The Transition Probability of the $q$-TAZRP ($q$-Bosons) with Inhomogeneous
Jump Rates

The Transition Probability of the $\boldsymbol{q}$-TAZRP
($\boldsymbol{q}$-Bosons) with Inhomogeneous Jump Rates\star\star$\star$This
paper is a contribution to the Special Issue on Asymptotics and Universality
in Random Matrices, Random Growth Processes, Integrable Systems and
Statistical Physics in honor of Percy Deift and Craig Tracy. The full
collection is available at http://www.emis.de/journals/SIGMA/Deift-Tracy.html

Dong WANG † and David WAUGH ‡

D. Wang and D. Waugh

† Department of Mathematics, National University of Singapore, 119076,
Singapore<EMAIL_ADDRESS>http://www.math.nus.edu.sg/~matwd/

‡ BNP Paribas, 787 7th Avenue, New York, NY, 10019, USA<EMAIL_ADDRESS>
Received January 03, 2016, in final form April 08, 2016; Published online
April 14, 2016

In this paper we consider the $q$-deformed totally asymmetric zero range
process ($q$-TAZRP), also known as the $q$-boson (stochastic) particle system,
on the $\mathbb{Z}$ lattice, such that the jump rate of a particle depends on
the site where it is on the lattice. We derive the transition probability for
an $n$ particle process in Bethe ansatz form as a sum of $n!$ $n$-fold contour
integrals. Our result generalizes the transition probability formula by
Korhonen and Lee for $q$-TAZRP with a homogeneous lattice, and our method
follows the same approach as theirs.

zero range process; transition probability; interacting particle systems;
Bethe ansatz

82B23; 82C22; 60J35

##  1 Introduction

The zero range process (ZRP) is a well studied model in non-equilibrium
stochastic mechanics. Since it was introduced by Spitzer [17], it has become a
popular model in both the physics and mathematics communities [1].

With properly chosen parameters, as in the $q$-TAZRP to be described below,
the $1$-dimensional ZRP model is integrable, and is solvable by the Bethe
ansatz [5, 13, 15]. The application of the Bethe ansatz to find the explicit
transition probabilities in $1$-dimensional particle models has come along two
decades since [8] and [16], and it has yielded the explicit transition
probability formulas for the asymmetric simple exclusion process (ASEP) in
Tracy and Widom’s seminal work [18]. These exact formulas enabled (after
further manipulation) Tracy and Widom to calculate the asymptotic distribution
of one particle as the number of particles approaches infinity with special
initial conditions [19], and greatly improve our understanding of the KPZ
universality [7, 14]. Since then, different methods have been developed and
exact formulas for the transition probabilities have been studied for various
integrable $1$-dimensional particle models [10], including the $q$-deformed
totally asymmetric simple exclusion process ($q$-TASEP), which is an example
of the general framework of Macdonald processes [2] and has duality with
$q$-TAZRP [5].

In our paper, we consider the $q$-deformed totally asymmetric zero range
process ($q$-TAZRP), of which a definition is given below. This model was
considered by Borodin and Corwin as a specialization of Macdonald processes,
and then by Borodin, Corwin and Sasamoto in [5] in the light of its stochastic
duality (in the sense of [11, Section 2.3]) to the $q$-TASEP model. Our main
inspiration is the paper [9] by Korhonen and Lee, where the transition
probability for the $q$-TAZRP with homogeneous jump rates is derived. In [9]
the authors used Tracy and Widom’s method rather than results from Macdonald
processes. Our paper is a generalization to [9], and we extend their method to
inhomogeneous jump rates. Let us note that, this integrable ZRP was first
introduced by Sasamoto and Wadati [15] under the name of $q$-Boson totally
asymmetric diffusion model, and this name is used in later publications like
[4], so we refer to this name in the title. But later in this paper we follow
the nomenclature in [9] and call the model $q$-TAZRP.

As mentioned above, $q$-TAZRP and $q$-TASEP have the stochastic duality. There
is a natural particle-spacing duality between zero range process and simple
exclusion process (SEP) models generally [17], such that if particles
$\\{x_{i}(t)\\}$ are in a simple exclusion process, then their spacings
$\\{y_{i}=x_{i}-x_{i+1}\\}$ are in a corresponding zero range process. While
in [5], a different stochastic duality that is associated to a duality
function is observed, and it is applied there to derive moment formulas of
$q$-TASEP. Therefore it is natural that the exact formulas enjoyed by
$q$-TASEP have a dual version for $q$-TAZRP, and vice versa, although the
relation between the formulas may not be direct. However, there is a certain
discrepancy between the results for $q$-TASEP and those for $q$-TAZRP. The
$q$-TASEP as a specialization of Macdonald process naturally contains
parameters, which are independent of $q$, for the jumping rates of particles,
so the particles may not be identical, but some are fast particles and some
are slow. In the study of $q$-TAZRP, Korhonen and Lee assume that the jump
rate of a particle only depends on the height of the stack where the particle
belongs, and the model has no parameter other than $q$.

The particle-spacing duality between ZRP and SEP models suggests if a SEP
model can have fast and slow particles and still be integrable, then the dual
ZRP model can both be integrable and have inhomogeneous jump rates, that is,
some lattice sites are easy to pass through and some are difficult. The
purpose of our paper is to generalize the result in [9] to the $q$-TAZRP model
with inhomogeneous jump rates. In this sense, we remove the discrepancy
between the existing results for $q$-TASEP and $q$-TAZRP.

Before going into detail, we remark that the $q$-TAZRP with homogeneous jump
rates can be analysed by spectral theory [4], and proving the transition
probability formula is equivalent to proving the direct Plancherel theory, see
[3, Section 7.5] for an explanation in the ASEP case. It would be very
interesting to see if our transition probability results likewise lead to a
direct Plancherel theory for the inhomogeneous rate model – though such a
theory has not yet been developed.

It is worth noting that very recently Borodin and Petrov investigated the
fully inhomogeneous stochastic higher spin six vertex model in a quadrant [6],
and the spectral theory was analysed. Our $q$-TAZRP model, together with the
above mentioned $q$-TASEP and ASEP models, can all be regarded as
degenerations of the general higher spin six vertex model.

Below we give a description of the $q$-TAZRP model. We keep the discussion on
the background to a minimum, since [9] has a detailed introduction to this
effect.

###  1.1 Description of $\boldsymbol{q}$-TAZRP

Throughout this paper, $q\in(0,1)$ is a constant. Let $n$ particles be on the
integer lattice $\mathbb{Z}$, and denote the position of the $k$-th particle
by $x_{k}$, or $x_{k}(t)$ if we specify the time $t\geq 0$. We also use
$x_{k}$ to denote the $k$-th particle itself by abuse of notation. We assume
that initially $(x_{1}(0),x_{2}(0),\dotsc,x_{n}(0))=:X(0)\in\mathbb{W}^{n}$,
where $\mathbb{W}^{n}=\\{(i_{1},\dotsc,i_{n})\in\mathbb{Z}^{n}\,|\,i_{1}\geq
i_{2}\geq\dotsb\geq i_{n}\\}$. From the dynamics given below, we know that for
all $t>0$ the order $x_{1}(t)\geq x_{2}(t)\geq\dotsb\geq x_{n}(t)$ is kept, or
equivalently, $X(t)\in\mathbb{W}^{n}$.

Each particle is at an integer site at any given time, and several consecutive
particles can share a site. Suppose
$x_{k-1}(t)>x_{k}(t)=x_{k+1}(t)=\dotsb=x_{k+j-1}(t)=m>x_{k+j}(t)$, then we say
that at time $t$, the particles $x_{k},x_{k+1},\dotsc,x_{k+j-1}$ are in a
stack at $m$ with height $j$, such that $x_{k}$ is at the top of the stack,
$x_{k+i}$ is below $x_{k+i-1}$ ($i=1,\dotsc,j-1$), and $x_{k+j-1}$ is at the
bottom. See Fig. 2.

Figure 1: Particles in $q$-TAZRP. The $k$-th particle is on the top of a stack
at $i-1$ with height $3$.

Figure 2: The $k$-th particle jumps to the bottom of the stack at $i$. Until
the $k$-th particle jumps, the $(k+1)$-th and $(k+2)$-th particles stay put.

On each site $i$, we assign a _conductance_ $a_{i}>0$. Throughout this paper,
we denote $b_{i}=a_{i}(1-q)$ for notational simplicity. The dynamics of the
particles is given as follows. If a particle is in a stack but is not on the
top of the stack, then it cannot move; otherwise if the particle is on the top
of a stack at $m$ with height $j$, it moves to site $m+1$ at rate
$a_{m}(1-q^{j})=b_{m}(1+q+\dotsb+q^{j-1})$. The movement occurs regardless of
whether the right neighbour site is occupied or not (hence the model is named
“zero range”). If the right neighbour site is occupied, then $x_{k}$ moves to
the bottom of the stack there, and the height of that stack increases by $1$.
See Fig. 2 for illustration.

The $q$-TAZRP is clearly a Markov process. Let
$X=(x_{1},x_{2},\dotsc,x_{n})\in\mathbb{W}^{n}$ denote a state of the model,
then $P(X;t)$, the probability that the model is at state $X$ at time $t$,
satisfies the master equation

$\displaystyle\frac{d}{dt}P(X;t)=HP(X;t)=\sum_{Y\in\mathbb{W}^{n}}P(Y;t)(c(Y,X)-c(X,Y)),$
(1.1)

where $H$ is the generator of the Markov process, and $c(X,Y)$ is the
transition rate for the Markov process from state $X$ to state $Y$. To express
the master equation more concretely, for
$X=(x_{1},\dotsc,x_{n})\in\mathbb{W}^{n}$ we denote

$\displaystyle
X^{k,-}=(x_{1},\dotsc,x_{k-1},x_{k}-1,x_{k+1},\dotsc,x_{n}),\qquad
k=1,\dotsc,n,$

and denote

$\displaystyle\begin{split}&\ell(X):=\text{$\\#$ of the distinct values
attained by $x_{1},\dotsc,x_{n}$},\\\ &v_{i}(X):=\text{the $i$-th largest
value attained by $x_{1},\dotsc,x_{n}$},\\\ &p_{k}(X):=\text{$\\#$ of
$x_{1},\dotsc,x_{n}$ whose value is $k$},\\\
&n_{i}(X):=p_{v_{i}}(X),\quad\text{and}\quad
N_{i}(X):=n_{1}(X)+\dotsb+n_{i}(X).\end{split}$ (1.2)

For instance, if

$\displaystyle
X=(\underbrace{s_{1},\dotsc,s_{1}}_{k_{1}},\underbrace{s_{2},\dotsc,s_{2}}_{k_{2}},\dotsc,\underbrace{s_{m},\dotsc,s_{m}}_{k_{m}}),\qquad
s_{1}>s_{2}>\dotsb>s_{m},$ (1.3)

and let $K_{i}=k_{1}+k_{2}+\dotsb+k_{i}$, then $\ell(X)=m$, $v_{i}(X)=s_{i}$,
$n_{i}(X)=k_{i}$, and $N_{i}(X)=K_{i}$. For $X$ expressed in (1.3), the master
equation (1.1) can be expressed concretely as

$\displaystyle\frac{d}{dt}P(X;t)=\sum^{m}_{i=1}\big{(}1-q^{p_{s_{i}-1}(X^{K_{i},-})}\big{)}a_{s_{i}-1}P\big{(}X^{K_{i},-};t\big{)}-\sum^{m}_{i=1}\big{(}1-q^{k_{i}}\big{)}a_{s_{i}}P(X;t),$
(1.4)

where $p_{s_{i}-1}(X^{K_{i},-})$ is the number of particles on site $s_{i}-1$
for the state $X^{K_{i},-}$ (as defined in (1.2)). Here we remark that if
$X\in\mathbb{W}^{n}$ and $k\in\\{1,\dotsc,n\\}$, $X^{k,-}$ may not be in
$\mathbb{W}^{n}$. But all $X^{K_{i},-}$ on the right-hand side of (1.4) are in
$\mathbb{W}^{n}$.

###  1.2 Main result

In this paper we compute $P_{Y}(X;t)$, the probability that the particles are
in state $X\\!=\\!(x_{1},\dotsc,x_{n})$ at time $t>0$, given the initial
condition that the particles are in state $Y=(y_{1},\dotsc,y_{n})$ at time
$0$. Equivalently, $P_{Y}(X;t)$ is the solution to the master equation (1.1)
with initial condition $P_{Y}(X;0)=1$ if $X=Y$ and $P_{Y}(X;t)=0$ if
$X\in\mathbb{W}^{n}{\setminus}\\{Y\\}$.

To state the main theorem, we introduce some notations. First, for
$X\in\mathbb{W}^{n}$ in the form of (1.3), denote

$\displaystyle
W(X)=\prod^{\ell(X)}_{i=1}\prod^{n_{i}(X)}_{j=1}\frac{1-q^{j}}{1-q}=\prod^{m}_{i=1}[k_{i}]_{q}!,$

where we use the $q$-deformed factorial
$[k]_{q}!=(1-q)(1-q^{2})\dotsb(1-q^{k})/(1-q)^{k}$. Next, let
$w_{1},\dotsc,w_{n}$ be variables. For each pair of integers $\alpha<\beta$,
denote

$\displaystyle
S_{(\beta,\alpha)}=-\frac{qw_{\beta}-w_{\alpha}}{qw_{\alpha}-w_{\beta}}.$

For a permutation $\sigma\in S_{n}$, we say $(\beta,\alpha)$ is an _inversion_
of $\sigma$ if $\alpha<\beta$ and $\sigma^{-1}(\alpha)>\sigma^{-1}(\beta)$.
Then let

$\displaystyle A_{\sigma}(w_{1},\dotsc,w_{n})=\prod_{(\beta,\alpha)\text{ is
an inversion of }\sigma}S_{(\beta,\alpha)}.$

We write $A_{\sigma}=A_{\sigma}(w_{1},\dotsc,w_{n})$ if there is no
possibility of confusion, but later we occasionally use different variables
for $A_{\sigma}$. Note that for $\sigma=\operatorname{id}\in S_{n}$, since the
identity permutation has no inversion, we let $A_{\operatorname{id}}=1$. For
$X=(x_{1},\dotsc,x_{n})\in\mathbb{Z}^{n}$,
$Y=(y_{1},\dotsc,y_{n})\in\mathbb{Z}^{n}$ and $\sigma\in S_{n}$, define

$\displaystyle\Lambda_{Y}(X;t;\sigma)=\left(\prod^{n}_{k=1}\frac{-1}{b_{x_{k}}}\right)\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}A_{\sigma}\prod^{n}_{j=1}\left[\sideset{}{{}^{\prime}}{\prod}^{x_{j}}_{k=y_{\sigma(j)}}\left(\frac{b_{k}}{b_{k}-w_{\sigma(j)}}\right)e^{-w_{j}t}\right],$
(1.5)

where the notation
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{i}}dw_{i}$
is a shorthand for $(2\pi i)^{-1}\oint_{C_{i}}dw_{i}$ and the notation
$\prod^{\prime}$ is an extension of the usual $\prod$ notation such that

$\displaystyle\sideset{}{{}^{\prime}}{\prod}^{n}_{k=m}f(k)=\begin{cases}\displaystyle\prod^{n}_{k=m}f(k)&\text{if
$n\geq m$},\\\ 1&\text{if $n=m-1$},\\\
\displaystyle\prod^{m-1}_{k=n+1}\frac{1}{f(k)}&\text{if $n<m-1$},\end{cases}$

for example,
$\prod\limits^{-3}_{k=0}\\!\\!{\vphantom{\prod}}^{\prime}f(k)=1/[f(-1)f(-2)]$.
Later we use the identity

$\displaystyle\left(\sideset{}{{}^{\prime}}{\prod}^{x}_{k=m+1}f(k)\right)\left(\sideset{}{{}^{\prime}}{\prod}^{m}_{k=y}f(k)\right)=\sideset{}{{}^{\prime}}{\prod}^{x}_{k=y}f(k).$

With respect to any $w_{j}$, the integrand in (1.5) has explicit poles in the
form of $w_{j}=b_{k}$ for some $k$, and the factor $A_{\sigma}$ may introduce
poles in the form of $w_{j}=qw_{i}$ and $w_{j}=q^{-1}w_{l}$ for some $i$ and
$l$ such that $i<j$ and $l>j$. We require that poles $b_{k}$ and $qw_{i}$ are
enclosed by $C_{j}$, but $q^{-1}w_{l}$ are not. To be concrete, we can take
that $C_{j}=\\{\lvert z\rvert=R\\}$ for all $j$ with a large enough $R$, and
choose the counter-clockwise orientation.

######  Theorem 1.1.

Let $X,Y\in\mathbb{W}^{n}$. Then

$\displaystyle P_{Y}(X;t)=\frac{1}{W(X)}\sum_{\sigma\in
S_{n}}\Lambda_{Y}(X;t;\sigma).$ (1.6)

######  Remark 1.2.

The prefactor $W(X)$ is the same $W(X)$ in [9, Theorem 2.1] in the homogeneous
case. It corresponds to the fact that the model is PT symmetric with respect
to the invariant measure, and $W(X)$, like the $C_{q}(\vec{n})$ in [4], gives
the invariant measure for the $q$-TAZRP process, up to a constant. See [4,
Remark 2.3] and [2, Corollary 3.3.12].

With special initial conditions, the transition probability formula (1.6) can
be simplified. Here we consider the initial condition
$y_{1}=y_{2}=\dotsb=y_{n}=0$. To state the result, we denote

$\displaystyle B(w_{1},\dotsc,w_{n})=\prod_{1\leq i<j\leq
n}\frac{w_{i}-w_{j}}{w_{i}-qw_{j}}.$ (1.7)

######  Corollary 1.3.

Let $X\in\mathbb{W}^{n}$. Then

$\displaystyle
P_{(0,0,\dotsc,0)}(X;t)=[n]_{q}!\frac{1}{W(X)}\left(\prod^{n}_{k=1}\frac{-1}{b_{x_{k}}}\right)$
$\displaystyle\hphantom{P_{(0,0,\dotsc,0)}(X;t)=}{}\times\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}B(w_{1},\dotsc,w_{n})\prod^{n}_{j=1}\left[\sideset{}{{}^{\prime}}{\prod}^{x_{j}}_{k_{j}=0}\left(\frac{b_{k_{j}}}{b_{k_{j}}-w_{j}}\right)e^{-w_{j}t}\right].$
(1.8)

We note that the transition probability formula (1.8) shares formal similarity
with the moment formulas in the $q$-TASEP with various initial conditions, see
[5, Theorem 2.11 and Corollary 2.12], where the parameters $a_{i}$ correspond
to the $a_{i}$ in our paper. This is not a coincidence: The duality between
the $q$-TASEP and $q$-TAZRP implies that the moment formula for $q$-TASEP
satisfies the true evolution equation for $q$-TAZRP, as explained in [5]. In
the special case where $a_{i}$ are identical, moment formulas in $q$-TASEP are
derived from properties of $q$-TAZRP in [4].

### Organization of the paper

The basic idea of our proof to Theorem 1.1 is the Bethe ansatz, following the
approach of [18] and [9]. In Section 2 we consider the one particle case where
the interaction between particles is missing as a warm-up. In Section 3 the
two particle case, the simplest non-trivial case, is analysed explicitly, and
in Section 4 the general case is worked out. The essence of the Bethe ansatz
is that the interaction among multiple particles is equivalent to the
superposition of two-particle interactions. We use the idea implicitly. At
last, we prove Corollary 1.3 in Section 5.

##  2 One particle case

In the simplest case that there are only one particle $x_{1}$, the master
equation (1.4) becomes (we write $X=(x)$ as $x$)

$\displaystyle\frac{d}{dt}P(x;t)=b_{x-1}P(x-1;t)-b_{x}P(x;t).$

Given the initial condition that the particle is at $y\in\mathbb{Z}$ at time
$t=0$, that is, $P(y;0)=1$ and $P(x;0)=0$ if $x\neq y$, we can compute
$P_{y}(x;t)$ by elementary probability. Since the particle jumps only to the
right, so $P_{y}(x;t)=0$ for all $x<y$. The time that the particle first jumps
away from $y$ is in the exponential distribution with rate $b_{y}$, and after
it jump away, it cannot be back, so $P_{y}(y;t)=e^{-b_{y}t}$. Next,
$P_{y}(y+1;t)$ is the probability that the particle jumps exactly once from
time $0$ to $t$, and it is

$\displaystyle\int^{t}_{0}b_{y}e^{-b_{y}s}e^{-b_{y+1}(t-s)}ds=b_{y}(b_{y+1}-b_{y})^{-1}\big{(}e^{-b_{y}t}-e^{-b_{y+1}t}\big{)},\qquad\text{if
$b_{y+1}\neq b_{y}$}.$

Inductively, we use

$\displaystyle
P_{y}(y+k;t)=\int^{t}_{0}b_{y}e^{-b_{y}s_{1}}\left(\int^{t}_{s_{1}}b_{y+1}e^{-b_{y+1}s_{2}}\left(\int^{t}_{s_{2}}\dotsb\vphantom{\int^{t}_{s_{k-1}}}\right.\right.$
$\displaystyle\hphantom{P_{y}(y+k;t)=}{}\times\left.\left.\left(\int^{t}_{s_{k-1}}b_{y+k-1}e^{-b_{y+k-1}s_{k}}e^{-b_{y+k}(t-s_{k})}ds_{k}\right)\dotsb
ds_{3}\right)ds_{2}\right)ds_{1}$
$\displaystyle\hphantom{P_{y}(y+k;t)}{}=\int^{t}_{0}b_{y}e^{-b_{y}s_{1}}P_{y+1}(y+k;t-s_{1})ds_{1},$

and derive

$\displaystyle P_{y}(x;t)=\begin{cases}0&\text{if $x<y$},\\\
e^{-b_{x}t}&\text{if $x=y$},\\\
\displaystyle\left(\prod^{x-1}_{k=y}b_{k}\right)\sum^{x}_{k=y}\left(\prod_{y\leq
j\leq x,\ j\neq k}\frac{1}{b_{j}-b_{k}}\right)e^{-b_{k}t}&\text{if
$x>y$}.\end{cases}$

Here and later, we assume $b_{k}$ are distinct, and understand the formulas by
l’Hôpital’s rule if some of them are identical. A more convenient way to
express $P_{y}(x;t)$ is ($w=1-z^{-1}$)

$\displaystyle
P_{y}(x;t)=\frac{-1}{b_{x}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C}\sideset{}{{}^{\prime}}{\prod}^{x}_{k=y}\frac{b_{k}}{b_{k}-w}e^{-wt}dw$
(2.1a)
$\displaystyle\hphantom{P_{y}(x;t)}{}=\frac{1}{b_{x}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\tilde{C}}\left(\sideset{}{{}^{\prime}}{\prod}^{x}_{k=y}\frac{b_{k}z}{1-z(1-b_{k})}\right)e^{(z^{-1}-1)t}\frac{dz}{z^{2}},$
(2.1b)

where the contour $C$ encloses all the poles $b_{k}$ in positive orientation,
and the contour $\tilde{C}$ encloses the pole $0$ in positive orientation. If
some $b_{k}\neq 1$, then $\tilde{C}$ should not enclose poles
$(1-b_{k})^{-1}$. Thus the $n=1$ case of Theorem 1.1 is proved.

######  Remark 2.1.

Although the expression (2.1b) of $P_{y}(x;t)$ seems unnatural, it reduces to
the $n=1$ case [9, Theorem 2.1] when $b_{k}=1$ for all $k\in\mathbb{Z}$.

##  3 Two particle case

In the $2$-particle case, where the two particles are at $x_{1}\geq x_{2}$,
the master equation (1.4) can be written as

$\displaystyle\frac{d}{dt}P((x_{1},x_{2});t)=\begin{cases}b_{x_{1}-1}P((x_{1}-1,x_{2});t)-b_{x_{1}}P((x_{1},x_{2});t)&\\\
{}+b_{x_{2}-1}P((x_{1},x_{2}-1);t)-b_{x_{2}}P((x_{1},x_{2});t),&\text{if
$x_{1}>x_{2}+1$},\\\ (1+q)b_{x}P((x,x);t)-b_{x+1}P((x+1,x);t)&\text{if
$x_{2}=x$}\\\ {}+b_{x-1}P((x+1,x-1);t)-b_{x}P((x+1,x);t),&\text{and
$x_{1}=x+1$},\\\ b_{x-1}P((x,x-1);t)-(1+q)b_{x}P((x,x);t)&\text{if
$x_{1}=x_{2}=x$}.\end{cases}\\!\\!\\!\\!$ (3.1)

and we want to solve it under the initial condition ($y_{1}\geq y_{2}$)

$\displaystyle P((y_{1},y_{2});0)=1,\qquad\text{and}\qquad
P((x_{1},x_{2});0)=0\qquad\text{for all
$(x_{1},x_{2})\in\mathbb{W}^{2}{\setminus}\\{(y_{1},y_{2})\\}$}.\\!\\!\\!$
(3.2)

To simplify the master equation, we consider functions
$u^{0}((x_{1},x_{2});t)$ that are differentiable in $t\geq 0$, for all
$(x_{1},x_{2})\in\mathbb{Z}^{2}$. By contrast, the probabilities
$P((x_{1},x_{2});t)$ can also be viewed as a differentiable function in $t$,
but it is only defined when $(x_{1},x_{2})\in\mathbb{W}^{2}$.

If the functions $u^{0}((x_{1},x_{2});t)$ satisfy the differential equations
(so-called free particle evolution equations)

$\displaystyle\frac{d}{dt}u^{0}((x_{1},x_{2});t)=b_{x_{1}-1}u^{0}((x_{1}-1,x_{2});t)-b_{x_{1}}u^{0}((x_{1},x_{2});t)$
$\displaystyle\hphantom{\frac{d}{dt}u^{0}((x_{1},x_{2});t)=}{}+b_{x_{2}-1}u^{0}((x_{1},x_{2}-1);t)-b_{x_{2}}u^{0}((x_{1},x_{2});t),$
(3.3)

and in addition, they satisfy the relations (so-called boundary condition)

$\displaystyle
b_{x-1}u^{0}((x-1,x);t)=b_{x-1}qu^{0}((x,x-1);t)+b_{x}(1-q)u^{0}((x,x);t)\qquad\text{for
all $x\in\mathbb{Z}$},\\!\\!\\!$ (3.4)

then we have that the functions with $t\geq 0$ and
$(x_{1},x_{2})\in\mathbb{W}^{2}$

$\displaystyle
u((x_{1},x_{2});t)=\frac{1}{W((x_{1},x_{2}))}u^{0}((x_{1},x_{2});t),$ (3.5)

where

$\displaystyle W((x_{1},x_{2}))=\begin{cases}1&\text{if $x_{1}>x_{2}$},\\\
1+q&\text{if $x_{1}=x_{2}$},\end{cases}$

satisfy the master equation (3.1) with
$P((x_{1},x_{2});t)=u((x_{1},x_{2});t)$. To see it, we note that if
$x_{1}>x_{2}+1$, then (3.1) is the same as (3.3) with
$P((x_{1},x_{2});t)=u^{0}((x_{1},x_{2});t)$; if $x_{1}=x+1$ and $x_{2}=x$,
then (3.1) is equivalent to (3.3) with $P((x+1,x);t)=u^{0}((x+1,x);t)$,
$P(x+1,x-1);t)=u^{0}((x+1,x-1);t)$, and
$P((x,x);t)=(1+q)^{-1}u^{0}((x_{1},x_{2});t)$; if $x_{1}=x_{2}=x$, and we let
$P((x,x);t)=u^{0}((x,);t)$ and $P((x,x-1);t)=u^{0}((x,x-1);t)$, then the
right-hand side of then (3.1) is the same as $(1+q)^{-1}$ times (3.3),
assuming the identity (3.4). Thus (3.3), (3.4) and (3.5) imply that
$u((x_{1},x_{2});t)$ satisfy (3.1).

Hence by solving (3.3) with the boundary condition (3.4) and the partial
initial condition on $\mathbb{W}^{2}$:
$u^{(0)}(y_{1},y_{2};0)=W((y_{1},y_{2}))$ and $u^{(0)}(x_{1},x_{2};0)=0$ if
$(x_{1},x_{2})\in\mathbb{W}^{2}{\setminus}\\{(y_{1},y_{2})\\}$, we solve the
master equation (3.1) with the initial condition (3.2).

######  Remark 3.1.

It is not obvious that from a solution $u((x_{1},x_{2});t)$ to (3.1) where
$(x_{1},x_{2})\in\mathbb{W}^{2}$, we can construct $u^{0}((x_{1},x_{2});t)$
for $(x_{1},x_{2})\in\mathbb{Z}^{2}$, that satisfy (3.5) if
$(x_{1},x_{2})\in\mathbb{W}^{2}$, as well as solve (3.3) and (3.4). Although
it is a digression of our main theorem, for its own interest, we describe the
construction is as follows:

$\displaystyle u^{0}((x_{1},x_{2});t)=u((x_{1},x_{2});t)\quad\text{if
$x_{1}>x_{2}$},\qquad u^{0}((x,x);t)=(1+q)u((x,x);t),$ $\displaystyle
u^{0}((x-1,x);t)=qu((x,x-1);t)+\frac{b_{x}}{b_{x-1}}\big{(}1-q^{2}\big{)}u((x,x);t),$

and then inductively with respect to $k\geq 1$, under the assumption that any
$u^{0}((x_{1},x_{2});t)$ is expressed as linear combinations of
$u((z_{1},z_{2});t)$ if $x_{1}\geq x_{2}-k$,

$\displaystyle
u^{0}((x-(k+1),x);t)=\frac{1}{b_{x-(k+1)}}\bigg{[}\frac{d}{dt}u^{0}((x-k,x);t)$
$\displaystyle\hphantom{u^{0}((x-(k+1),x);t)=}{}-b_{x-1}u^{0}((x-k,x-1);t)+(b_{x-k}+b_{x})u^{0}((x-k,x);t)\bigg{]}.$
(3.6)

Note that in (3.6), not only are $u^{0}((x-k,x-1);t)$ and $u^{0}((x-k,x);t)$
expressed by linear combinations of $u((z_{1},z_{2});t)$, but also
$\frac{d}{dt}u^{0}((x-k,x);t)$ is the linear combinations of
$\frac{d}{dt}u((z_{1},z_{2});t)$, and in turn the linear combinations of
$u((z_{1},z_{2});t)$. For example, for $k=1$,

$\displaystyle u^{0}((x-2,x);t)$
$\displaystyle=\frac{1}{b_{x-2}}\left[\frac{d}{dt}u^{0}((x-1,x);t)-b_{x-1}u^{0}((x-1,x-1);t)+(b_{x-1}+b_{x})u^{0}((x-1,x);t)\right]$
$\displaystyle=\frac{1}{b_{x-2}}\left[\frac{d}{dt}\left(qu((x,x-1);t)+\frac{b_{x}}{b_{x-1}}\big{(}1-q^{2}\big{)}u((x,x);t)\right)-b_{x-1}(1+q)u((x-1,x-1);t)\right.$
$\displaystyle\hphantom{=\frac{1}{b_{x-2}}}\left.\vphantom{\frac{d}{dt}}+(b_{x-1}+b_{x})\left(qu((x,x-1);t)+\frac{b_{x}}{b_{x-1}}\big{(}1-q^{2}\big{)}u((x,x);t)\right)\right]$
$\displaystyle=\frac{1-q^{2}}{b_{x-2}}\left[\left(b_{x}+q\frac{b^{2}_{x}}{b_{x-1}}\right)u((x,x);t)+b_{x}u((x,x-1);t)-b_{x-1}u((x-1,x-1);t)\right.$
$\displaystyle\hphantom{=\frac{1-q^{2}}{b_{x-2}}}+\left.\frac{q}{1-q^{2}}b_{x-2}u((x,x-2);t)\right].$

We note that for any meromorphic function $f(w_{1},w_{2})$, the integral with
parameters $x_{1}$, $x_{2}$, $t$

$\displaystyle
I_{f}((x_{1},x_{2});t)=\frac{1}{b_{x_{1}}b_{x_{2}}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}\\!dw_{1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{2}}\\!dw_{2}f(w_{1},w_{2})\left(\sideset{}{{}^{\prime}}{\prod}^{x_{1}}_{k=1}\frac{b_{k}}{b_{k}-w_{1}}\right)\\!\left(\sideset{}{{}^{\prime}}{\prod}^{x_{2}}_{k=1}\frac{b_{k}}{b_{k}-w_{2}}\right)e^{-w_{1}t}e^{-w_{2}t}$

satisfies the differential equations (3.3) for $u^{0}((x_{1},x_{2});t)$. To
see it, we compute

$\displaystyle\frac{d}{dt}I_{f}((x_{1},x_{2});t)=I_{g}((x_{1},x_{2});t),\qquad$
$\displaystyle\text{where}\quad g(w_{1},w_{2})=-(w_{1}+w_{2})f(w_{1},w_{2}),$
$\displaystyle I_{f}((x_{1}-1,x_{2});t)=I_{f_{1}}((x_{1},x_{2});t),\qquad$
$\displaystyle\text{where}\quad
f_{1}(w_{1},w_{2})=\frac{b_{x_{1}}-w_{1}}{b_{x_{1}-1}}f(w_{1},w_{2}),$
$\displaystyle I_{f}((x_{1},x_{2}-1);t)=I_{f_{2}}((x_{1},x_{2});t),\qquad$
$\displaystyle\text{where}\quad
f_{2}(w_{1},w_{2})=\frac{b_{x_{2}}-w_{2}}{b_{x_{2}-1}}f(w_{1},w_{2}).$

We then have that for $I_{f}((x_{1},x_{2});t)$ to satisfy (3.3), it suffices
to show $g=b_{x_{1}-1}f_{1}+b_{x_{2}-1}f_{2}-(b_{x_{1}}+b_{x_{2}})f$, which is
straightforward to verify.

In the same way as above, we have that $I_{f}((x_{1},x_{2});t)$ satisfies the
boundary condition (3.4) for $u^{0}((x_{1},x_{2});t)$ if and only if for all
$x\in\mathbb{Z}$

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{2}}dw_{2}f(w_{1},w_{2})(qw_{2}-w_{1})\left(\sideset{}{{}^{\prime}}{\prod}^{x}_{k=1}\frac{b_{k}}{b_{k}-w_{1}}\right)\left(\sideset{}{{}^{\prime}}{\prod}^{x}_{k=1}\frac{b_{k}}{b_{k}-w_{2}}\right)e^{-w_{1}t}e^{-w_{2}t}=0.\\!\\!\\!$
(3.7)

Then inspired by the constructions in [9], we take
$f(w_{1},w_{2})=F_{y_{1},y_{2}}(w_{1},w_{2})$, where
$(y_{1},y_{2})\in\mathbb{Z}^{2}$, and

$\displaystyle
F_{y_{1},y_{2}}(w_{1},w_{2})=\left(\sideset{}{{}^{\prime}}{\prod}^{0}_{k=y_{1}}\frac{b_{k}}{b_{k}-w_{1}}\right)\left(\sideset{}{{}^{\prime}}{\prod}^{0}_{k=y_{2}}\frac{b_{k}}{b_{k}-w_{2}}\right)$
$\displaystyle\hphantom{F_{y_{1},y_{2}}(w_{1},w_{2})=}{}-\frac{qw_{1}-w_{2}}{qw_{2}-w_{1}}\left(\sideset{}{{}^{\prime}}{\prod}^{0}_{k=y_{2}}\frac{b_{k}}{b_{k}-w_{1}}\right)\left(\sideset{}{{}^{\prime}}{\prod}^{0}_{k=y_{1}}\frac{b_{k}}{b_{k}-w_{2}}\right).$

It is straightforward to check that this
$I_{F_{y_{1},y_{2}}}((x_{1},x_{2});t)$ satisfies identity (3.7).

Now we consider the value of $I_{F_{y_{1},y_{2}}}((x_{1},x_{2});0)$ for
$(x_{1},x_{2})$ and $(y_{1},y_{2})\in\mathbb{W}^{2}$. First, in the double
integral

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{2}}dw_{2}\left(\sideset{}{{}^{\prime}}{\prod}^{x_{1}}_{k=y_{1}}\frac{b_{k}}{b_{k}-w_{1}}\right)\left(\sideset{}{{}^{\prime}}{\prod}^{x_{2}}_{k=y_{2}}\frac{b_{k}}{b_{k}-w_{2}}\right),$
(3.8)

if $x_{2}<y_{2}$, then the inner integral with respect to $w_{2}$ vanishes,
since the integrand is holomorphic with respect to $w_{2}$; if $x_{2}>y_{2}$,
then by enlarging the contour $C_{2}$ to $\\{\lvert w_{2}\rvert=M\\}$ with
$M\to\infty$, we still have that the inner integral vanishes. Next, in the
double integral

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{2}}dw_{2}\frac{qw_{1}-w_{2}}{qw_{2}-w_{1}}\left(\sideset{}{{}^{\prime}}{\prod}^{x_{1}}_{k=y_{2}}\frac{b_{k}}{b_{k}-w_{1}}\right)\left(\sideset{}{{}^{\prime}}{\prod}^{x_{1}}_{k=y_{1}}\frac{b_{k}}{b_{k}-w_{2}}\right),$
(3.9)

if $x_{2}<y_{2}$, then $x_{2}<y_{1}$, and we have that the inner integral with
respect to $w_{2}$ vanishes if $x_{2}<y_{1}$, since the integrand has no pole
within $C_{2}$ with respect to $w_{2}$; if $x_{2}>y_{2}$, then $x_{1}>y_{2}$,
and by enlarging the contour $C_{1}$ to $\\{\lvert w_{1}\rvert=M\\}$ with
$M\to\infty$, we have that the integral on $C_{1}$ with respect to $w_{1}$
vanishes for $x_{1}>y_{2}$. So the integral (3.8) vanishes unless
$x_{2}=x_{1}$, and the integral (3.9) vanishes unless
$x_{1}=x_{2}=y_{1}=y_{2}=y$. In the case that $x_{2}=y_{2}$, the integral
(3.8) is equal to $b_{y_{1}}b_{y_{2}}$. Similarly, if
$x_{1}=x_{2}=y_{1}=y_{2}$, the integral (3.9) is equal to $qb^{2}_{y}$. Since
$I_{F_{y_{1},y_{2}}}(x_{1},x_{2});0)$ is $(b_{x_{1}}b_{x_{2}})^{-1}$ times the
difference between the two integrals in (3.8) and (3.9), we have that

$\displaystyle
I_{F_{y_{1},y_{2}}}((x_{1},x_{2});0)=\begin{cases}W((x_{1},x_{2}))&\text{if
$(x_{1},x_{2})=(y_{1},y_{2})$},\\\ 0&\text{if
$(x_{1},x_{2})\in\mathbb{W}^{2}{\setminus}\\{(y_{1},y_{2})\\}$}.\end{cases}$

We conclude that $u^{0}((x_{1},x_{2});t)=I_{F_{y_{1},y_{2}}}((x_{1},x_{2});t)$
is a solution to (3.3) and (3.4), and by (3.5) it yields $u((x_{1},x_{2});t)$
that is a solution to the master equation (3.1) with the initial condition
(3.2). Note that our solution $u((x_{1},x_{2});t)$ is expressed by a double
contour integral formula, and it agrees with the $n=2$ case of (1.6) after a
simple change of variables. Thus the $n=2$ case of Theorem 1.1 is proved.

##  4 $\boldsymbol{n}$ particle case

Now we prove Theorem 1.1 in the general form. Our strategy is the same as in
the proof in the $n=2$ case in Section 3.

First we consider functions $u^{0}(X;t)$, where
$X=(x_{1},\dotsc,x_{n})\in\mathbb{Z}^{n}$, such that they are differentiable
functions in $t$, and satisfy the free particle evolution equation

$\displaystyle\frac{d}{dt}u^{0}(X;t)=\sum^{n}_{k=1}\big{(}b_{x_{k}-1}u^{0}\big{(}X^{k,-};t\big{)}-b_{x_{k}}u^{0}(X;t)\big{)}.$
(4.1)

We also consider the boundary conditions that generalize (3.4), such that for
all $k=1,\dotsc,n-1$

$\displaystyle b_{x-1}u^{0}((\dotsc,x_{k-1},\underbrace{x-1,x}_{\text{$x_{k}$
and
$x_{k+1}$}},x_{k+2},\dotsc);t)=qb_{x-1}u^{0}((\dotsc,x_{k-1},x,x-1,x_{k+2},\dotsc);t)$
$\displaystyle\qquad{}+(1-q)b_{x}u^{0}((\dotsc,x_{k-1},x,x,x_{k+2},\dotsc);t).$
(4.2)

Then let for all $X\in\mathbb{W}^{n}$

$\displaystyle u(X;t)=\frac{1}{W(X)}u^{0}(X;t).$ (4.3)

We have the following result:

######  Lemma 4.1.

Suppose $u^{0}(X;t)$ is a solution to the free particle evolution equation
(4.1) with boundary condition (4.2). The functions $u(X;t)$ defined by (4.3)
satisfy the master equation (1.4) with $P(X;t)=u(X;t)$.

This lemma is essentially equivalent to the (A) $\Leftrightarrow$ (B) part of
[5, Proposition 2.7]. For the sake of readability, we give the proof below. We
also remark that given functions $u(X;t)$ for $X\in\mathbb{W}^{n}$ such that
they satisfy (1.4) with $P(X;t)=u(X;t)$, then we can construct $u^{0}(X;t)$
inductively for all $X\in\mathbb{Z}^{n}$, such that $u^{0}(X;t)$ satisfy (4.1)
for all $X\in\mathbb{Z}^{n}$, and are related to $u(X;t)$ by (4.3) if
$X\in\mathbb{W}^{n}$. The $n=2$ case is discussed in detail in Remark 3.1, and
we do not give the proof for the general $n$ case since it is not related to
our main result.

###### Proof of Lemma 4.1.

Functions $u(X;t)$ satisfy (1.4) if and only if for all $X$ expressed in
(1.3),

$\displaystyle\frac{d}{dt}u^{0}(X;t)=\sum^{m}_{i=1}D_{i},$ (4.4)

where

$\displaystyle
D_{i}=\frac{W(X)}{W(X^{K_{i},-})}\big{(}1-q^{p_{s_{i}-1}(X^{K_{i},-})}\big{)}a_{s_{i}-1}u^{0}\big{(}X^{K_{i},-};t\big{)}-\big{(}1-q^{k_{i}}\big{)}a_{s_{i}}u^{0}(X;t).$

Below we show that for each $i$,

$\displaystyle
D_{i}=\sum^{K_{i}}_{j=K_{i-1}+1}\big{(}b_{s_{i}-1}u^{0}\big{(}X^{j,-};t\big{)}-b_{s_{i}}u^{0}(X;t)\big{)}.$
(4.5)

Plugging (4.5) into (4.4), we have that (4.4) is equivalent to (4.1), and then
prove the lemma.

In the case $i=m$ or $i<m$ and $s_{i}-1>s_{i+1}$, we have
$p_{s_{i}-1}(X^{K_{i},-})=1$ and $W(X)/W(X^{K_{i},-})$ $=(1-q^{k_{i}})/(1-q)$.
On the other hand, if $i<m$ and $s_{i}-1=s_{i+1}$, we have
$p_{s_{i}-1}(X^{K_{i},-})=k_{i+1}+1$ and
$W(X)/W(X^{K_{i},-})=(1-q^{k_{i}})/(1-q^{k_{i-1}+1})$. Then in both cases, we
have

$\displaystyle
D_{i}=(1-q^{k_{i}})\big{(}a_{s_{i}-1}u^{0}\big{(}X^{K_{i},-};t\big{)}-a_{s_{i}}u^{0}(X;t)\big{)}$
$\displaystyle\hphantom{D_{i}}{}=\frac{1-q^{k_{i}}}{1-q}\big{(}b_{s_{i}-1}u^{0}\big{(}X^{K_{i},-};t\big{)}-b_{s_{i}}u^{0}(X;t)\big{)}.$
(4.6)

If $k_{i}=1$, then (4.6) gives (4.5) directly. Otherwise, we use (4.2)
recursively with $j=K_{i}-1$, $K_{i}-2,\dotsc,K_{i-1}+2,K_{i-1}+1$ to derive

$\displaystyle
b_{s_{i}-1}u^{0}\big{(}X^{j,-};t\big{)}=q^{K_{i}-j}b_{s_{i}-1}u^{0}\big{(}X^{K_{i},-};t\big{)}+\big{(}1-q^{K_{i}-j}\big{)}b_{s_{i}}u^{0}(X;t),$

and then verify (4.5). ∎

Next, we use the idea of the Bethe ansatz to construct the solution to
equations (4.1) and (4.2) with initial condition

$\displaystyle u^{0}(X;0)=\begin{cases}W(Y)&\text{if
$X=Y\in\mathbb{W}^{n}$},\\\ 0&\text{if
$X\in\mathbb{W}^{n}{\setminus}\\{Y\\}$}.\end{cases}$ (4.7)

The existence of the solution is not obvious, and it depends on the
completeness of the Bethe ansatz for this model, which is so far only proved
in [4] in the homogeneous rate parameter case. Nevertheless, we imitate the
construction in [9] and guess the formula. Then we simply check that the
formula is correct.

######  Lemma 4.2.

Let $Y\in\mathbb{W}^{n}$. The function

$\displaystyle u^{0}_{Y}(X;t)=\sum_{\sigma\in S_{n}}\Lambda_{Y}(X;t;\sigma)$

satisfies equation (4.1), boundary condition (4.2) and initial condition
(4.7).

###### Proof.

Free particle evolution equations (4.1). It is straightforward to check that
for all $\sigma$, $\Lambda_{Y}(X;t;\sigma)$ satisfy (4.1), so $u^{0}_{Y}(X;t)$
also satisfies (4.1).

Boundary conditions (4.2). To check that $u^{0}_{Y}(X;t)$ satisfies (4.2), we
fix $k\in\\{1,\dotsc,n-1\\}$, and denote

$\displaystyle
X(a,b)=(x_{1},\dotsc,x_{k-1},\\!\\!\underbrace{a,b}_{x_{k}\text{ and
}x_{k+1}}\\!\\!,x_{k+2},\dotsc,x_{n}).$

We divide $S_{n}$ into $n!/2$ pairs, such that permutations $\mu$ and $\nu$
are in a pair if and only if $\mu=\nu\circ(k,k+1)$ and $\nu=\mu\circ(k,k+1)$,
that is, $\mu(i)=\nu(i)$ if $i\neq k,k+1$, and $\mu(k)=\nu(k+1)$,
$\mu(k+1)=\nu(k)$. Below we assume that $\mu$ and $\nu$ are in a pair such
that $\mu(k)=\nu(k+1)=\alpha<\beta=\mu(k+1)=\nu(k)$. It is not hard to check
that

$\displaystyle A_{\nu}=A_{\mu}S_{(\beta,\alpha)}.$

For notational simplicity, we denote

$\displaystyle
F(w_{1},\dotsc,w_{n})=\frac{1}{b_{x}}A_{\mu}\prod^{n}_{l=1}e^{-w_{l}t}\prod_{\begin{subarray}{c}j=1,\dotsc,n\\\
j\neq
k,k+1\end{subarray}}\frac{-1}{b_{x_{j}}}\sideset{}{{}^{\prime}}{\prod}^{x_{j}}_{l_{j}=y_{\mu(j)}}\frac{b_{l_{j}}}{b_{l_{j}}-w_{\mu(j)}},$
$\displaystyle
G(w_{1},\dotsc,w_{n})=F(w_{1},\dotsc,w_{n})\prod^{x}_{i=y_{\alpha}}\sideset{}{{}^{\prime}}{\prod}^{x}_{j=y_{\beta}}\frac{b_{i}b_{j}}{(b_{i}-w_{\alpha})(b_{j}-w_{\beta})}.$

Then we have

$\displaystyle
b_{x-1}\left(\Lambda_{Y}(X(x-1,x);t;\mu)+\Lambda_{Y}(X(x-1,x);t;\nu)\right)$
$\displaystyle\qquad{}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}F(w_{1},\dotsc,w_{n})\left(\sideset{}{{}^{\prime}}{\prod}^{x-1}_{l_{k}=y_{\alpha}}\frac{b_{l_{k}}}{b_{l_{k}}-w_{\alpha}}\sideset{}{{}^{\prime}}{\prod}^{x}_{l_{k+1}=y_{\beta}}\frac{b_{l_{k+1}}}{b_{l_{k+1}}-w_{\beta}}\right.$
$\displaystyle\left.\qquad\quad{}-\frac{qw_{\beta}-w_{\alpha}}{qw_{\alpha}-w_{\beta}}\sideset{}{{}^{\prime}}{\prod}^{x-1}_{l_{k}=y_{\beta}}\frac{b_{l_{k}}}{b_{l_{k}}-w_{\beta}}\sideset{}{{}^{\prime}}{\prod}^{x}_{l_{k+1}=y_{\alpha}}\frac{b_{l_{k+1}}}{b_{l_{k+1}}-w_{\alpha}}\right)$
$\displaystyle\qquad{}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}G(w_{1},\dotsc,w_{n})\left(\frac{b_{x}-w_{\alpha}}{b_{x}}-\frac{qw_{\beta}-w_{\alpha}}{qw_{\alpha}-w_{\beta}}\frac{b_{x}-w_{\beta}}{b_{x}}\right).$
(4.8)

Similarly,

$\displaystyle
b_{x}\left(\Lambda_{Y}(X(x,x);t;\mu)+\Lambda_{Y}(X(x,x);t;\nu)\right)$
$\displaystyle\qquad{}{}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}G(w_{1},\dotsc,w_{n})\left(1-\frac{qw_{\beta}-w_{\alpha}}{qw_{\alpha}-w_{\beta}}\right),$
(4.9) $\displaystyle
b_{x-1}\left(\Lambda_{Y}(X(x,x-1);t;\mu)+\Lambda_{Y}(X(x,x-1);t;\nu)\right)$
$\displaystyle\qquad{}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}G(w_{1},\dotsc,w_{n})\left(\frac{b_{x}-w_{\beta}}{b_{x}}-\frac{qw_{\beta}-w_{\alpha}}{qw_{\alpha}-w_{\beta}}\frac{b_{x}-w_{\alpha}}{b_{x}}\right).$
(4.10)

With the integral expressions (4.8), (4.9) and (4.10), and the algebraic
identity

$\displaystyle
q\left(\frac{b_{x}-w_{\beta}}{b_{x}}-\frac{qw_{\beta}-w_{\alpha}}{qw_{\alpha}-w_{\beta}}\frac{b_{x}-w_{\alpha}}{b_{x}}\right)+(1-q)\left(1-\frac{qw_{\beta}-w_{\alpha}}{qw_{\alpha}-w_{\beta}}\right)$
$\displaystyle\qquad{}=\frac{b_{x}-w_{\alpha}}{b_{x}}-\frac{qw_{\beta}-w_{\alpha}}{qw_{\alpha}-w_{\beta}}\frac{b_{x}-w_{\beta}}{b_{x}},$

it is clear that

$\displaystyle
b_{x-1}\sum_{\sigma\in\\{\mu,\nu\\}}\Lambda_{Y}(X(x-1,x);t;\sigma)$
$\displaystyle\qquad{}=qb_{x-1}\sum_{\sigma\in\\{\mu,\nu\\}}\Lambda_{Y}(X(x,x-1);t;\sigma)+(1-q)b_{x}\sum_{\sigma\in\\{\mu,\nu\\}}\Lambda_{Y}(X(x,x);t;\sigma).$

Since the identity holds for all the $n!/2$ pairs in $S_{n}$, we sum up them
together, and conclude that $u^{0}_{Y}(X;t)$ satisfies (4.2).

Initial condition (4.7). We check the initial condition inductively. The $n=1$
case can be done by direct computation, and if we assume that (4.7) holds when
the particle number is $n-1$, we show that it also holds when the particle
number is $n$.

For any $k=1,\dotsc,n$, we define two subsets of $S_{n}$:

$\displaystyle S_{n}(k)=\\{\sigma\in
S_{n}\,|\,\sigma(1)=k\\},\qquad\text{and}$ $\displaystyle
S^{-1}_{n}(k)=\\{\sigma\in
S_{n}\,|\,\sigma(k)=1\\}=\big{\\{}\sigma\,|\,\sigma^{-1}\in
S_{n}(k)\big{\\}}.$ (4.11)

Then we define two one-to-one correspondences $\varphi_{k}\colon S_{n}(k)\to
S_{n-1}$ and $\psi_{k}\colon S^{-1}_{n}(k)\to S_{n-1}$, such that for all
$i=1,\dotsc,n-1$

$\displaystyle\varphi_{k}(\sigma)=\tau\quad\text{if\/f}\quad\sigma(i+1)=\begin{cases}\tau(i)&\text{if
$\tau(i)<k$},\\\ \tau(i)+1&\text{if $\tau(i)\geq k$},\end{cases}$
$\displaystyle\psi_{k}(\sigma)=\lambda\quad\text{if\/f}\quad\sigma^{-1}(i+1)=\begin{cases}\lambda^{-1}(i)&\text{if
$\lambda^{-1}(i)<k$},\\\ \lambda^{-1}(i)+1&\text{if $\lambda^{-1}(i)\geq
k$}.\end{cases}$ (4.12)

As an example, $\sigma=(123)=\left(\begin{smallmatrix}1&2&3\\\
2&3&1\end{smallmatrix}\right)\in S_{3}$ is in $S_{3}(2)$ and $S_{3}(3)$, with
$\varphi_{2}(\sigma)=(12)\in S_{2}$ and $\psi_{3}(\sigma)=\operatorname{id}\in
S_{2}$.

We have for any $\sigma\in S_{n}$,

$\displaystyle A_{\sigma}(w_{1},\dotsc,w_{n})$
$\displaystyle\qquad{}=\begin{cases}\displaystyle\prod^{k-1}_{j=1}\frac{qw_{k}-w_{j}}{w_{k}-qw_{j}}A_{\tau}(w_{1},\dotsc,w_{k-1},w_{k+1},w_{n})&\text{if
$\sigma(1)=k$ and $\varphi_{k}(\sigma)=\tau$},\\\
\displaystyle\prod^{k-1}_{j=1}\frac{qw_{\sigma(j)}-w_{1}}{w_{\sigma(j)}-qw_{1}}A_{\lambda}(w_{2},\dotsc,w_{n})&\text{if
$\sigma(k)=1$ and $\psi_{k}(\sigma)=\lambda$}.\end{cases}$ (4.13)

We write

$\displaystyle
u^{0}_{Y}(X;0)=\sum^{n}_{k=1}I^{(k)}_{Y}(X),\qquad\text{where}\quad
I^{(k)}_{Y}(X)=\sum_{\sigma\in S_{n}(k)}\Lambda_{Y}(X;0;\sigma),$

and compute $I^{(k)}_{Y}(X)$ for $k=1,\dotsc,n$, under the assumption that
(4.7) holds when $n$ is replaced by $n-1$.

In the simplest case $k=1$, it is straightforward. Note that if $\sigma\in
S_{n}(1)$, by (4.13) $A_{\sigma}(w_{1},\dotsc,w_{n})$ does not depend on
$w_{1}$, and we have ($(x_{1})$ is represented by $x_{1}$ as in Section 2)

$\displaystyle\left((-1)^{n}\prod^{n}_{i=1}b_{x_{i}}\right)I^{(1)}_{Y}(X)$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\sideset{}{{}^{\prime}}{\prod}^{x_{1}}_{l_{1}=y_{1}}\frac{b_{l_{1}}}{b_{l_{1}}-w_{1}}\sum_{\sigma\in
S_{n}(1)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{2}}dw_{2}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}A_{\sigma}(w_{1},\dotsc,w_{n})\prod^{n}_{j=2}\sideset{}{{}^{\prime}}{\prod}^{x_{j}}_{l_{j}=y_{\sigma(j)}}\frac{b_{l_{j}}}{b_{l_{j}}-w_{\sigma(j)}}$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\sideset{}{{}^{\prime}}{\prod}^{x_{1}}_{l_{1}=y_{1}}\frac{b_{l_{1}}}{b_{l_{1}}-w_{1}}\sum_{\tau\in
S_{n-1}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{2}}dw_{2}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}A_{\tau}(w_{2},\dotsc,w_{n})\prod^{n-1}_{j=1}\sideset{}{{}^{\prime}}{\prod}^{x_{j+1}}_{l_{j}=y_{\tau(j)+1}}\frac{b_{l_{j}}}{b_{l_{j}}-w_{\tau(j)+1}}$
$\displaystyle=\left((-1)^{n}\prod^{n}_{i=1}b_{x_{i}}\right)u^{0}_{y_{1}}(x_{1};0)u^{0}_{(y_{2},\dotsc,y_{n})}((x_{2},\dotsc,x_{n});0).$

So by the inductive assumption,

$\displaystyle I^{(1)}_{Y}(X)=\begin{cases}W((x_{2},\dotsc,x_{n}))&\text{if
$X=Y$},\\\ 0&\text{if $X\in\mathbb{W}^{n}{\setminus}\\{Y\\}$}.\end{cases}$

In the general case $k>1$, for all $\sigma\in S_{n}(k)$, if
$S_{n-1}\ni\tau=\varphi_{k}(\sigma)$, we have

$\displaystyle\left((-1)^{n}\prod^{n}_{i=1}b_{x_{i}}\right)\Lambda_{Y}(X;0;\sigma)$
$\displaystyle\qquad{}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{k-1}}dw_{k-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{k+1}}dw_{k+1}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}A_{\tau}(w_{1},\dotsc,w_{k-1},w_{k+1},\dotsc,w_{n})$
$\displaystyle\qquad\quad{}\times\prod^{n}_{j=2}\sideset{}{{}^{\prime}}{\prod}^{x_{j}}_{l_{j}=y_{\sigma(j)}}\frac{b_{l_{j}}}{b_{l_{j}}-w_{\sigma(j)}}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{k}}\prod^{k-1}_{i=1}\frac{qw_{k}-w_{j}}{qw_{j}-w_{k}}\sideset{}{{}^{\prime}}{\prod}^{x_{1}}_{l_{1}=y_{k}}\frac{b_{l_{1}}}{b_{l_{1}}-w_{k}}dw_{k}\right).$
(4.14)

The inner integral vanishes if $x_{1}>y_{k}$, so $\Lambda_{Y}(X;0;\sigma)=0$
if $x_{1}>y_{k}$. To see it, we deform the contour $C_{k}$ into a circle
centred at $0$ with radius $R_{k}\to\infty$ while keeping the other $C_{j}$
($j\neq k$) in place, and use the estimate that the integrand of the inner
integral is $\mathcal{O}\big{(}w^{y_{k}-x_{1}-1}_{k}\big{)}$ as
$w_{k}\to\infty$. Note that this deformation does not change the integral on
the right-hand side of (4.14), since the only requirement for $C_{k}$ is that
it is big enough to enclose all possible poles $b_{l_{1}}$.

Next, we assume that $\sigma\in S^{-1}_{n}(m)$, and
$\psi_{m}(\sigma)=\lambda$. Then we have

$\displaystyle\left((-1)^{n}\prod^{n}_{i=1}b_{x_{i}}\right)\Lambda_{Y}(X;0;\sigma)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{2}}dw_{2}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}A_{\lambda}(w_{2},\dotsc,w_{n})\prod^{n}_{j=2}\sideset{}{{}^{\prime}}{\prod}^{x_{\sigma^{-1}(j)}}_{l_{j}=y_{j}}\frac{b_{l_{j}}}{b_{l_{j}}-w_{j}}$
$\displaystyle\hphantom{\left((-1)^{n}\prod^{n}_{i=1}b_{x_{i}}\right)\Lambda_{Y}(X;0;\sigma)=}{}\times\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}\prod^{k-1}_{j=1}\frac{qw_{\sigma(j)}-w_{1}}{w_{\sigma(j)}-qw_{1}}\sideset{}{{}^{\prime}}{\prod}^{x_{m}}_{l_{1}=y_{1}}\frac{b_{l_{1}}}{b_{l_{1}}-w_{1}}dw_{1}\right).$
(4.15)

If $x_{m}<y_{1}$, then the inner integral in (4.15) vanishes, for its
integrand has no pole within $C_{1}$. So $\Lambda_{Y}(X;0;\sigma)=0$ if
$x_{m}<y_{1}$.

Since $x_{1}\geq\dotsb\geq x_{m}$ and $y_{1}\geq\dotsb\geq y_{k}$, by argument
above, we see that $\Lambda_{Y}(X;0;\sigma)=0$ unless
$x_{1}=\dotsb=x_{m}=y_{1}=\dotsb=y_{k}$, and in particular,
$x_{1}=y_{1}=\dotsb=y_{k}$.

When $x_{1}=y_{1}=\dotsb=y_{k}$, by deforming the contour $C_{k}$ into a big
circle centred at $0$ with radius $R_{k}\to\infty$, the inner integral in
(4.14) can be explicitly computed as

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{k}}\prod^{k-1}_{i=1}\frac{qw_{k}-w_{j}}{w_{k}-qw_{j}}\frac{b_{x_{1}}}{b_{x_{1}}-w_{k}}dw_{k}=-q^{k-1}b_{x_{1}},$
(4.16)

since its integrand is
$-q^{k-1}b_{x_{1}}w^{-1}_{k}\big{(}1+\mathcal{O}\big{(}w^{-1}_{k}\big{)}\big{)}$.
Plugging in (4.16) into (4.14), we have

$\displaystyle\left((-1)^{n}\prod^{n}_{i=1}b_{x_{i}}\right)\Lambda_{Y}(X;0;\sigma)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{k-1}}dw_{k-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{k+1}}dw_{k+1}\dotsi$
$\displaystyle\qquad\quad{}\times\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}A_{\tau}(w_{1},\dotsc,w_{k-1},w_{k+1},\dotsc,w_{n})\prod^{n}_{j=2}\sideset{}{{}^{\prime}}{\prod}^{x_{j}}_{l_{j}=y_{\sigma(j)}}\frac{b_{l_{j}}}{b_{l_{j}}-w_{\sigma(j)}}q^{k-1}b_{x_{1}}$
$\displaystyle\qquad{}=q^{k-1}\left((-1)^{n}\prod^{n}_{i=1}b_{x_{i}}\right)\Lambda_{(y_{2},\dotsc,y_{n})}((x_{2},\dotsc,x_{n});0;\tau),$
(4.17)

where $\tau=\varphi_{k}(\sigma)$. Thus summing up with $\sigma\in S_{n}(k)$,
we have that $\tau$ runs over $S_{n-1}$ and

$\displaystyle
I^{(k)}_{Y}(X)=q^{k-1}u^{0}_{(y_{2},\dotsc,y_{n})}((x_{2},\dotsc,x_{n});0).$

If $X\neq Y$, then the assumption $x_{1}=y_{1}=\dotsb=y_{k}$ implies
$(x_{2},\dotsc,x_{n})\neq(y_{2},\dotsc,y_{n})$, and by inductive assumption
have that $I^{(k)}_{Y}(X)=q^{k-1}\cdot 0=0$. Otherwise, $X=Y$, and then the
assumption $x_{1}=y_{1}=\dotsb=y_{k}$ holds only if $k\leq k_{1}$, if $X$ is
expressed in the form of (1.3).

The result derived above can be summarized as follows: If $X$ is expressed by
(1.3), then by the inductive assumption,

$\displaystyle I^{(k)}_{X}(X)=\begin{cases}0&\text{if
$X\in\mathbb{W}^{n}{\setminus}\\{Y\\}$},\\\ 0&\text{if $X=Y$ and
$k>k_{1}$},\\\ q^{k-1}W((x_{2},\dotsc,x_{n}))&\text{if $X=Y$ and $k\leq
k_{1}$},\end{cases}$

and hence $u^{0}_{Y}(X)=0$ if $X\in\mathbb{W}^{n}{\setminus}\\{Y\\}$ and
$u^{0}_{X}(X)=\big{(}1+q+\dotsb+q^{k_{1}-1}\big{)}W((x_{2},\dotsc,x_{n}))=W(X)$,
and we prove (4.7). ∎

Combining Lemmas 4.2 and 4.1, we prove Theorem 1.1 in the general $n$ particle
case.

##  5 Proof of Corollary 1.3

By Theorem 1.1, we have

$\displaystyle
P_{(0,\dotsc,0)}(X;t)=\frac{1}{W(X)}\left(\prod^{n}_{k=1}\frac{-1}{b_{x_{k}}}\right)\sum_{\sigma\in
S_{n}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi$
$\displaystyle\hphantom{P_{(0,\dotsc,0)}(X;t)=}{}\times\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}A_{\sigma}(w_{1},\dotsc,w_{n})\prod^{n}_{j=1}\left[\sideset{}{{}^{\prime}}{\prod}^{x_{j}}_{k=0}\left(\frac{b_{k}}{b_{k}-w_{\sigma(j)}}\right)e^{-w_{\sigma(j)}t}\right]$
$\displaystyle\hphantom{P_{(0,\dotsc,0)}(X;t)}{}=\frac{1}{W(X)}\left(\prod^{n}_{k=1}\frac{-1}{b_{x_{k}}}\right)\sum_{\sigma\in
S_{n}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{1}}dw_{1}\dotsi$
$\displaystyle\hphantom{P_{(0,\dotsc,0)}(X;t)=}{}\times\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{C_{n}}dw_{n}A_{\sigma}(w_{\sigma^{-1}(1)},\dotsc,w_{\sigma^{-1}(n)})\prod^{n}_{j=1}\left[\sideset{}{{}^{\prime}}{\prod}^{x_{j}}_{k=0}\left(\frac{b_{k}}{b_{k}-w_{j}}\right)e^{-w_{j}t}\right],$

where we utilize that the contours $C_{j}$ can be taken as identical. Hence it
is clear that if we can prove

$\displaystyle\sum_{\sigma\in
S_{n}}A_{\sigma}(w_{\sigma^{-1}(1)},\dotsc,w_{\sigma^{-1}(n)})=[n]_{q}!B(w_{1},\dotsc,w_{n}),$
(5.1)

then (1.8) is a direct consequence of (1.6). We prove (5.1) by induction. For
$n=1,2$, (5.1) holds obviously. Suppose (5.1) holds with $n$ replaced by
$n-1$. Recall the notations $S^{-1}_{n}(k)$ defined in (4.11) and $\psi_{k}$
defined in (4.12) for $k=1,\dotsc,n$, and let $v^{(k)}_{i}=w_{i}$ for
$i=1,\dotsc,k-1$ and $v^{(k)}_{i}=w_{i+1}$ for $i=k,\dotsc,n-1$. We have

$\displaystyle\sum_{\sigma\in
S^{-1}_{n}(k)}A_{\sigma}(w_{\sigma^{-1}(1)},\dotsc,w_{\sigma^{-1}(n)})$
$\displaystyle=\prod^{k-1}_{j=1}\frac{qw_{j}-w_{k}}{w_{j}-qw_{k}}\sum_{\sigma\in
S^{-1}_{n}(k)}A_{\psi_{k}(\sigma)}(w_{\sigma^{-1}(2)},\dotsc,w_{\sigma^{-1}(n)})$
$\displaystyle=\prod^{k-1}_{j=1}\frac{qw_{j}-w_{k}}{w_{j}-qw_{k}}\sum_{\lambda\in
S_{n-1}}A_{\lambda}\big{(}v^{(k)}_{\lambda^{-1}(1)},\dotsc,v^{(k)}_{\lambda^{-1}(n-1)}\big{)}$
$\displaystyle=[n-1]_{q}!\prod^{k-1}_{j=1}\frac{qw_{j}-w_{k}}{w_{j}-qw_{k}}B\big{(}v^{(k)}_{1},\dotsc,v^{(k)}_{n-1}\big{)},$

and then

$\displaystyle\sum_{\sigma\in
S_{n}}A_{\sigma}(w_{\sigma^{-1}(1)},\dotsc,w_{\sigma^{-1}(n)})=\sum^{n}_{k=1}\left(\sum_{\sigma\in
S^{-1}_{n}(k)}A_{\sigma}(w_{\sigma^{-1}(1)},\dotsc,w_{\sigma^{-1}(n)})\right)$
$\displaystyle\qquad{}=[n-1]_{q}!\sum^{n}_{k=1}\prod^{k-1}_{j=1}\frac{qw_{j}-w_{k}}{w_{j}-qw_{k}}B\big{(}v^{(k)}_{1},\dotsc,v^{(k)}_{n-1}\big{)}$
$\displaystyle\qquad{}=[n-1]_{q}!\sum^{n}_{k=1}\prod^{k-1}_{j=1}\frac{qw_{j}-w_{k}}{w_{j}-qw_{k}}\left(\prod^{k-1}_{j=1}\frac{qw_{k}-w_{j}}{w_{k}-w_{j}}\prod^{n}_{l=k+1}\frac{qw_{l}-w_{k}}{w_{l}-w_{k}}B(w_{1},\dotsc,w_{n})\right)$
$\displaystyle\qquad{}=[n-1]_{q}!C(w_{1},\dotsc,w_{n})B(w_{1},\dotsc,w_{n}),$

where

$\displaystyle C(w_{1},\dotsc,w_{n})=\sum^{n}_{k=1}\prod_{j=1,\dotsc,n,\ j\neq
k}\frac{qw_{j}-w_{k}}{w_{j}-w_{k}}.$

To prove (5.1), it suffices to show

$\displaystyle
C(w_{1},\dotsc,w_{n})=\frac{[n]_{q}!}{[n-1]_{q}!}=\frac{1-q^{n}}{1-q}.$ (5.2)

Denoting the function $h(z)=z^{-1}\prod\limits^{n}_{k=1}(z-qw_{k})/(z-w_{k})$
that is meromorphic in $z$, we have

$\displaystyle(1-q)C(w_{1},\dotsc,w_{n})=\sum^{n}_{k=1}\operatorname*{Res}_{z=w_{k}}h(z)=-\operatorname*{Res}_{z=\infty}h(z)-\operatorname*{Res}_{z=0}h(z)=1-q^{n}.$

Thus we verify (5.2) and finish the proof of the corollary. (Inspired by an
anonymous referee’s suggestion, the authors find that (5.2) can be proved
algebraically by taking $z=0$ in [12, the first formula on p. 210].)

### Acknowledgements

The authors are indebted to Ivan Corwin for numerous comments, especially on
the relation to the spectral theory and on the proof of Corollary 1.3. We also
thank Eunghyun Lee for valuable suggestions and comments. At last, we thank
the anonymous referees for their comments, suggestions and corrections,
especially on a sign error in the manuscript. The first named author is
supported partially by the startup grant R-146-000-164-133.

## References

  * [1] Balázs M., Komjáthy J., Order of current variance and diffusivity in the rate one totally asymmetric zero range process, J. Stat. Phys. 133 (2008), 59–78, arXiv:0804.1397.
  * [2] Borodin A., Corwin I., Macdonald processes, Probab. Theory Related Fields 158 (2014), 225–400, arXiv:1111.4408.
  * [3] Borodin A., Corwin I., Petrov L., Sasamoto T., Spectral theory for interacting particle systems solvable by coordinate Bethe ansatz, Comm. Math. Phys. 339 (2015), 1167–1245, arXiv:1407.8534.
  * [4] Borodin A., Corwin I., Petrov L., Sasamoto T., Spectral theory for the $q$-Boson particle system, Compos. Math. 151 (2015), 1–67, arXiv:1308.3475.
  * [5] Borodin A., Corwin I., Sasamoto T., From duality to determinants for $q$-TASEP and ASEP, Ann. Probab. 42 (2014), 2314–2382, arXiv:1207.5035.
  * [6] Borodin A., Petrov L., Higher spin six vertex model and symmetric rational functions, arXiv:1601.05770.
  * [7] Corwin I., The Kardar–Parisi–Zhang equation and universality class, Random Matrices Theory Appl. 1 (2012), 1130001, 76 pages, arXiv:1106.1596.
  * [8] Gwa L.-H., Spohn H., Six-vertex model, roughened surfaces, and an asymmetric spin Hamiltonian, Phys. Rev. Lett. 68 (1992), 725–728.
  * [9] Korhonen M., Lee E., The transition probability and the probability for the left-most particle’s position of the $q$-totally asymmetric zero range process, J. Math. Phys. 55 (2014), 013301, 15 pages, arXiv:1308.4769.
  * [10] Lee E., Transition probabilities of the Bethe ansatz solvable interacting particle systems, J. Stat. Phys. 142 (2011), 643–656, arXiv:1011.2841.
  * [11] Liggett T.M., Interacting particle systems, Classics in Mathematics, Springer-Verlag, Berlin, 2005.
  * [12] Macdonald I.G., Symmetric functions and Hall polynomials, 2nd ed., Oxford Mathematical Monographs, The Clarendon Press, Oxford University Press, New York, 1995.
  * [13] Povolotsky A.M., On the integrability of zero-range chipping models with factorized steady states, J. Phys. A: Math. Theor. 46 (2013), 465205, 25 pages, arXiv:1308.3250.
  * [14] Quastel J., Introduction to KPZ, in Current Developments in Mathematics, 2011, Int. Press, Somerville, MA, 2012, 125–194.
  * [15] Sasamoto T., Wadati M., Exact results for one-dimensional totally asymmetric diffusion models, J. Phys. A: Math. Gen. 31 (1998), 6057–6071.
  * [16] Schütz G.M., Exact solution of the master equation for the asymmetric exclusion process, J. Stat. Phys. 88 (1997), 427–445, cond-mat/9701019.
  * [17] Spitzer F., Interaction of Markov processes, Adv. Math. 5 (1970), 246–290.
  * [18] Tracy C.A., Widom H., Integral formulas for the asymmetric simple exclusion process, Comm. Math. Phys. 279 (2008), 815–844, arXiv:0704.2633.
  * [19] Tracy C.A., Widom H., Asymptotics in ASEP with step initial condition, Comm. Math. Phys. 290 (2009), 129–154, arXiv:0807.1713.