= Solution
{c}
For the stated <Hamiltonian operator>, put $\boldsymbol\Omega=(V,0,-\delta)$ and $\Omega=\sqrt{V^2+\delta^2}$. The <Bloch vector> obeys
$$
\dot{\mathbf r}=\boldsymbol\Omega\mathbin\times\mathbf r,\qquad
U(t)=\cos\frac{\Omega t}{2}\,I-i\sin\frac{\Omega t}{2}\,
\frac{V\sigma_x-\delta\sigma_z}{\Omega}.
$$
Thus it rotates right-handedly about the fixed axis $\boldsymbol\Omega$, preserving its length and its projection on that axis. For example, on <resonance> $\delta=0$, a north-pole <Bloch vector> follows $(0,-\sin Vt,\cos Vt)$ and reaches the south pole at $t=\pi/V$.
Take $V>0$, with the field <phase> chosen accordingly. The <Hadamard gate> is $H=(\sigma_x+\sigma_z)/\sqrt2$. Align the axis with $(1,0,1)$ and rotate through $\pi$:
$$
\boxed{\delta=-V,\qquad \tau=\frac{\pi}{\sqrt2V},\qquad U(\tau)=-iH.}
$$
The factor $-i$ is an irrelevant common <global phase>. This realizes the whole <quantum gate>, rather than merely its action on one input <quantum state>.
There is a <basis>-ordering convention to make explicit: this $H$ is the <Hadamard gate> matrix in the ordered <basis> $(|1\rangle,|0\rangle)$ chosen in part (a). If instead its definition is the usual ket rules $H_{01}|0\rangle=(|0\rangle+|1\rangle)/\sqrt2$ and $H_{01}|1\rangle=(|0\rangle-|1\rangle)/\sqrt2$, its matrix in our order is $(\sigma_x-\sigma_z)/\sqrt2$. For that definition choose $\boxed{\delta=+V,\ \tau=\pi/(\sqrt2V)}$. Both conventions use the same pulse duration; the sign of the <Rabi detuning> selects the desired axis.
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