 Rabi pi pulse

For a constant-amplitude square pulse on resonance, a Rabi pi pulse lasts ${\tau }_{\pi }=\pi /\Omega$ and gives $|g⟩\mapsto -i{e}^{-i\varphi }|e⟩$ in the rotating frame. A half-duration pulse prepares $(|g⟩-i{e}^{-i\varphi }|e⟩)/\sqrt{2}$. The quantum pulse area $\Omega t$ determines these rotations only under the stated resonant, constant-axis assumptions; a detuned drive rotates about a different axis.

As an original numerical illustration, choose $\Omega =2\pi \times 1000\mathrm{rad}{\mathrm{s}}^{-1}$ and hence ${\tau }_{\pi }=0.500\mathrm{ms}$. These are chosen model parameters, not historical apparatus settings. Holding that duration fixed givesThus $P_{\pi }(0)=1$, $P_{\pi }(1)=\frac{1}{2}{\sin }^2(\pi /\sqrt{2})\simeq 0.3166$, and $P_{\pi }(\sqrt{3})=0$. The last zero comes from completing a full rotation about the tilted axis, rather than switching the drive off. At zero Rabi detuning, doubling the pulse duration restores the ground-state population; continuous driving does not keep transferring population irreversibly.