Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 10C (Quantum Information and Computation)

The controlled-NOT gate sends $|0x⟩$ to $|0x⟩$ and $|1\bar{x}⟩$ to $|1x⟩$. HenceSince the Hadamard gate satisfies $H^2=I$,

The required identity isthe first exponent in the scanned statement should be $z$. Indeed, $X^x$ changes the parity label from $0$ to $x$, while $Z^z$ multiplies the component whose first qubit is $1$ by $(-1{)}^z$, creating the phase label. Equivalently, the hint permits either one-qubit operator to be transferred to the second half of the Bell state by transposition.

This is superdense coding. To send the two classical bits $(z,x)$, Alice applies $Z^z{X}^x$ to her half of the shared Bell state $|{\beta }_{00}⟩$, producing $|{\beta }_{zx}⟩$, and sends that one qubit to Bob. Bob applies $\mathrm{CX}$ followed by $H$ on the first qubit, as in part (i), and measures both qubits in the computational basis. The outcome is $zx$ with certainty, so one transmitted qubit communicates two classical bits using the previously shared quantum entanglement.