text:
\section{Introduction}

Systems of particles governed by stochastic dynamics and exclusion interactions
have been studied for a long time \cite{Ligget,Mukamel,Derrida}.
One aspect of this problem is the motion of a tagged particle sometimes called
the tracer particle \cite{Ha,Jepsen,Le,alexander}.
The diffusion of a tagged particle,
in a  one dimensional system of Brownian particles, interacting
via hard core interaction is a model for the motion of a single
molecule in a crowded one dimensional environment such as  a biological
pore or channel \cite{karger,Burada,Kim,Golan}, and
in experimentally studied physical systems
such as zeolites \cite{Hahn1}, confined colloid particles
\cite{Wei,Lutz,LMCRD} and charged spheres in circular channels \cite{Saint}.
Since particles do not pass each other such diffusion processes are
called single file diffusion.

 Confinement of a tagged Brownian particle, due to its interaction with other
Brownian particles, leads to a slow-down of the diffusion of the
tagged particle $\langle (x_T)^2 \rangle \propto t^{1/2} $ instead of normal
diffusion $\langle (x_T)^2 \rangle \propto t$ when  the particles
in the system are uniformly distributed \cite{Ha,Le}.
Such many body problems
can be treated using methods
which exploit the relation between the dynamics of the interacting tagged
particle and the motion of a particle  free of interactions
 \cite{Jepsen,Le,Percus,Percus1,Leb,Kalinay}. In recent years
at-least six new directions  of research emerged. (i) The effect
of an external field acting on all the particles
\cite{Silbey}
or on the tagged particle only
\cite{Oshanin,Lizana}
has attracted attention since pores induce entropic barriers \cite{Burada}
and are generally
inhomogeneous. In this category the examples
of  single file motion in a periodic potential \cite{Mar}
or in a box \cite{LIAM}
 were considered in detail. (ii) Initial conditions may have
a profound impact on diffusion of the tagged particle. For example
if particles start as a narrow Gaussian packet
the diffusion of the tagged center particle  increases linearly in time
$\langle (x_T)^2 \rangle\propto
t$ \cite{Aslangul} (see details below).
Power law initial conditions induce $\langle (x_T)^2 \rangle \propto t^\xi$
and $\xi$ is  neither $1$ nor $1/2$ \cite{Ophir}
(see further details in text).
When an external
field is acting on the system, it is important to consider
 non uniform initial conditions where
the density of particles
is determined naturally from Boltzmann's distribution.
(iii) In some investigations
the underlying motion is not normal diffusion, instead the particles  may be non-Brownian,
and following anomalous kinetics \cite{Ophir,Band}. (iv) Usually hard core interactions are considered,
although the hard problem  of more general interactions has been recently
treated in \cite{Lizana,kollmann,peeters}.
Screened hydrodynamic interactions which seem important at short times
were investigated in \cite{Cui} both theoretically and experimentally.
Granular single file diffusion with inelastic collisions
shows the typical $t^{1/2}$ sub-diffusion \cite{Cecconi,Vulp}.
(v)  Interacting particles in systems with quenched disorder is yet another
challenge.
The motion of a tagged particle was recently treated in the context of single file diffusion in the Sinai model
\cite{Ben}. (vi) Finally, if the motion of the tagged particle is not
normal Brownian motion, what stochastic theory replaces the usual
Brownian-Langevin framework? In this direction interesting connections
to fractional calculus emerged \cite{Lizana,taloni,Lim}.
Roughly speaking and under certain
conditions half order time derivatives $(d^{1/2} / d t^{1/2})$
enter in the Langevin equation \cite{Lizana}
and fractional Brownian noise replaces white noise.

 Here we provide a very general theory of single file diffusion of the center tagged particle,
valid in the presence or the absence of an external potential field $V(x)$, as
well as for thermal and non-thermal initial conditions. Our main results
reproduce previously obtained formulas and many new ones, by mapping the
many particle problem onto a single particle model. Our  method,
explained in Sec. \ref{SecMethod}, exploits the
theoretical concept of the Jepsen line \cite{Jepsen},
is limited to hard core point particles,
but, as we show in Sec. \ref{SecBeyond}, is not limited to Brownian particles.
After providing general results in Sec. \ref{SecRes}
we limit our  attention to symmetric
initial conditions and potentials where the tagged particle has no
average drift. A general
relation between the mean square displacement
of the tagged particle and reflection probability of the non interacting
particle is given in Eq.
(\ref{eq21}).
Detailed calculations of the mean square displacement
of the tagged particle then
follow, for special choices of
force fields and initial conditions,
in Sec. \ref{secPI}.
  As we discuss  in sub-section
\ref{SecGumbel},
in certain limits our problem is related
to order statistics \cite{Gumbel}, a fact worth mentioning since it allows
us to solve our problem and related ones using known methods.
In Sec.
\ref{SecBeyond}, we discuss non-Brownian kinds of motion and the Percus
relation.
A brief report of part of our results was
recently published \cite{Silbey}.

\section{Model and Methods}
\label{SecMethod}

 In our model,
 $2 N + 1$ identical point particles with hard core particle-particle
interactions are undergoing Brownian motion in one dimension,
so particles cannot
pass each other. The diffusion constant of particles free of interaction
is $D$. As mentioned,
 towards the end of the paper we discuss the more general case
where the dynamics between collision events is not necessarily Brownian.
 An external potential
$V(x)$ acts on the particles.  The system stretches from
$-\overline{L}$ to $\overline{L}$; however, unless stated otherwise
we will let $\overline{L} \to \infty$ and obtain a thermodynamic
limit where $N /\overline{L}$ is fixed.
We tag the center particle, which clearly has $N$
particles to its left and $N$ to it right. Initially the tagged particle is
at the origin $x=0$. The motion of a single particle in the absence
of interactions with other particles is described by a single
particle Green function $g(x,x_0,t)$, with the initial conditions
$g(x,x_0,0)= \delta(x - x_0)$. In the case of over damped Brownian
motion the Green function is the solution of the Fokker-Planck equation
\cite{Risken}
\begin{equation}
{ \partial g(x,x_0,t) \over \partial t} = D \left[ {\partial^2 \over \partial x^2} - {1 \over k_b T} {\partial \over \partial x} F(x) \right] g(x,x_0, t),
\label{eqzzz}
\end{equation}
where $F(x) = - V'(x)$ is the force field, $T$ is the temperature and
$k_b$ is Boltzmann's constant.
The initial conditions of $N$ particles residing initially to the
right (left)  of the test  particle are drawn from
the probability density function (PDF)
$f_R(x_0)$ $(f_L(x_0))$ respectively. We consider
an ensemble of trajectories and average over trajectories
and initial
conditions (see details below).
Our goal is to obtain the PDF of the position $x_T$ of the
tagged center particle $P(x_T)$  in the large $N$
limit.

\subsection{The Jepsen Line}

 A schematic diagram of the problem is presented
 in Fig. \ref{fig1} for particles in a box.
 Initial positions of particles
are given by $x_{0} ^{j}$ where $j=-N, \cdots, 0, \cdots N$ where $j$
 is the label
of the interacting particles (see Fig. \ref{fig1}). The tagged
particle whose coordinate is denoted with $x_T (t)$, is the
center particle $j=0$ (bold line in Fig. \ref{fig1}).
Initially the tagged particle is at the origin $x_T (0) =0$.
  Since particles do not pass each
other, their order is clearly maintained, and the number of particles
to the left and right of the tagged particle  $N$ is  fixed.

In Fig. \ref{fig1} the straight line which starts at $x=0$ is called
the Jepsen line and follows $x(t) = v t$,
where $v$ is a test velocity \cite{Jepsen,Le}.
We label the interacting particles according to their initial position,
increasing to the right (see Fig. \ref{fig1}).
The tagged particle starts just to the right
of the Jepsen line so at $t=0$ the label of the particle
to the right of the Jepsen line is zero.
In this system we
have $2 N + 1$ particles.
Hence initially we have
$N$ particles to the left of the Jepsen line and including the tagged particle  $N+1$
particles to the right.

\begin{figure}
\begin{center}
\epsfig{figure=traj.eps,totalheight=0.34\textheight, width=0.45\textwidth}
\end{center}
\caption{
Schematic motion of Brownian particles in a box, where particles
cannot penetrate through each other.
 The center tagged particle label is $0$, its
trajectory is restricted by collisions with neighboring particles. The straight
line is the Jepsen line, it follows $v t$ as explained in the  text.
As explained in the text,  in
an equivalent non-interacting picture, we allow particles to pass through
each other, and at time $t$ we search for the position of the particle
which has $N$ particles to its right and $N$ to its left (i.e. the
center particle).
}
\label{fig1}
\end{figure}

 Let $\tilde{\alpha}(t)$ be the label number of the first particle
situated to the
right of the Jepsen line. According to our rules, at
$t=0$ we have $\tilde{\alpha} =0$, and then the random variable
$\tilde{\alpha}$ will increase or
decrease in steps of $+1$ or $-1$ according to:\\
i) if a particle crosses the Jepsen line from left to right $\tilde{\alpha} \to \tilde{\alpha} - 1$ \\
ii) if a particle crosses the Jepsen line from right to left
$\tilde{\alpha} \to \tilde{\alpha}+  1$.
Thus the counter $\tilde{\alpha}$ is performing a random walk
decreasing or increasing its value $+1$ or $-1$ at random times.

 A collision between two hard core particles is represented schematically in
Fig. \ref{fig2}. In one dimension a hard core
collision event is equivalent to two  particles
that pass through each other, i.e. non interacting particles,
 and then after the particles cross each other, the labels of
the pair of  particles are switched (see Fig. \ref{fig2}).
 Instead of relabeling
particles after each collision, we let particles pass through each
other, and then at time $t$ we label our particles (or if we are
interested only in tagged particle, locate the central particle).
Operationally this means that in the time interval
$(0,t)$ we view the particles
as non interacting, and then find the particle with $N$ particles
to its right and $N$ to its left, which is equivalent
to  the tagged particle
in the interacting system.
Hence  the problem is related to the mathematical topic of order statistics \cite{Gumbel}
as we will discuss briefly later.

  Following Jepsen \cite{Jepsen} and Levitt \cite{Le} we introduce the
stochastic process for the non interacting particles  $\alpha(t)$ where:\\
i) if a particle crosses the Jepsen line from left to right $\alpha \to \alpha - 1$ \\
ii) if a particle crosses the Jepsen line from right to left
$\alpha \to \alpha+  1$.
The process $\alpha(t)$
is the same as the process $\tilde{\alpha}(t)$,
in statistical sense.

\begin{figure}
\begin{center}
\epsfig{figure=arrow.eps, width=0.45\textwidth}
\end{center}
\caption{
In one dimension, the paths of a pair of particles in
a hard core collision event can be represented by
two non interacting particles which pass through each other,
and then their label numbers are switched.
}
\label{fig2}
\end{figure}

The event where $\alpha(t)=0$ implies that the number of particles
crossing the Jepsen line from right to left is the same
as those crossing from left to right.
Switching back to the interacting system, the event $\alpha(t)=0$
means  that the particle
to the right of the Jepsen line at time $t=0$ is also
there at time $t$. Hence the probability that $\alpha(t)=0$ is
the same as the probability of finding the tagged interacting
 particle
to be the first particle to  the right of the Jepsen line.
Therefore we will calculate the probability for $\alpha(t)=0$ and
with it obtain in the large $N$ limit,  the probability $P(x_T) {\rm d} x_T$
of finding
the tagged particle in a small interval $x_T,x_T+{\rm d} x_T$.

The random variable $\alpha$ is a sum of many random variables
\begin{equation}
\alpha = {\sum_{j=-N}^{N}}' \delta \alpha_j
\label{eq01}
\end{equation}
where $\delta \alpha_j$ is the number of times particle $j$
crossed the Jepsen line from right to left minus the number of
times it crossed the line from left to right. Clearly $\delta \alpha_j$
may attain the values $-1$ or $0$ if the particle started on the left,
 or $1$ or $0$ if it started on the right. Since we are interested in the
large $N$ limit we may neglect the contribution of $\delta \alpha_0$,
which is indicated by the prime in the sum \cite{remark0}.

 We calculate the probability
$P_N (\alpha)$
 of the random variable $\alpha$. For that we designate $P_{LL}(x_0 ^{-j})$,
the probability that particle $-j$, starting to the left of the
Jepsen line at $t=0$ at  $x_0 ^{-j}<0$, is found at time
$t$ to the left of the Jepsen line. Clearly for the corresponding trajectory
$\delta \alpha_{-j}=0$, since the particle crossed
the Jepsen line an even number of times, or did not cross it at all.
Similarly, $P_{LR} (x_0 ^{-j})$ is the probability to start to the
left of the line and end to its right, and $P_{RR}(x_0 ^j)$, $P_{RL}(x_0 ^j)$
are defined similarly for particles starting on $x_0 ^j>0$
to the right of the line.

In our non interacting picture, the motion of particles is
independent, hence we can use random walk theory \cite{Montroll} and
 Fourier series
to find $P_N (\alpha)$.
It is convenient to rewrite Eq. (\ref{eq01})
\begin{equation}
\alpha = \sum_{j=1}^N  \Delta \alpha_j
\label{eq01a}
\end{equation}
where $\Delta \alpha_j = \delta \alpha_j + \delta \alpha_{-j}$ may attain the values $1 , 0, -1$.
Each summand $\Delta \alpha_j$ takes into account
 one  particle starting to the
left of the Jepsen line and one to the right.

 First consider $N=1$, that is one particle that starts on the
left of the Jepsen line and one to its right.
Then $\alpha= \Delta \alpha_1= \delta \alpha_{1} + \delta \alpha_{-1}$ and
as mentioned $\alpha=1$ or $0$ or $-1$.
Examples for possible trajectories are shown in Fig. \ref{fig3}.
It is easy to see that
\begin{equation}
P_{N=1} \left( \alpha \right) = \left\{
\begin{array}{l l}
\alpha = 1 & P_{RL}(x_0 ^1 ) P_{LL} (x_0 ^{-1} ) \\
\alpha = 0 & P_{LL}(x_0 ^{-1} ) P_{RR} (x_0 ^{1} ) + P_{LR}(x_0 ^{-1} ) P_{RL} (x_0 ^{1})  \\
\alpha = -1 & P_{RR}(x_0 ^1 ) P_{LR} (x_0 ^{-1} ) .
\end{array}
\right.
\label{eq02}
\end{equation}
We define the structure function
\begin{widetext}
\begin{equation}
 \lambda(\phi, x_0 ^{-j},x_0 ^j) =
e^{ i \phi}
P_{LL}(x_0 ^{-j} )
 P_{RL} (x_0 ^j)
+ \left[ P_{LL} (x_0 ^{-j}) P_{RR} (x_0 ^j ) + P_{LR} (x_0 ^{-j}) P_{RL} (x_0 ^j ) \right] +
 e^{- i \phi}
P_{LR} (x_0 ^{-j} )
P_{RR}( x_0 ^j).
\label{eq03}
\end{equation}
\end{widetext}
The structure function describes a single step in the random walk
Eq. (\ref{eq01a})
in the following usual way: the coefficient of $\exp( i \phi)$
is the probability that $\Delta \alpha_j$ is equal one (i.e. $P_{LL} P_{RL})$,
the coefficient of $\exp( i \phi) =1$ (i.e. $\phi=0$)
yields
the probability $\Delta \alpha_j =0$, and similarly for $\exp( - i \phi)$.
Since the summands in Eq. (\ref{eq01a}) are independent,
Fourier analysis gives:
\begin{equation}
P_N (\alpha) = {1 \over 2 \pi} \int_{-\pi} ^\pi {\rm d} \phi  \Pi_{j = 1} ^N \lambda\left( \phi, x_0 ^{-j} , x_0 ^{j} \right) e^{ - i \alpha \phi}.
\label{eq04}
\end{equation}
We average Eq. (\ref{eq04}) with respect to the initial conditions
$x_0$ which are assumed independent identically distributed random
variable and we find
\begin{equation}
\langle P_N \left( \alpha \right) \rangle = { 1 \over 2 \pi} \int_{-\pi} ^\pi
{\rm d} \phi \langle \lambda\left( \phi\right) \rangle^N e^{ - i \alpha \phi}
\label{eq05}
\end{equation}
where from Eq. (\ref{eq03})
$$ \langle \lambda\left( \phi \right) \rangle = $$
\begin{equation}
e^{ i \phi } \langle P_{LL} \rangle \langle P_{RL} \rangle +
\langle P_{LL} \rangle \langle P_{RR} \rangle + \langle P_{LR} \rangle \langle P_{RL} \rangle + e^{ - i \phi}
\langle P_{LR} \rangle
\langle P_{RR} \rangle .
\label{eq06}
\end{equation}
Here $\langle P_{ij} \rangle$ is the probability of starting
in  $i=L,R$ (relative to the the Jepsen line) and
ending in $j=L,R$
and $\langle \cdots \rangle$ denotes an average over initial
condition.
The probability $\langle P_{LR} \rangle$ is given in terms
of the Green function of the non interacting particle $g(x,x_0,t)$
and the initial
density of particles situated initially to the left of the Jepsen line
$f_L(x_0)$
\begin{equation}
 \langle P_{LR} (vt) \rangle = \int_{-\overline{L}}  ^0 f_L (x_0) \int_{vt} ^{\overline{L}} g(x,x_0, t ) {\rm d} x {\rm d} x_0 .
\label{eq06g}
\end{equation}
We see that
for  $\langle P_{LR} \rangle$ we average over initial conditions in the
domain $(-\overline{L},0)$ weighted by $f_L(x_0)$
 (since the starting point is to the left of the
Jepsen line) and also integrate over $x$ in the domain
$(vt, \overline{L})$ with the weight $g(x,x_0,t)$
(since the end point is to the right of the line).
Similarly
\begin{equation}
\langle P_{RR} (vt) \rangle = \int_0 ^{\overline{L}} f_{R} (x_0) \int_{vt} ^{\overline{L}} g(x,x_0 , t ) {\rm d} x {\rm d} x_0.
\label{eq07}
\end{equation}
As mentioned $f_L(x_0)$ $[f_R(x_0)]$
 is the PDF of initial positions of particles that initially
are at $x_0<0$ $(x_0>0)$ respectively,
hence $f_L(x_0) = 0$ when $x_0>0$
and $\int_{-\overline{L}} ^0 f_L (x_0) {\rm d} x_0 = 1$,
while  $f_R(x_0)$ is normalized and  non zero only in $(0,\overline{L})$.
$\langle P_{LL} \rangle$ and $\langle P_{RR} \rangle$ are defined similarly,
and they satisfy
$\langle P_{LL} \rangle = 1 - \langle P_{LR} \rangle$,
and $\langle P_{RL} \rangle = 1 - \langle P_{RR} \rangle$.

\begin{figure}
\begin{center}
\epsfig{figure=ella.eps, width=0.45\textwidth}
\end{center}
\caption{
Trajectories crossing the Jepsen line. In the left panel one particle
started on $L$ and ended on $L$ the other on $R$ and ended on $R$.
These two paths are assigned the probability $P_{LL}(x_0 ^{-1}) P_{RR}
(x_0 ^1)$ and yield in Eq.
(\ref{eq02})
 $\alpha=0$. Similarly the trajectories on the right panel
correspond to $P_{LL}(x_0 ^{-1} ) P_{RL} (x_0 ^{1} ) $ and they
contribute $\alpha= 1$.
}
\label{fig3}
\end{figure}

\section{Dynamics of the Tagged Particle}
\label{SecRes}

We now
apply the central limit theorem  to analyze
the sum Eq. (\ref{eq01a}) using
Eq. (\ref{eq05})
when $N \to \infty$.
The small $\phi$ expansion of the structure function
Eq. (\ref{eq06})
is
\begin{equation}
\langle \lambda(\phi) \rangle = 1
+ i \langle \Delta \alpha \rangle \phi - {1 \over 2}\langle (\Delta \alpha)^2 \rangle \phi^2 + O(\phi^3)
\label{eq08}
\end{equation}
where
\begin{equation}
\langle \Delta \alpha \rangle  = \langle P_{RL} \rangle - \langle P_{LR} \rangle
\label{eq09}
\end{equation}
as expected,
and the variance
$\sigma_{\Delta \alpha} ^2 =  \langle (\Delta \alpha)^2 \rangle- \langle \Delta \alpha \rangle^2 $ is
\begin{equation}
\sigma_{\Delta \alpha}  ^2 = \langle P_{RL}\rangle \langle P_{RR} \rangle +
 \langle P_{LR}\rangle \langle P_{LL} \rangle.
\label{eq10}
\end{equation}
Using the central limit theorem
we have the probability of zero total crossing of
the Jepsen line, namely $\alpha = 0$ in the limit $N \to \infty$
\begin{equation}
 P_N (\alpha = 0 ) \sim
{1 \over \sqrt{ 2 \pi N} \sigma_{\Delta \alpha} }
 \exp\left( - { N \langle \Delta \alpha \rangle^2 \over 2 \sigma_{\Delta \alpha} ^2 } \right).
\label{eq11}
\end{equation}
This simple result, valid for a large class of Green functions
and initial conditions, is  suited for the investigation
of a large number of single file problems.

 The probability $P_N(\alpha =0)$ is according to our earlier
discussion equal to the probability that the tagged particle
is in the interval $v t < x_T < vt + \delta x$, where $\delta x$ is
the distance between the Jepsen line and the position of the
second particle to the right of the Jepsen line. We assume that
$\delta x$ is small relative to the typical length scale
the tagged particle traveled. For example if the
 density of interacting particles
$\rho$ is a constant, we assume $1/\rho << \sqrt{ \langle (x_T)^2 \rangle }$.
 Hence we may replace $vt \to x_T$ to transform
$P_N(\alpha=0)$ to the PDF of the tagged particle
$P(x_T)$ \cite{remark}.

 Making the replacement $v t \to x_T$, using Eq.
(\ref{eq11}) we have in the limit of large $N$
$$ P(x_T) \sim $$
\begin{equation}
C \exp\left\{ - { N \left[ \langle P_{LR} (x_T) \rangle - \langle P_{RL} (x_T) \rangle \right]^2 \over 2 \left[ \langle P_{LL} (x_T) \rangle \langle P_{LR} (x_T) \rangle + \langle P_{RR} (x_T) \rangle \langle P_{RL} (x_T) \rangle \right] } \right\}
\label{eq12}
\end{equation}
where $C$ is a normalization constant. In Eq. (\ref{eq12})
and what follows $\langle P_{ij} (x_T) \rangle$ is given by
Eqs.
(\ref{eq06g},\ref{eq07}) with $vt \to x_T$.
Eq. (\ref{eq12}) yields
the PDF of the center particle $P(x_T)$ in terms of
 $\langle P_{ij} (x_T) \rangle$ which according to Eq. (\ref{eq07})
depends on the free
particle green function and the initial conditions.
Thus the information contained in the non-interacting Green function
is sufficient for the determination of the single file diffusion
of the tagged particle.

\subsection{Thermal Equilibrium}

 In the long time limit, and in the presence of a binding potential field,
e.g. harmonic field or particles in a box, an equilibrium is reached.
 Then initial conditions do not play a role. For example
$\langle P_{LR} (x_T) \rangle = P^{eq}_{R} (x_T)$
with
\begin{equation}
P^{\rm eq}_R (x_T) = {1 \over Z} \int_{x_T} ^{\overline{L}}
\exp\left( - { V(x) \over k_b T } \right) {\rm d} x
\label{eq13a}
\end{equation}
where $Z$ is the normalizing partition function
\begin{equation}
Z= \int_{-\overline{L} } ^{\overline{L} } \exp\left[ - {V\left(x\right) \over k_b T }\right] {\rm d} x.
\label{eq13b}
\end{equation}
Similarly
\begin{equation}
P^{\rm eq}_L (x_T) = {1 \over Z} \int_{-\overline{L}} ^{x_T}
\exp\left( - { V(x) \over k_b T } \right) {\rm d} x.
\label{eq14a}
\end{equation}
In Eqs. (\ref{eq13a},\ref{eq14a}) we used the steady state solution,
 $\lim_{t \to \infty}  g(x,x_0,t) = \exp[ - V(x)/k_b T] / Z$, which is
Boltzmann's distribution suited for a system in thermal equilibrium.
Using Eq. (\ref{eq12})
the position PDF of the tagged  particle is
$\lim_{t \to \infty} P(x_T) = P^{\rm eq}(x_T)$
\begin{equation}
P^{\rm eq} (x_T) \sim  C \exp\left\{ - { N \left[ \langle P_{R}^{eq}  (x_T) \rangle - \langle P_{L} ^{eq}  (x_T) \rangle \right]^2 \over 4  \langle P_{L} ^{eq} (x_T) \rangle \langle P_{R} ^{eq}  (x_T) \rangle } \right\}.
\label{eq15a}
\end{equation}
If the potential is symmetric $V(x) = V(-x)$, e.g. particles in a box or
harmonic field, we have for
not too large $x_T$, $P_{L} ^{eq} (x_T) \simeq 1/2$,
$P_{R} ^{eq} (x_T) \simeq 1/2$ hence
from Eq. (\ref{eq12})
\begin{equation}
P^{\rm eq} (x_T) \sim C \exp \left\{ - N \left[ \langle P_R ^{eq} (x_T) \rangle -
\langle P_L (x_T) \rangle \right]^2 \right\}.
\label{eq16a}
\end{equation}
Expanding the expression in the exponent in $x_T$ (since $N$ is large) we find
using Eqs. (\ref{eq13a},\ref{eq14a},\ref{eq15a})
\begin{equation}
P^{\rm eq} ( x_T ) \sim {2 \sqrt{N} \over \sqrt{\pi} Z} \exp \left[ - { 4 N \over Z^2} \left( x_T \right)^2 \right]
\label{eq17a}
\end{equation}
where with out loss of generality we assigned $V(x=0)=0$.
Hence the standard deviation is
\begin{equation}
\langle (x_T)^2 \rangle \sim {Z^2 \over 8 N} .
\label{eq17aa}
\end{equation}
The same expression is found in Appendix A using the many body
Boltzmann distribution, and integrating over all the particles
except the tagged particle.

\subsection{Simple Illustration}
\label{secSI}

 We now consider the situation of particles free of a force $F(x)=0$ with open
boundary conditions $\overline{L} \to \infty$ where initially all the particles
are on the vicinity of  the origin.
More precisely, the tagged particle is initially situated at
$x_T =0$, $N$ particles to its right on $\epsilon\to 0^{+}$ and
$N$ particles on $-\epsilon$. This problem was solved already by Aslangul \cite{Aslangul},
and here we recover the known result using our formulas.
The Green function $g(x,x_0,t)$ of
a free particle is
\begin{equation}
g(x,x_0,t) = {\exp \left[ - { \left( x - x_0\right)^2 \over 4 D t} \right] \over \sqrt{ 4\pi D t} }
\label{eqAs01}
\end{equation}
where  as mentioned $D$ is the diffusion coefficient of the free particle.
With the specified initial conditions we
have
$$ \langle P_{RL} \left( x_T \right) \rangle = \lim_{\epsilon \to 0} \int_0 ^\infty \delta(x_0 - \epsilon) \int_{-\infty} ^{x_T} { \exp\left[ - { \left( x - x_0 \right)^2 \over 4 D t }  \right] \over \sqrt{4\pi D t} } {\rm d} x {\rm d} x_0 = $$
\begin{equation}
{1 \over 2} + \int_0 ^{x_T} { e^{ - { x^2 \over 4 D t}} \over \sqrt{ 4 \pi D t } } {\rm d} x.
\label{eqAs02}
\end{equation}
Similarly
\begin{equation}
 \langle P_{LR} \left( x_t \right) \rangle =
{1 \over 2} - \int_0 ^{x_T} { e^{ - { x^2 \over 4 D t}} \over \sqrt{ 4 \pi D t } } {\rm d} x.
\label{eqAs03}
\end{equation}
When $x_T\ll \sqrt{2 D t}$ we Taylor expand in $x_T$ to find $(\langle P_{LR} \rangle - \langle P_{RL} \rangle)^2 \sim
(x_T)^2/ \pi D t $, using Eq. (\ref{eq12})
we recover the result in \cite{Aslangul}
\begin{equation}
P(x_T) \sim { 1 \over \pi} \sqrt{ {N \over D t}} \exp\left[ - { N (x_T)^2 \over \pi D t } \right],
\label{eqAs03}
\end{equation}
hence
\begin{equation}
\langle (x_T)^2 \rangle \sim  {\pi D t \over 2 N }.
\label{eqAs03a}
\end{equation}
The diffusion is
normal, in the sense that the mean square displacement increases linearly
in time. However the diffusion of the tagged particle
is slowed down compared
with a free particle, by a factor of $1/N$ which is  due to the collisions
with all other Brownian particles in the  system. Clearly the approximation
breaks down if one is interested in the tails of $P(x_T)$, since we
used $x_T \ll \sqrt{ 2 D t}$. Though clearly
 when $N$ is large,
the probability of finding such a particle is extremely small (i.e.
use Eq. (\ref{eqAs03}) $P(x_T = \sqrt{ 2 D t}) \sim \exp( -N 2/\pi)$).

\subsection{Formula for $\langle (x_T)^2 \rangle$ }

 We now consider
 symmetric potential fields $V(x)=V(-x)$, and symmetric initial
conditions.
 The latter means that the density of
particles at time $t=0$ to the left of the tagged particle, i.e. those
residing in $x_0<0$, is the same as for those residing to the right,
$f_R(x_0) = f_L ( - x_0)$ (e.g. uniform initial conditions). In this
case the subscript $R$ and $L$ is redundant and we use $f(x_0)=f_R(x_0)=f_L(-x_0)$,
where $f(x_0)=0$ if $x_0<0$, $f(x_0)\ge 0$ and $\int_0 ^\infty f(x_0) {\rm d} x_0=1$.
From symmetry it is clear that the tagged particle is unbiased,
namely $\langle x_T \rangle = 0 $. Further, since $N$ is large we may
expand the expressions in the exponent in Eq. (\ref{eq12})
in $x_T$ to obtain
the leading term
$$ \langle P_{RL} \rangle - \langle P_{LR} \rangle = $$
\begin{equation}
{\partial \over \partial x_T} \left[ \langle P_{RL} \left( x_T \right) \rangle -
  \langle P_{LR} \left( x_T \right) \rangle\right]|_{x_{T}=0}
  x_T + O[(x_T)^2].
\label{eq12a}
\end{equation}
Similarly
for small $x_T$
\begin{equation}
\begin{array}{c}
 \langle P_{LL} \rangle \langle P_{LR} \rangle + \langle P_{RR} \rangle \langle P_{RL} \rangle= \\
2 \langle P_{RR} \left( x_T \right) \rangle \left[ 1 - \langle P_{RR}\left( x_T \right) \rangle \right] |_{x_T =0} + O(x_T)
\end{array}
\label{eq14}
\end{equation}
To derive Eq. (\ref{eq12a}) we used the symmetry of the problem,
 which means that
the probability of crossing the point $x_T=0$ from left to right
is the same as  the probability to cross from right to left
$\langle P_{LR} \rangle_{x_T =0} =\langle P_{RL} \rangle_{x_T = 0} $ and
similarly
$\langle P_{LL} \rangle_{x_T =0}  =
\langle P_{RR} \rangle_{x_T =0}$ by symmetry.
We designate the reflection coefficient,
\begin{equation}
{\cal R} = \langle P_{RR} \rangle_{x_T = 0}
\label{eq15}
\end{equation}
[or ${\cal R} = \langle P_{LL} \rangle_{x_T = 0}$]
since it is the probability that a particle starting at $x_0>0$, is found in $x>0$ at time
$t$ when an average over initial conditions is made
\begin{equation}
{\cal R} = \int_0 ^{\overline{L}} f(x_0) \int_0 ^{\overline{L}} g\left( x, x_0 , t \right) {\rm d} x {\rm d} x_0.
\label{eq16}
\end{equation}
A transmission coefficient
is defined through
${\cal T} = \langle P_{RL}  \rangle_{x_T = 0} =
 \langle P_{LR} \rangle_{x_T = 0} $
which is related to the reflection coefficient in the usual way
$ {\cal T} = 1 -{\cal R}$.

Turning our attention to Eq. (\ref{eq12a}), from left-right symmetry we have
\begin{equation}
{\partial \over \partial x_T } \langle P_{RL} \left( x_T \right) \rangle|_{x_T = 0}
= - {\partial \over \partial x_T } \langle P_{LR} \left( x_T \right) \rangle|_{x_T = 0}.
\label{eq18}
\end{equation}
Hence we define
\begin{equation}
{\it r} =
{\partial \over \partial x_T } \langle P_{RL} \left( x_T \right) \rangle|_{x_T = 0}
\label{eqj}
\end{equation}
where from  Eq.
(\ref{eq07})
\begin{equation}
{\it r}   = \int_0 ^{\overline{L}} f(x_0) g(0,x_0, t ) {\rm d} x_0 .
\label{eq19}
\end{equation}
So  ${\it r}$ is the density of non interacting particles at $x=0$
for an initial density $f(x_0)$.
Note that
since
$\langle P_{RL}(x_T) \rangle + \langle P_{RR}(x_T) \rangle = 1$
we have
\begin{equation}
{\it r} = - {\partial \over \partial x_T}  \langle P_{RR} (x_T ) \rangle|_{x_T = 0}.
\label{eq19u}
\end{equation}

Inserting Eq.
(\ref{eqj})
in  Eq. (\ref{eq12a}) using Eq.
(\ref{eq18})
we have $\langle P_{RL} \rangle - \langle P_{LR} \rangle = 2 r x_T + \cdots$.
Inserting Eq. (\ref{eq15}) in Eq. (\ref{eq14})
we find our main result: the
probability density function  of the position of the tagged central particle
\begin{equation}
P(x_T) \sim { 1 \over \sqrt{ 2 \pi \langle (x_T)^2 \rangle }}
\exp\left[ - {\left( x_T \right)^2 \over 2 \langle \left( x_T \right)^2 \rangle} \right],
\label{eq20}
\end{equation}
where
\begin{equation}
\langle (x_T)^2 \rangle = { {\cal R} \left( 1 - {\cal R} \right)\over 2 N {\it r} ^2 }
\label{eq21}
\end{equation}
is the  mean square displacement of the tagged  particle.
 The single particle probability $\langle P_{RR}(x_T) \rangle$
 gives ${\cal R}$ Eq.
(\ref{eq15})
and ${\it r}$ Eq.
(\ref{eq19u}) which in turn yield the mean square displacement of
the tagged particle Eq. (\ref{eq21}).
We will soon use this equation to demonstrate a variety of physical behaviors.
However first we establish a connection between our work
and the theory of order statistics.

\subsection{Order Statistics}
\label{SecGumbel}

 Generate $n$ independent identically distributed random variables
drawn from a PDF $\hat{r}(x)$ and arrange them in
increasing order.  Order statistics deals with
the m-th observable $m=1,\cdots,n$ among $n$ observations taken
in the increasing order, which is denoted $x_m$. The PDF of $x_m$, $\phi(x_m)$
depends on the PDF $\hat{r}(x)$, the sample size $n$ and
the order $m$. Let $\hat{R}(x)$ be the cumulative distribution
of $x$, e.g. if the domain of $x$ is $-\infty<x<\infty$, $\hat{R}(x)=
\int_{-\infty} ^x \hat{r}(x) {\rm d} x$ as usual.
Following well known result \cite{Gumbel}, define $\hat{x}$
with
\begin{equation}
\hat{R}(\hat{x} ) = { m\over n+1}
\label{eqOrder1}
\end{equation}
then when $n$ is large and $m/n $ is of the order $1/2$
\begin{equation}
\phi(x_m) = \mbox{const} \exp\left\{ - { n \left( x- \hat{x} \right)^2 \hat{r}^2 (\hat{x} ) \over 2 \hat{R}(\hat{x} ) \left[ 1 - \hat{R}(\hat{x} ) \right] } \right\}.
\label{eqorder2}
\end{equation}
Hence the variance of $x_m$
\begin{equation}
(\sigma_m)^2  = { \hat{R}(\hat{x}) \left[ 1 - \hat{R}\left( \hat{x} \right) \right] \over n \hat{r}^2 \left( \hat{x} \right) }
\label{eqorder3}
\end{equation}
which has some resemblance to Eq. (\ref{eq21}).

 The problem of the motion of a tagged particle is mathematically
identical to the problem of order statistics in two cases: (i)
in the presence of a binding potential and in the long
time limit and (ii) when all the particles start on the
same point. In both cases the single particle
 PDF $g(x,x_0,t)$ of all
the particles are identical since it is either  independent of $x_0$ (case (i))
or we have a unique initial condition (case treated in
sub section \ref{secSI}).
For example in the presence
of a binding field $\lim_{t \to \infty} g(x,x_0,t) = \exp[- V(x)/k_b T ] / Z$
which is independent of the initial position of the particle.
Hence in equilibrium, to find the center particle we may draw $(2 N+1)$
random variables from Boltzmann's distribution, and search for the center
particle (which will give the position of the interacting tagged particle).
Or, using the language of order statistics we have  $\hat{r}(x) =
\exp[- V(x)/k_b T ] / Z$.
Using  a symmetric potential $V(x)=V(-x)$
and Eq.
(\ref{eqOrder1})
we insert  $n=2 N+1$, and $m=N+1$; hence, when $N$ is large
we have
\begin{equation}
\hat{R}(\hat{x} ) = \int_{-\infty} ^{{\hat x}} { \exp\left[ - {V(\hat{x})\over  k_b T} \right] \over Z} {\rm d} x  = {1\over 2} .
\label{eqorder4}
\end{equation}
Thus, $\hat{x} = 0$. Then using Eq.
(\ref{eqorder3}) we find
\begin{equation}
\langle (x_T)^2 \rangle = {Z^2 \over 8 N}
\end{equation}
which is the same as Eq.
(\ref{eq17aa}) (recall $V(0) =0$).

While  Eq. (\ref{eqorder3}) has superficially a structure similar to Eq.
(\ref{eq21}) they are different. Our ${\cal R}$
Eq.
(\ref{eq15})
is generally not equal half, neither
must it be close to that value
[so Eq. (\ref{eqOrder1}) is not generally related to our problem].
In fact when $t\to 0$ we must have $\langle (x_T)^2 \rangle \to 0$ which
is found when $\lim_{t \to 0} {\cal R} = 1$ (see examples below).
In the problem of motion of a tagged particle, the number of particles
which can interact with the center particle is usually increasing
with time (hence $n$ is not fixed). For example consider the classical
case of  uniformly
distributed particles in infinite space and in the absence of forces.
Roughly speaking  particles  at distances of the order of
$l_{{\rm eff}} = \sqrt{ Dt}$ or shorter
can influence the tagged particle motion via collisions.
So in this  case roughly $ N_{{\rm eff}}=\rho \sqrt{D t}$
particles participate in the
process. In contrast, if all particles are initially at the vicinity
of the origin $N$ particles participate i.e. influence
the motion of the tagged particle. Interestingly
this gives an argument for the well known behavior
of the mean square displacement of the tagged particle, with uniform density of
particles,  namely use Eq.
(\ref{eqAs03a})  $\langle (x_T)^2 \rangle \sim \pi D t / 2 N$
and replace
$N$ with $N_{{\rm eff}} = \rho \sqrt{D t}$
to find $\langle (x_T)^2 \rangle \simeq
\sqrt{ D t} / \rho$
which of course misses the correct numerical prefactor (see Eq.
\ref{eqbox06}
below).

\section{Physical Illustrations}
\label{secPI}

\subsection{Particles in a Box}

 Consider particles in a finite box extending from $-\overline{L}$ to
$\overline{L}$ which was recently treated with the Bethe ansatz and
numerical simulations by Lizana  and Ambj\"ornson  \cite{LIAM}.
The tagged particle initially
at $x_T =0$ has $N$ particles to its right and $N$ to its left.
These particles are assumed uniformly distributed, hence
\begin{equation}
f(x_0)= \left\{
\begin{array}{l l}
{1\over \overline{L}} \  & \
0<x_0 < \overline{L} \\
\ & \ \\
0 \ & \ \mbox{otherwise} .
\end{array}
\right.
\label{eqbox0}
\end{equation}
In the limit $\overline{L} \to \infty$
and $N\to \infty$ in such a way that the density $\rho= N / \overline{L}$ is fixed,
we obtain single file diffusion in an infinite system,
a case well studied long ago \cite{Ha,Le}.

 The single particle Green function of a particle in a box,
with reflecting boundary conditions
$\partial g(x,x_0,t) / \partial x|_{x=\pm
\overline{L}}= 0$ is solved using an eigenfunction expansion
\cite{Risken}
$$ g(x,x_0,t) = {1 \over 2 \overline{L} } + $$
\begin{equation}
{1 \over \overline{L} }
\sum_{n=1} ^\infty \cos \left[ { n \pi \over 2\overline{L}} \left( x + \overline{L} \right) \right] \cos\left[ { n \pi \over 2 \overline{L} } \left( x_0 + \overline{L} \right) \right] \exp\left( - D { n^2 \pi^2 \over 4 \overline{L}^2 } t \right).
\label{eqbox01}
\end{equation}
With Eqs. (\ref{eq16},\ref{eqbox0},\ref{eqbox01}) we find
\begin{equation}
{\cal R}(t) = {1 \over 2} + { 4 \over \pi^2} \sum_{n=1, \mbox{Odd} } ^\infty
{ \exp\left( - D { n^2 \pi^2 \over 4 \overline{L}^2 } t \right) \over n^2}
\label{eqbox02}
\end{equation}
where the summation is over odd $n$.
At $t=0$, ${\cal R} =1$ since all particles
initially in $(0, \overline{L})$ did not have
time to move to the other side of the box, and
$\lim_{t \to \infty} {\cal R} = 1/2$ since in the
long time limit there is equal probability for
a non interacting particle to occupy each half of
the box.
Using Eqs. (\ref{eq19},\ref{eqbox0},\ref{eqbox01}) we find
\begin{equation}
r= {1 \over  2 \overline{L} }
\label{eqbox02aa}
\end{equation}
hence Eq. (\ref{eq21}) gives
the mean square displacement of the tagged particle
\begin{equation}
\langle \left( x_T \right)^2 \rangle \sim 2 {{\cal R}(t) \left[ 1 - {\cal R}(t) \right] \overline{L}^2 \over N}.
\label{eqbox03}
\end{equation}
The eigenvalues of the non-interacting particle determine the
multi exponential type of  decay of
${\cal R}(t)$ with time,  which in turn determines the
dynamics of the interacting tagged particle \cite{remark3}.
\begin{figure}
\begin{center}
\epsfig{figure=Nick1.eps,totalheight=0.34\textheight, width=0.45\textwidth}
\end{center}
\caption{ For particles in a box
we show $\langle (x_T)^2 \rangle N / \overline{L}^2$ of the tagged particle, in equilibrium versus $N$. Exact expression
obtained in \cite{LIAM} valid for all $N$ (dots) converges in the limit
of $N \to \infty$ to the behavior predicted by our theory (dashed line)
Eq. (\ref{eqbox07a}). For $N=70$ clear deviations between
our asymptotic theory and the exact result in \cite{LIAM}
are found.
}
\label{fig11}
\end{figure}

 Let $\delta^2 = D \pi^2 t / 4 \overline{L}^2$ hence the limit $\delta \ll 1$ gives
the short time dynamics before the particles interact with the walls,
or equivalently the limit of an infinite system.
The reflection coefficient is rewritten
\begin{equation}
{\cal R} = { 1 \over 2} + { 4 \over \pi^2} \left(
\delta \sum_{n=1, \mbox{Odd}} ^\infty { e^{ - \delta^2 n^2} - 1 \over \delta^2 n^2 } \delta + \sum_{n=1, Odd} ^\infty { 1 \over n^2} \right).
\label{eqbox04}
\end{equation}
When $\delta\ll 1$ we may replace the first summation with integration
\begin{equation}
\sum_{n=1,\mbox{Odd}} ^\infty { e^{- \delta^2 n^2 } - 1 \over \delta^2 n^2} \delta \simeq {1 \over 2} \int_0 ^\infty { e^{- y^2} - 1 \over y^2} {\rm d} y = { - \sqrt{\pi} \over 2}
\label{eqbox05}
\end{equation}
where the factor $1/2$ on the RHS  comes from  the summation over  only odd $n$ on the LHS.
Using $\sum_{n=1,\mbox{Odd} } ^\infty 1/n^2 = \pi^2/ 8$ we have
${\cal R} = 1 - { \sqrt{D t} \over \overline{L}  \sqrt{\pi} }+ \cdots$,
and hence
\begin{equation}
\langle \left( x_T \right)^2 \rangle \sim { 2 \over \sqrt{\pi} }
{\sqrt{ D t} \over \rho} .
\label{eqbox06}
\end{equation}
This result was obtained in \cite{Ha,Le}, and
it describes the dynamics of the tagged particle in a box,
before particles have time to interact with the walls,
namely when $\delta$ is small.

 In the opposite limit of long times and finite systems
we attain equilibrium, then $\lim_{t \to \infty} {\cal R} = 1/2$
and
\begin{equation}
P(x_T) \sim { \sqrt{N} \over \sqrt{\pi} \overline{L} } e^{ - N \left( x_T\right)^2 / \overline{L}^2 },
\label{eqbox07}
\end{equation}
hence
\begin{equation}
\lim _{t \to \infty} \langle (x_T)^2 \rangle = \overline{L}^2 / 2 N.
\label{eqbox07a}
\end{equation}
Eq. (\ref{eqbox07}) is a special case of the more general Eq.
(\ref{eq17a}) since the single particle normalizing
 partition function
for our example is $Z= 2 \overline{L}$.

 As mentioned, in \cite{LIAM} the Bethe ansatz was used to
solve the problem of tagged particle motion in a box.
Among other things an
 exact expression for the long time limit of $\langle (x_T)^2 \rangle$,
 valid for all
$N$ was found \cite{LIAM}
\begin{equation}
\lim _{t \to \infty} \langle (x_T)^2 \rangle = \overline{L}^2 \left( {1 \over 4}\right)^{N+1}  { \Gamma\left({1\over 2} \right) \Gamma\left[ 2\left( N + 1 \right) \right]  \over \Gamma\left( N + 1 \right) \Gamma\left( N + {5 \over 2} \right) }.
\label{eqbox07liz}
\end{equation}
Since our formalism is valid only in the large $N$ limit,
comparison of our solution to an exact result like Eq.
 (\ref{eqbox07liz}) provides
insight to the question of convergence.
In Fig. \ref{fig11} we plot
$   \lim_{t \to \infty } \langle (x_T)^2  \rangle N / \overline{L}^2$
versus $N$ using Eq. (\ref{eqbox07liz}) comparing it to our
Eq. (\ref{eqbox07a}) which gives
$  \lim_{t \to \infty } \langle (x_T)^2 \rangle N/ \overline{L}^2  =1/2 $.
Not surprisingly we see that the two results yield asymptotically
the same result. The figure illustrates
that even for $N=100$ deviations between exact results and the present
theory are observable.

 In \cite{LIAM} numerical simulation of systems with  $N=1,10,70$
(maximum  of  $141$
particles) were performed, and favorably
compared with the Bethe ansatz solution.
These simulations were made for finite size hard core particles
whose diameter is $\Delta$. To make comparison between our theory
and simulation we scale  the size of the box according to $2 \overline{L} \to
2 \overline{L} - (2 N ) \Delta$. In Fig. \ref{fig10}  the scaled
mean square displacement versus $t/\tau_{eq}$ is shown, where $\tau_{eq} = 4 \overline{L}^2 / D$.
 Some general features
of our theory can now be discussed. First, at short
times simulations show normal behavior $\langle (x_T)^2 \rangle = 2 D t$,
which is a trivial effect: particles did not have time to
collide and hence the tagged particle diffuses normally as
if it is free. After particles start to collide, deviations
from normal diffusion are observed
(for $N=70$ but  as expected not for $N=1$). These
are mainly due to collisions with other particles. Finally
saturation due to the finite size of the system is found. While
our theory shows the general trend of simulations (say for $N=70$
and for not too short times) it is clearly
not in perfect agreement. We argue that this is due to
the small number of particles $N=70$,
since as we showed in Fig. \ref{fig11}, at least for large
times and large $N$ our theory gives the exact result.
It would be nice to have simulations with point particles,
with larger $N$ and since there are three phases of the motion:
normal diffusion, single file diffusion without the  influence of
the walls
and finally saturation to equilibrium.
Separation of time scales is needed to demonstrate
these behaviors clearly.

\begin{figure}
\begin{center}
\epsfig{figure=long.eps,totalheight=0.34\textheight, width=0.45\textwidth}
\end{center}
\caption{ Motion of tagged particle in a box is shown for $N=1,10,70$.
Simulation results (squares) are taken from \cite{LIAM} (see text). At
short
time simulations  exhibit normal diffusion $\langle (x_T)^2 \rangle = 2 D t$
 as illustrated by the straight dashed line which is a guide to the eye.
Our theory is expected to work well in the large $N$ limit
and when many collision events between tagged particle and surrounding
Brownian particles took place.
Hence agreement between theory (solid line) and simulation
is reasonable at most only for $N=70$ and not for too short times.
}
\label{fig10}
\end{figure}

\subsection{Gaussian Packet}

 Consider particles without  external forces $V(x)=0$, in an infinite system.
Initially particles are spread with a Gaussian packet with width $\xi$:
\begin{equation}
f(x_0) = { \sqrt{2} \over \sqrt{\pi} \xi } \exp\left[ - { (x_0) ^2 \over 2 \xi^2} \right]
\label{eqGP1}
\end{equation}
for $x_0 >0$. As before we consider the tagged particle motion,
which is initially at $x_T=0$, with $N$ particles to its
left and $N$ to its right. With the free particle Green function
$g(x,x_0,t)$ Eq. (\ref{eqAs01}) we proceed to find $\langle (x_T )^2 \rangle$.

\begin{figure}
\begin{center}
\epsfig{figure=gauss1.eps,totalheight=0.34\textheight, width=0.45\textwidth}
\end{center}
\caption{ Scaled mean square displacement of the tagged particle
with Gaussian initial
conditions of the packet of particles exhibits a transition
between
short time $\langle (x_T)^2 \rangle\propto
t^{1/2}$ law to $\langle (x_T)^2 \rangle \propto t$ behavior.
Dashed lines are short and long time asymptotic behavior Eq.
(\ref{eqGPii}),
circles represent Eq. (\ref{eqGP6}).
}
\label{fig12a}
\end{figure}

  The reflection probability Eq. (\ref{eq16}) is
\begin{equation}
{\cal R} = \int_0 ^\infty {\rm d} x_0 \sqrt{{2 \over \xi^2  \pi}} e^{ - {(x_0)^2 \over 2 \xi^2 } }
\int_0 ^\infty { {\rm d} x \over \sqrt{ 4 \pi D t } } e^{ - { (x- x_0)^2 \over 4 D t} }  .
\label{eqGP2}
\end{equation}
Changing variables  according to $y^2/2 = (x-x_0)^2/ 4 Dt$
and using dimensionless parameter $\tilde{\xi} = \xi/ \sqrt{ 2 D t}$, we find
\begin{equation}
{\cal R} = {1 \over 2} + { 1 \over \tilde{\xi}} \sqrt{ { 2 \over \pi}} \int_0 ^\infty
{\rm d} \tilde{x}_0 e^{ - { (\tilde{x}_0)^2 \over 2 \tilde{\xi}^2 }} { \mbox{Erf}\left( \tilde{x}_0/\sqrt{2} \right)
\over 2}
\label{eqGP3}
\end{equation}
where
$\mbox{Erf}(\tilde{x}_0/\sqrt{2} )/2=\int_0 ^{\tilde{x}_0 } e^{ - y^2 /2 } {\rm d} y/\sqrt{ 2 \pi} $
is the error function \cite{Abr}.
Mathematica solves the integral
in Eq. (\ref{eqGP3}) and we find
\begin{equation}
{\cal R} = {1 \over 2} + {1 \over \pi} \mbox{arccot}\left( { \sqrt{ 2 D t} \over \xi } \right).
\label{eqGP4}
\end{equation}
For short times $\sqrt{D t}\ll \xi$ we have ${\cal R}\sim 1 - { \sqrt{ 2 D t} \over \pi \xi } $,
namely most particles did not have time to cross the origin, while in the
opposite limit $\lim_{t \to \infty} {\cal R} = 1/2$ due to the symmetry of
initial conditions.
The calculation of $r$ Eq. (\ref{eq19}) using Eqs. (\ref{eqAs01},\ref{eqGP1}) is straightforward
\begin{equation}
r = { 1  \over 2 \sqrt{ \pi D t} \sqrt{ 1 + \xi^2 / 2 D t}  },
\label{eqGP5}
\end{equation}
Inserting Eqs. (\ref{eqGP4},\ref{eqGP5}) in Eq.
(\ref{eq21}) we find
\begin{equation}
\langle ( x_T )^2 \rangle \sim \xi^2 {\pi \over N} \left( 1 + { 2 D t \over \xi^2} \right) \left[ {1 \over 4} -
{1 \over \pi^2 } \mbox{arccot}^2 \left( \sqrt{{ 2 D t \over \xi^2} } \right) \right],
\label{eqGP6}
\end{equation}
This solution is shown in Fig. \ref{fig12a}
with its two  limiting behaviors
\begin{equation}
\langle (x_T)^2 \rangle \sim \left\{
\begin{array}{l l}
\xi { \sqrt{ 2 D t} \over N} & \mbox{short times} \ \  2 D t \ll \xi^2 \\
{\pi D  \over 2 N} t & \mbox{long times} \ \  2 D t \gg \xi^2 .
\end{array}
\right.
\label{eqGPii}
\end{equation}
For short times the particles do not have time to disperse;
hence, the motion of the tagged particle is slower than normal, increasing
as $t^{1/2}$ which is similar to the single file diffusion
with a uniform density Eq.
(\ref{eqbox06}). Roughly speaking,
for short times
the tagged particle sees a uniform density of particles with $\rho=N/\xi$.
For long times we recover the behavior in Eq.
(\ref{eqAs03a}) since the scale of diffusion is much larger than $\xi$.
Hence if we start with   a Gaussian or delta function packet
we get in the long time limit similar behavior, as we showed.

\subsection{Particles in Harmonic Oscillator}

 Consider particles in a harmonic potential $V(x) = m \omega^2 x^2 / 2$ where
$\omega>0$ is the harmonic frequency. The single particle  undergoes
an Ornstein Uhlenbeck process \cite{Risken} and the corresponding
single particle Green
function is
$$ g(x,x_0, t) = $$
\begin{equation}
{ 1 \over \sqrt{ 2 \pi D \tau \left( 1 - e^{ - 2 t /\tau} \right)}} \exp\left[ - { \left( x - x_0 e^{ - t / \tau} \right)^2 \over 2 D \tau \left( 1 - e^{ - 2 t / \tau } \right) } \right],
\label{eqHO01}
\end{equation}
where $(\tau)^{-1} = D m \omega^2/ k_b T $ is the inverse relaxation time.

\begin{figure}
\begin{center}
\epsfig{figure=har1.eps,totalheight=0.34\textheight, width=0.45\textwidth}
\end{center}
\caption{ Scaled mean square displacement  of tagged particle in harmonic field
Eq.
(\ref{eqHO06})
exhibits a transition between a short time $\langle (x_T)^2 \rangle\propto
t^{1/2}$ law to saturation due to the binding field.
The horizontal dashed line is the long time $\langle (x_T)^2 \rangle = \pi \xi^{2}  _{th} /4 N$ behavior,
the dashed line is the  short time limit  Eq. (\ref{eqHO07}),
and the dotted dashed line is Eq.
(\ref{eqHO06}).
}
\label{fig12}
\end{figure}

 We assume thermal initial conditions
\begin{equation}
f(x_0) = { 2 \sqrt{ m \omega^2 } \over \sqrt{ 2 \pi k_b T} } \exp\left( - {m \omega^2 x^2 \over 2 k_b T} \right), \ \ \ x_0 > 0.
\label{eqHO02}
\end{equation}
Using Eq. (\ref{eq19}) it is easy to show that
\begin{equation}
r = { 1 \over \sqrt{ 2 \pi} \xi_{{\rm th}} },
\label{eqHO03}
\end{equation}
where the thermal length is $\xi_{{\rm th}} = \sqrt{ D \tau} = \sqrt{ m \omega^2 / k_b T} $.
 Note that one can write $r = 1/ Z$ which  as we will soon prove
is a general result
valid for all potential fields satisfying $V(x)=V(-x)$,
 provided that the initial condition
$f(x_0)$ is the thermal equilibrium .
The reflection coefficient [Eq.
(\ref{eq16})] is
\begin{equation}
{\cal R} = {1 \over 2} + \sqrt{{1\over 2 \pi} } \eta \int_0 ^\infty e^{ - {\eta^2 y^2 \over 2} } \mbox{Erf} \left( { y \over \sqrt{2} } \right) {\rm d} y,
\label{eqHO04}
\end{equation}
where $\eta = e^{t/\tau} \sqrt{ 1 - e^{ -2 t / \tau}}$. Using
Mathematica
\begin{equation}
{\cal R} = {1 \over 2} + { 1 \over \pi} \mbox{arccot} \left( \sqrt{ e^{ 2 t /\tau } - 1} \right).
\label{eqHO05}
\end{equation}
For short times $t/\tau<<1 $,
${\cal R} \sim 1 - \sqrt{ 2 t }/ (\pi \sqrt{\tau} )$,
and for long time ${\cal R} \sim  1/2 + e^{ - t/\tau} / \pi$.

Using Eqs.(\ref{eq21},\ref{eqHO03},\ref{eqHO05})  the mean square displacement
of the tagged particle is
\begin{equation}
\langle \left( x_T \right)^2 \rangle = { \pi \over N } \xi_{{\rm th}} ^2 \left[
{ 1 \over 4} - { 1 \over \pi^2} \mbox{arccot}^2 \left( \sqrt{ e^{ 2 t / \tau} - 1 } \right) \right] .
\label{eqHO06}
\end{equation}
For short times $t<<\tau$
\begin{equation}
\langle \left( x_T \right)^2 \rangle \sim {\xi_{{\rm th}} \over N} \sqrt{ 2 D t}
\label{eqHO07}
\end{equation}
Such a behavior is expected, as well known
 for short times  the diffusion process of the Ornstein Uhlenbeck process
$x\sim t^{1/2}$
 is faster
than the drift $x\sim t$  and dominates the process, i.e.
take $t<<\tau$ in Eq. (\ref{eqHO01}) and get the Green function of the
force free particle Eq.
(\ref{eqAs01}).
 Hence for short
times we can neglect the
external field, but use of course the Gaussian  initial condition
Eq. (\ref{eqHO02}).
Then the problem reduces to the short time
behavior of the Gaussian packet, free of external forces
(compare short time behavior in  Eq.
(\ref{eqGPii}) with Eq. (\ref{eqHO07}) when $\xi\to \xi_{{\rm th}}$).
The long time limit of Eq.
(\ref{eqHO06}) gives
$\langle \left( x_T \right)^2 \rangle \sim \pi \xi_{{\rm th}} ^2 / 4 N$
which is the thermal equilibrium behavior predicted more generally
in Eq.
(\ref{eq17a}).
Behavior of the scaled  mean square displacement and its asymptotic behaviors is
presented in Fig. \ref{fig12}.

\subsection{Thermal Initial Conditions}

 If we assume that initially the particles are in thermal
equilibrium, our simple formulas simplify even more.
If $f(x_0)= 2 \exp[ - V(x_0)/k_b T]/Z$ then $r = 1/Z$.
To see this notice that  for symmetric
potentials $V(x) = V(-x)$ we have $g(0,-x_0,t) = g(0,x_0,t)$ and
\begin{equation}
r= {2 \over Z}  \int_0 ^\infty \exp\left[ - {V(x_0) \over k_b T} \right]
g(0,x_0, t) {\rm d} x_0
\label{eqTI01}
\end{equation}
yields
\begin{equation}
r= \int_{-\infty} ^\infty { \exp\left[ - { V(x_0) \over k_b T} \right] \over Z} g(0,x_0,t) {\rm d} x_0,
\label{eqTI02}
\end{equation}
where we assumed that the potential is binding so a stationary solution
of the Fokker-Planck equation is reached; i.e. the free particle
is excluded. Therefore $r$ [Eq. (\ref{eqTI02})] is the probability of
finding a non interacting particle at the origin, with thermal initial
conditions. Since the thermal equilibrium density is
the stationary solution of the Fokker Planck operator,
$r$ is time independent and equal to
$r= \exp[ - V(0)/k_b T]/Z$. We can always take
$V(0)= 0$ and then $r = 1/ Z$.  Examples for this behavior
are the already analyzed cases of  particles in a box with uniform initial
density [Eq.
(\ref{eqbox02aa})] (since $Z= 2 \overline{L}$ for that case) and similarly for
particles in a harmonic potential with initial thermal density
 [Eqs.
(\ref{eqHO02},
\ref{eqHO03})].
Hence using Eq.
(\ref{eq21})
\begin{equation}
\langle (x_T)^2 \rangle = {{\cal R} \left( 1 - {\cal R} \right) Z^2 \over 2 N}.
\label{eqTI03}
\end{equation}
For  the symmetric and binding potentials under consideration,
 we have $\lim_{t \to \infty} {\cal R} = 1/2$ and hence Eq.
(\ref{eqTI03}) in the long time limit yields the equilibrium
behavior  Eq.
(\ref{eq17aa}).

\subsection{Power Law Type of Initial Conditions}

 Flomenbom and Taloni \cite{Ophir}
considered particles free of external forces,
hence the free particle Green function is
\begin{equation}
g(x,x_0, t ) = { 1 \over \sqrt{ 4 \pi D t } } \exp\left[ - { \left( x - x_0 \right)^2 \over 4 D t } \right],
\label{eqOP01}
\end{equation}
with initial conditions of power law type
\begin{equation}
f(x_0) = B | x_0 |^{- \beta}  \ \ \ \ 0<x_0 <L
\label{eqOP02}
\end{equation}
for $0<\beta<1$. Here the system size $\overline{L} \to \infty$ and
the reader should not confuse $\overline{L}$ with $L$.
The normalization of the PDF of Eq. (\ref{eqOP02}) yields
$B= (1 - \beta) L^{\beta - 1} $. In the limit $\beta \to 0$ and $L \to \infty$
in such a way that $\rho = N/L$ remains fixed, we anticipate the
classical case of single file diffusion in the  presence of
a uniform density of particles: $\langle x^2 \rangle \propto t^{1/2}$ [Eq.
(\ref{eqbox06})]. In the opposite limit of $\beta \to 1$ the particles
are initially concentrated at the origin, and we expect the
behavior $\langle x^2 \rangle \propto t$ [Eq.
(\ref{eqAs03a})]. Thus  $0<\beta<1$ bridges between these two known behaviors, indeed
one finds
$\langle (x_T)^2 \rangle \propto t^{(\beta + 1 )/2}$ as shown in \cite{Ophir}.
The latter is valid for times $\sqrt{4 Dt} \ll L$ as discussed below. This indicates
that specially chosen initial condition may control the qualitative
behavior of the diffusion of the tagged particle. Here we analyze
this case using our formalism finding analytical expressions for the
mean square displacement.

 To obtain $\langle (x_T )^2 \rangle$ all
we need to do is to find $r$ and ${\cal R}$. Using Eqs.
(\ref{eq19},\ref{eqOP01},\ref{eqOP02})
\begin{equation}
r = { \left( 1 - \beta \right) \over \sqrt{\pi} } { L^{\beta-1} \over \left( \sqrt{ 4 D t } \right)^\beta }
\int_0 ^{ {L \over \sqrt{ 4 D t}}} y^{-\beta} e^{ - y^2 } {\rm d} y,
\label{eqOP03}
\end{equation}
solving the integral
\begin{equation}
r = { \left( 1 - \beta \right) \over \sqrt{\pi} } { L^{\beta-1} \over \left( \sqrt{ 4 D t } \right)^\beta }
{1 \over 2} \left[ \Gamma\left( {1 - \beta \over 2} \right) -
\Gamma\left({1 - \beta\over 2} , {L^2 \over 4 D t } \right)\right],
\label{eqOP03a}
\end{equation}
where $\Gamma(a,z)$ is the incomplete Gamma function \cite{Abr}.
The following behaviors are found for short and long
times
\begin{equation}
r = \left\{
\begin{array}{l l}
c_r { L^{\beta - 1} \over \left( \sqrt{ 4 D t } \right)^\beta} & \ \sqrt{4 D t } \ll L \\
{ 1 \over \sqrt{ 4 \pi D t} } & \ \sqrt{4 D t} \gg L .
\end{array}
\right.
\label{eqOP04}
\end{equation}
In the short time limit we took the upper bound in integration
in Eq. (\ref{eqOP03}) to $\infty$ so
\begin{equation}
c_r = { (1 - \beta ) \over \sqrt{\pi} } \int_0 ^\infty y^{ - \beta} e^{ - y^2 } {\rm d} y = { 1 - \beta \over \sqrt{\pi} } { \Gamma\left( 1 - \beta/2 \right) \over 2}.
\label{eqOP05}
\end{equation}

Inserting Eqs. (\ref{eqOP01},\ref{eqOP02}) in Eq. (\ref{eq16}) we find
the reflection coefficient
\begin{equation}
 {\cal R} = {1 \over 2} +
{1 \over 2 } \left( 1 - \beta \right)\left( {  L \over \sqrt{ 4 D t} } \right)^{\beta -1}
\int_0 ^{L/ \sqrt{ 4 D t}} {\rm d} y y^{- \beta} \mbox{Erf} \left( y \right) .
\label{eqOP06}
\end{equation}
In the limit of long times we get the expected behavior $\lim_{t \to \infty} {\cal R} = 1/2$
since then half of the particles are to the left of the origin and half to the right (in statistical
sense).
Mathematica solves the integral in the last equation
in terms of tabulated functions
\begin{equation}
\begin{array}{l}
\int_0 ^x y^{- \beta} \mbox{Erf} \left(y \right) {\rm d} y =\\
 { x^{-\beta}  \over (1 -\beta) \sqrt{\pi} } \left\{ x \left[ \sqrt{\pi} \mbox{Erf} \left( x \right) + x \mbox{E}_{\beta/2} \left( x^2 \right)\right] - x^\beta \Gamma\left( 1 - { \beta \over 2} \right) \right\},
\end{array}
\label{eqOP07a}
\end{equation}
where $\mbox{E}_n (z)= \int_1^\infty {\rm d} t \exp(-z t) / t^n $
is the exponential integral function.

To analyze the short time behavior we use integration by parts in Eq.
(\ref{eqOP06}) and find
\begin{equation}
\begin{array}{c}
{\cal R} = {1 \over 2} +
{1\over 2} \left( { L \over \sqrt{ 4 D t} } \right)^{\beta - 1}\times\\
\left[
\left( { L \over \sqrt{ 4 D t} } \right)^{1 - \beta} \mbox{Erf}\left( { L \over \sqrt{ 4 D t } } \right)
- { 2 \over \sqrt{\pi} } \int_0 ^{L/\sqrt{ 4 D t} } y^{ 1 - \beta} e^{ - y^2 } {\rm d} y  \right].
\end{array}
\label{eqOP07}
\end{equation}
Using asymptotic properties of the $\mbox{Erf}$ function \cite{Abr}, neglecting
terms of the order of $\exp( - L^2 / 4 D t)$,
Eq. (\ref{eqOP07})
for $\sqrt{ 4 D t } \ll L$ gives
\begin{equation}
{\cal R} \sim 1 - c_{{\cal R}}  \left( {\sqrt{ 4 D t } \over L} \right)^{1 - \beta} ,
\label{eqOP08}
\end{equation}
with
\begin{equation}
c_{{\cal R}} = { 1 \over \sqrt{\pi}} \int_0 ^\infty y^{1 - \beta} e^{ - y^2 } {\rm d} y ={\Gamma \left( 1 - {\beta \over 2} \right) \over 2 \sqrt{\pi} } .
\label{eqOP09}
\end{equation}

Eqs.
(\ref{eqOP03a},\ref{eqOP06},\ref{eqOP07a})
yield the  mean square displacement
 Eq. (\ref{eq21}). The short time behavior
$\sqrt{4 D t} \ll L$
\begin{equation}
\langle \left( x_T \right)^2 \rangle
\sim {1 \over 2 N} {2 \sqrt{\pi} \over (1 - \beta)^2 \Gamma\left(1 - \beta/2\right) }
L^{1-\beta} \left( \sqrt{ 4 D t } \right)^{ 1 + \beta},
\label{eqOP10}
\end{equation}
while for long times $\sqrt{ 4 D t} \gg L$
\begin{equation}
\langle \left( x_{{\cal T}}\right)^2 \rangle \sim { \pi D t \over 2 N } .
\label{eqOP11}
\end{equation}
Thus after a long time the diffusion is normal $\langle (x_T)^2 \rangle \propto t$,
exactly like  the case where all particles started
initially on the origin Eq.
(\ref{eqAs03a}). While for short times
$\langle (x_T)^2 \rangle \propto t^{(1 + \beta)/2}$.

Taking  the limit $\beta\to 0$
in Eq. (\ref{eqOP10}),
we get the
well known result of single file motion in uniform density of particles
 \cite{Ha,Le} Eq.
(\ref{eqbox06}), with the density $\rho=N/L$.
In the opposite limit $\beta \to 1$ we have $\langle x^2 \rangle
=
{ \pi D t \over 2 N }$ for all times
which is the same as in Eq. (\ref{eqAs03a}).
Note that the limit $\beta\to 1$ must be treated with care,
the short time limit and the $\beta\to 1$ limit do not commute
due to the the $1/(1-\beta)^2$ divergence  in
Eq.
(\ref{eqOP10}). When $\beta\to 1$  all particles are centered on the origin
hence we get the same behavior as in  Eq.
(\ref{eqAs03a}).

\section{Percus Relation: Beyond Brownian Particles}
\label{SecBeyond}

 So far we considered the case where particles are diffusing according
to the laws of Brownian motion in between collision events.
What happens for other types of motion? For example what happens
when the underlying motion itself is anomalous \cite{Ophir,Band}.

Percus \cite{Percus1} (see also \cite{Hahn,Ophir})
investigated  a general relation between the diffusion
of the tagged particle
and motion in the absence of
interactions  (free motion)
$\langle (x_T)^2 \rangle \sim \langle |x| \rangle_{{\rm free}}/\rho$.
Such a relation was suggested for normal diffusion
where we have $\langle |x| \rangle_{{\rm free}}
\propto t^{1/2}$,  so $ \langle (x_T)^2 \rangle \propto t^{1/2}$
and for a particle moving ballistically between collision
events
hence $\langle |x| \rangle_{{\rm free}} \propto t$ and
therefore  when we turn on the interactions
$\langle (x_T)\rangle^2 \propto t$.
This simple relation between free particle motion and the
 mean square displacement
of the  tagged interacting particle,
is expected to work for an infinite system, with a uniform
density $\rho$ of particles, and when external forces are
zero $[F(x) = 0]$.
Here we will derive the Percus relation from our formalism, for more general
dynamics.

 Assume that the Green function of the non interacting particle
can be
written in the scaling form
\begin{equation}
g(x,x_0,t) = {1 \over \sqrt{K_\alpha} t^{\gamma/2} } G\left( { x - x_0 \over \sqrt{K_\gamma} t^{\gamma/2} } \right) .
\label{eqAno1}
\end{equation}
We assume $G(y) = G(-y)$ and from normalization $\int_{-\infty} ^ \infty G(y) {\rm d} y = 1$ .
Here the free particle motion is anomalous, so that
$\langle |x| \rangle_{{\rm free}}
\propto t^{\gamma/2} $ for $0<\gamma$ and not equal unity.
For example the underlying motion might be sub-diffusive continuous time
random walk (CTRW) \cite{SM,Bouchaud,Review} or fractional Brownian motion, where in the latter
case $G(y)$ is Gaussian and in the former $G(y)$ is expressed
in terms of L\'evy distributions (see some details below).
We assume that moments of the process
are finite. The constant $K_{\gamma} $ has units of $\mbox{m}^2/\mbox{sec}^\gamma$.

 To obtain the mean square displacement of the tagged interacting
particle we calculate
the reflection coefficient
Eq. (\ref{eq16})
\begin{equation}
{\cal R} = {1 \over \overline{L}} \int_0 ^{\overline{L}} {\rm d} x_0 \int_0 ^{\overline{L}} G\left({ x- x_0 \over \sqrt{K_\gamma} t^{\gamma/2}} \right) { {\rm d} x \over
\sqrt{K_\gamma} t^{\gamma/2} } .
\label{eqAno2}
\end{equation}
Here we used uniform density of particle $f(x_0) = 1/ \overline{L}$,
and we will consider the limit $\overline{L} \to \infty$ with the
density $\rho$ being kept fixed. Change of variables
$y = (x-x_0) / \sqrt{K_\gamma} t^{\gamma/2}$
and using $\overline{L} / \sqrt{K_\gamma} t^{\gamma/2} \to \infty$
we have
\begin{equation}
{\cal R} \sim {1 \over \overline{L} } \int_0 ^{\overline{L}} {\rm d} x_0
\int_{ {- x_0 \over \sqrt{K_\gamma} t^{\gamma/2}} } ^\infty  G(y) {\rm d} y.
\label{eqAno3}
\end{equation}
Integrating by parts, changing variables
 $y= x_0 / \sqrt{K_\gamma} t^{\gamma/2} $, and using the
normalization
condition,
we have
\begin{equation}
{\cal R} \sim
1 - {  \langle |x| \rangle_{{\rm free}} \over  2 \overline{L}}
\label{eqAno4}
\end{equation}
where the mean of the absolute value of the free particle
motion is by its definition
\begin{equation}
\langle |x| \rangle_{{\rm free}} = \int_{-\infty} ^\infty |x| G\left( { x \over \sqrt{K_\gamma}  t^{\gamma/2} } \right)
{{\rm d} x \over \sqrt{K_\gamma} t^{\gamma/2} }.
\label{eqAno5}
\end{equation}
Here we used the assumed symmetry $G(y) = G(-y)$ which means that
the underlying random walk is not biased, as a result
$\langle |x| \rangle = \sqrt{K_\gamma} t^{\gamma/2} 2 \int_0 ^\infty y G(y)
{\rm d} y $. It is easy to see that $r=1/2 \overline{L}$. In the limit
where $N \to \infty$ and $\overline{L} \to \infty$ we find using
Eqs. (\ref{eq21},\ref{eqAno4})
the Percus formula for a general class of stochastic dynamics
\begin{equation}
\langle \left( x_T\right)^2 \rangle = { \langle | x| \rangle_{\rm free} \over \rho} .
\label{eqAno6}
\end{equation}
Clearly this equation  gives
 a useful relationship between motion of a
free particle and the same particle moving in single file when it is
surrounded by identical particles whose density is $\rho$,

 As a simple example consider fractional Brownian motion where
$g(x,x_0,t) = ( \sqrt{ 4 \pi K_\gamma t^\gamma})^{-1} \exp \left[ - (x- x_0)^2 / 4 K_\gamma t^\gamma\right]$, where $0<\gamma<2$. Then Eq. (\ref{eqAno6})
gives
\begin{equation}
\langle \left( x_T \right)^2 \rangle = {2 \sqrt{K_\gamma t^\gamma} \over \rho \sqrt{\pi} }  .
\label{eqAno7}
\end{equation}
Hence if the underlying motion is ballistic $\gamma\to 2$ the motion of
the tagged particle undergoing single file dynamics is normal
with respect to time $\langle (x_T)^2 \rangle \propto t$.
For a CTRW particle in the continuum approximation the
non-interacting single particle green function is governed
by the fractional diffusion equation \cite{Schneider,MBK,Review}
\begin{equation}
{ \partial^\gamma \over \partial t^\gamma} g(x,x_0,t) = \overline{K}_\gamma {\partial^2 \over \partial x^2} g(x,x_0,t)
\label{eqAno8}
\end{equation}
with $0<\gamma<1$.
From this equation it is easy to find \cite{BarkaiPRE}
$\langle|x|\rangle_{{\rm free}} =
\sqrt{ \overline{K}_\gamma t^\gamma} / \Gamma(1 + \gamma/2)$.
 Hence
using Eq. (\ref{eqAno6})
\begin{equation}
\langle \left( x_T \right)^2 \rangle ={ \sqrt{\overline{K}_\gamma} t^{\gamma/2} \over \rho \Gamma\left(1 + \gamma/2\right) } .
\label{eqAno9}
\end{equation}
We see that independent of the mechanism of the underlying anomalous
diffusion (i.e. fractional Brownian motion, or CTRW) we get
for the tagged particle $\langle (x_{{\cal T}})^2 \rangle \sim t^{\gamma/2}$.
Further our general results show that the Green function of the
tagged particle is Gaussian even though the Green function
of non-interacting CTRW particles  is highly non-Gaussian.

  {\bf Warning:} One should be careful  when applying our
results to the CTRW model, namely Eq. (\ref{eqAno9}) should not
be abused.
One mechanism of anomalous diffusion are the power law waiting times
of Scher and Montroll \cite{SM}  which yield slow dynamics, as captured by the
CTRW  \cite{Bouchaud,Review} and the fractional Eq.
(\ref{eqAno8}) \cite{BarkaiPRE}.
Single file diffusion
with such sub-diffusive motion as the starting point was considered
recently theoretically and with simulations in \cite{Ophir,Band}.
It should be noted that the meaning of collision in such a model
should be taken with care. For a random walk on a lattice with
waiting times on each lattice point, one can envision several
collision rules. For example one might allow two particles
to occupy the same site at a given time, or one may consider a mechanism
were a particle once hopping into a trap already occupied will eject
the particle previously residing in the trap.  Or a particle
is allowed to jump only into an empty site, as is usually assumed.
These type of collision
rules might yield behaviors different than ours. For example
a particle stuck with a very large sojourn time, might be ejected
by another particle, hence one can imagine a situation where
some form of interaction causes the particles to move faster.
 In our work the collision implies that we can
let particles go past one another as if they were non interacting
(so on a lattice two particles may occupy the same point at the same time)
and eventually we look for the center particle.
Even more interesting will be to investigate interacting particles
in systems with quenched disorder, since the latter,
when strong enough,
is known to lead
to non Gaussian sub-diffusion \cite{SM,Bouchaud}.  There the simple formula
Eq. (\ref{eqAno6}) is generally not expected to hold.
Indeed in \cite{Ben} single file motion of a tagged particle in the Sinai
model was considered, the results are much richer when compared
with Eq. (\ref{eqAno6}).

\section{Discussion}

 The one dimensional problem of motion of a tagged particle
interacting via hard core interactions with other particles
was solved using the Jepsen line. The formalism we developed
treats both Brownian and non Brownian motion in between collision
events, is suited for
rather  general external fields acting on the particles,
for open and closed system,  and
handles also different types of initial conditions. Following
others we have mapped the problem onto a non interacting
problem using the Jepsen line. The motion of the tagged particle
belongs to the general problem of order statistics.
The problem reduces to considering a list of $2 N +1 $ random
variables and finding the distribution of
the  variable which has $N$ variables
smaller than it and $N$ larger corresponding to center particle
(note that  the right most particle we will have an
extreme value problem). Classical theory of order statistics deals however
with the case where all the random variables have identical distributions.
In contrast in the exclusion process under consideration, particles
have non identical distribution. Thus except for two cases:
i) all the particles initially on the same position and ii) equilibrium
state, the problem deals with non identically distributed random variables
(since the initial condition are non identical).
 While we
treated the problem of symmetric potential and symmetric initial
condition for the center particle in detail, it is left for
future work to consider non-symmetric potential fields,
non symmetric initial conditions, and the dynamics of the particle
in the tails of the packet. We believe that the methods developed here
with some modifications can treat these cases too.

{\bf Acknowledgment} EB is supported by the Israel Science Foundation, and
RS by the NSF. We thank Ludvig Lizana and
 Tobias Ambj\"ornson for sharing with us their data in
Fig. \ref{fig10} and for providing helpful insight. We also
thank M. Lomholt and A. Taloni for useful discussions.

\section{Appendix A}

In this Appendix we use Boltzmann's distribution for the interacting
system to find the PDF of the tagged center particle in equilibrium.
The multi dimensional PDF for $2 N + 1$ interacting
particles, in the presence of an external binding field $V(x)$, with $V(x) = V(-x)$, acting
on all of them is
\begin{widetext}
\begin{equation}
P\left( x_{-N}, \cdots,x_{-1} , x_{0} , x_1, \cdots, x_N \right) =
{1 \over Z_{2 N+ 1} } \exp\left[ - \sum_{j=-N} ^N {V(x_j) \over k_b T}\right]
\theta\left(x_{-N+1} - x_{-N}\right)\theta\left( x_{-N + 2 } - x_{-N + 1} \right) \cdots
\theta\left(x_{N} - x_{N-1}\right)
\label{eqAp1}
\end{equation}
where $Z_{2 N + 1}$ is a normalizing factor, and $\theta(x)$ is a step function:
$\theta(x)=0$ if $x<0$, $\theta(x)=1$ for  $x\ge 0$.
 The  center tagged particle is $x_0=  x_T$.
To find the PDF  of $x_T$ in equilibrium, which we call
$P^{{\rm eq}}(x_T)$,  we must integrate Eq. (\ref{eqAp1})  over
all coordinates besides $x_0 \to x_T$
\begin{equation}
\begin{array}{l}
P^{{\rm eq}} \left(x_T\right) =  {\exp\left[ - { V\left( x_T \right) \over k_b T } \right] \over Z_{2 N + 1} } \times \\
 \ \\
\int_{-\infty}^{x_T} {\rm d } x_{-N} \int_{x_{-N}} ^{x_T} {\rm d} x_{-N + 1} \cdots \int_{x_{-2}} ^{x_T} {\rm d} x_{-1}
\exp\left[ - { \sum_{j=-N} ^{-1} V\left(x_j\right) \over k_b T }  \right]
\int_{x_{T}}^{\infty} {\rm d } x_{1}
\int_{x_1}^{\infty} {\rm d } x_{2}
\cdots \int_{x_{N-1} } ^\infty {\rm d} x_N
\exp\left[ - { \sum_{j=1} ^{N} V\left(x_j\right) \over k_b T }  \right] .
\end{array}
\label{eqAp2}
\end{equation}
We rearrange the integration limits,
as explained in
Fig. \ref{fig4}
\begin{equation}
\int_{-\infty} ^{x_T} {\rm d} x_{-N} \int_{x_{-N}} ^{x_T} {\rm d} x_{-N + 1}  \cdots =
\int_{-\infty} ^{x_T} {\rm d} x_{-N} \int_{-\infty} ^{x_{-N} } {\rm d} x_{-N + 1}  \cdots .
\label{eqApnn}
\end{equation}
 Hence we can use
\begin{equation}
 \int_{-\infty} ^{x_T} {\rm d} x_{-N} \int_{x_{-N}} ^{x_T} {\rm d} x_{-N + 1}  \cdots =
{1 \over 2} \left[ \int_{-\infty} ^{x_T} {\rm d} x_{-N} \int_{x_{-N}} ^{x_T} {\rm d} x_{-N + 1}  \cdots  +
\int_{-\infty} ^{x_T} {\rm d} x_{-N} \int_{-\infty} ^{x_{-N}  } {\rm d} x_{-N + 1}  \cdots \right] = {1 \over 2} \int_{-\infty} ^{x_T} {\rm d} x_N \int_{-\infty} ^{x_T} {\rm d} x_{-N + 1} \cdots .
\label{eqApnn1}
\end{equation}
Repeating this procedure  we rewrite Eq.  (\ref{eqAp2}) as
\begin{equation}
P^{{\rm eq}} \left( x_T \right) = \mbox{Nor} \left\{ \int_{-\infty} ^{x_T} {\rm d} x {\exp\left[ - {V\left( x \right) \over k_B T } \right]\over Z} \right\}^N
 \left\{ \int_{x_T} ^{\infty} {\rm d} x {\exp\left[ - {V\left( x \right) \over k_B T } \right]\over Z}  \right\}^N
{\exp\left[ - {V \left( x_T \right) \over k_B T} \right]\over Z}.
\label{eqAp3}
\end{equation}
\end{widetext}
where $\mbox{Nor}$ is a normalization constant
and $Z$ is defined in Eq. (\ref{eq13b}).

\begin{figure}
\begin{center}
\epsfig{figure=area.eps, width=0.45\textwidth}
\end{center}
\caption{
Left panel integration in the domain
$-\infty < x_{-N} < x_{T}$, $x_{-N} < x_{-N +1}<x_T$
is equivalent to integration $-\infty <x_{-N} < x_T$,
$-\infty <x_{-N +1} <x_{-N}$ (right panel).
}
\label{fig4}
\end{figure}

 Thus Eq. (\ref{eqAp3}) describes a problem
of order statistics which is extensively investigated by mathematicians,
as mentioned  in Sec.
\ref{SecGumbel}.
 $P^{{\rm eq}}(x_T)$
Eq. (\ref{eqAp3})
is the PDF of the random variable
which has exactly $N$ random variable  larger than it and $N$ smaller.
In this sense we have transformed
the problem to a non-interacting system, similar to the non interacting
picture in the main text.
 The information contained in the
single non-interacting particle, i.e. the single particle
Boltzmann distribution $\exp[ - V(x)/ k_b T]/Z$ is all what is
needed
for the calculation of the position of the tagged particle.

  Using the  symmetry $V(x)=V(-x)$ we have
\begin{equation}
\int_{-\infty} ^{x_T} { e^{- { V(x) \over k_B T} } \over Z} {\rm d} x = { 1 \over 2} + \int_0 ^{ x_T}
 { e^{- { V(x) \over k_B T} } \over Z} {\rm d} x,
\label{eqAp4}
\end{equation}
and
\begin{equation}
\int_{x_T} ^\infty { e^{- { V(x) \over k_B T} } \over Z} {\rm d} x = { 1 \over 2} - \int_0 ^{ x_T}
 { e^{- { V(x) \over k_B T} } \over Z} {\rm d} x.
\label{eqAp5}
\end{equation}
Using Eqs. (\ref{eqAp4},\ref{eqAp5}), we rewrite Eq. (\ref{eqAp3})
\begin{equation}
\begin{array}{c}
P^{{\rm eq}} \left( x_T \right)  = \mbox{Nor} { e^{ - V(x_T)/k_b T } \over Z}e^{N \ln {1 \over 4}}  \\
 \exp\left\{ N \ln \left[ 1
- 4 \left( { \int_0 ^{x_T} e^{ - V(x)/k_b T } {\rm d} x \over Z} {\rm d} x \right)^2 \right] \right\}.
\end{array}
\label{eqAp6}
\end{equation}
In the limit $N \to \infty$ only $x_T$ with $\int_0 ^{x_T} \exp[ - V(x)/k_b T ] {\rm d} x /Z \ll 1$
will have a measurable contribution to $P^{{ \rm eq}} (x_T)$,
since if this condition is not satisfied,
the value of $P^{{\rm eq}} (x_T)$
is exponentially small in $N$. Expanding the $\ln$ in Eq.
(\ref{eqAp6}) we have
\begin{equation}
P^{{\rm eq}} (x_T) \propto \exp\left[  - {V(x_T) \over k_b T } \right] \exp\left[  - 4 N \left({ \int_0 ^{x_T} {\rm d} x
e^{ - {V(x) \over k_b T}}  \over Z} \right)^2\right].
\label{eqAp7}
\end{equation}
Since $N>>1$ we expand $\int_0 ^{x_T} \exp( - V(x) /k_b T) {\rm d} x = x_T$
where we
used $V(0)=0$.
We use  $V(x_T)/k_b T \ll N (x_T)^2 /Z^2$ which holds in the center part
of the PDF of the tagged particle (since $N>>1$, $V(0) = 0$, and
$V(x)$ is analytic)
$P^{{\rm eq}} (x_T ) \sim C \exp \left[ - { 4 N \left( x_T \right)^2 \over Z^2} \right] $
where $C$ is a normalization constant.
This final approximate result is the same as Eq.
(\ref{eq17a}), justifying the tricks used to derive our  main results.

redpajama_set_name: RedPajamaArXiv


Your videos have been a source of widespread enjoyment - as you probably know all too well. Thanks for posting them.
Honored that others find some of the same joy we felt at the time, I wish there were more I could share!
Tony, I do not believe it, are you the master who rode her?
Congratulations for yours video Tony!
(09-20-2018, 11:48 PM)Mike Wrote: Tony, I do not believe it, are you the master who rode her?
I have a question! Are you writing about this ponygirl?
According to an email from a friend, it is 1.50 m large and the man (rider) weighs 110 kg.
the complete movie material (http://iporntv.net/download/video/ph56c9...up-(funny) - Are only a handful of them).
I would like to have met myself once!
But you just can not have everything in life.
Although I ride two ponies, but this well-shaped body I would like to have had between the thighs.
Glad I was able to catch and ride that comet before she retired!


GAA talking points: Can Kerry make a statement?
Ahead of the weekend's GAA action, we take a look at the main talking points.
Can Kerry make a statement?
With two wins from two, Kerry are off to a flying start under Peter Keane. However, it's set to go up a level in Tralee on Saturday night as All-Ireland champions Dublin come to town.
The Kingdom's youngsters have passed the two tests thus far with flying colours but won't have things all their own way against a Dublin team quickly adjusting to the pace after a late start to their preseason.
Nonetheless, a win at Austin Stack Park would represent a major milestone for this young team as they aim to bounce back following a disappointing 2018. In Kerry, All-Ireland titles are generally the only metric regarded in end of year accounts, but this is a coming team and trumping the Dubs would signal real progress.
Can Mayo continue winning start?
James Horan is well and truly up and running in his second stint as Mayo boss. The Connacht side are two wins from two ahead of the visit of Cavan to Castlebar on Saturday, and will fancy their chances of accounting for the Breffni.
Horan has opted for five changes from the side that demolished Tyrone in Omagh, with David Clarke replacing Rob Hennelly between the sticks.
Young players are continuing to burst through, with the experienced contingent picking up game-time. In all, Mayo are motoring along nicely as Horan looks to build a squad capable of challenging in the summer.
Will the real Tyrone please stand up?
The Red Hands were touted as contenders ahead of the Division 1 campaign, after looking sharp on their way to the McKenna Cup title.
However, since then they've failed to raise a glove, scoring an aggregate of just 17 points in losses to Kerry and Mayo.
While the league is a time for learning about your team before the summer, Mickey Harte will know the importance of delivering a result when they travel to Roscommon on Sunday. A loss to the Rossies would mean last year's All-Ireland finalists are in a real relegation scrap, and a run of losses would unsettle a team looking to build on a promising 2018.
Who can seize momentum on moving weekend in Division 2?
The second tier is widely hailed as perhaps the most fiercely contested of all the divisions. After this weekend, it's likely that it will become clear which sides are in the hunt for relegation, and who will be battling relegation.
Donegal are hot favourites to go straight back up, following a perfect start with wins over Clare and Meath. They head to Semple Stadium to take on Tipperary, who are struggling with an injury crisis. The three Munster counties lie bottom of the table with one point each, and all three will be looking to secure a much-needed win this weekend. Cork travel to Ennis to face Clare.
Meanwhile, Kildare are the latest side attempting to break down a resolute Fermanagh defence when they make the trip to Enniskillen, as Meath and Armagh, after promising starts, aim to keep pace with the leaders when they meet in Navan.
It looks like a huge round of games.
Who will reach the St Patrick's Day decider?
The race for the All-Ireland club hurling title is wide open. The Thurles meeting of Ballyhale Shamrocks and Ballygunner is the standout tie, with both sides fancying their chances after impressive provincial titles.
2013 All-Ireland champs St Thomas are favourites to atone for their semi-final loss two years ago and overcome Cushendall at Parnell Park, but the Antrim side won't be lacking in confidence given how they overcame Galway opposition upon their last foray outside of Ulster.
Both of Saturday's encounters are tricky to call.
We'll send you push notifications so you'll receive all the big GAA news!


text:
Q: Based On First Checkbox list value 2nd checkbox should come dynamically Using angularjs? i am new to this technology please any one help me out.i have two different checkbox list 1st one is Select Stores 2nd one is Select your device Company name .baesd on first checkbox checked value 2nd(Select your device Company name) checkbox list should come achievements: achievement :Based on first check box list(Select Stores) values 2nd checkbox(Select your device Company name) list should come
Example1: if i select Sai Stores in Select Stores list .Select your device Company name list should show only Sai Stores
Example2 : if i select Sai Stores,Sri Ram Stores in Select Stores list .Select your device Company name list should show Sai Stores and Sri Ram Stores
Example3 :
if nothing is select in that Select Stores list . Select your device Company name list should be hide or not visible
Example 4:now i have already selected Sai Stores in Select Stores list.Select your device Company name list should show only Sai Stores .here instead of that its showing both in Select your device Company name list
if matched then only second list of checkbox should show other wise it should be hidden
https://jsfiddle.net/bq5c1txd/11/

var app = angular.module("myApp", []);
    app.controller("Test1Controller", function($scope) {
    var serverData = ["Sai Stores", "Sri Ram Shop", "Ragava Stores","Baba DHl Shop"];
    $scope.items = [];
 var selectedStore = 'Sai Stores';
 $scope.itemChecked = false;
                for (var i = 0; i < serverData.length; i++) {
                    var modal = {
                        name: serverData[i],
                        selected: false

                    };
                  if (selectedStore.indexOf(serverData[i]) >= 0 ) {
                        modal.selected = true;

                    }

                    $scope.items.push(modal);

                }

        $scope.itemsChecked = function() {
            $scope.itemChecked = false;
            for (var i = 0; i < $scope.items.length; i++) {
                if ($scope.items[i].selected) {
                    $scope.itemChecked = true;
                }
            }
        }
//-----------------------------------------------------------------------------------------------------
$scope.stores = [{

        id: "1",
        storeid: "86754634",
        storename: "Sai Stores",
  servicedevice:"Htc"

    }, {
         id: "2",
        storeid: "3746579",
        storename: "Sri Ram Stores",
  servicedevice:"Samsung"
    }

    ]
//-----------------------------------------------------------------------------------------------------
        $scope.check = function() {
            var checkedItems = [];
            for (var i = 0; i < $scope.items.length; i++) {
                if ($scope.items[i].selected) {
                    checkedItems.push($scope.items[i].name);
                }
            }
            console.log(checkedItems);
        }
  $scope.check1 = function() {
            var checkedItems1 = [];
            for (var i = 0; i < $scope.stores.length; i++) {
                if ($scope.stores[i].selected) {
                    checkedItems1.push($scope.stores[i].servicedevice);
                }
            }
            console.log(checkedItems1);
        }
    });

    <form name="projectForm">

             Select Stores

                <input type="checkbox" ng-model="item.selected" ng-change="itemsChecked()" required="" />{{item.name}}

            Select your device Company name

                <input type="checkbox" ng-model="store.selected"  required="" />{{store.storename}}

* Mandatory

            <input type="button" name="submit" value="submit" ng-disabled="!itemChecked" ng-click="check();check1" />

    </form>

redpajama_set_name: RedPajamaStackExchange


Sunday, October 1st was a beautiful day in our community. A morning of life-changing declarations. With friends from Compass Housing Alliance, the neighborhood, and our church community we were able to see folks take a next step in their faith journeys. To tell the world that they love Jesus, have been saved by his grace, and will live for Him all the days of their life.


What Can Students Do to Protect Themselves?
The Consumer Financial Protection Bureau has developed a suite of tools to help students manage college debt and improve their financial literacy.
Americans owe more than a trillion dollars in student loan debt. That's more than we owe on credit cards, more than we owe on car loans – and it's still growing.
So, if you're going to invest in a college degree, we want you to be able to choose the best deal for your situation. Students are overwhelmed with options and aren't sure how to compare them. In the absence of apples-to-apples comparisons, they're left to their own devices when making a choice that will have significant consequences for their financial future.
From start to finish, we can help you make informed financial decisions about paying for college.
While we can't give you advice for your exact situation, we can point you in the right direction.
Three steps that can help you get the right loan for you.
We'll forward your issue to the company, give you a tracking number, and keep you updated on the status of your complaint. Even if you have federal loans, we can help make sure you get to the right place to submit a complaint.
Choosing your first bank account is an important decision. Unlike that first school ID photo, your first banking relationship could last long after you graduate. Making a smart decision now will mean fewer surprise fees that can add up later.
We're building a tool that will let you compare financial aid offers so you can see how all those numbers impact your payments down the road.


Are you motivated by what others think about you? Do you donate to charity because it's a good thing to do, or because to be seen to donate by others feels good? If your mental self-protection strategies are working, you'll definitely think it's the former but you may feel an unconscious itch which worries about it being the latter. In this post, I consider what really motivates human actions, look through research into human behaviour and see how little words like 'thank you' can have a large effect.
Acting in a selfless manner certainly makes us feel good, but research shows that if we happen to be seen doing it, it makes us feel even better. We thrive off this validation. Our propensity to decide how we act through the lens of the perception of others is called image motivation, or signalling motivation.
To test this, Professor of Psychology and Behavioural Economics at Duke University, Dan Ariely and his colleagues, devised an experiment in which participants were recruited to click an awkward combination of computer keys over and over again. The task was deliberately mundane and soul destroying, but the number of times the participant completed it determined how much money was donated on their behalf to the American Red Cross. Since 92% of participants thought highly of the Red Cross, performing the dreary task could be assumed to make people look good in the eyes of their peers. People were randomly assigned to either a private group, where only the participant knew the amount of the donation, or a public group, where the participant had to stand up at the end of the session and share this information with the group. Consistent with the hypothesis that image mattered, participants exerted much greater effort in the public case; with the average number of public clicks (900) nearly doubling the average of private clicks (517).
Think about how this finding could be an influence beyond the experimenter's lab. In the context of social media, when you're deciding to post a photo, do you find your finger hovering over the 'Post' button just before you share it with the world? In this hesitation, you're really asking yourself "Is this activity good?". Your subconscious pauses because it wants others to see you as a good person and if the answer is positive, you post the pic. The number of likes you receive validates if your activity was 'good' and by extension, you're now a good person in the eyes of your peers. This will influence future behaviour, leading you to post more pictures similar to that last one.
"You're about to buy a new car, it has two models. The first is environmentally friendly but more expensive than the second standard petrol equivalent model.
There are many ingenious pieces of psychological research like above, all of which lean towards the idea that we're highly motivated by how our image will be perceived by others. So do companies use this knowledge when attempting to promote engagement with their products or services?
How are companies rewarding customer behaviour?
When companies want to reward their customers, many ignore intrinsic motivations and think that showing gratitude is best done by offering money as a reward for their custom. Why is this? It can partly be down to competing with their competitors and needing to be seen as a good company?
Xbox offered points on Xbox Live member's birthday, but it equated to about 25p. Is that really how much Microsoft values someone that is spending £35 a year to use their platform? Perhaps even more if the customer is buying expansion packs and perhaps even introducing more players to join Xbox Live.
Offering money as a reward for custom or another interaction might not go down as well as you'd think. It may feel like an intuitive idea to acknowledge loyalty with a financial gift, but here it changed the gamer's perception of the relationship. They changed from being a loyal customer who identifies themselves as an Xbox devotee, to simply engaging in a monetary transaction — and where the transaction was insulting.
When money is introduced, it adds a transactional element to the relationship. When a value is attached to the relationship it acts as a reason to consider what your actions are worth. Suddenly you equate your actions to payment, rather than considering yourself as prosocial, or 'good'. This is known as the "Trivialisation Effect".
Okay, if money doesn't always work, how can businesses show they care without money? What else can you do to show someone you appreciate them?
How can companies help us feel like we're good people?
We want to be seen as and feel like a good person. Think of a time you opened a door for someone. How did you feel when they said "thanks"? I'm not saying it's as exciting as meeting your favourite football player but I bet it was enough to put a smile on your face!
What if, instead of a thank you, the person handed you 10 pence for your trouble? You probably wouldn't react positively. It reduces your polite behaviour to a mere exchange of goods or services.
It's basic manners, but a simple "thank you" can go a loooong way. It's such a simple thing to do, and so many business (and individuals) forget to say it and it can add so much value to your brand, by simply being polite!
Studies have shown customers prefer a "thank you" over a 5–10% discount. Just try to make sure it sounds natural. You could increase the monetary reward to offset the trivialisation effect, but if you're offering a 40% discount where a simple "thank you" would suffice, you're missing a trick.
An example of showing appreciation without the use of money is Monzo's golden ticket. Monzo, "a bank that lives on your smartphone and built for the way you live today" got it right. Whether intentionally or unintentionally, they combined an understanding of the trivialisation effect with another cognitive bias called 'the entourage effect'. Customers feel valued for their custom, and their status is elevated because they could share their VIP treatment with their friends and family. Nice!
Golden Tickets were introduced for existing customers to invite their friends, giving them the chance to skip the waiting list. When a lucky customer received a golden ticket, they now felt pretty good because not only had they been rewarded, they were also given the opportunity to act prosocially by helping a new business grow and invite their friends to what they were doing.
When thinking about how to 'reward' customers, try to understand what works by doing the right kind of research. Asking outright what a customer wants can get you into trouble (ever heard of Coke vs Classic Coca Cola? or the 185 billion dollar mistake by Walmart?) This is because typically, customers don't know and may tell you the intuitive answer of money or some other tangible reward. We're all flawed when it comes to knowing what motivates us, so it's better to observe customers in context to get closer to understanding what makes them tick.
So next time you're thinking of offering a reward, consider the trivialisation effect and ask yourself, would a simple heartfelt 'thank you' be enough?
Thanks to The Design Psych.


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The BlackBerry Classic is a return to basics for BlackBerry. After the lukewarm reception to the BlackBerry Passport, they're likely looking for a critical win. Their recent struggles are widely known and for those of us who carry a nostalgic love for the brand, it's been disheartening to watch.
Now they're hoping to win back the customers by offering something similar to the BlackBerry of old: less bells and whistles, more simple, focused functionality. Did it work? Yes and no. Read on to check out some of the reactions folks around the web had to BlackBerry's new design of an old idea.
"Unfortunately, the Classic reminds me of early "bricklike" BlackBerrys because of its weight. The BlackBerry "app gap" may be smaller than it once was, but it still exists, and Android apps just don't run as well as they do on Android phones.
Theres a huge amount to like in the BlackBerry Classic. It will make you more productive. It feels serious. And it doesnt look too bad either. But in part youre more productive because there are fewer distractions, and therein is the root of the problem. While this may be a tool for getting things done it is not really a tool of the modern world.
The consensus seems to be that the BlackBerry Classic is exactly what its name implies. It's faithful to the BlackBerry name in that it excels in productivity but doesn't offer much beyond that. Half a decade ago, that wasn't much of a problem. Unfortunately, the 2014 concept of a smartphone is lightyears ahead of what BlackBerry is offering. If you're looking for a distraction-free business phone, this seems to be the one to get, but everyone else should stay away.


Here's a wonderfully easy Chicken Cordon Bleu casserole that is great for a pot luck or a delicious meal any day of the week!
This is great served over egg noodles, rice, mashed potatoes or just over some toast points.
Whisk together the eggs and hot sauce in a medium bowl. Toss in the chicken chunks.
Put your Panko bread crumbs in a dish and toss a few of the chicken chunks in at a time and press the bread crumbs into the chicken. Just use maybe a half cup at a time of bread crumbs so you don't waste them.
Heat about 1/2 inch of oil in a frying pan until hot. Fry the nuggets until golden brown.
Turn and fry the other side until golden brown. Drain on paper towels.
Spray a 9 x 13 glass baking dish with non-stick cooking spray. Arrange the nuggets in the bottom of the baking dish.
Sprinkle the cheese on top of the nuggets. I didn't have quite enough shredded cheese so I used some provolone cheese too.
Arrange the diced ham on top of the cheese.
In a medium bowl, whisk together the soups and milk.
Pour over the top of the casserole.
Bake in a preheated 350° F oven for 30 minutes or until hot and bubbly. Serve over egg noodles, rice, mashed potatoes or even toast!
I hope you enjoy this easy chicken cordon bleu recipe!