Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 15D (Quantum Information and Computation)

Let $|\alpha ⟩=a|0⟩+b|1⟩$ be Alice's unknown qubit, and label her half of the shared Bell state by $A$ and Bob's by $B$. Alice applies a Controlled-NOT gate from the unknown qubit to $A$, then a Hadamard gate to the unknown qubit. The three-qubit state becomesAlice measures her two qubits in the computational basis and sends the two classical outcomes $(j,k)$ to Bob. Bob applies $Z^j{X}^k$, recovering $|\alpha ⟩$ with certainty. This is quantum teleportation; Alice never needs to know the state.

For the shared GHZ state, Charlie first applies a Hadamard gate to his qubit. ThenCharlie measures in the computational basis. Outcome $0$ leaves Alice and Bob with $|{\varphi }^{+}⟩$; outcome $1$ leaves $|{\varphi }^{-}⟩$, which either party converts to $|{\varphi }^{+}⟩$ by applying $Z$ after Charlie reports the outcome. Thus

Quantum no-signalling says that, without communicating a measurement outcome, any local trace-preserving operation performed on subsystem $A$ leaves all outcome probabilities for measurements on subsystem $B$ unchanged; equivalently, it leaves the reduced density matrix ${\rho }_B$ unchanged.

The two Bell states are orthogonal. Applying the inverse Bell-state preparation circuit, namely a Controlled-NOT gate from $A$ to $B$ followed by a Hadamard gate on $A$, givesA computational-basis measurement therefore distinguishes them perfectly, soIf only $B$ is accessible, however,Every measurement on $B$ consequently has the same outcome distribution under both hypotheses. With equal prior probabilities no information is available, andthe error probability of a random guess.