 Identical-pulse Ramsey return probability

For ideal identical $R_x(\pi /2)$ pulses separated by $R_z(-\delta T)$, multiply the unitary operators to obtain $U=-i[\cos (\delta T/2)X-\sin (\delta T/2)Z]$. An input at either pole of the Bloch sphere returns to that same pole with probability ${\sin }^2(\delta T/2)$. In particular zero delay gives a full NOT, so its return probability is zero. Opposite-sign pulses instead give ${\cos }^2(\delta T/2)$. The distinction matters when identifying the measured quantum state population.