What is the Von Neumann Entropy of the GHZ State?

[neu]

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:

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

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

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

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

 

[neu]

oui ou non?

 

[genneth]

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

 

[neu]

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

Yeah I realize 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