1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Von Neumann Entropy of GHZ state

  1. Apr 15, 2008 #1

    neu

    User Avatar

    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. Apr 17, 2008 #2

    neu

    User Avatar

    oui ou non?
     
  4. Apr 17, 2008 #3
    No. The density matrix has off-diagonal terms as well.
     
  5. Apr 17, 2008 #4

    neu

    User Avatar

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?