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Von Neumann entropy in terms of the tangle

  1. May 4, 2010 #1
    The Von Neumann entropy is [tex]\mathcal{S}(|\psi\rangle) = -Tr[\rho_a ln \rho a] [/tex]. The linear entropy [tex]S_L = \frac{l}{l-1}(1 - Tr[\rho_a^2])[/tex] For l =2 the linear entropy is written [tex]4Det(\rho_A)[/tex] which is also called the tangle [tex]\tau[/tex]. I understand this just fine, I can show that. Now it says the Von Neumann can be written:

    [tex]\mathcal{S}(|\psi\rangle) = -xln_{2}x - (1-x)ln_{2}(1-x) [/tex] where [tex] x = (1+\sqrt{1-\tau})/2[/tex]

    I don't know how to show this last step? Anyone offer any insight? This is for a 2-dimensional case if that isn't clear from the above.
    Last edited: May 4, 2010
  2. jcsd
  3. May 4, 2010 #2
    Nobody know?
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