 Solution

For the stated Hamiltonian operator, put $\mathbf{\Omega }=(V,0,-\delta )$ and $\Omega =\sqrt{V^2+{\delta }^2}$. The Bloch vector obeysThus it rotates right-handedly about the fixed axis $\mathbf{\Omega }$, 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 $H=({\sigma }_x+{\sigma }_z)/\sqrt{2}$. Align the axis with $(1,0,1)$ and rotate through $\pi$: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 $(|1⟩,|0⟩)$ chosen in part (a). If instead its definition is the usual ket rules $H_{01}|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H_{01}|1⟩=(|0⟩-|1⟩)/\sqrt{2}$, its matrix in our order is $({\sigma }_x-{\sigma }_z)/\sqrt{2}$. For that definition choose $\boxed{\delta =+V,\tau =\pi /(\sqrt{2}V)}$. Both conventions use the same pulse duration; the sign of the Rabi detuning selects the desired axis.