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Consider a situation in which each of Alice, Bob and Charlie holds a qubit, where the three qubits are acted on by a 3-qubit gate $U$. Assume that the gate $U$ can be implemented by first performing a $CNOT$ gate on Bob’s and Alice’s qubits, with the control on Bob’s qubit, and then performing a CNOT gate on Bob’s and Charlie’s qubits, with the control again on Bob’s qubit. Alice, Bob and Charlie can each prepare an initial state of their qubit as input to the gate $U$. They can perform basic measurements on their qubits after the gate $U$ is applied (that is, after both steps in the implementation of the gate U are complete).

Is it possible for Charlie to send a message $m ∈ {0, 1}$ to Alice and vice versa?

Charlie to Alice seems impossible to me since Charlie's qubit gets involved in the computation after Alice's and there's no way his qubit can affect Alice's measurement as Alice's computation is finished by the time Charlie's starts.

For Alice to Charlie, my initial idea was if m=0, let Alice prepare $|+⟩$ else $|-⟩$.

Let Bob prepare $|+⟩$

Let Charlie prepare $|0⟩$

Then applying $U$ results in the Bell state $|\phi ^+⟩$ if m=0 and $|\phi^-⟩$ if m=1, but this is an entangled state hence there's no way for Charlie to make a measurement independent of Bob's measurement to get an outcome.

Is there any way to prove/disprove these 2 problems?

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The density matrix of Alice's final qubit state, after tracing out Bob and Charlie, can be written as a function of things not involving Charlie. Therefore Charlie cannot communicate a bit to Alice. Another way of seeing this is to realize everything that Alice does after her CNOT commutes with anything Charlie does, therefore whatever the final result she measures it must be independent of Charlie's actions. If Alice measures before Charlie acts, the density matrix must be identical; due to the commutation.

That being said, if classical communication is free, this setup can be used to create entanglement between Alice and Charlie which can then be used to send a qubit to Alice by using teleportation. So it won't let you send bits, but it'll let you upgrade sending bits into sending a qubit.

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  • $\begingroup$ What about the opposite direction? (Alice to Charlie?) $\endgroup$ Apr 29 at 20:54

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