Greenberger-Horne-Zeilinger versus<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mi>W</mml:mi></mml:math>states: Quantum teleportation through noisy channels

Which state loses less quantum information between Greenberger-Horne-Zeilinger (GHZ) and $W$ states when they are prepared for two-party quantum teleportation through a noisy channel? We address this issue by solving analytically a master equation in the Lindblad form with introducing the noisy channels that cause the quantum channels to be mixed states. It is found that the answer to this question is dependent on the type of noisy channel. If, for example, the noisy channel is $({L}_{2,x},{L}_{3,x},{L}_{4,x})$ type, where the $L$'s denote the Lindblad operators, the GHZ state is always more robust than the $W$ state, i.e., the GHZ state preserves more quantum information. In, however, the $({L}_{2,y},{L}_{3,y},{L}_{4,y})$-type channel the situation becomes completely reversed. In the $({L}_{2,z},{L}_{3,z},{L}_{4,z})$-type channel, the $W$ state is more robust than the GHZ state when the noisy parameter $(\ensuremath{\kappa})$ is comparatively small while the GHZ state becomes more robust when $\ensuremath{\kappa}$ is large. In isotropic noisy channel we found that both states preserve an equal amount of quantum information. A relation between the average fidelity and entanglement for the mixed state quantum channels are discussed.