 Pulse-and-precession spin one-half map

Multiplying the two matrix exponentials with the Pauli matrix multiplication law givesOne has $C^2+|\mathbf{v}{|}^2=1$. If $|\mathbf{v}|\ne 0$, put $\theta =2\arccos C$ and $\mathbf{n}=\mathbf{v}/|\mathbf{v}|$. The exponential of a Pauli vector gives $U_F^N=I\cos (N\theta /2)-i\mathbf{n}\cdot \mathbf{\sigma }\sin (N\theta /2)$. Its action on a Bloch vector is a rotation through $N\theta$ about $\mathbf{n}$, and its quasienergies are $\pm \hslash \theta /(2T)$ modulo $2\pi \hslash /T$. When $U_F=\pm I$, its axis is arbitrary and the Bloch vector is unchanged.