5 A. Thermal Equilibrium B. Simple Illustration C. Formula for 〈 ( x$_{T}$ ) $^{2}$〉 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 〈 P$_{LR}$ ( x$_{T}$ ) 〉 = P eq R ( x$_{T}$ ) with N particles on − ϵ . This problem was solved already by Aslangul [26], and here we recover the known result us- ing our formulas. The Green function g ( x, x$_{0}$, t ) of a free particle is where Z is the normalizing partition function where as mentioned D is the diffusion coefficient of the free particle. With the specified initial conditions we have Similarly In Eqs. (16,18) we used the steady state solution, $^{lim$_{t}$}$→∞ g ( x, x$_{0}$, t ) = exp[ − V ( x ) /k$_{b}$T ] /Z , which is Boltz- mann’s distribution suited for a system in thermal equi- librium. Using Eq. (15) the position PDF of the tagged $^{particle is lim$_{t}$}$→∞ P ( x$_{T}$ ) = P $^{eq}$( x$_{T}$ ) Similarly When x$_{T}$ ≪ √$_{2}$$_{Dt}$ we Taylor expand in x$_{T}$ to find ( 〈 P$_{LR}$ 〉 − 〈 P$_{RL}$ 〉 ) 2 ∼ ( x$_{T}$ ) $^{2}$/πDt , using Eq. (15) we re- cover the result in [26] If the potential is symmetric V ( x ) = V ( − x ), e.g. parti- cles in a box or harmonic field, we have for not too large x$_{T}$ , P eq L ( x$_{T}$ ) ≃ 1 / 2, P eq R ( x$_{T}$ ) ≃ 1 / 2 hence from Eq. (15) Expanding the expression in the exponent in x$_{T}$ (since N is large) we find using Eqs. (16,18,19) where with out loss of generality we assigned V ( x = 0) = 0. Hence the standard deviation is The same expression is found in Appendix A using the many body Boltzmann distribution, and integrating over all the particles except the tagged particle. We now consider the situation of particles free of a force F ( x ) = 0 with open boundary conditions L → ∞ where initially all the particles are on the vicinity of the origin. More precisely, the tagged particle is initially sit- uated at x$_{T}$ = 0, N particles to its right on ϵ → 0 + and We now consider symmetric potential fields V ( x ) = V ( − x ), and symmetric initial conditions. The latter means that the density of particles at time t = 0 to the left of the tagged particle, i.e. those residing in x$_{0}$ < 0, is the same as for those residing to the right, f$_{R}$ ( x$_{0}$ ) = f$_{L}$ ( − x$_{0}$ ) (e.g. uniform initial conditions). In this case the subscript R and L is redundant and we use f ( x$_{0}$ ) = f$_{R}$ ( x$_{0}$ ) = f$_{L}$ ( − x$_{0}$ ), where f ( x$_{0}$ ) = 0 if The diffusion is normal, in the sense that the mean square displacement increases linearly in time. However the dif- fusion of the tagged particle is slowed down compared with a free particle, by a factor of 1 /N which is due to the collisions with all other Brownian particles in the sys- tem. Clearly the approximation breaks down if one is in- terested in the tails of P ( x$_{T}$ ), since we used x$_{T}$ ≪ √$_{2}$$_{Dt}$$_{.}$ Though clearly when N is large, the probability of find- ing such a particle is extremely small (i.e. use Eq. (26) P ( x$_{T}$ = √$_{2}$$_{Dt}$$_{)}$ ∼ exp( − N 2 /π )). hence P eq R ( x$_{T}$ ) = 1 Z ∫ L x$_{T}$ exp ( − V ( x ) k$_{b}$T ) d x (16) g ( x, x$_{0}$, t ) = exp [ − ( $^{x}$−x$_{0}$ ) 2 4 Dt ] √$_{4}$$_{πDt}$ (23) 〈 P$_{RL}$ ( x$_{T}$ ) 〉 = lim $^{ϵ}$→ 0 ∫ ∞ 0 δ ( x$_{0}$ − ϵ ) ∫ x$_{T}$ −∞ exp [ − ( $^{x}$−x$_{0}$ ) 2 4 Dt ] √$_{4}$$_{πDt}$ d x d x$_{0}$ = 1 2 + ∫ x$_{T}$ 0 e − x 2 4 Dt √$_{4}$$_{πDt}$ d x. (24) Z = ∫ L − L exp [ − V ( x ) k$_{b}$T ] d x. (17) P eq L ( x$_{T}$ ) = 1 Z ∫ x$_{T}$ − L exp ( − V ( x ) k$_{b}$T ) d x. (18) 〈 P$_{LR}$ ( x$_{t}$ ) 〉 = 1 2 − ∫ x$_{T}$ 0 e − x 2 4 Dt √$_{4}$$_{πDt}$ d x. (25) P $^{eq}$( x$_{T}$ ) ∼ C exp { − N [ 〈 P eq R ( x$_{T}$ ) 〉 − 〈 P eq L ( x$_{T}$ ) 〉 ] 2 4 〈 P eq L ( x$_{T}$ ) 〉〈 P eq R ( x$_{T}$ ) 〉 } . (19) P ( x$_{T}$ ) ∼ 1 π √ N Dt exp [ − N ( x$_{T}$ ) 2 πDt ] , (26) 〈 ( x$_{T}$ ) $^{2}$〉 ∼ πDt 2 N . (27) P $^{eq}$( x$_{T}$ ) ∼ C exp { − N [ 〈 P eq R ( x$_{T}$ ) 〉 − 〈 P$_{L}$ ( x$_{T}$ ) 〉 ] 2 } . (20) P $^{eq}$( x$_{T}$ ) ∼ $_{2}$√$_{N}$ √$_{πZ}$ exp [ − 4 N Z 2 ( x$_{T}$ ) 2 ] (21) 〈 ( x$_{T}$ ) $^{2}$〉 ∼ Z 2 8 N . (22)