Integral Formulas of ASEP and q-TAZRP on a Ring

Abstract

In this paper, we obtain the transition probability formulas for the asymmetric simple exclusion process and the q-deformed totally asymmetric zero range process on the ring by applying the coordinate Bethe ansatz. We also compute the distribution function for a tagged particle with general initial condition.

A. Convergence Lemmas for ASEP and q-TAZRP on a Ring

We need to make sure that the right side of (1.16) and (1.26) are well-defined before we prove Theorems 1 and 3. That is, we need to show that the sum of $\Lambda ^{\mathbb {L}}_Y(X; t; \sigma )$ over the infinite lattice $\mathbb {Z}^N(0)$ converges. Also, we need to show that the integrands of the z-integral on the right side of (1.17) and (1.27) converge in Theorems 2 and 4. For these integrands, the problem is more serious since we need to deal with convergence of other series in the proofs of Theorems 2 and 4. In this appendix, we show that all the required convergences hold.

We establish some notation and a basic estimate to facilitate the upcoming proofs in this section. Recall the notation set up in (4.9). It is straightforward to see that given any k and n, there are only finitely many $\mathbb {L}\in \mathbb {Z}^N(n)$ such that $\max (\mathbb {L}) < k$. We have a very rough estimate.

Lemma A.1

The number of $\mathbb {L}\in \mathbb {Z}^N(n)$ with $\max (\mathbb {L}) < k$ is no more than $N^N (k - \lfloor n/N \rfloor )^N$ if $n < Nk$, and there is no $\mathbb {L}\in \mathbb {Z}^N(n)$ with $\max (\mathbb {L}) < k$ if $n \ge Nk$.

Proof

The statement for $n \ge Nk$ is obvious. For $n < Nk$, we assume that $n = lN$ with l an integer without loss of generality. Note that, if $\mathbb {L}= (\ell _1, \dotsc , \ell _N) \in \mathbb {Z}^N(n)$, then $\mathbb {L}' (\ell _1 - l, \ell _2 - l, \dotsc , \ell _N - l) \in \mathbb {Z}^N(0)$. Then, the inequality $\max (\mathbb {L}) < k$ is equivalent to $\max (\mathbb {L}') < k - l$. Hence, the lemma is reduced to the $n = 0$ case.

Suppose $n = 0$ and $\mathbb {L}= (\ell _1, \dotsc , \ell _N) \in \mathbb {Z}^N(0)$ satisfies $\max (\mathbb {L}) < k$. Then, if there is an $\ell _i$ with $\ell _i < (1 - N)k$, the sum of the other $(N - 1)$ components is greater than $(N - 1)k$, and at least one among them is greater than k, a contradiction. So, the value of each $\ell _i$ is less than k and no less than $(1 - N)k$. Hence, each component $\ell _i$ has no more than Nk possible values and $\mathbb {L}$ has no more than $(Nk)^N$ possible choices. $\quad \square$

1.1 A.1 q-TAZRP case

In this subsection, we assume $C = \{ |z |= R \}$ is a circular contour with positive orientation and R a large constant. Recall that $b_{[1]}, \cdots , b_{[L]}$ are positive constants defined in Sect. 1.1.2. We assume $g(w_1, \dotsc , w_N)$ is an analytic function on $\mathbb {C}$ and

$$\begin{aligned} \tilde{g}(w_1, \dotsc , w_N) = g(w_1, \dotsc , w_N) \prod ^N_{j = 1} \prod ^L_{i = 1} (b_{[i]} - w_j)^{-m} \end{aligned}$$

(8.29)

for some $m > 0$. In this subsection, we take $D_{\mathbb {L}}$ as defined by (1.23).

Lemma A.2

Denote

(A.1)

Then $\Lambda ^{\mathbb {L}} = 0$ if $\max (\mathbb {L}) > m$.

Proof

Consider the integral over $w_{m(\mathbb {L})}$. We have that the contour C encircles no poles with respect to $w_{m(\mathbb {L})}$. Then, the integral of $w_{m(\mathbb {L})}$ over $C_{m(\mathbb {L})}$ vanishes, which implies that $\Lambda ^{\mathbb {L}}$ vanishes. $\quad \square$

Next, we consider the special case with $b_{[1]} = \cdots = b_{[N]} = 1$. Then, the function $\tilde{g}(w_1, \dotsc , w_N)$ defined in (A.1) becomes $g(w_1, \dotsc , w_N) \prod ^N_{j = 1} (1 - w_j)^{-mL}$. Furthermore, we assume that the radius R of the contour C is bigger than N. We denote

$$\begin{aligned} M_g = \max _{w_i \in C,\ i = 1, \dotsc , N} |g(w_1, \dotsc , w_N) |. \end{aligned}$$

(A.2)

Lemma A.3

Denote

(A.3)

and $\tilde{\Lambda }_s = \sum _{\mathbb {L}\in \mathbb {Z}^N(s)} |\tilde{\Lambda }^{\mathbb {L}} |$ for some $s \in \mathbb {Z}$. Furthermore, assume the radius R of the contour C satisfies

$$\begin{aligned} \frac{R^{2LN}}{(R - 1)^{L(2N - 1)}} \left( \frac{1 + q}{1 - q} \right) ^{2N^2} < 1. \end{aligned}$$

(A.4)

Then, $\tilde{\Lambda }_s < C M_g$ for some constant C that depends on m, s but not g.

Proof

We partition $\mathbb {Z}^N(s)$ into disjoint subsets $Z_0 \cup Z_1 \cup Z_2 \cup \cdots$, with

$$\begin{aligned} Z_0&= \{ \mathbb {L}\in \mathbb {Z}^N(s) \mid L \max (\mathbb {L})< m \}\nonumber \\ Z_n&= \{ \mathbb {L}\in \mathbb {Z}^N(s) \mid m + (n - 1)L \le L\max (\mathbb {L}) < m + nL \} \end{aligned}$$

(A.5)

for $n = 1, 2, \dotsc$. By Lemma A.1, each $Z_n$ is a finite set and $|Z_n |< N^N (n + \lceil m/L \rceil - \lfloor s/N \rfloor )^N$ if $n + m/L - s/N > 0$. Without loss of generality, we assume $m = s = 0$ below for notational convenience. Then, in this case, $Z_0 = \emptyset$ and we only need to consider $n \ge 1$.

For $\mathbb {L}= (\ell _1, \dotsc , \ell _N) \in Z_n$, we have that the integrand in (A.4) has no pole with respect to $w_{m(\mathbb {L})}$ within C and has a zero of order at least $(n-1)L$ at 1. So, by the identity

$$\begin{aligned} \frac{1}{w} = \frac{1}{w(1 - w)^{(n-1)L}} - \sum ^{(n-1)L}_{k = 1} \frac{1}{(1 - w)^k}, \end{aligned}$$

(A.6)

we have

(A.7)

Also, we have $\sum ^N_{i = 1} |\ell _i |< 2nN$ since $\max ^N_{i = 1} (\ell _i) < n$ and $\sum ^N_{i = 1} \ell _i = 0$. Then, it is not hard to see that

$$\begin{aligned} |D_{\mathbb {L}}(w_1, \dotsc , w_N) |< \left( \left( \frac{R}{R - 1} \right) ^L \left( \frac{1 + q}{1 - q} \right) ^N \right) ^{2nN}, \end{aligned}$$

(A.8)

if $w_i \in C$ for all $i = 1, \dotsc , N$. Additionally, we have

$$\begin{aligned} \left|\frac{ g(w_1, \dotsc , w_N)}{(1 - w_{m(\mathbb {L})})^{(n-1)L}} \prod _{1 \le i< j \le N} (qw_i - w_j)^{-1} \right|< \frac{M_g}{(R - 1)^{(n-1)L}} ((1 + q)R)^{N(N - 1)/2}. \end{aligned}$$

(A.9)

Thus, we obtain the following estimate,

$$\begin{aligned} \sum _{\mathbb {L}\in Z_n} |\tilde{\Lambda }^{\mathbb {L}} |< (N n)^N M_g (R-1) ((1 + q)R)^{N(N - 1)/2} \left( \frac{R^{2N L}}{(R - 1)^{(2N+1) L}} \left( \frac{1 + q}{1 - q} \right) ^{2N^2} \right) ^n, \end{aligned}$$

(A.10)

by combining the previous two estimates. Hence, we obtain the result by summing the last estimate over $n>1$. $\quad \square$

1.2 A.2 ASEP case

Later in this subsection, we assume that $C = \{ |z |= p^2 \}$ is a circular contour with positive orientation. Let $f(\xi _1, \dotsc , \xi _N)$ be a meromorphic function such that it is analytic on the multi-cylinder $\{ |\xi _i |\le p^2 \mid i = 1, \dotsc , N \}$ with $\max _{|\xi _i |= r,\ i = 1, \dotsc , N} |f(\xi _1, \dotsc , \xi _N) |= M_f$. Let $t \in \mathbb {R}_{>0}$, $m \in \mathbb {Z}_{>0}$ and $s \in \mathbb {Z}$ be constants. Then, we have the following estimate.

Lemma A.4

Denote

(A.11)

We have $\Lambda _s < C_1 M_f$ for a constant $C_1$ that depends on m, s but not on $f(\xi _1, \dotsc , \xi _N)$.

For the proof of Lemma A.4, we need to estimate $\Lambda ^{\mathbb {L}}$ for each $\mathbb {L}$. We decompose the exponential factor $e^{\epsilon (\xi _{m(\mathbb {L})})t}$ into the sum of two terms: one is the partial sum of the Taylor expansion of the exponential function, and the other is the remainder term. To be precise, we write

$$\begin{aligned} e^z = P_n(z) + E_n(z), \quad \text {so that} \quad P_n(z) = \sum ^n_{k = 0} \frac{z^k}{k!}, \quad E_n(z) = \sum ^{\infty }_{k = n + 1} \frac{z^k}{k!}. \end{aligned}$$

(A.12)

Then, we write

$$\begin{aligned} \Lambda ^{\mathbb {L}} = \Lambda ^{\mathbb {L}; P}(n) + \Lambda ^{\mathbb {L}; E}(n), \end{aligned}$$

(A.13)

for any $n \ge 0$ with

(A.14)

for $\square = P$ or E.

Lemma A.5

Suppose $\max (\mathbb {L})L - m = M \ge 0$. Then

• 1. $\Lambda ^{\mathbb {L}; P}(M) = 0$.

• 2. If $\mathbb {L}\in \mathbb {Z}^N(s)$, we have $|\Lambda ^{\mathbb {L}; E}(M) |< C_2 M_f e^{c_3 M + c'_3 L} /M!$ for some constant $C_2, c_3, c'_3 > 0$ depending on s, t, m but independent of f and L.

Proof

The proof of statement 1 is similar to the proof of Lemma A.2. We first note that the contour C encloses no pole with respect to $\xi _{m(\mathbb {L})}$. In particular, there is no pole at $\xi _{m(\mathbb {L})} =0$ due to the assumption that $\max (\mathbb {L}) = \ell _{m(\mathbb {L})} \ge m$. Hence, the integral with respect to $\xi _{m(\mathbb {L})}$ over $C_{m(\mathbb {L})}$ vanishes for the multiple integral that defines $\Lambda ^{\mathbb {L}; P}_Y(X; t; \sigma ; M)$, and so does $\Lambda ^{\mathbb {L}; P}_Y(X; t; \sigma ; M)$.

For statement 2, we note that $|\xi _i |= p^2$ if $\xi _i$ is on the contour C. Then,

$$\begin{aligned} \left|\prod ^N_{j = 1} \xi ^{-m}_j \prod _{j \ne m(\mathbb {L})} \xi ^{L \ell _j}_j \right|= p^{2(-mN + L(s - m(\mathbb {L})L))} \end{aligned}$$

(A.15)

for all $\mathbb {L}\in \mathbb {Z}^N(s)$. Also, we have

$$\begin{aligned} \left|e^{t(q \xi _{m(\mathbb {L})} - 1)} \prod _{j = 1, \dotsc , N, \, j \ne m(\mathbb {L})} e^{\epsilon (\xi _j)t} \right|\le e^{N c_4 t} \end{aligned}$$

(A.16)

with $c_4 = (1 + p^2)(p + p^{-2}q)$ since $\mathfrak {R}\epsilon (\xi _j) \le c_1$ and $\mathfrak {R}(q \xi _j - 1) \le c_4$ if $|\xi _j |= p^2$. Next, we have that

$$\begin{aligned} \prod ^N_{j = 1} \prod ^N_{k = 1} \left|\frac{p + q\xi _k \xi _j - \xi _k}{p + q\xi _k \xi _j - \xi _j} \right|^{\ell _j} \le \prod ^N_{j = 1} c^{|\ell _j |}_5 \le c^{N \max (\mathbb {L})}_5 = c^{N(M + m)}_5 \end{aligned}$$

(A.17)

with $c_5 = (1 + p + p^3q)/(q - p^3q)$ for all $i = 1, \dotsc , N$ since

$$\begin{aligned} p(q - p^3q) \le |p + q\xi _i \xi _j - \xi _i |\le p(1 + p + p^3q). \end{aligned}$$

(A.18)

At last, we have an adequate upper bound of the integrand in (A.15) for $\Lambda ^{\mathbb {L}; E}_Y(X; t; \sigma ; M)$ by using the estimate

$$\begin{aligned} |E_M(tp \xi ^{-1}_{m(\mathbb {L})}) |\le \frac{2}{(M + 1)!} (tp^{-1})^{M + 1}, \end{aligned}$$

(A.19)

which corresponds to the estimate of the error term in a Taylor expansion. Then, we have

$$\begin{aligned} |\Lambda ^{\mathbb {L}; E}(M) |\le M_f p^{2(-m(N+1) + Ls)} c^{N(M + m)}_5 \frac{2}{(M + 1)!} (tp^{-1})^{M + 2} e^{N c_4 t} \end{aligned}$$

(A.20)

after evaluating the integral. This implies statement 2 of the lemma. $\quad \square$