- #201

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I'm not. Rather my point is precisely that we have to be careful how we apply models to experiment. You can use a model that contradicts the experiment you are modelling.I think you are misunderstanding the distinction between what is observed, and a model intended to EXPLAIN those observations.

This is all true and irrelevant to the point I'm making.Yes, what is observed is that Alice measures the spin on one particle, and Bob measures the spin on another particle. The class of models that Bell is interested in (the class that Einstein, Podolsky and Rose were interested in) are models of the following form:

So in general, Alice's result ##R_A## can be written as a function ##R_A(\lambda, \alpha, \lambda_A)##, where ##\alpha## is the choice of which measurement to perform, and ##\lambda_A## is other facts about Alice's device. Bob's result ##R_B## is similarly a function ##R_B(\lambda, \beta, \lambda_B)## where ##\beta## is his choice of measurement, and ##\lambda_B## is other facts about Bob's device.

- When a twin-pair is produced, there is an associated state variable ##\lambda## describing the pair.
- When Alice measures her particle, the result depends on (1) facts about her measuring device, and (2) the value of ##\lambda##
- Similarly, when Bob measures his particle, the result depends on facts about his measuring device, and the value of ##\lambda##.

So now, we have to take into account a stark fact about these twin-particles: There is PERFECT anti-correlation (or correlation, depending on the exact type of EPR experiment performed). That means that for each ##\lambda##, if it happens to be that ##\alpha = \beta## (that is, if Alice and Bob perform the same measurement), they always get opposite result: No matter what ##\lambda_A## and ##\lambda_B## are, we always have:

##R_A(\lambda, \alpha, \lambda_A) = - R_B(\lambda, \alpha, \lambda_B)##

I'm sure you agree that it is wrong to assume perfect anti-correlation between one particle of a pair and another particle of a different pair even if it is prepared similarly. In the same way as it is wrong to assume perfect anti-correlation between heads of one toss and tails of a different toss, even of the exact same coin.

Again this is all trivially true and irrelevant so long as ##A_n, B_n, C_n## arise from the same context ##n##. The three averages ##\langle A B \rangle, \langle A C \rangle, \langle B C \rangle## are not simply independent averages without any relationship with each other. Based on the way you derived the expression, the inequality relationship embodies all the assumptions you used in it's derivation, including the fact that they are all based on the same context ##n##. Note that by dropping the ##n## subscripts, you are being a little careless and perhaps that is why you are not getting the point. This will be crucial when you apply this relationship to experimental data.This implies that ##R_A## and ##R_B## don't actually depend on ##\lambda_A## and ##\lambda_B## at all. If Alice's result is NOT determined by ##\lambda## and ##\alpha##, then sometimes she would get a result that would not be anti-correlated with what Bob gets.

So we pick three possible measurements for Alice, ##\alpha_1, \alpha_2, \alpha_3##. For each ##\lambda##, let ##A(\lambda)## be ##R_A(\lambda, \alpha_1)##, let ##B(\lambda)## be ##R_A(\lambda, \alpha_2)## and let ##C(\lambda)## be ##R_A(\lambda, \alpha_3)##.

Now, if we create a sequence of twin-pairs, each twin pair is associated with some value of ##\lambda##. So we let

##A_n = A(\lambda_n)##

##B_n = B(\lambda_n)##

##C_n = C(\lambda_n)##

where ##\lambda_n## is the value of ##\lambda## for the ##n^{th}## twin pair. These are not measurement results, they are just numbers, unknown functions of ##\lambda## evaluated at ##\lambda = \lambda_n##. But we're ASSUMING that the significance of these numbers is that

##A_n## is the result that Alice WOULD get, if she chose to measure her ##n^{th}## particle using device setting ##\alpha_1##.

##B_n## is the result that Alice WOULD get, if she chose to measure her ##n^{th}## particle using device setting ##\alpha_2##.

##C_n## is the result that Alice WOULD get, if she chose to measure her ##n^{th}## particle using device setting ##\alpha_3##.

Under the assumption of perfect anti-correlation, Bob would get ##-A_n, -B_n, -C_n## for the corresponding measurements.

So now, the numbers ##A_n, B_n, C_n## are just three numbers, each are assumed to be ##\pm 1##. So we can do manipulations as real numbers to come to the conclusion that:

##|\langle A B \rangle + \langle A C \rangle | \leq 1 + \langle B C \rangle##

where ##\langle A B \rangle = \frac{1}{N} \sum_n A_n B_n## and ##\langle A C \rangle = \frac{1}{N} \sum_n A_n C_n## and ##\langle B C \rangle = \frac{1}{N} \sum_n B_n C_n##, and where ##N## is the number of twin pairs produced.

This is simply a mathematical fact about ANY sequence of triples of numbers ##A_n, B_n, C_n## where each number is ##\pm 1##.

This is the key. You have not clearly stated the assumption. It is not simply that ##A_n B_n## over some of the ##n## is approximately the same as the average over all ##n##.Now, the question is whether it is possible to measurement the quantities ##\langle A B \rangle##, ##\langle A C \rangle##, ##\langle B C \rangle##. We can't, actually, because the definition of (for example) ##\langle A B \rangle## is that it is the average of ##A_n B_n## over all values of ##n##. But we don't measure ##A_n## and ##B_n## over all possible values of ##n##. We only measure it on for some of the ##n##. So to compare theory with experiment, we have to assume that the average of ##A_n B_n## over some of the ##n## is approximately the same as the average over all ##n##.

The assumption is in fact that the relationship between three averages ##\langle A_n B_n \rangle, \langle A_n C_n \rangle, \langle B_n C_n \rangle## from the same context ##n## is the same relationship as that between ##\langle A_i B_i \rangle## from one context ##i## and ##\langle A_j C_j \rangle## from a different context ##j## and ##\langle B_k C_k \rangle## from a yet another context ##k## with ##i, j, k## disjoint.

At the very least, you have to agree that this assumption is implied. Do you disagree? Bell's realism assumption definitely includes this "sub-assumption" if you will, as soon as the inequality relationship is applied to any experiment in which simultaneous measurement of ##\langle A_n B_n \rangle, \langle A_n C_n \rangle, \langle B_n C_n \rangle## was not performed (eg EPRB).

If this assumption is true. It should be possible to start from the variables ##A_i, B_i, A_j, C_j, B_k, C_k## and derive the same relationship as what you derived for ##A_n, B_n, C_n## and ask the question, what additional assumptions will be required in that case. It turns out it will be required to assume that ##A_i = A_j, B_i = B_k, C_j = C_k## and ##B_i B_k = 1## which is the same as assuming perfect correlation between heads of one coin toss and tails of another coin toss.