Two-particle Bethe ansatz for periodic TASEP

How does exclusion change two independently moving particles into a Bethe eigenvector? For continuous-time TASEP on a finite ring, the answer is a contact relation: an attempted hop into an occupied site is removed from the generator. Two plane waves can satisfy that relation if their amplitudes have a particular ratio; carrying a particle around the ring then quantizes their spectral parameters. We derive this construction for two particles and verify a four-site mode exactly. The result is a decay mode of a probability generator, with no claim that this calculation supplies every eigenvector at arbitrary size.

Required background. Multiply a matrix by a vector, work with complex powers, and distinguish a probability distribution from an observable; the Markov-generator lesson supplies the last distinction. Helpful background. TASEP on a finite ring derives the stationary measure and current. The model record and stochastic conventions fix the regime and normalizations.

Two particles on a directed ring

Fix an integer $N\geq3$, a ring of sites $1,\ldots,N$, and two indistinguishable particles. An occupied site attempts a jump to its right neighbor at rate $r\gt0$; the jump occurs only if that neighbor is empty. Site $N$ has right neighbor $1$. The state space contains $\binom N2$ unordered occupied pairs, represented by

$$1\leq x\lt y\leq N.$$

Probability columns evolve by $\dot p=Qp$. Thus $Q_{\eta',\eta}$ is the rate from configuration $\eta$ to a different configuration $\eta'$, and the diagonal is minus the total exit rate. We seek right eigenvectors

$$Q\psi=E\psi.$$

The component $\psi(x,y)$ is generally signed or complex. It is an eigenvector coefficient, not the probability of the pair $(x,y)$. The rate $r$ and eigenvalue $E$ have units of inverse time; setting $r=1$ means measuring time by $rt$.

This is the probability-column and coordinate-eigenvector convention of Golinelli and Mallick 2004, § 2.1, p. 3, equations (1)–(2), arXiv v1 PDF. Their ring length $L$, particle number $n$, and unit rate become $N$, $2$, and $r$ here.

Coordinates across the closing bond

For the derivation, lift the coordinates to integers satisfying $x\lt y\lt x+N$. The same physical configuration has two cyclically ordered representatives, so impose

$$\psi(x,y)=\psi(y,x+N).$$

Applying this relation twice also gives $\psi(x+N,y+N)=\psi(x,y)$. It is cyclic order, rather than independent periodicity of each plane wave, that will generate the interacting Bethe equations.

A pair is separated from contact on both sides when

$$1\lt y-x\lt N-1.$$

The configuration $(1,N)$ is adjacent across the closing bond, even though its displayed coordinates are far apart. Ignoring that second contact would solve an open-coordinate problem rather than the ring.

The free equation and the contact relation

For a pair separated on both sides, both particles can jump out. The two incoming configurations have the first or second particle one site to the left. The eigenvector equation is therefore

$$E\psi(x,y) =r\psi(x-1,y)+r\psi(x,y-1)-2r\psi(x,y).$$

The left shifts appear because this is an equation for the coefficient at the destination configuration. Replacing them by right shifts would describe the transpose convention.

At $y=x+1$, only the right particle can jump out. The only incoming move comes from $(x-1,x+1)$, so the physical equation is

$$E\psi(x,x+1)=r\psi(x-1,x+1)-r\psi(x,x+1).$$

Extend a trial function formally to coincident coordinates and compare this equation with the free equation evaluated at $(x,x+1)$:

$$\begin{aligned} E\psi(x,x+1) &=r\psi(x-1,x+1)+r\psi(x,x)\\ &\quad-2r\psi(x,x+1). \end{aligned}$$

The two expressions agree precisely when

$$\boxed{\psi(x,x)=\psi(x,x+1).}$$

There is no physical state with two particles at $x$. The diagonal value is an auxiliary value of the trial function that cancels the forbidden hop and its extra loss term. It does not relax the exclusion rule.

The closing bond requires no new scattering rule. Write its adjacent configuration as $(N,N+1)$, equivalent to $(1,N)$. Its equation is

$$\begin{aligned} E\psi(N,N+1)&=r\psi(N-1,N+1)-r\psi(N,N+1),\\ E\psi(1,N)&=r\psi(1,N-1)-r\psi(1,N). \end{aligned}$$

The second line follows from cyclic identification. It is exactly the incoming move $(1,N-1)\to(1,N)$ and the outgoing move $(1,N)\to(2,N)$. A trial function satisfying the free equation, contact relation, and cyclic identification therefore satisfies every physical equation. For $N=3$, every physical pair is adjacent: the free equation is then an auxiliary construction, not an additional equation at a separated physical configuration.

Contact scattering of two plane waves

Take dimensionless complex parameters $z_1,z_2$ and a trial function

$$\psi(x,y)=A_{12}z_1^xz_2^y+A_{21}z_2^xz_1^y.$$

For the regular construction in this section, assume

$$z_1z_2\ne0,\qquad z_1\ne z_2,\qquad z_1\ne1,\quad z_2\ne1, \qquad A_{12}\ne0.$$

These assumptions permit the inverse powers and amplitude ratios below. We will still check that the resulting vector on physical configurations is nonzero.

Each plane wave obeys the free equation with the same eigenvalue,

$$\boxed{E=r\left(z_1^{-1}+z_2^{-1}-2\right).}$$

Substitution into the contact relation gives

$$(z_1z_2)^x \bigl[A_{12}(1-z_2)+A_{21}(1-z_1)\bigr]=0.$$

Consequently the amplitude ratio is

$$\boxed{S_{12}\equiv\frac{A_{21}}{A_{12}} =-\frac{1-z_2}{1-z_1}.}$$

This is a scattering amplitude in the coordinate Bethe ansatz. It is not a transition probability: it can be complex, and it need not have modulus one. The stochastic generator is not a Hermitian quantum Hamiltonian.

Periodic Bethe equations

Insert the same two-wave expression into the cyclic condition:

$$\begin{aligned} \psi(y,x+N) &=A_{12}z_1^yz_2^{x+N} +A_{21}z_2^yz_1^{x+N}\\ &=(A_{21}z_1^N)z_1^xz_2^y +(A_{12}z_2^N)z_2^xz_1^y. \end{aligned}$$

Matching the two distinct plane-wave coefficients imposes

$$A_{12}=A_{21}z_1^N, \qquad A_{21}=A_{12}z_2^N.$$

Thus the regular two-particle Bethe equations are

$$\boxed{z_1^N=S_{12}^{-1},\qquad z_2^N=S_{12}.}$$

Multiplication yields $(z_1z_2)^N=1$. The total spectral product is on a free ring grid, but the two individual parameters are coupled by exclusion. In particular, imposing $z_1^N=z_2^N=1$ independently would generally discard the contact scattering.

An equivalent form is

$$\begin{aligned} (z_1-1)^2z_1^{-N}&=-(1-z_1)(1-z_2),\\ (z_2-1)^2z_2^{-N}&=-(1-z_1)(1-z_2). \end{aligned}$$

These are the $n=2$ specialization of Golinelli and Mallick 2004, § 2.2, p. 4, equations (4)–(6), arXiv v1 PDF. Their lowercase $z_i$ agrees with ours; their later variable $Z_i=2/z_i-1$ does not. Their dimensionless eigenvalue is $E/r$.

The reasoning establishes a sufficient construction: regular parameters obeying these relations give a right eigenvector provided its reconstructed coefficients are not all zero. It does not count the independent vectors, settle exceptional solutions, or prove completeness.

Exceptional parameters and the stationary state

The exclusions matter before any denominators are cleared. A zero $z_i$ invalidates the inverse powers used in the free equation. A root at $1$ requires returning to the undivided contact equation. If $z_1=z_2=z\ne1$, that equation forces $A_{21}=-A_{12}$, so the displayed two-wave function vanishes identically.

The constant function $\psi(x,y)=1$ satisfies the physical equation with $E=0$. It also satisfies the free and contact equations formally with $z_1=z_2=1$ and $A_{12}+A_{21}=1$. But $S_{12}$ is then the undefined ratio $0/0$: the regular scattering formula has not constructed this state. Normalizing the constant function gives

$$p_{\mathrm{st}}(x,y)=\binom N2^{-1}.$$

Its stationarity has a direct counting proof, given in the finite-ring lesson. An elementary stationary distribution and a complete Bethe eigenbasis are different results.

A four-site mode checked against the generator

Set $N=4$ and use the ordered configuration basis

$$\mathcal B=(12,13,14,23,24,34),$$

where $12$ abbreviates the pair $(1,2)$. Listing right jumps gives

$$\begin{gathered} 12\to13,\qquad13\to14,23,\qquad14\to24,\\ 23\to24,\qquad24\to12,34,\qquad34\to13. \end{gathered}$$

Every arrow has rate $r$. In the stated probability-column convention,

$$\frac Qr= \begin{pmatrix} -1&0&0&0&1&0\\ 1&-2&0&0&0&1\\ 0&1&-1&0&0&0\\ 0&1&0&-1&0&0\\ 0&0&1&1&-2&0\\ 0&0&0&0&1&-1 \end{pmatrix}.$$

Columns sum to zero. Rows also sum to zero for this periodic homogeneous process, which explains the uniform stationary vector; it does not make $Q$ symmetric. For example, the move $12\to13$ has no reverse move.

Choose

$$z_1=z=e^{2\pi i/3},\qquad z_2=z^{-1},\qquad A_{12}=1.$$

Since $z^3=1$ and $1+z+z^2=0$, the scattering amplitude is $S_{12}=z^{-1}$. Then $z_1^4=z=S_{12}^{-1}$ and $z_2^4=z^{-1}=S_{12}$, so both periodic equations hold. The eigenvalue is

$$E=r(z+z^{-1}-2)=-3r.$$

Reconstructing the six coefficients gives

$$\psi=-zv,\qquad v=\begin{pmatrix}1&-2&1&1&-2&1\end{pmatrix}^{\mathsf T}.$$

This vector is nonzero. Direct multiplication by the independently assembled matrix yields

$$\frac Qr v= \begin{pmatrix}-3&6&-3&-3&6&-3\end{pmatrix}^{\mathsf T} =-3v.$$

This check includes the seam and makes no use of a numerical root finder. The computational laboratory reconstructs both the generator and the Bethe vector and compares finite-state evolution with trajectories.

Turning a mode into a probability perturbation

For any eigenvector with $E\ne0$, probability conservation implies

$$0=\mathbf1^{\mathsf T}Q\psi =E\mathbf1^{\mathsf T}\psi, \qquad \mathbf1^{\mathsf T}\psi=0.$$

Thus a nonzero real decay mode necessarily has both signs. Here $\sum_\eta v_\eta=0$, and the family

$$p(t)=\frac16\mathbf1+a e^{-3rt}v$$

is an exact probability solution for

$$-\frac16\leq a\leq\frac1{12}.$$

The bounds make every initial component nonnegative; for later times the coefficient moves toward zero and remains within those bounds. This describes the specified one-mode initial family, not arbitrary initial data.

The mode is also not the slowest decay. A direct determinant calculation gives

$$\det\left(qI-\frac Qr\right) =q(q+1)^2(q+3)(q^2+3q+4).$$

The nonzero eigenvalues are $-r$ (twice), $-3r$, and $r(-3\pm i\sqrt7)/2$. The smallest nonzero decay rate measured by $-\operatorname{Re}E$ is therefore $r$, while this example decays at $3r$. A chosen initial distribution can have zero projection onto the slowest modes, as the displayed family does.

Exercises

Reverse the convention deliberately

Starting from $p(0)=\delta_{12}$ in the four-site basis, compute $\dot p(0)$ for the right-moving generator and for its transpose. Explain why checking only stationarity and the vector $v$ would miss this mistake.

Solution. Read the first column of each matrix:

$$\begin{aligned} Q\delta_{12}&=r(-1,1,0,0,0,0)^{\mathsf T},\\ Q^{\mathsf T}\delta_{12}&=r(-1,0,0,0,1,0)^{\mathsf T}. \end{aligned}$$

The first evolves from $12$ toward $13$, as right jumps require; the second evolves toward $24$. Yet both matrices annihilate $\mathbf1$, and direct multiplication shows $Q^{\mathsf T}v=-3rv$ too. A direction-sensitive check is essential. In general, $Q^{\mathsf T}$ is the generator acting on observables; its accidental interpretation here as another probability generator uses the additional zero row sums.

Change the ring size while retaining a conjugate pair

Set $z_1=z$, $z_2=z^{-1}$, with $z\ne0,1$ and $z\ne z^{-1}$. Reduce the Bethe equations to one power equation. Determine whether this restricted root family can produce a regular mode for $N=3$.

Solution. The contact ratio reduces to $S_{12}=z^{-1}$, so the first periodic equation requires $z^{N-1}=1$; the second is equivalent. For $N=3$, the only candidates are $z=1,-1$. The first is excluded by contact division, and the second gives equal roots $z=z^{-1}$. Thus this restricted regular conjugate-pair construction supplies no $N=3$ mode. The task has no regular solution; that is not a claim that the three-state generator lacks nonstationary eigenvectors. It illustrates why a special root family need not cover a spectrum.

Read the current carried by the four-site perturbation

For the family $p(t)=\mathbf1/6+a e^{-3rt}v$, compute the expected total rate of jumps and the expected rate through a single bond.

Solution. The configuration exit rates are $r(1,2,1,1,2,1)$. Their pairing with $\mathbf1/6$ is $4r/3$, and their pairing with $v$ is $-4r$. Hence

$$\mathbb E[\text{total jump rate at }t] =\frac{4r}{3}-4ra e^{-3rt}.$$

The vector $v$, and thus the whole probability family, is invariant under rotating all sites once. Every bond has the same expected rate, one quarter of the total:

$$j(t)=\frac r3-ra e^{-3rt}.$$

The decay mode therefore changes a concrete observable, not just an abstract matrix coefficient. The stationary limit is $r/3$, agreeing with the fixed-particle-number current in the model record.

Scope of the construction

The contact and cyclic equations establish regular two-particle eigenvectors of the finite, homogeneous, continuous-time ring generator. The explicit four-site calculation verifies one such mode and its time evolution. It does not establish a complete Bethe basis at all sizes, solve singular root families, or derive thermodynamic relaxation exponents. Changing to open boundaries, adding reverse jumps, using different rates at different bonds, or replacing continuous-time updates changes the problem and its equations.

For an independent local-operator definition of the generator, see the model record. The corresponding source matrices are in Golinelli and Mallick 2006, § III.A, p. 5, equations (20)–(21), arXiv v1 PDF; the source uses unit right rate and the local order $11,10,01,00$.

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, PDF. Locators above refer to the printed pages of that 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, PDF.