= Solution
Put $\alpha=\Delta\tau/\hbar$ and $\beta=Ex_0\tau/\hbar$. The <Pauli matrix multiplication law> gives the <pulse-and-precession spin one-half map>
$$
U_\Delta U_E=CI-i\mathbf v\cdot\boldsymbol\sigma,\quad C=\cos\alpha\cos\beta,\quad
\mathbf v=(\cos\alpha\sin\beta,-\sin\alpha\sin\beta,-\sin\alpha\cos\beta).
$$
Because $C^2+|\mathbf v|^2=1$, a suitable <angle> and <unit vector> for the <rotation> are
$$
\boxed{\theta=2\arccos(\cos\alpha\cos\beta),\qquad
\mathbf n=\frac{(\cos\alpha\sin\beta,-\sin\alpha\sin\beta,-\sin\alpha\cos\beta)}{\sqrt{1-\cos^2\alpha\cos^2\beta}}.}
$$
If the denominator vanishes, $U=\pm I$ and the axis is arbitrary. The <Bloch vector> $\mathbf r=\langle\boldsymbol\sigma\rangle$ rotates through $\theta$ about $\mathbf n$ per period; after $N$ periods the angle is $N\theta$. The eigenphases equivalently give <quasienergies> $\pm\hbar\theta/(2\tau)$ modulo $2\pi\hbar/\tau$.
In the two-orbital subspace the position operator is $X=x_0\sigma_x$. The <Heisenberg equation of motion> between pulses gives $P/m=(i/\hbar)[H_0,X]=(2\Delta x_0/\hbar)\sigma_y$, hence
$$
\langle X\rangle=x_0r_x,\qquad \langle P\rangle=\frac{2m\Delta x_0}{\hbar}r_y.
$$
Thus 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 $\langle X\rangle$ unchanged at the kick, while free evolution mixes the position and momentum components. Starting in $|1s\rangle$, one complete period gives
$$
\mathbf r=(-\sin2\alpha\sin2\beta,-\cos2\alpha\sin2\beta,\cos2\beta),
$$
which exhibits the induced motion explicitly. These are projected two-orbital <observables>.
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