Von Neumann Entropy of GHZ state

  1. I just wanted to run this working by some of you.

    Simplest Greenberger-Horne-Zeilinger state (entagled) state is:

    [tex]\mid GHZ \rangle = \frac{1}{\sqrt{2}}\left(\mid 0 \rangle_{A}\mid 0 \rangle_{B}\mid 0 \rangle_{C}+\mid 1 \rangle_{A}\mid 1 \rangle_{B}\mid 1 \rangle_{C}\right)[/tex]

    density matrix is:
    [tex] \rho = \frac{1}{2} \left( \mid 0 \rangle \langle 0 \mid_{A}\mid 0 \rangle \langle 0 \mid_{B}\mid 0 \rangle \langle 0 \mid_{C} + \mid 1 \rangle \langle 1 \mid_{A}\mid 1 \rangle \langle 1 \mid_{B}\mid 1 \rangle \langle 1 \mid_{C} \right) [/tex]

    reduced density matrix of qubit A:

    [tex] \rho_{A} = Tr_{B}\left(Tr_{C}\rho\right) = \frac{1}{2} \left( \mid 0 \rangle \langle 0 \mid_{A}Tr\left(\mid 0 \rangle \langle 0 \mid_{B}\right)Tr\left(\mid 0 \rangle \langle 0 \mid_{C}\right) + \mid 1 \rangle \langle 1 \mid_{A}Tr\left(\mid 1 \rangle \langle 1 \mid_{B}\right)Tr\left(\mid 1 \rangle \langle 1 \mid_{C}\right) \right) [/tex]

    [tex] \rho_{A} = \frac{1}{2}\left( \mid 0 \rangle \langle 0 \mid_{A} + \mid 1 \rangle \langle 1 \mid_{A}\right) = \frac{1}{2}
    \left[\left(
    \begin{array}{ c c }
    1 & 0 \\
    0 & 0
    \end{array}\right) +
    \left(
    \begin{array}{ c c }
    0 & 0\\
    0 & 1
    \end{array}\right)\right]
    [/tex]

    So the eigenvalue equation of [tex]\rho_{A}[/tex] is :
    [tex]
    \mid
    \begin{array}{ c c }
    \frac{1}{2}-\lambda & 0\\
    0 & \frac{1}{2}-\lambda
    \end{array}\mid = 0
    [/tex]

    so [tex]\lambda = \frac{1}{2}[/tex] and Von neumann entropy [tex] S(\rho_{A}) = - \Sigma_{i} \lambda_{i} log_{2} \lambda_{i} [/tex] is:

    [tex] 2^{-2S(\rho_{A})} = \frac{1}{2} [/tex]

    So [tex] S(\rho_{A}) = \frac{1}{2}[/tex]
     
  2. jcsd
  3. oui ou non?
     
  4. No. The density matrix has off-diagonal terms as well.
     
  5. Yeah I realise this, but they cancel when finding the reduced matrix from the tracing.

    So get same result.

    Thanks I get it anyway now; I've gone over it a few times
     
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