Motional sideband (${\omega }_{\mathrm{drive}}={\omega }_0,{\omega }_0-\nu ,{\omega }_0+\nu$)

A motional sideband is a transition that changes internal excitation and a motional quantum number together. Define the internal energy gap as $\hslash {\omega }_0>0$. A carrier drives $|g,n⟩\leftrightarrow |e,n⟩$; the first red sideband drives $|g,n⟩\leftrightarrow |e,n-1⟩$; the first blue sideband drives $|g,n⟩\leftrightarrow |e,n+1⟩$. Their resonant drive angular frequencies follow by subtracting these energy levels. For a Raman transition, the drive is a beam-difference frequency with a sign fixed by the selected internal transition.

Choose a real carrier Rabi angular frequency $\Omega$ defined so that resonant carrier excitation is ${\sin }^2(\Omega t/2)$. In the Lamb-Dicke regime, isolated first-sideband Hamiltonians in a rotating frame can be written, after choosing the pulse phase, asThese formulas use a rotating-wave approximation: neglected carrier and other-sideband terms oscillate rapidly compared with the retained coupling. The drive's effective width and sideband coupling must be small compared with $\nu$; sufficient suppression of off-resonant carrier excitation must also be checked. The red interaction is a Jaynes-Cummings model with a motional quantum replacing a cavity photon. The blue interaction creates internal excitation and a motional quantum together; the external laser supplies energy. The driven-system rotating-frame Hamiltonian alone is not the total laboratory energy.

Consider an original ideal calculation with no heating, decoherence, detuning or spectator modes. In the ordered basis $(|g,n⟩,|e,n-1⟩)$ for $n\ge 1$, the red motional sideband has $H_r=\hslash \eta \Omega \sqrt{n}{\sigma }_x/2$. Here ${\sigma }_x$ is the Pauli matrix exchanging the two basis states. Since ${\sigma }_x^2=I$, its unitary time evolution isThus $U_n^{†}{U}_n=I$, and excitation from $|g,n⟩$ has probability $P_r(n,t)={\sin }^2(\theta \sqrt{n}/2)$. For $n=0$ it vanishes. The blue motional sideband similarly gives $P_b(n,t)={\sin }^2(\theta \sqrt{n+1}/2)$. A red Rabi pi pulse calibrated for $n=1$ has $\theta =\pi$; it is not a Rabi pi pulse for every occupied $n$.

Choose $\eta =0.08$, $\Omega /(2\pi )=100\mathrm{kHz}$ and $\nu /(2\pi )=2\mathrm{MHz}$. The $n=1$ red Rabi pi pulse lasts $t_{\pi }=\pi /(\eta \Omega )=62.5\mu \mathrm{s}$. Let motion have the Boltzmann distribution of a quantum harmonic oscillator, with chosen mean occupation $\bar{n}=0.25$:Here $\sum p_n=1$, $\sum n{p}_n=\bar{n}$ and $p_0=0.8$. Reindex the red sum using $n=m+1$ and $p_{m+1}=r{p}_m$: $P_r(t)=r{P}_b(t)$ for equal resonant pulse parameters, even outside the short-pulse limit. At times with $P_b>0$, the ratio $R=P_r/P_b$ therefore gives $\bar{n}=R/(1-R)$. A nonthermal distribution, unequal drive strengths, detuning or state-detection bias can invalidate that thermometer. Ground-state population cannot generally be deduced from the mean alone without specifying the distribution.

The chosen pulse gives $P_r\simeq 0.181393$ and $P_b\simeq 0.906967$. These are internal excitation probabilities, not quantum gate fidelities or cooling-cycle efficiencies. An independent matrix exponentiation of each two-state block agrees with the sine formula and preserves normalization. Summing $n=0,\dots ,79$ omits probability $r^{80}<1.21\times 10^{-56}$; because every sine squared lies between zero and one, this also bounds each omitted response. The thermal relation supplies a separate dimensional check:Here $k_B$ is the Boltzmann constant. Substitution recovers $\bar{n}=0.25$; the thermal root-mean-square expansion parameter is $\eta \sqrt{2\bar{n}+1}\simeq 0.098$. This checks the populated part of the Lamb-Dicke regime, while exponentially rare large $n$ still exceed any fixed small-parameter criterion. The pulse is a leading-order model; it does not bound corrections from the full coupling or experimental noise.