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

[lugita15]

You admitted that your "probability" equation was for scenario (c). And you also admit that QM and real experiments are for (b). So you still have to account for that disconnect. Why does it make sense to compare apples (b) with oranges (c) as you do in your arguments?

I thought I accounted for that in my post #187.

Surely you understand that local causality does not give you that justification since both (b) and (c) mean completely different things even when local causality is true.

Certainly they mean different things. I'm just saying the probabilities in (b) happen to be equal to the probabilities in (c) due to counterfactual definiteness and the no-conspiracy condition.

When I asked "what physical or logical basis do you have to expect the result of measurement on one set of photons to restrain the results of measurements on a different set of photons?". Unless you can answer this question, you can not link (c) to (b). And you MUST be able to link (c) to (b) in order use the QM predictions and experimental results the way you are doing.

I already gave you my attempt at answering this question. If the probability that the result that you would get at -30° differs from the result you would get at 0° is 25% for those particle pairs for which you actually orient the polarizers at -30° and 0°, then the no-conspiracy condition allows you to conclude that the probability that the result you would get at -30° differs from the result you would get at 0° even for those particle pairs for which you DON'T orient the polarizers at -30° and 0°. So as I said, I think this dispute boils down to a disagreement about what constitutes a valid use of the no-conspiracy condition.

Now, in a recent post in response to the triangle inequality you suggested that the triangle was the whole set of photons and that for some yet specified or justified reason one set of photons was congruent with a different set. This claim is the same as simply stating that (c) and (b) are the same without any justification. This claim is the same as saying measurements performed on one set of photons should be able to restrain the results of measurements performed on a different set of photons. How is that possible? You do not explain or even attempt a justification, yet you continue to think that your argument is logical.

I did attempt a justification, first in post #182 and then again in post #187. Could you tell me what you found to be unsatisfactory in post #187?

To see how ridiculous this argument is consider the following example:

We have a die which we throw on a table with little square depressions which exactly fit the die, point being that when the die settles it will always have one of it's vertical sides facing north, another facing south and the others east and west. The actual numbers on those sides will vary randomly for a fair die. We then throw the die and read of the number facing north (say a N6). For the same die, the outcome N6 (6 is facing North), now restraints the other possibilities for that same die. For example, S6 is impossible. There is no logical or physical reason to expect a different die, *congruent* with the first one, thrown at a different time, from being restrained by the outcome we obtained on the first one.

Let me use your analogy. Suppose, for some strange reason, you could only find out which way two of the six faces on the die were located, after which the die would spontaneously self-destruct. You roll a whole bunch of these dice, sometimes observing where 1 and 2 are located, sometimes observing (say) where 1 and 6 are located. Now let me invent a property called coolness. A face on a die is called cool if it's oriented in one of the cardinal directions, and is called lame otherwise (i.e. if it's oriented up or down). You're interested in whether they two faces you pick on the die have the same coolness or different coolness. You find that the probability that faces 1 and 2 have the same coolness is always p (I'm too lazy to calculate it). Now you're assuming the no-conspiracy condition, so which way the dice lands does not depend on the choice you're going to make as to which faces you're going to choose to locate. Thus you reason that even for those dice for which you happen to choose to determine the location of other faces instead of 1 and 2, it is still true that the probability that you would have found faces 1 and 2 to have the same coolness if you HAD determines the location of 1 and 2 is still p. So in this case I am linking the probabilities in (b) and the probabilities in (c). Do you think my reasoning in this example is invalid. If you think it's valid, in what way does it differ from my linking of (b) and (c) in the Bell's theorem argument?

Anyway, I look forward to hearing your response to my post #187. I think that should crystalize the source of our disagreement.

 

[lugita15]

Sorry, your question doesn't make sense to me. Experiments (and thus experimenters) exploit loopholes by their design.

I'm sorry if I'm quibbling over semantics, but in my terminology it doesn't make sense to say that an experimenter or experiment "exploits" a loophole. Rather, experimenters either leave a loophole open, meaning that they did not definitively prove what they wanted to prove with their experiment, or they close a loophole.

To give an illustration from SR: an experimenter wants to prove that the speed of light is truly isotropic relative to the laboratory. For that purpose he synchronises two clocks next to each other and slowly transports one of them to a far distance. Next he sends a light pulse in two directions and measures the time intervals with the two clocks, showing that they are equal. Of course there is a loophole which he exploits.

I don't see what loophole he is exploiting when he demonstrates the isotropy of the one-way speed of light with respect to slow transport synchronization. I don't even know what it means to exploit a loophole.

Other experiments such as the Michelson interferometer exploit again other loopholes if performed for that purpose.

Again, I don't know what loopholes you're talking about. To my mind, the only manner in which you could use the word "loophole" in the context of the Michelson-Morley experiment is to say that the experiment would have proven that the Earth is at rest with respect to the luminiferous aether, except that Lorentz and Fitzgerald pointed out a loophole that the experiment left open: namely, that the apparatus could have shrunk parallel to the motion of the Earth with respect to the aether, and thus we can't know for sure that the speed of light is really isotropic with respect to the earth.

Thus you could ask, how is it that the kind of particles for which we have closed the length contraction loophole just happen to be the kind of particles that exploit the synchronisation loophole, and vice versa - thus suggesting that SR is wrong.

I know this is just a strawman argument you're using to illustrate a point, but I really don't understand what you're saying here.

However, that would merely be due to a lack of insight in how SR works, and it seems reasonable to assume that it's similar with QM.

I still don't see the analogy you're making, but perhaps you're making the point "a conspiracy of nature is a law of nature" (Poincare) i.e. if it seems like nature continually conspires to make something appear true, maybe it really is. That's certainly a useful principle in science. When repeated attempts to detect the aether failed, perhaps there's a reason it's so hard to detect the aether - either it's undetectable, or it doesn't exist. When repeated attempts to measure with certainty both the position and momentum of a particle at time t fails, maybe there's a reason for that too - the particle may not possesses both attributes at the same time, or attempts to measure one attribute disturbs the other attribute dramatically. Well, there's several possibly ways to apply this principle to the issue of Bell tests:

• Maybe it seems really hard to close all the loopholes at once because it's impossible
• Maybe it's hard to definitively show that local models are incorrect because they're not.
• Maybe the fact that quantum mechanics doesn't get disproved no matter what loopholes we close means that quantum mechanics is true, or at any rate that is right about Bell tests

 

[harrylin]

I'm sorry if I'm quibbling over semantics, but in my terminology it doesn't make sense to say that an experimenter or experiment "exploits" a loophole. [..]

Well, it was first of all me who was sorry because it doesn't make sense to me that a particle could "exploit" a loophole! :tongue2:

[..] I don't even know what it means to exploit a loophole.

If you don't know what it means, then why did you bring it up as argument in posts #203+208 ?

[..] To my mind, the only manner in which you could use the word "loophole" in the context of the Michelson-Morley experiment is to say that the experiment would have proven that the Earth is at rest with respect to the luminiferous aether, [..]

No no no, I wasn't talking about MMX at all (nor did I mention or suggest it!). Instead I gave an example of an often seen misunderstanding here at physicsforums of people who think that synchronisation is somehow absolute, and who come with fake experimental evidence to back up their claims. Do you find the established "conspiracy"* of relativistic effects "unlikely" to be true?

[..] it seems like nature continually conspires to make something appear true, maybe it really is. [..]

To the contrary, that seems to be your argument and in my example which is based on the PoR that argument is wrong (thus, and contrary to what you seem to think, the PoR is not equal to that argument!).

[..]

• Maybe it seems really hard to close all the loopholes at once because it's impossible
• Maybe it's hard to definitively show that local models are incorrect because they're not.
• Maybe the fact that quantum mechanics doesn't get disproved no matter what loopholes we close means that quantum mechanics is true, or at any rate that is right about Bell tests

That's much better!

*it may look like a conspiracy if you don't understand how and why SR works, just as loopholes seem to conspire in QM experiments

 

[lugita15]

Well, it was first of all me who was sorry because it doesn't make sense to me that a particle could "exploit" a loophole! :tongue2:

I mean, a person analyzing an experiment can exploit a loophole to claim that a particle isn't behaving as the experimenter claims the particle is behaving. I just find it a convenient shorthand to say that, e.g. the particle is exploiting this or that loophole to escape Bell's theorem.

If you don't know what it means, then why did you bring it up as argument in post #203 ?

I mean, I don't know what it means for an experimenter to exploit this or that loophole.

No no no, I wasn't talking about MMX at all (nor did I mention or suggest it!).

You mentioned that the Michelson interferometer exploits loopholes.

Instead I gave an example of an often seen misunderstanding here at physicsforums of people who think that synchronisation is somehow absolute, and who come with fake experimental evidence to back up their claims

I think you would have to specify their fallacious argument in greater detail for me to see the fallacy.

. Do you find the established "conspiracy"* of relativistic effects "unlikely" to be true?

Are you referring to the fact that at first glance special relativity seems like a conspiracy to make the aether frame undetectable?

To the contrary, that seems to be your argument and in my example which is based on the PoR that argument is wrong (thus, and contrary to what you seem to think, the PoR is not equal to that argument!).

I think the Einstein's application of the principle of relativity IS equal to that argument. For instance, you see a magnet and a conductor moving relative to each other, and although the descriptions in different frames are diffferent, somehow the electric and magnetic fields conspire to produce the same basic phenomenon in every frame. So applying the argument, maybe there's a reason why the electromagnetic force shows a greater degree of frame invariance than seems evident from Maxwell's equations.

That's much better!

Hold your horses! I was just giving possible applications of the argument in the case of Bell tests.

 

[harrylin]

I mean, a person analyzing an experiment can exploit a loophole to claim that a particle isn't behaving as the experimenter claims the particle is behaving. I just find it a convenient shorthand to say that, e.g. the particle is exploiting this or that loophole to escape Bell's theorem.
I mean, I don't know what it means for an experimenter to exploit this or that loophole.

An experimenter is the first person analysing an experiment. I meant that an experimenter can just as well exploit a flaw by ignoring or downgrading its existence; thus he uses (exploits) the flaw in his experiment to reach a wrong conclusion. But I agree that that's an unusual way of putting it, I should have written that he denigrates the experimental flaw as "loophole". As a matter of fact, I used "loophole" as if it were a neutral word because I don't know a neutral replacement. "Flaw" won't do either as it is just as biased. Any suggestions?

You mentioned that the Michelson interferometer exploits loopholes.

Such an apparatus can be used for many purposes. Therefore I mentioned "if performed for that purpose" (emphasis mine) - thus for the purpose of demonstrating true isotropy relative to the apparatus.

I think you would have to specify their fallacious argument in greater detail for me to see the fallacy. [..]

If you're not familiar with it then it's quite useless as example, however I suppose that you are familiar with it as you participated in one such a discussion:

physicsforums.com/showthread.php?t=584743

Thus for an experimenter who would try to prove "true isotropy" relative to the laboratory, possible clock slowdown is the loophole for relativity in the first experiment but possible length contraction is the loophole in the experiment with the interferometer. That may look like an unreasonable conspiracy between very different things.

somehow the electric and magnetic fields conspire to produce the same basic phenomenon in every frame. [..]

That's quite similar to my example, although less striking (perhaps we misunderstood of each other who argues what?). Such a perceived "conspiracy" is not evidence that SR is wrong. Instead, such "conspiracies" may indicate deep and not yet fully understood connections in the workings of nature. (Reminder: this was an elaboration of posts #207,208).

 

[billschnieder]

I'm just saying the probabilities in (b) happen to be equal to the probabilities in (c) due to counterfactual definiteness and the no-conspiracy condition.

Well let us examine that claim to see if at all CFD or no-conspiracy allow you to equate (b) and (c) as you claim. Here are the (b) and (c) scenarios again for reference:

(b) Just like in QM and EPR-Experiments, the probabilities are calculated for different sets of independent photons.
(c) The probabilities are calculated for the same set of photons by imagining what could have been measured even if they are not in fact measured.

And here again are the probabilities for reference:

P(A) = The probability of mismatches between -30, and 0
P(B) = The probability of mismatches between 0 and 30
P(C) = The probability of mismatches between -30 and 30

And here again is your disputed claim (3) for reference

(3) If there is a mismatch for (-30, 30) then there must also be a mismatch for either (0, 30) or (-30, 0).

So for a given set of particles under (c) we have the following possible set of triples of outcomes:

Photon Set 1:
-30 0 30
+ + +
+ + -
+ - +
- + +
- - +
- + -
+ - -
- - -

These are the only 8 possible outcomes for scenario (c). Clearly, CFD allows us to make this list because we can think hypothetically of what would be observed for each angle without any experimental limitations. That is what scenario (c) is all about anyway. Of course your claim (3) is trivially true for this scenario because as can be seen, whenever there is a mismatch for (-30,30), there is necessarily a mismatch for either (-30,0) or (30, 0) (for example, see all outcomes except the first and the last). The transitive relationship holds trivially. Remember, the only reason justifying your claim (3) as stated in your own words, is the fact that the outcome at -30 for the (-30, 0) pair is exactly the same as the outcome at -30 for the (-30, 30) pair! Same thing for the other pairs! We have exactly 3 outcomes each time from which we select paired combinations! This is the most important reason why your claim (3) is valid for scenario (c) as stated in your own words:

If the results at -30° and 0° differ, then either the results at -30° and 0° differ or the results at 0° and 30° differ

Do you understand this? Do you understand that you do not and can not have any other possible reason why your above quote makes logical sense? Here is another quote from you which says this:

Step 3 is a trivial consequence of the transitive property of equality. If the results at -30° and 0° are the same and the results at 0° and 30° are the same, then clearly the results at -30° and 30° must be the same.

Do you understand that your only justification for making such a statement is the fact that for scenario (c) we have exactly 3 outcomes from which we pick combinations of pairs? Clearly, saying if X = Y and Y = Z, then X = Z can not be true unless X is in fact always equal to Y and Y is in fact always equal to Z.

However, for scenario (b), you do not have 3 outcomes but 3 pairs of independent outcomes. For each pair, we have the following possibilities


Photon set A:
-30 0
1: + +
2: + -
3: - +
4: - -

Photon set B:
0 +30
1: + +
2: + -
3: - +
4: - -

Photon set C:
-30 +30
1: + +
2: + -
3: - +
4: - -

Your transitive relationship which allowed you to state claim (3) now does not apply to this scenario for the following reasons:

* The three photon sets are disjoint!
* Every possibility in each set is also a possiblity for every entry in the other set so we have a total of 4*4*4 = 64 distinct possible outcome combinations.
* For some of the posibilities (a quarter of the total) the outcome at an angle for one set is different from the outcome at the same angle for a different set therefore it is very naive to even think of transitive relationships in this case.
* There is ONLY one scenario under which we may be able to apply your transitive relationship to the scenario (b). It can be done IF and only IF we can rearange the rows of the three distinct sets of outcome possibilities such that after placing them side by side, on each row the same angles have exactly the same outcomes no matter the set of photons. So let us try that.

Looking at the sets, we observe that the outcomes for the two -30 columns already match, and the outcomes for the two +30 columns already match. So we can place the outcomes side by side already. However, we also find that the outcomes for the 0 columns do not match. While those in set A alternate +-+-, those in set B do not ++--. In order to make the 0 columns match we will have to rearragne the rows but unfortunately it is impossible to find a rearrangement for the 0° column without at the same time causing a mismatch in the other columns which already match. For example we could change the order of rows for set A to (1324) but now the -30 column of A will not match the -30 column of set C any more.

Therefore it is impossible to satisfy the key requirement of the transitive relationship (equal outcomes at equal angles for each possibility) in scenario (b) and as a result it is naive and foolish to think that anything such as CFD let alone no-conspiracy will allow your claim (3) to succeed. You claim that CFD and no-conspiray allows you to apply your transitive realationship to scenario (b) just as in scenario (c). As I've explained above, it is not even wrong. To the contrary, the mere act of trying to force the relationship to scenario (b) is a conspiracy on your part as explained above. CFD doesn't even come into scenario (b) it is something completely within the domain of scenario (c).

Therefore, contrary to your claims you haven't even attempted to show how your inequality derived under scenario (c) is applicable to scenario (b).

 

[ThomasT]

Here's my current understanding of lugita15's logic:

Where,
A = different results at (30_0)
B = different results at (-30_0)
C = different results at (-30_30)

Given the assumptions of locality and determinism (in steps 1 and 2 in no required order), then it follows that, for any given individual entangled pair:

3. If C, then either A OR B (but NOT A AND B),

because the result at 0 can't be both detection and nondetection at the same time.

Then lugita15 brings in probablilty, to wit:

4. P(A OR B) ≤ P(A) + P(B)

Since, in step 4, we're dealing with probability, then we're no longer dealing with any given individual entangled pair (wrt which we can apply counterfactual definite reasoning via the assumption of determinism) but with accumulations of many results at settings A and B.

I'm not sure what step 4 refers to, but insofar as I suppose P(A) and P(B) to refer to certain numbers of mismatches (wrt any given set of runs), then I suppose step 4 to mean that the number of mismatches for A OR the number of mismatches for B must be less than or equal the sum of the mismatches for A AND B. So, wrt that supposition, then step 4 is just trivially correct arithmetic, but I'm not sure what it has to do with (that is, that it necessarily follows from) step 3.

So, @ lugita15, I understand step 3 (I had asked in a previous post where step 3 came from), but I'm now not sure how, or if, step 4 is necessitated by step 3.

 

[lugita15]

Well let us examine that claim to see if at all CFD or no-conspiracy allow you to equate (b) and (c) as you claim.

Just to clarify, I'm just equating the probabilities calculated in (b) and (c), not equating them conceptually. And counterfactual definiteness just allows me to talk about (c) at all. It's the no-conspiracy condition that allows me to equate the probabilities.

Here are the (b) and (c) scenarios again for reference:

(b) Just like in QM and EPR-Experiments, the probabilities are calculated for different sets of independent photons.
(c) The probabilities are calculated for the same set of photons by imagining what could have been measured even if they are not in fact measured.

Yes, and I'm saying that the photons do not "know" in advance which set they're going to end up being in, so if photons in one set behave a certain way (i.e with a certain probability of mismatch) when the polarizers are oriented in a certain way then the no-conspiracy condition allows me to conclude that the photons in the other set would have behaved in the same way (i.e. with the same probability of mismatch) if the polarizers had been oriented in the same way. I don't know what else to say.

There is ONLY one scenario under which we may be able to apply your transitive relationship to the scenario (b).

I'm not applying the transitive property of equality to (b). I'm applying it to (c), using it to draw certain conclusions about the probabilities in (c), and then using the no-conspiracy condition to equate the probabilities in (b) and the probabilities in (c).

 

[billschnieder]

Just to clarify, I'm just equating the probabilities calculated in (b) and (c), not equating them conceptually.

Huh? You take a valid relationship between probabilities calculated under (c) and you are expecting the relationship to hold for probabilities under (b) how is that not equating them conceptually? Or are you now admitting by the above that you do not expect the relationship to hold for scenario (b).

It's the no-conspiracy condition that allows me to equate the probabilities.
...
Yes, and I'm saying that the photons do not "know" in advance which set they're going to end up being in, so if photons in one set behave a certain way (i.e with a certain probability of mismatch) when the polarizers are oriented in a certain way then the no-conspiracy condition allows me to conclude that the photons in the other set would have behaved in the same way (i.e. with the same probability of mismatch) if the polarizers had been oriented in the same way.

Your argument boils to to the following: Since the photons in each set do not know whether they will be measured at (-30,30) or (-30,0) or (30,0), the no-conspiracy condition allows you to conclude that if there is a mismatch for (-30, 30) then there must also be a mismatch for either (0, 30) or (-30, 0). Is this really your argument? If this is not the argument you are making, please explain to me why you would think the inequality you obtained for (c) is at all applicable to (b).

The central relationship on which your argument hangs is your claim (3). Without claim (3) you do not and can not have your probability inequality, period. Your claim (3) is valid *because* of the transitive relationship. Without the transitive relationship your claim (3) fails. Without equal outcomes at equal angles, your transitive relationship dies and so does your claim (3). As explained in my previous post, your transitive relationship can not survive in scenario (b).

If you are trying to say you rely only on the equality of probabilities then why do you need claim (3) at all? I have explained to you in my last post by listing all the possible outcomes, that it is impossible to fulfill the requirements of the transitive relationship under scenario (b). Yet you continue to think for some yet unexplained reason the no-conspiracy condition will allow you to use the probability inequality which you obtained by relying on the transitive relationship, in scenario (b) where the transitive relationship is impossible.

I made a simple diagram to illustrate the folly in that line of argument (see attachment), since my last illustration apparently was not clear to you. From the diagram, your argument pretty much boils to the suggestion that since the areas (cf "probabilities") A,B,C on the left and on the right are the same, the no-conspiracy condition allows you to say that the total area occupied by A,B,C on the left is also the same as the total area occupied by A,B,C on the right.

Do you see how wrong that is? Just because the individual probabilities P(A), P(B), and P(C) are the same does not mean all probabilities or relationships between the two scenarios are the same. For example let us calculate P(A or B or C). We know from probability theory that
P(A or B or C) = P(A) + P(B) + P(C) – P(A and B and C). For scenario (c), P(A and B and C) is non-zero. However for scenario (b) in which all three outcome sets are disjoint, P(A and B and C) MUST be zero. Therefore, even though the individual probabilities are the same, their relationships are clearly different for the two scenarios. Where is the conspiracy in this?

You keep saying that the no-conspiracy allows you to equate the probabilities. Please explain how the no-conspiracy assumption allows you to escape this predicament, or explain the conspiracy in the above straight-forward application of probability theory.

We do not even begin to talk about what photons may know in advance or not, etc. Your argument fails long before we get to that point so it is mind-boggling why you would think no-conspiracy bails you out.

I'm not applying the transitive property of equality to (b).

Does the inequality apply to scenario (b) or not. Please answer this question explicitly without equivocating. If you say it does, then you are contradicting yourself here. If you say it doesn't then why have we been wasting time with all this. Why are we even discussing this at all if the inequalities do not apply to QM results or experimental outcomes ?

 

[lugita15]

Huh? You take a valid relationship between probabilities calculated under (c) and you are expecting the relationship to hold for probabilities under (b)

Yes, because I am saying the probabilities in (b) are equal to the probabilities in (c).

how is that not equating them conceptually?

All I meant was that they are two different concepts. The probabilities in (b) and the probabilities in (c) do not have to be equal as a matter of logic. They just happen to be equal due to the no-conspiracy condition. Just like how general relativity doesn't conceptually equate inertial mass and gravitational mass, it just says they happen to be equal due to the equivalence principle.

Or are you now admitting by the above that you do not expect the relationship to hold for scenario (b).

I definitely do expect the relationship to hold for (b), at least the final relationship P(C)≤P(A)+P(B).

Your argument boils to to the following: Since the photons in each set do not know whether they will be measured at (-30,30) or (-30,0) or (30,0), the no-conspiracy condition allows you to conclude that if there is a mismatch for (-30, 30) then there must also be a mismatch for either (0, 30) or (-30, 0). Is this really your argument? If this is not the argument you are making, please explain to me why you would think the inequality you obtained for (c) is at all applicable to (b).

No, that's not the argument I'm making. Rather, I'm saying that in (c) a mismatch at (-30,30) implies that there is a mismatch at either (-30,0) or (0,30), from which I am concluding that P(C)≤P(A)+P(B) where P(A), P(B), and P(C) are the probabilities in (c). And then I am arguing that due to the no-conspiracy condition, P(A), P(B), and P(C) are the same in (b) and (c).

My application of the no-conspiracy condition goes as follows: since the photons in each set do not know whether they will be measured at (-30,30) or (-30,0) or (30,0), the probability of mismatch any of the three sets would have if the photons in the set were measured at, say, (-30,30), is the same for all three sets, regardless of what angles you actually set the polarizers for each of the three sets.

The central relationship on which your argument hangs is your claim (3). Without claim (3) you do not and can not have your probability inequality, period. Your claim (3) is valid *because* of the transitive relationship. Without the transitive relationship your claim (3) fails. Without equal outcomes at equal angles, your transitive relationship dies and so does your claim (3). As explained in my previous post, your transitive relationship can not survive in scenario (b).

The point is that the transitive relationship stated in step 3 IS valid for (c), so the Bell inequality derived in step 6 IS valid for (c), and then the no-conspiracy condition implies that the inequality is also valid for (b), because the probabilities are equal.

If you are trying to say you rely only on the equality of probabilities then why do you need claim (3) at all?

I need the transitive relationship in step 3 in order to derive the inequality for (c). I rely on the equality of probabilities in (b) and (c) in order to conclude that the inequality holds for (b).

I made a simple diagram to illustrate the folly in that line of argument (see attachment), since my last illustration apparently was not clear to you. From the diagram, your argument pretty much boils to the suggestion that since the areas (cf "probabilities") A,B,C on the left and on the right are the same, the no-conspiracy condition allows you to say that the total area occupied by A,B,C on the left is also the same as the total area occupied by A,B,C on the right.

No, that's not what I'm doing at all. I'm saying that areas A, B, and C on the left are equal to those on the right, therefore if area C is less than or equal to area A + area B on the left, area C must also be less than or equal to area A + area B on the right.

Do you see how wrong that is? Just because the individual probabilities P(A), P(B), and P(C) are the same does not mean all probabilities or relationships between the two scenarios are the same.

I certainly agree that just because P(A), P(B), and P(C) are the same in (b) and (c) doesn't mean ALL relationships in (b) and (c) are the same. However, SOME relationships are the same, and P(C)≤P(A)+P(B) is clearly one of them. I don't know how you can possibly dispute that. If P(A) and P(B) are the same, then P(A)+P(B) is the same. If P(A)+P(B) and P(C) are the same, then P(C)≤P(A)+P(B) is the same.

Does the inequality apply to scenario (b) or not. Please answer this question explicitly without equivocating. If you say it does, then you are contradicting yourself here. If you say it doesn't then why have we been wasting time with all this. Why are we even discussing this at all if the inequalities do not apply to QM results or experimental outcomes ?

Yes, the inequality applies to (b) because of the no-conspiracy condition and the fact that it applies to (c).

 

[billschnieder]

My application of the no-conspiracy condition goes as follows: since the photons in each set do not know whether they will be measured at (-30,30) or (-30,0) or (30,0), the probability of mismatch any of the three sets would have if the photons in the set were measured at, say, (-30,30), is the same for all three sets, regardless of what angles you actually set the polarizers for each of the three sets.

What do you mean by probability of mismatch please state clearly what that means in scenario (c) and in the experiment of scenario (b). For both scenarios, please state clearly what is the numerator and what is the denominator in your probability calculation. Please state clearly what is measured in the experiment and how you calculate the above probability from it. Once you do that, you will see that you are comparing apples and oranges.

Sure the photons do not know at which angle they will be measured but how does that translate to the probability of measurement at (-30,30) being the same as the probability for measurement at (-30,0). We know from classical physics that the probability of detecting a the pair for the (-30,30) angles is different from the probability of detecting the pair at (-30,0). Are you saying there is a conspiracy in Malus law?

The essence of your argument is the following:

- The trivial inequality is true because of the transitive property of scenario (c).
- But even though the transitive property is false for scenario (b), the inequality is allegedly still true because the probabilities under (b) are allegedly always equal to the probabilities under (c) due to no-conspiracy.
- But really, the QM experimental probabilities under (b) are not really equal to those under (c), yet the inequality allegedly still applies and we get a violation.
- Therefore we must interpret the violation to mean there is something "awry" with scenario (b).

You have not provided a convincing reason why the probabilities measured under (b) should be same as the hypothetical ones imagined under (c). Your no-conspiracy argument doesn't make sense. A photon can very well be predetermined to not be detected at certain angles without any conspiracy even with a perfect experimental setup. Malus law tells us that. In that case, when calculating probabilities for the hypothetical scenario (c), you are dividing the number of mismatches by the total number of photons which you know since it is hypothetical. However for scenario (b) you do not and can not know the total of all photons produced. You only know what you can measure in which case all the undetected photons have been filtered off. And we know from Malus law that the fraction of photons will change with angular difference. So the frequencies you calculate are not the same as the probabilities from (c).

So the probabilities in (b) which you claim are equal to those in (c) are in fact not measurable and are not the same ones predicted by QM and experiments. The QM and experimental probabilities pertain to measurable outcomes ONLY. You have fallen victim to DrC's "full universe" fallacy. You have full universe for (c) you can not and do not have full universe for QM or experiments. You are comparing apples to oranges.

 

[lugita15]

Given the assumptions of locality and determinism (in steps 1 and 2 in no required order)

I still think that it's better to say that the assumptions are identical behavior at identical angles and local determinism, but this is largely a matter of taste.

I'm not sure what step 4 refers to, but insofar as I suppose P(A) and P(B) to refer to certain numbers of mismatches (wrt any given set of runs), then I suppose step 4 to mean that the number of mismatches for A OR the number of mismatches for B must be less than or equal the sum of the mismatches for A AND B. * So, wrt that supposition, then step 4 is just trivially correct arithmetic, but I'm not sure what it has to do with (that is, that it necessarily follows from) step 3.

. Well, technically probability has to a number between 0 and 1, so the probability would be the number of mismatches divided by the number of runs. But that's OK, I can phrase step 4 in terms numbers of mismatches instead: the number of particle pairs for which either A or B is true is less than or equal to the number of particle pairs for which A is true plus the number of particle pairs for which B is true. Dividing this by the number of runs, so we return to probabilities: the percentage of particle pairs for which A or B is true is less than or equal to the percentage of particle pairs for which A is true plus the percentage of particle pairs for which B is true. Do you understand that?

So, @ lugita15, I understand step 3 (I had asked in a previous post where step 3 came from), but I'm now not sure how, or if, step 4 is necessitated by step 3.

To clarify the logic, steps 3 and 4 follow from steps 1 and 2, step 5 folloas from step 3, and step 6 follows from steps 4 and 5.

 

[billschnieder]

I still think that it's better to say that the assumptions are identical behavior at identical angles and local determinism, but this is largely a matter of taste.

Do you know that for the 64 possible outcome combinations in scenario (b), in 1/4 of them you do not have identical behaviour at identical angles even though local causality is true ?! (see https://www.physicsforums.com/showpost.php?p=3856772&postcount=125)

A good highly recommended article which discusses this is the following:
http://edelstein.huji.ac.il/staff/pitowsky/#Publications
George Boole's 'Conditions of Possible Experience' and the Quantum Puzzle
Brit. J. Phil. Sci. 45 (1994). 95-125
http://edelstein.huji.ac.il/staff/pitowsky/papers/Paper23.pdf

In the mid-nineteenth century George Boole formulated his 'conditions of possible
experience'. These are equations and inequalities that the relative frequencies of
(logically connected) events must satisfy. Some of Boole's conditions have been
rediscovered in more recent years by physicists, including Bell inequalities, Clauser
Horne inequalities, and many others.
...
CAN BOOLE'S CONDITIONS BE VIOLATED?
One thing should be clear at the outset: none of Boole's conditions of possible
experience can ever be violated when all the relative frequencies involved have been
measured in a single sample. The reason is that such a violation entails a logical
contradiction.
...
In case we deal with relative frequencies in a single sample, a violation of any of the relevant Boole's
conditions is a logical impossibility.
...
But sometimes, for various reasons, we may choose or be forced to measure
the relative frequencies of (logically connected) events, in several distinct
samples.In this case a violation of Boole's conditions may occur. There are
various possible reasons for that, and they are listed below in an increasing
order of abstractness:
(a) Failure of randomnes...
(b) Measurement biases...
(c) No distribution...
(d) Mathematical oddities...
...
These are the cases, of which I am aware, where Boole's conditions might be
violated. Another possibility, which has been neglected, is the case where we
erroneously believe that some logical relation among the events obtains, and
thus, wrongly expect some condition to be satisfied. Strictly speaking, this case
does not represent a violation of Boole's conditions, but rather an error of
judgement.

 

[lugita15]

What do you mean by probability of mismatch please state clearly what that means in scenario (c) and in the experiment of scenario (b). For both scenarios, please state clearly what is the numerator and what is the denominator in your probability calculation. Please state clearly what is measured in the experiment and how you calculate the above probability from it. Once you do that, you will see that you are comparing apples and oranges.

All right, in (b), P(A) for example is the number of mismatches among the particle pairs for which you orient the polarizers at -30 and 0, divided by the number for particle pairs for which you orient the polarizers at -30 and 0. In (c), P(A) is the number of mismatches you would get if you oriented the polarizers at -30 and 0 for all the particle pairs, divided the total number of particle pairs. And I am claiming that the P(A) in (b) and the P(A) in (c) are equal due to the no-conspiracy condition. Is that clear enough?

Sure the photons do not know at which angle they will be measured but how does that translate to the probability of measurement at (-30,30) being the same as the probability for measurement at (-30,0).

I don't know what you mean by the probability of measurement. I'm assuming that 100% of the particle pairs emitted by the source are sent through the polarization detectors, and that these detectors output either a 1 or a 0, 1 indicating that the photon is polarized in the direction of the detector and 0 indicating that the photon is polarized perpendicular to the direction of the detector. I'm doing the Bell inequality here (in the form Herbert presents it), not the CHSH inequality, so I'm not allowing for nondetection, in the sense of a particle not even reaching the detector.

If you meant "probability of mismatch" instead, I certainly don't think that the probability of mismatch at -30 and 30 must be the same as the probability of mismatch at -30 and 0. Rather, I'm saying that the no-conspiracy condition implies that the probability that you would get a mismatch at (say) -30 and 30 is the same for the particle pairs for which you DO set the polarizers at -30 and 30, and the particle pairs for which you don't.

The essence of your argument is the following:

- The trivial inequality is true because of the transitive property of scenario (c).
- But even though the transitive property is false for scenario (b), the inequality is allegedly still true because the probabilities under (b) are allegedly always equal to the probabilities under (c) due to no-conspiracy.
- But really, the QM experimental probabilities under (b) are not really equal to those under (c), yet the inequality allegedly still applies and we get a violation.
- Therefore we must interpret the violation to mean there is something "awry" with scenario (b).

Here's my summary of the argument, paraphrasing you:

- The Bell inequality in step 6 holds for (c) because the transitive property in step 3 holds for (c)
- But even though the transitive property in step 3 does not make sense for (b), the Bell inequality in step 6 holds for (b) because the probabilities P(A), P(B), and P(C) in (b) are equal to the probabilities in (c) due to the no-conspiracy condition.
- But really the QM experimental probabilities in (b) do not satisfy the inequality in step 6
- We interpret the violation as there being something awry with the assumption used in deriving the inequality for (c), specifically the assumption of local determinism

You have not provided a convincing reason why the probabilities measured under (b) should be same as the hypothetical ones imagined under (c). Your no-conspiracy argument doesn't make sense.

The argument seems so clear to me. If the particle pairs are randomly divided into two sets, one set for which you WILL orient the polarizers at -30 and 0, and the other set for which you WON'T, and the particles do not "know" which set they're going to end up in, then the no-conspiracy condition says that the probability of mimatch you would get if you oriented the polarizers at -30 and 0 in one set must be the same as for the other set.

A photon can very well be predetermined to not be detected at certain angles without any conspiracy even with a perfect experimental setup. Malus law tells us that. In that case, when calculating probabilities for the hypothetical scenario (c), you are dividing the number of mismatches by the total number of photons which you know since it is hypothetical. However for scenario (b) you do not and can not know the total of all photons produced. You only know what you can measure in which case all the undetected photons have been filtered off. And we know from Malus law that the fraction of photons will change with angular difference. So the frequencies you calculate are not the same as the probabilities from (c).

I think we're talking about different things. You're talking about polarizing filters that simply absorbs and "forgets" about photons that are polarized perpendicular, and pass through photons that are polarized parallel. I'm talking about polarization detectors that let you know no matter what that a photon has arrived at the detector, and then tells you whether the photon is polarized parallel or perpendicular.

 

[billschnieder]

All right, in (b), P(A) for example is the number of mismatches among the particle pairs for which you orient the polarizers at -30 and 0, divided by the number for particle pairs for which you orient the polarizers at -30 and 0.

The numerator still very ambiguous. "Particle pairs for which you orient the polarizers at -30 and 0 ..." is not clear. Are you referring to:
- mismatches observed for particle pairs going through the polarizers oriented at -30 and 0
- or mismatches existing in particle pairs sent to the polarizers oriented at -30 and 0.
You need to be precise or state clearly if you are assuming the two are the same. Do you understand that the two are not necessarily the same?
Same thing applies for the denominator. Are you deviding by:
- total number of particles pairs going through the polarizers oriented at -30 and 0
- or total number of particle pairs sent to the polarizers oriented at -30 and 0.
Again, be precise or state clearly if you are assuming the two are the same.

And I am claiming that the P(A) in (b) and the P(A) in (c) are equal due to the no-conspiracy condition. Is that clear enough?

No. See above.

I don't know what you mean by the probability of measurement. I'm assuming that 100% of the particle pairs emitted by the source are sent through the polarization detectors, and that these detectors output either a 1 or a 0, 1 indicating that the photon is polarized in the direction of the detector and 0 indicating that the photon is polarized perpendicular to the direction of the detector.
...
so I'm not allowing for nondetection

You have a polarized beam leaving a calcite crystal and going throug a polarizer oriented at an arbitrary angle and you are assuming 100% detection? How is that not a violation of Malus law? What does 1 mean and what does 0 mean then. Doesn't 0 mean that the photon was not detected by the given detector? So a paired outcome (1,1) means both detectors detected photons, (1,0) and (0,1) means only one detected photons and (0,0) means none of them detected anything. Am I misunderstanding the experiment? So when you talk of mismatches are you are referring to just (1,0) and (0,1) right?
Here is what Herbert says:

For instance, light completely polarized along the calcite axis will excite only one of the detectors. Completely unpolarized light will excite both detectors to the same extent.
...
We point SPOT at a source of light and SPOT puts out a sequence of pulses which we label "1" or "0" depending on which of the two detectors fires. We calibrate SPOT by aiming it at a source of Vertically polarized light, turning the tube till one detector fires all the time, calling that detector "1" and moving SPOT's little red arrow so that it points vertically. In this calibration mode SPOT's output looks like this:
...1, 1, 1, 1, 1, 1, 1, 1, 1...
which we interpret to mean that SPOT is looking at a beam of light made up solely of Vertically-polarized photons. If we turn SPOT by 90 degrees while viewing the same Vertically polarized light beam, its output will look like this:
...0, 0, 0, 0, 0, 0, 0...

Please clearify that you agree that nondetection IS involved. Now say in a given experiment we had ONLY 0,0,0,... for both detectors. Using your answers to my earlier questions, how would you calculate the probability of mismatches. I'm not interested in the answer but the method of calculation. Precisely how you obtain the numerator and how you obtain the denominator. And please while explaining this, do not use any information that would not be available/measurable by an experimenter doing an actual experiment.

 

[lugita15]

The numerator still very ambiguous. "Particle pairs for which you orient the polarizers at -30 and 0 ..." is not clear. Are you referring to:
- mismatches observed for particle pairs going through the polarizers oriented at -30 and 0
- or mismatches existing in particle pairs sent to the polarizers oriented at -30 and 0.
You need to be precise or state clearly if you are assuming the two are the same. Do you understand that the two are not necessarily the same?

All right, I am stating clearly that I am assuming the two are the same. I am talking about polarization detectors, for instance consisting of a calcite crystal and two photon counters as in Herbert, that split the beam of photons into two beams, one of which is polarized in the polarization direction of the detector and one of which is polarized perpendicular to the polarization direction of the detector. But regardless, there are no photon pairs we don't know about.

Same thing applies for the denominator. Are you deviding by:
- total number of particles pairs going through the polarizers oriented at -30 and 0
- or total number of particle pairs sent to the polarizers oriented at -30 and 0.
Again, be precise or state clearly if you are assuming the two are the same.

Again, I am assuming the two are the same. Note that I am often sloppy with my terminology, saying "going through" when I really mean going through in the "1" or "main" beam.

You have a polarized beam leaving a calcite crystal and going throug a polarizer oriented at an arbitrary angle and you are assuming 100% detection? How is that not a violation of Malus law?

Because Malus' Law is about the probability that a photon goes through or is absorbed by a polarizing filter. I'm talking here not about polarizing filters, but about an apparatus that does not absorb photons that are polarized perpendicular. Rather, it sends photons that are polarized perpendicular on a separate path from those that are polarized parallel, so that the photons going on the two paths hit separate photon counters. This can be done via a calcite crystal like Herbert does.

What does 1 mean and what does 0 mean then. Doesn't 0 mean that the photon was not detected by the given detector? So a paired outcome (1,1) means both detectors detected photons, (1,0) and (0,1) means only one detected photons and (0,0) means none of them detected anything. Am I misunderstanding the experiment? So when you talk of mismatches are you are referring to just (1,0) and (0,1) right?

A 1 means that the photon was detected by the 1 photon counter, and a 0 means that the photon was detected by the 0 photon counter. There is a calcite crystal, a 0 photon counter, and a 1 photon counter at each end.

Please clearify that you agree that nondetection IS involved.

I'm sorry, I don't agree with that. Every photon is either detected by the "1" photon counter or the "0" photon counter.

Now say in a given experiment we had ONLY 0,0,0,... for both detectors. Using your answers to my earlier questions, how would you calculate the probability of mismatches. I'm not interested in the answer but the method of calculation. Precisely how you obtain the numerator and how you obtain the denominator. And please while explaining this, do not use any information that would not be available/measurable by an experimenter doing an actual experiment.

If I have 0,0,0,... for both detectors, then the numerator is the number of observed mismatches, which is 0, and the denominator is the number of particles observed, so I conclude that the probability of mismatch is 0.

 

[billschnieder]

All right, I am stating clearly that I am assuming the two are the same. I am talking about polarization detectors, for instance consisting of a calcite crystal and two photon counters as in Herbert, that split the beam of photons into two beams, one of which is polarized in the polarization direction of the detector and one of which is polarized perpendicular to the polarization direction of the detector. But regardless, there are no photon pairs we don't know about.

OK I went back and re-read Herbert's description and I was indeed misunderstanding the setup. He does appear to claim 100% detection of all photons. I do not believe such a device exists but I'll have to ponder it some more but it seems like a waste of time discussing a such a setup when we have actual setups of experiments which have been performed.

In any case, I still do not think it is legitimate to mix and match probability spaces the way you are doing. Besides violation of the relationship P(A or B) <= P(A) + P(B) in standard probability theory simply means the events A and B do not belong to the same probability space and the QM probabilities you are using do not belong to the same probability space, neither does the ones from the experiment. It seems circular to try to argue that the relationship should apply to QM and experiments because the probabilities are the same but at the same time argue that the probabilities are not the same. Why not derive the relationship directly from scenario (b). Why go through the winded way through scenario (c)?

 

[lugita15]

OK I went back and re-read Herbert's description and I was indeed misunderstanding the setup. He does appear to claim 100% detection of all photons. I do not believe such a device exists but I'll have to ponder it some more but it seems like a waste of time discussing a such a setup when we have actual setups of experiments which have been performed.

The problem is that actual experimental setups allow for all sorts of experimental loopholes that tend to muddy the waters. I'm less interested in the question of whether currently practical Bell tests definitively disprove local determinism. What I'm trying to do here is to show is that an ideal experiment would be able adjudicate the differences between the predictions of quantum mechanics and those of local determinism.

There's a lot of ways to implement what's required for the proof. If you don't like birefringent calcite crystals, we can instead conduct a position measurement before it reaches a polarizing filter. This way, regardless of whether it succeeds in passing through the filter, we at least know that it was about to hit the filter, so that there are no photon pairs we don't know about.

In any case, I still do not think it is legitimate to mix and match probability spaces the way you are doing. Besides violation of the relationship P(A or B) <= P(A) + P(B) in standard probability theory simply means the events A and B do not belong to the same probability space and the QM probabilities you are using do not belong to the same probability space, neither does the ones from the experiment.

But in (c), statements A, B, and C all have well-defined truth-values for all particle pairs. So there's no issue of separate probability spaces.

It seems circular to try to argue that the relationship should apply to QM and experiments because the probabilities are the same but at the same time argue that the probabilities are not the same.

Let me be clear what I'm doing, which is a proof by contradiction. I am assuming that there is a local deterministic reality that underlies quantum mechanical phenomena. Counterfactual definiteness allows me to meaningfully talk not merely the results I did get for a particular particle pair, but also about the results I would have gotten if I had made different measurement decisions. I then get P(C)≤P(A)+P(B) for (c), and then I use the no-conspiracy condition to get that P(A), P(B), and P(C) are the same for (b) and (c), so the inequality is valid for (b). But the probabilities I actually get from the experimental results of quantum mechanics do not conform to the inequality. Thus the inequality in (b) is wrong, so the inequality in (c) must also be wrong, and thus we can reject the assumption that local determinism underlies quantum mechanical phenomena.

Why not derive the relationship directly from scenario (b). Why go through the winded way through scenario (c)?

Let's review what (b) and (c) are:
"(b) Just like in QM and EPR-Experiments, the probabilities are calculated for different sets of independent photons.
(c) The probabilities are calculated for the same set of photons by imagining what could have been measured even if they are not in fact measured."

To put it another way, (b) means we restrict ourselves only to factual outcomes, whereas (c) means we also consider counterfactual outcomes, outcomes we would have gotten if our measurement decisions had been different. So by asking me why the proof has to invoke (c), you're asking my why the proof involves counterfactual reasoning. Well, that is just a general feature of Bell's theorem proofs. Restricting ourselves to factual reasoning, all we can do is the EPR argument, not Bell. That's why counterfactual definiteness is an assumption of the theorem. But you seem to accept both counterfactual definiteness and the no-conspiracy condition. So how can you reject the validity of Bell's theorem?

 

[billschnieder]

The problem is that actual experimental setups allow for all sorts of experimental loopholes that tend to muddy the waters.

On the contrary it is Herberts (and your) approach that muddies the waters because it take a gedanken experiment which has never been performed and then midway brings in results from an actual experiment and it's QM prediction to make gradiose claims based on the alledged disagreement between actual experimental results and the predictions of the never performed gedanken experiment. Not only that, it makes use of many assumptions about the physical systems which have not be verified.

I'm less interested in the question of whether currently practical Bell tests definitively disprove local determinism. What I'm trying to do here is to show is that an ideal experiment would be able adjudicate the differences between the predictions of quantum mechanics and those of local determinism.

The best you can do with this approach is show that your toy theory disagrees with QM in the context of your toy experiment. Nothing more. How useful is that? Besides the probabilities you calculate in your example are global variables which by itself is a "loophole". It is impossible to calculate your probabilities without using joint information from the two stations at the same time after the fact so why would you even think that what you are doing tells you anything about local determinism?

There's a lot of ways to implement what's required for the proof. If you don't like birefringent calcite crystals, we can instead conduct a position measurement before it reaches a polarizing filter. This way, regardless of whether it succeeds in passing through the filter, we at least know that it was about to hit the filter, so that there are no photon pairs we don't know about.

So long as the thought experiment you are basing your "proof" on is not the exact same experiment that was actually done, and for which you have QM results, it makes no sense to even proceed.

But in (c), statements A, B, and C all have well-defined truth-values for all particle pairs. So there's no issue of separate probability spaces.

They all have well defined values in (b) too, just not simultaneously for the same population. The validity of the "proof" for scenario (c) is not and was never in question so I'm not sure what you mean by the above.

Let me be clear what I'm doing, which is a proof by contradiction. I am assuming that there is a local deterministic reality that underlies quantum mechanical phenomena. Counterfactual definiteness allows me to meaningfully talk not merely the results I did get for a particular particle pair

But you will not obtain a different inequality by assumint non-local causality. All you need is a single list of triples of results at -30, 0 and 30. It doesn't matter how you obtain the list. It could be results of spooky action, or any other non-local effect you choose. You will still obtain the inequality you obtained. If you disagree, find me a which violates the inequality. So contrary to what you may think, the inequality has nothing to do with local determinism. It is simply a mathematical tautology for lists of triples. You even recognized this yourself when you said:

Strictly speaking, none of my steps constitute a locality condition. Step 2 describes a condition that is satisfied by ALL local deterministic theories, but not ONLY local deterministic theories.

How come you do not conclude that all theories are ruled out?

But the probabilities I actually get from the experimental results of quantum mechanics do not conform to the inequality. Thus the inequality in (b) is wrong, so the inequality in (c) must also be wrong, and thus we can reject the assumption that local determinism underlies quantum mechanical phenomena.

See previous point. Why focus on local determinism when you know that non-local theories are also ruled out? I've been trying to tell you all along that you should be looking elsewhere for the reason of the violation.

To put it another way, (b) means we restrict ourselves only to factual outcomes, whereas (c) means we also consider counterfactual outcomes, outcomes we would have gotten if our measurement decisions had been different. So by asking me why the proof has to invoke (c).

I'm asking you why you are unable to directly derive the inequality from (b).

you're asking my why the proof involves counterfactual reasoning. Well, that is just a general feature of Bell's theorem proofs.

Do you what is another general feature of Bell's inequality? It is also violated by real experiments.

That's why counterfactual definiteness is an assumption of the theorem. But you seem to accept both counterfactual definiteness and the no-conspiracy condition. So how can you reject the validity of Bell's theorem?

I accept that Bell's inequality is iron-clad valid. However Bell's theorem involves comparing an unperformable experiment with results from actual experiments and QM predictions for actual experiment. I assume you are familiar with the modal fallacy? There is something not too unlike it in play here. The elephant in the room is the fact that unperformed experiments have no results. There is a difference between an algebraic manipulation and an empirical manipulation.

 

[billschnieder]

Permit me to reproduce an explanation I gave previously about CFD/Realism etc. I believe it is relevant here.
https://www.physicsforums.com/showpost.php?p=3333711&postcount=186

"Everything that is *actual* is *possible*, but not everything that is *possible* is actual"
A prediction about an experimental outcome such as P(a,b) is a conditional statement, premised on the exact experimental conditions -- of the form: If Alice and Bob measure along a and b, the result will be P(a,b)
Such statements are not *actualities*, but *possibilities*. All *possibilities* are simultaneously true (CFD). Again remember that to say a *possibility* is true, means the relationship between the antecedent and the consequent is correct and valid. It does not mean the antecendent alone is true or the consequent alone is true. It is easy to get confused about this if you erroneously strip of conditioning clauses. This is fully consistent with realism. Since the prediction is relying on the existence of hidden properties in the particles and the instrument, which together with the settings will give the results.

However, you run into problems when you confuse such predictions with properties. And expect each particle to *possess* an outcome. This is obviously what most Bell proponents are doing. Having naively confused such *possiblities* to be *actualities*, you start wondering why they violate Bell's inequality which is also based on "actualities". You then naively conclude that since these "actualities" (according to you) violate Bell's inequality it means realism is false or CFD is false, since realism requires that *actualities* be simultaneously actual.

But the violation is not due to the failure of the realistic statement that: "All actualities are simultaneously actual" or the CFD statement that "All possibilities are simultaneously true". The violation is due to the faulty and naive understanding of realism or CFD to be equivalent to: "All possibilities are simultaneously actual", where in your mind you haven't made the clear distinction between an actuality and a possibility. I find that this mistake is made mostly by those who do not have a clear grasp of the difference between ontology and epistemology. An actuality is ontological, a possiblities is epistemological.

NOTE! All possibilities ARE simultaneously TRUE, but that does not mean all possibilities are simultaneously *actual*. Realism implies simultaneous actuality, and since everything that is simultaneously actual is simultaneously possible, everything that is real is simultaneously possible as well. However, this does not mean everything that is simultaneously "possible" is simultaneously "actual". This is the syllogistic fallacy often committed by Bellists when they try to interpret the EPR paper. In the case of Bell's inequalities, we just happened to have an expression in which all the "possible" terms were not and can NEVER be simultaneously "actual". So when a violation is obtained, Bellists fallaciously proclaim the failure of realism/CFD/Locality etc.

The terms in Bell's inequality are actualities to an omniscient observer who does not need to make any measurements. But to QM and Experiments, those terms can NEVER be realized in any experiment because only two measurements can be made for any particle pair. Therefore Bell's inequalities can NEVER be tested in any experiment which is possible. It is limited to the realm of impossible gedanken experiments and omniscient beings. The results from real experiments and predicted by QM, therefore correspond to a different experiment than that modeled by Bell, hence the violation. All attempts to derive inequalities compatible with these experiments have resulted in inequalities which are never violated by QM.

 

[harrylin]

Permit me to reproduce an explanation I gave previously about CFD/Realism etc. I believe it is relevant here.
https://www.physicsforums.com/showpost.php?p=3333711&postcount=186

"Everything that is *actual* is *possible*, but not everything that is *possible* is actual"[..]

Hi Bill, here you seem to be arguing against an IMHO superfluous assumption of lugita15, which was discussed in the thread on Herbert's proof. As far as I can see, you never commented on my posting https://www.physicsforums.com/showthread.php?p=3829811 there. The resulting consensus (in which you didn't share) was that the flaw in Herbert's proof is that experimental results so far are not as he pretends, but that for an "ideal" experiment his proof would be valid. But it appears that you think that even if his experimental assumptions were correct, still Herbert's proof is faulty. So, it may be useful for everyone if you could clarify in that thread, which is still open, what according to you the flaw is in Herbert's proof if his idealised experimental assumptions were true.

PS in view of post #26 there, it's useless to explain the inefficient detection flaw; just consider Herbert's proof for 100% detection efficiency.

 

[billschnieder]

Hi Bill, here you seem to be arguing against an IMHO superfluous assumption of lugita15, which was discussed in the thread on Herbert's proof. As far as I can see, you never commented on my posting https://www.physicsforums.com/showthread.php?p=3829811 there. The resulting consensus (in which you didn't share) was that the flaw in Herbert's proof is that experimental results so far are not as he pretends, but that for an "ideal" experiment his proof would be valid. But it appears that you think that even if his experimental assumptions were correct, still Herbert's proof is faulty. So, it may be useful for everyone if you could clarify in that thread, which is still open, what according to you the flaw is in Herbert's proof if his idealised experimental assumptions were true.

PS in view of post #26 there, it's useless to explain the inefficient detection flaw; just consider Herbert's proof for 100% detection efficiency.

Hi Harrylin,
I do not have time now to read the whole of that thread to know what was or wasn't addressed. But I believe the main points are addressed as follows:

- The main claim that "Starting with two completely identical binary messages, if A's 30 degree turn introduces a 25% mismatch and B's 30 degree turn introduces a 25% mismatch, then the total mismatch (when both are turned) can be at most 50%." is not necessarily true. It is only true if you assume that the mismatch is linearly produced from the angle. This is false if you consider what we already know about polarization of light, from Malus law.

- The second issue which I have discussed in this thread is that the inequality is derived for possibilities which can never be simultaneously realized (actualized). In principle it is impossible to test experimentally, so trying to take experimental results on the basis that probabilities are the same doesn't make sense. The probabilies may be the same but not simultaneously.

 

[harrylin]

Hi Harrylin,
I do not have time now to read the whole of that thread to know what was or wasn't addressed.

I fully agree, which is why I referred you to a specific post of mine which I ask you to answer there - you only have to click on the link, and next to click on Reply!

But I believe the main points are addressed as follows: [..]

If you agree that it apples to the aforementioned linked post of mine (it seems that it does), then I'll add your comment on the topic of Herbert's proof to that topic - and no doubt I'll consider it and comment on it. :tongue2:

 

[billschnieder]

I fully agree, which is why I referred you to a specific post of mine which I ask you to answer there - you only have to click on the link, and next to click on Reply!

If you agree that it apples to the aforementioned linked post of mine (it seems that it does), then I'll add your comment on the topic of Herbert's proof to that topic - and no doubt I'll consider it and comment on it. :tongue2:

Please go ahead.

 

[lugita15]

On the contrary it is Herberts (and your) approach that muddies the waters because it take a gedanken experiment which has never been performed and then midway brings in results from an actual experiment and it's QM prediction to make gradiose claims based on the alledged disagreement between actual experimental results and the predictions of the never performed gedanken experiment.

I am not midway bringing in actual experimental results. I am describing a thought experiment, which indeed has not been performed but could in principle be performed, and I am determining what local determinism and quantum mechanics would each predict for this experiment. I conclude that their predictions differ, thus demonstrating that the question of whether local determinism underlies observed quantum mechanical phenomena is in principle experimentally testable.

Not only that, it makes use of many assumptions about the physical systems which have not be verified.

I freely admit to that. But it is not my goal here to show that existing experimental data is sufficient to refute local determinism. Rather, it is that there EXISTS an experiment which can be performed for which QM predicts one result and local determinism predicts a different result. If I can get you (and others) to agree on just this much, that's enough for me.

The best you can do with this approach is show that your toy theory disagrees with QM in the context of your toy experiment. Nothing more. How useful is that?

Well, I certainly agree that I am reasoning in the context of a "toy experiment", i.e. one that so far people haven't performed and thus for which we don't definitively know what the results would be. But we DO know what different theories would predict the results to be.

Besides the probabilities you calculate in your example are global variables which by itself is a "loophole". It is impossible to calculate your probabilities without using joint information from the two stations at the same time after the fact so why would you even think that what you are doing tells you anything about local determinism?

I agree that in order to determine the probabilities, we need to compare the results of distant polarization detectors. But why is that a "loophole"? Local determinism leads to certain conclusions as to the relationship between the probabilities, even if the probabilities are "global" in the sense that locally available data (at the time of measurement) is not sufficient to compute them. How does that present a contradiction in any way?

So long as the thought experiment you are basing your "proof" on is not the exact same experiment that was actually done, and for which you have QM results, it makes no sense to even proceed.

But even if it wasn't actually done, we can still talk about what quantum mechanics and local determinism predict for it.

They all have well defined values in (b) too, just not simultaneously for the same population. The validity of the "proof" for scenario (c) is not and was never in question so I'm not sure what you mean by the above.

OK, so then if you agree that P(C)≤P(A)+P(B) in (c), do you disagree with me concerning the three probabilities being equal in (b) and (c) due to the no-conspiracy condition?

But you will not obtain a different inequality by assumint non-local causality. All you need is a single list of triples of results at -30, 0 and 30. It doesn't matter how you obtain the list. It could be results of spooky action, or any other non-local effect you choose. You will still obtain the inequality you obtained. If you disagree, find me a which violates the inequality. So contrary to what you may think, the inequality has nothing to do with local determinism. It is simply a mathematical tautology for lists of triples.

But the thing is, in a nonlocal theory you need not have a pre-existing triple that both particles have in common that is determining their behavior. Instead, the result the particle 1 gives when sent through polarization detector 1 oriented at angle x may depend directly on the angle setting y of the distant polarization detector 2, through which particle 2 is being sent. That is how things work in Bohmian mechanics, for example.

You even recognized this yourself when you said:

Strictly speaking, none of my steps constitute a locality condition. Step 2 describes a condition that is satisfied by ALL local deterministic theories, but not ONLY local deterministic theories.

How come you do not conclude that all theories are ruled out?

Notice what I said. I said that ALL local deterministic theories and SOME nonlocal theories are ruled out by the argument. Bohmian mechanics is an example of a theory that is not ruled out.

See previous point. Why focus on local determinism when you know that non-local theories are also ruled out?

Because most nonlocal theories are not ruled out. The ones that ARE ruled out by the argument don't really have adherents, and they're not really interesting. If someone's interested in going the nonlocal route, why would they specifically choose a theory that is ruled out by Bell when they can choose many others that are still viable?

I'm asking you why you are unable to directly derive the inequality from (b).

Because counterfactual definiteness is an assumption of Bell's theorem. If you were a local determinist who refuses to accept counterfactual definiteness, i.e. the kind of person who thinks that reasoning in (b) is valid and reasoning in (c) is meaningless, this argument wouldn't be able to persuade you.

I accept that Bell's inequality is iron-clad valid. However Bell's theorem involves comparing an unperformable experiment with results from actual experiments and QM predictions for actual experiment. I assume you are familiar with the modal fallacy? There is something not too unlike it in play here. The elephant in the room is the fact that unperformed experiments have no results.

Certainly unperformed experiments don't have results, but by counterfactual definiteness there are definite results they WOULD have had if they HAD been performed, and by the no-conspiracy condition the probabilities P(A),P(B), and P(C) you WOULD have gotten if you had performed those experiments are the same as the probabilities that you DID in fact get. This latter point seems clearly to be the locus of our disagreement.

 

[lugita15]

The main claim that "Starting with two completely identical binary messages, if A's 30 degree turn introduces a 25% mismatch and B's 30 degree turn introduces a 25% mismatch, then the total mismatch (when both are turned) can be at most 50%." is not necessarily true. It is only true if you assume that the mismatch is linearly produced from the angle. This is false if you consider what we already know about polarization of light, from Malus law.

But the linearity (or rather sublinearity) is NOT an arbitrary assumption of an argument. It is the conclusion of an argument, an argument I have tried to lay out more carefully than Herbert did in this thread.

The second issue which I have discussed in this thread is that the inequality is derived for possibilities which can never be simultaneously realized (actualized). In principle it is impossible to test experimentally, so trying to take experimental results on the basis that probabilities are the same doesn't make sense. The probabilies may be the same but not simultaneously.

What do you mean "the probabilities may be the same but nor simultaneously"? For each set of particle pairs, we are considering three scenarios, one factual scenario and two counterfactual scenarios. So if we orient the polarization detectors at -30 and 0 for one set of particle pairs, then the factual scenario is actually orienting the detectors at -30 and 0, whereas the counterfactual scenarios are orienting the detectors at -30 and 30 and at 0 and 30. We only directly know what happens in the factual scenario, so we know P(A) for the factual scenario, but we don't directly know P(B) and P(C) respectively in the two counterfactual scenarios. Similarly for another set of particle pairs, we know what P(B) is in the factual scenario, but not P(A) and P(C) respectively in the two counterfactual scenarios. But the no-conspiracy condition says that since the particles are randomly put into the three sets, we can conclude that P(B) we found in the factual scenario in the B-set is the same as the P(B) in the counterfactual scenarios in the A-set and C-set.

 

[billschnieder]

I agree that in order to determine the probabilities, we need to compare the results of distant polarization detectors. But why is that a "loophole"? Local determinism leads to certain conclusions as to the relationship between the probabilities, even if the probabilities are "global" in the sense that locally available data (at the time of measurement) is not sufficient to compute them. How does that present a contradiction in any way?

Global probabilities depend on both sides. So if the joint probability is not separable, it does not mean non-locality. Knowing the results on one side changes the proability space you use when considering the results at the other side. Of cause moving one detector will affect the the "results" considered for the other. The whole concept of "match" is global. If the goal is to truly prove non-locality rather than just pulling wool over eyes, why use "match". why not use an entirely local statistic such as the one-sided ratio of detector 1 to detector 0. So indeed, using global probabilies and global phenomena such as "matches" IS a loophole.

 

[lugita15]

Global probabilities depend on both sides.

I certainly agree with that. The probabilities P(A), P(B), and P(C) depend on the results of both photons.

So if the joint probability is not separable, it does not mean non-locality.

What do you mean separable? Are you referring to the factorization of conditional probability that is done in Bell's original proof? Well, the proof by Herbert that I am outlining here doesn't invoke factorization arguments, so that is certainly not what I'm using to refute local determinism.

Knowing the results on one side changes the proability space you use when considering the results at the other side.

Certainly knowing the result for one photon changes your expectations about what the result for the other photon will be. But I'm not using conditional probabilities at all. I'm not asking "Given the result of photon 1, what is the conditional probability that the result of photon 2 will be the same?" I'm asking the question "What is the probability that photon 1 will give the same result as photon 2?"

And I already gave told you the definition of this probability, for both (b) and (c), back in post #224. Now you can definitely say that different probability spaces are being considered in (b) and (c). Indeed, there are three probability spaces in (b) for the three different factual scenarios. In each of these three probability spaces, only one of the three probabilities P(A), P(B), and P(C) is well-defined. And in (c), there is yet another probability space, a bigger one that includes all the particle pairs because in (c) the three statements A, B, and C are all simultaneously well-defined for all particle pairs, so that P(A), P(B), and P(C) are all well-defined in this probability space. To repeat, my contention is that no-conspiracy implies that P(A) in (c) is equal to the P(A) in the A-space of (b), etc.

Of cause moving one detector will affect the the "results" considered for the other.

I think we're using the word "result" in different ways. What I mean by result is whether the photon "goes through" (i.e. is measured as polarized parallel to the orientation of the polarizer) or "does not go through" (i.e. is measured as polarized perpendicular to the orientation of the polarizer). So in that sense, I hope you agree that in a local deterministic theory that conforms to the no-conspiracy condition, the result of the polarization measurement of photon 1 cannot possibly depend on the setting of the distant polarizer 2.

The whole concept of "match" is global.

I agree with this wholeheartedly.

If the goal is to truly prove non-locality rather than just pulling wool over eyes, why use "match".

I'm certainly not trying to pull the wool over your eyes. Let me describe exactly what local determinism has to do with the global probabilities.

We start out with prediction of quantum mechanics that there are identical results (i.e. zero probability of mismatch) at identical polarizer settings whenever we DO orient the polarizers at identical angle settings, i.e. in (b). Then we use the no-conspiracy condition to conclude that the probability of mismatch at identical angle settings is also zero in (c). But how is identical results at identical angle settings in (c) possible in a local deterministic universe? Assuming the no-conspiracy condition, it is only possible if for (c) it is determined in advance both photons would go through and what photons would not go through. And this implies that for each particle pair, the three statements A, B, and C have pre-existing truth values in (c), independent of what angles the polarizers actually happen to be oriented at for that particle pair. From this, twe show that P(C)≤P(A)+P(B) in (c), and then a final application of no-conspiracy allows us to conclude that P(C)≤P(A)+P(B) in (b).

Does that make it clear where I'm invoking local determinism?

So indeed, using global probabilies and global phenomena such as "matches" IS a loophole.

I still don't see how this is a loophole.

 

[San K]

Understanding/checking Bell's theorem via scenario analysis

Theoretical simulation (of a case(s)) can be another way to understand/validate Bell's theorem.

So let's take a scenario:

If the entangled photons were oriented at say 0 degrees(see note below) and one polarizer (A) was at 0 degrees and the other (B) at 15 degrees.

What is the logic, and calculations, we use to calculate the match prediction per LHV hypothesis? and per QM/actual?

Note: per QM/QE - it is impossible to tell what degree the photons are oriented at because they are in an indeterminate state and as soon as we measure it, the entanglement breaks. That said... let's assume hypothetically/theoretically.

 

[billschnieder]

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

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

And I already gave told you the definition of this probability, for both (b) and (c), back in post #224. Now you can definitely say that different probability spaces are being considered in (b) and (c). Indeed, there are three probability spaces in (b) for the three different factual scenarios. In each of these three probability spaces, only one of the three probabilities P(A), P(B), and P(C) is well-defined.

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.

And in (c), there is yet another probability space, a bigger one that includes all the particle pairs because in (c) the three statements A, B, and C are all simultaneously well-defined for all particle pairs, so that P(A), P(B), and P(C) are all well-defined in this probability space.

But P(ABC) is well defined in this scenario (c) contrary to the (b) scenario.

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). 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.

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. But that is not very credible. The inequality is a deeper relationship between probabilities not just specific values of the probabilities. This is evident in the fact that P(ABC|w) is well defined and P(ABC|xyz) is not.

There is nothing new here, this same issue was addressed by Boole 150 years ago. The article by Itamar Pitowsky I cited earlier goes into great details about this. Violation of inequality simply means the data are not from the same population. How does no-conspiracy enable you to recombine separate incompatible samples into a single one? It can't, they are not called incompatible for no reason.[/QUOTE]

 

[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.