A scheme is presented for the long-distance teleportation of an unknown atomic state between two separated cavities. The scheme is based on the Raman coupling and cavity decay. In the scheme, the effective atom-cavity coupling strength is much smaller than the cavity decay rate and thus cavities of high quality factor are unnecesssary.
A scheme is presented for the teleportation of an unknown atomic state between two separated cavities. The scheme involves two interaction-detection cycles and uses resonantly coupled atoms with an additional ground state not coupled to the cavity field. Remarkably, the damping of one basis state is balanced by that of the other basis state and the state with photon loss in the first interaction-detection cycle is eliminated by the second cycle. Therefore, the fidelity of teleportation is independent of the teleported state and insensitive to the atomic spontaneous emission, cavity decay, and detection inefficiency, which is obviously in contrast to the original scheme by Bose et al. [Phys. Rev. Lett. 83, 5158 (1999)].
We present a physical scheme to teleport an unknown atomic entangled state via cavity decay. In the teleportation process, four-particle Greenberger–Horne–Zeilinger (GHZ) state is used as quantum channel, and two unknown entangled atoms and two of four atoms in the four-particle GHZ state are trapped in four leaky cavities, respectively. Based on the joint detection of the photons leak out from the four cavities, we can teleport an unknown entangled state to two other remote atoms with certain probability and high fidelity.