Understanding bell's theorem: why hidden variables imply a linear relationship?

[lugita15]

The probabilities P(A), P(B), and P(C) depend on the results of both photons.

Then those probabilities are conditional probabilities.

They don't seem like conditional probabilities to me. P(A) is not "the probability of mismatch when the polarizers are oriented at -30 and 0, given that the result of photon 1 was such-and-such and the result of photon 2 was so-and-so." Rather, it is just "the probability of mismatch when the polarizers are oriented at -30 and 0, given no information as to what the results of the two photons are".

Besides, when you look deeply enought, all probabilities are conditional.

Well, I suppose that's trivially true. Any regular probability can be written as "the probability that such-and-such is the case, given nothing."

So what in your opinion is P(ABC) for scenario (b)? Unless P(ABC) is well defined, an expression involving P(A), P(B) and P(C) does not make any practical sense.

Sorry, what does P(ABC) mean? In fact, what does ABC mean? Does it mean "A and B and C" or does it mean something else?

Le us represent the sets of photons as one set "w" for scenario (c), and three different sets ("x", "y","z") for scenario (b). The the probabilities for (c) are more accurately represented as

P(A|w), P(B|w), P(C|w).

I don't think this is the right notation. Usually a conditional probability is of the form "the probability that statement X is true, given statement Y." But you're putting a set of photons on the other side of the bar. I think you mean "The probability that A is true for the photons in w", not "The probability that A is true given w", which doesn't make sense to me. But anyway, this is a minor point, since it is just a notation.

Surely we can combine these probabilities into the same expression and obtain the trivial inequality P(C|w) <= P(A|w) + P(B|w). This makes sense because we are dealing with the same space "w" (aka, the same context, condition, etc). There is no question here.

Good, at least we're agreed on that.

However the probabilities for scenario (b) are more accurately represented as P(A|x), P(B|y), P(C|z). You are saying that because for the specific physical scenario you described, P(A|x) = P(A|w) and P(B|y) = P(B|w) and P(C|z) = P(C|w), therefore P(C|z) <= P(A|x) + P(B|y) should be valid.

Yes, that is exactly what I'm saying. If we know that p≤q+r, and we also know that p=p',q=q', and r=r', can't we conclude that p'≤q'+r'? How can you possibly disagree with that?

But that is not very credible. The inequality is a deeper relationship between probabilities not just specific values of the probabilities.

This makes absolutely no sense to me. It's just an inequality. It just says that this number is less than or equal to this number plus this number. Then we're just using the substiution property of equality.

It seems so obvious to me. P(A|x)=.25, P(B|y)=.25, and P(C|z)=.75. Thus P(A|w)=.25, P(B|w)=.25, and P(C|w)=.75. And you agreed that P(C|w) ≤ P(A|w) + P(B|w). Then we reach the conclusion .75≤.25+.25. What do you think I'm doing wrong here?

 

[San K]

And you agreed that P(C|w) ≤ P(A|w) + P(B|w). Then we reach the conclusion .75≤.25+.25.

The mismatch is 0.25 more than expected (by the additive law of probability)

What is the explanation for that per QM/QE?

Is it something like -

Since the two particles are entangled and act as one prior to interaction (with the polarizer):

the angle/orientation at both the polarizers will be "taken into account" and this causes the increase in mismatch.

side note: Do the calculation involve some sort of "joint probabilities" ?

 

[billschnieder]

Well, I suppose that's trivially true. Any regular probability can be written as "the probability that such-and-such is the case, given nothing."

"Given nothing", you have nothing. But you always have something by which to define you probability space otherwise any calculation is meaningless.

Sorry, what does P(ABC) mean?

That is the joint probability for the events A, B, C.

But you're putting a set of photons on the other side of the bar I think you mean "The probability that A is true for the photons in w", not "The probability that A is true given w", which doesn't make sense to me.

If A = "mismatch at -30 and 0". P(A|w) is the probability of mismatch at -30 and 0 for the probability space "w". If "w" was all we were ever talking about, it will make sense to simply leave it out and write P(A) but that will not mean P(A) is not a conditional probability. However your argument is not such a case; therefore to to be precise and avoid mistakes, it is recommended to write P(A|w).

How can you possibly disagree with that?This makes absolutely no sense to me. It's just an inequality. It just says that this number is less than or equal to this number plus this number. Then we're just using the substiution property of equality.

I can disagree with that because I'm not limitting my thinking to an expression on a sheet of paper or on screen. I can disagree because I understand the difference between an empirical manipulation and an algebraic manipulation. You are focusing on the algebraic and ignoring the empirical. This is what I've been trying to tell you about Boole's work.

And if you attempt to tell me what the joint probability P(ABC) is for both scenario (b) and (c) then you may start having a paradigm shift to understand the issue I have with your argument. I suppose your no-conspiracy condition implies that they two are equal. But that is impossible because P(ABC) is undefined for scenario (b). Let me even make it easier: your expression P(C|w) <= P(A|w) + P(B|w) is actually derived from the equality P(C|w) = P(A|w) + P(B|w) - P(AB|w) so let us stay with the equality for a moment. Please use your so called "no-conspiracay condition" to write down the equivalent equality for scenario (b) with appropriate notation. As soon as you start writing it you will realize that after the second term on the RHS you are stuck, there is no P(AB|..) term because A and B are incompatible experiments. So contrary to appearances, your "no-conspiracy" inequality P(C|z) <= P(A|x) + P(B|y) is nonsense. By writing it as an inequality, you have hidden the error of the impossible P(AB|..) term.

And that means you are not allowed, ever, to use P(A|x), P(B|y) and P(C|z) in the same expression the way you do. You can only do that if the three are defined over the exact same probability space where the joint probabilities are well defined. Since you insist on doing the forbidden, don't blame anything else for violations. Going back to the inequality for a bit, if you insist on using the inquality and obtain a violation for scenario (b), such a violation means P(AB|..) is negative which is impossible! But of cause that is because you insisted on using an impossible probability P(AB|..) to start with so "garbage-in", "garbage out".

This is what Boole published 150 years ago. Please read the article I cited earlier.

 

[lugita15]

Sorry, what does P(ABC) mean?

That is the joint probability for the events A, B, C.

I assume you mean P(A and B and C), more compactly written as P(A & B & C).

If A = "mismatch at -30 and 0". P(A|w) is the probability of mismatch at -30 and 0 for the probability space "w". If "w" was all we were ever talking about, it will make sense to simply leave it out and write P(A) but that will not mean P(A) is not a conditional probability. However your argument is not such a case; therefore to to be precise and avoid mistakes, it is recommended to write P(A|w).

It's fine if you want to use that notation, but my only quibble was that usually you put a statement on each side of the bar, you don't put a probability space.

I can disagree with that because I'm not limitting my thinking to an expression on a sheet of paper or on screen. I can disagree because I understand the difference between an empirical manipulation and an algebraic manipulation. You are focusing on the algebraic and ignoring the empirical.

Well, we are talking about numbers, so naturally algebra is where I turn.

And if you attempt to tell me what the joint probability P(ABC) is for both scenario (b) and (c) then you may start having a paradigm shift to understand the issue I have with your argument.

Sorry, I'm not having any paradigm shift. Of course A is only a meaningful statement in x, B is only a meaningful statement in y, C is only a meaningful statement in z, and all three statements are meaningful in w. So of course "A & B & C" is a meaningless statement in any of the three probability spaces in (b), and thus P(A & B & C) is not well-defined in any of them.

I suppose your no-conspiracy condition implies that they two are equal.

No, it doesn't. It's a conspiracy if the probability of mismatch you would get if you measured at -30 and 0 depends on whether you actually measure at -30 and 0, because the particles don't "know" whether they're going to be measured at -30 and 0. But it's not a conspiracy if C, and thus P(C), is well-defined for w but not for x. That just means that (c) allows for counterfactual reasoning and (b) doesn't.

Let me even make it easier: your expression P(C|w) <= P(A|w) + P(B|w) is actually derived from the equality P(C|w) = P(A|w) + P(B|w) - P(AB|w) so let us stay with the equality for a moment.

Actually, you forgot a factor of 2. The correct equation is P(C|w)=P(A XOR B|w)=P(A|w)+P(B|w)-2P(A & B|w). That's because if both A and B are true, then C is false.

Please use your so called "no-conspiracay condition" to write down the equivalent equality for scenario (b) with appropriate notation. As soon as you start writing it you will realize that after the second term on the RHS you are stuck, there is no P(AB|..) term because A and B are incompatible experiments.

Yes, the statement "A & B" is meaningless in any of the three probability spaces in (b), and thus P(A & B) is not well-defined in any of three probability spaces in (b), the equation doesn't make any sense using probabilities in (b).

So contrary to appearances, your "no-conspiracy" inequality P(C|z) <= P(A|x) + P(B|y) is nonsense. By writing it as an inequality, you have hidden the error of the impossible P(AB|..) term.

Look, I see no contradiction in saying the following:
1. The equation is meaningless in (b).
2. The equation is meaningful and correct in (c).
3. The equation in (c) implies the inequality in (c).
4. The inequality in (c) implies the inequality in (b)
5. The inequality is meaningful and correct in (b).

Why do you see a contradiction in this? It's true that the equation doesn't apply to (b), but that is irrelevant. The reason the inequality applies to (b) is not because the equation applies to (b), but rather because the inequality applies to (c), which is ultimately because the equation applies to (c).

And that means you are not allowed, ever, to use P(A|x), P(B|y) and P(C|z) in the same expression the way you do. You can only do that if the three are defined over the exact same probability space where the joint probabilities are well defined.

But they're just numbers. Why am I not allowed to use numbers in whatever expression I want to?

Since you insist on doing the forbidden, don't blame anything else for violations.

But how in the world is it forbidden?

Going back to the inequality for a bit, if you insist on using the inquality and obtain a violation for scenario (b), such a violation means P(AB|..) is negative which is impossible! But of cause that is because you insisted on using an impossible probability P(AB|..) to start with so "garbage-in", "garbage out".

Yes, a violation of the inequality in (b) implies a violation of the inequality in (c), which implies that P(A & B|w) is negative, which is impossible. But I am never invoking the ill-defined probabilities P(A & B|x), P(A & B|y), or P(A & B|z), so I don't see any garbage-in, garbage-out here.

 

[billschnieder]

Your argument is that no-conspirary allows you to conclude that the inequality P(C|w) <= P(A|w) + P(B|w) which is valid for scenario (c) must be valid for scenario (b), ie P(C|z) <= P(A|x) + P(B|y) because according to you the no-conspiracy implies that P(C|z) = P(C|w) and P(A|x)=P(A|w) and P(B|w)= P(B|y). This is your argument.
My argument is is the following:
The inequality P(C|w) <= P(A|w) + P(B|w) is obtained from P(C|w) = P(A|w) + P(B|w) - 2P(AB|w). So if you are claiming that the inequality should apply to scenario (b) it means the equality should also apply. You can not take the inequality and reject the equality, the inequality is just a different view into the what is already there in the equality.

Yes, the statement "A & B" is meaningless in any of the three probability spaces in (b), and thus P(A & B) is not well-defined in any of three probability spaces in (b), the equation doesn't make any sense using probabilities in (b).

You agree that the *equality* does not make sense for scenario (b) yet you insist on using the inequality. You claim that 3 of the 4 terms in the equality are the same for both (b) and (c) due to your so called "no-conspiracy condition". However, you conveniently hide the 4th term by removing it and changing the equality sign to an inequality. By doing that, you are assuming that the meaningless undefined probability which does not exist has a value. Yet you do not see the problem. This is why I asked you earlier to derive the inequality directly from scenario (b) using any assumptions you like without going through (c). But you didn't and can't.
What more do you want me to say if you refuse to see the fallacy.

Look, I see no contradiction in saying the following:
1. The equation is meaningless in (b).
2. The equation is meaningful and correct in (c).
3. The equation in (c) implies the inequality in (c).
4. The inequality in (c) implies the inequality in (b)
5. The inequality is meaningful and correct in (b).

Then I can't help you.

Why do you see a contradiction in this?

Because there is one.

It's true that the equation doesn't apply to (b), but that is irrelevant.

It is very relevant as I have explained. The inequality does not exist in the ether. You can not have a valid inequality if the equality is invalid. This much is obvious from you inability to derive the inequality directly under scenario (b) without invoking scenario (c) at all.

The reason the inequality applies to (b) is not because the equation applies to (b), but rather because the inequality applies to (c), which is ultimately because the equation applies to (c).

This is precisely the fallacy you continue to make. You can not put P(A|x) and P(B|y) and P(C|z) together in a single expression and calculate anything empirically meaningful. There is no alternate logic, or alternate probability theory which allows you do do that. Yet you insist on doing just that.

But they're just numbers. Why am I not allowed to use numbers in whatever expression I want to?

You are free to use them in whatever expression you like. But don't fool yourself to think the result will be meaningful in any way. This is what Boole, the father of Boolean logic, showed 150 years ago, which I've been encouraging you to read.

Yes, a violation of the inequality in (b) implies a violation of the inequality in (c), which implies that P(A & B|w) is negative, which is impossible. But I am never invoking the ill-defined probabilities P(A & B|x), P(A & B|y), or P(A & B|z), so I don't see any garbage-in, garbage-out here.

As soon as you write down your inequality and associate it with scenario (b), you have hidden those terms under a rug. They are embedded inside the inequality sign. So contrary to your claim that you are not using them, you are infact using them. I take back my suggestion that anyting was "forbidden". You are ofcourse free to do what you want but there is nothing more I can say. I have better things to do with my time.

 

[lugita15]

Your argument is that no-conspirary allows you to conclude that the inequality P(C|w) <= P(A|w) + P(B|w) which is valid for scenario (c) must be valid for scenario (b), ie P(C|z) <= P(A|x) + P(B|y) because according to you the no-conspiracy implies that P(C|z) = P(C|w) and P(A|x)=P(A|w) and P(B|w)= P(B|y). This is your argument.

Yes, it is.

My argument is is the following:
The inequality P(C|w) <= P(A|w) + P(B|w) is obtained from P(C|w) = P(A|w) + P(B|w) - 2P(AB|w). So if you are claiming that the inequality should apply to scenario (b) it means the equality should also apply. You can not take the inequality and reject the equality, the inequality is just a different view into the what is already there in the equality.

It is true that the equation in (c) implies the inequality in (c). But the inequality is its own statement, even if it's implied by another statement. Let me write things more explicitly so we can key in on our disagreement:
1. (P(A|w)+P(B|w)-P(C|w))/2=P(A & B|w)
2. P(A & B|w)≥0
3. (P(A|w)+P(B|w)-P(C|w))/2≥0
4. (P(A|x)+P(B|y)-P(C|z))/2≥0
Now I'm open in my "hiding" of a term. What I'm saying is that 1 and 2 imply 3, and even though 1 and 2 have no analogue for (b), it is nevertheless true that 3 implies 4.

However, you conveniently hide the 4th term by removing it and changing the equality sign to an inequality. By doing that, you are assuming that the meaningless undefined probability which does not exist has a value. Yet you do not see the problem.

No, I'm not assuming that P(A & B) is defined in any of the three probability spaces in (b). The equation implies the inequality, but the inequality need not imply the equation.

This is why I asked you earlier to derive the inequality directly from scenario (b) using any assumptions you like without going through (c). But you didn't and can't.

But there's a fundamental reason for that: counterfactual definiteness is a crucial assumption for Bell's theorem. If you were the kind of person who rejected counterfactual reasoning, in other words rejected reasoning in (c) as invalid, then this argument can't persuade you. If you restrict yourself to reasoning in (b), i.e. if you don't believe in counterfactual definiteness, then you can only carry out the argument for EPR, not Bell.

Look, I see no contradiction in saying the following:
1. The equation is meaningless in (b).
2. The equation is meaningful and correct in (c).
3. The equation in (c) implies the inequality in (c).
4. The inequality in (c) implies the inequality in (b)
5. The inequality is meaningful and correct in (b).

Then I can't help you.

What is the logical contradiction you see?

This much is obvious from you inability to derive the inequality directly under scenario (b) without invoking scenario (c) at all.

I explained my inability to do so.

This is precisely the fallacy you continue to make. You can not put P(A|x) and P(B|y) and P(C|z) together in a single expression and calculate anything empirically meaningful. There is no alternate logic, or alternate probability theory which allows you do do that. Yet you insist on doing just that.

OK, I still don't understand your point, since I'm just doing valid operations with numbers, but what if I side-stepped the issue and reasoned as follows:
1. P(C|w)≤P(A|w)+P(B|w)
2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75
3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w)
4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75
5. .75 ≤ .25 + .25

What step do you think I'm going wrong in?

 

[billschnieder]

1. P(C|w)≤P(A|w)+P(B|w)
2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75
3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w)
4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75
5. .75 ≤ .25 + .25

What step do you think I'm going wrong in?

1. P(C|w)=P(A|w)+P(B|w) - 2P(AB|w)
2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75, P(AB|x?y?)=?
3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w), P(AB|x?y?)=?
4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75, P(AB|w)=??
5. .75 = .25 + .25 + ?


Steps (1), (2), (3), (4) and (5). You are hiding the dirt under the inequality "rug". Enough said. If you still do not understand my argument that's your problem.

 

[San K]

A short note that:

Bell's theorem, though fairly convincing, is not, as many of you are aware, the only "proof" of non-local action

Other experiments, for example those listed below, also provide evidence of quantum entanglement

- Two-photon interference - where interference is demonstrated (at a distance) without the entangled photons meeting at the beam-splitter which is not replicable via two (un-entangled) photons

- Quantum entangled swapping - where photons are entangled (via their "twins") without ever meeting

- improvements in Bell's test detection efficiency

 

[lugita15]

You are hiding the dirt under the inequality "rug".

OK, let me be more explicit in my logic and not use inequalities at all.
1. P(C|w)=P(A|w)+P(B|w)-2P(A & B)|w)
2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75
3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w) (From no-conspiracy.)
4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75 (From 2 and 3)
5. .75 = .25 + .25 -2P(A & B|w) (From 1 and 4)
6. P(A & B|w) = -.125 (From 5)

Now what step is in error? I am no longer "hiding" ill-defined probabilities like P(A & B|x) behind inequality signs.

 

[billschnieder]

OK, let me be more explicit in my logic and not use inequalities at all.
1. P(C|w)=P(A|w)+P(B|w)-2P(A & B|w)
2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75
3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w) (From no-conspiracy.)
4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75 (From 2 and 3)
5. .75 = .25 + .25 -2P(A & B|w) (From 1 and 4)
6. P(A & B|w) = -.125 (From 5)

Now what step is in error? I am no longer "hiding" ill-defined probabilities like P(A & B|x) behind inequality signs.

Your argument is the same and the criticism is the same. P(AB|w) disappears in Step (2) and reappears in step (5). What happened to it in steps (2), (3) and (4)? In step (5) you have effectively the frankenstein equation P(C|z) = P(A|x) + P(B|y) - 2P(AB|w) from which you conclude in step (6) that 2P(AB|w) is negative! Why do you leave out the P(AB|..) terms from (2), (3) and (4)? Because you know that by including them, it reveals your error. Your equation (1) contains the term 2P(AB|w). How come your no-conspiracy condition can make a statement about P(C|w), P(A|w), and P(B|w) but conveniently ignores P(AB|w)? Isn't it curious that your "no-conspiracy" condition "just happens" to be silent about P(AB|w), the same probability on which your conclusion (6) hinges? That is the conspiracy of your "no-conspiracy".

How many different ways do I have to explain this?

 

[lugita15]

billschnieder and I continued this discussion in another thread, but on his suggestion I'm bringing it back here.

Your no-conspiracy condition is essentially that Scenario X and Scenario Y (from above) are exactly the same,

Yes, the no-conspiracy condition says that if a statement is meaningful in both scenarios, then the probability is equal for both scenarios.

in other words, your no-conspiracy condition is equivalent to saying, the QM result from a single wavefunction must be the same as the QM result from three different wavefunctions.

Where in the world did you get that from? We're talking about different possible measurements we could perform on a system with the same wavefunction. We're not talking about different wavefunctions.

And I showed you that step (3) was incomplete, Step (3) What does no-conspiracy say about P(AB|w). According to your logic, no-conspiracy also implies that P(AB|w)=P(AB|x,y).

No-conspiracy states that if a statement S is meaningful in both x and w, then P(S|w)=P(S|x) (and similarly for y and z). But A & B is not meaningful in x, so no-conspiracy doesn't tell you anything in this case.

Also, what do you mean by P(A & B|x,y)? Do you mean a combined space which is the union of x and y? Well, my reasoning doesn't talk about combined spaces like that. It only discusses x, y, z, and w.

But x and y are two different sets of photons, which means P(AB|x,y) is undefined/meaningless.

Again, I didn't say anything about P(A & B|x,y).

All you have proven is the triviality that the joint probablity distribution P(ABC|x,y,z) for outcomes from three different sets of photons (x,y,z) is undefined, although the joint probability distribution P(ABC|w) from the single set of photons (w) is well defined.

I didn't say anything about P(A & B & C|x,y,z). And again, since A & B & C is meaningless in x, y, or z, the no-conspiracy condition says nothing in this case either.

 

[billschnieder]

Yes, the no-conspiracy condition says that if a statement is meaningful in both scenarios, then the probability is equal for both scenarios.

But then, such a condition is seriously flawed.

Where in the world did you get that from? We're talking about different possible measurements we could perform on a system with the same wavefunction. We're not talking about different wavefunctions.

Please review post #240 above:

Le us represent the sets of photons as one set "w" for scenario (c), and three different sets ("x", "y","z") for scenario (b). Then the probabilities for (c) are more accurately represented as P(A|w), P(B|w), P(C|w) ... the probabilities for scenario (b) are more accurately represented as P(A|x), P(B|y), P(C|z).

So clearly in the first scenario, we are talking about different measurements that could be done on the same real system, but since we can only measure the system once, the other two terms are counter-factual AND clearly, in the second scenario we are talking about three different measurements on three different systems "x", "y", "z" (hence 3 wavefunctions) and there is nothing counterfactual in it.

Your argument, at least as I understand it, is that:
- the probabilities from the first scenario should be the same as the probabilities from the second scenario, because according to what you call the "no-conspiracy" condition, if we can talk meaningfully of P(A|w), P(B|w), P(C|w) in the first scenario, and we can talk meaningfully of P(A|x), P(B|y), P(C|z) in the second scenario, then those probabilities must be equal, ie P(A|x) = P(A|w) and P(B|y) = P(B|w) and P(C|z) = P(C|w), and if P(C|w) <= P(A|w) + P(B|w), then it must also be true that under "no-conspiracy", P(C|z) <= P(A|x) + P(B|y). From which you argue that since P(A|x) = P(B|y) = 0.25 and P(C|z) = 0.75, the inequalty P(C|z) <= P(A|x) + P(B|y) will be violated. .75≤.25+.25.

Now, is the above not exactly your argument. Please, let me know if I've misunderstood your argument in any way. Once we are clear on what your argument is, I will proceed to show why such a "no-conspiracy" condition is seriously flawed.

 

[lugita15]

So clearly in the first scenario, we are talking about different measurements that could be done on the same real system, but since we can only measure the system once, the other two terms are counter-factual AND clearly, in the second scenario we are talking about three different measurements on three different systems "x", "y", "z" (hence 3 wavefunctions) and there is nothing counterfactual in it.

Well, before measurement the system is described by the same wavefunction in each of the three cases x, y, and z.

Your argument, at least as I understand it, is that:
- the probabilities from the first scenario should be the same as the probabilities from the second scenario, because according to what you call the "no-conspiracy" condition, if we can talk meaningfully of P(A|w), P(B|w), P(C|w) in the first scenario, and we can talk meaningfully of P(A|x), P(B|y), P(C|z) in the second scenario, then those probabilities must be equal, ie P(A|x) = P(A|w) and P(B|y) = P(B|w) and P(C|z) = P(C|w), and if P(C|w) <= P(A|w) + P(B|w), then it must also be true that under "no-conspiracy", P(C|z) <= P(A|x) + P(B|y). From which you argue that since P(A|x) = P(B|y) = 0.25 and P(C|z) = 0.75, the inequalty P(C|z) <= P(A|x) + P(B|y) will be violated. .75≤.25+.25.

Now, is the above not exactly your argument. Please, let me know if I've misunderstood your argument in any way. Once we are clear on what your argument is, I will proceed to show why such a "no-conspiracy" condition is seriously flawed.

Yes, that is exactly my argument.

EDIT: In fact, your statement of my argument is unnecessarily cumbersome. We just need to say that P(C|w) <= P(A|w) + P(B|w), and by no-conspiracy P(A|x) = P(A|w), P(B|y) = P(B|w), P(C|z) = P(C|w), and according to QM we have P(A|x)=P(B|y)=.25 and P(C|z)=.75, so we have P(A|w)=P(B|w)=.25 and P(C|w)=.75, so we have .75 <= .25 + .25.

So we don't even need the intermediate step P(C|z) <= P(A|x) + P(B|y).

 

[billschnieder]

Well, before measurement the system is described by the same wavefunction in each of the three cases x, y, and z.

I guess this is a key point of disagreement. There are three different systems of photons "x", "y" and "z". They can not be described by the same wavefunction. It is only in "w" that we have a single system.

Yes, that is exactly my argument.

Let me explain then what your problem is:

1) Theoretically: An inequality such as P(C|w) <= P(A|w) + P(B|w), is not an independent mathematical truth, but just a simplification of the equality P(C|w)= P(A|w)+P(B|w)-2P(AB|w) from which it is derived. Then, you claim that because of "no-conspiracy", P(A|x) = P(A|w) and P(B|y) = P(B|w) and P(C|z) = P(C|w). Now, if that is the case, then it must also be the case that P(C|z) = P(A|x) + P(B|y) - ?. What in your opinion is the missing term, there must be one, otherwise your no-conspiracy condition is inconsistent. I would guess that it should be 2P(AB|xy) = 2P(AB|w) because of "no-conspiracy". If that is the case, let us remind ourselves at this point what A,B all mean:

Where A denotes statement "The result at -30° differs from the result at 0°", B denotes the statement "The result at 0° differs from the result at 30°", and C denotes the statement "The result at -30° differs from the result at 30°".

P(AB|w) is therefore the probability that the set of photon pairs "w" were destined to mismatch at both (-30°, 0°) AND (30°, 0°). Note here, that it is the same set that must satisfy both conditions.

P(AB|xy) is the probability that the set of photon pairs "x" was destined to mismatch at (-30°, 0°) and a completely different set of pairs "y" was also destined to mismatch at (30°, 0°). Note here that each set has just a single condition to fulfill.

 

[billschnieder]

Then, your no conspiracy condition boils down to the suggestion that a requirement for a single system to fulfill two conditions simultaneously, is the same as a requirement for two different systems to each fulfill only one of the two conditions. In simpler terms. The suggestion is akin to the suggestion that the probability of finding a tall girl in a class, is the same as the probability of finding a tall person in one class and a girl of any height in a different class.

2) Practically: When talking of measurements, since P(C|w), P(A|w), and P(B|w) are all derived from a single set of photon pairs, on which we can only ever measure one, two of the three must be counter-factual and unmeasurable. Since P(C|z), P(A|x), and P(B|y) are each derived from a different set of photon pairs, they are each directly measurable without any counter-factuals. Your no-conspiracy condition then becomes the suggestion that it is okay to substitute the results measured from three different systems into an inequality based on unmeasurable counterfactuals describing a single system.

Here is what Bell says about this idea:

"It was not the objective measurable predictions of quantum mechanics which ruled out hidden variables. It was the arbitrary assumption of a particular (and impossible) relation between the results of incompatible measurements either of which might be made on a given occasion but only one of which can in fact be made."

P(C|z), P(A|x), and P(B|y) are each objectively measurable.
P(C|w) <= P(A|w) + P(B|w), is an impossible relation between results of incompatible measurements either of which might be made on a given occasion but only one of which can in fact be made.

So it is the mixing of "w", with "x,y,z" that results in the violation. In otherwords, it is the so-called "no-conspiracy assumption" itself that has to be abandoned.

 

[billschnieder]

P(C|z) = P(A|x) + P(B|y) - ?

On the other hand, if you refuse to provide the last term, let us leave it as ? and follow the calculation through.

0.75 = 0.25 + 0.25 - ?

and

? = - 0.25

Where is the violation? Why do you have a problem with ? = -0.25.

 

[lugita15]

I guess this is a key point of disagreement. There are three different systems of photons "x", "y" and "z". They can not be described by the same wavefunction. It is only in "w" that we have a single system.

I think this is just a minor point. x is the set of photon pairs for which we're going to measure polarization at -30 and 0, y is the set of photon pairs for which we're going to measure 0 and 30, and z is the set of photon pairs for which we're going to measure -30 and 30. All I was saying is that according to quantum mechanics, the wave function of a photon pair in any of these three sets is exactly the same. Do you really disagree with that?

Let me explain then what your problem is:

1) Theoretically: An inequality such as P(C|w) <= P(A|w) + P(B|w), is not an independent mathematical truth, but just a simplification of the equality P(C|w)= P(A|w)+P(B|w)-2P(AB|w) from which it is derived.

That's why I rewrote things in terms of equations in post #249.

Then, you claim that because of "no-conspiracy", P(A|x) = P(A|w) and P(B|y) = P(B|w) and P(C|z) = P(C|w). Now, if that is the case, then it must also be the case that P(C|z) = P(A|x) + P(B|y) - ?. What in your opinion is the missing term, there must be one, otherwise your no-conspiracy condition is inconsistent.

Just refer to my post #249: .75 = .25 + .25 - 2P(A & B|w). Remember, the no-conspiracy condition only applies to statements S which are meaningful in both x and w (and similarly for y and z). It does not apply to A & B, which is not meaningful for x, y, or z.

I would guess that it should be 2P(AB|xy) = 2P(AB|w) because of "no-conspiracy".

No, I'm not claiming anything about P(A & B|x,y).

 

[lugita15]

Then, your no conspiracy condition boils down to the suggestion that a requirement for a single system to fulfill two conditions simultaneously, is the same as a requirement for two different systems to each fulfill only one of the two conditions. In simpler terms. The suggestion is akin to the suggestion that the probability of finding a tall girl in a class, is the same as the probability of finding a tall person in one class and a girl of any height in a different class.

I don't see how the no-conspiracy condition boils down to that at all, because I see P(A & B|x,y) as irrelevant.

2) Practically: When talking of measurements, since P(C|w), P(A|w), and P(B|w) are all derived from a single set of photon pairs, on which we can only ever measure one, two of the three must be counter-factual and unmeasurable. Since P(C|z), P(A|x), and P(B|y) are each derived from a different set of photon pairs, they are each directly measurable without any counter-factuals.

Yes, that's a fair summary,

Your no-conspiracy condition then becomes the suggestion that it is okay to substitute the results measured from three different systems into an inequality based on unmeasurable counterfactuals describing a single system.

Yes, and I think that it is okay in this case.

Here is what Bell says about this idea:

"It was not the objective measurable predictions of quantum mechanics which ruled out hidden variables. It was the arbitrary assumption of a particular (and impossible) relation between the results of incompatible measurements either of which might be made on a given occasion but only one of which can in fact be made."

That quote isn't about the assumptions of Bell's theorem, it's about Von Neumann's earlier hidden variable theorem that Bell critiqued as being too demanding in its assumptions. I don't think Bell thought of his own theorem the same way.

P(C|z), P(A|x), and P(B|y) are each objectively measurable.
P(C|w) <= P(A|w) + P(B|w), is an impossible relation between results of incompatible measurements either of which might be made on a given occasion but only one of which can in fact be made.

I don't know why you call it an impossible relation. It's a relationship between unmeasurable things, but the no-conspiracy condition allows us to say that certain unmeasurable quantities are equal to certain measurable quantities.

So it is the mixing of "w", with "x,y,z" that results in the violation.

What is wrong with an equation that contains x, y, z, and w? What is wrong with putting any numbers you want in an equation? That's fundamental to mathematics. Even if two numbers were taken from vastly different sources, if they have the same value, then they can be freely substituted into any statement without changing the truth value. That's Leibniz's famous Identity of Indiscernibles: if F(a) is true, and a=b, then F(b) must always be true. Is that what you're disputing?

In otherwords, it is the so-called "no-conspiracy assumption" itself that has to be abandoned.

OK, if you want to abandon it, you can, but at least address the argument I gave earlier in the thread for why it should be true:

I am talking about the result I would get if I perform a particular experiment on a particle, regardless of whether I actually perform that experiment. Now how do I connect this to real experiments, which are obviously concerned with possibility b)? I make the crucial assumption, which I expect that you disagree with or think is misleading, that the following two probabilities are always equal:
1. The probability that this photon would go through a polarizer if it is oriented at angle x, given that the polarizer is actually oriented at angle x.
2. The probability that this photon would go through a polarizer if it is oriented at angle x, given that the polarizer is NOT actually oriented at angle x, but instead some different angle y.

Now why do I assume that these two probabilities are equal? Because I am assuming that the answers to the following two questions are always the same:
1. What result would you get if you send this photon through a polarizer oriented at angle x, given that the polarizer is actually oriented at angle x?
2. What result would you get if you send the photon through a polarizer oriented at angle x, given that the polarizer is NOT oriented at angle x, but instead some different angle y?

And why do I assume that these questions have the same answer? That seems to me to be a consequence the no-conspiracy condition: the properties (or answers to questions) that are predetermined cannot depend on the specific measurement decisions that are going to be made later, since by assumption those decisions are free and independent.

What do you disagree with here? Do you disagree that questions 1 and 2 always have the same answer?

 

[lugita15]

P(C|z) = P(A|x) + P(B|y) - ?

On the other hand, if you refuse to provide the last term, let us leave it as ? and follow the calculation through.

0.75 = 0.25 + 0.25 - ?

and

? = - 0.25

Where is the violation? Why do you have a problem with ? = -0.25.

I have a problem with .75 = .25 + .25 - 2P(A & B|w) as shown in post #249.

 

[billschnieder]

All I was saying is that according to quantum mechanics, the wave function of a photon pair in any of these three sets is exactly the same. Do you really disagree with that?

Of course, they can not be exactly the same, there are 3 different systems with different space-time coordinates. You probably mean that the QM predictions for the three different systems
"x", "y", "z" as concerns this particular experiment are the same, which I will agree with.

.75 = .25 + .25 - 2P(A & B|w). Remember, the no-conspiracy condition only applies to statements S which are meaningful in both x and w (and similarly for y and z). It does not apply to A & B, which is not meaningful for x, y, or z.No, I'm not claiming anything about P(A & B|x,y).

But that is a nonsensical equation .75 = .25 + .25 - 2P(A & B|w). By your own argument, if your no-conspiracy condition can not say anything about P(A & B|x,y) then it can not say anything about 2P(A & B|w) either. Or do you want to explain how A&B is meaningful in both x and w (and similarly for y and z). By writing .75 = .25 + .25 - 2P(A & B|w), you are backhandedly trying to use it to say something about 2P(A & B|w). You say you are not making any claim about P(A & B|x,y) but I'm telling you that you are, indirectly.

How come you do not want to say anything about the equality for the second scenario only. In other words, why did you need w, why didn't you just write down the inequality using ONLY x, y, and z? What I'm asking you is, is there a genuine expression of the form

P(C|z) = P(A|x) + P(B|y) - ?

involving ONLY x,y,z and no w? If not, why not, and if so what is it?

 

[billschnieder]

That quote isn't about the assumptions of Bell's theorem, it's about Von Neumann's earlier hidden variable theorem that Bell critiqued as being too demanding in its assumptions.

Of course Bell did not intend to argue against his own theorem in a paper written before his theorem. But the point is that his argument is sound, and obviously applies just as well to his theorem whether he thought about it or not.

I don't know why you call it an impossible relation. It's a relationship between unmeasurable things

You've answered your own question. As far as measurements are concerned, it is an impossible relation.

What is wrong with an equation that contains x, y, z, and w? What is wrong with putting any numbers you want in an equation?

Nothing. Until you start drawing physical inferences from the equation. As I explained previously, if the time to put on just shoes is Ta, and the time to put on just socks is Tb, then you can write any equation you like which includes Ta and Tb. But you can not conclude that the Ta + Tb = Time to put on shoes and then socks.

Even if two numbers were taken from vastly different sources, if they have the same value, then they can be freely substituted into any statement without changing the truth value.

That's just not correct. Two numbers do not have the same value just because you claim they do. It is the physical situation producing the two values that determines whether they have the same value or not, not lugita. And I've already explained to you that the probability of finding a tall girl in a class is not the same as the probability of finding a tall person in one class and finding a girl of any height in another class.

That's Leibniz's famous Identity of Indiscernibles: if F(a) is true, and a=b, then F(b) must always be true. Is that what you're disputing?OK, if you want to abandon it, you can

What are you talking about?

Continuing in the next post ...

 

[billschnieder]

I make the crucial assumption, ... that the following two probabilities are always equal:
1. The probability that this photon would go through a polarizer if it is oriented at angle x, given that the polarizer is actually oriented at angle x.
2. The probability that this photon would go through a polarizer if it is oriented at angle x, given that the polarizer is NOT actually oriented at angle x, but instead some different angle y.

To illustrate the error, I will rephrase it as follows:
*) The expected result R1, if the photon is measured at angle x
*) The expected result R3, if the photon were instead measured at angle x, given that we already measured it at angle y and obtained result R2.

Your crucial point is that R1 must be equal to R3 because of no-conspiracy. Of course that is false. I will give you a clear counterexample example using Bernouli's urn.

We have an urn with 2 balls, 1 white and 1 red, we pick two balls in sequence without replacement.
*) The probability of the first ball being Red, R1
*) The probability of the first ball being Red, R3, given that the second ball was revealed to be Red R2.
According to your no-conspiracy condition, R1 = R3 which is false. R1 = 1/2 and R3 = 0. As I hope you appreciate, the underlined statements (R2) place restrictions on what we are allowed to say about R3. By ignoring this part of your argument, it has obscured your ability to see the error.

Not only do we have results for the actually measured angle y, we are also using it jointly in the same expression as the results at angle x.

the properties (or answers to questions) that are predetermined cannot depend on the specific measurement decisions that are going to be made later, since by assumption those decisions are free and independent.

Obviously false or naive. In the Bernouli example above, the result of the first pick was dependent on the result of the second pick which was made later.

So I will appreciate answers to my own questions too:

What I'm asking you is, is there a genuine expression of the form

P(C|z) = P(A|x) + P(B|y) - ?

involving ONLY x,y,z and no w? If not, why not, and if so what is it?

 

[lugita15]

Of course, they can not be exactly the same, there are 3 different systems with different space-time coordinates. You probably mean that the QM predictions for the three different systems
"x", "y", "z" as concerns this particular experiment are the same, which I will agree with.

The wavefunction of a system can be written as a product of a positional part and a spin part. All I meant is that the spin state of each photon pair is the same.

But that is a nonsensical equation .75 = .25 + .25 - 2P(A & B|w).

Why is it nonsensical?

By your own argument, if your no-conspiracy condition can not say anything about P(A & B|x,y) then it can not say anything about 2P(A & B|w) either.

Whether the no-conspiracy condition says anything about P(A & B|x,y) is not really something that concerns me, because it's completely irrelevant to my argument. But yes, the no-conspiracy condition does not allow you to conclude that P(A & B|w) = P(A & B|x), for instance, because A & B is not meaningful in x. To repeat, the no-conspiracy condition states that if S is meaningful in both x and w, then P(S|w)=P(S[x).

Or do you want to explain how A&B is meaningful in both x and w (and similarly for y and z).

No, it's definitely not meaningful in x, y, or z.

By writing .75 = .25 + .25 - 2P(A & B|w), you are backhandedly trying to use it to say something about 2P(A & B|w).

All I'm saying is that you can't directly apply the no-conspiracy condition to P(A & B|w).

You say you are not making any claim about P(A & B|x,y) but I'm telling you that you are, indirectly.

How am I doing that?

How come you do not want to say anything about the equality for the second scenario only. In other words, why did you need w, why didn't you just write down the inequality using ONLY x, y, and z?

Well, you can write the inequality in terms of only x, y, and z, but you can't write the equation in terms of only x, y, and z. How is that a problem?

In a deeper sense, if you're asking why we need w at all in this proof, that's because one of the assumptions of Bell's theorem is counterfactual definiteness.

What I'm asking you is, is there a genuine expression of the form

P(C|z) = P(A|x) + P(B|y) - ?

involving ONLY x,y,z and no w? If not, why not, and if so what is it?

No, there isn't. (I mean, of course there are equations of that form, but nothing relevant.) And the reason there isn't is that the no-conspiracy condition applies to P(A|w), P(B|w), and P(C|w), but it doesn't apply to P(A & B|w), because A & B is only meaningful in w.

 

[lugita15]

Of course Bell did not intend to argue against his own theorem in a paper written before his theorem. But the point is that his argument is sound, and obviously applies just as well to his theorem whether he thought about it or not.

OK, fair enough, it's just that the way you've been using the quote makes it seem like Bell was acknowledging a shortcoming in his own theorem.

You've answered your own question. As far as measurements are concerned, it is an impossible relation.

Well, if all you mean is that P(C|w) <= P(A|w) + P(B|w) is a relationship between quantities that are not directly measurable, then yes I agree with that.

Nothing. Until you start drawing physical inferences from the equation.

I think this is where we fundamentally disagree. If the mass of a stick and the mass of a stone are equal, and the mass of the stick is great than the mass of an ant, then the mass of the stone is also greater than the mass of the ant. Numbers which are equal can be freely substituted for each other in any statement without changing the truth value of that statement, whether that statement is about the physical world or not. Do you really disagree with that? That's Leibniz's law.

As I explained previously, if the time to put on just shoes is Ta, and the time to put on just socks is Tb, then you can write any equation you like which includes Ta and Tb. But you can not conclude that the Ta + Tb = Time to put on shoes and then socks.

But all your example shows is that the time to do a then b is not equal to the time to do a + the time to do b, or in symbols, T_(a then b) does not equal T_a + T_b. In other words, the "then" operation of actions does not correspond to the addition operation. But we're not dealing with that kind of issue at all here.

That's just not correct. Two numbers do not have the same value just because you claim they do. It is the physical situation producing the two values that determines whether they have the same value or not, not lugita.

I didn't say that any two numbers are equal if I say so. All I said is IF they'e equal, then they can be freely substituted. So if you want to disagree with me, you have to disagree with P(A|w)=P(A|x). But if I'm right about that, then after that you can't disagree, can you, with me substituting P(A|x) into the equation P(C|w)= P(A|w)+P(B|w)-2P(AB|w), and drawing valid physical inferences from a correct equation?

And I've already explained to you that the probability of finding a tall girl in a class is not the same as the probability of finding a tall person in one class and finding a girl of any height in another class.

I agree with that, but I don't see how I'm doing anything analogous to that.

What are you talking about?

I'm talking about Leibniz's Law, known as the Identity of Indiscernibles. It says that if F(a) is a true statement (like "a is an apple"), and a=b is a true statement, then F(b) ("b is an apple") must always be a true statement. Do you disagree with that?

 

[lugita15]

To illustrate the error, I will rephrase it as follows:
*) The expected result R1, if the photon is measured at angle x
*) The expected result R3, if the photon were instead measured at angle x, given that we already measured it at angle y and obtained result R2.

Your crucial point is that R1 must be equal to R3 because of no-conspiracy.

I am NOT asserting that the probability that a given photon would have yielded a certain result R1 at angle x, given that it was measured at y and yielded R2, is equal to the probability that it that will yield R1 if we measure it at x. Instead I'm saying that the probability that a given photon have yielded R1 at x, given that it was measured at y, but NOT given the result of the measurement, is equal to the probability that it will yield R1 if we measure it at x.

That was a long string of words, so let me write it in fraction form. This is what I am NOT saying, but what you seem to think I'm saying:
(number of photons yielded R2 at y but would have yielded R1 at x)/(number of photons which yielded R2 at y) = (number of photons which yielded R1 at x)/(number of photons measured at x).

This is what I AM saying:
(number of photons which were measured at y but would have yielded R1 at x)/(number of photons which were measured at y) = (number of photons which yielded R1 at x)/(number of photons measured at x).

If you disagree with this statement, you can, but please tell me your thoughts on the argument I presented for it:

"Now why do I assume that these two probabilities are equal? Because I am assuming that the answers to the following two questions are always the same:
1. What result would you get if you send this photon through a polarizer oriented at angle x, given that the polarizer is actually oriented at angle x?
2. What result would you get if you send the photon through a polarizer oriented at angle x, given that the polarizer is NOT oriented at angle x, but instead some different angle y?

And why do I assume that these questions have the same answer? That seems to me to be a consequence the no-conspiracy condition: the properties (or answers to questions) that are predetermined cannot depend on the specific measurement decisions that are going to be made later, since by assumption those decisions are free and independent."
Do you agree or disagree that these two questions have the same answer?

Of course that is false. I will give you a clear counterexample example using Bernouli's urn.

We have an urn with 2 balls, 1 white and 1 red, we pick two balls in sequence without replacement.
*) The probability of the first ball being Red, R1
*) The probability of the first ball being Red, R3, given that the second ball was revealed to be Red R2.
According to your no-conspiracy condition, R1 = R3 which is false. R1 = 1/2 and R3 = 0. As I hope you appreciate, the underlined statements (R2) place restrictions on what we are allowed to say about R3. By ignoring this part of your argument, it has obscured your ability to see the error.

This is not at all what I'm doing.

 

[billschnieder]

I am NOT asserting that the probability that a given photon would have yielded a certain result R1 at angle x, given that it was measured at y and yielded R2, is equal to the probability that it that will yield R1 if we measure it at x. Instead I'm saying that the probability that a given photon have yielded R1 at x, given that it was measured at y, but NOT given the result of the measurement, is equal to the probability that it will yield R1 if we measure it at x.

I can only assume then that you do not understand what you are saying. Do you at least agree now, that the two probabilities would not be the same if R2 were given? If you can agree to this, then you have understood the argument and all I have to do is to convince you that in your argument, R2 is indeed given.

If you disagree with this statement, you can, but please tell me your thoughts on the argument I presented for it

What do you think I've been doing? You are just repeating the same post and ignoring my argument against it. Do you even understand the argument at all, let-alone agreeing with it?

Now why do I say R2 must be given. Because you have both results in the same expression from which you are trying to draw physical conclusions. You are arguing as if you only had a single probability written on one sheet of paper, and another probability written on another sheet of paper. Rather you have an expression P(A|x) + P(B|y) which includes BOTH the results at x and the results at y, so how can the results at y not be given?! You have "given" it, by having it in the expression. You can't have your cake and not have it.

This is not at all what I'm doing.

It is indeed, as I've explained now, maybe 5 different ways. Each time you simply say it is not. Just because you phrase your argument in a way that obscures the argument does not mean I can not see through it and go right through to the core issue. That is what I have done, many times already.

 

[billschnieder]

Numbers which are equal can be freely substituted for each other in any statement without changing the truth value of that statement ... Do you really disagree with that?

But the numbers are not equal, isn't that the argument? Otherwise why are you asking me to explain why I claim the numbers are not equal.

But all your example shows is that the time to do a then b is not equal to the time to do a + the time to do b

You completely missed the point which was to show you that the value of T_a, T_b in the expression T_a + T_b, will be different from the value of just T_a or T_b separately. The point being that the physical situation of in the expression "T_a + T_b" (putting on shoes and then Socks) determines which value T_a or T_b will have in the expression and you can not simply take values from considering T_a separately and plug them in expecting things to be meaningful. Of course you are free to do it, but do not be surprised when nonsensical paradoxes result.

All I said is IF they'e equal, then they can be freely substituted.

And I'm arguing that the free standing T_b is not the same as the T_b when considering it jointly with T_a. I'm arguing that they are not equal.

So if you want to disagree with me, you have to disagree with P(A|w)=P(A|x).

No I don't. I disagree with you going from

P(C|w)= P(A|w)+P(B|w)-2P(AB|w)
to P(C|z)= P(A|x)+P(B|y)-2P(AB|w)

And apparently, you too disagree with it when if agree that there is no genuine expression of the form P(C|z) = P(A|x) + P(B|y) - ? involving only x, y, z.

I disagree with "P(A|w)=P(A|x) AND P(B|w)=P(B|y) AND P(C|w)=P(C|z) simultaneously"

I mean, of course there are equations of that form, but nothing relevant.

Please, please it is very relevant. This is at the core of the disagreement, how can it not be relevant. Please provide the equation. Unless you just want to wiggle out of admitting that there is none. The question did not include any w, so why even bring no-conspiracy into it. Do you have a complete expression including just x, y, z for of the form P(C|z) = P(A|x) + P(B|y) - ? or not? If you do please provide it. If none exists, please explain why none exists. This is a crucial question.

 

[lugita15]

I can only assume then that you do not understand what you are saying. Do you at least agree now, that the two probabilities would not be the same if R2 were given?

I agree that the percentage of photons yielding R2 for y, for which x would have yielded R1, is different from the percentage of photons measured at x, for which x yielded R1. But I also believe that the percentage of photons measured at y, for which x would have yielded R1, is the SAME as the percentage of photons measured at x, for which x yielded R1. Do you agree or disagree with this second statement?

What do you think I've been doing? You are just repeating the same post and ignoring my argument against it. Do you even understand the argument at all, let-alone agreeing with it?

I think I understand my argument, but I don't know what your counterargument is. At least so far as I can tell, you haven't told me whether my 2 questions always have the same answer. In post #262, you responded to that post by giving the example of Bernoulli's urn, which was about probabilities. But my two questions have absolutely nothing to do with probabilities.

Now why do I say R2 must be given. Because you have both results in the same expression from which you are trying to draw physical conclusions. You are arguing as if you only had a single probability written on one sheet of paper, and another probability written on another sheet of paper. Rather you have an expression P(A|x) + P(B|y) which includes BOTH the results at x and the results at y, so how can the results at y not be given?!

First of all, we have some ambiguity as to what x and y mean. The x and y in my two questions are angle settings of a polarizer. The x and y in P(A|x) + P(B|y) are probability spaces, x being the set of photons for which you measure at -30 and 0, and y being the set of photons for which you measure at 0 and 30. To minimize confusion, in future why don't we use θ1 and θ2 to refer to angle settings, and continue to use x, y, z, and w to refer to probability spaces?

You have "given" it, by having it in the expression. You can't have your cake and not have it.

Instead of arguing about what's been "given" or not, why don't you simply tell me whether you agree or disagree that the following two fractions are equal?
1. (Number of photons pairs which yielded different results when measured at -30 and 0)/((Number of photon pairs which were measured at -30 and 0).
2. (Number of photon pairs which were measured at 0 and 30 but would have yielded different results at -30 and 0)/(Number of photon pairs which were measured at 0 and 30)

It is indeed, as I've explained now, maybe 5 different ways. Each time you simply say it is not.

Your example of Bernoulli's urn dealt with different probabilities than the ones I'm dealing with. I wrote the probabilities that I'm dealing with in fraction form above.

 

[lugita15]

But the numbers are not equal, isn't that the argument? Otherwise why are you asking me to explain why I claim the numbers are not equal.

If you'r only claim is that P(A|w)=P(A|x) is wrong for some reason, that's one thing that we can discuss. But you also seem to claim that "it is the mixing of "w", with "x,y,z" that results in the violation". What is the reason why equations involving x, y, z, and w are not allowed, or physically meaningless, or whatever your contention is?

You completely missed the point which was to show you that the value of T_a, T_b in the expression T_a + T_b, will be different from the value of just T_a or T_b separately.

I have no idea what this sentence means. Can you explain it different words?

The point being that the physical situation of in the expression "T_a + T_b" (putting on shoes and then Socks) determines which value T_a or T_b will have in the expression and you can not simply take values from considering T_a separately and plug them in expecting things to be meaningful. Of course you are free to do it, but do not be surprised when nonsensical paradoxes result.

Again, I don't quite understand what you're saying here, and I don't know why you're associating T_a + T_b with putting on shoes then socks, rather than just the time to put on shoes plus the time to put on socks. Can you use a different example to illustrate your point, whatever it is?

And I'm arguing that the free standing T_b is not the same as the T_b when considering it jointly with T_a. I'm arguing that they are not equal.

T_b is just a fixed number, just like 5. Whether you put it on its own in a statement (with physical significance or not), or whether you put it in a statement (with physical significance or not) which also has T_a in it, doesn't it still retain its value?

No I don't. I disagree with you going from

P(C|w)= P(A|w)+P(B|w)-2P(AB|w)
to P(C|z)= P(A|x)+P(B|y)-2P(AB|w)

And apparently, you too disagree with it when if agree that there is no genuine expression of the form P(C|z) = P(A|x) + P(B|y) - ? involving only x, y, z.

I definitely don't disagree with going from P(C|w)= P(A|w)+P(B|w)-2P(AB|w) to P(C|z)= P(A|x)+P(B|y)-2P(AB|w). What do you see wrong with that step?

I disagree with "P(A|w)=P(A|x) AND P(B|w)=P(B|y) AND P(C|w)=P(C|z) simultaneously"

This is a rather strange statement. What does it mean for three statements to be true, versus three statement being true simultaneously? Are you operating in some non-classical logic or something?

Please, please it is very relevant. This is at the core of the disagreement, how can it not be relevant. Please provide the equation. Unless you just want to wiggle out of admitting that there is none. The question did not include any w, so why even bring no-conspiracy into it. Do you have a complete expression including just x, y, z for of the form P(C|z) = P(A|x) + P(B|y) - ? or not? If you do please provide it. If none exists, please explain why none exists. This is a crucial question.

There aren't any useful equations of that form which involve just x, y, and z. There's lots of useless and unimportant equations of that form, but they're of zero relevance to what we're talking about. (For instance P(C|z) = P(A|x) + P(B|y) - (1 - P(not A|x) - P(not B|y) + P(not C|z)) or something silly like that.) Anyway, the point is, why do we need to express the Bell inequality in terms of an equation that only involves x, y, and z? What is wrong with an equation that involves x, y, z, and w? And why can't we derive physical inferences from such an equation?

And also, why can't we just sidestep this whole issue, like I did in post #249, and just directly get to .75 = .25 + .25 - 2P(A & B|w) from P(C|w)= P(A|w)+P(B|w)-2P(AB|w), P(A|w)=P(A|x)=.25, P(B|w)=P(B|y)=.25, and P(C|w)=P(C|z)=.75?

 

[billschnieder]

What is wrong with an equation that involves x, y, z, and w? And why can't we derive physical inferences from such an equation?

Because the expression is incorrect to start with. It is incorrect because you have assumed that terms are equal, which are not. The fact that a genuine expression exists for w, and none exists for xyz, confirms that. In other words, due to the physical situation "w", there is a single joint probability distribution p(ABC|w) from which you can extract P(A|w), P(B|w), and P(C|w) but there is no joint probability distribution p(ABC|xyz) in the physical situation "xyz" from which you can extract P(A|x),P(B|y), and P(C|z). Because of the physical situation, "w" the probabilities P(A|w), P(B|w), and P(C|w) are mutually dependent having originated from the same system, but the physical situation "xyz" mandates that the probabilities P(A|x), P(B|y), and P(C|z) are independent, having originated from three separate systems.

After substituting the probabilities from the physical situation "xyz" into an expression derived for the physical situation "w", the violation you obtain tells you clearly that the probabilities were not the same as you naively assumed. And the impossibility of deriving a similar expression directly in physical situation "xyz" without first going through "w" is a big hint that your assumptions are wrong.

To summarize:
1) There are two scenarios/physical sittuations involved in this discussion:

Scenario "xyz", involving the three probabilities:

P(A|x) = what we would get if we measure system "x" at angles (a,b)
P(B|y) = what we would get if we measure system "y" at angles (b,c)
P(C|z) = what we would get if we measure system "z" at angles (a,c)​

Scenario "w", involving the three probabilities:

P(A|w) = what we would get if we measure system "w" at angles (a,b)
P(B|w) = what we would have gotten had we measured system "w" at angles (b,c) instead of (a,b)
P(C|w) = what we would have gotten had we measured system "w" at angles (a,c) instead of (a,b)​

2) Scenario "xyz" is different from Scenario "w". The three probabilities in "w" are not equal to the three probabilities from "xyz".
3) Bell's inequalities are derived for scenario "w" ONLY. A similar inequality cannot be derived for scenario "xyz".
4) The predictions of QM are for scenario "xyz" only, not "w"
5) Experimental results are for scenario "xyz" only
6) Using probabilities from "xyz" in an expression derived for "w", amounts to introducing an assumption that the probabilities are all equal. It is this assumption that should be rejected when a violation results.

So what part of this argument don't you understand, or do understand but disagree with and why?