 Solution

Put $\alpha =\Delta \tau /\hslash$ and $\beta =E{x}_0\tau /\hslash$. The Pauli matrix multiplication law gives the pulse-and-precession spin one-half mapBecause $C^2+|\mathbf{v}{|}^2=1$, a suitable angle and unit vector for the rotation areIf the denominator vanishes, $U=\pm I$ and the axis is arbitrary. The Bloch vector $\mathbf{r}=⟨\mathbf{\sigma }⟩$ rotates through $\theta$ about $\mathbf{n}$ per period; after $N$ periods the angle is $N\theta$. The eigenphases equivalently give quasienergies $\pm \hslash \theta /(2\tau )$ modulo $2\pi \hslash /\tau$.

In the two-orbital subspace the position operator is $X=x_0{\sigma }_x$. The Heisenberg equation of motion between pulses gives $P/m=(i/\hslash )[H_0,X]=(2\Delta x_0/\hslash ){\sigma }_y$, henceThus two components of the Bloch vector directly describe the mean position and momentum; its third gives the orbital population imbalance. The electric kick rotates about the position axis, leaving $⟨X⟩$ unchanged at the kick, while free evolution mixes the position and momentum components. Starting in $|1s⟩$, one complete period giveswhich exhibits the induced motion explicitly. These are projected two-orbital observables.