# Proof that no non-local variable can exist ?

1. Jan 3, 2014

### jk22

Suppose we consider the measurement of $$A\otimes B-A\otimes B'+A'\otimes B+A'\otimes B'$$at angles 0, 45, 90, 135 degrees.

If there exist a non-local variable that determine the result of the pair of result, then one gets for result of measurement $$0, 4, -4$$

Whereas in quantum mechanics, the total Bell operator is $$\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 2 &0&0&2\\0&-2&2&0\\0&2&-2&0\\2&0&0&2\end{array}\right)$$ which has as possible measurement outcomes : $$0, 2\sqrt{2},-2\sqrt{2}$$ which are not the same as the one given by a nonlocal variable since the eigenvalue of a sum of operator is not equal to the sum of the eigenvalues.

Since the measurement outcomes are not the same, does this indicates that there can be no non-local variable can exist ?

2. Jan 3, 2014

### Demystifier

I think you are doing the same mistake von Neumann have done a long time ago. Measuring A+B is not the same as measuring A, measuring B, and then adding results.

3. Jan 3, 2014

### jk22

Can we say that the addition is not part of the measurement process, since it's made by a computer, so that we really measure and add +1 and -1 in an experiment ? I made another mistake, it's that the eigevalues of the sum of operators is the sum of the eigenvalues if both operator are diagonal.

Last edited: Jan 3, 2014
4. Jan 4, 2014

### Staff: Mentor

If anything (computer, human, animal, vegetable, or mineral) interacts with the system in a way that extracts a value of A, that's a measurement of A and it will disturb the system in such a way that (the distribution of subsequent) measurements of B will be affected.

5. Jan 4, 2014

### jk22

Ok.
But does the way we prepare an experiment change : if we measure AB we set the experimental apparatus to give 1 or -1 as result, whereas if we measure AB-AB'+A'B+A'B' at the given angle we should set the experimental apparatus to give -2sqrt(2) as result ? Or does the experimental apparatus in some way know we measure the sum ? In the case of the Bell operator the singlet state is an eigenstate and is not degenerate, so it's not an average, we know the result with certainty.

6. Jan 7, 2014

Yes we can.