So far these smart people are only successful in theory. In practice their success is still very, very limited. Only time will tell where they will end up.
1/root(2) [a|000>+a|011>+b|100>+b|111>] applying cnot where q1 is target and q2 & q3 are targets , we get 1/root(2) [a|000>+a|011>+b|111>+b|100>] but how are you getting 1/root(2) [a|000>+a|011>+b|101>+b|110>]
Thanks for sharing, good explanation of the principle. However, I don' get what the purpose of this is: if we have to transfer a qBit of the entangled pair to Bob, why don't we just transfer |Ψ> directly? Is is just because we can share the entangled pair in advance and come up / compute |Ψ> after that and then teleport? Even if that is the case we still have to wait for the classical information to be transferred, which likely takes the same amount of time as transferring a qBit.
Way above my head..... thank god we have such smart people
So far these smart people are only successful in theory. In practice their success is still very, very limited. Only time will tell where they will end up.
At 3:03, inside the box labeled "STATE 2" , why did the CNOT flip the 3rd bit (Bob's bit) instead of the 2nd bit?
Well technically the flip should have occurred with Alice's pair but since these are entangled the result comes out to be the same
she just interchanged the third and fourth term so still the same
1/root(2) [a|000>+a|011>+b|100>+b|111>]
applying cnot where q1 is target and q2 & q3 are targets , we get
1/root(2) [a|000>+a|011>+b|111>+b|100>]
but how are you getting
1/root(2) [a|000>+a|011>+b|101>+b|110>]
Thanks for sharing, good explanation of the principle. However, I don' get what the purpose of this is: if we have to transfer a qBit of the entangled pair to Bob, why don't we just transfer |Ψ> directly?
Is is just because we can share the entangled pair in advance and come up / compute |Ψ> after that and then teleport? Even if that is the case we still have to wait for the classical information to be transferred, which likely takes the same amount of time as transferring a qBit.
Okay nevermind. My question is answered by the part starting at 08:46
A bit more energy and enthusiasm in the presentation wouldn't have hurt this video.
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