TASEP on a finite ring

How can particles keep flowing when their probability distribution has stopped changing? On a homogeneous TASEP ring, all configurations with the same particle number have equal stationary weight, yet the directed current is positive. You will prove that stationarity by counting interfaces, derive the exact finite-size current, and see why fixed particle number prevents independent occupations. The worked ring has four sites and two particles; the counting argument applies to any finite ring in the stated regime.

Required background. Probability and Markov generators explains transition rates, probability columns, waiting times and observables. You need elementary counting and finite matrices, with no quantum prerequisite.

Helpful background. The TASEP model fixes the physical process and boundaries. The stochastic conventions reference distinguishes per-bond current, total jump rate and per-particle jump rate.

Right-moving particles with exclusion

Put $M$ indistinguishable particles on $N$ labelled sites around a ring, with $N\geq2$ and $0\lt M\lt N$. A configuration is a binary string

$$\eta=(\eta_1,\ldots,\eta_N), \qquad \eta_n\in\{0,1\}, \qquad \sum_n\eta_n=M.$$

Here $1$ means occupied and $0$ empty. Site $N+1$ is site $1$. At each bond $(n,n+1)$, the change $10\to01$ occurs at rate $r\gt0$ when that pattern is present. A particle cannot jump into an occupied site. The rate is per eligible bond, and the process uses continuous time rather than simultaneous updates.

This is the totally asymmetric simple exclusion process, or TASEP. “Totally asymmetric” specifies the absence of left jumps; “exclusion” specifies at most one particle per site. Time has units such that $r$ is an inverse time, and $rt$ is dimensionless. The source uses $L,n$ for our $N,M$ and unit jump rate; multiplying its generator by $r$ restores our time scale. See Golinelli and Mallick 2004, § 2.1, printed p. 3, equations (1)–(2), PDF.

The number of configurations is $\binom NM$. Translating a configuration around the ring usually gives another state: site labels are retained. Indistinguishable particles remove particle labels, not site labels.

Entry check and repair

On a four-site ring, compare $1100$ with $1010$. List their permitted jumps, including the bond $4\to1$. Do these configurations have the same total exit rate?

Repair. From $1100$, only the particle at site $2$ can move, giving $1010$. Its exit rate is $r$. From $1010$, particles at sites $1$ and $3$ can move, giving $0110$ and $1001$. Its exit rate is $2r$. The holding times therefore have means $1/r$ and $1/(2r)$. A common rate per eligible bond does not mean a common rate per configuration.

To check the seam explicitly, $0101$ permits $4\to1$, giving $1100$, as well as $2\to3$, giving $0011$.

Build the six-state generator

For $N=4,M=2$, abbreviate a configuration by its occupied pair and order the probability column as

$$p=(p_{12},p_{13},p_{14},p_{23},p_{24},p_{34})^{\mathsf T}.$$

The complete transition list is short enough to check:

Starting state	Binary string	Destinations, each at rate $r$	Exit rate
$12$	$1100$	$13$	$r$
$13$	$1010$	$14$, $23$	$2r$
$14$	$1001$	$24$	$r$
$23$	$0110$	$24$	$r$
$24$	$0101$	$12$, $34$	$2r$
$34$	$0011$	$13$	$r$

Place each outgoing rate in its destination row and starting-state column. Subtract the exit rate on the diagonal:

$$\frac Qr= \begin{pmatrix} -1&0&0&0&1&0\\ 1&-2&0&0&0&1\\ 0&1&-1&0&0&0\\ 0&1&0&-1&0&0\\ 0&0&1&1&-2&0\\ 0&0&0&0&1&-1 \end{pmatrix}, \qquad \frac{dp}{dt}=Qp.$$

Every column sums to zero, as required for probability conservation. For example, starting certainly in $12$ gives

$$p'(0)=r(-1,1,0,0,0,0)^{\mathsf T}.$$

This derivative loses probability from $12$ and gains it in $13$. Transposing $Q$ would send the gain to $24$ instead. Because both the rows and columns of this particular matrix sum to zero, a sum test alone would miss that reversal; a known initial transition detects it.

Prove uniform stationarity by counting interfaces

Let $n_{10}(\eta)$ count occupied-empty bonds in a ring configuration and $n_{01}(\eta)$ count empty-occupied bonds, including the seam. Their difference telescopes:

$$\begin{aligned} n_{10}(\eta)-n_{01}(\eta) &=\sum_{n=1}^N \bigl[\eta_n(1-\eta_{n+1})-(1-\eta_n)\eta_{n+1}\bigr]\\ &=\sum_{n=1}^N(\eta_n-\eta_{n+1})=0. \end{aligned}$$

There are $n_{10}(\eta)$ outgoing jumps. Each $01$ bond in $\eta$ identifies an incoming jump: replace that $01$ by $10$ to obtain the predecessor configuration. Thus there are $n_{01}(\eta)$ incoming jumps, all with rate $r$.

If every configuration has probability $1/\binom NM$, the incoming and outgoing probability rates at each configuration agree. Therefore

$$\pi(\eta)=\frac1{\binom NM}, \qquad Q\pi=0.$$

For the six-state example, $\pi=\mathbf1/6$ and multiplication by the displayed matrix checks this immediately. The general proof explains why that computation works: a closed ring has equally many $10$ and $01$ interfaces, and homogeneity gives them equal rates. Uniform stationarity is also stated in Golinelli and Mallick 2006, § II.A, printed pp. 2–3, PDF.

This argument proves that a process started in $\pi$ remains in $\pi$. It does not say that every trajectory stays in one configuration. Individual particles continue to jump.

Stationarity is weaker than detailed balance

Detailed balance would require equality of stationary probability flow for every pair of configurations:

$$Q_{\eta',\eta}\pi(\eta) =Q_{\eta,\eta'}\pi(\eta').$$

For the four-site transition $12\to13$, the left side is $r/6$, while the reverse transition has rate zero. Detailed balance fails. Stationarity balances the total incoming and outgoing flow at each configuration; it does not require a reverse arrow for every allowed jump.

This is why a time-independent distribution is compatible with a nonzero directed particle current. The persistent flow is a physical observable, even though configuration probabilities are constant.

Count the exact stationary current

Let $K_n(t)$ count successful jumps across bond $n\to n+1$ during $[0,t]$. Its instantaneous mean rate is

$$\frac d{dt}\mathbb E[K_n(t)] =r\,\mathbb E[\eta_n(t)(1-\eta_{n+1}(t))].$$

At stationarity, define the per-bond current $j$ as this constant rate. The uniform fixed-$M$ distribution is translation invariant, so $j$ is the same for every bond.

To count the favorable configurations, fix site $n$ occupied and site $n+1$ empty, then place the remaining $M-1$ particles among $N-2$ sites. Hence

$$\begin{aligned} \Pr_\pi(\eta_n=1,\eta_{n+1}=0) &=\frac{\binom{N-2}{M-1}}{\binom NM}\\ &=\frac{M(N-M)}{N(N-1)},\\ j&=r\frac{M(N-M)}{N(N-1)}. \end{aligned}$$

The same result follows from conditioning: the first site is occupied with probability $M/N$; once it is occupied, $N-M$ of the remaining $N-1$ sites are empty. Multiplying these two probabilities uses the fixed-particle constraint exactly.

For four sites and two particles,

$$j=\frac r3,\qquad \text{total mean jump rate}=Nj=\frac{4r}{3},\qquad \text{mean rate per particle}=\frac{Nj}{M}=\frac{2r}{3}.$$

These quantities have different normalizations. A bond sees the current $j$; counting every successful jump anywhere on the ring gives $Nj$. Dividing by particle number gives the average jump frequency per particle. With unit lattice spacing that last quantity is the mean displacement rate per particle on an unwrapped ring.

A direct six-state check gives total rate $r(1+2+1+1+2+1)/6=4r/3$, agreeing with the counting formula.

Why independent occupations give the wrong finite answer

Let $\rho=M/N$. Uniform weight over fixed-$M$ configurations does not make the sites independent. The exact current is

$$j=r\rho(1-\rho)\frac{N}{N-1},$$

whereas multiplying the separate one-site probabilities would give $r\rho(1-\rho)$. For $N=4,M=2$, these are $r/3$ and $r/4$, respectively.

The reason is conditional: knowing one site is occupied leaves slightly fewer particles for the other sites. In fact, for distinct sites $i,j$,

$$\mathbb E_\pi[\eta_i\eta_j] =\frac{M(M-1)}{N(N-1)}, \qquad \operatorname{Cov}_\pi(\eta_i,\eta_j) =-\frac{\rho(1-\rho)}{N-1}.$$

This negative correlation increases the probability of an occupied-empty pair relative to the independent estimate. The factor $N/(N-1)$ approaches one as the ring grows, but it should not be dropped in a finite calculation.

Two limiting checks are useful. With one particle, $j=r/N$ and the particle’s mean jump rate is $r$, since it is never blocked. With one hole, $j=r/N$ again, but the total jump rate is only $r$: exactly one particle can enter the hole at a time.

Exercises

Guided practice: count across the periodic seam

In the four-site, two-particle stationary state, identify the configurations permitting a jump $4\to1$. Compute the current across this bond and the mean number of such jumps during a stationary observation interval of length $T$. Explain why the transition $12\to13$ violates detailed balance even though $\pi$ is stationary.

Hint

For a seam jump, site $4$ must be occupied and site $1$ empty. Each of the six configurations has probability $1/6$. A constant mean event rate integrates to its rate times elapsed time; this does not assert that the entire counting process is Poisson.

Solution

The configurations are $24$ and $34$. Each permits one seam jump at rate $r$, so

$$j=\frac{2r}{6}=\frac r3, \qquad \mathbb E_\pi[K_4(T)]=\frac{rT}{3}.$$

This is a mean; exclusion produces time-dependent availability of the bond, so a Poisson count distribution does not follow from the calculation. For detailed balance, $Q_{13,12}\pi_{12}=r/6$, while $Q_{12,13}\pi_{13}=0$. Other incoming flows restore the total stationary balance at each configuration.

Independent practice: retain the fixed-particle correlation

On a six-site ring with two particles, derive the probability that two specified distinct sites are both occupied. Find their covariance. Compute the per-bond current, total jump rate and per-particle rate, and compare the current with the independent-occupation estimate.

Hint

There are $\binom62=15$ equally weighted configurations and exactly one occupies both specified sites. The density is $1/3$. For the current, first occupy one specified site and then condition on the other being empty.

Solution

The joint probability is $1/15$, whereas the product of marginals is $1/9$. Thus

$$\operatorname{Cov}(\eta_i,\eta_j) =\frac1{15}-\frac19=-\frac2{45}.$$

Conditioned on one occupied site, four of the other five sites are empty. Therefore

$$j=r\frac13\frac45=\frac{4r}{15}, \qquad Nj=\frac{8r}{5}, \qquad \frac{Nj}{M}=\frac{4r}{5}.$$

The independent estimate is $r(1/3)(2/3)=2r/9$. The exact current is larger by the factor $6/5$. This is the finite fixed-number correction, not a numerical discrepancy or a modification of the microscopic rate.

Transfer: give one bond a different rate

Change the model to one particle on four sites, with the rightward exit rates $(r_1,r_2,r_3,r_4)=(r,r,r,r/2)$. Find the stationary probability at each site and the per-bond stationary current. Which step in the uniform-stationarity proof has failed?

Hint

For one particle, stationarity at site $n$ says $r_{n-1}\pi_{n-1}=r_n\pi_n$, with periodic indices. All bond currents are equal, but the waiting times differ.

Solution

Write the common bond current as $j$. Then $\pi_n=j/r_n$, and normalization gives

$$j=\left(\sum_{n=1}^4\frac1{r_n}\right)^{-1} =\frac r5, \qquad \pi=\frac15(1,1,1,2)^{\mathsf T}.$$

The particle spends twice as much stationary probability at site $4$, just before the slower bond. The total jump rate is $4r/5$. The ring still has equally many $10$ and $01$ interfaces, but their rates can no longer be factored out as one common $r$. Counting arrows without their weights therefore does not prove uniform stationarity. The homogeneous current formula cannot be applied unchanged to this modified process.

From stationary counting to time evolution

The stationary law and mean current followed from probability conservation and finite counting. Those arguments alone do not establish an integrable structure or solve relaxation from an arbitrary initial configuration. The Library derivation constructs a two-particle Bethe mode and imposes periodicity. The exclusion-process laboratory compares finite-state evolution with trajectories sampled at fixed times and tests that mode against the six-state generator.

The ring, fixed particle number, homogeneous rates and continuous-time update rule are all consequential. Reservoir boundaries, site-dependent rates, simultaneous updates, or additional left jumps define different problems; their stationary laws and currents require their own derivations.

References

• Golinelli, Olivier, and Kirone Mallick. “Bethe Ansatz calculation of the spectral gap of the asymmetric exclusion process.” Journal of Physics A: Mathematical and General 37 (2004), 3321–3331. DOI. Author version arXiv:cond-mat/0312371v1, submitted 2003; Open PDF. The model locator refers to printed p. 3 of this version.
• Golinelli, Olivier, and Kirone Mallick. “The asymmetric simple exclusion process: an integrable model for non-equilibrium statistical mechanics.” Journal of Physics A: Mathematical and General 39 (2006), 12679–12705. DOI. Author version arXiv:cond-mat/0611701v1; Open PDF. Locators refer to its printed page labels.