Rabi oscillations

Intuition

Shine light on an atom, or microwaves on a qubit, at exactly the frequency of its two levels — on resonance — and the atom does not just absorb: it goes up and comes down again, periodically, as long as the drive lasts. The probability of the upper level swings between 0 and 1 as $\sin^{2}(\Omega t/2)$ , where $\Omega$ , the Rabi frequency, grows with the strength of the drive. On the Bloch sphere the swing is a rotation about an axis in the equator. A pulse lasting half a period, a $\pi$ pulse, turns the lower level into the upper; a quarter period, a $\pi/2$ pulse, makes an equal superposition. These pulses are the gates of a quantum computer.

Push a child on a swing exactly at its own rhythm and each push adds; yet in this quantum version the pushes, once the swing is full, start taking energy out again, and the cycle repeats. The rhythm of filling and emptying is the Rabi frequency.

The probability of the upper level of a two-level atom driven exactly at its own frequency, started in the lower level, against $\Omega t$ : $\sin^{2}(\Omega t/2)$ . It reaches 1 at $\Omega t=\pi$ , the end of a pulse, and returns to 0 at .

Write the atom’s states as $\ket{e}$ , the upper, and $\ket{g}$ , the lower. Driven exactly at its own frequency, which is called driving on resonance, and seen in a frame that turns with the drive (the next lesson), the atom has the Hamiltonian

\hat{H}=\frac{\hbar\Omega}{2}\,\sigma_{x},\qquad P_{e}(t)=\sin^{2}\frac{\Omega t}{2}

Properties

$\Omega$ , the Rabi frequency, is proportional to the drive: $\hbar\Omega=|\bra{e}\hat{V}\ket{g}|$ for a drive — for an atom in light, the dipole element times the field’s amplitude.

Rabi oscillation on resonance

\hat{H}=\frac{\hbar\Omega}{2}\sigma_{x},\ \ \ket{\psi(0)}=\ket{g}\ \Longrightarrow\ P_{e}(t)=\sin^{2}\frac{\Omega t}{2}

The evolution operator is the exponential of $-i\Omega t\sigma_{x}/2$ . Because $\sigma_{x}$ squares to one, its series splits into a cosine times the identity and a sine times , which swaps the two states. Acting on the lower level, the sine term puts amplitude into the upper one.

Proof steps

\hat{U}(t)=e^{-i\Omega t\sigma_{x}/2}

The Hamiltonian does not depend on time.

e^{-i\varphi\sigma_{x}/2}=\cos\tfrac{\varphi}{2}\,\hat{1}-i\sin\tfrac{\varphi}{2}\,\sigma_{x}

$\sigma_{x}^{2}=\hat{1}$ splits the exponential series into even and odd powers.

\hat{U}(t)\ket{g}=\cos\tfrac{\Omega t}{2}\ket{g}-i\sin\tfrac{\Omega t}{2}\ket{e}

$\sigma_{x}$ swaps $\ket{g}$ and $\ket{e}$ .

P_{e}(t)=\sin^{2}\frac{\Omega t}{2}=\tfrac12\bigl(1-\cos\Omega t\bigr)

The squared amplitude of $\ket{e}$ .

Applications

Practice

The Swing

Driven at its own frequency, a two-level atom started in the lower level has the upper level with probability $\sin^{2}(\Omega t/2)$ .

P_{e}(t)=\sin^{2}\frac{\Omega t}{2}

Try it

What is $P_{e}$ at $\Omega t=\pi/2$ ?