Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

An inverse Bell theorem

  1. Apr 23, 2015 #1
    if we consider a separable measurement operator $$(A+A')\otimes (B-B')$$ then quantum mechanics predict the result is in [-2;2]

    Whereas going to classical results would give in [-4;4]

    This could indicate that going from measurement operator in the quantum realm to measurement results is maybe more complicated than just retranscripting the terms ?
     
  2. jcsd
  3. Apr 24, 2015 #2

    zonde

    User Avatar
    Gold Member

    jk22, your post is not very coherent
    Measurement operator represents measurements and in Bell test two measurement angles are always considered independent.
    But measured pair of particles can be in separable or entangled state.

    Classical results can't give such values for CHSH inequality. Correlation values in inequality are not just arbitrary numbers but maximum numbers you can get when using shared initial variable for particle pair.
     
  4. May 6, 2015 #3
    Well this operator is not Chsh but it shows a discrepancy between eigenvalues and the measurement results since the value of each operator A,A',B,B' is in {-1;1} whereas if i consider the operators like in a bell test they cannot reach such eigenvalues but not in general this is true my post is not very precise.
     
    Last edited: May 6, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: An inverse Bell theorem
  1. Bell's theorem (Replies: 264)

Loading...