= Solution
With $\Omega=\gamma B_x/2$, the <Born rule> applied to the supplied down-spin <probability amplitude> gives
$$
P_\downarrow(t)=\frac{\Omega^2}{\delta^2+\Omega^2}
\sin^2\left(\frac{\sqrt{\delta^2+\Omega^2}\,t}{2}\right).
$$
For nonzero transverse field, the maxima occur at
$$
\boxed{t_j=\frac{(2j+1)\pi}{\omega_R},\quad j=0,1,\ldots,
\qquad P_{\max}(\omega)=\frac{\Omega^2}{(\omega-\omega_z)^2+\Omega^2}.}
$$
The first maximum is reached at $\pi/\omega_R$; on resonance this is a <Rabi pi pulse>. The <resonance envelope of a Rabi transition> is a <Lorentzian line shape>, centred on $\omega_{\mathrm{res}}=\omega_z$ under the orientation convention above. It has half its peak value at $|\omega-\omega_z|=|\Omega|$, so its <full width at half maximum> is
$$
\boxed{\Delta\omega=2|\Omega|=|\gamma B_x|,
\qquad\frac{\omega_{\mathrm{res}}}{\Delta\omega}=\frac{|B_z|}{|B_x|}.}
$$
If resonance width is instead defined as the half-width, the ratio differs by a factor of two but has the same requested proportionality. The <quality factor> is large precisely in the regime where the <rotating-wave approximation> is justified. The envelope is obtained by choosing the best time separately at each drive frequency; a measurement at one fixed pulse duration has a different line profile.
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2021-q4-resonance-profile.png]
{title=Maximum spin-flip probability versus drive detuning}
{description=Original <rotating-wave approximation> envelope, with the two half-height detunings marked. The maximum is over interaction time independently at each drive frequency.}
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