- #1
kelly0303
- 579
- 33
Hello! 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