Resonant frequency shift for cavity superconducting circuits For superconducting circuits coupled to a resonator, we get a shift in the transmission spectrum (vacuum Rabi splitting) because of the interaction of the qubit and the resonator.
Then one goes to the dispersive limit where we have large detuning $\Delta$. Then you can approximate the Jaynes-Cummings Hamiltonian and see the interaction term as perturbation. Afterwards the Hamiltonian shows a cavity frequency shift of
$$\pm g^2/\Delta \, .$$
With this we can determine the state of the qubit.
Why do we have to detune the system when we already see a splitting because of the interaction of the resonator and the qubit? Can we perform non-demolition measurement only in the detuned case?
The splitting for the detuned case is really small, when $\Delta$ is big but is it still bigger than for the not detuned case?
 A: The resonator$^{[a]}$ is usually connected to a transmission line so that we can collect the dispersed readout pulse and detect it's phase/amplitude.
If the qubit is near in frequency to the resonator, then energy from the qubit goes into the resonator and leaves via the transmission line, i.e the qubit has a decay channel.
In quantum mechanics, that decay process is known as the "Purcell effect" or "Fermi's Golden rule", but in the case of nearly harmonic qubits (like the tramsmon), it can be understood entirely through conventional circuit analysis.
For large enough detuning, the decay rate imparted by the resonator on the qubit is often approximated as
$$\kappa_\text{qubit} = \kappa_\text{resonator} (g^2/ \Delta^2) \, .$$
Beware that it's not a great approximation when the qubit anharmonicity is modest.
The large detuning is needed to have reasonably fast $\kappa_\text{resonator}$ without imparting too much decay in the qubit.
$[a]$: The resonator is a cavity only in the case of 3D  devices.
