\section{Cumulants at arbitrary density}
\subsection{Law of the distance between two particles}
We consider two tagged particles (TPs) in the SEP with density $\rho$. The initial distance
between them is $L$ and we derive the equilibrium distribution of the distance
(in particular we show that the distribution of the distance is time-independant at large time).
We proceed as follow: (i) We write the law of the number of particles $N_p$ between the TPs.
This number is fixed initially and does not evolve. (ii) We write the law of the number
of vacancies $N_v$ between the tracers at equilibrium; this law depends on $N_p$.

Initially there are $L-1$ sites between the TPs. Each site is occupied with probability $\rho$. This gives us a binomial law for $N_p$:
\begin{equation}
 \mathbb{P}(N_p = k) = \binom{L-1}{k} \rho^k (1-\rho)^{L-1-k}
\end{equation}

At large time, the number of vacancies between the two TPs, knowing that there
are $k$ particles between them, is given by
the law of the number $m$ of failures before $k+1$ successes in a game in which the probability of a success is $\rho$.
It is a negative binomial law:
\begin{equation}
 \mathbb{P}(N_v = m|N_p = k) = \binom{m+k}{m} (1-\rho)^m \rho^{k+1}
\end{equation}

The distance $D$ between the tracers is given by $D=N_p + N_l + 1$, its law is:
\begin{align}
 \mathbb{P}(D=\delta) &= \sum_{k=0}^{L-1} \mathbb{P}(N_p = k)\mathbb{P}(N_v = \delta-k-1|N_p = k) \\
 &= \sum_{k=0}^{L-1}\binom{L-1}{k} \rho^k (1-\rho)^{L-1-k}
 \binom{\delta-1}{\delta-k-1} (1-\rho)^{\delta-k-1} \rho^{k+1} \\
 &= \sum_{k=0}^{L-1}\binom{L-1}{k} \binom{\delta-1}{k} \rho^{2k+1} (1-\rho)^{L+\delta-2k-2}
 \label{si:eq:predDist}
\end{align}
This law is in very good agreement with the numerical simulations (Fig.~\ref{si:fig:dist}).

\subsection{Large deviations of the law of the distance}
From the law of the distance \eqref{si:eq:predDist}, one can derive the generating function
$G_D(z)$.
\begin{align}
 G_D(z) &\equiv \sum_{\delta=1}^\infty \mathbb{P}(D=\delta) z^\delta \\
 &= \sum_{\delta=1}^\infty z^\delta \sum_{k=0}^{L-1} \mathbb{P}(N_p = k)\mathbb{P}(N_v = \delta-k-1|N_p = k) \\
 &= z \sum_{k=0}^{L-1} z^k \mathbb{P}(N_p = k) \sum_{m=0}^\infty z^m \mathbb{P}(N_v = m|N_p = k) \\
 &= z \sum_{k=0}^{L-1} z^k \binom{L-1}{k} \rho^k (1-\rho)^{L-1-k}
 \sum_{m=0}^\infty  \binom{m+k}{m} (1-\rho)^m \rho^{k+1} \\
 &= z \sum_{k=0}^{L-1} z^k \binom{L-1}{k} \rho^k (1-\rho)^{L-1-k}
 \left(\frac{\rho}{1-(1-\rho)z}\right)^{k+1} \\
 &= \left(\frac{\rho z}{1-(1-\rho)z}\right) \left(\frac{\rho^2 z}{1-(1-\rho)z} + 1-\rho\right)^{L-1}
\end{align}

We can then derive a large deviation scaling in the limit $L\to\infty$:
\begin{equation}
 \frac{1}{L} \ln G_D(e^t) \xrightarrow[N\to\infty]{} \ln\left(\frac{\rho^2 z}{1-(1-\rho)z} + 1-\rho\right)
 \equiv \phi(t)
\end{equation}
From the Gärtner-Ellis theorem \cite{Touchette:2009}, this implies:
\begin{align}
 P(D=L(1+\tilde d)) &\asymp e^{-LI(\tilde d)} \\
 I(\tilde d) &= \text{sup}_{t\in\mathbb{R}} \(t(1+\tilde d) - \phi(t)\)
 \label{si:eq:SI_extremum}
\end{align}

We now consider the high-density limit: $\rho = 1-\rho_0$ with $\rho_0 \ll 1$. We obtain:
\begin{equation}
 \phi(t) = t + 2\rho_0 \cosh
\end{equation}
Thus, the supremum in \eqref{si:eq:SI_extremum} is at $t^\ast$ such that:
\begin{equation}
 \sinh t^\ast = \frac{\tilde d}{2\rho_0}
\end{equation}
At the end of the day,
\begin{equation}
 I(\tilde d) = 2\rho_0\left\{
 1-\sqrt{1+\(\frac{\tilde d}{2\rho_0}\)^2}
 + \frac{\tilde d}{2\rho_0} \log\(\frac{\tilde d}{2\rho_0} + \sqrt{1+\(\frac{\tilde d}{2\rho_0}\)^2}\)
 \right\}
\end{equation}
This is exactly what is found with our high-density approach: Eq.~\eqref{eq:probDist}
of the main text.

\subsection{Large time behavior of the cumulants of $N$ particles}
We consider $N$ TPs having displacements $Y_1, \dots Y_N$.
We know that the moments of a single particle scale as $t^{1/2}$ \cite{Imamura:2017}
while the moments of the distance scale as $t^0$ (previous section).
\begin{align}
 \langle Y_1^{2p} \rangle &= \mathcal{O}(t^{1/2})\qquad \forall p \in \mathbb{N} \label{si:eq:SI_X1} \\
 \langle (Y_i - Y_1)^{2p} \rangle &= \mathcal{O}(t^0)\qquad \forall i \leq N, \forall p \in \mathbb{N}
 \label{si:eq:SI_dist}
\end{align}

From this we want to show that
\begin{equation} \label{si:eq:SI_HR}
 A_{p_1, \dots p_N}^{(N)} \equiv \langle Y_1^{p_1}\dots Y_N^{p_N} \rangle
 - \langle Y_1^{p_1 + \dots + p_N} \rangle = \mathcal{O}(t^{1/4}) \qquad
 \forall p_1, \dots, p_N
\end{equation}

We proceed by induction: the case $N=1$ is straightforward. Now assuming that
(\ref{si:eq:SI_HR}) holds for a given $N$, we want to prove it for $N+1$.
As we have $A_{p_1, \dots p_N, 0}^{(N+1)} = A_{p_1, \dots p_N}^{(N)} = \mathcal{O}(t^{1/4})$,
we prove by induction that
$A_{p_1, \dots p_N, q}^{(N+1)}= \mathcal{O}(t^{1/4})\ \forall q\geq 0$.
Indeed, if $A_{p_1, \dots p_N, q'}^{(N+1)}= \mathcal{O}(t^{1/4})\ \forall q'<q$, we can
write:
\begin{align}
 &A_{p_1, \dots p_N, q}^{(N+1)} =
 \langle Y_1^{p_1}\dots Y_N^{p_N}X_{N+1}^q \rangle - \langle Y_1^{p_1 + \dots + p_N + q} \rangle \\
  &= \left\langle Y_1^{p_1}\dots Y_N^{p_N}\left[(Y_{N+1}-Y_1)^q
  -\sum_{r=1}^q \binom{q}{r} (-1)^r Y_1^r Y_{N+1}^{q-r}\right] \right\rangle
  - \langle Y_1^{p_1 + \dots + p_N + q} \rangle \\
  &= \langle Y_1^{p_1}\dots Y_N^{p_N}(Y_{N+1}-Y_1)^q \rangle
  -\sum_{r=1}^q \binom{q}{r} (-1)^r \langle Y_1^{p_1+r} Y_2^{p_2} \dots Y_{N+1}^{q-r}\rangle
  - \langle Y_1^{p_1 + \dots + p_N + q} \rangle \\
  &= \langle Y_1^{p_1}\dots Y_N^{p_N}(Y_{N+1}-X_1)^q \rangle
  -\sum_{r=1}^q \binom{q}{r} (-1)^r \left[ \langle Y_1^{p_1+r} Y_2^{p_2} \dots Y_{N+1}^{q-r}\rangle
  - \langle Y_1^{p_1 + \dots + p_N + q}\rangle \right]
\end{align}
All the terms in the sum are of order $\mathcal{O}(t^{1/4})$ from (\ref{si:eq:SI_HR})
and the first term can be bounded by the Cauchy–Schwarz inequality:
\begin{equation}
 \left|\langle Y_1^{p_1}\dots Y_N^{p_N}(Y_{N+1}-Y_1)^q \rangle\right| \leq
 \sqrt{\langle \(Y_1^{p_1}\dots Y_N^{p_N}\)^2 \rangle\langle (Y_{N+1}-Y_1)^{2q} \rangle}
 = \sqrt{\mathcal{O}(t^{1/2})\mathcal{O}(t^0)} = \mathcal{O}(t^{1/4})
\end{equation}
using (\ref{si:eq:SI_X1}), (\ref{si:eq:SI_dist}) and (\ref{si:eq:SI_HR}).
This ends the proof of (\ref{si:eq:SI_HR}).

This implies that if $p_1 + \dots + p_N$ is even:
\begin{equation}
  \langle Y_1^{p_1}\dots Y_N^{p_N} \rangle \equi{t\to\infty} \langle Y_1^{p_1 + \dots + p_N}
  \rangle = \mathcal{O}(t^{1/2})
\end{equation}
The moments of $N$ particles are equal, in the large time limit, are given by
the moments of a single particle.
This extends to the cumulants $\kappa^{(N)}_{p_1, \dots, p_N}$ defined from the second
characteristic function.
\begin{equation}
 \psi(k_1, \dots k_N) \equiv \sum_{p_1, \dots p_N} \kappa^{(N)}_{p_1, \dots, p_N}
  \frac{(ik_1)^{p_1} \dots (ik_N)^{p_N}}{p_1!\dots p_N!}
  \equiv \log\left[\sum_{p_1, \dots p_N} \langle Y_1^{p_1}\dots Y_N^{p_N} \rangle
  \frac{(ik_1)^{p_1} \dots (ik_N)^{p_N}}{p_1!\dots p_N!}\right]
\end{equation}
\begin{equation}
  \kappa^{(N)}_{p_1, \dots, p_N} \equi{t\to\infty} \kappa^{(1)}_{p_1 + \dots + p_N}
  \equi{t\to\infty} B_{p_1+\dots + p_N} \sqrt{t}
\end{equation}
The coefficient $B_p$ are computed in Ref. \cite{Imamura:2017}.

\begin{figure}
 \includegraphics[width=15cm]{figDist}
 \caption{Comparison of the numerical probability distribution of the distance of two TPs
 (dots)
 with the prediction \eqref{si:eq:predDist} (lines) for $\rho = 0.25, 0.5, 0.75$ and
 $L=5, 10, 20$. The average is performed other $10^6$ simulations at final time
 $t=2\cdot 10^4$ (we checked that this is enough for the convergence).}
 \label{si:fig:dist}
\end{figure}

\section{Detailed calculations in the high-density limit}
\subsection{Approximation and thermodynamic limit}
Let us consider a system of size $\mathcal{N}$ with $M$ vacancies
and denote by $\vec Y(t) = (X_i(t)-X_i^0)_{i=1}^N$
the vector of the displacements of the TPs.
The probability $P^{(t)} (\vec Y|\{Z_j\})$ of having displacements $\vec Y$ at
time $t$ knowning that the $M$ vacancies started at sites $Z_1\dots Z_M$ is exactly given by~:
\begin{equation}
 P^{(t)} (\vec Y|\{Z_j\}) =
  \sum_{\vec Y_1,\dots,\vec Y_M} \delta_{\vec Y, \vec Y_1+\dots+\vec Y_M}
  \mathcal{P}^{(t)} (\{\vec Y_j\}|\{Z_j\})
\end{equation}
where $\mathcal{P}^{(t)} (\{\vec Y_j\}|\{Z_j\})$ is the probability of displacement $\vec Y_j$ due to the vacancy $j$
for all $j$ knowing the initial positions of all the vacancies.

Assuming that in the large density large ($\rho\to 1$) the vacancies interact
independantly with the TPs, we can
link it to the probability $p^{(t)}_Z(\vec Y)$ that the tracers have moved by $\vec Y$ at time $t$
due to a single vacancy that was initially at site $Z$:
\begin{equation}
 \mathcal{P}^{(t)} (\{\vec Y_j\}|\{Z_j\}) \underset{\rho\to 1}{\sim}
  \prod_{j=1}^M p^{(t)}_{Z_j}(\vec Y_j)
\end{equation}
so that
\begin{equation}\label{si:eq:approx}
 P^{(t)} (\vec Y|\{Z_j\}) \underset{\rho\to 1}{\sim}
  \sum_{\vec Y_1,\dots,\vec Y_M}
  \delta_{\vec Y, \vec Y_1+\dots+\vec Y_M}
  \prod_{j=1}^M p^{(t)}_{Z_j}(\vec Y_j)
\end{equation}

We take the Fourier transform and we average over the initial positions of the vacancies:
\begin{equation}
\tilde p^{(t)}(\vec k) \equiv \frac{1}{\mathcal{N}-N}
\sum_{Z \notin \{X_i^0\}} \sum_{\vec Y} p^{(t)}_Z(\vec Y) e^{i\vec k\vec Y}
\end{equation}
and mutatis mutandis for $\tilde P^{(t)} (\vec k)$.
Furthermore we write $\tilde p_Z^{(t)}(\vec k) = 1 + \tilde q_Z^{(t)}(\vec k)$ ($q_Z$ corresponds to the deviation
from a Dirac centered in $0$).
\begin{equation}
 \tilde P^{(t)} (\vec k) \underset{\rho\to 1}{\sim} \left[\tilde p^{(t)}(\vec k)\right]^M
 = \left[\frac{1}{\mathcal{N}-N}\sum_{Z\notin \{X_i^0\}}\tilde p_Z^{(t)}(\vec k)\right]^M
 = \left[1 + \frac{1}{\mathcal{N}-N}\sum_{Z\notin \{X_i^0\}} \tilde q_Z^{(t)}(\vec k)\right]^M
\end{equation}

We now take the limit $\mathcal{N},M\to\infty$
with $\rho_0 \equiv 1-\rho = M / \mathcal{N}$ (density of vacancies) remaining constant.
The second characteristic function reads:

\begin{equation} \label{si:eq:psi1}
 \lim_{\rho_0\to 0} \frac{\psi^{(t)}(\vec k)}{\rho_0} \equiv
 \lim_{\rho_0\to 0} \frac{\ln \left[\tilde P^{(t)}(\vec k)\right]}{\rho_0}
 = \sum_{Z \notin \{X_i^0\}} \tilde q_Z^{(t)}(\vec k)
\end{equation}

\subsection{Expression of the single-vacancy propagator}
One can partition over the first passage of the vacancy
to the site of one of the tracers to get an expression for $\tilde q_Z$ which is the main quantity involved
in (\ref{si:eq:psi1}):
\begin{align}
 p_Z^{(t)}(\vec Y) &= \delta_{\vec Y, \vec 0}  \(1 - \sum_{j=0}^t \sum_{\nu = \pm 1, \pm 2} F^{(j)}_{\nu, Z}\)
 + \sum_{j=0}^t \sum_{\nu = \pm 1, \pm 2} p_{-\nu}^{(t-j)}(\vec Y) F^{(j)}_{\nu, Z} \\
 \tilde p_Z^{(t)}(\vec k) &= 1 - \sum_{j=0}^t \sum_{\nu = \pm 1, \pm 2}
 F^{(j)}_{\nu, Z} + \sum_{j=0}^t \sum_{\nu = \pm 1, \pm 2}
 \tilde p_{-\nu}^{(t-j),\zeta} (\vec k) F^{(j)}_{\nu, Z} \\
\label{si:eq:qAll}
 \tilde q_Z^{(t)}(\vec k) &= - \sum_{j=0}^t \sum_{\nu = \pm 1, \pm 2}
 \left[1 - \(1+\tilde q_{-\nu}^{(t-j), \zeta} (\vec k)\) F^{(j)}_{\nu, Z}\right]
\end{align}
An exponant $\zeta$ to a quantity means that this quantity
is computed taking into account $\zeta = \zeta(Z)$.

We now need an expression for $\tilde q_\eta^\zeta$ where $\eta$ is a special site.
To do so we decompose the propagator of the displacements over the successive passages of the vacancy to
the position of one of the tracers:
\begin{multline} \label{si:eq:decompo}
 p^{(t),\zeta}_\eta (\vec Y) =
 \delta_{\vec Y,\vec 0} \(1 - \sum_{j=0}^t \sum_\mu F^{(j),\zeta}_{\mu, \eta}\) \\
  + \sum_{p=1}^\infty \sum_{m_1,\dots,m_p=1}^{\infty} \sum_{m_{p+1}=0}^\infty
 \delta_{t, \sum_i\! m_i} \sum_{\nu_1, \dots, \nu_p}
  \delta_{\vec Y, \sum_i\! \vec e_{\nu_i}}
   \(1- \sum_{j=0}^{m_{p+1}} \sum_\mu F^{(j),\zeta}_{\mu, -\nu_p}\)
  F^{(m_p),\zeta}_{\nu_p, -\nu_{p-1}} \dots F^{(m_2),\zeta}_{\nu_2, -\nu_1}
  F^{(m_1),\zeta}_{\nu_1, \eta}
\end{multline}
the sums on $\mu$ and $\nu_i$ run over the special sites ($\pm 1, \dots \pm N$).

The discrete Laplace transform (power series) of a function of time $g(t)$ is
$\hat g(\xi)\equiv \sum_{t=0}^\infty g(t) \xi^t$.
We can now take both the Laplace and Fourier transforms of (\ref{si:eq:decompo}) to get:
\begin{equation}
 \hat p_{\eta}^\zeta(\vec Y, \xi) = \frac{1}{1-\xi}\left\{ \delta_{\vec Y,\vec 0}
 \(1 - \sum_\mu \hat F_{\mu,\nu}^\zeta\)
 + \sum_{p=1}^\infty \sum_{\nu_1, \dots, \nu_p} \delta_{\vec Y, \sum_i\! \vec e_{\nu_i}}
 \sum_\mu \(1-\hat F_{\mu,-\nu_p}^\zeta\) \hat F_{\nu_p, -\nu_{p-1}}^\zeta
 \dots \hat F_{\nu_2, -\nu_1}^\zeta \hat  F_{\nu_1, \eta}^\zeta
 \right\}
\end{equation}

\begin{equation} \label{si:eq:qBorder}
 \hat{\tilde q}_\eta^\zeta (\vec k, \xi)
 \equiv \hat{\tilde p}_\eta^\zeta(\vec k, \xi) - \frac{1}{1-\xi}
 = \frac{1}{1-\xi} \sum_{\mu,\nu}
 \{[1-T^\zeta(\vec k, \xi)]^{-1}\}_{\nu\mu}
 \times \(1-e^{-i\,\vec k\vec e_\nu}\) e^{i\,\vec k\vec e_\mu}
 \hat F^\zeta_{\mu\eta}(\xi)
\end{equation}

The matrix $T$ is defined by
$T^\zeta(\vec k,\xi)_{\nu\mu} = \hat F^\zeta_{\nu,-\mu} (\xi) e^{i\,\vec k \vec e_{\nu}}$.

\subsection{Expression of the characteristic function}
Introducing (\ref{si:eq:qAll}) into (\ref{si:eq:psi1}) directly gives:
\begin{align}
 \lim_{\rho_0\to 0}\frac{\hat \psi (\vec k, \xi)}{\rho_0} &= - \sum_\nu \bigg[
 \frac{1}{1-\xi}
 - \(\frac{1}{1-\xi} + \hat{\tilde q}^{(\zeta)}_{-\nu} (\vec k, \xi)\)
 e^{i\,\vec k\vec e_\nu} \bigg] h_\zeta(\xi) \\
 h_\zeta(\xi) &= \sum_{Z\notin \{X_i^0\}} \hat F_{\zeta,Z} (\xi)
 = \sum_{Z=X_\zeta^0 + 1}^{X_{\zeta+1}^0 - 1} \hat F_{\zeta,Z} (\xi)
\end{align}
with $\hat{\tilde q}_{\nu}^\zeta$ given by (\ref{si:eq:qBorder}).

\subsection{Expression of the quantities of interest
($\hat F_{\nu, \eta}^{(\zeta)}$ and $h_\mu$)}
The two results \cite{Hughes:1995} that we recalled in the article are:
\begin{itemize}
 \item The Laplace transform of the first passage density at the origin at time $t$
 of a symmetric 1d Polya walk starting from site $l$ is:
 \begin{equation} \label{si:eq:walk1}
  \hat f_l(\xi) = \alpha^{|l|} \mbox{ with } \alpha = \frac{1-\sqrt{1-\xi^2}}{\xi}
 \end{equation}
  \item The probability of first passage at site $s_1$ starting from $s_0$
  and considering $s_2$ as an absorbing site is given by
\begin{equation} \label{si:eq:walk2}
 \hat F^\dagger (s_1|s_0,\xi) =
 \frac{\hat f_{s_1-s_0}(\xi) - \hat f_{s_1-s_2}(\xi) \hat f_{s_2-s_0}(\xi)}
 {1-\hat f_{s_1-s_2}(\xi)^2}
\end{equation}
\end{itemize}

From (\ref{si:eq:walk1}), we have:
\begin{equation}
\hat F_{-1,-1} = \hat F_{+N, +N} = \alpha
\end{equation}

From (\ref{si:eq:walk1}, \ref{si:eq:walk2}), and recalling that the distance between
TP $\mu$ and TP $\mu+1$ is $L_\mu^{(\zeta)}$, we have
\begin{align}
 \hat F_{\mu,\mu}^{(\zeta)} = \hat F_{-\mu+1,-\mu-1}^{(\zeta)} &=
 \frac{\alpha -\alpha^{2L_\mu^{(\zeta)}-1}}{1-\alpha^{2L_\mu^{(\zeta)}}} \\
 \hat F_{\mu,-\mu+1}^{(\zeta)} = \hat F_{-\mu+1,\mu}^{(\zeta)} &=
 \frac{\alpha^{L_\mu^{(\zeta)}-1} -\alpha^{L_\mu^{(\zeta)} + 1}}{1-\alpha^{2L_\mu^{(\zeta)}}}
\end{align}

And the sums are:
\begin{align}
 h_0 = h_N &= \sum_{Z=-\infty}^{-1} \alpha^{|Z|} = \frac{\alpha}{1-\alpha} \\
 h_\mu &= \sum_{Z=1}^{L_\mu-1} \frac{\alpha-\alpha^{2L_\mu-Z}}{1-\alpha^{2L_\mu}}
 = \frac{\alpha (1-\alpha^{L_\mu-1})(1-\alpha^{L_\mu})}{(1-\alpha)(1-\alpha^{2L_\mu})}
 \label{si:eq:exprEnd}
\end{align}

\section{Results in the high density limit}
\subsection{Scaling of the characteristic function}
Using a numerical software (Mathematica)
we obtain the following result for the discrete Laplace tranform
of the second characteristic function is:
\begin{multline} \label{si:eq:psiResSI}
 \lim_{\rho_0\to 0}\frac{\hat \psi(\vec k, \xi)}{\rho_0} = \frac{1}{(1-\alpha^2)(1-\xi)}
\sum_{n=0}^{N-1} \sum_{i=1}^{N-n}
 \alpha^{\mathcal{L}_i^n}
 \bigg\{
 2\alpha (1-\alpha^{L_{i-1}})(1-\alpha^{L_{i+n}}) \cos(k_i + \dots + k_{i+n}) \\
 + (1-\alpha) Q_n(k_i, \dots, k_{i+n}) + C
 \bigg\}
\end{multline}
with $C$ a constant enforcing $\hat\psi(\vec k =\vec 0) = 0$ and
\begin{align}
 \mathcal{L}_i^n &= L_i + \dots + L_{i+n-1} \\
 Q_2(k_1, k_2) &= \alpha^{L_1}\(e^{ik_1} + e^{-ik_2}\) \label{si:eq:SI_Q2} \\
 Q_3(k_1, k_2, k_3) &= \alpha^{L_1}\(e^{ik_1} + e^{-ik_2}\)
 + \alpha^{L_2}\(e^{ik_2} + e^{-ik_3}\)
 + \alpha^{L_1+L_2}\(e^{i(k_1+k_2)} + e^{-i(k_2+k_3)} + 2\cos k_2\)
\end{align}
Similar expressions exist for $Q_n, n>3$.

We define the total length $L = L_1 + \dots + L_{N-1}$ and the rescaled variables
$l_i^n = \mathcal{L}_i^n/L$.

We define $p=1-\xi$ and $\tilde p = p L^2$. We take the limit
$L\to\infty$ keeping $\tilde p$ constant. This is a limit of large time.

The asymptotic behavior of $\alpha$ is given by~:
\begin{equation}
 \alpha = \frac{1-\sqrt{1-\xi^2}}{\xi}
 = 1-\sqrt{2p} + \mathcal{O}_{p\to 0}(p)
\end{equation}
so that
\begin{equation}
 \alpha^{rL} \underset{L\to\infty}{\sim} e^{-r\sqrt{2\tilde p}}
\end{equation}

(\ref{si:eq:psiResSI}) becomes~:
\begin{multline} \label{si:eq:psiResSI2}
 \lim_{\rho_0\to 0}\frac{\hat \psi(\vec k, p=L^2\tilde p)}{\rho_0}
 \underset{L\to\infty}{\sim}
 \frac{L^3}{\sqrt{2} \tilde p^{3/2}} \sum_{n=0}^{N-1} \sum_{i=1}^{N-n}
 \(e^{-\sqrt{2\tilde p}\, l_i^n} - e^{-\sqrt{2\tilde p}\, l_{i-1}^{n+1}}
 - e^{-\sqrt{2\tilde p}\, l_i^{n+1}} + e^{-\sqrt{2\tilde p}\, l_{i-1}^{n+2}}
 \) \\
 \times \(\cos(k_i + \dots + k_{i+n}) - 1\)
\end{multline}

The following continuous inverse Laplace tranform is known:
\begin{equation}
 \hat h(p) = \frac{e^{-r\sqrt{2p}}}{p^{3/2}} \Leftrightarrow
 h(t) = 2\sqrt\frac{t}{\pi} g\(\frac{r}{\sqrt{2t}}\)
\end{equation}
\begin{equation}
 g(u) = e^{-u^2} - \sqrt\pi\, u\, \erfc(u)
\end{equation}

(\ref{si:eq:psiResSI2}) can then be inverted and we find:
\begin{multline}
 \lim_{\rho_0\to 0}\frac{\psi(\vec k, t)}{\rho_0}
 \underset{t\to\infty}{\sim} \sqrt\frac{2t}{\pi}
\sum_{n=0}^{N-1} \sum_{i=1}^{N-n}
 \left[
 g\(\frac{L_i^n}{\sqrt{2t}}\) - g\(\frac{L_{i-1}^{n+1}}{\sqrt{2t}}\)
 - g\(\frac{L_i^{n+1}}{\sqrt{2t}}\) + g\(\frac{L_{i-1}^{n+2}}{\sqrt{2t}}\)
 \right] \\
 \times \(\cos(k_i + \dots + k_{i+n}) - 1\)
\end{multline}

\subsection{Effects of the initial conditions on the odd cumulants}
We consider two TPs at an initial distance $L$. We show that, due to the fact that
we impose their initial positions, the two TPs separate slightly from one another.

From (\ref{si:eq:psiResSI}) and (\ref{si:eq:SI_Q2}) we obtain:
\begin{equation}
 \lim_{\rho_0\to 0} \frac{\langle X_2\rangle (\xi)}{\rho_0}
 = \frac{\alpha^L}{(1+\alpha)(1-\xi)} \equi{L\to\infty} L^2
 \frac{e^{-\sqrt{2\tilde p}}}{2\tilde p}
\end{equation}
with $\tilde p = (1-\xi)/L^2$.

The inversion of the Laplace transform gives:
\begin{equation}
 \lim_{\rho_0\to 0} \frac{\langle X_2\rangle (t)}{\rho_0}
 \equi{t\to\infty} \frac{1}{2} \text{erfc}\(\frac{L}{\sqrt{2t}}\)
\end{equation}
and similarly
\begin{equation}
 \lim_{\rho_0\to 0} \frac{\langle X_1\rangle (t)}{\rho_0}
 \equi{t\to\infty} -\frac{1}{2} \text{erfc}\(\frac{L}{\sqrt{2t}}\)
\end{equation}
We indeed see that the two tracers separate a little bit. This effect is obvious
when $L=1$: initially the two TPs are on neighboring sites, and at large time
there is a probability $\rho_0$ that there is a vacancy inbetween: $\langle X_2-X_2\rangle(t\to\infty) = \rho_0$.

Similar effects are expected on all the odd cumulants in the $N$-tag problem.
These effect also add a term of order $t^0$ in the even cumulants.

\subsection{Large deviation function for a single TP}
For a single tracer at high density our approach (which coincides with \cite{Illien:2013}) gives:
\begin{equation}
 \psi(k, t) = \mu_t \(\cos k - 1\)
\end{equation}
with $\mu_t = \rho_0 \sqrt\frac{2t}{\pi}$.
The G\"artner-Ellis theorem gives:
\begin{align} \label{si:eq:SM_ld1}
 P\(Y_i=\mu_t\tilde y\) &\asymp e^{-\mu_t I(\tilde y)} \\
 I(\tilde y) &= \sup_{q\in\mathbb{R}}
 \(q \tilde y - (\cosh q - 1)\)
\end{align}

Solving for the extremum gives:
\begin{equation}
 e^{\pm q} = \pm \tilde y + \sqrt{1 + \tilde y^2}
\end{equation}
and finally:
\begin{equation}
I(\tilde y) = 1 - \sqrt{1+\tilde y^2}
 + \tilde y\ln\left[\tilde y + \sqrt{1+\tilde y^2}\right]
\end{equation}

\subsection{Large deviation function for $N$ TPs}
We assume that the limits $\rho_0\to 0$ and $t\to\infty$ can be exchanged
(this would need to be proven) and we write
\begin{multline} \label{si:eq:psiResSI3}
 \psi(i\vec q, t)
 \underset{\substack{\rho_0\to 0\\ t\to\infty}}{\sim} \mu_t
\sum_{n=0}^{N-1} \sum_{i=1}^{N-n}
 \left[
 g\(\frac{\Lambda_i^n}{\sqrt\pi}\) - g\(\frac{\Lambda_{i-1}^{n+1}}{\sqrt\pi}\)
 - g\(\frac{\Lambda_i^{n+1}}{\sqrt\pi}\) + g\(\frac{\Lambda_{i-1}^{n+2}}{\sqrt\pi}\)
 \right] \\
 \times \(\cosh(q_i + \dots + q_{i+n}) - 1\)
\end{multline}
with $\mu_t = \rho_0\sqrt\frac{2t}{\pi}$ and $\Lambda_i^n = L_i^n \sqrt\frac{\pi}{2t}$.
In the following, the limits will be implicit.

The Gärtner-Ellis theorem \cite{Touchette:2009} for $\mu_t\to\infty$
then gives us the probability distribution (at exponential order, denoted $\asymp$).
\begin{align}
 P\(\{Y_i = \mu_t\tilde y_i\}\) &\asymp e^{-\mu_t J(\{\tilde y_i\})} \\
 J(\{\tilde y_i\}) &= \sup_{\{q_i\}\in\mathbb{R}}
 \(\sum_{i=1}^N q_i \tilde y_i - \mu_t^{-1} \psi(-i\vec q)\) \label{si:eq:funJSI1}
\end{align}

To simplify the problem we define the following variables:
\begin{equation}
 Y = \frac{Y_1 + Y_N}{2} \qquad D_i = \frac{Y_{i+1}-Y_i}{2}\quad (i=1,\dots N-1)
\end{equation}
\begin{equation}
 u = q_1+\dots+q_N \qquad v_i = -q_1 - \dots - q_i + q_{i+1} + \dots + q_N\quad (i=1,\dots N-1)
\end{equation}

\textbf{Small rescaled length.}
\begin{equation}
 g(\Lambda/\sqrt{\pi}) = 1 - \Lambda + \mathcal{O}_{\Lambda\to 0}(\Lambda^2)
\end{equation}
At the first order in the rescaled lengths, only the ``boundary'' terms contribute in
(\ref{si:eq:psiResSI3}). Using the variables defined above, one checks that (\ref{si:eq:funJSI1})
becomes simpler:
\begin{multline}
 J(\tilde y, \{\tilde d_i\})
 \underset{\Lambda_i \ll 1}{\sim}
 \sup_{u,\{v_i\}\in\mathbb{R}}
 \bigg\{
 u\tilde y + \sum_{i=1}^{N-1} v_i\tilde d_i - \(\cosh(u) - 1\)
 - \sum_{i=1}^{N-1} \Lambda_i\(2\cosh\frac{u}{2} \cosh\frac{v_i}{2} - \cosh(u) - 1\) \bigg\}
\end{multline}

This extremum can only be solved for in the special cases defined in the main text:
$\Lambda_i = 0$, marginal probability of the distances (this corresponds to $u=0$)
and Gaussian limit ($\tilde y \ll 1$, $\tilde v_i \ll 1$).

\textbf{Two tracers, arbitrary length.}
One should note that for two tracers, (\ref{si:eq:funJSI1}) is already rather simple
if written with the right variables: we don't need to assume $\Lambda\ll 1$.
\begin{equation}
 J(\tilde y, \tilde d) =
 \sup_{u,v\in\mathbb{R}}
 \bigg\{
 u\tilde y + v\tilde d - \(\cosh(u) - 1\)
 - \(1-g\(\frac{\Lambda}{\sqrt{\pi}}\)\)\(2\cosh\frac{u}{2} \cosh\frac{v}{2} - \cosh(u) - 1\) \bigg\}
\end{equation}
This expression is used to provide a numerical predition in the main text.

\subsection{Numerical verification of the expression of $\kappa^{(2)}_{11}$ at arbitrary
density}

From the Edwards-Wilkinson equation, one expects~:
\begin{equation}\label{si:eq:cumEW}
 \kappa^{(2)}_{11} = \kappa^{(1)}_{2} g\(\frac{1}{\sqrt{2\tau}}\)
 = \frac{1-\rho}{\rho} \sqrt\frac{2t}{\pi} g\(\frac{1}{\sqrt{2\tau}}\)
\end{equation}
This behavior is in good agreement with numerical simulations (see Fig.~\ref{si:fig:cumInterm}, left).

For the other cumulants, we showed that
\begin{equation} \label{si:eq:equal1N}
\lim_{t\to\infty}\frac{\kappa^{(N)}_{p_1, \dots, p_N}}{\sqrt{t}}
= \lim_{t\to\infty}\frac{\kappa^{(1)}_{p_1 + \dots + p_N}}{\sqrt{t}}
= B_{p_1+\dots+p_N},
\end{equation}
where the constants $B_{k}$ characteristic of a single tracer have been determined in Ref.
\cite{Imamura:2017}.

\begin{figure}
 \includegraphics[width=15cm]{figCumInterm2}
 \caption{Evolution of the cumulants associated to two TPs
 at different densities and different distances. The cumulants are
 rescaled by the single-tag cumulants, following \eqref{si:eq:cumEW}.
 The dashed line corresponds to $g((2\tau)^{-1/2})$. }
 \label{si:fig:cumInterm}
\end{figure}

\subsection{Argument for the breaking of our scaling shape at arbitrary density}
In analogy with \eqref{si:eq:cumEW}, one would like to be able to make the following
bold conjecture~:
\[\kappa^{(2)}_{p, 2n-p} = B_{2n} \sqrt{t}\, g\((2\tau)^{-1/2}\right)\]
Unfortunately we show that this is incompatible with the law of the distance \eqref{si:eq:predDist}.

Let us focus on the 4th cumulant of the distance (we denote $\langle\rangle_c$ the cumulants)~:
\begin{equation}
  \langle (X_2-X_1)^4 \rangle_c = \langle X_1^4\rangle_c + \langle X_2^4\rangle_c
  -4\langle X_1^3 X_2 \rangle_c - 4\langle X_1 X_2^3 \rangle_c + 6\langle X_1^2 X_2^2 \rangle_c
\end{equation}
We \textit{assume} that~:
\begin{align}
 \langle X_1^4\rangle_c = \langle X_2^4\rangle_c &= B_4\sqrt{t} + o(1) \\
 \langle X_1^3 X_2\rangle_c = \langle X_1 X_2^3\rangle_c = \langle X_1^2 X_2^2\rangle_c
 &= B_4\sqrt{t} g\(\frac{1}{\sqrt{2\tau}}\) = B_4\sqrt{t} - B_4\sqrt\frac{\pi}{2} L + o(1)
\end{align}
This leads us to:
\begin{equation} \label{si:eq:assumeEq}
  \langle (X_2-X_1)^4 \rangle_c = \sqrt{2\pi} B_4 L
\end{equation}

From Ref.~\cite{Imamura:2017},
\begin{equation}
B_4 = \sqrt\frac{2}{\pi} \frac{1-\rho}{\rho^3}
\left[1 - (4-(8-3\sqrt{2})\rho)(1-\rho) + \frac{12}{\pi} (1-\rho)^2\right]
\end{equation}
while from \eqref{si:eq:predDist},
\begin{equation}
  \langle (X_2-X_1)^4 \rangle_c \equi{L\to\infty} 2L \frac{1-\rho}{\rho^3}
  \(12 - 24\rho + 13\rho^2\)
\end{equation}
At an arbitrary density, this is inconsistent with \eqref{si:eq:assumeEq},
thus our conjecture must be wrong. Note that \eqref{si:eq:assumeEq} does hold as expected when $\rho\to 1$.

In Fig.~\ref{si:fig:cumInterm} center (resp. right), we tried the following guess:
$\kappa^{2}_{22} = B_4\sqrt{t}\, g\((2\tau)^{-1/2}\right)$ (resp.
$\kappa^{2}_{31} = B_4\sqrt{t}\, g\((2\tau)^{-1/2}\right)$). We see that
it is valid only at high density, as expected.

\section{Description of numerical simulations}
\subsection{Continuous time simulations on a lattice (for the cumulants)}
$N_\text{parts}$ particles are put uniformly at random on the line of size $N_\text{size}$,
except the $N$ tagged particles which are put deterministically on their initial positions.
We used $N_\text{size}=5000$

Each particle has an exponential clock of time constant $\tau = 1$. Thus, the whole
system has an exponential clock of time constant $\tau_\text{all} = \tau / N_\text{parts}$.
When it ticks, a particle is chosen at random and tries to move either to the left or
to the right with probability $1/2$. If the arrival site is already occupied the
particle stays where it was.

The cumulants of the $N$ TPs are averaged over 100\,000 to 500\,000 simulations to obtain
their time dependence.

\subsection{Vacancy-based simulations (for the probability distribution)}
The previous approach does not enable one to get sufficient statistics
to investigate the probability law.

In the case of ponctual brownian particles, Ref.~\cite{Sabhapandit:2015} was able to
use the propagator of the displacement to directly obtain the state of the system at
a given time and investigate the probability distribution.

Here we used a numerical scheme close to out theoretical approach:
at high density and in discrete time, we simulate the behavior of the vacancies
considered as independant random walker. The displacement $\Delta x$ of a vacancy
at (discrete) time $t$ is given by a binomial law:
\begin{align}
 \Delta x &= 2 n_\text{right} - t \\
 P(n_\text{right}, t) &= \frac{1}{2^t} \binom{t}{n_\text{right}}
\end{align}
One is able to recover the final positions of the TPs from the final positions of
the vacancies.

For two TPs at distance $L$, we put a vacancy at each site between the TPs with
probability $\rho_0$ (density of vacancies). We consider a number of sites
$N_\text{sites}$ ($N_\text{sites} = 100\, 000$) on the left of the first TP and
on the right of the second TP and we put a deterministic number of vacancies
$N_\text{vac} = \rho_0 N_\text{sites}$ at random positions on these sites.

We make $10^8$ repetition of the simulation before outputting the probability law.

\end{widetext}
\end{document}


\section{Introduction}
The relativistic heavy-ion collider experiments show that human beings may have produced quark-gluon plasma (QGP) in the laboratory \cite{torres2015flavor}.
It is generally believed that some quark bound states can survive in QGP.
In thermal Quantum chromodynamics (QCD), the thermal
properties of QGP can be determined by studying the behavior of these quark bound states in hot medium.
In 1986, Satz pointed out that the suppression of $J/\psi$ could be a signature of QGP formation
in the relativistic heavy ion collisions \cite{matsui1986j}. Since then, many people have systematically studied
the melting of Charmonium and Bottomonium.
But the dissociation temperature of nucleon, the lightest baryon, has not been studied systematically.
This is due to the difficulties of solving three-body system and obtaining the temperature dependent quark-quark interaction potential within the nucleon.

QCD is the fundamental theory of strong interaction. It works well in the
perturbative region but it does not work in the non-perturbative region. It is difficult
for us to use QCD to study the thermal properties of quark bound states directly.
In this case, people have to use the model \cite{karsch1988color,satz2006colour,alberico2005heavy,zhen2012dissociation} to study the dissociation of
quarkonium states. In high temperatures and density, the interaction between quarks
is screened \cite{digal2005heavy} and the binding energy will be decreased. As a result, the nucleon
will start to melt when the binding energy become low enough. The melting of nucleon can be solved by Schr\"odinger equation of three body.
For the calculation, we need the interacting potentials among
quarks of the nucleon in the hot medium, which is temperature-dependent. Unfortunately, the potentials
are not yet well understood up to now. The free energy of a static heavy three-quark system $F_{qqq}(r,T)$ can be
calculated in lattice QCD and the internal energy can be obtained by using thermodynamic relation.
In the present approach, the needed potential is assumed to be the internal energy, i.e. $V=F+sT$ with
$s$ being the entropy density $s=-\partial F/\partial T$. The temperature-dependent form of
$F_{q\bar q}(r,T)$ can be constructed based on Debye-H\"uckel theory \cite{dixit1990charge},
as having been done in Ref. \cite{digal2005heavy}. Then, we can determine the T-dependent parameter
in the free energy by fitting it with the lattice data. According to the relation between
$F_{q\bar q}(r,T)$ and $F_{qqq}(r,T)$, we can obtain the free energy of heavy three-quark system.
We assume the conclusions are applicable to light quark system due to the flavor independence of the strong interaction. After constructing the potential
of nucleon system at finite temperature, we can obtain the temperature dependence of binding energies
and radii by solving the corresponding Schr\"odinger equation. The dissociation temperature is
the point where the binding energy decreases to zero. The Gaussian expansion method (GEM), which
is an efficient and powerful method in few-body system \cite{hiyama2003gaussian}, is employed
to calculate dissociation temperature of nucleon in this paper.

This paper is organized as follows. In Sec.\ref{sec:quakonium}, we show the rationality of GEM
on studying the melting of Charmonium and Bottomonium
by comparing our results  with others. In Sec.\ref{sec:forma}, we construct the potential of nucleon
and apply GEM to solve corresponding Schr\"odinger equation. In Sec.\ref{sec:res}, we show the
results at quenched and 2-flavor QCD, respectively. Sec.\ref{sec:summ} contains summary and conclusion.

\section{The Results on Dissociation Temperature of Quarkonium}\label{sec:quakonium}

 Before studying the dissociation of nucleon, we test the rationality of GEM on studying dissociation temperatures of quarkonium
 by comparing our results with others. To compare with Satz's results, the potential of quarkonium at
 finite temperature we use is the same as Satz's work \cite{satz2006colour}. The results on dissociation
 temperature of Charmonium, Bottomonium in Ref.\cite{satz2006colour} and our calculation results are listed
 in Table\ref{tab:temp}, Table\ref{tab:temp2},
\begin{table}[!h]
\begin{center}
\caption{\label{tab:temp}Dissociation Temperature $T_d/T_c$ of Charmonium in Ref.\cite{satz2006colour} and our results.}
\renewcommand\arraystretch{1.8}
\begin{tabular}{cccc}
\hline
Charmonium   &  1S  &  1P  &  2S \\
\hline
Ref.\cite{satz2006colour} & 2.10 & 1.16 & 1.12 \\
\hline
Our Results & 2.08 & 1.16 & 1.14   \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{table}[!h]
\begin{center}
\caption{\label{tab:temp2}Dissociation Temperature $T_d/T_c$ of Bottomonium in Ref.\cite{satz2006colour} and our results.}
\renewcommand\arraystretch{1.8}
\begin{tabular}{cccccc}
\hline
Bottomonium   &  1S  &  1P  &  2S & 2P & 3S \\
\hline
Ref.\cite{satz2006colour} & $>$ 4.0 & 1.76 & 1.60 & 1.19 & 1.17 \\
\hline
Our Results & 5.83 & 1.72 & 1.59 & 1.18 & 1.17\\
\hline
\end{tabular}
\end{center}
\end{table}
which show our results are consistent with that in Ref.\cite{satz2006colour}. So GEM can give accurate results
on the dissociation temperature of quarkonium. Giving accurate binding energy and wave function \cite{hiyama2003gaussian}
makes GEM very suitable for studying dissociation temperature of quark bound states (more detail can be found in Appendix A). In the following, we will use this method to
calculate the dissociation temperature of nucleon.

\section{Formalism}\label{sec:forma}

\subsection{Constituent Quark Model}

The constituent quark model is a non-relativistic quark model \cite{yang2008dynamical}. In the constituent quark model,
baryons are formed by three constituent quarks, which are confined by a confining potential and interact with each
other \cite{yoshida2015spectrum}. The potential of baryon can be described by a sum of the potential of corresponding
two-quark system. In Kaczmarek's work \cite{PhysRevD.75.054504}, it has been calculated in lattice QCD that the
potential of diquark system is about half of that of corresponding quark-antiquark system, i.e.
$V_{q q}=\frac 12 V_{q \bar q}$. The simplest and most frequently used potential for a $q \bar q$ system is
the Cornell potential \cite{digal2005heavy},
\begin{equation}
    V_{q \bar q}(r) = -\frac{\alpha}{r} + \sigma r
\label{eq:cornel}
\end{equation}
where $\alpha$ is the coupling constant, and $\sigma$ is the string tension. In the present work, we neglect the
spin-dependent part of potential here. Thus our Hamiltonian is written as
\begin{equation}
H = \sum_{i=1}^3 (m_i + \frac{\boldsymbol{p}_i^2}{2m_i}) - T_{cm} + \sum_{1=i<j}^3 \frac 12 V(r_{ij})
\label{eq:hamil}
\end{equation}
\begin{equation}
V(r_{ij}) = \sigma r_{ij} - \frac{\alpha}{r_{ij}}
\label{eq:cor1}
\end{equation}
where $m_i$ is the constituent quark mass of the $i$-th quark, and $T_{cm}$ is the kinetic energy of center-of-mass frame (cm). $\boldsymbol{r}_{ij}=\boldsymbol{r}_i-\boldsymbol{r}_j$ is the relative coordinate between $i$-th quark and
$j$-th quark. In this model, the mass of light quark (u and d quark) we use is $300$ MeV. The parameters of
Cornell potential we use are: $\alpha=1.4$, $\sqrt{\sigma}=0.131$ GeV. Solving the corresponding Schr\"odinger
equation, $ H\Psi_{total}=E_m \Psi_{total}$, with GEM, we can get the mass $E_m$ and corresponding wave function
of nucleon $\Psi_{total}$. We define the radii of nucleon as
\begin{equation}
R=\frac 13 \sum_{i=1}^3 \sqrt{\langle r_i^2 \rangle}
\label{eq:radii}
\end{equation}
with
\begin{equation}
\langle r_i^2 \rangle=\int \Psi^*_{total} r_i^2 \Psi_{total} d\tau
\label{eq:radii1}
\end{equation}
where $r_i$ is the distance between the center of nucleon and $i$-th quark. Using the calculated wave function,
we can calculate the radii of nucleon. The calculating mass and radii of nucleon are $939$ MeV and $0.83933$ fm,
respectively. While the corresponding experimental data are about $939$ MeV and $0.841$ fm. We can see this model
gives a good estimation of the properties of nucleon even if the spin-dependent part is neglected.
So it is reasonable for us to use this potential model to study the dissociation of nucleon. Of course,
we need notice that the spin-dependent part plays an important role in the baryon spectrum.

\subsection{Wave Function}

Here, we solve the Schr\"odinger equation with GEM. In this method, three sets of Jacobi
coordinates (Fig.\ref{fig:jacobi}) are introduced to express three-quark wave function. The Jacobi coordinates in
each channel $c (c=1,2,3)$ are defined as
\begin{figure}[!htbp]
\centering
\includegraphics[width=0.4\textwidth]{jacobi.png}
\caption{\label{fig:jacobi} Three sets of Jacobi coordinates for a three-body system \cite{hiyama2003gaussian}.}
\end{figure}
\begin{equation}
\boldsymbol{r}_i=\boldsymbol{x}_j-\boldsymbol{x}_k
\label{eq:recoor1}
\end{equation}
\begin{equation}
\boldsymbol{R}_i=\boldsymbol{x}_i-\frac {m_j\boldsymbol{x}_j+m_k\boldsymbol{x}_k}{m_j+m_k}
\label{eq:recoor2}
\end{equation}
where $\boldsymbol{x}_i$ is the coordinate of the $i$-th quark and $(i,j,k)$ are given by Table~\ref{tab:jacobi}.
\begin{table}[!h]
\begin{center}
\caption{\label{tab:jacobi}The quark assignments $(i,j,k)$ for Jacobi channels.}
\renewcommand\arraystretch{1.8}
\begin{tabular}{cccc}
\hline
channel   &  i  &  j  &  k \\
\hline
1 & 1 & 2 & 3 \\
\hline
2 & 2 & 3 & 1 \\
\hline
3 & 3 & 1 & 2 \\
\hline
\end{tabular}
\end{center}
\end{table}
The total wave function is given as a sum of three rearrangement channels ($c=1-3$)
\begin{equation}
\Psi_{total}^{JM} = \sum_{c,\alpha} C_{c,\alpha} \Psi_{JM,\alpha}^{(c)}(\boldsymbol{r}_c,\boldsymbol{R}_c)
\label{eq:totalwave}
\end{equation}
where the index $\alpha$ represents ($s,S,l,L,I,n,N$). Here $s$ is the spin of the $(i,j)$ quark pair,
$S$ is the total spin, $l$ and $L$ are the orbital angular momentum for the coordinate $\boldsymbol{r}$
and $\boldsymbol{R}$, respectively, and I is the total orbital angular momentum. The wave function for
channel $c$ is given by
\begin{equation}
\Psi_{JM,\alpha}^{(c)}(\boldsymbol{r}_c,\boldsymbol{R}_c) = \phi_c \otimes [X_{S,s}^{(c)}
\otimes \Phi_{l,L,I}^{(c)}]_{JM} \otimes H_{T,t}^{(c)}
\end{equation}
as given in Ref. \cite{yoshida2015spectrum}. The orbital wave function $\Phi_{l,L,I}^{(c)}$ is given
in terms of the Gaussian basis functions written in Jacobi coordinates $\boldsymbol{r}_c$ and $\boldsymbol{R}_c$
\begin{equation}
\Phi_{l,L,I}^{(c)} = [\phi_l^{(c)}(\boldsymbol{r}_c) \psi_L^{(c)}(\boldsymbol{R}_c)]_I
\label{eq:orbital}
\end{equation}
\begin{equation}
\phi_{lm}^{(c)}(\boldsymbol{r}_c) = N_{nl} r_c^l e^{-\nu_n r_c^2} Y_{lm}(\hat{\boldsymbol{r}}_c)
\label{eq:orbital1}
\end{equation}
\begin{equation}
\psi_{LM}^{(c)}(\boldsymbol{R}_c) = N_{NL} R_c^L e^{-\lambda_N R_c^2} Y_{LM}(\hat{\boldsymbol{R}}_c)
\label{eq:orbital2}
\end{equation}
where the range parameters, $\nu_n$ and $\lambda_N$, are given by
\begin{gather}
\nu_n = 1/{r_n^2}, r_n = r_1 a^{n-1} (n=1,...,n_{max}), \notag \\
\lambda_N = 1/{R_N^2}, R_N = R_1 A^{N-1} (N=1,...,N_{max});
\label{eq:rangepara}
\end{gather}
In Eqs. (\ref{eq:orbital1}) and (\ref{eq:orbital2}), $N_{nl}$($N_{NL}$) \cite{hiyama2003gaussian} denotes
the normalization constant of Gaussian basis. The coefficients $C_{c,\alpha}$ of the variational
wave function, Eq. (\ref{eq:totalwave}), are determined by Rayleight-Ritz variational principle.

\subsection{Potential model for nucleon}
The potential of nucleon at zero temperature has been discussed above and its parameters have been determined by
fitting the properties of nucleon. To determine the dissociation temperature of nucleon, we need the potential in
hot medium, i.e. $V_{qqq}(\boldsymbol{r},T)$ (the index q represents u or d quark). Here, we assume that
the potential is just the internal energy
\begin{equation}
\begin{split}
V_{qqq}(\boldsymbol{r},T) &= U_{qqq}(\boldsymbol{r},T) \\
                          &= F_{qqq}(\boldsymbol{r},T) + sT
\end{split}
\label{eq:poten}
\end{equation}
where $s$ is the entropy density $s=-\partial F_{qqq}/\partial T$.
In Refs.~\cite{hubner2008heavy,hubner2005free,huebner2005heavy}, Kaczmarek's works show that the color singlet
free energies of the heavy three-quark system ($F_{qqq}^1$) can be described by the sum of antitriplet
free energies of the corresponding diquark system ($F_{qq}^{\overline 3}$) plus self energy contributions
when the temperature is above $T_c$. It can be expressed as
\begin{equation}
F_{qqq}^1(P,T) \simeq \sum_{i<j} F_{qq}^{\overline 3}(R_{ij},T)-3F_q(T)
\label{eq:free}
\end{equation}
where $P=\sum_{i<j}R_{ij}$ and the self energy $F_q(T)=\frac 12 F_{qq}^{\overline 3}(\infty,T)$. In Ref.~\cite{PhysRevD.75.054504}, O.Kaczmarek's work suggests a simple relation between free energies of anti-triplet $qq$ states and color singlet $q\bar{q}$
\begin{equation}
F_{q\bar{q}}^1(r,T) \simeq 2(F_{qq}^{\overline 3}(r,T)-F_q(T))
\label{eq:free1}
\end{equation}
The form of $F_{q\bar{q}}^1$ can be obtained based on studies of screening in Debye-H\"uckel theory.
It can be written as \cite{digal2005heavy}
\begin{eqnarray}
F_{q\bar{q}}^1(r,T)=-\frac{\alpha}{r}\left[e^{-\mu r}+\mu r\right]+\frac{\sigma}{\mu}[\frac{\Gamma\left(1/4\right)}{2^{3/2}
    \Gamma\left(3/4\right)}            \notag \\
    -\frac{\sqrt{\mu r}}{2^{3/4}\Gamma\left(3/4\right)}K_{1/4}\left[\left(\mu r\right)^2+\kappa \left(\mu r\right)^4\right]]
    \label{eq:free2}
\end{eqnarray}
where screening mass $\mu$ and the parameter $\kappa$ are temperature-dependent, and $K_{1/4}[x]$ is the modified
Bessel function. We can determine the T-dependent $\mu$ and $\kappa$ by fitting $F_{q\bar{q}}^1(r,T)$ to the
lattice result obtained in quenched \cite{kaczmarek2002heavy} and 2-flavor \cite{kaczmarek2005static} QCD.
At $r=\infty$, the free energy $F_{q\bar{q}}^1(T)$ is wirtten as
\begin{equation}
F_{q\bar{q}}^1(T) = \frac {\sigma}{\mu\left(T\right)} \frac {\Gamma\left(1/4\right)}{2^{3/2} \Gamma\left(3/4\right)} - \alpha \mu\left(T\right)
\label{eq:inffree}
\end{equation}
Thus, the form of $\mu(T)$ is given as function of $F(T)$
\begin{equation}
\mu\left(T\right) = \frac {\left[\sqrt{{F_{q\bar{q}}^1(T)}^2+4\sigma\alpha \frac {\Gamma\left(1/4\right)}{2^{3/2} \Gamma\left(3/4\right)}}-F_{q\bar{q}}^1(T)\right]}{2\alpha}
\label{eq:fmu}
\end{equation}
Once we obtain the temperature dependence of $\mu(T)$, we fit Eq.~(\ref{eq:free2}) to the lattice data to obtain
$\kappa(T)$. The results for $\mu(T)$ and $\kappa(T)$ are shown in Fig. \ref{fig:miu} and Fig. \ref{fig:kai},
respectively. In Fig.~\ref{fig:lattfree}, we show our fit curves (solid lines) together with the lattice results.
We can see that the resulting $F_{q\bar q}^1(r,T)$ fits the lattice results quite well for all $r$ and in a broad
range of temperatures from $T_c$ to $4T_c$ in the two cases. For higher temperature, the resulting $F_{q\bar q}^1(r,T)$
cannot be fitted quite well to the lattice results in quenched QCD. There can be higher order corrections to
Poisson equation \cite{digal2005heavy}.

\begin{figure}[!htbp]
\centering
\includegraphics[width=0.4\textwidth]{miu.png}
\includegraphics[width=0.4\textwidth]{2flavormiu.png}
\caption{\label{fig:miu} Results for $\mu(T)$ in quenched (upper figure) and 2-flavor (lower figure) QCD.}
\end{figure}

\begin{figure}[!htbp]
\centering
\includegraphics[width=0.4\textwidth]{kai.png}
\includegraphics[width=0.4\textwidth]{2flavorkai.png}
\caption{\label{fig:kai}Results for $\kappa(T)$ in quenched (upper figure) and 2-flavor (lower figure) QCD.}
\end{figure}

\begin{figure}[!htbp]
\centering
\includegraphics[width=0.4\textwidth]{free.png}
\includegraphics[width=0.4\textwidth]{2flavorfree.png}
\caption{\label{fig:lattfree}Results for free energy ($F_{q\bar{q}}^1$) in quenched (upper figure) and 2-flavor (lower figure) QCD.}
\end{figure}
To obtain the binding energies of nucleon, we define a effective potential as
\begin{equation}
\tilde V_{qqq}(\boldsymbol{R} ,T) = V_{qqq}(\boldsymbol{R} ,T) - V_{qqq}(\infty,T)
\label{eq:effpoten}
\end{equation}
Combining Eqs.(\ref{eq:poten}-\ref{eq:free1},\ref{eq:effpoten}), we can get a relation between effective potential
and free energies of $q\bar q$
\begin{equation}
\tilde V_{qqq}(\boldsymbol{R},T) = \sum_{i<j} \frac 12 (\tilde F_{q\bar{q}}^1(R_{ij},T) - T\frac {\partial \tilde F_{q\bar{q}}^1(R_{ij},T)}{\partial T})
\label{eq:effpoten1}
\end{equation}
where
\begin{equation}
\tilde F_{q\bar{q}}^1(R_{ij},T) = F_{q\bar{q}}^1(R_{ij},T) - F_{q\bar{q}}^1(\infty,T)
\label{eq:efffree}
\end{equation}
Replacing the potential term, $\sum_{1=i<j}^3 \frac 12 V(r_{ij})$, in Eq.~(\ref{eq:hamil}) with this effective potential
$\tilde V_{qqq}(\boldsymbol{R},T)$, we can get a new Hamiltonian for nucleon at finite temperature written as
\begin{equation}
H_{new} = \sum_{i=1}^3 \frac{\boldsymbol{p}_i^2}{2m_i} - T_{CM} + \tilde V_{qqq}(\boldsymbol{R},T)
\label{eq:hamil1}
\end{equation}
Solving corresponding Schr\"odinger euqation,
$$
H_{new}\Psi_{total}^{JM}=\epsilon(T) \Psi_{total}^{JM},
$$
with GEM, we can get the binding energies $\Delta E(T)(=-\epsilon(T))$ and corresponding wave function at finite temperature.
Using the wave function, we can calculate the T-dependent radii according to Eq.~(\ref{eq:radii}).

\section{Numerical Results}\label{sec:res}
In Fig. \ref{fig:bind}, we show the resulting binding energies behaviour for nucleon in quenched and 2-flavor QCD.
We can see there is little difference between the two lines. When they vanish, the nucleon no longer exists.
So $\Delta E(T)=0$  determines the dissociation temperature. From Fig. \ref{fig:bind}, we get the dissociation
temperature in quenched and 2-flavor QCD are about $3.0T_c$ and $3.3T_c$, respectively; in Fig. \ref{fig:ra},
we show the corresponding nucleonic radii. The dissociation temperature determined from Fig. \ref{fig:ra} is consistent
with that determined from Fig. \ref{fig:bind}. It is seen that the divergence of the radii defines quite well
the different dissociation points in the two cases. The resulting dissociation temperatures have a little
difference between the two cases.

\begin{figure}[!htbp]
\centering
\includegraphics[width=0.4\textwidth]{bindingenergy1.png}
\caption{\label{fig:bind}T-dependent of binding energy in quenched and 2-flavor QCD, respectively.}
\end{figure}
\begin{figure}[!htbp]
\centering
\includegraphics[width=0.4\textwidth]{radii1.png}
\caption{\label{fig:ra}T-dependent of radii. in quenched and 2-flavor QCD, respectively}
\end{figure}
\section{Summary and Conclusion}\label{sec:summ}
The free energies of quark-antiquark system we construct based on Debye-H\"uckel theory at finite temperature
fit the lattice results quite well from $T_c$ to $4T_c$, but not well for higher temperature. According to
Kaczmarek's works, we can get a relation between color singlet free energy of heavy $qqq$ system $F_{qqq}^1$
and color singlet free energy of heavy $q\bar q$ system $F_{q\bar q}^1$, written as $F_{qqq}^1 \simeq \sum_{i<j}
\frac 12 F_{q\bar q}^1$. The dissociation temperature of nucleon in quenched and 2-flavor QCD we calculate are
about $3.0T_c$ and $3.3T_c$, respectively. There are a little difference between the two results. Comparing with
$J/\psi$, the dissociation temperature of nucleon is higher. So, nucleon is more difficult to melt than charmonium.
For the potential, we neglect the spin-dependent part in this work which may has some effects to the resulting
dissociation temperature. The effects arising from spin-dependent part deserve further studies.

\section*{Acknowledgement}

We are grateful to Emiko Hiyama, Pengfei Zhuang, Yunpeng liu, Min He and Fan Wang for their work and helpful discussions.

This work is supported in part by the National Natural Science Foundation of China (under Grants Nos. 11475085, 11535005,
11690030 and 11775118), the Fundamental Research Funds for the Central Universities (under Grant No. 020414380074),
and the International Science \& Technology Cooperation Program of China (under Grant No. 2016YFE0129300).


\section{Introduction}
Let $\tau(n) = \sum_{d|n} 1$ be the divisor function, and
\begin{align*}
S(\alpha) = \sum_{n\le x} \tau(n)e(n\alpha),  \qquad  e(\alpha) = e^{2\pi i \alpha}.
\end{align*}
In 2001 Brudern \cite{Brudern} considered the $L^1$ norm of $S(\alpha)$ and claimed to prove
\begin{equation}   \sqrt{x} \ll \int_0^1 |S(\alpha)| d\alpha \ll \sqrt{x}. \end{equation}
However there is a mistake in the proof given there which depends on a lemma which is false.
In this note we prove the following result.
\begin{theorem}
We have
\label{thm}
\begin{equation}
\sqrt{ x} \ll \int_0^1 |S(\alpha)|d\alpha\ll \sqrt{x}\log x.
\end{equation}
\end{theorem}
The upper bound here is obtained by following Brudern's proof with corrections. The lower bound is based on the method Vaughan introduced to study the $L^1$ norm for exponential sums over primes \cite{Va}, and also makes use of a more recent result of Pongsriiam and Vaughan \cite{PV} on the divisor sum in arithmetic progressions. We do not know whether the upper bound or the lower bound reflects the actual size of the $L^1$ norm here.

\section{Proof of the upper bound}
Let $u$ and $v$ always be positive integers. Following Br\"udern, we have
\[ \begin{split}
S(\alpha) &=  \sum_{n\le x} \left( \sum_{uv=n}1\right) e(n\alpha) \\&=\sum_{uv\le x}e( uv\alpha) \\&=  2\sum_{u\le\sqrt{x}}\sum_{u < v\le x/u} e( uv\alpha) + \sum_{u\le\sqrt{x}} e( u^2\alpha)\\ &
:= 2T(\alpha) + V(\alpha) .
\end{split} \]
By  Cauchy-Schwarz and Parseval
\[ \int_0^1 |V(\alpha)|\, d\alpha \le \left( \int_0^1 |V(\alpha)|^2\, d\alpha \right)^{\frac12} = \sqrt{ \lfloor \sqrt{x}\rfloor} \le x^{\frac14} ,\]
and therefore by the triangle inequality
\[ \int_0^1 |S(\alpha)|\, d\alpha = 2 \int_0^1 |T(\alpha)|\, d\alpha + O(x^{\frac14}). \]
Thus to prove the upper bound in Theorem 1 we need to establish
\begin{equation} \label{integral}
\int_0^1 |T(\alpha)| d\alpha\ll\sqrt{x}\log x.
\end{equation}

We proceed as in the circle method. Clearly in \eqref{integral} we can replace the integration range $[0,1]$  by $[1/Q, 1+1/Q]$. By Dirichlet's theorem for any $\alpha \in [1/Q, 1+1/Q]$ we can  find a fraction $\frac{a}{q}$, $1\le q \le Q$,  $1\leq a \le q$, $(a,q)=1$, with $|\alpha - \frac{a}{q}| \le 1/(qQ)$. Thus the intervals $ [\frac{a}{q}- \frac{1}{qQ}, \frac{a}{q} +\frac{1}{qQ} ] $ cover the interval $[1/Q,1+1/Q]$.  Taking
\begin{equation}  \label{Qequation}  2\sqrt{x} \le Q \ll \sqrt{x},  \end{equation}
we obtain
\[
\int_0^1 |T(\alpha)|d\alpha
\le \sum_{q\le Q}\sum_{\substack{a = 1\\ (a,q)=1}}^q \int_{\frac{a}{q}-1/(2q\sqrt{x})}^{\frac{a}{q}+1/(2q\sqrt{x})} \left\vert T(\alpha)\right\vert d\alpha.
\]
On each interval $ [\frac{a}{q}- \frac{1}{2q\sqrt{x}}, \frac{a}{q} +\frac{1}{2q\sqrt{x}} ] $ we decompose $T(\alpha)$
into
\[  T(\alpha) = F_q(\alpha) + G_q(\alpha) \]
where
\[  F_q(\alpha) = \sum_{\substack{u\le\sqrt{x}\\ q|u}}\sum_{u < v\le x/u} e( uv\alpha) \]
and
\[  G_q(\alpha) = \sum_{\substack{u\le\sqrt{x}\\ q\nmid u}}\sum_{u < v\le x/u} e( uv\alpha), \]
and have
\[ \begin{split} \int_0^1 |T(\alpha)|d\alpha
&\le  \sum_{q\le Q}\sum_{\substack{a = 1\\ (a,q)=1}}^q \int_{\frac{a}{q}-1/(2q\sqrt{x})}^{\frac{a}{q}+1/(2q\sqrt{x})} \left\vert F_q(\alpha)\right\vert d\alpha + \sum_{q\le Q}\sum_{\substack{a = 1\\ (a,q)=1}}^q \int_{\frac{a}{q}-1/(2q\sqrt{x})}^{\frac{a}{q}+1/(2q\sqrt{x})} \left\vert G_q(\alpha)\right\vert d\alpha\\ &
:=  I_F + I_G. \end{split}
\]
The upper bound in Theorem 1 follows from the following two lemmas.
\begin{lemma} [Br\"udern]   We have
\[  I_F \ll \sqrt{x}. \]
\end{lemma}

\begin{lemma}   We have
\[  I_G \ll \sqrt{x}\log x . \]
\end{lemma}

In what follows we always assume $(a,q)=1$, and  define the new variable $\beta$ by
\begin{equation} \label{alpha} \alpha = \frac{a}{q} +\beta .  \end{equation}
\begin{proof}[Proof of Lemma 2]  The proof follows from the estimate
\begin{equation} \label{Festimate}
F_q(\alpha)
 \ll \left\{ \begin{array}{ll}
      {\min\left(x,\frac{1}{|\beta|}\right)\frac{\log
\frac{2\sqrt{x}}{q}}{q} ,} & \mbox{if $q\le \sqrt{x}$, $ |\beta|\le \frac{1}{2q\sqrt{x}}$;} \\
      0, & \mbox{if $q>\sqrt{x}$;} \\
\end{array}
\right.
\end{equation}
since this implies
\[   \begin{split}   I_F &\ll \sum_{q\le \sqrt{x}} q \int_0^{1/(2q\sqrt{x})}\min\left(x, \frac{1}{|\beta|}\right) \frac{\log
\frac{2\sqrt{x}}{q}}{q}\, d\beta \\&
\ll \sum_{q\le \sqrt{x}} \log
\frac{2\sqrt{x}}{q}\left(\int_0^{1/(2x)} x \, d\beta +  \int_{1/(2x)}^{1/(2q\sqrt{x})} \frac{1}{\beta} \, d\beta \right) \\& \ll \sum_{q\le \sqrt{x}} \left(
\log
\frac{2\sqrt{x}}{q}\right)^2 \quad \ll \sqrt{x}.
\end{split} \]
To prove \eqref{Festimate}, we first note that the conditions $q|u$ and $u \le \sqrt{x}$ force $F_q(\alpha)=0$ when $q>\sqrt{x}$. Next, when $q\le \sqrt{x}$ we write $u=jq$ and have
\[ F_q(\alpha) = \sum_{j\le \frac{\sqrt{x}}{q}}\sum_{jq\le v\le \frac{x}{jq}} e( jqv\beta).\]
Making use of the estimate
\begin{equation} \label{basic} \sum_{N_1 < n \le N_2} e(n\alpha)  \ll  \min\left( N_2-N_1, \frac{1}{\Vert \alpha \Vert}\right)\end{equation}
we have
\[ F_q(\alpha) \ll \sum_{j\le \frac{\sqrt{x}}{q}}\min\left(\frac{x}{jq}, \frac{1}{\Vert jq\beta \Vert}\right).\]
In this sum $jq\le \sqrt{x}$ so that $|jq\beta | \le |\beta| \sqrt{x}$, and hence the condition $ |\beta|\le \frac{1}{2q\sqrt{x}}$ implies $|jq\beta | \le \frac{1}{2q} \le \frac12$. Hence $ \Vert jq\beta \Vert = jq|\beta|$, and we have
\[ F_q(\alpha) \ll \sum_{j\le \frac{\sqrt{x}}{q} } \frac{1}{jq} \min\left(x, \frac{1}{|\beta|}\right)\ll \min(x, \frac{1}{|\beta|})\frac{\log\frac{2\sqrt{x}}{q}}{q}  .\]
\end{proof}

\begin{proof}[Proof of Lemma 3] The proof follows from the estimate,
\begin{equation} \label{Gestimate}
G_q(\alpha) \ll (\sqrt{x} + q)\log q , \quad \text{for} \  \alpha = \frac{a}{q} + \beta, \quad  |\beta| \le \frac{1}{2q\sqrt{x}}, \end{equation}
since this implies
\[   \begin{split}   I_G &\ll \sum_{q\le Q} q \int_0^{1/(2q\sqrt{x})}(\sqrt{x} +q) \log q\,  d\beta \\&
\ll  \frac{1}{\sqrt{x}}( Q(\sqrt{x} +Q)\log Q \ll \sqrt{x}\log x
\end{split} \]
by \eqref{Qequation}.
To prove \eqref{Gestimate}, we apply \eqref{basic} to the sum over $v$ in $G_q(\alpha)$ and obtain
\[  G_q(\alpha) \ll  \sum_{\substack{u\le\sqrt{x}\\ q\nmid u}} \min\left( \frac{x}{u},\frac{1}{\Vert u\alpha \Vert}\right).  \]
Recalling $\Vert x \Vert =  \Vert -x \Vert $ and the triangle inequality $\Vert  x+y\Vert \le \Vert x\Vert +\Vert y\Vert$, and using the conditions $1\le u\le \sqrt{x}$, $q\nmid u$, $ |\beta| \le \frac{1}{2q\sqrt{x}}$, we have
\[  \begin{split}\left \Vert u\alpha\right \Vert &\ge \left\Vert \frac{au}{q}\right \Vert - \Vert u \beta \Vert \\&
\ge \left \Vert \frac{au}{q} \right\Vert - u|\beta| \\&
\ge\left \Vert \frac{au}{q} \right\Vert -  \frac{u} {2q\sqrt{x}} \\&
\ge \left \Vert \frac{au}{q} \right \Vert -  \frac{1} {2q}   \\&
\ge \frac12\left \Vert \frac{au}{q} \right \Vert , \end{split} \]
and therefore
 \[  G_q(\alpha) \ll  \sum_{\substack{u\le\sqrt{x}\\ q\nmid u}} \frac{1}{\left\Vert \frac{au}{q} \right \Vert}.  \]
Here $\left\Vert \frac{au}{q} \right \Vert = \frac{b}{q}$ for some integer $1\le b \le \frac{q}{2}$ and since the  integers $\{au: 1\le u \le \frac{q}{2}\}$ are distinct modulo $q$ since $(a,q)=1$, we see
\[ \sum_{1\le u \le \frac{q}{2} }  \frac{1}{\left\Vert \frac{au}{q}\right \Vert} = \sum_{1\le b \le \frac{q}{2}}\frac{q}{b} \ll  q\log q .\]
If  $q> \sqrt{x}$  then
\[  \sum_{\substack{u\le\sqrt{x}\\ q\nmid u}} \frac{1}{\left\Vert \frac{au}{q} \right \Vert} \le 2   \sum_{1\le u \le \frac{q}{2} }  \frac{1}{\left\Vert \frac{au}{q}\right \Vert} \ll q\log q ,\]
while if $q\le \sqrt{x}$ then
 the sum bounding $G_q(\alpha)$ can be split into  $\ll \frac{ \sqrt{x}}{q}$ sums of this type and
 \[ G_q(\alpha) \ll  \frac{\sqrt{x}}{q} (q\log q )\ll \sqrt{x}\log q. \]
\end{proof}

\section{Proof of the lower bound}
Following Br\"udern, consider the intervals  $|\alpha - \frac{a}{q}| \le 1/(4x)$ for $1\le a\le q\le Q $, where we take $\frac12 \sqrt{x} \le Q \le \sqrt{x}$. These intervals are pairwise disjoint because for two distinct fractions
$|a/q - a'/q'| \ge 1/(qq') \ge 1/x$. (We will see later why these intervals have been chosen shorter than required to be disjoint.) Hence, using \eqref{alpha}
\[ \int_0^1 |S(\alpha)|\, d\alpha = \int_{1/Q}^{1+1/Q} |S(\alpha)|\, d\alpha  \ge \sum_{q\le \frac12\sqrt{x} }\sum_{\substack{a = 1\\ (a,q)=1}}^q \int_{-1/(4x)}^{1/(4x)} \left\vert S\left(\frac{a}{q} +\beta\right)\right\vert d\beta.
\]
Next we follow Vaughan's method  \cite{Va} and apply the triangle inequality to obtain the lower bound
\[ \int_0^1 |S(\alpha)|\, d\alpha  \ge \sum_{q\le \frac12\sqrt{x} } \int_{-1/(4x)}^{1/(4x)} \left\vert \sum_{\substack{a = 1\\ (a,q)=1}}^qS\left(\frac{a}{q} +\beta\right)\right\vert d\beta.
\]
Letting
\begin{equation} \label{Uq}  U_q(x;\beta) := \sum_{\substack{a = 1\\ (a,q)=1}}^qS\left(\frac{a}{q} +\beta\right) = \sum_{n\le x} \tau(n) c_q(n)e(n\beta) , \end{equation}
where
\[ c_q(n) = \sum_{\substack{a = 1\\ (a,q)=1}}^q e\left(\frac{an}{q}\right) \]
is the Ramanujan sum,  our lower bound may now be written as
\begin{equation} \label{lowerbound} \int_0^1 |S(\alpha)|\, d\alpha  \ge \sum_{q\le \frac12\sqrt{x} } \int_{-1/(4x)}^{1/(4x)} \left\vert U_q(x;\beta)\right\vert d\beta. \end{equation}
To complete the proof of the lower bound we need the following lemma, which we prove at the end of this section.
\begin{lemma}
\label{Usum}
For $	q\ge 1$ we have
\begin{align}
U_q(x; 0) = \frac{\varphi(q)}{q}x(\log (x/q^2) + 2\gamma - 1) + O(q\tau(q) (x^{\frac13} + q^{\frac12})x^\epsilon),
\end{align}
where $\gamma$ is Euler's constant.
\end{lemma}

\begin{proof}[Proof of the lower bound in Theorem 1]  For any exponential sum
$T(x; \beta) =\sum_{n\le x} a_n e(n\beta)$ we have by partial summation or direct verification
\[ T(x; \beta) =  e(\beta x) T(x;0) - 2\pi i \beta \int_1^x e(\beta y)  T(y;0) \, dy.\]
Taking $T(x;\beta) = U_q(x;\beta)$ we thus obtain from \eqref{lowerbound} and the triangle inequality
\begin{equation} \label{lower} \int_0^1 |S(\alpha)|\, d\alpha  \ge \sum_{q\le \frac12\sqrt{x} } \int_{-1/(4x)}^{1/(4x)} \left( |U_q(x;0)| - 2\pi |\beta | \int_1^x |U_q(y; 0)| \, dy \right) d\beta.\end{equation}
By Lemma 4,  with $q\le \frac12\sqrt{x}$,
\[ \begin{split}    \int_1^x |U_q(y; 0)| \, dy & \le \frac{\varphi(q)}{q} \left( \int_1^x  y | \log(y/q^2)| +(2\gamma -1)y \, dy\right)  +   O(xq\tau(q) (x^{\frac13} + q^{\frac12})x^\epsilon) \\&
\le  \frac{\varphi(q)}{q} \left( \int_1^{q^2}  y \log(q^2/y) \, dy + \int_{q^2}^x  y \log(y/q^2) \, dy+ \frac12 x^2 (2\gamma -1) \right) \\ & \hskip 2.5in +   O(xq\tau(q) (x^{\frac13} + q^{\frac12})x^\epsilon) \\&
= \frac{x}{2}\left( \frac{\varphi(q)}{q}\left( x(\log (x/q^2)+ 2\gamma -1) - \frac{x}{2}  + \frac{q^4}{x} \right) +   O(q\tau(q) (x^{\frac13} + q^{\frac12})x^\epsilon)\right)\\&
\le \frac{x}{2} U_q(x,0)  +   O(q\tau(q) (x^{\frac13} + q^{\frac12})x^\epsilon).
\end{split}\]
Using $|\beta | \le 1/(4x)$, we have
\[  |U_q(x;0)| - 2\pi \beta \int_1^x |U_q(y; 0)| \, dy  \ge \left( 1 - \frac{\pi}{4} \right)|U_q(x;0)|  -  O(q\tau(q) (x^{\frac13} + q^{\frac12})x^\epsilon). \]
We conclude, returning to \eqref{lower} and making use of Lemma 4 again,
\[ \begin{split}  \int_0^1 |S(\alpha)|\, d\alpha & \ge \frac{4-\pi}{8 x} \sum_{q\le \frac12\sqrt{x} } \left(|U_q(x;0)| - O(q\tau(q) (x^{\frac13} + q^{\frac12})x^\epsilon)\right) \\&
\ge \frac{4-\pi}{8} \sum_{q\le \frac12\sqrt{x} } \frac{\varphi(q)}{q}(\log (x/q^2) + 2\gamma - 1) -O(x^{\frac13+\epsilon}) .\end{split} \]
It is easy to see that the sum above is $\gg  \sqrt{x} $ which suffices to proves the lower bound.  More precisely, using
\[ \frac{\varphi(n)}{n} = \sum_{d|n} \frac{ \mu(d)  }{d} \]
a simple calculation gives the well-known result
\[ \sum_{ n\le x} \frac{\varphi(n)}{n} = \frac{6}{\pi^2} x +O(\log x), \]
and then by partial summation we find
\[ \sum_{q\le \frac12\sqrt{x} } \frac{\varphi(q)}{q}(\log (x/q^2) + 2\gamma - 1)\sim \frac{6}{\pi^2}(\log 2 +\gamma -1)\sqrt{x} .\]
\end{proof}
\begin{proof}[Proof of Lemma 4]  Pongsriiam and Vaughan \cite{PV} recently proved the following very useful result on the divisor
function in arithmetic progressions.  For inteqer $a$ and $d\ge 1$ and real $x\ge 1$ we have
\[  \sum_{\substack{ n\le x \\ n\equiv a(\text{mod}\, d)}}\tau(n) = \frac{x}{d} \sum_{r | d}\frac{c_r(a)}{r}\left(\log\frac{x}{r^2} + 2\gamma -1\right) +O( (x^{\frac13} + d^{\frac12})x^\epsilon),\]
where $\gamma$ is Euler's constant and $c_r(a)$ is the Ramanujan sum. We need the special case when $a=0$ which along with the situation $(a,d)>1$ is explicitly allowed in this formula.  Hence we have
\begin{equation} \label{PV}
  \sum_{\substack{ n\le x \\ d|n}} \tau(n) = \frac{x}{d}f_x(d) + O( (x^{\frac13} + d^{\frac12})x^\epsilon)
\end{equation}
where
\begin{equation}\label{f-g}
f_x(d) = \sum_{r|d} g_x(r),    \quad   g_x(r) = \frac{\varphi(r)}{r}(\log(x/r^2) + 2\gamma - 1).
\end{equation}
Making use of
\[
c_q(n) = \sum_{d|(n, q)}d\mu\left(\frac{q}{d}\right),
\]
and  \eqref{PV} we have
\[ \begin{split} U_q(x;0) &= \sum_{n\le x}\tau(n)c_q(n)\\&
= \sum_{d|q}d\mu\left(\frac{q}{d}\right)\underset{d|n}{\sum_{n\le x}}\tau(n) \\& =
x\sum_{d|q}\mu\left(\frac{q}{d}\right)f_x(d) +  O( q\tau(q)(x^{\frac13} + d^{\frac12})x^\epsilon).
 \end{split} \]
We evaluate the sum above using  Dirichlet convolution and the identity $1*\mu  = \delta$ where $\delta(n)$ is the identity for Dirichlet convolution defined to be 1 if $n=1$ and zero otherwise.  Hence
\[ \sum_{d|q}\mu\left(\frac{q}{d}\right)f_x(d) =  (f_x * \mu)(q)   = ( (g_x*1)*\mu)(q) =   (g_x * \delta)(q)  = g_x(q),
\]
and Lemma 4 is proved.
\end{proof}