- #1
jarekduda
- 82
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In page 9 of http://www.theory.caltech.edu/people/preskill/ph229/notes/chap4.pdf we can find simple form of Bell inequalities for three binary variables:
$$ P(A=B) + P(A=C) + P(B=C) \geq 1 $$
which is kind of obvious: "This is satisfied by any probability distribution for the three coins because no matter what the values of the coins, there will always be two that are the same."
This lecture also contains example of state for which QM gives P(A=B) + P(A=C) + P(B=C) =3/4 violation.
For Bell violation by MERW, I wanted to construct an example of its violation using only real non-negative amplitudes, getting this candidate (|ABC>):
$$\psi=(|001\rangle + |010\rangle +|100\rangle +|011\rangle +|101\rangle +|110\rangle)/\sqrt{6}$$
The question is: what is the probability distribution when measuring two out of three such variables?
The interesting basis for measuring first two variables (AB, taking trace over C), is:
$$n_{00}=(|000\rangle+|001\rangle) /\sqrt{2}\qquad
n_{01}=(|010\rangle+|011\rangle) / \sqrt{2}$$
$$n_{10}=(|100\rangle+|101\rangle ) / \sqrt{2}\qquad
n_{11}=(|110\rangle+|111\rangle) / \sqrt{2}$$
getting
$$|\langle n_{00}|\psi\rangle|^2=1/12\quad
|\langle n_{01}|\psi\rangle|^2=4/12 \quad
|\langle n_{10}|\psi\rangle|^2=4/12 \quad
|\langle n_{11}|\psi\rangle|^2=1/12 $$
However, as this basis is not complete, instead of 1, these four possibilities sum to 10/12.
Can we normalize by this 10/12?
Getting probability of 00 as 1/10, and finally violation to P(A=B) + P(A=C) + P(B=C) = 6/10?
$$ P(A=B) + P(A=C) + P(B=C) \geq 1 $$
which is kind of obvious: "This is satisfied by any probability distribution for the three coins because no matter what the values of the coins, there will always be two that are the same."
This lecture also contains example of state for which QM gives P(A=B) + P(A=C) + P(B=C) =3/4 violation.
For Bell violation by MERW, I wanted to construct an example of its violation using only real non-negative amplitudes, getting this candidate (|ABC>):
$$\psi=(|001\rangle + |010\rangle +|100\rangle +|011\rangle +|101\rangle +|110\rangle)/\sqrt{6}$$
The question is: what is the probability distribution when measuring two out of three such variables?
The interesting basis for measuring first two variables (AB, taking trace over C), is:
$$n_{00}=(|000\rangle+|001\rangle) /\sqrt{2}\qquad
n_{01}=(|010\rangle+|011\rangle) / \sqrt{2}$$
$$n_{10}=(|100\rangle+|101\rangle ) / \sqrt{2}\qquad
n_{11}=(|110\rangle+|111\rangle) / \sqrt{2}$$
getting
$$|\langle n_{00}|\psi\rangle|^2=1/12\quad
|\langle n_{01}|\psi\rangle|^2=4/12 \quad
|\langle n_{10}|\psi\rangle|^2=4/12 \quad
|\langle n_{11}|\psi\rangle|^2=1/12 $$
However, as this basis is not complete, instead of 1, these four possibilities sum to 10/12.
Can we normalize by this 10/12?
Getting probability of 00 as 1/10, and finally violation to P(A=B) + P(A=C) + P(B=C) = 6/10?
