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In studying Mach-Zehnder and Ramsey interferometers, I came across the expression "$\pi/2$ pulse". What does it mean exactly? I am working with a Bloch vector representation $(u,v,w)$ of a 2 state system. We have a Rabi frequency $\Omega_0$ and a detuning parameter $\delta$ to the $|1\rangle\rightarrow|2\rangle$ transition frequency. In those conditions, I think the "$\pi/2$ pulse" is a rotation around the $(-\Omega_0,0,-\delta)$ axis for a duration $\tau=\frac{\pi/2}{\sqrt{\Omega_0^2+\delta^2}}$. Is that correct?

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$\frac{π}{2}$ pulse means that all the particles in the system have gone to the higher level. $π$ pulse excites all particles in the first half time and de-exites in the second, so all particles are in lower level.

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  • $\begingroup$ We use mathjax here. $\endgroup$ – Anubhav Goel Feb 17 '16 at 4:03
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    $\begingroup$ Despite being the accepted answer, this is not correct. A $\pi$ pulse sends $|0\rangle\to|1\rangle$. A $\pi/2$ pulse sends $|0\rangle\to(|0\rangle+i|1\rangle)/\sqrt{2}$. The most intuitive reason for this naming is how the corresponding Bloch sphere vectors transform (spoilers: rotation by $\pi$ and $\pi/2$, respectively.) Source: en.wikipedia.org/wiki/Rabi_problem $\endgroup$ – Daniel Apr 1 '19 at 15:39

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