How does Bell measurement work in the teleportation?

Quantum Computing Asked by Souroy on December 24, 2020

I’m a complete beginner and one of the first things I was taught was the teleportation protocol. In the protocol, the party sending its state (which we call say $|phirangle$) makes a Bell measurement on a Bell state it has from before along with $|phirangle$. From this, it finds the indices of its Bell state which it sends to the receiver. But my question is why does the fact that it is measuring $|phirangle$ together with its Bell state have any bearing on the indices of the Bell state, considering $|phirangle$ is not involved in the Bell measurement?

2 Answers

In the teleportation protocol, the two parties share the entangled Bell State and it is implemented via a CNOT gate between the state to be sent (suppose Alice is sending the state) and the part of entangled Bell state Alice has. The CNOT gate creates another entangled state whose measurement Alice will send to the second party, say it's Bob, to perform the measurements accordingly. So if you assume the state to be sent is 1 qubit state after CNOT you will have a 3 qubit state (as Bell sate is 2 qubit state). Now Alice will measure the middle qubit which will be the part of classical information she will communicate to Bob.

So, you see the controlled-NOT acts as an entangling operator without affecting the first qubit (here $|phirangle$) and changing the entangled Bell state and hence its measurement outcome.

Correct answer by aditikatoch on December 24, 2020

Perhaps the following diagram helps to show what acts where. I think you've got a little muddled. enter image description here

Answered by DaftWullie on December 24, 2020

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