# Von Neumann Entropy of GHZ state

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

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

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

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

reduced density matrix of qubit A:

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

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

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

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

$$2^{-2S(\rho_{A})} = \frac{1}{2}$$

So $$S(\rho_{A}) = \frac{1}{2}$$

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oui ou non?

No. The density matrix has off-diagonal terms as well.

No. The density matrix has off-diagonal terms as well.
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