Thank you both, DrChinese and Gordon Watson, for your feedback.
Let me restate where my doubts are specifically located, so that my point will become more clear and you will be able to give me more "targeted" help. First, let me repeat the necessary formulas:
JK423 said:
S_j = A_j\left( {{a_1}} \right)B_j\left( {{b_1}} \right) + A_j\left( {{a_1}} \right)B_j\left( {{b_2}} \right) + A_j\left( {{a_2}} \right)B_j\left( {{b_1}} \right) - A_j\left( {{a_2}} \right)B_j\left( {{b_2}} \right),
where A_j\left( {{a_i}} \right) = \pm 1 and B_j\left( {{b_i}} \right) = \pm 1, and j denoting a particular photon pair,
is always {S_j} = \pm 2, for any measurement result A and B.
When we take the mean value over all photon pairs, \,\left\langle S \right\rangle = \frac{1}{N}\sum\limits_{i = 1}^N {{S_j}} we find it to be bounded, i.e.
- 2 \le \,\left\langle S \right\rangle \le 2.
This quantity is bounded whatever the values of A and B for the photon pairs.
Now, the
reason why the CHSH inequality holds is because S_j is always S_j=±2 for every photon pair separately. And this is due to the locality assumption, i.e. that the outcome in Alice's side does not depend on what Bob measures, etc. Mathematically this assumption is expressed by the fact that {A_j}\left( {{a_1}} \right) is the same in both {A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _1}} \right) and {A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _2}} \right), i.e. whether Bob measures \beta_1 or \beta_2 is irrelevant, the outcome {A_j}\left( {{a_1}} \right) will be the same. The same reasoning applies to {A_j}\left( {{a_2}} \right).
The
important thing here is that this is always true because {A_j}\left( {{a_1}} \right) corresponds to the
same photon in these two expressions {A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _1}} \right) and {A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _2}} \right), so it cannot be different if locality is assumed.
Now, look what happens if you consider the measureable edition of \left\langle S \right\rangle, where each of the quantities \left\langle {A\left( {{a_i}} \right)B\left( {{b_j}} \right)} \right\rangle are mean values over the measurements. For simplicity let me consider only
one of the measured values (instead of the whole mean value) in order to make my point clear. So assume just one run:
\left\langle {A\left( {{a_1}} \right)B\left( {{b_1}} \right)} \right\rangle = {A_1}\left( {{a_1}} \right){B_1}\left( {{b_1}} \right) , corresponding to the measured value of the photon pair "1",
\left\langle {A\left( {{a_1}} \right)B\left( {{b_2}} \right)} \right\rangle = {A_2}\left( {{a_1}} \right){B_2}\left( {{b_2}} \right), corresponding to the measured value of the photon pair "2".
Now take the sum of these in order to form the first half part of the quantity S:
\left\langle {A\left( {{a_1}} \right)B\left( {{b_1}} \right)} \right\rangle + \left\langle {A\left( {{a_1}} \right)B\left( {{b_2}} \right)} \right\rangle = {A_1}\left( {{a_1}} \right){B_1}\left( {{b_1}} \right) + {A_2}\left( {{a_1}} \right){B_2}\left( {{b_2}} \right).
(1)
I told you previously that CHSH holds because {A_j}\left( {{a_1}} \right) has the same value in these two quantities, since it corresponds to the same photon.
But now that we have considered the mean values over measurements, {A_1}\left( {{a_1}} \right) and {A_2}\left( {{a_1}} \right) are, generally, different since they correspond to
different photons "1" and "2", and that way they could be mimicking non-locality, since it looks as if {A}\left( {{a_1}} \right) depends on what Bob measures.
You can generalize (1) for N photon pairs and take a more appropriate mean value. The moral in this story is that the photons are different in each quantity, so there is no obvious reason why CHSH would not be violated.
I hope that i made my point clear..
I'm looking forward to your feedback!
Giannis