Von Neumann Entropy of GHZ state

neu

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]

neu

oui ou non?

genneth

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

neu

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