On the KPZ Scaling and the KPZ Fixed Point for TASEP

Abstract

We consider all totally asymmetric simple exclusion processes (TASEPs) whose transition probabilities are given by the Schütz-type formulas and which jump with homogeneous rates. We show that the multi-point distribution of particle positions and the KPZ scaling are described using the probability generating function of the rightmost particle’s jump. For all TASEPs satisfying certain assumptions, we also prove the pointwise convergence of the kernels appearing in the joint distribution of particle positions to those appearing in the KPZ fixed point formula. Our result generalizes the result of Matetski, Quastel, and Remenik [18] in the sense that we provide the KPZ fixed point formulation for a class of TASEPs, instead of for one specific TASEP.

Appendices

Appendix A The KPZ fixed point for the continuous time TASEP with jump rate $\beta$

In this section, we use our method to show that the KPZ fixed point is obtained in the continuous time TASEP with jump rate $\beta \in (0, \infty )$. From (2.3) and Proposition 2, we get

$$\begin{aligned} {\mathcal {M}}(w)=e^{\beta (w-1)}. \end{aligned}$$

Besides, by (2.25), we have

$$\begin{aligned} \gamma (w)=e^{-\frac{\beta }{2}w}. \end{aligned}$$

If $\gamma (w)$ satisfies Assumption 5 and 7, we can show Theorem 8 using Theorem 4 and Proposition 9. Therefore, we prove that $\gamma (w)$ satisfies Assumption 5 and Assumption 7.

First, we show that $\gamma (w)$ satisfies Assumption 5. Calculating the derivatives of $\gamma (w)$ up to the third order, we get

$$\begin{aligned} \gamma ^{'}(w)= & {} -\frac{\beta }{2}e^{-\frac{\beta }{2}w}, ~\gamma ^{(2)}(w)=\frac{\beta ^2}{4}e^{-\frac{\beta }{2}w},\nonumber \\ ~\gamma ^{(3)}(w)= & {} -\frac{\beta ^3}{8}e^{-\frac{\beta }{2}w}. \end{aligned}$$

(A1)

Substituting $w=0$ in (A1) gives

$$\begin{aligned} \gamma ^{'}(0)= & {} -\frac{\beta }{2}, ~\gamma ^{(2)}(0)=\frac{\beta ^2}{4},\nonumber \\ ~\gamma ^{(3)}(0)= & {} -\frac{\beta ^3}{8}. \end{aligned}$$

(A2)

Thus, we obtain

$$\begin{aligned} \gamma ^{(3)}(0)-3\gamma ^{(2)}(0)\gamma ^{'}(0)+2\{\gamma ^{'}(0)\}^3- 2\gamma ^{'}(0)=\beta >0. \end{aligned}$$

Next we prove that $\gamma (w)$ satisfies Assumption 7. By (2.38), (2.39), (2.40), and (A2), we have

$$\begin{aligned} D=\frac{2}{\beta }, ~ E=\frac{1}{2}, ~ F=\frac{1}{2}. \end{aligned}$$

(A3)

Note that $\log (1+x)< x$ for $x\in (-1, \infty )\setminus \{0\}$. For $\theta \in [-\pi , -\frac{\pi }{3})\cup (\frac{\pi }{3}, \pi ]$, by (A3), we obtain

$$\begin{aligned} E\log |2-e^{i\theta }|+D\log |\gamma (1-e^{i\theta })| <\frac{1}{4}(4-4\cos \theta )+\cos \theta -1=0 \end{aligned}$$

and

$$\begin{aligned} F\log |2-e^{i\theta }|-D\log |\gamma (e^{i\theta }-1)|<\frac{1}{4}(4-4\cos \theta )+\cos \theta -1=0. \end{aligned}$$

This completes the proof.

Appendix B The KPZ fixed point for the discrete time Bernoulli TASEP with sequential update

In this section, we use our method to prove that the KPZ fixed point is derived in the discrete time Bernoulli TASEP with sequential update. This proof can be shown in a similar manner to Appendix A. By (2.4) and Proposition 2, we have

$$\begin{aligned} {\mathcal {M}}(w)=1-p+pw. \end{aligned}$$

Furthermore, by (2.25), we get

$$\begin{aligned} \gamma (w)=1-\frac{p}{2-p}w. \end{aligned}$$

If $\gamma (w)$ satisfies Assumption 5 and Assumption 7, we can prove Theorem 8 using Theorem 4 and Proposition 9. Thus, we show that $\gamma (w)$ satisfies Assumption 5 and Assumption 7.

First, we prove that $\gamma (w)$ satisfies Assumption 5. Calculating the derivatives of $\gamma (w)$ up to the third order, we obtain

$$\begin{aligned} \gamma ^{'}(w)=-\frac{p}{2-p}, ~\gamma ^{(2)}(w)=0, ~\gamma ^{(3)}(w)=0. \end{aligned}$$

(B1)

Substituting $w=0$ in (B1) gives

$$\begin{aligned} \gamma ^{'}(0)=-\frac{p}{2-p}, ~\gamma ^{(2)}(0)=0, ~\gamma ^{(3)}(0)=0. \end{aligned}$$

(B2)

Therefore, we get

$$\begin{aligned} \gamma ^{(3)}(0)-3\gamma ^{(2)}(0)\gamma ^{'}(0)+2 \{\gamma ^{'}(0)\}^3-2\gamma ^{'}(0)=\frac{8p(1-p)}{(2-p)^3}>0. \end{aligned}$$

Next we show that $\gamma (w)$ satisfies Assumption 7. By (2.38), (2.39), (2.40), and (B2), we obtain

$$\begin{aligned} D=\frac{(2-p)^3}{4p(1-p)}, ~ E=\frac{2-p}{4}, ~ F=\frac{2-p}{4(1-p)}. \end{aligned}$$

(B3)

Note that $\log (1+x)< x$ for $x\in (-1, \infty )\setminus \{0\}$. For $\theta \in [-\pi , -\frac{\pi }{3})\cup (\frac{\pi }{3}, \pi ]$, by (B3), we have

$$\begin{aligned} E\log |2-e^{i\theta }|+D\log |\gamma (1-e^{i\theta })|< \frac{2-p}{2}\left( 1-\cos \theta \right) + \frac{2-p}{2}\left( \cos \theta -1\right) =0 \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} F\log |2-e^{i\theta }|&-D\log |\gamma (e^{i\theta }-1)|\\&=\frac{2-p}{8(1-p)}\log \left( 1+\frac{4(4-p)(1-p)}{(2-p)^2+4p(1-\cos \theta )}(1-\cos \theta )\right) \\&\quad +\frac{(2-p)(4-p)}{8p}\log \left( 1+\frac{4p}{(2-p)^2+4p(1-\cos \theta )}(\cos \theta -1)\right) \\&\quad < \frac{(2-p)(4-p)}{2\left\{ (2-p)^2+4p(1-\cos \theta )\right\} }\left( 1-\cos \theta \right) \\&\quad +\frac{(2-p)(4-p)}{2\left\{ (2-p)^2+4p(1-\cos \theta )\right\} }\left( \cos \theta -1\right) \\&=0. \end{aligned} \end{aligned}$$

Note that we used the expression transform

$$\begin{aligned} \begin{aligned} \frac{2-p}{8(1-p)}&\log (5-4\cos \theta )-\frac{(2-p)^3}{8p(1-p)}\log \left( 1+\frac{4p(1-\cos \theta )}{(2-p)^2}\right) \\&=\frac{2-p}{8(1-p)}\left[ \log (5-4\cos \theta )-\log \left( 1+\frac{4p(1-\cos \theta )}{(2-p)^2}\right) \right] \\&\quad -\frac{(2-p)(4-p)}{8p}\log \left( 1+\frac{4p(1-\cos \theta )}{(2-p)^2}\right) \\&=\frac{2-p}{8(1-p)}\log \left( 1+\frac{4(4-p)(1-p)}{(2-p)^2+4p(1-\cos \theta )}(1-\cos \theta )\right) \\&\quad +\frac{(2-p)(4-p)}{8p}\log \left( 1+\frac{4p}{(2-p)^2+4p(1-\cos \theta )}(\cos \theta -1)\right) . \end{aligned} \end{aligned}$$

(B4)

This completes the proof.

Appendix C The KPZ fixed point for the discrete time geometric TASEP with parallel update

In this section, we use our method to show that the KPZ fixed point is obtained in the discrete time geometric TASEP with parallel update. This proof can be proved in a similar manner to Appendix A and Appendix B. By (2.7) and Proposition 2, we obtain

$$\begin{aligned} {\mathcal {M}}(w)=\frac{1-\alpha }{1-\alpha w}. \end{aligned}$$

Moreover, by (2.25), we have

$$\begin{aligned} \gamma (w)=\frac{1}{1+\frac{\alpha }{2-\alpha }w}. \end{aligned}$$

If $\gamma (w)$ satisfies Assumption 5 and Assumption 7, we can show Theorem 8 using Theorem 4 and Proposition 9. Therefore, we prove that $\gamma (w)$ satisfies Assumption 5 and Assumption 7.

First, we show that $\gamma (w)$ satisfies Assumption 5. Calculating the derivatives of $\gamma (w)$ up to the third order, we obtain

$$\begin{aligned} \begin{aligned} \gamma ^{'}(w)&=-\frac{\alpha }{2-\alpha }\left( \frac{1}{1+\frac{\alpha }{2-\alpha }w}\right) ^2, ~\gamma ^{(2)}(w)=2\left( \frac{\alpha }{2-\alpha }\right) ^2\left( \frac{1}{1+\frac{\alpha }{2-\alpha }w}\right) ^3,\\ \gamma ^{(3)}(w)&=-6\left( \frac{\alpha }{2-\alpha }\right) ^3\left( \frac{1}{1+\frac{\alpha }{2-\alpha }w}\right) ^4. \end{aligned} \end{aligned}$$

(C1)

Substituting $w=0$ in (C1) gives

$$\begin{aligned} \gamma ^{'}(0)= & {} -\frac{\alpha }{2-\alpha }, ~\gamma ^{(2)}(0)=2\left( \frac{\alpha }{2-\alpha }\right) ^2,\nonumber \\ ~\gamma ^{(3)}(0)= & {} -6\left( \frac{\alpha }{2-\alpha }\right) ^3. \end{aligned}$$

(C2)

Thus, we have

$$\begin{aligned} \gamma ^{(3)}(0)-3\gamma ^{(2)}(0)\gamma ^{'}(0) +2\{\gamma ^{'}(0)\}^3-2\gamma ^{'}(0)=\frac{8\alpha (1-\alpha )}{(2-\alpha )^3}>0. \end{aligned}$$

Next we prove that $\gamma (w)$ satisfies Assumption 7. By (2.38), (2.39), (2.40), and (C2), we obtain

$$\begin{aligned} D=\frac{(2-\alpha )^3}{4\alpha (1-\alpha )}, ~ E=\frac{2-\alpha }{4(1-\alpha )}, ~ F=\frac{2-\alpha }{4}. \end{aligned}$$

(C3)

We remark that $\log (1+x)< x$ for $x\in (-1, \infty )\setminus \{0\}$. For $\theta \in [-\pi , -\frac{\pi }{3})\cup (\frac{\pi }{3}, \pi ]$, by (C3), we get

$$\begin{aligned} \begin{aligned}&E\log |2-e^{i\theta }|+D\log |\gamma (1-e^{i\theta })|\\&\quad =\frac{2-\alpha }{8(1-\alpha )}\log \left( 1+\frac{4(4-\alpha )(1-\alpha )}{(2-\alpha )^2+4\alpha (1-\cos \theta )}(1-\cos \theta )\right) \\&\qquad +\frac{(2-\alpha )(4-\alpha )}{8\alpha }\log \left( 1+\frac{4\alpha }{(2-\alpha )^2+4\alpha (1-\cos \theta )}(\cos \theta -1)\right) \\&\qquad < \frac{(2-\alpha )(4-\alpha )}{2\left\{ (2-\alpha )^2+4\alpha (1-\cos \theta )\right\} }\left( 1-\cos \theta \right) \\&\qquad +\frac{(2-\alpha )(4-\alpha )}{2\left\{ (2-\alpha )^2+4\alpha (1-\cos \theta )\right\} }\left( \cos \theta -1\right) \\&\quad =0. \end{aligned} \end{aligned}$$

and

$$\begin{aligned} F\log |2-e^{i\theta }|-D\log |\gamma (e^{i\theta }-1)|< \frac{2-\alpha }{2}\left( 1-\cos \theta \right) + \frac{2-\alpha }{2}\left( \cos \theta -1\right) =0 \end{aligned}$$

Note that we used a transformation similar to (B4). This completes the proof.