= Trapped-ion sideband pulse model
{title2=$P_r(n,t)=\sin^2(\eta\Omega t\sqrt n/2)$}
Consider an original ideal calculation with no heating, <decoherence>, detuning or spectator modes. In the ordered basis $(|g,n\rangle,|e,n-1\rangle)$ for $n\ge1$, the red <motional sideband> has $H_r=\hbar\eta\Omega\sqrt n\,\sigma_x/2$. Here $\sigma_x$ is the <Pauli matrix> exchanging the two basis states. Since $\sigma_x^2=I$, its <unitary time evolution> is
$$
U_n(t)=\cos\frac{\theta\sqrt n}{2}\,I-i\sin\frac{\theta\sqrt n}{2}\,\sigma_x,\qquad
\theta=\eta\Omega t.
$$
Thus $U_n^\dagger U_n=I$, and excitation from $|g,n\rangle$ 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 $t_\pi=\pi/(\eta\Omega)=62.5\ \mathrm{\mu s}$. Let motion have the <Boltzmann distribution> of a <quantum harmonic oscillator>, with chosen mean occupation $\bar n=0.25$:
$$
p_n=(1-r)r^n,\qquad r=\frac{\bar n}{1+\bar n}=0.2,\qquad
P_r(t)=\sum_{n=0}^{\infty}p_n\sin^2\frac{\theta\sqrt n}{2},\qquad
P_b(t)=\sum_{n=0}^{\infty}p_n\sin^2\frac{\theta\sqrt{n+1}}{2}.
$$
Here $\sum p_n=1$, $\sum np_n=\bar n$ and $p_0=0.8$. Reindex the red sum using $n=m+1$ and $p_{m+1}=rp_m$: $P_r(t)=rP_b(t)$ 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.
\Image[/trapped-ion-thermal-sideband-pulse.png]
{title=Chosen thermal motion and ideal red/blue sideband pulse responses}
{description=Original calculation with stipulated parameters, not experimental data. Both responses average the same initial thermal number-state distribution. The vertical line marks a red pi pulse for the $n=1$ component; a thermal mixture has no single common pi-pulse time.}
The chosen pulse gives $P_r\simeq0.181393$ and $P_b\simeq0.906967$. 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 $n=0,\ldots,79$ omits <probability> $r^{80}<1.21\times10^{-56}$; because every sine squared lies between zero and one, this also bounds each omitted response. The thermal relation supplies a separate dimensional check:
$$
\bar n=\frac{1}{e^{\hbar\nu/(k_BT)}-1},\qquad
T=\frac{\hbar\nu}{k_B\log(1+1/\bar n)}\simeq59.6\ \mathrm{\mu K}.
$$
Here $k_B$ is the <Boltzmann constant>. Substitution recovers $\bar n=0.25$; the thermal root-mean-square expansion parameter is $\eta\sqrt{2\bar n+1}\simeq0.098$. 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.
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