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\title{Supplementary online material to the manuscript:\\`` Resolved atomic interaction sidebands in an optical clock transition'' }

\author
{ M. Bishof$^{1}$, Y. Lin$^{1}$, M. D. Swallows$^{1}$, M. J. Martin$^{1}$,  A. V. Gorshkov$^{2}$, J. Ye$^{1}$ and A. M. Rey$^{1}$} %\ast}$ }
\affiliation{$^{1}$JILA, NIST, Department of Physics, University of Colorado, 440, UCB, Boulder, CO 80309, USA}
 \affiliation{$^{2}$Institute for Quantum Information, California Institute of Technology, Pasadena, CA 91125, USA}
\maketitle

 %\affiliation{$^\ast$To whom correspondence should be addressed; E-mail: arey@jilau1.colorado.edu}
\section{Interaction parameters}
We consider    nuclear-spin polarized fermionic atoms with two different electronic degrees of freedom ($e,g$) confined in a harmonic trap with trapping frequencies $\omega_{X,Y,Z}$ and assume the atoms interact only via s-wave interactions. The  matrix element that describes  collisions between two atoms ($i = 1, 2$) in the $\textbf{n}_{i} = (n_{iX},n_{iY},n_{iZ})$ harmonic oscillator eigenmodes $\phi_{n_{iX}}^{X}(X) \phi_{n_{iY}}^{Y}(Y) \phi_{n_{iZ}}^{Z}(Z)$ is given by
%The  matrix element that describes  collisions between two atoms in the  $\{\textbf{n}_1,\textbf{n}_2\}$  harmonic oscillator eigenmodes $\phi_{n_{iX}}^{X}(X)$, $\phi_{n_{iY}}^{Y}(Y)$, and $\phi_{n_{iZ}}^{Z}(Z)$ [here $i=1, 2$ and $\textbf{n}_{i} = (n_{iX},n_{iY},n_{iZ})$] is given by
\begin{eqnarray}
%U_{\bf{n}_1,\bf{n}_2} &=&\frac{8 \pi \hbar a_{eg}^-}{m}\int  d^3\mathbf{R} \prod_{i=X,Y,Z} [\phi_{n_{1\mathbf{R}_i}}^{\mathbf{R}_i}(\mathbf{R}_i)]^2 [\phi_{n_{2\mathbf{R}_i}}^{\mathbf{R}_i}(\mathbf{R}_i)]^2 \notag\\
U_{\bf{n}_1,\bf{n}_2} &=&\frac{8 \pi \hbar a_{eg}^-}{m}\int  d^3\mathbf{R} \prod_{\mu =X,Y,Z} [\phi_{n_{1\mu}}^{\mu}(\mu)]^2 [\phi_{n_{2\mu}}^{\mu}(\mu)]^2 \notag\\
& = & u I_{n_{1X},n_{2X}}I_{n_{1Y},n_{2Y}} I_{n_{1Z},n_{2Z}},
\end{eqnarray}
%with
where $\mathbf{R} = (X,Y,Z)$, $u = 4  a_{eg}^- \sqrt{\frac{m \omega_X \omega_Y \omega_Z}{h}}$, %$I_{n_1,n_2}=\frac{\sqrt{2 \pi} \int e^{-2\xi^2 }H_{n_1}^2(\xi) H_{n_2}^2(\xi) d\xi}{2^{n_1+n_2} (n_1!) (n_2!)}$
$I_{n_1,n_2}=\frac{\sqrt{2} \int e^{-2 z^2 }H_{n_1}^2(z) H_{n_2}^2(z) d z}{\sqrt{\pi} 2^{n_1+n_2} (n_1!) (n_2!)}$,
%$I_{n_1,n_2}=\sqrt{2} \int e^{-2 z^2 }H_{n_1}^2(z) H_{n_2}^2(z) d z/\left[\sqrt{\pi} 2^{n_1+n_2} (n_1!) (n_2!)\right]$,
and %$H_{n}(\xi)$
$H_{n}(z)$  are Hermite polynomials.

At finite temperature, expectation values need to be calculated by averaging  over all possible  combinations of modes
weighted  according to their  Boltzmann factor:  $\langle U\rangle_{T} \approx u  \vartheta(\frac{\hbar\omega_X }{ k_B T_X})\vartheta(\frac{\hbar\omega_Y }{ k_B T_Y})\tilde \vartheta(\frac{\hbar\omega_Z }{ k_B T_Z})$ with
%$\vartheta(\alpha)\equiv\left( \frac{\sum_{n_{1},n_{2}}I_{{n_1},{n_2}} e^{-\alpha (n_1+n_2 )}}{\sum_{n_{1},n_{2}} e^{-\alpha(n_1+n_2) }}\right)$
$\vartheta(\alpha)\equiv \frac{\sum_{n_{1},n_{2}}I_{{n_1},{n_2}} e^{-\alpha (n_1+n_2 )}}{\sum_{n_{1},n_{2}} e^{-\alpha(n_1+n_2) }}$
and %$\tilde \vartheta(\alpha)\equiv\left( \frac{\sum_{n_{1}>n_{2}}I_{{n_1},{n_2}} e^{-\alpha (n_1+n_2 )}}{\sum_{n_{1}>n_{2}} e^{-\alpha(n_1+n_2) }}\right)$.
$\tilde \vartheta(\alpha)\equiv \frac{\sum_{n_{1}>n_{2}}I_{{n_1},{n_2}} e^{-\alpha (n_1+n_2 )}}{\sum_{n_{1}>n_{2}} e^{-\alpha(n_1+n_2) }}$. The $\tilde \vartheta$ function is introduced since in our experiment  atoms start polarized in $g$, and  even at $T=0$  they  are  required to occupy  different modes  along $\hat Z$ to satisfy the Pauli exclusion principle.

For $\frac{\hbar\omega_Z }{ k_B T_Z} \ll 1$, eigenmodes with  $|n_{1Z}- n_{2Z}|\gg 1$ dominate, so that
%making Eq.\ (S2) valid for $I_{n_{1Z},n_{2Z}}$. For $|n_{1}- n_{2}| > 1$, to a good approximation, %to a very good approximation, After a close analysis of   $I_{n_1,n_{2}}$ one can show that for $|n_{1}- n_{2}| \gg 1$, to a very good approximation,
\begin{eqnarray}
I_{n_{1Z},n_{2Z}}\approx 2 \frac{K\left[\frac{1}{2} \left(1-\frac{n_{1Z}+n_{2Z}}{|n_{1Z}-n_{2Z}|}\right)\right]}{\pi\sqrt{ \pi |n_{1Z}-n_{2Z}|}},
\end{eqnarray}
where $K$ is the complete elliptic integral of the first kind. %EllipticK  Integral function.
Furthermore, we checked numerically that, for our parameters, ISB is not significantly perturbed by approximating $K$ with $K(0)$, so that %$I_{|n_1- n_2|\gg 1}\approx \frac{1}{\sqrt{\pi }  \sqrt{|n_1-n_2|}}$
 $I_{n_{1Z},n_{2Z}} \approx \frac{1}{\sqrt{\pi }  \sqrt{|n_{1Z}-n_{2Z}|}}$.
 %Since for $\frac{\hbar\omega_Z }{ k_B T_Z}<1$ terms with $|n_1- n_2|\gg 1$ are dominant one can use this approximation to show that
%Using this approximation for $I_{n_{1Z},n_{2Z}}$,
Using this approximation, one can show that
$\vartheta(\alpha \ll 1)\sim\tilde \vartheta(\alpha \ll 1)\sim \sqrt{\alpha}$.

The (absolute value of the) mean interaction energy per particle $E_{\rm int}$ is
$(N-1)|\langle U\rangle_{T}|/2$, where
$N$ is the number of atoms in the trap.


\section{Interaction sidebands (ISB) at finite Temperature  for two atoms (N=2)}
Here, we derive Eq.\ (3) in the main text by performing a thermal average over different axial modes.  First, consider a given set of populated modes along $\hat Z$, $\vec{n}=(n_1,n_2)$. If the atoms are in the strongly interacting regime,
$|{U}_{\vec{n}}| \gg \bar{\Omega}_{\vec{n}}$, the ISB  is given by
\begin{eqnarray}
 N^{e}_{\vec{n}}(t,\delta) &\approx&  f(t,\delta -U_{\vec{n}}, \Delta{\Omega}_{\vec{n}}) \label{R}
\end{eqnarray}
with $f(t,\delta, y)\equiv  \frac{y^2}{y^2+\delta^2} \sin^2\left (\frac{t \sqrt{ y^2+\delta^2}}{2}\right)$.

At finite temperature, we need to evaluate $\langle N^e \rangle_T = \frac{\sum_{n_{1}>n_{2}}N^{e}_{\vec{n}}(t,\delta)e^{-\alpha (n_1+n_2 )}}{\sum_{n_{1}>n_{2}} e^{-\alpha(n_1+n_2) }}$ with $\alpha= \frac{\hbar\omega_Z }{ k_B T_Z}$. %\hbar \omega_Z/(k_B T_Z)$.
 %is a complicated sum but
We evaluate $\langle  N^{e} \rangle_T$ under the approximations $
\Delta{\Omega}_{\vec{n}}\approx \frac{\Omega_B \eta_Z^2 (n_1-n_2)}{\sqrt 2}$ and $U_{\vec{n}}\approx \vartheta(\frac{\hbar\omega_X }{ k_B T_X})\vartheta(\frac{\hbar\omega_Y }{ k_B T_Y})\frac{u}{ \sqrt{\pi|n_1-n_2|}}\equiv  \frac{\xi}{\sqrt{\alpha|n_1-n_2|}} $.


To carry out the sum,  we  treat $\alpha (n_1-n_2)\equiv x$ and $\alpha (n_1+n_2) \equiv y$ as continuous variables  (valid if  $\alpha \ll 1$). We also define $\kappa(x)=\frac{\Omega_B \eta_Z^2 x}{\alpha\sqrt 2}\equiv b x$ and $U(x)=\frac{\xi}{\sqrt{x}}$. Then  $\langle N^e \rangle_T\approx \int_0^{\infty} N^e(x/\alpha) e^{-x} dx$ with $N^e(x/\alpha)=\frac{\kappa^2( x)}{\kappa^2( x)^2+\delta^2} \sin^2\left (\frac{t \sqrt{ \kappa^2( x)+(\delta-U(x)) ^2}}{2}\right)$. Noticing that  $ N^{e}$ is sharply peaked at $x_0=(\xi/\delta)^2$, we evaluate $e^{-x}$ and $\kappa$ at $x = x_0$. We write $x = x_0 + x'$ and expand $U$ in $x'$ to first order.  We then define $p = |\xi| x'/(2 b x_0^{5/2})$, so that
\begin{eqnarray}
\langle N^e \rangle_T & \approx &e^{-x_0} \frac{2 b x_0^{5/2}}{|\xi|}\times \notag\\
&&\int_{-\infty}^{\infty} d p \frac{1}{1 + p^2} \sin^2 \left(c \sqrt{1 + p^2}\right),
\end{eqnarray}
where $c = t b x_0/2$. For $c \ll 1$, the integral in Eq.\ (S4) is approximately equal to $\pi c$. This yields  Eq.\ (3) in the main text.







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