
#1
Apr1508, 06:43 PM

P: 245

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



#2
Apr1708, 10:20 AM

P: 245

oui ou non?




#3
Apr1708, 11:14 AM

P: 981

No. The density matrix has offdiagonal terms as well.




#4
Apr1708, 03:31 PM

P: 245

Von Neumann Entropy of GHZ stateSo get same result. Thanks I get it anyway now; I've gone over it a few times 


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