For the <controlled-NOT gate> $U=\operatorname{CNOT}_{12}$, propagation of the Pauli generators gives
$$
\boxed{
UX_1U^\dagger=X_1X_2,
\qquad UX_2U^\dagger=X_2,
\qquad UZ_1U^\dagger=Z_1,
\qquad UZ_2U^\dagger=Z_1Z_2}.
$$
Thus an $X$ on the control propagates forward to the target, while a $Z$ on the target propagates backward to the control.

Suppose $\widetilde V$ has the same four conjugation rules and put $W=U^\dagger\widetilde V$. Then $W$ commutes with $X_1,X_2,Z_1,Z_2$. These generators span the full two-qubit operator algebra, so its <commutant> consists only of scalar multiples of the identity. Hence $W=e^{i\phi}I$ and
$$
\boxed{\widetilde V=e^{i\phi}U}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 324 / 4 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-324/4/a/ii/solution