Three outcome systems in Bell/CHSH

[gespex]

Hello everybody,

In Bell's theorem and CHSH, A(a, \lambda) is defined as the outcome for detector A with setting a and hidden variable \lambda. The outcome of this can be either -1, 0 or 1, so three outcomes.

It is clear what outcomes -1 and 1 refer to. But what about 0? Two relevant examples:
- In spin measurement, using a stern-gerlach device, -1 and 1 would be opposing spin directions (what about 0? The particle not being measured at all?)
- In polarization measurement, -1 would be "blocked" by the filter, and 1 would be "passed" through the filter (and I've got no idea at all what 0 could possible mean?)

So what is this "0" outcome?


Thanks in advance

 

[billschnieder]

Hello everybody,

In Bell's theorem and CHSH, A(a, \lambda) is defined as the outcome for detector A with setting a and hidden variable \lambda. The outcome of this can be either -1, 0 or 1, so three outcomes.

This is not correct. According to Bell, A(a,λ)= ±1, B(a,λ)= ±1. Zero is not an allowed outcome.

 

[gespex]

Okay, fair enough. But here:
http://en.wikipedia.org/wiki/Bell's_theorem#Original_Bell.27s_inequality

It states "This inequality is not used in practice. For one thing, it is true only for genuinely "two-outcome" systems, not for the "three-outcome" ones (with possible outcomes of zero as well as +1 and −1) encountered in real experiments."
(So yes, it actually mentions the *lack* of a third outcome)

But what do they refer to as an outcome of zero then, in the two given experiments?Thanks for your answer!

 

[billschnieder]

Okay, fair enough. But here:
http://en.wikipedia.org/wiki/Bell's_theorem#Original_Bell.27s_inequality

It states "This inequality is not used in practice. For one thing, it is true only for genuinely "two-outcome" systems, not for the "three-outcome" ones (with possible outcomes of zero as well as +1 and −1) encountered in real experiments."
(So yes, it actually mentions the *lack* of a third outcome)

But what do they refer to as an outcome of zero then, in the two given experiments?


Thanks for your answer!

Sorry, I miseed the CHSH part. For CHSH zero is "not-detected".

For a, given analyzer setting a and emission λ, there are three possible results at apparatus 1: a count in the + detector, a count in the —detector, or no count in either detector.

 

[billschnieder]

Bill, Correct me if I'm wrong, please:

I understand that CHSH (strictly) refers to the joint paper by CHSH (1969): Phys. Rev. Lett. 23, 15, 880-884 (1969).

That is correct. Though it also generally refers to the inequalities of the same form. There is no direct mention of three outcomes in the 1969 paper although it is implied since they consider a case (∞) in which the polarizer is taken out of the beam on one side (not unlike an undetected particle). In any case, you do not need a third outcome to derive the inequality because they still eliminate it from the equations by assuming that

P(A^+B^∞) = P(A^+B^+) + P(A^+B^-).

So just to clarify the answer to the OP, there are two subtly different views about what the "0" might mean:

1. The functions A(a,λ) = (+1, -1), B(b,λ) = (+1, -1), ±1 are the only alowed outcomes for the functions, but to facilitate comparison with experiments, we may use "0" to represent the cases in which, due to experimental imperfections, we do not know if the true value is +1 or -1.

2. A(a,λ) = (+1, 0, -1), B(b,λ) = (+1, 0, -1). Where "0" is a genuine value. In other words, non-detection is a valid outcome just as much as +1 or -1.

The above equation from the CHSH 1969 paper imples view (1) since it assumes that the undetected photon would have resulted in only one of (+1 or -1). It does not consider "0" as valid outcome but rather as lack of knowledge about what the valid outcome would have been had the photon been detected. Therefore their functions A(a,λ), and B(b,λ) still agree with Bell's restriction of only ±1 as outcomes and the "0" does not affect the derivation of the inequality.

 

[gespex]

Thanks a lot for your help guys!