= Commuting-involution unitary combination
{title2=$U=(A+iB)/\sqrt2$}
If A and B are commuting <self-adjoint operators> with $A^2=B^2=I$, then the displayed combination is a <unitary operator>. Indeed,
$$
U^\dagger U=\tfrac12(A-iB)(A+iB)=\tfrac12(A^2+B^2+iAB-iBA)=I,
$$
and the reverse product is also I. The factors are <involutions>, and their commutation is essential to cancel the cross terms. Operators $A=U_A\otimes I$ and $B=I\otimes U_B$ on separate tensor factors provide a natural example. More generally, $\cos\theta\,A+i\sin\theta\,B$ is unitary for real $\theta$ by the same proof. Replacing i by a real coefficient usually loses this property.
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