Past exam of the computer science course of the University of Oxford / 2021 / Quantum information / 1 / c

Prepare Bob in $|0{⟩}_B$ and encode the bit using Alice's preparation $|m{⟩}_A$. The controlled-NOT gate givesBob measures in the computational basis and reads $m$ with probability one. Alice's final measurement is unnecessary. The interaction transmits a classical bit; it does not clone an arbitrary unknown qubit.

Reverse communication is also possible. Prepare Alice in $|+{⟩}_A$. Bob encodes $m=0$ as $|+{⟩}_B$ and $m=1$ as $|-{⟩}_B$. Since $X|\pm ⟩=\pm |\pm ⟩$, phase kickback giveswhere $|(-{)}^0⟩=|+⟩$ and $|(-{)}^1⟩=|-⟩$. Alice measures in the Hadamard basis: $+$ means zero and $-$ means one, deterministically.

The target's Pauli X gate eigenvalue becomes a relative phase on the control. This is the operational content of CNOT control reversal in the Hadamard basis; the computational basis control arrow alone does not determine every possible direction of communication.

Charlie cannot send a bit to Alice with the stated preparation, fixed interaction and final local measurements. Write $V={\mathrm{CNOT}}_{B\to A}$ and $W={\mathrm{CNOT}}_{B\to C}$, so the complete gate is $U=WV$ with the rightmost gate acting first. Alice has finished interacting before the gate involving Charlie acts.

For arbitrary product input density matrices, put ${\tau }_{AB}=V({\rho }_A\otimes {\rho }_B)V^{†}$. The final reduced density matrix of Alice isThe equality follows because a unitary operator on the traced-out systems leaves the partial trace unchanged. The right-hand side is independent of Charlie's preparation. Therefore every outcome probability of every local measurement by Alice is independent of Charlie's message. Quantum no-signalling applies after Alice's last interaction, even though phase kickback can allow communication between the two qubits of an individual CNOT gate.

Alice cannot send a bit to Charlie either. An argument based only on the apparent chronological chain would be insufficient: Alice interacts with Bob before Bob interacts with Charlie. The extra fact is that these particular gates commute.

In Bob's computational basis the complete gate isBoth controlled-NOT gates preserve Bob's control label, and their target operations act on different systems. Thus $WV=VW$. We may consequently regard Bob–Charlie as the first interaction and Bob–Alice as the last one. The final operation on $AB$ cannot change Charlie's reduced density matrix, and Charlie's earlier state cannot depend on Alice's initially separate preparation.

The target no-signalling in a common-control unitary formula makes the independence explicit:Alice's density matrix does not appear. The formula includes arbitrary complex initial amplitudes and every final measurement on Charlie. This conclusion uses the specified pair of gates; a general intervening operation on Bob could change the communication possibilities.