# Proof that no non-local variable can exist ?

#### 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 ?

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#### Demystifier

2018 Award
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.

#### 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.

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#### Nugatory

Mentor
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 ?
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.

#### 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.

#### Demystifier

2018 Award
Can we say that the addition is not part of the measurement process, since it's made by a computer
Yes we can.

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