Recently, bidirectional quantum teleportation has attracted a great deal of research attention. However, existing bidirectional teleportation schemes are normally discussed on the basis of perfect quantum environments. In this paper, we first put forward a bidirectional teleportation scheme to transport three-qubit Greenberger–Horne–Zeilinger (GHZ) states based on controled-not (CNOT) operation and single-qubit measurement. Then, we generalize it to the teleportation of multi-qubit GHZ states. Further, we discuss the influence of quantum noise on our scheme by the example of an amplitude damping channel, then we obtain the fidelity of the teleportation. Finally, we utilize the weak measurement and the corresponding reversing measurement to protect the quantum entanglement, which shows an effective enhancement of the teleportation fidelity.Recently, bidirectional quantum teleportation has attracted a great deal of research attention. However, existing bidirectional teleportation schemes are normally discussed on the basis of perfect quantum environments. In this paper, we first put forward a bidirectional teleportation scheme to transport three-qubit Greenberger–Horne–Zeilinger (GHZ) states based on controled-not (CNOT) operation and single-qubit measurement. Then, we generalize it to the teleportation of multi-qubit GHZ states. Further, we discuss the influence of quantum noise on our scheme by the example of an amplitude damping channel, then we obtain the fidelity of the teleportation. Finally, we utilize the weak measurement and the corresponding reversing measurement to protect the quantum entanglement, which shows an effective enhancement of the teleportation fidelity.

Chinese Physics BApr 1, 201749 citations

Greenberger-Horne-Zeilinger versus W states: Quantum teleportation through noisy channels

Eylee Jung, Mi-Ra Hwang, You Hwan Ju +7 more

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.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.

arXiv:0801.1433Physical Review AJul 9, 2008137 citations

Quantum Teleportation of a Two Qubit State Using GHZ-Like State

Kaushik Nandi, Chandan Mazumdar

International Journal of Theoretical PhysicsNov 26, 201368 citations

Quantum Teleportation of an Arbitrary Three-Qubit State Using GHZ-Like States

Wen Yuan

International Journal of Theoretical PhysicsAug 3, 201416 citations

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.