# Question on Bell's theorem

1. Dec 29, 2013

### jk22

Hi all,

in Bell's theorem we consider the measurement of correlation $A\otimes B$ and $A\otimes B'$

Then the theorem states that the "result" for A in both quantities is the same. But does QM not only gives probabilities of the result, hence the result could be in principle different for the measurement with the A operator in those 2 correlation, so that no factorization is possible for the result of measurement ?

2. Dec 29, 2013

### Staff: Mentor

The distributions for A are the same, and you can select measurements with specific results at A afterwards if you like.

3. Dec 31, 2013

### jk22

I thought about selection and rearranging, but I could not achieve to get Bell's theorem : suppose the S value is

S=AB-A1B'+A'B1+A'1B'1

suppose we fix the A datas, since the datas have same length and have the same average, we can rearrange A1 so that A=A1

but if we rearrange A1 we rearranged B' because the datas are in couples (else we change the value of the correlation), now B'1 should be rearranged like B', so that we rearranged A'1

Since A'1=A' we shall rearrange B1, but then B1=B we should rearrange A which contradicts the hypothesis, so we never get AB-AB'+A'B+A'B' with the respective equivalence of the results.

4. Jan 2, 2014

### billschnieder

You can find similar observations in :
"What's wrong with this rebuttal,"Found. Phys. Lett. 19 (6) 625-629 (2006). http://arxiv.org/abs/quant-ph/0604124 (see section IV)

and
"Bell's inequalities I: An explanation for their experimental violation" Optics Communications 170 (1999) 55-60
http://arxiv.org/abs/quant-ph/0101087

5. Jan 2, 2014

### DrChinese

Both of the above citations by Bill are inappropriate and should be disregarded. They do not meet forum guidelines.

Bill, please stop now or I will report you for the Nth time. I won't argue with you further on this matter.

6. Jan 3, 2014

### billschnieder

This is false.

7. Jan 3, 2014

### Staff: Mentor

Closed for moderation