= Solution
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. Hence
$$
M_3^\dagger M_3=\frac12(A-iB)(A+iB)=\frac12(A^2+B^2+iAB-iBA)=I.
$$
The same calculation in the opposite order gives $M_3M_3^\dagger=I$. Thus \b[$M_3$ 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>.
Back to article page
