GHZ versus W : Quantum Teleportation through Noisy Channels
Which state does lose less quantum information between GHZ and W states when they are prepared for two-party quantum teleportation through noisy channel? We address this issue by solving analytically a master equation in the Lindbald form with introducing the noisy channels which makes the quantum channels to be mixed states. It is found that the answer of the question is dependent on the type of the noisy channel. If, for example, the noisy channel is ($L_{2,x}$, $L_{3,x}$, $L_{4,x}$)-type where $L's$ denote the Lindbald operators, GHZ state is always more robust than W state, i.e. GHZ state preserves more quantum information. In, however, ($L_{2,y}$, $L_{3,y}$, $L_{4,y}$)-type channel the situation becomes completely reversed. In ($L_{2,z}$, $L_{3,z}$, $L_{4,z}$)-type channel W state is more robust than GHZ state when the noisy paramter ($κ$) is comparatively small while GHZ state becomes more robust when $κ$ is large. In isotropic noisy channel we found that both states preserve equal amount of quantum information. A relation between the average fidelity and entanglement for the mixed state quantum channels are discussed.