Past exam of the computer science course of the University of Oxford / 2019 / Quantum information / 1 / b

The columns are $i|+⟩$ and $-i|-⟩$. Their norms are one and their inner product is zero, since $|+⟩,|-⟩$ are orthonormal. Thus $M_1$ is a unitary matrix. Equivalently,using the Hadamard gate and Pauli Z gate. The different column phases do not destroy orthonormality.

The Pauli Y gate satisfies $Y^{†}=Y$ and $Y^2=I$. With $c=\cos (\pi /6)=\sqrt{3}/2$ and $s=\sin (\pi /6)=1/2$, the given matrix is Hermitian, andTherefore $M_2$ is not a unitary matrix. Its eigenvalues are $c+s$ and $c-s$, neither of which has modulus one. An expression $\cos \theta I+i\sin \theta Y$ would behave differently, but the printed matrix has no such factor of $i$.

Set $A=U_A\otimes I_B$ and $B=I_A\otimes U_B$. They are self-adjoint operators and satisfy $A^2=B^2=I$, because the original factors are both self-adjoint and unitary. They also commute: each acts on a different tensor factor. HenceThe same calculation in the opposite order gives $M_3{M}_3^{†}=I$. Thus $M_3$ is a unitary matrix for every allowed pair of factors. Commutation cancels the cross terms, while their relative imaginary coefficient supplies the opposite signs. This is the commuting-involution unitary combination.