Rabi Oscillations of Spin

A static field makes a spin precess but cannot flip it between energy levels. To drive transitions we add a weak oscillating field perpendicular to the static one, tuned near the Larmor frequency. The result is Rabi oscillation: the population sloshes coherently back and forth between spin-up and spin-down.

The driven Hamiltonian

With a static field $B_0\hat{z}$ and a transverse drive of amplitude $B_1$ oscillating at frequency $\omega$ , the Hamiltonian (after moving to the rotating frame and dropping fast-oscillating terms — the rotating-wave approximation) is

H_{\text{rot}} = \frac{\hbar}{2}\bigl(\Delta\,\sigma_z + \Omega\,\sigma_x\bigr),

where $\Delta = \omega_0 - \omega$ is the detuning from resonance ( $\omega_0 = \gamma B_0$ ) and $\Omega = \gamma B_1$ is the Rabi frequency set by the drive strength.

On-resonance dynamics

Exactly on resonance, $\Delta = 0$ , so $H_{\text{rot}} = \tfrac{\hbar\Omega}{2}\sigma_x$ . A state started in $|0\rangle$ then evolves as a rotation about the $x$ -axis,

|\psi(t)\rangle = e^{-i\Omega t\,\sigma_x/2}\,|0\rangle = R_X(\Omega t)\,|0\rangle,

and the probability of finding the spin in the excited state is

P_{\downarrow}(t) = \sin^2\!\left(\frac{\Omega t}{2}\right).

The population oscillates fully between $0$ and $1$ at the Rabi frequency $\Omega$ . A pulse with $\Omega t = \pi$ (a $\pi$ -pulse) flips $|0\rangle \to |1\rangle$ completely; a pulse with $\Omega t = \pi/2$ creates an equal superposition.

Off resonance

With nonzero detuning the oscillation is faster but incomplete. The generalized Rabi frequency is

\Omega_R = \sqrt{\Omega^2 + \Delta^2}, \qquad P_{\downarrow}(t) = \frac{\Omega^2}{\Omega^2 + \Delta^2}\,\sin^2\!\left(\frac{\Omega_R t}{2}\right).

The maximum excited-state probability is $\Omega^2/(\Omega^2 + \Delta^2) < 1$ : only on resonance can the spin be fully inverted. Sweeping $\omega$ and watching the peak excitation is exactly how an experiment finds the qubit's transition frequency.

Try it

Apply a resonant drive with accumulated angle $\Omega t = \pi/2$ to a spin in $|0\rangle$ , i.e. $R_X(\pi/2)$ , and let the grader read the excited-state probability. It should be $\sin^2(\pi/4) = \tfrac{1}{2}$ .

Run your code to see the quantum state.