#### DrChinese

Gold Member
1. Actually, it is perfect anti-correlation since the product of the two coins is always -1. Perfect correlation would be a product of +1 every time... If the question was changed to "does a perfectly anti-correlated outcome violate a Bell Inequality?", the is YES. To verify replace each of the expectation values with a -1 and you get 3 on the LHS.

2. Again as I explained to DK, focusing on angles just muddies the water and takes us off topic. There are no angles involved in the example. "a", "b", "c" are just coins. In case you are tempted to think angles are relevant for "Bell's inequalities" as opposed to "Bell's theorem", remember than "a", "b", "c" in the mathematical derivation performed by Bell are just symbols...

3. Now that is an interesting claim. You think the example as described in the openning post is non-local? Please elaborate...

- The outcomes are non-contextual (not obeserver dependent) -- there is nothing the observers can do to change the results except refuse to toss.

4. Yet the inequality is violated. Why?
1. I know that in your particular example, perfect correlation is a -1 result. So I am not questioning the arithmetic.

2. Well, this does matter if you are setting up a comparison to Bell. There must be 3 angles, usually labelled a/b/c or A/B/C, so that Alice can pick 1 and Bob can pick 1. The selection is done independently by each. This is necessary so we can have condition that the result will be local.

I am not questioning that you have the right to specify the outcomes in advance; in fact to be realistic, you MUST do this for all 3. Your example meets this criteria sufficiently.

3. Here is a key issue: you change the results according to which 2 of a/b/c are selected. This violates one of the Bell conditions which is necessary to get the inequality. You cannot bypass this and expect to convince anyone that this is a local realistic simulation. The rule is: you must specify the possible results independent of what attribute (a/b/c) the observers Alice & Bob freely choose to examine.

4. The root rule is not the CHSH inequality. The root requirements are:
i) the possibility of any permutation must be within the range of 0 to 1 (realism);
ii) the outcome cannot change based on what the observers choose to look at (locality or separability or whatever you want to call it).

In your example, i) is satisfied but ii) is not. Note that QM predicts values outside the range of i). You can see that negative probabilities are such a prediction at:

http://drchinese.com/David/Bell_Theorem_Negative_Probabilities.htm

To convince yourself that your example, if properly respecting ii), would not violate any type of Bell inequality, just write down a set of data points for your a/b/c (we have had this discussion previously of course). To make things work out for you, I will make it simple: I will always select a for one of the two, and will randomly pick between b and c. That way, you can respect ii) (since the outcome b does not change). Below are the only 2 permutations:

a b c
+ - -
- + +

This has <ab>=-1, <ac>=-1, <bc>=+1, and the inequality is not violated (as it was when requirement i was violated). QED.

#### billschnieder

2. Well, this does matter if you are setting up a comparison to Bell. There must be 3 angles, usually labelled a/b/c or A/B/C, so that Alice can pick 1 and Bob can pick 1. The selection is done independently by each. This is necessary so we can have condition that the result will be local.
Again I ask you, if Alice and Bob are given their devices already fixed to a given angle, will you then argue that violation of the inequality is not significant? If not why do you think that line of argument is relevant here?
3. Here is a key issue: you change the results according to which 2 of a/b/c are selected.
You forget that the experimenters do not select which coins they get. Whatever they get, that is what they measure. So it is wrong to say the results are "changed" based on what is selected. The coins they get is the coins they get. It is not up to them to "select" any thing so the results are not changed based on the situation. The results for each toss are predetermined, not contextual.
This violates one of the Bell conditions which is necessary to get the inequality.
This is the crux of the matter. Could you elaborate what condition exactly which is violated. Please do not tell me about angles because the proof in the opening post does not use any angles yet it is a valid proof. So what condition required to obtain the proof in the OP is violated by the experiment in the OP.
You cannot bypass this and expect to convince anyone that this is a local realistic simulation.
This is common sense. You do not have to look at any inequality to see that the experiment is local and realistic. There is no spooky action at a distance, there is no instantaneous influence. So it is surprising that you continue to claim that the experiment is non-local. In fact this experiment can be performed by humans. Just place a human in the box give him a stop-watch and a hand calculator, and give him 3 coins labelled "a", "b", "c". It is clearly obvious that everything is local and realistic and local and you know it.
The rule is: you must specify the possible results independent of what attribute (a/b/c) the observers Alice & Bob freely choose to examine.
4. The root rule is not the CHSH inequality. The root requirements are:
i) the possibility of any permutation must be within the range of 0 to 1 (realism);
ii) the outcome cannot change based on what the observers choose to look at (locality or separability or whatever you want to call it).
Again, there is no free chosing of coins in the example in the OP. Yet the inequality is proven (see the first part of the OP). We calculate the LHS for every possible combination and observe that none of them ever violates the inequality. That is sufficient proof that the inequality is valid for three coins labelled "a", "b", "c", freedom of choice is just a red-herring in this example. Secondly as already explained, the outcomes are predetermined and the observers have no say in choosing what they look at, so it is wrong to suggest that the outcome in this example changes based on what observers choose to look at. The predetermined outcome in this example simply changes with time and that is sufficient to violate the inequality maximally.
In your example, i) is satisfied but ii) is not. Note that QM predicts values outside the range of i). You can see that negative probabilities are such a prediction at:
http://drchinese.com/David/Bell_Theorem_Negative_Probabilities.htm
Negative probabilities do not make sense, unless you redefine what is mean by "probability". But that is fodder for a different thread.
To convince yourself that your example, if properly respecting ii), would not violate any type of Bell inequality, just write down a set of data points ...
No need to, I have already proven the inequality in the first part of the OP by calculating for every possibility. Yet we have here an obviously local and realistic macroscopic experiment which violates the proven inequalities for a specific reason which has nothing to do with locality or realism or freedom of choice. Why.

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