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 &amp; 0 \\<br /> 0 &amp; 0<br /> \end{array}\right) +<br /> \left(<br /> \begin{array}{ c c }<br /> 0 &amp; 0\\<br /> 0 &amp; 1<br /> \end{array}\right)\right]<br />

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

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