Driven Two-Level Systems

The driven two-level system is the simplest exactly-solvable time-dependent problem and the prototype for every qubit gate. Here we connect the perturbative results of this module to the exact Rabi solution, and see precisely where perturbation theory breaks down.

Setup

Take levels $|0\rangle$ (energy $0$ ) and $|1\rangle$ (energy $\hbar\omega_0$ ) driven by a harmonic perturbation $\hat H'(t) = \hbar\Omega_R\cos(\omega t)\,\hat\sigma_x$ , where $\Omega_R$ — the Rabi frequency — measures the drive strength via the matrix element $\hbar\Omega_R = V_{10} = \langle 1|\hat V|0\rangle$ (so $\Omega_R = |V_{10}|/\hbar$ ). Writing $|\Psi\rangle = c_0(t)|0\rangle + c_1(t)|1\rangle$ , the interaction-picture equations couple the two amplitudes.

Perturbative result and its limit

The first-order (perturbative) RWA result of the previous lessons, written in terms of $\Omega_R = |V_{10}|/\hbar$ , is

P_{0\to 1}^{(1)}(t) = \frac{|V_{10}|^2}{4\hbar^2}\,\frac{\sin^2(\Delta t/2)}{(\Delta/2)^2} = \frac{\Omega_R^2}{\Delta^2}\,\sin^2\!\Big(\tfrac{\Delta t}{2}\Big) \;\xrightarrow{\;\Delta=0\;}\; \Big(\tfrac{\Omega_R t}{2}\Big)^2 ,

reproducing the quadratic short-time growth. It is trustworthy only for weak coupling, $\Omega_R \ll |\Delta|$ , and short times. But for a two-level system we can solve the RWA equations exactly — there is no continuum to escape into, so probability is conserved between just two states.

The exact Rabi solution

Solving the RWA two-level problem exactly gives the Rabi formula, with population oscillating at the generalized Rabi frequency $\Omega = \sqrt{\Delta^2 + \Omega_R^2}$ :

P_{0\to 1}(t) = \frac{\Omega_R^2}{\Delta^2 + \Omega_R^2}\, \sin^2\!\Big(\tfrac{1}{2}\sqrt{\Delta^2+\Omega_R^2}\,t\Big) .

The first-order result above is exactly this with $\Omega_R^2$ neglected next to $\Delta^2$ . On resonance ( $\Delta = \omega_0 - \omega = 0$ ), the exact populations oscillate cleanly:

\boxed{\;P_{0\to 1}(t) = \sin^2\!\Big(\frac{\Omega_R t}{2}\Big),\qquad P_{0\to 0}(t) = \cos^2\!\Big(\frac{\Omega_R t}{2}\Big)\;}

This is Rabi flopping: population sloshes back and forth between the two levels, fully and reversibly. The perturbative $(\Omega_R t/2)^2$ is just the small-angle expansion of $\sin^2(\Omega_R t/2)$ — perturbation theory is the leading edge of the true sinusoid, valid only while $\Omega_R t \ll 1$ .

Pulse area and qubit gates

The quantity $\theta = \Omega_R t$ is the pulse area. On resonance the drive rotates the Bloch vector by $\theta$ about an axis in the equatorial plane — exactly an $R_Y(\theta)$ (or $R_X(\theta)$ , depending on the drive phase) rotation:

R_Y(\theta)\,|0\rangle = \cos\tfrac{\theta}{2}\,|0\rangle + \sin\tfrac{\theta}{2}\,|1\rangle .

A $\pi$ -pulse ( $\theta=\pi$ ) fully inverts the qubit: $|0\rangle\to|1\rangle$ . This is the NOT gate.
A $\pi/2$ -pulse ( $\theta=\pi/2$ ) creates the equal superposition $(|0\rangle+|1\rangle)/\sqrt 2$ — a half-flop. This is how real hardware builds Hadamard-like operations.

Why perturbation theory fails for a strong drive

First-order theory predicts $P\propto t^2$ without bound, which is nonsense once $P$ nears 1. The exact solution saturates at $P=1$ and turns around — the population returns to $|0\rangle$ after a full period. Perturbation theory simply cannot see the turnaround because it keeps only the lowest power of the drive. The two-level system is the cleanest place to witness this: short pulses agree with perturbation theory, long pulses reveal the full, bounded Rabi oscillation.

Try it

The simulator represents the qubit as a single-qubit statevector starting in $|0\rangle$ . Apply a resonant $\pi/2$ -pulse using the $R_Y$ rotation to produce the half-flopped state $(|0\rangle+|1\rangle)/\sqrt 2$ , then press Check.

Run your code to see the quantum state.