Confused about Ramsey technique Assume we have a 2 level system with the frequency between the 2 levels $\omega_0$ and ignore the lifetime of the upper state. In order to measure the transition frequency using Ramsey technique, you apply 2 $\pi/2$ pulses separated by a time $T$. We have that the probability of finding an atom in the excited sate under the interaction with light (assuming strong field i.e. not using perturbation theory) is given by: $\frac{\Omega^2}{W^2}sin^2(Wt/2)$ where $\Omega$ is the Rabi frequency and $W^2=\Omega^2+(\omega-\omega_0)^2$ with $\omega$ being the laser frequency (which we scan in order to find $\omega_0$). If we are on resonance, i.e. $W=\Omega$, a $\pi/2$ pulse is a pulse of duration $t=\frac{\pi}{2\Omega}$ such that, starting in the ground state, the atom becomes an equal superposition of upper and lower states. However in the Ramsey technique, we don't know $\omega_0$ beforehand (this is what we want to measure). So I am not sure I understand how we can still create an equal superposition of upper and lower level using a $\pi/2$ pulse. Assuming we use a pulse of duration $t=\frac{\pi}{2W}$, from the equation above we get that the population of the upper state is $\frac{\Omega^2}{2W^2}$ which is not $1/2$ as in the resonance case. How do we still get equal superposition of upper and lower case when we are not on resonance? Thank you
 A: You are correct that the Ramsey sequence should be applied close to resonance, but it does not need to be perfectly on resonance -- it only needs to be close to resonance, relative to the Rabi frequency used for the $\pi/2$ pulses.
The Ramsey technique gives a measurement signal that oscillates as a function of the gap time $T$ between $\pi/2$ pulses, where the oscillation frequency is given by the detuning $\omega-\omega_0$ and the contrast is determined by the weights of the superposition produced by the $\pi/2$ pulse. In the ideal case (maximal contrast), the $\pi/2$ pulse would transfer 50% of the population to the second state; however, the signal is still usable when the $\pi/2$ is imperfect. In practice, as long as $|\omega - \omega_0| \lesssim  \Omega$, then you find from the expression that you wrote $\frac{\Omega^2}{W^2} \sin^2(Wt/2)$ that close to 50% of the population will be transferred.
If you know approximately where the resonance is to within a bandwidth of $\sim \Omega$, then Ramsey can be used to more precisely find the resonance (ultimately limited by the longest gap time $T$ that can be used). If you don't have any idea where the resonance is, then you can first find it by applying an approximate $\pi$ pulse over a wider range of frequencies and finding which frequency gives the largest population transfer.
