Resonant driving and the rotating frame · Quantum Mechanics

Intuition

A real drive is an oscillating field, $\hbar\Omega\cos\omega t\,\sigma_{x}$ , on an atom whose levels $\hbar\omega_{0}$ apart already make its state turn fast about $z$ . Go to a frame that turns about $z$ at the drive’s own frequency $\omega$ , as a camera on a carousel would. In that frame the atom’s own turning is reduced to the difference $\omega_{0}-\omega$ , and the drive becomes a steady part plus a part turning at twice the drive’s frequency. When the drive is weak against the frequencies involved, that fast part averages to nothing and can be dropped — the rotating-wave approximation. What is left is a two-level Hamiltonian that does not depend on time: the drive’s strength as a coupling, the detuning as a bias.

Seen from a merry-go-round turning at the same speed, a rider opposite you seems to sit still and anything slightly faster drifts slowly. The rotating frame turns with the drive, so only the slow mismatch and the drive’s steady push remain.

The upper level’s probability for an atom driven at $\omega=\omega_{0}$ with $\Omega=\omega_{0}/10$ , against . The solid curve is computed exactly, with the drive’s fast part kept; the dashed one is the rotating-wave approximation, . The exact curve wiggles about the smooth one at twice the drive frequency, never more than 0.03 away.

The rotating frame

For $\hat{H}=\tfrac12\hbar\omega_{0}\sigma_{z}+\hbar\Omega\cos\omega t\,\sigma_{x}$ , in the basis , set :

\begin{aligned}\hat{H}_{R}&\approx\frac{\hbar}{2}\bigl(-\delta\,\sigma_{z}+\Omega\,\sigma_{x}\bigr)\\ \delta&=\omega-\omega_{0}\end{aligned}

Properties

$\delta$ , the detuning, is how far the drive is from the atom’s own frequency.
The rotating-wave approximation drops terms turning at $2\omega$ ; it holds when $\Omega$ and $|\delta|$ are small against $\omega$ .
The leftover fast terms shift the resonance slightly, by about (the Bloch–Siegert shift, stated).

The rotating-wave Hamiltonian

\begin{aligned}\hat{H}&=\tfrac12\hbar\omega_{0}\sigma_{z}+\hbar\Omega\cos\omega t\,\sigma_{x}\\ \Longrightarrow\ \hat{H}_{R}&\approx\frac{\hbar}{2}\bigl(-\delta\sigma_{z}+\Omega\sigma_{x}\bigr)\end{aligned}

Substitute the turned state into Schrödinger’s equation: the frame’s turning subtracts $\tfrac12\hbar\omega\sigma_{z}$ , and conjugating $\sigma_{x}$ by the turning gives . Multiplied by the drive’s cosine this is half of plus terms at , which are dropped.

Proof steps

\hat{H}_{R}=\hat{U}^{\dagger}\hat{H}\hat{U}-i\hbar\,\hat{U}^{\dagger}\dot{\hat U},\qquad \hat{U}=e^{-i\omega t\sigma_{z}/2}

Schrödinger’s equation for $\ket{\psi_{R}}=\hat{U}^{\dagger}\ket{\psi}$ .

-i\hbar\,\hat{U}^{\dagger}\dot{\hat U}=-\tfrac12\hbar\omega\,\sigma_{z}

The frame’s own turning.

\hat{U}^{\dagger}\sigma_{x}\hat{U}=\sigma_{x}\cos\omega t-\sigma_{y}\sin\omega t

$\sigma_{x}$ seen from the turning frame.

\begin{aligned}&\cos\omega t\,(\sigma_{x}\cos\omega t-\sigma_{y}\sin\omega t)\\ &\quad=\tfrac12\sigma_{x}+\tfrac12(\sigma_{x}\cos2\omega t-\sigma_{y}\sin2\omega t)\end{aligned}

Products of cosines and sines at $\omega$ .

\hat{H}_{R}\approx\tfrac12\hbar(\omega_{0}-\omega)\sigma_{z}+\tfrac12\hbar\Omega\,\sigma_{x}

Drop the terms at $2\omega$ , which average to nothing when $\Omega\ll\omega$ .

Applications

Practice

Turning With the Drive

In a frame that turns about $z$ at the drive’s frequency, the atom’s own fast turning is reduced to the mismatch, and the drive looks almost steady.

\ket{\psi}=e^{-i\omega t\sigma_{z}/2}\ket{\psi_{R}}

Try it

Why go to a frame that turns with the drive?