Detuning · Quantum Mechanics

Intuition

Drive the atom a little off its frequency and the Rabi oscillation changes in two ways. It no longer reaches the upper level: the largest probability is $\Omega^{2}/(\Omega^{2}+\delta^{2})$ . And it swings faster, at the generalised Rabi frequency $\sqrt{\Omega^{2}+\delta^{2}}$ . On the Bloch sphere the state turns about a tilted axis, so it never reaches the south pole. Swept through the resonance, the largest probability traces a peak of half-width $\Omega$ : a strong drive broadens the line it drives.

Push a swing slightly out of step and the pushes help for a while, then hinder: the swing never gets as high as with pushes in step, and its rhythm of rising and falling is quicker. A detuned drive does the same to an atom.

The probability of the upper level against $\Omega t$ , on resonance and with the detuning $\delta=\Omega$ . Detuned, the largest value falls to $\tfrac12$ and the swing quickens to .

Rabi oscillations with detuning

With $\hat{H}_{R}=\tfrac12\hbar(-\delta\sigma_{z}+\Omega\sigma_{x})$ and the atom started in :

P_{e}(t)=\frac{\Omega^{2}}{\Omega^{2}+\delta^{2}}\,\sin^{2}\frac{\Omega't}{2},\qquad \Omega'=\sqrt{\Omega^{2}+\delta^{2}}

Properties

The largest probability, $\Omega^{2}/(\Omega^{2}+\delta^{2})$ , is 1 only on resonance and $\frac{}{}$ at .

The detuned Rabi formula

\begin{aligned}\hat{H}_{R}&=\tfrac12\hbar(-\delta\sigma_{z}+\Omega\sigma_{x})\\ \Longrightarrow\ P_{e}(t)&=\frac{\Omega^{2}}{\Omega'^{2}}\sin^{2}\frac{\Omega't}{2}\end{aligned}

Write the Hamiltonian as $\tfrac12\hbar\Omega'$ times $\mathbf{n}\cdot\vec{\sigma}$ with a unit vector . Its evolution operator is a cosine times the identity and a sine times . Only the sine term connects to , through the component of along .

Proof steps

\hat{H}_{R}=\tfrac12\hbar\Omega'\,\mathbf{n}\cdot\vec{\sigma},\qquad \mathbf{n}=\frac{(\Omega,0,-\delta)}{\Omega'}

Factor out the length of the field.

\hat{U}(t)=\cos\tfrac{\Omega't}{2}\,\hat{1}-i\sin\tfrac{\Omega't}{2}\,\mathbf{n}\cdot\vec{\sigma}

$(\mathbf{n}\cdot\vec{\sigma})^{2}=\hat{1}$ splits the exponential.

\bra{e}\hat{U}(t)\ket{g}=-i\sin\tfrac{\Omega't}{2}\,\bra{e}\mathbf{n}\cdot\vec{\sigma}\ket{g}=-i\sin\tfrac{\Omega't}{2}\,\frac{\Omega}{\Omega'}

Only $\sigma_{x}$ connects $\ket{g}$ to $\ket{e}$ .

P_{e}(t)=\frac{\Omega^{2}}{\Omega'^{2}}\sin^{2}\frac{\Omega't}{2}

Square the modulus.

Applications

Practice

Not All the Way

Off resonance the upper level is never reached completely: the largest probability is $\Omega^{2}$ over $\Omega^{2}+\delta^{2}$ .

P_{e}^{\max}=\frac{\Omega^{2}}{\Omega^{2}+\delta^{2}}

Try it

A drive has detuning $\delta=2\Omega$ . What is the largest probability of the upper level?