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How to differentiate GHZ state and W state?

  1. Oct 24, 2016 #1
    • OP warned about not using the homework template
    Good day,

    I try to differentiate GHZ-state and W-state using three tangle. Suppose The value three tangle for GHZ-state equal to 1 while W-state equal to 0.

    I used three tangle formula,

    $$\tau_{ABC}=\tau_{A(BC)}-\tau_{AB}-\tau_{AC}=2(\lambda^{AB}.\lambda^{AB}+\lambda^{AC}.\lambda^{AC})$$

    where the the

    $$ \lambda^{AB}.\lambda^{AB}$$ and $$\lambda^{AC}.\lambda^{AC}$$ are eigenvalues of

    $$\rho_{AB}.\rho^{\sim}_{AB}=(\sigma_y\otimes\sigma_y\rho^*_{AB}\sigma_y\otimes\sigma_y)$$

    where the asterisk denotes complex conjugation in the standard basis and $$\sigma_y=\begin{pmatrix} 0 &-i &
    \\i & 0
    \end{pmatrix}
    $$


    I measure the W-state and the density matrix of W-state:

    $$\rho_w=\begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & \frac{1}3 & \frac{1}3& 0 & \frac{1}3 & 0 & 0 & 0
    \\0 & \frac{1}3 & \frac{1}3& 0 & \frac{1}3 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & \frac{1}3 & \frac{1}3& 0 & \frac{1}3 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    \end{pmatrix}$$


    and the

    $$\rho_{AB}=\rho_{AC}= \begin{pmatrix} \frac{1}3 & 0 & 0 & 0
    \\0 & \frac{1}3 & \frac{1}3 & 0 \\0 & \frac{1}3 & \frac{1}3 & 0\\0 & 0& 0 & 0
    \end{pmatrix}$$


    Since it real matrix I assume it a multiplication matrix between $$\rho_{AB}.\rho^{\sim}_{AB}$$.

    I get the eigen values= 2/3 and 1/3

    and the value of three tangle for W-state = 0.8888... not 0

    Where the wrong I did? I'm really stuck. Please help me to show the way to calculate three tangle correctly.

    Thank you
     
  2. jcsd
  3. Oct 29, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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