I am searching for an elementary proof of the first order approximation of the current of particles in ASEP (asymmetric simple exclusion process) and TASEP (totally asymetric). To avoid technical details unuseful for the intuition, I will focus on the current at the site $0$.

A simple case : step initial condition

It has been known for a long time (see for exemple theorem 5.12 of the section devoted to exclusion processes in "Interacting particle systems", Liggett) that the current of particles at site $0$ in the ASEP model converges almost surely to $t\gamma/4$ when $t$ tends to infinity, where $\gamma=q-p$, when the initial condition is the so called step initial condition. (i.e. $\eta_0(x)=\mathbf{1}_{x>0}$).

One can explain this result using the density (renormalized) profile of particles in the window $[-\gamma t, \gamma t]$ (or $[-\gamma , \gamma ]$ after renormalization) which is actually a stronger result, but more natural. The fact is that this density profile is linear, hence a number of particles on the left of $0$ like $t\gamma/4$.

Step Bernoulli initial condition

When the initial condition is step Bernoulli, i.e., $\eta_0(x)=\mathbf{1}_{x>0}Ber_x(\rho)$, where $Ber_x(\rho)$ is a Bernoulli random variable with parameter $\rho$, the approximation for the current stays the same for $\rho\geq 1/2$, and the fluctuations follow tracy-widom law on a $t^{1/3}$ scale. For $\rho< 1/2$, the current in $0$ is like $t\gamma\rho(1-\rho)$ (see for example Tracy-Widom, "On ASEP with Step Bernoulli Initial Condition"), and the fluctuations are Gaussian.

In the proof of the fluctuations, the phase transition appears clearly in the calculations (when approximating the fredholm determinant for $t\rightarrow +\infty$), but I would like to find a simple physical interpretation ( I mean a simple and/or direct proof) explaining why the approximation does not depend on $\rho$ as soon as $\rho >1/2$.


For me the "physical intuition" is the fact that exclusion processes tend to clog up for large densities and hence you cannot expect a faster current than in the $\rho = \frac{1}{2}$ case, even when you add more particles.

PS: to answer your question in the comments, the phase transition should occur at $\rho = \frac{1}{2}$ by symmetry with respect to interchanging the notion of "particles" and "holes". Indeed (heuristically speaking) if the process has $\rho = \frac{1}{2}$, there are just enough holes that each particle basically behaves like a free particle. If you now increase the density, the front of particles still can only move as fast as the free particle and all the other ones block each other behind it until the first particle in front has moved away a bit. Only then the next one can follow and in this case again it moves like a free particle. This is (my) intuition why the particle current cannot be higher than in the "free" case.

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  • $\begingroup$ Yes, but it doesn't really explain why it strictly doesn't depend on rho, and why the transition is at rho = 1/2. $\endgroup$ – Guillaume Nov 18 '12 at 1:36
  • $\begingroup$ Thank you for your answer and this idea of symmetry. However, I am not sure about the idea of freeness. For example in the case $\rho=1/2$, if the particles were basically free, each particle would move at speed $\gamma$ (averagely), and the currrent would be like $t\gamma/2$ and not $t\gamma/4$. $\endgroup$ – Guillaume Nov 20 '12 at 18:17
  • $\begingroup$ You're right. So I guess one should replace "free" by "as free as it can possibly be". However, here my own intuition ends. $\endgroup$ – Daniel Nov 21 '12 at 11:12

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