Does realism imply locality or vice versa?

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SUMMARY

This discussion centers on the relationship between realism and locality in quantum mechanics, particularly through the lens of Bell's inequality. It establishes that while local realism assumes realism, the violation of realism does not necessarily imply the violation of locality, as non-local hidden variable theories can exist. The conversation highlights that if realism is absent, it does not automatically negate locality, and various interpretations of quantum mechanics can preserve locality even without predetermined outcomes. Key references include Bell's inequality and the implications of entanglement tests.

PREREQUISITES
  • Understanding of Bell's inequality and its implications in quantum mechanics.
  • Familiarity with the concepts of local realism and signal locality.
  • Knowledge of quantum entanglement and measurement processes.
  • Basic grasp of quantum mechanics interpretations and their philosophical implications.
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  • Explore the implications of Bell's inequality in quantum mechanics.
  • Research non-local hidden variable theories and their compatibility with quantum mechanics.
  • Study the concept of entanglement swapping and its relation to time symmetry.
  • Read Bell's original papers for foundational insights into local realism and quantum correlations.
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Physicists, quantum mechanics researchers, philosophy of science scholars, and anyone interested in the foundational questions of quantum theory and its interpretations.

  • #121
stevendaryl said:
where
C_\lambda(\alpha, \alpha', \beta, \beta') \equiv X_A(\alpha, \lambda) X_B(\beta, \lambda) + X_A(\alpha', \lambda) X_B(\beta, \lambda) + X_A(\alpha, \lambda) X_B(\beta', \lambda) - X_A(\alpha', \lambda) X_B(\beta', \lambda)

It might not be obvious, but since each of the Xs are between +1 and -1, it follows that C_\lambda(\alpha, \alpha', \beta, \beta') must lie in the range [-2, +2].

Letting
  • A \equiv X_A(\alpha, \lambda)
  • A' \equiv X_A(\alpha', \lambda)
  • B \equiv X_B(\beta, \lambda)
  • B' \equiv X_B(\beta', \lambda)
  • F(A,A', B, B') \equiv AB + A'B + AB' -A'B'
We want to show that -2 \leq F(A,A',B,B') \leq +2, under the assumption that all 4 variables lie in the range [-1, +1].

Let A_{max}, B_{max}, A'_{max}, B'_{max} be a choice of the variables that maximizes F. Then:
  • Either A_{max} = \pm 1 or \frac{\partial F}{\partial A}|_{A = A_{max}, B=B_{max}, A'=A'_{max}, B' = B'_{max}} = 0
  • Either A'_{max} = \pm 1 or \frac{\partial F}{\partial A'}|_{A = A_{max}, B=B_{max}, A'=A'_{max}, B' = B'_{max}} = 0
Suppose \frac{\partial F}{\partial A} = 0. That implies that B_{max} + B'_{max} = 0. So in this case: F(A_{max}, A'_{max}, B_{max}, B'_{max}) = A'_{max} (B_{max} - B'_{max}) \leq 2

Suppose \frac{\partial F}{\partial A'} = 0. That implies that B_{max} - B'_{max} = 0. So in this case: F(A_{max}, A'_{max}, B_{max}, B'_{max}) = A_{max} (B_{max} + B'_{max}) \leq 2

So we conclude that if F > 2, it must be when A_{max} = \pm 1 and A'_{max} = \pm 1. We have two cases: they have the same sign, or they have opposite signs.

Suppose A_{max} = \pm 1 and A'_{max} = +A_{max}. Then F(A_{max}, A'_{max}, B_{max}, B'_{max}) = \pm (B_{max} + B'_{max}) \pm (B_{max} - B'_{max}) = \pm 2 B_{max} \leq 2

Suppose A_{max} = \pm 1 and A'_{max} = -A_{max}. Then F(A_{max}, A'_{max}, B_{max}, B'_{max}) = \pm (B_{max} + B'_{max}) \mp (B_{max} - B'_{max}) = \pm 2 B'_{max} \leq 2

So in all cases, F(A,A',B, B') \leq 2. We can similarly prove -2 \leq F(A,A',B,B'). So we conclude:

-2 \leq F(A,A',B,B') \leq +2
 

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