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Bell's inequality

  1. Jul 8, 2008 #1
    I'm trying to understand a part of the text where they prove QM doesn't satisfy Bell's inequality. I get how he derives the inequality. Apparently it's same as Sakuri (1985) and Townsend (2000). Problem is I lose him as soon as he starts the main part. Quoting almost directly,

    Consider a qubit oriented in an arbitrary direction. Consider a unit vector [tex]\vec{n} = \sin \theta \cos\phi \hat{x} + \sin \theta \sin\phi \hat{y} + \cos \theta \hat{z}[/tex]. The eigenvectors of [tex]\sigma \cdot \vec{n} [/tex] are..

    [tex]|+_n \rangle = \cos \frac{\theta }{2} |0 \rangle + e^{i\phi }\cos \frac{\theta }{2} |1 \rangle[/tex]

    [tex]|-_n \rangle = \cos \frac{\theta }{2} |0 \rangle - e^{i\phi }\cos \frac{\theta }{2} |1 \rangle[/tex]

    He then goes on to interpret [tex]|\langle 0|+_n\rangle |^2[/tex] as a probability & i can follow from there. I just don't get the bit I posted. The hint is to consider the x & y axes, eg to get the eigenvectors of [tex]\sigma \cdot \hat{x} [/tex] set [tex]\theta =\pi /2[/tex] & [tex]\phi =0[/tex]

    Last edited: Jul 8, 2008
  2. jcsd
  3. Jul 11, 2008 #2
    The key is to understand what [tex]\sigma \cdot \vec{n}[/tex] means. This is a dot product of matrices (Pauli matrices) times scalars (components of [tex]\vec{n}[/tex]. Let's just look at the first component. You will take the x Pauli matrix times by the x component of [tex]\vec{n}[/tex] to get [tex]\sigma_{x} n_{x}=\left(\begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right) \sin \theta \cos\phi=\left(\begin{array}{cc}0 & \sin\theta\cos\phi \\ \sin\theta\cos\phi & 0\end{array} \right)[/tex]. You do the same for the y and z components and add them all up to get your matrix, [tex]\sigma \cdot \vec{n}[/tex]. Find the eigenvectors of it and you should get what you gave from the book.
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