# Passive Particle Driven by Brownian Motion

Urna Basu  S. N. Bose National Centre for Basic Sciences, Kolkata 700106,
India  P. L. Krapivsky  Department of Physics, Boston University, Boston,
Massachusetts 02215, USA  Santa Fe Institute, Santa Fe, New Mexico 87501, USA
Satya N. Majumdar  LPTMS, CNRS, Univ. Paris-Sud, Université Paris-Saclay,
91405 Orsay, France

###### Abstract

We investigate the overdamped dynamics of a ‘passive’ particle driven by
nonreciprocal interaction with a ‘driver’ Brownian particle. When the
interaction between them is short-ranged, the long-time behavior of the driven
particle is remarkably universal—the mean-squared displacement (MSD) and the
typical position of the driven particle exhibits the same qualitative
behaviors independent of the specific form of the potential. In particular,
the MSD grows as $t^{1/2}$ in one dimension and $\log t$ in two spatial
dimensions. We compute the exact scaling functions for the position
distribution in $d=1$ and $d=2$. These functions are universal when the
interaction is short-ranged. For long-ranged interactions, the MSD of the
driven particle grows as $t^{\phi}$ with exponent $\phi$ depending on the tail
of the potential.

##  I Introduction

The stochastic dynamics of $N$ interacting particles is a subject of long-
standing interest in statistical physics. This is a hard problem, except for a
few solvable cases mostly in one dimension. Even the simple two body problem
$(N=2)$ is often hard. A two-particle system, with reciprocal interactions
between them that depend only on their relative coordinate, can be reduced to
a simpler one-body problem in an external potential in their relative co-
ordinate. However, when the interaction is non-reciprocal, this simple change
of coordinates does not work and the problem is nontrivial even for $N=2$. In
this paper, we investigate the stochastic dynamics of two particles with a
non-reciprocal interaction between them in all dimensions and derive exact
results for the position fluctuations at asymptototically late times. We find
rich and rather universal behaviors that are independent of the details of the
interactions but depend only on the dimension and on whether the nonreciprocal
interaction is short-ranged or long-ranged.

Our setup consists of two particles: we call one of them the ‘driver’ particle
and the other a ‘passive’ or ‘driven’ particle. We consider a non-reciprocal
(one-sided) interaction affecting only the passive particle. We assume that
the driver particle performs a standard Brownian motion Lévy (1965); Itô and
McKean (1996); Mörters and Peres (2010), while the motion of the passive
particle is overdamped and caused by a central force depending on the
separation from the Brownian particle. In one dimension, the position $x(t)$
of the driver particle and the position $y(t)$ of the passive particle evolve
according to

$\displaystyle\dot{x}$ $\displaystyle=\eta(t),$ (1a) $\displaystyle\dot{y}$
$\displaystyle=f[x(t)-y(t)]\,.$ (1b)

Here $\eta(t)$ is a Gaussian white noise obeying standard relations,

$\langle\eta(t)\rangle=0,\qquad\langle\eta(t_{1})\eta(t_{2})\rangle=2D\delta(t_{1}-t_{2})\,.$
(2)

The force acting on the passive particle depends only on the separation
between the particles. The force is necessarily potential in one dimension,
$f(z)=-\partial_{z}V(z)$. We make a few assumptions about the force, although
many results remain valid in more general situations. We assume that the force
$f(z)$ is an odd function of $z$. Equivalently, the potential $V(z)$ is an
even function of $z$. However, there is no restriction on the sign of the
potential, i.e., it can be either attractive or repulsive.

In higher dimensions, we assume that the force is central with the potential
$V(\boldsymbol{r})=V(r)$ that depends only on the separation
$\boldsymbol{r}=\boldsymbol{x-y}$ between the driver and the driven particles.

The behavior of the position of the driver particle is well-known since it
performs standard Brownian motion. The goal is to determine the statistics of
the position of the passive particle. For instance, we will explore how the
mean-squared displacement (MSD) grows with time and show that the behavior of
the MSD is very different depending on whether the potential is short-ranged
or long-ranged. One can thus distinguish between two types of forces $f(r)$
where $r$ represents the between the two particles in general dimensions:

  1. 1.

Short-ranged forces. A short-ranged force $f(r)$ is finite everywhere and
quickly decays with distance, e.g., faster than any power of $1/r$ for $r\gg
R$, the typical range of the force.

  2. 2.

Long-ranged forces. A long-ranged force is finite everywhere, but it decays
algebraically,

$f(r)\simeq\frac{A}{r^{a}}\quad\text{as}\quad r\to\infty,$ (3)

with $a>0$. For sufficiently large $a$, one anticipates the same behavior as
for short-ranged forces. Note that the force is repulsive if $A>0$, while it
is attractive if $A<0$.

We focus on short-ranged regular forces (no singular behavior as $r\to 0$) and
show that the emerging behaviors of the passive particle at late times are
universal, e.g., independent of the details of the potential and whether the
force is repulsive or attractive. For long-ranged forces, the behavior may
depend on the decay exponent $a$ in (3) and also on the sign of the amplitude
$A$ in (3), i.e., on whether the force is asymptotically repulsive or
attractive.

Before going into the details of computations, we first summarize our main
results. We start with the case where the force $f(r)$ on the driven particle
is short-ranged and characterize its position fluctuations at late times. For
simplicity we assume that the driven particle is initially at the origin
($\boldsymbol{y}(0)=0$). The symmetry of the force then ensures that the
average position of the driven particle does not change with time, i.e.,
$\langle\boldsymbol{y}(t)\rangle=0$. We find that the effect of the
nonreciprocal interaction is captured by the mean-squared displacement (MSD)
which shows a remarkable universal behavior. Irrespective of the specific form
of the interaction, the MSD of the passive particle increases indefinitely
with time in $d\leq 2$ dimensions and saturates when $d>2$:

$\sigma_{d}^{2}(t)\equiv\langle
y^{2}(t)\rangle\sim\left(\frac{V_{0}R}{D}\right)^{2}\times\begin{cases}\sqrt{Dt/R^{2}}&d=1\\\
\ln(Dt/R^{2})&d=2\\\ O(1)&d>2\end{cases}$ (4)

Here $V_{0}$ is the typical strength of the short-range potential, $R$ is its
range, and $y=|\boldsymbol{y}|$. The dimensionless quantity $V_{0}/D$ plays
the role of the Péclet number.

To appreciate the sub-diffusive growth laws (4), let us consider a lattice
version in which a driver particle performs a random walk. In one dimension,
in the late time limit, the random walker visits $\sim\sqrt{t}$ sites Feller
(1977); Weiss (1994) and hence it frequently returns to the visited sites and,
on an average, spends $\sqrt{t}$ time in each such site. The passive particle
is affected when it is close to the random walker. This leads to a heuristic
estimate for the typical displacement of the passive particle

$\sum_{1}^{\sqrt{t}}\pm 1\sim t^{1/4},$ (5)

which, in turn leads to the $\sqrt{t}$ growth of the MSD of the driven
particle.

In two dimensions, the random walker visits $t/\ln t$ sites Feller (1977);
Weiss (1994); Popov (2021), so a visited site is visited on average $\ln t$
times. The typical displacement is therefore

$\sum_{1}^{\ln t}\pm 1\sim\sqrt{\ln t},$ (6)

leading to the $\log t$ growth of the MSD of the driven particle. The behavior
of the MSD in $d=1$ and $2$ are shown in Figs. 1 and 2. For $d>2$, the random
walker visits a site a few times and eventually wanders away and never returns
Feller (1977); Weiss (1994), so the typical spread of the driven particle
remains finite.

It turns out that the complete position distribution of the driven particle is
also universal and depends only on the dimension and not on the specific
short-ranged potential. In one dimension, the MSD of the passive particle
exhibits the same $\sqrt{t}$ growth as that of a tagged particle in single-
file diffusion Harris (1965); Levitt (1973); Richards (1977); Alexander and
Pincus (1978). The distribution of the displacement of the tagged particle in
single-file diffusion is Gaussian Arratia (1983). One could expect the same
simple behavior for the displacement of the passive particle, but this is not
so. The distribution of the passive particle position $y(t)$ acquires a
scaling form

$P(y,t)\simeq\frac{1}{c_{1}\sigma_{1}(t)}\,{\mathcal{F}}_{1}(z),\qquad
z=\frac{y}{c_{1}\sigma_{1}(t)}$ (7a) in the long time limit, with scaling
function, $\displaystyle{\mathcal{F}}_{1}(z)$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi\sqrt{2}}\Bigg{[}\Gamma\left(\frac{1}{4}\right)\,_{0}F_{2}\left(\frac{1}{2},\frac{3}{4};-\frac{z^{4}}{256}\right)$
(7d)
$\displaystyle-\sqrt{2\pi}\,|z|\,_{0}F_{2}\left(\frac{3}{4},\frac{5}{4};-\frac{z^{4}}{256}\right)$
$\displaystyle-\frac{z^{2}}{2}\Gamma\left(\frac{3}{4}\right)\,_{0}F_{2}\left(\frac{5}{4},\frac{3}{2};-\frac{z^{4}}{256}\right)\Bigg{]}.$
Here
${}_{0}F_{2}(a,b;z)=\sum_{k=0}^{\infty}\frac{1}{(a)_{k}(b)_{k}}\,\frac{z^{k}}{k!}\,,$
(7e)

is a hypergeometric function where $(a)_{k}=\frac{\Gamma(a+k)}{\Gamma(a)}$ is
the Pochhammer symbol Graham  _et al._ (1994). The scaling function (7d) is
rather complicated but universal, the specific form of the interaction
potential only affects the constant $c_{1}$. The function
${\mathcal{F}}_{1}(z)$ is symmetric in $z$ and has the asymptotic behaviors

${\mathcal{F}}_{1}(z)\simeq\begin{cases}\displaystyle\frac{1}{2\sqrt{2}\pi}\Gamma\left[\frac{1}{4}\right]-\frac{|z|}{2\sqrt{\pi}}\quad&z\to
0\\\\[20.00003pt]
\displaystyle\frac{2^{1/6}}{\sqrt{3\pi}\,z^{1/3}}\,\exp\\!\left[-3\left(\frac{z}{4}\right)^{4/3}\right]\quad&z\to\infty.\end{cases}$
(8)

Thus the scaling function has a cusp at $z=0$ and decays super-exponentially
as $z\to\infty$. A plot of this scaling function is shown in Fig. 4.

In two dimensions, the distribution of the passive particle position is
asymptotically isotropic, i.e., depends only on $y=|\boldsymbol{y}|$, and has
a scaling form

$P(y,t)\simeq\frac{1}{[c_{2}\sigma_{2}(t)]^{2}}\,{\mathcal{F}}_{2}(z),\qquad
z=\frac{y}{c_{2}\sigma_{2}(t)}.$ (9a) The scaling function reads,
$\displaystyle{\mathcal{F}}_{2}(z)=\frac{1}{2\pi}\,K_{0}(z),$ (9b)

where $K_{0}(z)$ is the modified Bessel function of the second kind of order
zero. The constant $c_{2}$ is non-universal, viz., it depends on the specific
form of the interaction potential. The scaling function (9b) is again
universal and, amusingly, simpler than the scaling function (7d) in one
dimension. It has the asymptotic behaviors

${\mathcal{F}}_{2}(z)\simeq\begin{cases}\displaystyle-\frac{1}{2\pi}\,\ln
z+\frac{\ln 2-\gamma_{E}}{2\pi}\quad&z\to 0\\\\[10.00002pt]
\displaystyle\frac{1}{2\sqrt{2\pi z}}\,\exp(-z)\quad&z\to\infty,\end{cases}$
(10)

where $\gamma_{E}=0.5772\dots$ is the Euler constant. Thus the scaling
function diverges logarithmically as $z\to 0$, while it decays exponentially
for large $z$. A plot of the scaling function is shown in Fig. 5.

We next investigate the scenario where the passive particle is driven by a
long-ranged force of the form (3). We show that in one dimension, the MSD of
the driven particle grows algebraically at late times,

$\langle y^{2}(t)\rangle\sim t^{\phi}\,.$ (11)

The exponent $\phi$ is chiefly determined by the decay exponent $a>0$ of the
force [cf. Eq. (3)]

$\phi=\begin{cases}\frac{1}{2}&a>\frac{3}{2}\\\ 2-a&1<a<\frac{3}{2}\\\
\frac{2}{a+1}&0<a<1\quad(\text{repulsive})\\\
1&0<a<1\quad(\text{attractive})\,.\end{cases}$ (12)

The exponent $\phi$ undergoes a freezing transition at $a=\frac{3}{2}$: for
all $a>\frac{3}{2}$, the exponent has the same value $\phi=\frac{1}{2}$ as for
short-ranged potentials, while for $1<a<\frac{3}{2}$, the exponent depends on
$a$ as $\phi=2-a$. In the latter case, the driven particle still moves sub-
diffusively but faster than in the case of short-ranged potentials. The scaled
position distribution also depends on $a$ in this case. At the transition
point $a=3/2$, the growth of the MSD of the driven particle is logarithmically
faster than that for short-ranged potentials,

$\langle y^{2}(t)\rangle\sim\sqrt{t}\,(\log t)^{2}\quad\textrm{for}\quad
a=3/2.$ (13)

The behavior in the $a>1$ range remains largely universal, e.g., changing the
sign of the potential does not affect the leading asymptotic behavior.
However, in the $0<a<1$ range, the nature of the potential plays a role, and
qualitatively different behaviors emerge depending on whether the force is
repulsive or attractive, i.e., whether the amplitude in Eq. (3) is $A>0$ or
$A<0$.

The rest of the paper is organized as follows. We start with the case of
short-ranged interactions (Sec. II). First, we establish the asymptotic
behavior of the MSD in different dimensions (Sec. II.1). In Sec. II.2, we
analyze the position distribution employing a heuristic treatment. We argue
that this method leads to exact results for the scaled position distribution,
and support our claims numerically. In Sec. II.3, we present a few analytical
calculations for explicitly tractable one-dimensional short-ranged potentials.
In Sec. III, we study the one-dimensional passive particle driven by long-
ranged potentials. In Sec. IV, we present concluding remarks. The details of
some calculations are relegated to the Appendices.

##  II Short-ranged interaction

In this section, we explore the position fluctuations of the passive particle
when the interaction potential is short-ranged. Without loss of generality, we
set $x(0)=0$ and $y(0)=0$. Integrating Eq. (1b), we obtain

$\displaystyle y(t)=\int_{0}^{t}d\tau\,f[x(\tau)-y(\tau)]$ (14)

For simplicity, we use the one-dimensional notation; it is straightforward to
generalize to higher dimensions [see Appendix A for details]. For short-ranged
potentials, we anticipate that $|x(t)|\gg|y(t)|$. Hence, Eq. (14) reduces to a
Brownian functional,

$\displaystyle y(t)\simeq\int_{0}^{t}d\tau\,f[x(\tau)].$ (15)

Brownian functionals appear in various contexts both in physics and
mathematics, and they are well studied Majumdar (2005); Radice (2023). The
long-time approximation of the driven particle position being given by a
Brownian functional allows us to investigate the nature of fluctuations of
$y(t)$.

Figure 1: The MSD of the driven particle for short-ranged interaction in
$d=1$: Plot of $\langle y^{2}(t)\rangle$ versus $t$ measured from numerical
simulations using (a) tent potential [see Eq. (29)] and (b) Gaussian potential
[see Eq. (30)]. Different curves correspond to different values of the range
of interaction $R$, and the red dashed curves indicate the analytically
predicted $\sqrt{t}$ behavior. Here, we have taken $V_{0}=1$ and $D=1$.

###  II.1 Mean-squared displacement

We start with computation of the second moment $\langle y^{2}(t)\rangle$,
which is the same as the MSD since $\langle y(t)\rangle=0$ due to our
conventions about the intial condition. We are interested in the long time
behavior, so we employ Eq. (15) and obtain

$\displaystyle\langle
y^{2}(t)\rangle=2\int_{0}^{t}dt_{2}\int_{0}^{t_{2}}dt_{1}\,\Big{\langle}f(x(t_{2}))f(x(t_{1}))\Big{\rangle},$
(16)

where we have used the symmetry of the two-point correlation of the force
$\langle f(x(t_{1}))f(x(t_{2}))\rangle$ under the exchange of $t_{1}$ and
$t_{2}$. The two-point correlation, for $t_{2}>t_{1}$, reads

$\displaystyle\Big{\langle}f(x(t_{2}))f(x(t_{1}))\Big{\rangle}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}dx_{2}\int_{-\infty}^{\infty}dx_{1}\,f(x_{2})f(x_{1})$
(17) $\displaystyle\times$
$\displaystyle\mathcal{P}(\boldsymbol{2}|\boldsymbol{1})\mathcal{P}(\boldsymbol{1}|\boldsymbol{0}).$

We use the shorthand notations 2=$(x_{2},t_{2})$, 1=$(x_{1},t_{1})$, and
0=$(0,0)$. We denote by $\mathcal{P}(\boldsymbol{2}|\boldsymbol{1})$ the
probability density for the Brownian particle to move from $\boldsymbol{1}$ to
$\boldsymbol{2}$; similarly, $\mathcal{P}(\boldsymbol{1}|\boldsymbol{0})$ is
the probability density to move from $\boldsymbol{0}$ to $\boldsymbol{1}$.
These propagators are Gaussian; e.g.,

$\displaystyle\mathcal{P}(\boldsymbol{2}|\boldsymbol{1})=\frac{1}{\sqrt{4\pi
D(t_{2}-t_{1})}}\,\exp\\!{\left[-\frac{(x_{2}-x_{1})^{2}}{4D(t_{2}-t_{1})}\right]}\,.$
(18)

To compute the two-point correlation (17), it is convenient to express the
above propagator in Fourier space and substitute it in Eq. (17). Then,
performing the $x$-integrals, we get [see Appendix A for the details]

$\displaystyle\Big{\langle}f(x(t_{1}))f(x(t_{2}))\Big{\rangle}$
$\displaystyle=$
$\displaystyle\int\frac{dk_{1}}{2\pi}\int\frac{dk_{2}}{2\pi}\tilde{f}(k_{1}-k_{2})\tilde{f}(k_{2})$
(19) $\displaystyle\times$
$\displaystyle\exp{[-D(k_{1}^{2}t_{2}+k_{2}^{2}(t_{1}-t_{2})]},\qquad$ (20)

where $\tilde{f}(k)=\int_{-\infty}^{\infty}dz\,e^{ikz}f(z)$ is the Fourier
transform of the force $f(z)$. Substituting (20) in (16), and performing the
integrals over $t_{1}$ and $t_{2}$, we get

$\displaystyle\langle y^{2}(t)\rangle$ $\displaystyle=$
$\displaystyle\frac{2}{D^{2}}\int_{-\infty}^{\infty}\frac{dk_{1}}{2\pi}\int_{-\infty}^{\infty}\frac{dk_{2}}{2\pi}\frac{\tilde{f}(k_{1}-k_{2})\tilde{f}(k_{2})}{k_{1}^{2}k_{2}^{2}(k_{1}^{2}-k_{2}^{2})}$
(22)
$\displaystyle\times\left[k_{1}^{2}(1-e^{-Dk_{2}^{2}t})-k_{2}^{2}(1-e^{-Dk_{1}^{2}t})\right].~{}~{}$

The above expression gives the exact MSD for the process (15), but it cannot
be evaluated without knowing the explicit form of the force. However, we are
primarily interested in the long-time behavior where the process (15) provides
an asymptotically exact description of the actual evolution of the passive
particle. Therefore, we focus on the large-time behavior of Eq. (22) where we
can advance further. To this end, let us consider the Laplace transform of the
second moment, $M_{2}(s)=\int_{0}^{\infty}dt\,e^{-st}\langle y^{2}(t)\rangle$.
From Eq. (22) we then have

$\displaystyle M_{2}(s)$ $\displaystyle=$
$\displaystyle\frac{2}{s}\int_{-\infty}^{\infty}\frac{dk_{1}}{2\pi}\int_{-\infty}^{\infty}\frac{dk_{2}}{2\pi}\frac{\tilde{f}(k_{1}-k_{2})\tilde{f}(k_{2})}{(Dk_{1}^{2}+s)(Dk_{2}^{2}+s)}.\qquad$
(23)

The large-time behavior of the MSD is controlled by the small $s$ behavior of
its Laplace transform $M_{2}(s)$. The small $s$ behavior of the Laplace
transform $M_{2}(s)$ can be extracted systematically from Eq. (23), and it
turns out that it is independent of the specific form of $\tilde{f}(k)$. See
Appendix A for the details of the computation. Here, we only sketch the main
steps.

For small $s$, the dominant contribution to the integrals in Eq. (23) comes
from the region $-\Lambda\leq(k_{1},k_{2})\leq\Lambda$ where $\Lambda\sim
O(1)$ is an ultraviolet cut-off. Consequently, it suffices to consider the
small $k$ behavior of $\tilde{f}(k)$, which is analytic for a short-ranged
force. Hence, for small $k$, we must have, $\tilde{f}(k)=-ia_{1}k+O(k^{2})$
where $a_{1}=2\int_{0}^{\infty}dz\,zV^{\prime}(z)$ contains the information
about the specific interaction potential. Using this approximation and
performing the integrals in Eq. (23), we get

$\displaystyle M_{2}(s)\simeq\frac{a_{1}^{2}\Lambda}{\pi(Ds)^{3/2}}$ (24)

for small $s$ [see Appendix A for the details]. The above equation immediately
leads to

$\displaystyle\langle y^{2}(t)\rangle\simeq\frac{2a_{1}^{2}\Lambda}{(\pi
D)^{3/2}}\,\sqrt{t}$ (25)

for short-ranged forces in one dimension. Clearly, the specific form of the
potential only affects the numerical factor $a_{1}$, and the asymptotic
$\sqrt{t}$ growth of the MSD of the driven particle is universal for short-
ranged interactions. This sub-diffusive growth law is one of the main results
of this paper.

It is straightforward to generalize the above analysis to higher dimensions.
The $k_{1}$ and $k_{2}$ integrals in Eq. (23) become $d$-dimensional [see Eq.
(124)]. For small $s$, the dominating contribution to the $k$-integrals comes
again from the small $|k|$ regime. For short-ranged rotationally symmetric
interactions in two dimensions, we get
$\tilde{\boldsymbol{f}}(\boldsymbol{k})=-ia_{2}{\boldsymbol{k}}+O(k^{2})$
where $a_{2}=\frac{\pi}{2}\int_{0}^{\infty}dz\,z^{2}V^{\prime}(z)$ contains
the information about the interaction. Using the linear approximation for
$\tilde{\boldsymbol{f}}(\boldsymbol{k})$, the $k$ integrals can be performed
[see Appendix A]. Expanding the resulting expression for small $s$ we arrive
at

$\displaystyle
M_{2}(s)\simeq\frac{a_{2}^{2}}{8\pi^{2}D^{2}}\frac{\log(D/s)}{s},$ (26)

which can be inverted to obtain the long-time behavior of the MSD in two
dimensions:

$\displaystyle\langle
y^{2}(t)\rangle\simeq\frac{a_{2}^{2}}{8\pi^{2}D^{2}}(\gamma_{E}+\log Dt),$
(27)

where $\gamma_{E}$ denotes the Euler constant. Once again, we find a universal
behavior, namely a $\log t$ growth of the MSD, in $d=2$.

Following the same analysis, in $d=3$ we obtain

$\displaystyle
M_{2}(s)\simeq\frac{a_{3}^{2}\Lambda^{4}}{6\pi^{4}D^{2}s}+O(s^{-3/2}),$ (28)

which, in turn, implies that the MSD eventually reaches a constant value in
three dimensions. The same holds for all $d\geq 3$.

Figure 2: The behavior of the MSD of the driven particle in $d=2$ for (a) tent
potential [see Eq. (29)] and (b) Gaussian potential [see Eq. (30)], obtained
from numerical simulations with $D=1$ and the same $V_{0}=1$ for both
potentials and three different values of the range $R$. We have taken
$x_{0}=0.1$ for (a) and $x_{0}=0$ for (b). The red dashed lines on both panels
indicate the $\log t$ behavior.

We illustrate the universal behavior of the MSD at late times numerically in
one and two dimensions in Fig. 2, for two different short-ranged interaction
potentials. First is a tent potential, which is nothing but a truncated linear
potential,

$\displaystyle
V(\boldsymbol{r})=\begin{cases}V_{0}\Big{(}1-\frac{r}{R}\Big{)}~{}\text{for}~{}r<R\cr
0\qquad\qquad\quad\text{otherwise}.\end{cases}$ (29)

The driven particle in the tent potential feels a constant force $V_{0}/R$
whenever its distance from the Brownian particle is less than $R$, the range
of the potential (29). Our second example is a Gaussian interaction potential,

$\displaystyle V(\boldsymbol{r})=V_{0}\,e^{-(r/R)^{2}},$ (30)

which has a typical range $R$. The potentials (29) and (30) are attractive
when $V_{0}<0$ and repulsive when $V_{0}>0$. In one dimension, we performed
exact analutical calculations for the tent and Gaussian potential, and for the
square potential (Sec. II.3 and Appendixes). Schematic representations of
these potentials are shown in Fig. 3.

We performed numerical simulations of the two-particle dynamics (1a)–(1b)
using interaction potentials (29) and (30) in one and two spatial dimensions.
Figure 1 illustrates that, for $d=1$, the MSD asymptotically grows as
$\sqrt{t}$ for both interaction potentials. The results from the $d=2$
simulations [Fig. 2] illustrate the logarithmic growth, $\langle
y^{2}(t)\rangle\sim\log t$, for both interaction potentials.

Figure 3: The square, Gaussian, and tent potentials in one dimension (top to
bottom in the $|x|<R$ range).

###  II.2 Position distribution

The universal nature of the MSD of the driven particle raises the obvious
question of whether similar universal features are manifest in the full
position distribution of the driven particle. To answer this question, we
investigate the position distribution $P(y,t)$ at large-times $t$ using two
different approaches. Both approaches apply in the long time limit and give
exact scaled distributions. Their status is different, however. A more
rigorous approach based on the Feynman-Kac formalism is challenging, and we
implemented it only in some specific examples of short-ranged potentials in
one dimension where we succeeded in computing the distribution explicitly
(Sec. II.3). The second approach is less rigorous, and we call it a heuristic
argument. This approach relies on the first-passage properties of the Brownian
particle, and its obvious advantage is that the computations do not depend on
the specific form of the short-ranged potential that affects only a non-
universal constant. We implemented the second approach both in one and two
dimensions. Whenever both approaches are applicable, they give the same
universal scaling behavior. Here, we show how to determine the scaling
functions in the realm of the heuristic approach that is less rigorous but
applies to arbitrary short-ranged potentials in one and two dimensions.

The position distribution $P(y,t)$ of the driven particle is affected when it
is close to the position $x(t)$ of the Brownian particle. Hence, the
displacement of the driven particle can be estimated as

$\displaystyle y=\sum_{j=1}^{N}\eta_{j},$ (31)

where $N$ denotes the number of times the Brownian particle comes close to the
driven particle, and $\eta_{j}$ denotes the displacement of the driven
particle at each such instance. We have already used a simpler version of the
above equation, viz. Eq. (5), for a heuristic estimate of the typical
displacement of the driven particle. Equation (31) also looks like a heuristic
estimate. However, the discrete formulation implied by Eq. (31) becomes
asymptotically exact, and the instances when the interaction is non-negligible
become independent. Indeed, the driven particle is much ‘slower’ than the
Brownian particle, so it does not move far from its initial position. Hence,
one can think about $N$ as the number of returns of the Brownian particle to
the origin. The potential must be short-ranged to justify the discrete
treatment of interaction instances. Thus the goal is to improve the naive
estimate in Eq. (5) where we replaced $N$ by a typical number of return
events. In Eq. (5), we also used the simplest estimate $\eta_{j}=\pm 1$. This
simplification only misses the non-universal constant $c_{1}$, but we can
restore it at the end using the calculation of the MSD presented in Sec. II.1.

The above arguments imply that position distribution of the driven particle
can then be expressed as

$\displaystyle P(y,t)=\sum_{N}P(y|N)P(N|t)$ (32)

where $P(y|N)$ denotes the probability that the driven particle has a
displacement $y$, given $N$ return events, and $P(N|t)$ denotes the
conditional probability that there are $N$ returns of the Brownian particle to
the origin, during the interval $[0,t]$. Note that, for the sake of
simplicity, we have used the same letter $P$ to denote all probability
distributions.

To compute $P(N|t)$ we adopt a trajectory based approach. Let us consider a
trajectory of the Brownian particle with $N$ returns to the origin and let
$\tau_{i}$ denote the interval between $i-1$-th and $i$-th returns. The
probability of such a trajectory is

$\displaystyle
P(N,\\{\tau_{i}\\}|t)=\prod_{i=1}^{N}g(\tau_{i})q(\tau_{N+1})\delta(t-\sum_{i=1}^{N+1}\tau_{i})$
(33)

where $g(\tau)$ denotes the first passage probability of the Brownian motion
and

$\displaystyle
q(\tau_{N+1})=\int_{\tau_{N+1}}^{\infty}d\tau^{\prime}\,g(\tau^{\prime})$ (34)

denotes the probability that there were no returns during the final interval
$\tau_{N+1}$. Combining the contributions from all such trajectories with $N$
returns we arrive at

$\displaystyle
P(N|t)=\int\prod_{i=1}^{N}d\tau_{i}\,g(\tau_{i})\,d\tau_{N+1}\,q(\tau_{N+1})\,\delta\Big{(}t-\sum_{i=1}^{N+1}\tau_{i}\Big{)}.$
(35)

It is convenient to consider its Laplace transform,

$\displaystyle\tilde{P}(N,s)=\int_{0}^{\infty}P(N|t)e^{-st}dt=[\tilde{g}(s)]^{N}\tilde{q}(s),$
(36)

where $\tilde{g}(s)$ and $\tilde{q}(s)$ denote the Laplace transforms of
$g(t)$ and $q(t)$, respectively. These two quantities are related. Indeed,
performing the Laplace transform of Eq. (34) and using $q(0)=1$ we obtain

$\displaystyle\tilde{q}(s)=\frac{1-\tilde{g}(s)}{s}\,.$ (37)

In the following, we use Eq. (35) along with the known results for $P(y|N)$ to
compute $P(y,t)$ in $d=1$ and $d=2$.

Figure 4: Scaled position distribution of the driven particle in $d=1$
measured from numerical simulations with tent potential (open symbols) and
Gaussian potential (filled symbols). We have taken $D=1,R=0.5$, $V_{0}=1$, and
$x_{0}=0$. The solid black line indicates the analytically predicted scaling
function (7d). The non-universal scaling factor, $c$ in the figure, is
different for the two interaction potentials.

In one spatial dimension, the driven particle can only move in two directions.
The displacements, which can be set to unity as we argued before as along as
we are interested only in the functional form of the scaled position
distribution, are therefore $\eta_{i}=\pm 1$. The distribution of $y$ for a
given number, $N$, of interactions with the Brownian particle is
asymptotically Gaussian:

$\displaystyle P(y|N)=\frac{1}{\sqrt{4\pi
N}}\exp{\left(-\frac{y^{2}}{4N}\right)},$ (38)

for $N\gg 1$. Next, we need to compute $\tilde{P}(N,s)$ which requires
determination of $\tilde{g}(s)$ [cf. Eqs. (36) and (37)]. The first passage
properties of one-dimensional Brownian motion are well studied and the exact
form of $g(\tau)$ is known Weiss (1994),

$\displaystyle g(\tau)=\frac{|\Delta x|}{\sqrt{4\pi
D\tau^{3}}}\exp{\left[-\frac{(\Delta x)^{2}}{4D\tau}\right]},$ (39)

where $\Delta x$ denotes the initial distance of the Brownian and driven
particles. The corresponding Laplace transform is
$\tilde{g}(s)=e^{-b\sqrt{s}}$ with $b=\Delta x/\sqrt{D}$. We can set $b=1$ as
the precise value only affects the non-universal constant $c_{1}$ which we fix
otherwise [see Sec. II.1 and Appendix A]. Using $\tilde{g}(s)=e^{-\sqrt{s}}$
and Eq. (37) we have $\tilde{q}(s)=s^{-1}\big{(}1-e^{-\sqrt{s}}\big{)}$ and
deduce

$\tilde{P}(N,s)=\frac{1-e^{-\sqrt{s}}}{s}\,e^{-N\sqrt{s}}$ (40)

from Eq. (36). We are primarily interested in the large $t$ behavior, hence,
it suffices to consider the small $s$ behavior of the Laplace transform,

$\displaystyle\tilde{P}(N,s)\simeq\frac{e^{-N\sqrt{s}}}{\sqrt{s}}\,,$ (41)

which we invert to find

$\displaystyle P(N,t)\simeq\frac{1}{\sqrt{\pi
t}}\exp{\left(-\frac{N^{2}}{4t}\right)}.$ (42)

Replacing the summation in Eq. (32) by integration we obtain

$\displaystyle
P(y,t)\simeq\int_{0}^{\infty}\frac{dN}{\sqrt{4N\pi^{2}t}}\,\exp\\!\left[-\frac{y^{2}}{4N}-\frac{N^{2}}{4t}\right].$
(43)

Let us remark that this heuristic argument esentially maps the effective
motion of the passive particle to a continuous-time random walk (CTRW) problem
Montroll and Scher (1973). CTRW refers to a walker that performs $\pm 1$
random walk of $N$ steps, but before every jump the walker waits at a lattice
site for a random time $\tau$ drawn, independently for each jump, from a
waiting time distribution $g(\tau)$.

The distribution (43) can be expressed in a scaling form

$\displaystyle
P(y,t)\simeq\frac{1}{t^{1/4}}{\mathcal{F}}_{1}\left(\frac{y}{t^{1/4}}\right)$
(44)

where the scaling function is given by

$\displaystyle{\mathcal{F}}_{1}(z)=\frac{1}{\pi\sqrt{2}}\int_{0}^{\infty}\frac{du}{\sqrt{u}}\,\exp\\!\left[-\frac{z^{2}}{8u}-u^{2}\right]\,.$
(45)

The integral can be evaluated exactly leading to the announced scaling
function (7d). This is one of the central results of this work—for short-
ranged interactions, the distribution of typical position fluctuations of the
driven particle in the large-time regime is universal and independent of the
specific form of the interaction potential. The asymptotic behaviors of the
scaling function $\mathcal{F}_{1}(z)$ are given in (8). Note that the large
$z$ behavior in (8) is easier to extract by performing a saddle point analysis
of the integral representation (45), rather than from the formal expression in
(7d).

The universal nature of the position fluctuations of the driven particle is
illustrated in Fig. 4 comparing the scaled position distribution obtained from
numerical simulations with interaction potential (29) and (30) with the
analytical prediction Eq. (7d). The match is excellent. We used a numerical
factor to obtain the collapse for the two different interaction potentials.
This non-universal factor $c_{1}$ appearing in Eq. (7a) depends on the
specific form of the interaction and cannot be determined from the heuristic
argument presented above. In Sec. II.3, we discuss some specific examples
where we deduce this factor from an exact computation of the distribution
$P(y,t)$.

Figure 5: Scaled position distribution of the driven particle in $d=2$ for (a)
tent potential and (b) Gaussian potential. The different curves correspond to
the distribution of the scaled position $z=|y|/(c_{2}\sqrt{\log t})$ obtained
from numerical simulations; the non-universal factor $c_{2}$ is different for
the two potentials. The solid black lines correspond to the analytical
prediction (9b). We have taken $D=1,V_{0}=1$ and $R=0.5$ for both the cases.

This heuristic argument can be extended to two dimensions. In this case, the
approximation (31) describes a two-dimensional random walker implying a
Gaussian distribution of $y$ for a given $N$:

$\displaystyle P(y|N)=\frac{1}{4\pi
N}\,\exp\\!\left[-\frac{|y|^{2}}{4N}\right].$ (46)

The next step is to obtain $P(N|t)$, i.e., the probability that the two-
dimensional Brownian particle $x$ has $N$ return events to the origin within a
duration $t$, which is given by Eq. (35). A convenient way to find $P(N|t)$ is
again via its Laplace transform [see Eq. (36)]. Thus, we need $g(s)$, the
Laplace transform of the first-passage time distribution. For a two-
dimensional Brownian motion, the first-passage time is the time to return
within a radial distance $a$ of the origin, starting from a radial distance
$b>a$. Let $S(t,a,b)$ be the corresponding survival probability, i.e., the
probability that, starting from a distance $b$, the Brownian particle has not
returned within a distance $a$ of the origin up to time $t$. It is known that
the Laplace transform of $S(t,a,b)$ is given by Spitzer (1958),

$\displaystyle\int_{0}^{t}dt\,e^{-st}S(t,a,b)=\frac{K_{0}(b\sqrt{2s})}{sK_{0}(a\sqrt{2s})}\,,$
(47)

where $K_{0}(z)$ is the modified Bessel function of the second kind of order
zero. The corresponding first-passage time distribution is given by
$g(t)=-\partial_{t}S(t,a,b)$, so its Laplace transform reads

$\displaystyle\tilde{g}(s)=\frac{K_{0}(b\sqrt{2s})}{K_{0}(a\sqrt{2s})}\,.$
(48)

Since we are interested in the long-time regime, it suffices to consider the
small $s$ behavior of $\tilde{g}(s)$,

$\displaystyle\tilde{g}(s)=1+\frac{2\log(b/a)}{\log s}+O(s\log s).$ (49)

Consequently

$\displaystyle\tilde{q}(s)\simeq-\frac{c}{s\log s},$ (50)

where $c=2\log(b/a)$ is a non-universal constant. Using this result together
with Eqs. (36)–(37) we can write

$\displaystyle\tilde{P}(N,s)\simeq\frac{c}{s\log s}\exp{\bigg{(}\frac{cN}{\log
s}\bigg{)}}$ (51)

for small $s$ and large $N$. From the above equation, one can get the large
time behavior of $P(N,t)$ by inverting the Laplace transform using Bronwich
integral,

$\displaystyle P(N,t)=-\frac{c}{2\pi i}\int_{\Gamma}\frac{ds}{s\log
s}e^{st}\exp{\bigg{(}\frac{cN}{\log s}\bigg{)}}$ (52)

where $\Gamma$ denotes the standard Bromwich contour which encloses all the
singularties of the integrand. It is convenient to use a change of variable
$z=st$ which yields

$\displaystyle P(N,t)=-\frac{c}{2\pi i}\int_{\Gamma}\frac{dz}{z(\log z-\log
t)}e^{z}\exp{\bigg{(}\frac{cN}{\log z-\log t}\bigg{)}}.~{}$

This integral is dominated by the region where $z=O(1)$, i.e., $\log z\ll\log
t$. Thus

$\displaystyle P(N,t)\simeq\frac{c}{\log t}\exp\left(-\frac{cN}{\log
t}\right)\frac{1}{2\pi i}\int_{\Gamma}\frac{dz}{z}e^{z}$ (53)

from which

$\displaystyle P(N|t)\simeq\frac{1}{\log t}\exp{\left(-\frac{N}{\log
t}\right)},$ (54)

where, once again, we have put the non-universal constant $c=1$.

Substituting (46) and (54) into Eq. (32) and replacing summation by
integration we arrive at

$\displaystyle P(y,t)$ $\displaystyle\simeq$ $\displaystyle\frac{1}{4\pi\log
t}\int_{0}^{\infty}\frac{dN}{N}\exp{\Big{(}-\frac{y^{2}}{4N}-\frac{N}{\ln
t}\Big{)}}$ (55) $\displaystyle=$ $\displaystyle\frac{1}{\ln
t}{\mathcal{F}}_{2}\left(\frac{y}{\sqrt{\ln t}}\right)$ (56)

with the scaling function given by

$\displaystyle{\mathcal{F}}_{2}(z)=\frac{1}{4\pi}\int_{0}^{\infty}\frac{du}{u}\,\exp\\!\left[-\frac{z^{2}}{4u}-u\right]=\frac{1}{2\pi}\,K_{0}(z)\,,$
(57)

as announced in (9b).

Figure 5 shows a comparison of the distribution of the scaled radial distance
$z=|y|/(c_{2}\sqrt{\log t})$ of the driven particle, obtained from numerical
simulations for interaction potentials (29) and (30), with the predicted
scaling function (9b). By adjusting the non-universal factor $c_{2}$ an
excellent match is observed in both cases. The only exception is the small $z$
behavior where the ultimate logarithmic divergence [cf. Eq. (10)] is difficult
to see at a large but finite time.

Figure 6: Plot of $P(y,t)$ obtained from numerical simulations in $d=1$ with
repulsive ($V_{0}=1$) and attractive $(V_{0}=-1)$ Gaussian interaction
potential (30). We have taken $t=400$, $R=0.5$ and $D=1$.

The logarithmic growth of the MSD of the driven particle resembles an
anomalously slow diffusion on a two-dimensional lattice mediated by a single
vacancy Brummelhuis and Hilhorst (1988). Specifically, a symmetric exclusion
process on the square lattice with a single empty site (vacancy) was
investigated in Refs. Brummelhuis and Hilhorst (1988); Toroczkai (1997). The
vacancy performs a random walk, and whenever it occupies a site neighboring to
a tagged particle, the tagged particle can hop, exchanging its position with
the vacancy. The MSD of the tagged particle grows logarithmically and the
scaled position distribution follows (9b). The space is discrete (the square
lattice) and the analysis Brummelhuis and Hilhorst (1988); Toroczkai (1997) is
much more complicated than our analysis that also has the advantage of leading
to universal results independent of the short-ranged interaction potential,
and even on the nature of the potential, i.e., whether it is attractive,
repulsive, or a combination of both. An interesting attempt Newman (1999) of
developing a continuum theory of vacancy-mediated diffusion relied on a mean-
field approximation, and was also aimed at modeling the original lattice
problem Brummelhuis and Hilhorst (1988); Toroczkai (1997). Our setting is
continuous from the beginning, and we also treated arbitrary short-ranged
interaction potentials. As often happens in exact sciences, a more general
framework is frequently more tractable than a specific one.

For the one-dimensional symmetric exclusion process with a single empty site,
the motion of the tagged particle is trivial—it first hops to the right
(assuming that the vacancy was originally to the right of the tagged
particle), then to the left, then to the right, etc. Thus in this case, the
tagged particle is essentially confined only to two sites. A non-trivial
vacancy-meditated diffusion is feasible on quasi-one-dimensional lattices,
e.g., on the ladders. The emerging behavior is essentially the same as the
behavior of our driven particle in one dimension. The analysis Brummelhuis and
Hilhorst (1988) of the vacancy-mediated diffusion on the ladders is lengthy
and cumbersome, so the advantage of our continuous approach (in addition to
its applicability to arbitrary short-ranged interaction potentials) is even
more apparent.

The heuristic argument presented in this section does not depend on whether
the interaction potential is attractive or repulsive. Consequently, we expect
the results to hold irrespective of the sign of the force on the driven
particle. This is shown in Fig. 6 for $d=1$, where we compare $P(y,t)$
measured from numerical simulations for attractive ($V_{0}<0$) and repulsive
($V_{0}>0$) Gaussian interaction (30). The identical distribution in both the
cases illustrates that the universal scaling function is independent of the
sign of the interaction, i.e., whether the interaction is repulsive or
attractive.

Although the heuristic approach used in this subsection predicts the exact
scaling functions in $d=1$ and $d=2$, the non-universal factors $c_{1}$ and
$c_{2}$ remain unknown. We now present asymptotically exact computations in a
few explicit examples which allow the determination of these non-universal
factors. The emerging scaling functions are the same, supporting the validity
of the heuristic argument presented in this subsection.

###  II.3 Exact computations: Explicit examples

For short-ranged interactions in $d=1$, the Brownian functional form (15)
allows for the exact computation of the position distribution $P(y,t)$, using
the Feynman-Kac formalism. In this subsection, we derive the scaling function
for a few simple interaction potentials.

To compute the full distribution $P(y,t)$ for $y$ at fixed time $t$, we
consider the ‘Laplace transform’ of $P(y,t|x_{0})$, the distribution of $y$ at
time $t$, given that $x(0)=x_{0}$, with respect to $y$

$\displaystyle Q_{p}(x_{0},t)$ $\displaystyle=$
$\displaystyle\Big{\langle}e^{-p\int_{0}^{t}f(x(\tau))d\tau}\Big{\rangle}$
(58) $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}dy\,e^{-py}P(y,t|x_{0}),$

where $p$ denotes the conjugate variable. Note that the integration is over
$-\infty<y<+\infty$, as opposed to $0$ to $\infty$ used for the standard
Laplace transform. One can show Majumdar (2005) that $Q_{p}(x_{0},t)$ in Eq.
(58) satisfies the backward Feynman-Kac equation (where the starting point
$x_{0}$ is treated as a variable):

$\displaystyle\partial_{t}Q_{p}(x_{0},t)=D\partial_{x_{0}}^{2}Q_{p}(x_{0},t)-pf(x_{0})Q_{p}(x_{0},t).$
(59)

The function $Q_{p}(x_{0},t)$ must satisfy the initial condition
$Q_{p}(x_{0},0)=1$. This assertion follows from the definition in (58) and the
boundary conditions

$\displaystyle Q_{p}(x_{0}\to\pm\infty,t)=1.$ (60)

Indeed, if the particle starts at $\pm\infty$, at any finite time $t$, the
value of $f(x)$ in Eq. (58) is essentially zero since we assume that $f(x)\to
0$ as $x\to\pm\infty$. To find $P(y,t)$, we first need to solve Eq. (59) for
arbitrary $x_{0}$ and then set $x_{0}=0$ and invert the Laplace transform in
Eq. (58).

To solve Eq. (59), it is appropriate to take first the Laplace transform with
respect to $t$ by defining

$\displaystyle\tilde{Q}_{p}(x_{0},s)=\int_{0}^{\infty}dt\,e^{-st}Q_{p}(x_{0},t)$
(61)

Taking the Laplace transform of Eq. (59) and using the initial condition
$Q_{p}(x_{0},0)=1$ gives

$\displaystyle\mathcal{L}\tilde{Q}_{p}(x_{0},s)=-1$ (62a)
$\displaystyle\mathcal{L}=D\partial_{x_{0}}^{2}-s-pf(x_{0})$ (62b)

The boundary conditions

$\displaystyle\tilde{Q}_{p}(x_{0}\to\pm\infty,s)=\frac{1}{s}$ (63)

follow from (60). In general, Eq. (62) is hard to solve for arbitrary
$f(x_{0})$. So, we consider a couple of specific examples where
$\tilde{Q}_{p}(x_{0},s)$ can be computed explicitly.

####  II.3.1 Square potential

For the square potential

$\displaystyle V(x)=\left\\{\begin{split}V_{0}&\quad\text{for}\quad-R\leq
x\leq R\cr 0&\quad\text{for}\quad|x|>R\end{split}\right.$ (64)

the force is $f(x)=-V_{0}\delta(x+R)+V_{0}\delta(x-R)$. We thus need to solve
Eq. (62a) with

$\displaystyle\mathcal{L}=D\partial_{x_{0}}^{2}-s-pV_{0}[\delta(x_{0}-R)-\delta(x_{0}+R)].$
(65)

The solution is presented in Appendix B. In particular, defining

$\mathcal{Q}_{p}(s)=\tilde{Q}_{p}(x_{0}=0,s)\,,$ (66)

we find the exact expression

$\displaystyle\mathcal{Q}_{p}(s)=\frac{1}{s}\left[1-\frac{2\,p^{2}\,V_{0}^{2}\,\sinh(R\,\sqrt{s/D}\,)}{p^{2}\,V_{0}^{2}\,\sinh(2R\sqrt{s/D}\,)-2sD\,e^{2R\sqrt{s/D}}}\right]\,.$
(67)

Expanding $\mathcal{Q}_{p}(s)$ in powers of $p$ yields

$\displaystyle\mathcal{Q}_{p}(s)=\frac{1}{s}+\sum_{n\geq
1}\rho_{n}(s)\,p^{2n}$ (68a)
$\displaystyle\rho_{n}(s)=\frac{V_{0}^{2n}}{s(sD)^{n}}\,\left(\frac{\sinh(2R\sqrt{s/D})}{2}\right)^{n-1}$
(68b)
$\displaystyle\qquad\qquad\times\sinh\left(R\sqrt{\frac{s}{D}}\right)\,e^{-2nR\sqrt{s/D}},$
(68c)

where $(2n)!\,\rho_{n}(s)$ represents the Laplace transform of the $(2n)$-th
moment of $y(t)$. All odd moments vanish due to symmetry. The coefficient
$\rho_{1}(s)$ of the first term of the expansion in $p^{2}$ has the small $s$
behavior

$\rho_{1}(s)=\frac{V_{0}^{2}}{s^{2}D}\,\,\sinh(R\sqrt{s/D})e^{-2R\sqrt{s/D}}\simeq\frac{V_{0}^{2}R}{(sD)^{3/2}}$
(69)

from which we deduce (performing the inverse Laplace transform) the large-time
asymptotic of the MSD of the passive particle driven by the square potential

$\displaystyle\langle y^{2}(t)\rangle\simeq\frac{4RV_{0}^{2}}{\sqrt{\pi
D^{3}}}\sqrt{t}\,.$ (70)

To derive the full distribution $P(y,t)$ in the large time regime, we consider
the limit $s\to 0$ and $p\to 0$ in (67). To the leading order we get,

$\displaystyle\mathcal{Q}_{p}(s)\simeq\frac{1}{\sqrt{s}(\sqrt{s}-c_{1}^{2}p^{2})}.$
(71)

where $c_{1}^{2}=V_{0}^{2}R/D^{3/2}$. The large-time position distribution is
given by the double inverse Laplace transform of $\mathcal{Q}_{p}(s)$. This
can be done using a set of integral representations [see Appendix B for the
details]. First, we invert the Laplace transform with respect to $p$ and find,

$\displaystyle\hat{P}(y,s)\simeq\int_{0}^{\infty}\frac{dN}{\sqrt{4\pi
c_{1}^{2}Ns}}\,\,\exp{\Big{[}-N\sqrt{s}-\frac{y^{2}}{4c_{1}^{2}N}\Big{]}}.$
(72)

Then, we invert the Laplace transform with respect to $s$ to get,

$\displaystyle
P(y,t)\simeq\int_{0}^{\infty}\frac{dN}{\sqrt{4\pi^{2}c_{1}^{2}Nt}}\,\,\exp{\Big{[}-\frac{y^{2}}{4c_{1}^{2}N}-\frac{N^{2}}{4t}\Big{]}}.$
(73)

Making a change of variable $N=u\sqrt{4t}$, we get the scaling form for the
position distribution

$\displaystyle
P(y,t)\simeq\frac{1}{c_{1}t^{1/4}}{\mathcal{F}}_{1}\Big{(}\frac{y}{c_{1}t^{1/4}}\Big{)}.$
(74)

The scaling function ${\mathcal{F}}_{1}(z)$ is the same as predicted by the
heuristic argument, viz. it is given by the integral in Eq. (45) that reduces
to (7d). The exact solution also provides the non-universal scale factor
$c_{1}=V_{0}\sqrt{R}/D^{3/4}$.

####  II.3.2 Tent potential

For the tent potential (29), the corresponding force in $d=1$ is

$\displaystyle
f(x)=\left\\{\begin{split}\text{sgn}(x)\frac{V_{0}}{R}&\quad\text{for}~{}~{}|x|<R\cr
0&\quad\text{otherwise}.\end{split}\right.$ (75)

In this case, the backward Fokker-Planck equation is also solvable, see
Appendix C. As before, we compute the second moment by expanding
$\mathcal{Q}_{p}(s)$ around $p=0$, and inverting the Laplace transform of the
coefficient of $p^{2}$, we obtain

$\displaystyle\langle y^{2}(t)\rangle\simeq\frac{4RV_{0}^{2}}{3\sqrt{\pi
D^{3}}}\,\sqrt{t}\,.$ (76)

To derive the position distribution $P(y,t)$ of the driven particle we use the
same procedure as before [see Appendix C for details] and find

$\displaystyle\mathcal{Q}_{p}(s)\simeq\frac{1}{\sqrt{s}(\sqrt{s}-c_{1}^{2}p^{2})}$
(77)

with $c_{1}^{2}=RV_{0}^{2}/(3D^{3/2})$ when $s\to 0$ and $p\to 0$. This
formula differs from Eq. (71) only by the nonuniversal numerical factor
$c_{1}$. Thus, as expected, the scaling form (74) is the same for both
potentials.

Figure 7: Distribution of the scaled variable $z=y/(c_{1}t^{1/4})$ for the
tent potential (29) with $c_{1}=V_{0}\sqrt{R}/D^{3/4}$. The symbols denote the
data obtained from numerical simulations with $R=0.5$ and $V_{0}=1.0$, and the
solid line indicates the scaling function (7d). Here we have taken $D=1$.

Figure 7 shows a comparison of the scaled position distribution measured from
numerical simulations of the passive particle driven by the tent potential
with the theoretical prediction (7d) for the scaling function.

##  III Long-Ranged interaction

In this section, we focus on long-ranged interaction between the driver and
the driven particle, as in (3). For simplicity, we limit ourselves here to
$d=1$, and consider long-ranged forces that decay as

$f(z)\simeq\text{sgn}(z)\frac{A}{|z|^{a}}\quad\text{as}\quad|z|\to\infty,$
(78)

with the exponent $a>0$. The force can be either repulsive $(A>0)$ or
attractive $(A<0)$. Below we focus only on the MSD of the driven particle and
show that at late times it grows algebraically as $\sim t^{\phi}$ where the
exponent $\phi$ depends on the range of $a$, as announced in (12). Hence, we
consider the different ranges of $a$ separately below.

####  III.0.1 $a>1$: sub-diffusive growth.

To see whether the MSD of the driven particle shows any universal behavior for
long-ranged interactions, we start with Eq. (23), which gives the Laplace
transform of the MSD. We can rely on Eq. (23) only when the typical
displacement of the Brownian particle greatly exceeds a typical displacement
of the driven particle, $\langle y^{2}(t)\rangle\ll t$. In other words, the
driven particle exhibits sub-diffusive growth at late times. We will justify
this condition a posteriori.

We employ the same arguments as in the short-ranged interaction case. To
extract from Eq. (23) the small $s$ behavior of the Laplace transform
$M_{2}(s)$ of the MSD of the driven particle, we need the small $k$ behavior
of the Fourier transform of the force, $\tilde{f}(k)$. For long-ranged
interaction of the type in (78), the Fourier transform $\tilde{f}(k)$ is non-
analytic around $k=0$ for $a<2$ and its leading order behavior is

$\displaystyle\tilde{f}(k)\sim\begin{cases}k&\text{for}~{}~{}a>2\cr|k|^{a-1}&\text{for}~{}~{}a\leq
2.\end{cases}$ (79)

In this case, the small $s$ behavior of $M_{2}(s)$ is given by [see Appendix A
for the details]

$\displaystyle M_{2}(s)\sim\begin{cases}s^{a-3}&\text{for}\quad 1\leq
a\leq\frac{3}{2}\cr s^{-3/2}&\text{for}\quad a>\frac{3}{2}.\end{cases}$ (80)

Correspondingly, the long-time asymptotic behavior of the MSD of the driven
particle is given by

$\displaystyle\langle y^{2}(t)\rangle\sim t^{\phi}\quad{\textrm{a}s}\quad
t\to\infty\,,$ (81)

with the exponent

$\displaystyle\phi=\max\big{(}2-a,\tfrac{1}{2}\big{)}.$ (82)

The exponent $\phi$ characterizing the growth of the MSD of the driven
particle undergoes a ‘freezing transition’– $\phi=2-a$ is a strictly
decreasing function of $a$ in the range $1<a<\frac{3}{2}$, and it freezes at
the value $\phi=\frac{1}{2}$, the same as for short-ranged interactions, when
$a$ crosses the critical value $a_{c}=\frac{3}{2}$. The motion of the driven
particle is thus sub-diffusive, $\phi<1$, for any $a>1$, so our assumption
$\langle y^{2}(t)\rangle\ll t$ is valid in the long time limit justifying the
usage of Eq. (23). This proves the result announced in the first two lines of
(12).

Figure 8: Long-ranged interactions. The plot of the MSD of the passive
particle driven by the power-law interaction potential (83). The symbols
denote the data from numerical simulations, and the red dashed line indicates
the predicted asymptotic behavior (82) for $\mu<1/4$. Here we have taken
$V_{0}=1=D$.

We now illustrate this general result for $a>1$ by focusing on a class of
interaction potentials

$\displaystyle V(z)=\frac{V_{0}}{(1+z^{2})^{\mu}}\,,$ (83)

parametrized by the exponent $\mu>0$. The force associated to this potential
is long-ranged,

$f(z)=\frac{2\mu V_{0}z}{(1+z^{2})^{(1+\mu)}}\,,$ (84)

which decays for large $|z|$ as $|z|^{-(1+2\mu)}$. Hence, it corresponds to an
example of a force of the type in (78) with exponent

$a=1+2\mu\,.$ (85)

Since $\mu>0$, this corresponds to $a>1$ where we expect the result for MSD in
(81) and (82) to hold. It turns out that the Fourier transform of the force in
(84) can be written explicitly as

$\displaystyle\tilde{f}(k)=\frac{i\,V_{0}\,2^{\frac{3}{2}-\mu}\,\sqrt{\pi}}{\Gamma(\mu)}\,\text{Sgn}(k)\,|k|^{\mu+\frac{1}{2}}\,K_{\mu-\frac{1}{2}}(|k|)\,,$
(86)

where $K_{\nu}(z)$ denotes the modified Bessel function of the second kind of
order $\nu$.

Figure 9: Long-ranged interactions: Data collapse of the position distribution
of the passive particle driven by the power-law interaction potential (83)
obtained from numerical simulations for (a) $\mu<1/4$ and (b) $\mu>1/4$. (a)
Distribution of the scaled position $z=yt^{\mu-\frac{1}{2}}$ obtained from
numerical simulation with $\mu=0.1$. (b) Distribution of the scaled position
$z=y/(ct^{1/4})$, with an arbitrary scaling factor $c$, for $\mu=0.45$. The
solid line shows the scaling function (7d). Here we have taken $V_{0}=1$ and
$D=1$.

As discussed before, it suffices to look at the behavior of $\tilde{f}(k)$ for
small $k>0$. To extract this asymptotic, we need the small $k$ behavior of the
modified Bessel function of the second kind Bender and Orszag (1978)

$\displaystyle
K_{\nu}(k)\simeq\frac{1}{2}\left[\left(\frac{k}{2}\right)^{-\nu}\Gamma(\nu)+\left(\frac{k}{2}\right)^{\nu}\Gamma(-\nu)\right].$
(87)

This equation implies that the leading small $k$ behavior of the modified
Bessel function $K_{\mu-\frac{1}{2}}(k)$ is different depending on whether
$\mu>1/2$ or not. Substituting the asymptotic (87) into Eq. (86) we obtain the
anticipated algebraic behavior,

$\displaystyle\tilde{f}(k)\sim\begin{cases}|k|^{2\mu}&\text{for}~{}~{}\mu<\frac{1}{2}\cr
k&\text{for}~{}~{}\mu>\frac{1}{2}\end{cases}$ (88)

which is consistent with Eq. (79) with $a=1+2\mu$ [see Eq. (85)].
Consequently, the MSD of the passive particle driven by the long-ranged
potential (83) scales as

$\displaystyle\frac{\langle
y^{2}(t)\rangle}{V_{0}^{2}}\sim\begin{cases}t^{1-2\mu}&\text{for}~{}~{}\mu<\frac{1}{4}\cr\sqrt{t}&\text{for}~{}~{}\mu>\frac{1}{4}\end{cases}$
(89)

in the long time limit, with the ‘freezing transition’ occurring at
$\mu=\frac{1}{4}$. Using $a=1+2\mu$, this translates precisely into the result
in (82). This behavior is illustrated in Fig. 8 where this analytical
prediction is compared with the data obtained from numerical simulations,
which shows an excellent agreement.

To explore the signature of the freezing transition on the full position
distribution, we studied the position distribution $P(y,t)$ numerically. The
data support the emergence of the scaling in the long time limit:

$\displaystyle P(y,t)\simeq\frac{1}{\sqrt{\langle
y^{2}(t)\rangle}}\,{\mathcal{G}}_{\mu}\left(\frac{y}{\sqrt{\langle
y^{2}(t)\rangle}}\right).$ (90)

The freezing transition at $\mu=\frac{1}{4}$ is associated with a shape
transition of the scaling function. For $\mu>1/4$, as already indicated by Eq.
(89), the behavior of the driven particle is similar to that with a short-
ranged interaction and hence ${\mathcal{G}}_{\mu}(z)={\mathcal{F}}_{1}(z)$,
i.e., the scaling function is given by Eq. (7d). For $\mu<\frac{1}{4}$, the
distribution depends continuously on $\mu$ and has a bimodal shape; the
scaling function in this regime is unknown. We illustrate the shape transition
in Fig. 10, where we plot the scaled position distribution for two different
values of $\mu$ in the two regimes.

So far, we have tacitly assumed that $\mu\neq\frac{1}{4}$ and $\mu>0$. We now
analyze the behavior in the marginal cases $\mu=\frac{1}{4}$ and $\mu=0$, and
also the behavior when $\mu<0$. This range may seem unphysical, but recalling
that the force scales as $z^{-(1+2\mu)}$ when $z\to\infty$ we conclude that
the force asymptotically vanishes when $\mu>-\frac{1}{2}$. Below, we use the
potential (83) characterized by the exponent $\mu$ and a much more general
class of potentials with force decaying as $z^{-a}$. The qualitative results
are identical if we choose $a=1+2\mu$. In some calculations, we need the
entire potential, not only the tail, and we use the potential (83) in these
situations.

The marginal case $a=3/2$: For $\mu=\frac{1}{4}$ or equivalently $a=3/2$ in
(85), the leading small $k$ behavior of the modified Bessel function
$K_{0}(k)$ reads

$\displaystyle K_{0}(k)\simeq\log\tfrac{2}{k}-\gamma_{E}$ (91)

and Eq. (86) becomes

$\displaystyle\tilde{f}(k)\simeq
2\,i\,V_{0}\,k\,\big{[}\log\tfrac{2}{k}-\gamma_{E}\big{]}.$ (92)

The same analysis as before [see Appendix A] gives the small $s$ behavior
$M_{2}\sim s^{-3/2}(\log s)^{2}$ of the Laplace transform of the MSD of the
driven particle, from which the large-time behavior is

$\langle y^{2}(t)\rangle\sim V_{0}^{2}\,\sqrt{t}\,(\log t)^{2}\,.$ (93)

In addition to the announced temporal scaling (13), we included the dependence
on $V_{0}$ but omitted an exact amplitude as we omitted an exact
$\mu-$dependent amplitude in Eq. (89).

Figure 10: Long-ranged interaction with $a=3/2$: Plot of the MSD of the
passive particle driven by the power-law interaction potential (83) for the
marginal case $\mu=1/4$ (i.e., $a=3/2$), for different values of $V_{0}$. The
symbols denote the data from numerical simulations, and the red dashed lines
indicate the predicted asymptotic behavior (93). Here we have taken $D=1$.

Thus summarizing, for the passive particle driven by a long-ranged force
decaying as $z^{-a}$, we can restate growth laws (89) and (93) as

$\langle y^{2}(t)\rangle\sim\begin{cases}\sqrt{t}&a>\frac{3}{2}\\\
\sqrt{t}\,(\log t)^{2}&a=\frac{3}{2}\\\ t^{2-a}&1<a<\frac{3}{2}\end{cases}$
(94)

The exponent $\phi$ cited in (12) for $a>1$ immediately follows from (94).

####  III.0.2 $0<a<1$

To complete the growth laws of the MSD, we now probe the behavior in the
$0<a\leq 1$ range. Unfortunately, we can no longer rely on the methods
employed so far. Indeed, the driven particle is expected to be as mobile as
the Brownian particle, or even super-diffusive, so the chief starting point of
our analysis, the Brownian functional (15), is no longer applicable. We can
advance, however, once we recognize the crucial distinction between the
attractive $A<0$ and the repulsive $(A>0)$ long-ranged forces in (78) with
exponent $a$ in the range $0<a<1$.

If the force is repulsive $(A>0)$, the passive particle eventually settles on
one of the two half-lines, $y>0$ or $y<0$, and remains there. Both outcomes
are equiprobable if the starting positions coincide. In the former case, we
anticipate that the passive particle is pushed to a distance $y\gg\sqrt{Dt}$.
In this case, the force $A/z^{a}$ [cf. Eq. (78)] simplifies to $A/y^{a}$.
Therefore, the average position satisfies the deterministic equation

$\dot{y}=\frac{A}{y^{a}}$ (95)

in the leading order, from which

$y=[A\,(a+1)\,t]^{\frac{1}{a+1}}$ (96)

where $A>0$ ince the force is repulsive) and $0<a<1$. This provides the third
line of the result announced in (12). Thus in this case, the motion of the
driven particle is asymptotically deterministic, albeit the direction of the
motion is random. The scaling exponent obeys $\frac{1}{2}<\frac{1}{a+1}<1$
when $0<a<1$, so the motion is super-diffusive and sub-ballistic. Let us now
specialize to the potential in (83) or equivalently to the force in (84)
which, for large $z$, decays as $f(z)\simeq 2\mu V_{0}\,z^{-(1+2\mu)}$,
indicating that $A=2\mu V_{0}$ and $a=1+2\mu$ in (95). Since $0<a<1$, this
indicates that $-1/2<\mu<0$. In order that the force is repulsive, i.e.,
$A>0$, we must have $2\,\mu,V_{0}>0$. Since $-1/2<\mu<0$ is negative, we must
choose $V_{0}<0$ to have a repulsive force. The result in (96) then states

$y=[4\,V_{0}\,\mu\,(\mu+1)t]^{\frac{1}{2\,(\mu+1)}}\,,$ (97)

applicable when $0>\mu>-\frac{1}{2}$ and $V_{0}<0$.

Different behaviors of the driven particle emerge when the force is
attractive. Subtracting (1b) from (1a) we obtain a stochastic differential
equation

$\dot{z}=f(z)-\eta,$ (98)

for the separation $z=y-x$. The probability distribution $P(z,t)$ satisfies
the Fokker-Planck equation

$\frac{\partial P(z,t)}{\partial t}=-\frac{\partial}{\partial
z}\left[f(z)P(z,t)\right]+D\,\frac{\partial^{2}P(z,t)}{\partial z^{2}}\,.$
(99)

The crucial observation is that for slowly decaying attractive forces (3) with
exponent satisfying $0<a<1$, the driven particle is ‘slaved’ to the Brownian
particle which is mathematically reflected by the fact that the solution of
the Fokker-Planck equation becomes stationary in the long-time limit. Seeking
the stationary solution we reduce (99) to

$\frac{dP(z)}{dz}=\frac{f(z)}{D}\,P(z)$ (100)

which is solved to yield

$P(z)=C\,\exp\\!\left[\int_{0}^{z}dw\,\frac{f(w)}{D}\right]$ (101)

with amplitude $C$ fixed by normalization. Specializing (101) to the potential
(83) we obtain

$P(z)=C\,\exp\\!\left[-\frac{V_{0}}{D(1+z^{2})^{\mu}}\right]$ (102)

applicable when $0>\mu>-\frac{1}{2}$ and $V_{0}>0$ (since the force is
attractive).

Since the driven particle is slaved to the Brownian particle, the MSD of the
driven particle is

$\langle y^{2}(t)\rangle\simeq 2Dt$ (103)

in the leading order. The asymptotic behaviors (96) and (103) imply the
predictions for the exponent $\phi$ in the $0<a<1$ range announced in Eq.
(12).

####  III.0.3 $a=1$

Finally, we discuss the marginal case $a=1$. The entire force distribution is
necessary if the force is attractive [cf. Eq. (101)], and as an illustration
we take

$f(z)=-\frac{2\,A\,D\,z}{1+z^{2}}\,.$ (104)

The corresponding potential

$V(z)=A\,D\,\ln(1+z^{2})$ (105)

is a regularized $\mu=0$ version of the potential (83). Alternatively, this
potential is a regularized two-dimensional Coulomb potential, although we use
it in one dimension. We put the diffusion coefficient $D$ into (104), so the
amplitude $A$ is dimensionless. The force (104) is used in numerous
applications, particularly in modeling the spreading of the momentum of two-
level atoms in optical lattices Castin  _et al._ (1991); Marksteiner  _et al._
(1996); Lutz (2004) where $z$ is the momentum of an atom, $D$ describes
momentum fluctuations which lead to heating, and $f(z)$ the cooling force.
Hence, the Fokker-Planck equation with force (104) was the subject of several
articles Bray (2000); Castin  _et al._ (1991); Marksteiner  _et al._ (1996);
Lutz (2004); Chavanis and Lemou (2005); Levine  _et al._ (2005); Douglas  _et
al._ (2006); Kessler and Barkai (2010); Hirschberg  _et al._ (2011); Dechant
_et al._ (2011); Ray and Reuveni (2020); see Ref. Afek  _et al._ (2023) for a
review. In applications to laser cooling, $A>0$, so in our terminology, the
force is attractive. Hence, we can use known results in this case. A discrete
lattice model of a particle moving in a logarithmic potential was introduced
by Gillis in the probability literature Gillis (1956). For more recent studies
on the Gillis model and its generalizations, see Refs. Onofri  _et al._
(2020); Pozzoli  _et al._ (2020); Radice  _et al._ (2020); Artuso  _et al._
(2022); Zodage  _et al._ (2023). Here we focus on the continuous-space version
of the model, briefly re-derive a few simple results using our terminology and
notations and cite more complicated results.

The force (104) is attractive when $A>0$, and the stationary solution becomes

$P(z)=C(1+z^{2})^{-A}$ (106)

This solution is normalizable when $A>\frac{1}{2}$. In this case we fix the
amplitude from the normalization requirement,
$\int_{-\infty}^{\infty}dz\,P(z)=1$, and arrive at

$P(z)=\frac{\Gamma(A)}{\Gamma\big{(}\frac{1}{2}\big{)}\,\Gamma\big{(}A-\frac{1}{2}\big{)}}\,(1+z^{2})^{-A}$
(107)

The precise form (107) depends on the detailed form (104) of the force, but
the threshold value $A_{*}=\frac{1}{2}$ is universal. This assertion readily
follows from the general stationary solution (101).

The stationary solution (107) was noticed in Lutz (2004); Douglas  _et al._
(2006), and it was observed in optical lattice experiments Douglas  _et al._
(2006). Notice, however, that the stationary solution (107) predicts finite
second moment $\langle z^{2}\rangle$ only when $A>\frac{3}{2}$. The divergence
of the second moment was noticed in Kessler and Barkai (2010) where it was
shown that when $\frac{1}{2}<A\leq\frac{3}{2}$, the stationary solution (107)
is valid only in the central region $|z|\ll\sqrt{t}$; in the tail regions the
solution is time-dependent. The behavior of the second moment can be deduced
from a more comprehensive analysis Kessler and Barkai (2010) of the Fokker-
Planck with force (104):

$\langle z^{2}\rangle=\begin{cases}(2A-3)^{-1}&A>\frac{3}{2}\\\ \sim
t^{\frac{3}{2}-A}&\frac{1}{2}<A<\frac{3}{2}\\\
2D(1-2A)t&0<A<\frac{1}{2}\end{cases}$ (108)

We observe that when the force (104) is weakly attractive, $0<A<\frac{1}{2}$,
the separation between the driver and the passive particle grows diffusively.
This suggests to seek a scaling solution of the Fokker-Planck equation with
force (104):

$P(z,t)=\frac{1}{\sqrt{4Dt}}\,\mathcal{P}(\zeta),\qquad\zeta=\frac{z}{\sqrt{4Dt}}\,.$
(109a) In the range $0<A<\frac{1}{2}$, the scaled distribution
$\mathcal{P}(\zeta)=\frac{1}{\Gamma\big{(}\frac{1}{2}-A\big{)}}\,\zeta^{-2A}\,e^{-\zeta^{2}}$
(109b)

was found in Ref. Kessler and Barkai (2010). The same scaling solution applies
repulsive forces, $A<0$.

Using (109a)–(109b) one finds the diffusive growth

$\langle z^{2}\rangle=2\,D\,(1-2A)\,t$ (110)

asymptotically valid for $A<\frac{1}{2}$, i.e., for sufficiently weak
attractive forces (104) as it was already stated in Eq. (108) and for all
repulsive forces (104). In this range, the MSD of the driven particle is also
expected to grow diffusively

$\langle y^{2}(t)\rangle\simeq C(A)\,D\,t$ (111)

The amplitude $C(A)$ is unknown, apart from the obvious case of vanishing
force when $y\equiv 0$ and hence $C(0)=0$. The identity $\langle
z^{2}(t)\rangle=\langle x^{2}(t)\rangle-2\langle x(t)y(t)\rangle+\langle
y^{2}(t)\rangle$ and the knowledge of $\langle z^{2}(t)\rangle$, Eq. (110),
and $\langle x^{2}(t)\rangle=2Dt$ is insufficient since in this marginal case
the stochastic variables $x(t)$ and $y(t)$ are correlated and, in particular,
$\langle x(t)y(t)\rangle\neq 0$. The correlator cannot vanish because this
would lead to $C(A)=-4A$ which is clearly impossible for weakly attractive
forces: $0<A<\frac{1}{2}$. Only for very strong repulsive forces (104) the
naive prediction is asymtotically exact: $C(A)\simeq-4A$ when $A\to-\infty$.

##  IV Concluding Remarks

We have studied the motion of a passive particle driven by a non-reciprocal
interaction with a Brownian particle. We find that the late-time position
fluctuations of the driven particle are remarkably universal when the
interaction potential is short-ranged. The MSD of the driven particle scales
as $\sqrt{t}$ in $d=1$ and as $\log t$ in $d=2$, and the position
distributions are also universal and independent of the specific form of the
potential that only affects the numerical scale factor. In $d>2$, the typical
displacement of the passive particle from its original location remains
finite. We calculate the scaling functions in $d=1$ and $d=2$ using heuristic
arguments. We argue that our calculation is exact, modulo the scale factor,
and support our claims by making rigorous calculations for a few specific
interaction potentials and by numerical simulations.

We have also explored the motion of a passive particle in one dimension driven
by long-ranged interaction potentials. We have found a rich set of behaviors
chiefly determined by the exponent $a$ describing the decay of the force,
$f\sim z^{-a}$ when $z\to\infty$. The MSD of the driven particle grows as
$t^{\phi}$, with the exponent $\phi$ undergoing a freezing transition at
$a=\frac{3}{2}$: The same value $\phi=1/2$ as for short-ranged potentials
holds for $a>\frac{3}{2}$, while the exponent varies continuously in the
$1<a<\frac{3}{2}$ range where $\phi=2-a$. The behavior remains universal in
the $a>1$ range. The scaled position distribution is known only when
$a>\frac{3}{2}$, where it coincides with the scaled distribution in a short-
ranged potential. The scaled distributions in the $1<a<\frac{3}{2}$ are
unknown. The behavior in the $0<a<1$ range is less universal but simpler than
when $a\geq 1$. The behavior crucially depends on whether the potential is
attractive or repulsive. If the potential is repulsive, the driven particle
exhibits asymptotically self-averaging behavior—one of the two possible
directions is eventually selected (this varies from realization to
realization), and the motion is asymptotically deterministic. If the potential
is attractive, the passive particle is effectively slaved to the Brownian
particle. We studied the effect of long-ranged potentials in one dimension.
The extension to higher dimensions is left for future work.

In this paper, we restricted ourselves to a force $f(r)$ that is regular near
$r=0$. It would be interesting to study a singular force that diverges at
short distance as a power law $f(r)\simeq B\,r^{-b}$ with a positive exponent
$b>0$.

We mentioned the connection between our passive particle driven by a short-
ranged non-reciprocal interaction with a Brownian particle and diffusion of a
tagged particle mediated by a single vacancy Brummelhuis and Hilhorst (1988);
Toroczkai (1997); Newman (1999); Bénichou and Oshanin (2002). The latter
problem is traditionally studied in the realm of simple symmetric exclusion
process where particles can hop only to nearest neighbors, so in the case of a
single vacancy the number of possible moves is $2d$ in the case of the hyper-
cubic lattice $\mathbb{Z}^{d}$. Symmetric exclusion processes in which long-
ranged hops are allowed could be a counterpart of our problem with long-ranged
potential. In a supplementary direction, a charged tagged particle mediated by
a single vacancy exhibits intriguing behaviors Bénichou and Oshanin (2002)
suggesting a similar model in our setting, perhaps just considering a force
that is not an odd function of the separation.

Arcsine laws are old but still striking characteristics of Brownian motion
Lévy (1965); Itô and McKean (1996); Mörters and Peres (2010). One can ask
about analogues of arcsine laws for the passive particle. For long-ranged
forces, the answers could be easier than for short-ranged forces. For
instance, for forces decreasing as $z^{-a}$ when $z\to\infty$ with exponent
$0<a<1$, the distribution of the fraction $\tau$ of time spent by the passive
particle on the right half-line, $y>0$, is
$\frac{1}{2}\delta(\tau)+\frac{1}{2}\delta(\tau-1)$ when the force is
repulsive, and the same as for Brownian particle when the force is attractive.

In this work, we studied the most basic few-body system: A single driver
particle and a single passive particle, symbolically a (1,1) system.
Similarly, an $(m,n)$ system consists of $m$ Brownian particles and $n$
passive particles supplemented by non-reciprocal forces exerted by every
driver particle on every passive particle. The $(\infty,1)$ system in which
non-interacting Brownian particles are uniformly distributed in the
$\mathbb{R}^{d}$ space with density $\rho$ is an amusing microscopic model of
Langevin dynamics. The passive particle should behave diffusively. The
diffusion coefficient $\mathcal{D}$ of the passive particle depends on the
diffusion coefficient $D$ of Brownian particles, their density $\rho$, the
characteristic scale $V_{0}$ of the potential, and its range $R$. On
dimensional grounds, $\mathcal{D}=D\,\Phi(\rho R^{d},V_{0}/D)$, i.e., it
depends on two dimensionless parameters, the typical number $\rho R^{d}$ of
Brownian particles exerting the force on the passive particle, and $V_{0}/D$
playing the role of the Péclet number. Computing $\Phi(\rho R^{d},V_{0}/D)$,
e.g., for the Gaussian potential, is a promising avenue for future work.

###### Acknowledgements.

PLK is grateful to Y. Kafri for a useful discussion. UB acknowledges support
from Science and Engineering Research Board (SERB), India under the MATRICS
scheme (No. MTR/2023/000392). SNM acknowledges supports from ANR Grant No.
ANR-23-CE30-0020-01 EDIPS and the Science and Engineering Research Board
(SERB, Government of India), under the VAJRA faculty scheme (No.
VJR/2017/000110) during a visit to Raman Research Institute, where part of
this work was carried out.

##  Appendix A Large-time behavior of the MSD

In this Appendix we provide the detailed derivation of the long-time behavior
of the mean-squared displacement. We start with the $d$-dimensional version of
(15):

$\displaystyle\boldsymbol{y}(t)\simeq\int_{0}^{t}d\tau\,\boldsymbol{f}(x(\tau)),$
(112)

where $\boldsymbol{f}(\boldsymbol{z})=-\nabla V(\boldsymbol{z})$ denotes the
vector force derived from the central potential $V(|z|)$. The corresponding
mean-squared displacement is

$\displaystyle\langle
y^{2}(t)\rangle=2\int_{0}^{t}dt_{2}\int_{0}^{t_{2}}dt_{1}\left\langle\boldsymbol{f}(\boldsymbol{x}(t_{2}))\cdot\boldsymbol{f}(\boldsymbol{x}(t_{1}))\right\rangle$
(113)

where we have used the fact that the two-point correlation is symmetric under
the exchange $t_{1}$ and $t_{2}$. Hence, it suffices to compute the two-point
correlation for $t_{2}>t_{1}$, which is given by

$\displaystyle\Big{\langle}\boldsymbol{f}(\boldsymbol{x}(t_{2}))\cdot\boldsymbol{f}(\boldsymbol{x}(t_{1}))\Big{\rangle}=$
$\displaystyle\int d\boldsymbol{x}_{2}\int
d\boldsymbol{x}_{1}\,f(\boldsymbol{x}_{2})f(\boldsymbol{x}_{1})$ (114)
$\displaystyle\times\mathcal{P}(\boldsymbol{2}|\boldsymbol{1})\mathcal{P}(\boldsymbol{1}|\boldsymbol{0})$
(115)

where 2=$(\boldsymbol{x}_{2},t_{2})$, 1=$(\boldsymbol{x}_{1},t_{1})$,
0=$(\boldsymbol{0},0)$, and the propagator is given by

$\displaystyle\mathcal{P}(\boldsymbol{2}|\boldsymbol{1})=\frac{1}{[4\pi
D(t_{2}-t_{1})]^{\frac{d}{2}}}\exp{\left[-\frac{(\boldsymbol{x}_{2}-\boldsymbol{x}_{1})^{2}}{4D(t_{1}-t_{2})}\right]}$
(116)

in $d$ dimensions where we have taken $t_{2}>t_{1}$. To compute the two-point
correlation, we express the propagator as a Fourier integral and use it in Eq.
(115) to yield

$\displaystyle\left\langle\boldsymbol{f}(\boldsymbol{x}(t_{2}))\cdot\boldsymbol{f}(\boldsymbol{x}(t_{1}))\right\rangle=\int
d\boldsymbol{x_{1}}\int
d\boldsymbol{x_{2}}\,\boldsymbol{f}(\boldsymbol{x}_{2})\cdot\boldsymbol{f}(\boldsymbol{x}_{1})$
(117)
$\displaystyle\times\int\frac{d\boldsymbol{k}_{1}}{(2\pi)^{d}}\int\frac{d\boldsymbol{k}_{2}}{(2\pi)^{d}}\,\exp{[-D\\{k_{2}^{2}(t_{2}-t_{1})+k_{1}^{2}t_{1}\\}]}$
(118)
$\displaystyle~{}~{}~{}\times\exp{[-i\boldsymbol{k}_{2}\cdot(\boldsymbol{x}_{2}-\boldsymbol{x}_{1})-i\boldsymbol{k}_{1}\cdot\boldsymbol{x}_{1}]}.$
(119)

where $k_{i}^{2}=|\boldsymbol{k}_{i}|^{2}$. Performing the $\boldsymbol{x}$
integrals gives

$\displaystyle\Big{\langle}\boldsymbol{f}(\boldsymbol{x}(t_{2}))\cdot\boldsymbol{f}(\boldsymbol{x}(t_{1}))\Big{\rangle}=\int\frac{d\boldsymbol{k}_{1}}{(2\pi)^{d}}\int\frac{d\boldsymbol{k}_{2}}{(2\pi)^{d}}$
(120)
$\displaystyle\times\tilde{\boldsymbol{f}}(\boldsymbol{k}_{1}-\boldsymbol{k}_{2})\cdot\tilde{\boldsymbol{f}}(\boldsymbol{k}_{2})e^{-D[k_{1}^{2}t_{1}+k_{2}^{2}(t_{2}-t_{1})]}$
(121)

where $\tilde{\boldsymbol{f}}(\boldsymbol{k})=\int
d\boldsymbol{z}\,e^{i\boldsymbol{k}\cdot\boldsymbol{z}}\boldsymbol{f}(\boldsymbol{z})$
denotes the $d$-dimensional Fourier transform of the force. Substituting Eq.
(121) in (113) and performing the integrals over $t_{1}$ and $t_{2}$, we get,

$\displaystyle\langle y^{2}(t)\rangle$ $\displaystyle=$
$\displaystyle\frac{2}{D^{2}}\int\frac{d\boldsymbol{k}_{1}}{(2\pi)^{d}}\int\frac{d\boldsymbol{k}_{2}}{(2\pi)^{d}}\frac{\tilde{\boldsymbol{f}}(\boldsymbol{k}_{1}-\boldsymbol{k}_{2})\cdot\tilde{\boldsymbol{f}}(\boldsymbol{k}_{2})}{k_{1}^{2}k_{2}^{2}(k_{1}^{2}-k_{2}^{2})}$
(123)
$\displaystyle\times\left[k_{1}^{2}(1-e^{-Dk_{2}^{2}t})-k_{2}^{2}(1-e^{-Dk_{1}^{2}t})\right].$

The $k$-integrals are performed over the $d$-dimensional space. The above
expression gives the exact MSD for the process (112), but it cannot be
evaluated without knowing the explicit form of the force. However, since we
are primarily interested in the long-time behavior, we can proceed a bit
further. To this end, we consider the Laplace transform of the MSD with
respect to time, $M_{2}(s)=\int_{0}^{\infty}dt\,e^{-st}\langle
y^{2}(t)\rangle$. Performing the Laplace transform of Eq. (123) gives

$\displaystyle M_{2}(s)$ $\displaystyle=$
$\displaystyle\frac{2}{s}\int\frac{d\boldsymbol{k}_{1}}{(2\pi)^{d}}\int\frac{d\boldsymbol{k}_{2}}{(2\pi)^{d}}\frac{\tilde{\boldsymbol{f}}(\boldsymbol{k}_{1}-\boldsymbol{k}_{2})\cdot\tilde{\boldsymbol{f}}(\boldsymbol{k}_{2})}{(Dk_{1}^{2}+s)(Dk_{2}^{2}+s)}.\qquad$
(124)

We are interested in the large $t$ behavior of the MSD, which is controlled by
the small $s$ behavior of its Laplace transform $M_{2}(s)$. The small $s$
behavior of $M_{2}(s)$ cannot be extracted by naively putting $s=0$ in Eq.
(124) since the resulting integral diverges. To extract the leading divergence
of this double integral as $s\to 0$, we first split the integral into two

$\displaystyle M_{2}(s)=\frac{2}{s}[I_{1}(s)+I_{2}(s)],$ (125)

where $I_{1}(s)$ and $I_{2}(s)$ denote the contributions from the integral
over the region where $(|\boldsymbol{k}_{1}|,|\boldsymbol{k}_{2}|)\leq\Lambda$
with a constant cut-off $\Lambda\sim O(1)$, and outside this regime where
$(\boldsymbol{k}_{1}|,|\boldsymbol{k}_{2}|)>\Lambda$, respectively.

In the $s\to 0$ limit, $I_{2}(s)$ converges to a constant. This can be seen by
taking the limit in the integrand itself. Therefore

$\displaystyle M_{2}(s)=\frac{2}{s}I_{1}(s)+O\left(\frac{1}{s}\right).$ (126)

On the other hand, $I_{1}(s)$ is expected to diverge in the $s\to 0$ limit,
and consequently the small $s$ behavior of $M_{2}(s)$ is determined by the
contribution from $I_{1}(s)$. In the following we extract the small $s$
behavior of $M_{2}(s)$, by analysing the asymptotic behavior of $I_{1}(s)$,
for short and long-ranged forces, in different dimensions.

One dimension. In one dimension

$I_{1}(s)=\int_{-\Lambda}^{\Lambda}\frac{d{k}_{1}}{2\pi}\int_{-\Lambda}^{\Lambda}\frac{d{k}_{2}}{2\pi}\frac{\tilde{f}({k}_{1}-{k}_{2})\cdot\tilde{f}({k}_{2})}{(Dk_{1}^{2}+s)(Dk_{2}^{2}+s)}\,.$
(127)

To extract the small $s$ behavior, we first note that the symmetric nature of
the potential ensures that it suffices to consider the only the region
$(k_{1},k_{2})>0$. Using the scaled variables $q_{1}=k_{1}\sqrt{D/s}$ and
$q_{2}=k_{2}\sqrt{D/s}$, we recast Eq. (127) into

$\displaystyle I_{1}(s)$
$\displaystyle=\frac{4}{Ds}\int_{0}^{\overline{\Lambda}}\frac{dq_{1}}{2\pi}\int_{0}^{\overline{\Lambda}}\frac{dq_{2}}{2\pi}$
(128)
$\displaystyle\times\frac{\tilde{f}(\sqrt{\frac{s}{D}}(q_{1}-q_{2}))\tilde{f}(\sqrt{\frac{s}{D}}q_{2})}{(q_{1}^{2}+1)(q_{2}^{2}+1)}\,.$
(129)

Hereinafter we use the shorthand notation
$\overline{\Lambda}\equiv\Lambda\sqrt{D/s}$.

We now analyze the short-ranged and long-ranged forces separately. For short-
ranged forces, the Fourier transform of the force,
$\tilde{f}(k)=-2i\int_{0}^{\infty}dz\sin(kz)V^{\prime}(z)$, is analytic, and
to the leading order in $k$, $\tilde{f}(k)\simeq-ika_{1}$, where
$a_{1}=2\int_{0}^{\infty}dz\,zV^{\prime}(z)$. Therefore

$\displaystyle
I_{1}(s)\simeq\frac{4a_{1}^{2}}{D^{2}}\int_{0}^{\overline{\Lambda}}\frac{dq_{1}}{2\pi}\int_{0}^{\overline{\Lambda}}\frac{dq_{2}}{2\pi}\frac{q_{2}(q_{2}-q_{1})}{(q_{1}^{2}+1)(q_{2}^{2}+1)}.$
(130)

The $q_{2}$-integral is dominated by the contributions from large $q_{2}$, and
hence

$\displaystyle
I_{1}(s)\simeq\frac{4a_{1}^{2}}{D^{2}}\int_{0}^{\overline{\Lambda}}\frac{dq_{1}}{2\pi}\frac{1}{q_{1}^{2}+1}\int_{0}^{\overline{\Lambda}}\frac{dq_{2}}{2\pi}\frac{q_{2}^{2}}{q_{2}^{2}+1}\,.$
(131)

Evaluating the integrals we obtain

$\displaystyle
I_{1}(s)\simeq\frac{a_{1}^{2}}{\pi^{2}D^{2}}\tan^{-1}\overline{\Lambda}\left\\{\overline{\Lambda}-\tan^{-1}\overline{\Lambda}\right\\}.$
(132)

Recalling that $\overline{\Lambda}\equiv\Lambda\sqrt{D/s}$, taking the small
$s$ limit, and keeping only the leading small $s$ behavior of $M_{2}(s)$ we
arrive at Eq. (24).

For long-ranged forces the situation is a bit different since $\tilde{f}(k)$
cannot be expanded in a Taylor series. The expected small $k$ behavior is
$\tilde{f}(k)\simeq b\,\text{Sgn}(k)|k|^{\alpha}$ where $\alpha>0$ and $b$ is
a constant dependent on the specific potential. In that case, we have, from
Eq. (129),

$\displaystyle
I_{1}(s)\simeq\frac{4b^{2}s^{\alpha-1}}{D^{1+\alpha}}\int_{0}^{\overline{\Lambda}}\frac{dq_{1}}{2\pi}\frac{1}{q_{1}^{2}+1}\int_{0}^{\overline{\Lambda}}\frac{dq_{2}}{2\pi}\frac{q_{2}^{2\alpha}}{q_{2}^{2}+1}$
(133)

where we have used the fact that, similar to the short-ranged case, the
$q_{2}$ integral is dominated by the contributions from large $q_{2}$.
Performing the above integrals and taking the small $s$ limit yields

$\displaystyle I_{1}(s)\simeq\frac{b^{2}s^{\alpha-1}}{\pi
D^{\alpha+1}}[A+Bs^{\frac{1}{2}-\alpha}+O(s^{\frac{3}{2}-\alpha})],$ (134)

where $A$ and $B$ are $s$-independent constants. If $\alpha<1/2$, we have
$I_{1}(s)\sim s^{\alpha-1}$; for $\alpha>1/2$, $I_{1}(s)\sim s^{-1/2}$,
similar to the short-ranged scenario. Ignoring the sub-leading contributions
we get Eq. (80) in the main text.

Two dimensions. The Fourier transform of the rotational symmetric force reads

$\tilde{\boldsymbol{f}}(\boldsymbol{k})=-i\pi\frac{{\boldsymbol{k}}}{k}\int_{0}^{\infty}dz\,zJ_{1}(kz)V^{\prime}(z)$

where $J_{1}(z)$ is the Bessel function of the first kind of order one. Hence,
$\tilde{\boldsymbol{f}}(\boldsymbol{k})\simeq-ia_{2}{\boldsymbol{k}}$ with
$a_{2}=\frac{\pi}{2}\int_{0}^{\infty}dz\,z^{2}V^{\prime}(z)$ for short-ranged
forces in the leading order in $k$. The same arguments as in $d=1$ lead to

$\displaystyle
I_{1}(s)\simeq\frac{a_{2}^{2}s}{D^{3}}\int_{0}^{\overline{\Lambda}}\frac{dq_{1}}{2\pi}\frac{q_{1}}{q_{1}^{2}+1}\int_{0}^{\overline{\Lambda}}\frac{dq_{2}}{2\pi}\frac{q_{2}^{3}}{q_{2}^{2}+1}.$
(135)

Performing the integrals, we get

$\displaystyle
I_{1}(s)\simeq\frac{a_{2}^{2}s}{4\pi^{2}D^{3}}\log\left(1+\overline{\Lambda}^{2}\right)\Big{[}\overline{\Lambda}^{2}-\log\left(1+\overline{\Lambda}^{2}\right)\Big{]}.$

Taking the leading small $s$ behavior of the above expression we get the small
$s$ behavior of $M_{2}(s)$ quoted in Eq. (26) where we have put $\Lambda=1$
for simplicity.

Three dimensions. In the leading order in $k$, we have
$\tilde{\boldsymbol{f}}(\boldsymbol{k})\simeq-ia_{3}{\boldsymbol{k}}$ with
$a_{3}=\frac{4\pi}{3}\int_{0}^{\infty}dz\,z^{3}V^{\prime}(z)$. Consequently,
the $3d$ version of Eq. (131) is given by,

$\displaystyle I_{1}(s)\simeq\frac{a_{3}^{2}s}{(\pi
D)^{4}}\int_{0}^{\overline{\Lambda}}\frac{dq_{1}}{2\pi}\frac{q_{1}^{2}}{q_{1}^{2}+1}\int_{0}^{\overline{\Lambda}}\frac{dq_{2}}{2\pi}\frac{q_{2}^{4}}{q_{2}^{2}+1}.$
(136)

Performing the integrals, taking the small $s$ limit, and using it in Eq.
(126), we get Eq. (28) in the main text.

##  Appendix B Square potential

In this Appendix we provide the details for the computation of the MSD and
position distribution for the square-well interaction potential. We start from
the backward Feynman-Kac equation in the Laplace space (65), which is a second
order homogeneous differential equation except at $x_{0}=\pm R$. To solve this
equation, it is convenient to consider separately three regions and then match
the solutions in the different regimes.

  * •

Region I ($x_{0}\geq R$): Here the solution of (65) satisfying the boundary
condition $\tilde{Q}_{p}(x_{0}\to\infty,s)=1/s$ is

$\displaystyle\tilde{Q}_{p}(x_{0})=\frac{1}{s}+A_{1}e^{-x_{0}\,\sqrt{s/D}}$
(137)

where $A_{1}$ is an arbitrary constant to be determined. Hereinafter we
usually suppress $s$ variable, e.g.,
$\tilde{Q}_{p}(x_{0})\equiv\tilde{Q}_{p}(x_{0},s)$.

  * •

Region II ($-R\leq x_{0}\leq R$): In this region, the general solution reads,

$\displaystyle\tilde{Q}_{p}(x_{0})=\frac{1}{s}+A_{2}e^{x_{0}\,\sqrt{s/D}}+B_{2}e^{-x_{0}\,\sqrt{s/D}}$
(138)

where $A_{2}$ and $B_{2}$ are arbitrary constants.

  * •

Region III ($x_{0}\leq-R$): the general solution satisfying the boundary
condition $\tilde{Q}_{p}(x_{0})\to-\infty,s)=\frac{1}{s}$

$\displaystyle\tilde{Q}_{p}(x_{0})=\frac{1}{s}+A_{3}e^{x_{0}\,\sqrt{s/D}}$
(139)

where $A_{3}$ is an arbitrary constant.

To determine the unknown constants $A_{1}$, $A_{2}$, $B_{2}$ and $A_{3}$, we
use the fact that the solution $\tilde{Q}_{p}(x_{0},s)$ is continuous at
$x_{0}=R$ and $x_{0}=-R$. Additionally, by integrating Eq. (65) across a small
interval around $x=R$ and $x=-R$, we find two jump conditions to be satisfied
by the derivatives. Thus we have four conditions:

$\displaystyle\tilde{Q}_{p}(R^{+})$ $\displaystyle=$
$\displaystyle\tilde{Q}_{p}(R^{-})$ $\displaystyle
D\partial_{x_{0}}\\!\big{[}\tilde{Q}_{p}(R^{+})-\tilde{Q}_{p}(R^{-})\big{]}$
$\displaystyle=$ $\displaystyle pV_{0}\tilde{Q}_{p}(R)$
$\displaystyle\tilde{Q}_{p}(-R^{+})$ $\displaystyle=$
$\displaystyle\tilde{Q}_{p}(-R^{-})$ $\displaystyle
D\partial_{x_{0}}\\!\big{[}\tilde{Q}_{p}(-R^{+})-\tilde{Q}_{p}(-R^{-})\big{]}$
$\displaystyle=$ $\displaystyle-pV_{0}\tilde{Q}_{p}(-R)$

By inserting the explicit solutions (137)–(139) into the above conditions we
deduce four linear equations for four unknowns. After simplifications, they
read,

$\displaystyle A_{1}-A_{2}\rho^{2}-B_{2}$ $\displaystyle=0$ (140)
$\displaystyle(pV_{0}+\sqrt{sD})A_{1}+\sqrt{sD}(\rho^{2}A_{2}-B_{2})$
$\displaystyle=-\frac{pV_{0}}{s}\rho$ (141) $\displaystyle
A_{2}+B_{2}\rho^{2}-A_{3}$ $\displaystyle=0$ (142)
$\displaystyle\sqrt{sD}(A_{2}-\rho^{2}B_{2})+(pV_{0}-\sqrt{sD})A_{3}$
$\displaystyle=-\frac{pV_{0}}{s}\,\rho.$ (143)

where $\rho=e^{R\sqrt{\frac{s}{D}}}$. Using Eq. (138) we find

$\displaystyle\mathcal{Q}_{p}(s)\equiv\tilde{Q}_{p}(x_{0}=0,s)=\frac{1}{s}+A_{2}+B_{2}.$
(144)

Solving the linear equations (140)–(143) and substituting them in (144) gives
the result announced in Eq. (67).

To obtain the position distribution in the large time regime, we start from
Eq. (71). To invert the double Laplace transform, we use the integral
representation

$\displaystyle\frac{1}{\sqrt{s}-c_{1}^{2}p^{2}}\equiv\int_{0}^{\infty}dN\,\exp{[N(c_{1}^{2}p^{2}-\sqrt{s})]}$

and the identity

$\displaystyle\frac{1}{\sqrt{4\pi\sigma^{2}N}}\int_{-\infty}^{\infty}dy\,\exp{\Big{[}-\frac{y^{2}}{4\sigma^{2}N}-py\Big{]}}\equiv
e^{p^{2}\sigma^{2}N}$

to rewrite Eq. (71) as

$\displaystyle\mathcal{Q}_{p}(s)\simeq\int_{-\infty}^{\infty}dye^{-py}\int_{0}^{\infty}\frac{dN}{\sqrt{4\pi
c_{1}^{2}Ns}}\,e^{-N\sqrt{s}-\frac{y^{2}}{4c_{1}^{2}N}}\,.$ (145)

The Laplace transform $\hat{P}(y,s)={\mathcal{L}}_{t\to s}P(y,t)$ can then be
read off from the above equation, which is given in Eq. (72). Finally, the
Laplace transform with respect to $s$ can be inverted using the identity

$\displaystyle{\mathcal{L}}^{-1}_{s\to
t}\bigg{[}\frac{e^{-N\sqrt{s}}}{\sqrt{s}}\bigg{]}=\frac{1}{\sqrt{\pi
t}}\exp\\!\bigg{(}-\frac{N^{2}}{4t}\bigg{)}.$ (146)

Completing these steps we arrive at Eq. (73).

##  Appendix C Tent potential

To solve the backward Fokker-Planck equation (62) for the tent potential (29)
in one dimension, we consider separately four regions: (I) $x_{0}\geq R$, (II)
$0<x_{0}<R$, (III) $-R<x_{0}<0$ and (IV) $x_{0}<-R$. The solutions in
neighboring regions can then be matched using the appropriate boundary
conditions.

  * •

$|x_{0}|>R$: The force vanishes in these regions, and Eq. (62) reduces to

$\displaystyle D\partial_{x_{0}}^{2}\tilde{Q}_{p}-s\tilde{Q}_{p}=-1$ (147)

which is solved to yield

$\tilde{Q}_{p}(x_{0})=\begin{cases}C_{1}e^{x_{0}\sqrt{s/D}}+C_{2}e^{-x_{0}\sqrt{s/D}}+\frac{1}{s}&x_{0}<-R\\\
G_{1}e^{x_{0}\sqrt{s/D}}+G_{2}e^{-x_{0}\sqrt{s/D}}+\frac{1}{s}&x_{0}>R\end{cases}$

The boundary conditions (63) imply that $C_{2}=0=G_{1}$ and therefore

$\tilde{Q}_{p}(x_{0})=\begin{cases}\displaystyle
C_{1}e^{x_{0}\sqrt{s/D}}\,+\frac{1}{s}&x_{0}<-R\\\ \displaystyle
G_{2}e^{-x_{0}\sqrt{s/D}}+\frac{1}{s}&x_{0}>R\end{cases}$ (148)

  * •

$0<x_{0}<R$: In this region, the force is non-zero and Eq. (62) reduces to

$\displaystyle\partial_{x_{0}}^{2}\tilde{Q}_{p}-\lambda_{+}^{2}\tilde{Q}_{p}=-1,\quad\lambda_{+}=\sqrt{\frac{s+pV_{0}/R}{D}}$
(149)

which has the general solution

$\displaystyle\tilde{Q}_{p}(x_{0})=\frac{1}{D\lambda_{+}^{2}}+A_{1}e^{x_{0}\lambda_{+}}+A_{2}e^{-x_{0}\lambda_{+}}\,.$
(150)

  * •

$-R<x_{0}<0$: In this region, Eq. (62) reduces to

$\displaystyle\partial_{x_{0}}^{2}\tilde{Q}_{p}-\lambda_{-}^{2}\tilde{Q}_{p}=-1,\quad\lambda_{-}=\sqrt{\frac{s-pV_{0}/R}{D}}$
(151)

which has the general solution,

$\displaystyle\tilde{Q}_{p}(x_{0})=\frac{1}{D\lambda_{-}^{2}}+B_{1}e^{x_{0}\lambda_{-}}+B_{2}e^{-x_{0}\lambda_{-}}\,.$
(152)

To determine the $s$-dependent constants $A_{1,2}$, $B_{1,2}$, $C_{1}$ and
$G_{2}$ we note that $\tilde{Q}_{p}(x_{0})\equiv\tilde{Q}_{p}(x_{0},s)$ must
be a continuous function of $x_{0}$. Moreover, by integrating (62) for the
$f(x)$ given by (75) across $x_{0}=\pm R$ and $x_{0}=0$, we find that
$\partial_{x_{0}}\tilde{Q}_{p}(x_{0})$ must also be continuous. Thus

$\displaystyle\tilde{Q}_{p}(0^{+})$ $\displaystyle=$
$\displaystyle\tilde{Q}_{p}(0^{-}),~{}~{}~{}~{}~{}\partial_{x_{0}}\tilde{Q}_{p}(0^{+})=\partial_{x_{0}}\tilde{Q}_{p}(0^{-}),$
$\displaystyle\tilde{Q}_{p}(R^{+})$ $\displaystyle=$
$\displaystyle\tilde{Q}_{p}(R^{-}),~{}~{}~{}~{}\partial_{x_{0}}\tilde{Q}_{p}(R^{+})=\partial_{x_{0}}\tilde{Q}_{p}(R^{-}),$
$\displaystyle\tilde{Q}_{p}(-R^{+})$ $\displaystyle=$
$\displaystyle\tilde{Q}_{p}(-R^{-}),~{}~{}\partial_{x_{0}}\tilde{Q}_{p}(-R^{+})=\partial_{x_{0}}\tilde{Q}_{p}(-R^{-}).$

Inserting the above solutions into these conditions we get six linear
equations. We are particularly interested in the behavior of
$\mathcal{Q}_{p}(s)\equiv\tilde{Q}_{p}(0,s)$, so we eliminate $C_{1}$ and
$G_{2}$ leaving us with two long equations

$\displaystyle
A_{1}\left(1+\lambda_{+}\sqrt{\tfrac{D}{s}}\right)e^{\lambda_{+}R}$
$\displaystyle+$ $\displaystyle
A_{2}\left(1-\lambda_{+}\sqrt{\tfrac{D}{s}}\right)e^{-\lambda_{+}R}$
$\displaystyle=$ $\displaystyle\frac{pV_{0}/R}{s(s+pV_{0}/R)},$ $\displaystyle
B_{1}\left(1-\lambda_{-}\sqrt{\tfrac{D}{s}}\right)e^{-\lambda_{-}R}$
$\displaystyle+$ $\displaystyle
B_{2}\left(1+\lambda_{-}\sqrt{\tfrac{D}{s}}\right)e^{-\lambda_{+}R}$
$\displaystyle=$ $\displaystyle-\frac{pV_{0}/R}{s(s+pV_{0}/R)}$

and two shorter equations

$\begin{split}A_{1}+A_{2}-B_{1}-B_{2}&=\frac{2pV_{0}/R}{s^{2}-p^{2}V_{0}^{2}/R^{2}},\\\
A_{1}-A_{2}&=\frac{\lambda_{-}}{\lambda_{+}}(B_{1}-B_{2}).\end{split}$

These linear equations can be readily solved, but we omit long explicit
formulae for $A_{1,2}$ and $B_{1,2}$. We are primarily interested in the long-
time and large $y$ behavior, so it suffices to consider the small $s$ and
small $p$ limiting behavior of $\mathcal{Q}_{p}(s)$. As before, we first
compute the second moment. Expanding $\mathcal{Q}_{p}(s)$ around $p=0$ and
taking the coefficient of $p^{2}$ we obtain ${\mathcal{L}}_{t\to s}\langle
y^{2}(t)\rangle\simeq\frac{2RV_{0}^{2}}{3}(Ds)^{-\frac{3}{2}}$. Performing the
inverse Laplace transform we arrive at the announced result (76).

To compute the full distribution $P(y,t)$, we consider $\mathcal{Q}_{p}(s)$,
keep the leading order contributions in the limit of $s\to 0$ and $p\to 0$,
and deduce the announced Eq. (77) which leads to the scaling form of the
distribution.

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