Exploring Bell's Disproof: Fundamental Theoretical Arguments

[PeterDonis]

if you have full knowledge of $λ$, then Y becomes Y*

No it doesn't. Full knowledge of $\lambda$ does not mean all probabilities are 1.

 

[Nicky665]

I am not the one asserting there are counterfactual cases. Everything I see points to an observer dependent universe, one that lacks local realism. So when someone asks about a particular local realistic theory, one which is excluded by Bell but that the author claims is not, I always ask: Please describe those counterfactual cases.

" Let us break down the situation fully:

Monday:
If Alice picks setting "a" on Tuesday she will obtain result "A"
If Alice picks setting "b" on Tuesday she will obtain result "B"
If Alice picks setting "c" on Tuesday she will obtain result "C"

Tuesday:
Alice picks setting "a" and obtains result "A"

Wednesday:
Alice picked setting "a" on Tuesday and obtained result "A"
If Alice had picked setting "b" on Tuesday she would have obtained result "B"
If Alice had picked setting "c" on Tuesday she would have obtained result "C"

The original question then becomes clear: Do you agree that if Alice had picked a different setting on Tuesday than the one she picked, she would have obtained a result ?

Note that I don't care what the value of the actual result is, just the fact that there is a result. Also note that A, B, C are not actual values, but just labels for whatever result it is that Alice obtains for the corresponding setting. "

 

[stevendaryl]

" The original question then becomes clear: Do you agree that if Alice had picked a different setting on Tuesday than the one she picked, she would have obtained a result ?

In discussions of Bell's inequality, "counterfactual" is mentioned in the context of "counterfactual definiteness". CFD says that there is a definite answer to questions such as "What result would Alice have gotten if she had measured X instead of Y?"

 

[Nicky665]

It's not Bell's problem. Bell proved a theorem along the lines of: "All theories of type X have property Y. QM does not have property Y. Therefore, QM is not a theory of type X."

What are you disagreeing about?

"Perhaps if you spell out exactly what X and Y mean mathematically, I we will show you what we are disagreeing about.

For example, please describe to me a physical situation for which the relationship P(AB|λ) = P(A|λ)P(B|λ) is true
But the relationship P(AB|λ) = P(A|λ)P(B|Aλ) is false."

In discussions of Bell's inequality, "counterfactual" is mentioned in the context of "counterfactual definiteness". CFD says that there is a definite answer to questions such as "What result would Alice have gotten if she had measured X instead of Y?"

"And you think, QM disagrees with that?

Bell in 1964 Paper wrote:Measurements can be made, say by Stern-Gerlach magnets, on selected components of the spins

and

. If measurement of the component

, where

is some unit vector, yields the value +1 then, according to quantum mechanics, measurement of

must yield the value -1 and vice versa.Bell says according to QM, Alice would have gotten a definite answer if she had measured at the same angle as Bob. Is that CFD or not?

BTW, you didn't answer the first question. Would Alice have definitely obtained a result (any result at all) had she measured at a different setting?"

 

[DrChinese]

Would Alice have definitely obtained a result (any result at all) had she measured at a different setting?"

Obviously Alice literally obtains a result when she measures at any setting. But the relevant point is whether there is any dependency on what Bob does. It always comes back to the same place: The answer is interpretation dependent.