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Correlation coefficient of two-particle half-spin entangled state

  1. Apr 5, 2010 #1
    Hi, I can't get my head around this question.

    1. The problem statement, all variables and given/known data
    The Bell state:
    [tex]|\psi\rangle = \frac{1}{\sqrt{2}}\left(|\uparrow_1\rangle|\downarrow_2\rangle +|\downarrow_1\rangle |\uparrow_2\rangle \right)[/tex]

    Find the correlation coefficient of the measurement of the spins in the directions [tex]z,\phi[/tex].

    2. Relevant equations
    [tex] C(\phi) = \langle S_{z1} S_{\phi 2} \rangle [/tex]

    S takes the values [tex]\pm 1[/tex]

    3. The attempt at a solution
    I know that I need to calculate the probability of the states being parallel and subtract the probability that the states are anti-parallel. But how do you calculate those?

    [tex]Pr(\uparrow_{z1}\uparrow_{\phi 2}) = \left|\langle\uparrow_{z1}\uparrow_{\phi 2} \left |\frac{1}{\sqrt{2}}\left(|\uparrow_1\rangle|\downarrow_2\rangle +|\downarrow_1\rangle |\uparrow_2\rangle \right)\right| \uparrow_{z1}\uparrow_{\phi 2} \rangle \right|^2[/tex]

    Where do I go from here?

    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2
    I think I might be on to something with this, but am not quite there yet:

    [tex] \hat{S}_z =\frac{\hbar}{2} \begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}[/tex]

    Rotate by angles [tex]\theta, \phi[/tex] to get

    [tex] \hat{S}_\phi = \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta e^{-i\phi} \\
    \sin \theta e^{i\phi} & -\cos \theta \end{pmatrix}[/tex]

    Can I use this somehow to get the correlation coefficient?

    Unfortunately the notation is getting a bit mixed up because in this post [tex]\phi = 0[/tex] and in the post above [tex]\phi[/tex] is used in place of [tex]\theta[/tex].
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