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

A computational-basis vector $|x_1{x}_2{x}_3⟩$ has eigenvalue $(-1{)}^{x_i+x_j}$ under $Z_i{Z}_j$. Simultaneous eigenvalue $+1$ for $Z_1{Z}_2$, $Z_2{Z}_3$, and $Z_1{Z}_3$ therefore requiresThe stabilizer subspace isIt is two-dimensional because the three displayed nonidentity stabilizers contain only two independent generators; indeed $(Z_1{Z}_2)(Z_2{Z}_3)=Z_1{Z}_3$.

For the controlled-NOT gate $U={\mathrm{CNOT}}_{12}$, propagation of the Pauli generators givesThus an $X$ on the control propagates forward to the target, while a $Z$ on the target propagates backward to the control.

Suppose $\tilde{V}$ has the same four conjugation rules and put $W=U^{†}\tilde{V}$. 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 $W=e^{i\varphi }I$ and

Since the Hadamard gate conjugates $X$ to $Z$,For $n=1$, each exponential is $-I$, soFor $n=2$,Using $XZ=-iY$ givesso one convenient logarithm isBoth products are Clifford operations: the first is the identity and the second is a one-qubit Pauli Y gate up to global phase.