Gull and Gill's theory

[gill1109]

Another point Jaynes makes is: According to QM
for all
ie, knowledge of the setting for the A measurement does not change any predictions at B. Yet simultaneously, according to QM

Why do you think . Do you think this is due to non-locality or any spooky reason? If you do, then you've misunderstood even more of what Jaynes is saying in which case I would ask you to again describe step-by-step how you would calculate/estimate from the same experimental data above obtained from your distributed computing experiment. And please be careful to explain why you would use a different procedure to calculate from the one used to calculate .

It is perfectly unsurprising that . What goes on at the two detectors is influenced by what happened at the source. Given a and b, B helps us predict A.

I estimate P(B|Aab) by #ABab/#Aab and P(B|a) by #Ba/#a, how would you do it?

[Heinera]

[...]
Why do you think .

This is true not only for the QM predictions, but for most LHV models as well, and is perfectly unsurprising as Richard says. Why do you think it has any relevance for Bell's theorem?

[gill1109]

[...]
Why do you think .

This is true not only for the QM predictions, but for most LHV models as well, and is perfectly unsurprising as Richard says. Why do you think it has any relevance for Bell's theorem?

Indeed.

Of course P(B|b) = 1/2 doesn’t have to be true. But P(B|ab)= P(B|a) is called no-signalling and it needs to be true by the principles of relativity theory.

Amusingly, Pearle’s detection loophole model does not satisfy another kind of no-signalling: the rate of double detections depends on both settings. Pearle knew this very well. Gisin & Gisin’s model does not have that defect.

[minkwe]

I estimate P(B|Aab) by #ABab/#Aab and P(B|a) by #Ba/#a, how would you do it?

You did not explain why you calculate them differently. Practically, the conditioning is used to select the sub-sample within which the calculation is done. It implies nothing about no-signaling or non-locality. There is no difference in how you calculate that from an experiment, irrespective of whether you assume a local or a non-local theory. You do it exactly the same by comparing the data from both sides, and selecting the subset within which outcome A was obtained while settings (a,b) were in force, and then within that subset, you calculate the relative frequency of the specific B outcome. You must always use the A outcome to select the subset for P(B|Aab). It is this information about the A outcome that is "non-local", not anything physical happening in the long-completed experiment.

Of course P(B|b) = 1/2 doesn’t have to be true. But P(B|ab)= P(B|a) is called no-signalling and it needs to be true by the principles of relativity theory.

You think P(B|ab)= P(B|b) is no-signalling, therefore QM forbids it because according to QM P(B|ab)= P(B|b) = 1/2.

Do you think P(B|Ab) = P(B|b) is also no-signalling? Because according to QM P(B|Ab) =/= P(B|b) which according to you should imply signalling no?

[minkwe]

[...]
Why do you think .

This is true not only for the QM predictions, but for most LHV models as well...

So "most LHV" theories also allow signaling?

[minkwe]

[...]
Why do you think .

... Why do you think it has any relevance for Bell's theorem?

Jaynes explains why by making another the crucial point that all the inequalities are therefore not required and we can avoid half a century of arguments about inequalities by doing just this simple experiment to test QM directly, using



And a simple photon-correlation experiment in which is 90 degrees.

[Heinera]

[...]
Why do you think .

This is true not only for the QM predictions, but for most LHV models as well...

So "most LHV" theories also allow signaling?

But your inequality does not imply signaling. Just consider any LHV model that gives opposite results for Alice and Bob when they use identical settings, e.g. Peres' "exploding balls" model. Anyone who knows Alice's result can predict Bob's result with certainty. If they don't know, it's 50/50. So . This has nothing to do with signaling.

[Heinera]

[...]
Why do you think .

... Why do you think it has any relevance for Bell's theorem?

Jaynes explains why by making another the crucial point that all the inequalities are therefore not required and we can avoid half a century of arguments about inequalities by doing just this simple experiment to test QM directly, using



And a simple photon-correlation experiment in which is 90 degrees.

This doesn't make sense. You have a dependency on A in the lhs, yet it is nowhere to be found in the rhs.

[minkwe]

This doesn't make sense. You have a dependency on A in the lhs, yet it is nowhere to be found in the rhs.

It makes perfect sense. Maybe you need to relearn probability theory.

shouldn't make sense to you then because there is no on the rhs.

[minkwe]

So . This has nothing to do with signaling.

Does have anything to do with signalling?

[gill1109]

So . This has nothing to do with signaling.

Does have anything to do with signalling?

Yes. Because it means that Bob can learn something about Alice's setting by seeing whether or not B happens (while, of course, knowing b). The probability of what he sees varies, depending on Alice's setting a. So he can rightly make different bets on what b is, depending on whether or not B happens (while knowing, of course, b).

[Heinera]

This doesn't make sense. You have a dependency on A in the lhs, yet it is nowhere to be found in the rhs.

It makes perfect sense. Maybe you need to relearn probability theory.

shouldn't make sense to you then because there is no on the rhs.

But then your expression

is simply wrong, if you mean that the lhs is the same value for any A. Do you want me to write the correct expression for you?

[gill1109]

This doesn't make sense. You have a dependency on A in the lhs, yet it is nowhere to be found in the rhs.

It makes perfect sense. Maybe you need to relearn probability theory.

shouldn't make sense to you then because there is no on the rhs.

But then your expression

is simply wrong, if you mean that the lsh is the same value for any A. Do you want me to write the correct expression for you?

I think that Michel means "Alice's outcome is +1" by the event A. Similarly, "B" means Bob's outcome is +1. And AB is the event "both outcomes are +1".

[Heinera]

AB? He writes ab. I thought lower case letters indicated detector settings.

[gill1109]

AB? He writes ab. I thought lower case letters indicated detector settings.

He writes both. "AB" means, for him, I think, both outcomes are +1. "ab" means, for him, both settings. Michel can tell us what he means. Everyone gets used to their own notation.

[Heinera]

He writes both. "AB" means, for him, I think, both outcomes are +1. "ab" means, for him, both settings.

Yes, that's how I understand it, and then everything I wrote is correct.

[Heinera]

So . This has nothing to do with signaling.

Does have anything to do with signalling?

Of course it does. Alice can control the setting a and thus can use it for signaling. But Alice cannot control the outcome A, so that can't be used to signal.

[minkwe]

This doesn't make sense. You have a dependency on A in the lhs, yet it is nowhere to be found in the rhs.

It makes perfect sense. Maybe you need to relearn probability theory.

shouldn't make sense to you then because there is no on the rhs.

But then your expression

is simply wrong, if you mean that the lsh is the same value for any A. Do you want me to write the correct expression for you?

I think that Michel means "Alice's outcome is +1" by the event A. Similarly, "B" means Bob's outcome is +1. And AB is the event "both outcomes are +1".

Correct! I thought it was clear from the mention of Jaynes what this is all about? See https://bayes.wustl.edu/etj/articles/cmystery.pdf pages 10-15. Gull is mentioned on Page 15.

[Heinera]

Correct! I thought it was clear from the mention of Jaynes what this is all about? See https://bayes.wustl.edu/etj/articles/cmystery.pdf pages 10-15. Gull is mentioned on Page 15.

Yes, we have sorted that out. I have used this meaning throughout in all my replies. Hopefully we can now get back to the topic.

[minkwe]

So . This has nothing to do with signaling.

Does have anything to do with signalling?

Of course it does. Alice can control the setting and thus can use it for signaling. But Alice cannot control the outcome , so that can't be used to signal.

Mathematically, there is no difference between and since A and "a" are just random variables. But according to you, one represents signaling and the other does not, which means it is not the mathematical expression that conveys signaling but your interpretation of it.

Does Alice have any influence over the outcome? If not then why does she bother changing at all? If she does, influence outcome by changing the setting then it would appear your distinction is superficial.

But first, what do you understand by "signaling"?