 Solution

For the axis-angle composition of orthogonal spin rotations, set $A=\Omega T/2$ and $B=\Phi /2$. Multiplication, including ${\sigma }_z{\sigma }_x=i{\sigma }_y$, givesOne checks that $C^2+|\mathbf{v}{|}^2=1$. For $s=|\mathbf{v}|\ne 0$, chooseThis choice retains the full spinor sign and gives $U_{\Omega }{U}_{\Phi }=e^{-i\vartheta \hat{\mathbf{n}}\cdot \mathbf{S}}$. Repeated periods have the same axis:Thus the spin expectation vector rotates through $N\vartheta$, interpreted modulo $2\pi$ as a spatial rotation. The positive $y$ component comes from the stated ordering; reversing the two rotations changes that component's sign.

When $s=0$, the one-period operator is $+I$ or $-I$, and no unique rotation axis exists. The latter is a $2\pi$ spatial rotation with a minus sign on a spin-$1/2$ state. It is safer to retain this central group element than to replace the angle by zero in a spinor calculation.