### Why the upper bound on CHSH is 2\/2 and not 4 ?

Posted:

**Sat Sep 05, 2015 8:06 am*****

After Fred's recent post about my derivation of the upper bound on CHSH, we have been trying to better understand why the upper bound on CHSH is 2\/2 and not 4 ?

In fact I have already answered this question much more rigorously in this paper, but perhaps a simplified explanation is called for, especially because the underlying geometrical reason for the bound 2\/2 instead of 4 is quite straightforward. In particular, it has nothing to do with any mystical notion such as "quantum non-locality."

As in Fred's post linked above, we begin with the square of the CHSH-type string of the (geometric) products of the standardized variables (or standard scores):

,

where , , , and are unit bivectors about the directions , , , and , with being the volume form.

Now in Fred's post as well as in my original derivation, the upper bound of 2\/2 on CHSH is derived by using the assumption , justified on the physical grounds that and are space-like separated, and there does not exist a "third" particle in the EPR-B type experiment that can be detected along the exclusive direction in addition to the detections of the pair of particles along the directions and . This permits some rearrangement in the 16 terms, leading to the maximum value of Consequently, the upper bound on the CHSH in this case turns out to be for the four Bell test directions (or angles).

So far so good. But can we do better? Is it possible to increase the upper bound on CHSH to its apparent full potential of 4 ? And if not, then why not? Evidently, this seems to be possible if we can arrange a maximum value of to be 16. Then we would have as desired. Now from the above expansion of it is clear that we can have = 16 if each of its 16 terms equals to +1. Is that possible? A little reflection will convince you that it is possible provided we can have all four of the directions , , , and orthogonal to each other. But one cannot have four directions orthogonal to each other in a three dimensional space such as R^3.

So there. One cannot have all four directions , , , and orthogonal to each other in R^3. And therefore we cannot extend the upper bound on CHSH to 4.

The maximum value of the upper bound on CHSH within R^3 is 2\/2. Therefore the observed upper bound on CHSH is 2\/2. It has nothing to do with any mystical idieas like quantum entanglement or non-locality. It is simply a numerical constraint arising form the geometrical and topological properties of the physical space we live in.

In fact, |CHSH| 4 is physically not possible even within S^3, which is locally (or tangentially) just R^3. The only way to get the bound |CHSH| 4 is to have most of the observation directions orthogonal to each other, i.e., have a topology stronger than S^3, as shown in the last plot of this simulation: http://rpubs.com/jjc/84238.

Joy Christian

After Fred's recent post about my derivation of the upper bound on CHSH, we have been trying to better understand why the upper bound on CHSH is 2\/2 and not 4 ?

In fact I have already answered this question much more rigorously in this paper, but perhaps a simplified explanation is called for, especially because the underlying geometrical reason for the bound 2\/2 instead of 4 is quite straightforward. In particular, it has nothing to do with any mystical notion such as "quantum non-locality."

As in Fred's post linked above, we begin with the square of the CHSH-type string of the (geometric) products of the standardized variables (or standard scores):

,

where , , , and are unit bivectors about the directions , , , and , with being the volume form.

Now in Fred's post as well as in my original derivation, the upper bound of 2\/2 on CHSH is derived by using the assumption , justified on the physical grounds that and are space-like separated, and there does not exist a "third" particle in the EPR-B type experiment that can be detected along the exclusive direction in addition to the detections of the pair of particles along the directions and . This permits some rearrangement in the 16 terms, leading to the maximum value of Consequently, the upper bound on the CHSH in this case turns out to be for the four Bell test directions (or angles).

So far so good. But can we do better? Is it possible to increase the upper bound on CHSH to its apparent full potential of 4 ? And if not, then why not? Evidently, this seems to be possible if we can arrange a maximum value of to be 16. Then we would have as desired. Now from the above expansion of it is clear that we can have = 16 if each of its 16 terms equals to +1. Is that possible? A little reflection will convince you that it is possible provided we can have all four of the directions , , , and orthogonal to each other. But one cannot have four directions orthogonal to each other in a three dimensional space such as R^3.

So there. One cannot have all four directions , , , and orthogonal to each other in R^3. And therefore we cannot extend the upper bound on CHSH to 4.

The maximum value of the upper bound on CHSH within R^3 is 2\/2. Therefore the observed upper bound on CHSH is 2\/2. It has nothing to do with any mystical idieas like quantum entanglement or non-locality. It is simply a numerical constraint arising form the geometrical and topological properties of the physical space we live in.

In fact, |CHSH| 4 is physically not possible even within S^3, which is locally (or tangentially) just R^3. The only way to get the bound |CHSH| 4 is to have most of the observation directions orthogonal to each other, i.e., have a topology stronger than S^3, as shown in the last plot of this simulation: http://rpubs.com/jjc/84238.

Joy Christian