Quantum teleportation
User comments:
[2011-10-23 14:44:38] mnielsencourses - @IslandHermit As ...
@IslandHermit As discussed in earlier videos, it's not possible to determine the value of the unknown state \psi by measurement. So the strategy of measure-then-send doesn't work.
[2011-10-23 08:25:35] IslandHermit - I should have ...I should have waited for the next video, as it answers my question. First, A *can't* measure \psi. Second, even if she could it would require an infinite amount of information to be transmitted along the classical channel to B.
[2011-10-23 08:14:21] IslandHermit - I don't get the ...I don't get the point here. Given the presence of the classical communication channel, A could have just measured the \psi state and sent the result of that measurement to B over the classical channel. B could then have reproduced the \psi state. All of that could happen without the need of a shared Bell state.
[2011-08-01 16:12:52] mnielsencourses - I just recomputed, ...I just recomputed, and it seems that ZX \psi is correct. I can't say why you're getting XZ \psi without seeing working. Note that to compute ZX\psi you first have to apply X to \psi, then Z, not the other way round (which is an easy mistake to make, since we read from left to right).
[2011-08-01 04:28:31] jrupac - How come at 11:07 ...How come at 11:07 the last possible posterior state is ZX\psi and not XZ\psi?