On the TASEP with Second Class Particles
In this paper, we study some conditional probabilities for the totally asymmetric simple exclusion processes (TASEP) with second-class particles. To be more specific, we consider a finite system with one first-class particle and N−1 second-class particles, and we assume that the first-class particle...
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Інститут математики НАН України
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| Цитувати: | On the TASEP with Second Class Particles / E. Lee // Symmetry, Integrability and Geometry: Methods and Applications. — 2018. — Т. 14. — Бібліогр.: 18 назв. — англ. |
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| citation_txt | On the TASEP with Second Class Particles / E. Lee // Symmetry, Integrability and Geometry: Methods and Applications. — 2018. — Т. 14. — Бібліогр.: 18 назв. — англ. |
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| container_title | Symmetry, Integrability and Geometry: Methods and Applications |
| description | In this paper, we study some conditional probabilities for the totally asymmetric simple exclusion processes (TASEP) with second-class particles. To be more specific, we consider a finite system with one first-class particle and N−1 second-class particles, and we assume that the first-class particle is initially at the leftmost position. In this case, we find the probability that the first class particle is at x and it is still the leftmost particle at time t. In particular, we show that this probability is expressed by the determinant of an N×N matrix of contour integrals if the initial positions of particles satisfy the step initial condition. The resulting formula is very similar to a known formula in the (usual) TASEP with the step initial condition, which was used for asymptotics by Nagao and Sasamoto [Nuclear Phys. B 699 (2004), 487-502].
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 14 (2018), 006, 17 pages
On the TASEP with Second Class Particles
Eunghyun LEE
Department of Mathematics, Nazarbayev University, Kazakhstan
E-mail: eunghyun.lee@nu.edu.kz
URL: https://sites.google.com/a/nu.edu.kz/eunghyun-lee-s-homepage/
Received August 08, 2017, in final form January 08, 2018; Published online January 12, 2018
https://doi.org/10.3842/SIGMA.2018.006
Abstract. In this paper we study some conditional probabilities for the totally asymmetric
simple exclusion processes (TASEP) with second class particles. To be more specific, we
consider a finite system with one first class particle and N − 1 second class particles, and
we assume that the first class particle is initially at the leftmost position. In this case, we
find the probability that the first class particle is at x and it is still the leftmost particle at
time t. In particular, we show that this probability is expressed by the determinant of an
N ×N matrix of contour integrals if the initial positions of particles satisfy the step initial
condition. The resulting formula is very similar to a known formula in the (usual) TASEP
with the step initial condition which was used for asymptotics by Nagao and Sasamoto
[Nuclear Phys. B 699 (2004), 487–502].
Key words: TASEP; Bethe ansatz; second class particles
2010 Mathematics Subject Classification: 60J25; 60K35; 82B23
1 Introduction
In this paper we study some conditional probabilities in the totally asymmetric simple exclusion
processes (TASEP) with second class particles. So far, there have been many works on the
coordinate Bethe Ansatz applicable stochastic particle models, but most of them are on the
models of a single species such as the asymmetric simple exclusion processes (ASEP) [14, 15], the
q-totally asymmetric simple exclusion processes (q-TASEP) [2], the q-Hahn asymmetric exclusion
processes [1, 7] and the q-totally asymmetric zero range processes (q-TAZRP) [6, 9, 13, 18].
Our interests in these models include but are not limited to the transition probabilities, the
probability distribution of a tagged particle’s position for some special initial conditions, and
their asymptotics that may confirm that the models belong to the KPZ universality class.
However, in this direction of studies, the TASEP with second class particles have not been
explored much, and only a limited number of papers were published [3, 16, 17] because of
the complexity in computations. The transition probabilities of the TASEP with second class
particles were studied by Chatterjee and Schütz [3] and also studied in more general setting (the
ASEP with multiple species particles) by Tracy and Widom [17]. A motivation of this paper
starts from naive questions that “Can we extend the results in the TASEP (such as the one
point distributions for some initial conditions and their asymptotics) to the TASEP with second
class particles?” “Furthermore, what observables are meaningful and should be considered in
connection to the KPZ universality class?” In this paper, we do not provide complete answers
to these questions but would like to study some explicit and exact formulas that can serve as
a starting point for further studies. The integrability of the TASEP with second class particles
has been confirmed in [3, 17], hence as the next step, we are interested in finding a neat
probability formula of a certain event from which we hope to obtain a meaningful asymptotic
result. We provide a probability formula (1.4) whose form is very similar to a known formula
in the TASEP without second class particles. But, it is not clear at this moment whether this
mailto:eunghyun.lee@nu.edu.kz
https://sites.google.com/a/nu.edu.kz/eunghyun-lee-s-homepage/
https://doi.org/10.3842/SIGMA.2018.006
2 E. Lee
high similarity in the appearances will imply easy asymptotics or not. Although we consider the
TASEP with second class particles for algebraic simplicity, it seems that the techniques used in
this paper can be also used to extend the results in this paper to the ASEP with second class
particles. The author confirmed this extension for some small systems of the ASEP with second
class particles.
The definition of the TASEP with second class particles is as follows. Each site on Z can
be occupied by at most one particle and each particle belongs to one of two different species,
labeled 1 or 2. A particle of species i at x ∈ Z tries to jump to x + 1 after a waiting time
exponentially distributed with rate 1. If the target site x + 1 is empty, the particle jumps
to x+ 1. But, if x+ 1 is not empty, one of the following cases occurs: (i) if x+ 1 is occupied by
another particle of the same species i, the particle of species i at x cannot jump to x + 1, and
the waiting time is reset, (ii) if x is occupied by a particle of species 2 and x+ 1 is occupied by
a particle of species 1, then the particle of species 2 at x can jump to x+ 1 by interchanging the
positions with the particle of species 1 at x + 1, (iii) if x is occupied by a particle of species 1
and x+ 1 is occupied by a particle of species 2, then the particle of species 1 cannot jump
to x+ 1 and the waiting time is reset (see Fig. 1).
1 2
2 1 1 2
1 1 2 2
=⇒
×
× ×
(i)
(ii)
(iii)
Figure 1.
In other words, the particles of species 2 have a priority over those of species 1. The particles
of species 1 are called second class particles and the particles of species 2 are called first class
particles.
1.1 Statement of the main results
In this paper we consider a finite system with N particles. The state space of the process is
a countable set of pairs (X,π) where X = (x1, . . . , xN ) ∈WN with
WN =
{
(x1, . . . , xN ) : (x1, . . . , xN ) ∈ ZN and x1 < · · · < xN
}
and π = (π1 · · ·πN ) is a finite sequence of length N whose elements are 1 or 2. A state(
(x1, . . . , xN ), (π1 · · ·πN )
)
implies that the ith particle from the left is at xi and this particle belongs to species πi. We will
sometimes omit the parentheses in the expressions of (x1, . . . , xN ) and (π1 · · ·πN ). If we would
like to express a state at time t, we write(
(x1, . . . , xN ), (π1 · · ·πN ); t
)
We denote by P(Y,ν)(X,π; t) a transition probability, that is, the probability that the system is
at state (X,π) at time t, given that the initial state is (Y, ν). We denote by P(Y,ν) the probability
measure of the process with initial state (Y, ν). Let Et = {(X, 21 . . . 1; t) : x1 = x}, the event
that the leftmost particle is the first class particle and it is at x at time t.
On the TASEP with Second Class Particles 3
Remark 1.1. One of the motivations of this paper is the fact that if π = ν, then the transition
probabilities P(Y,ν)
(
(x1, . . . , xN ), ν; t
)
can be expressed as a determinant of an N×N matrix [3].
In the TASEP with N particles of the same kind, the transition probabilities are also expressed
as a determinant [14] and the distribution of the leftmost particle’s position is obtained by
summing the determinants over all possible x2, . . . , xN with x1 < x2 < · · · < xN for fixed x1.
This multiple sum of the determinants can be expressed as a closed form of a multiple contour
integral when the initial positions of particles are given by (1, 2, . . . , N) [12, 15]. This was
possible due to some special properties of the determinants [12, Theorem 4] or equivalently due
to the algebraic identity which was found in more general setting in [15, equation (1.6)]. Hence,
it is natural to ask if there is a similar algebraic structure originated from the determinantal
form of P(Y,ν)(X, ν; t) in the TASEP with second class particles, and (if so) whether or not it is
possible to express a multiple sum of P(Y,ν)(X, ν; t) over x2, . . . , xN with x1 < x2 < · · · < xN for
fixed x1 as a closed form of a multiple contour integral.
Throughout this paper, we will use the notation −
∫
for 1
2πi
∫
. The first result provides an
answer to the question “if the first class particle is initially the leftmost particle, what is the
probability that the first particle moves to x during time t without exchanging positions with
second class particles?”
Theorem 1.2. Let C be a counterclockwise circle centered at the origin with radius less than 1.
Then,
P(Y,21...1)(Et) = −
∫
C
· · · −
∫
C
(1− ξ1)
∏
1≤i<j≤N
ξj − ξi
1− ξi
×
N∏
i=1
1
1− ξi
N∏
i=1
(
ξx−yi−1i eε(ξi)t
)
dξ1 · · · dξN , (1.1)
where ε(ξi) = 1/ξi − 1.
This formula is comparable with that of the TASEP (with a single species). The formula for
the TASEP corresponding to (1.1) is
P(Y,i...i)(Ft) = −
∫
C
· · · −
∫
C
(1− ξ1 · · · ξN )
∏
1≤i<j≤N
ξj − ξi
1− ξi
×
N∏
i=1
1
1− ξi
N∏
i=1
(
ξx−yi−1i eε(ξi)t
)
dξ1 · · · dξN ,
where Ft = {(X, i . . . i; t) : x1 = x} [15].
Remark 1.3. In order to obtain (1.1), we need to find an expression of P(Y,21...1)(X, 21 . . . 1; t).
One novel point of this paper is that we provide an explicit and exact formula of the transition
probability P(Y,21...1)(X, 21 . . . 1; t). Chatterjee and Schütz [3], and Tracy and Widom [17] dis-
cussed a method to obtain P(Y,ν)(X,π; t) but did not provide an explicit formula for specific ν
and π. The transition probabilities of an N -particle system (with or without second class par-
ticles) are expressed in a compact matrix form of contour integrals as in (2.31), and each matrix
element is a specific transition probability. Finding an explicit formula of a matrix element
of PY (X; t) in (2.31) means finding an explicit formula of a matrix element of Aσ in (2.31). But
this may require a tedious matrix multiplication of several 2N by 2N matrices. To the best of the
author’s knowledge, there are no known shortcut methods to find each matrix element of Aσ.
Lemma 3.3 is an important result to provide an explicit formula of P(Y,21...1)(X, 21 . . . 1; t).
4 E. Lee
The second result is obtained by applying a special initial condition for the positions of
particles to (1.1). Let Y = (y1, . . . , yN ) be the initial positions of particles given by
yi =
{
1 if i = 1,
i+ l if i > 1
(1.2)
for some nonnegative integer l. When l = 0, this condition is called the step initial condition.
Let us recall that
hl(ξ1, . . . , ξN ) =
∑
1≤ii≤···≤il≤N
ξi1ξi2 · · · ξil
is called the lth complete symmetric polynomial of ξ1, . . . , ξN , and h0(ξ1, . . . , ξN ) = 1.
Theorem 1.4. If Y is given by (1.2),
P(Y,21...1)(Et) =
(−1)N(N−1)/2
N !
−
∫
C
· · · −
∫
C
hl(ξ1, . . . , ξN )
∏
1≤i<j≤N
(ξj − ξi)2
N∏
i=1
1
(ξi − 1)N−1
×
N∏
i=1
(
ξx−N−l−1i eε(ξi)t
)
dξ1 · · · dξN . (1.3)
If Y is given by the step initial condition, then (1.3) is very similar to a known formula in
the TASEP [15, Remark on p. 839]. (Also, see [4, 12] for similar results.)
Corollary 1.5. If Y is given by (1.2) with l = 0, then
P(Y,21...1)(Et) =
(−1)N(N−1)/2
N !
−
∫
C
· · · −
∫
C
∏
1≤i<j≤N
(ξj − ξi)2
N∏
i=1
1
(ξi − 1)N−1
×
N∏
i=1
(
ξx−N−1i eε(ξi)t
)
dξ1 · · · dξN . (1.4)
Also, (1.4) can be written as
P(Y,21...1)(Et) = (−1)N(N−1)/2 det
[
−
∫
C
ξi+j+x−N−1(ξ − 1)−(N−1)e(1/ξ−1)tdξ
]N−1
i,j=0
by the same method as in [15, Remark on p. 839].
Remark 1.6. It would be interesting to see if P(Y,ν)(E
π
t ) where Eπt = {(X,π; t) : x1 = x} for
arbitrary ν and π can be expressed as a determinant. To do so, at first, we need to find an
explicit formula of P(Y,ν)(X,π; t). However, as mentioned in Remark 1.3, there are no known
methods to find P(Y,ν)(X,π; t) except a tedius matrix computation. A partial generalization of
P(Y,21...1)(Et) was made a few months after the first version of this paper was posted [8].
1.2 Previous results
There are some previous results on the TASEP (or the ASEP) with a different initial configura-
tion. Suppose that initially all negative integers are occupied by first class particles, the origin is
occupied by a second class particle, and all positive integers are empty. In this case, the position
of the second class particle at time t is nontrivial because it is affected by first class particles’
On the TASEP with Second Class Particles 5
movements. Tracy and Widom obtained the exact expression of the distribution of the position
of the second class particle at time t in the ASEP [16]. Mountford and Guiol proved the strong
law of large numbers for the position of the second class particle in the TASEP [11]. On the other
hand, for the initial condition considered in this paper, the probability distribution of the position
of the (only) first class particle at time t can be trivially obtained because it is a free particle if
we do not care about the order of particles. But, the probability for the non-intersecting paths of
the particles from the initial configuration to any configurations such that the leftmost particle
(the first class particle) is at a specific position at time t is nontrivial as in Theorem 1.2.
Remark 1.7. The initial configuration (Y, ν) for Y = (. . . ,−2,−1) and ν = (. . . 221 . . . 1), and
the probability formula of the position of the mth rightmost first class particle at time t are of
particular interest. If we do not care about the order of the first particles and the second class
particle at time t, the system can be thought of as the TASEP without second class particles
because the first class particles see the second class particles as holes. Hence, the distribution of
the mth rightmost first class particle’s position is trivially obtained from the previous results of
the TASEP and the Tracy–Widom distribution should appear in the asymptotic studies. If we
consider an event that the initial order of particles does not change over time, the probability
of the mth rightmost first class particle’s position at time t confined in this even is nontrivial.
Organization of the paper. In Section 2, we provide some preliminary background of the
integrability and the transition probabilities of the TASEP with second class particles. Although
the majority of Section 2 is essentially overlapped with [3, 17], we introduce this background for
the notations in this paper and self-containedness. In Section 3, we derive (1.1), (1.3), and (1.4).
2 Preliminary
Since the state space of the process is countable, we may view P(Y,ν)(X,π; t) as elements of an
infinite matrix. This infinite matrix is denoted by P(t) and we assume that P(Y,ν)(X,π; t) is an
element at the (Y, ν)th column and at the (X,π)th row. We denote a submatrix of P(t) with
elements P(Y,ν)(X,π; t) for fixed X and Y as PY (X; t). Hence, PY (X; t) is a 2N × 2N matrix.
We assume that the elements of PY (X; t) are listed in the reverse lexicographic order of finite
sequences ν and π from left to right and from top to bottom, respectively (See (2.2) below for
an example). If G is the generator of the process, then P(t) satisfies the forward equation
d
dt
P(t) = GP(t) (2.1)
and it is subject to the initial condition P(0) = I where I is the identity matrix. The definition
of the TASEP with second class particles implies that G = [ajk]j,k∈Z is a band matrix with
sup
j,k
|ajk| = N < ∞ (so, G induces a bounded operator on l2(Z)), and the solution of (2.1) is
simply given by P(t) = etG. We want to find an explicit formula of each matrix element of etG,
that is, the transition probabilities P(Y,ν)(X,π; t) by using the standard method, the coordinate
Bethe Ansatz [14, 15].
2.1 Two-particle system
By the definition of the TASEP with second class particles, (2.1) implies that a submatrix
PY (x1, x2; t) given by
P(Y,11)(x1, x2, 11; t) P(Y,12)(x1, x2, 11; t) P(Y,21)(x1, x2, 11; t) P(Y,22)(x1, x2, 11; t)
P(Y,11)(x1, x2, 12; t) P(Y,12)(x1, x2, 12; t) P(Y,21)(x1, x2, 12; t) P(Y,22)(x1, x2, 12; t)
P(Y,11)(x1, x2, 21; t) P(Y,12)(x1, x2, 21; t) P(Y,21)(x1, x2, 21; t) P(Y,22)(x1, x2, 21; t)
P(Y,11)(x1, x2, 22; t) P(Y,12)(x1, x2, 22; t) P(Y,21)(x1, x2, 22; t) P(Y,22)(x1, x2, 22; t)
(2.2)
6 E. Lee
satisfies one of the following equations, depending on (x1, x2):
d
dt
PY (x1, x2; t) = PY (x1 − 1, x2; t) + PY (x1, x2 − 1; t)
− 2PY (x1, x2; t), x1 < x2 − 1, (2.3)
d
dt
PY (x1, x2; t) = PY (x1 − 1, x2; t)−
1 0 0 0
0 1 −1 0
0 0 2 0
0 0 0 1
PY (x1, x2; t), x1 = x2 − 1. (2.4)
Let U(x1, x2; t) be a 4 × 4 matrix whose elements are functions defined on Z2 × [0,∞). If we
suppose that U(x1, x2; t) is a solution of
d
dt
U(x1, x2; t) = U(x1 − 1, x2; t) + U(x1, x2 − 1; t)− 2U(x1, x2; t) (2.5)
for all (x1, x2) ∈ Z2 and is subject to
U(x, x; t)− 2U(x, x+ 1; t) = −
1 0 0 0
0 1 −1 0
0 0 2 0
0 0 0 1
U(x, x+ 1; t) for all x ∈ Z,
so that
U(x, x; t) =
1 0 0 0
0 1 1 0
0 0 0 0
0 0 0 1
U(x, x+ 1; t) := BU(x, x+ 1; t), (2.6)
then it is obvious that U(x1, x2; t) for x1 < x2 − 1 satisfies (2.3), and both (2.5) and (2.6)
imply that U(x1, x2; t) for x = x1 = x2 − 1 satisfies (2.4). An ansatz of variables separation,
U(x1, x2; t) = X(x1, x2)T (t) where X(x1, x2) is a 4× 4 matrix and T (t) is a scalar function of t,
reduces the matrix equation (2.5) to a linear matrix recurrence relation of two variables
εX(x1, x2) = X(x1 − 1, x2) + X(x1, x2 − 1)− 2X(x1, x2) (2.7)
and T (t) = eεt where ε is to be determined. By an ansatz from the theory of the linear recurrence
relation we see that I4ξ
x1
1 ξx22 is a solution of (2.7) where I4 is the 4× 4 identity matrix (we will
write In for the n × n identity matrix), and ξ1 and ξ2 are complex numbers with 0 < |ξ1| < 1
and 0 < |ξ2| < 1 (the reason for this restriction on ξ1 and ξ2 will be clear soon), and in this case,
ε is given by
ε =
1
ξ1
+
1
ξ2
− 2. (2.8)
In fact, by the linearity of the equation, for any 4 × 4 matrix A12, independent of x1 and x2,
A12ξ
x1
1 ξx22 is also a solution of (2.7). Also, we observe that I4ξ
x1
2 ξx21 is a solution of (2.7) with
the same ε in (2.8). Thus, for a general solution of (2.5), we put
U(x1, x2; t) =
(
A12ξ
x1
1 ξx22 + A21ξ
x1
2 ξx21
)
eεt, (2.9)
where A21 is another 4× 4 matrix, independent of x1 and x2. Substituting (2.9) into (2.6), we
obtain
(I4 − ξ2B)A12 = −(I4 − ξ1B)A21,
On the TASEP with Second Class Particles 7
and
A21 = −(I4 − ξ1B)−1(I4 − ξ2B)A12 =
−1−ξ2
1−ξ1 0 0 0
0 −1−ξ2
1−ξ1
ξ2−ξ1
1−ξ1 0
0 0 −1 0
0 0 0 −1−ξ2
1−ξ1
A12. (2.10)
Definition 2.1.
Sβα :=
− 1−ξβ
1−ξα 0 0 0
0 − 1−ξβ
1−ξα
ξβ−ξα
1−ξα 0
0 0 −1 0
0 0 0 − 1−ξβ
1−ξα
and Sβα := −
1− ξβ
1− ξα
. (2.11)
Hence, (2.9) with (2.10) satisfies (2.3) and (2.4). Next, we put A12 = I4ξ
−y1−1
1 ξ−y2−12 for
Y = (y1, y2) in (2.9) and integrate componentwise over counterclockwise circles C centered at
the origin with radii less than 1, to form a matrix of contour integrals written
−
∫
C
−
∫
C
(
ξx1−y1−11 ξx2−y2−12 I4 + ξx1−y2−12 ξx2−y1−11 S21
)
eεtdξ1dξ2. (2.12)
Recall that the matrix (2.2) satisfies the initial condition which states that at t = 0, if x1 = y1
and x2 = y2, then the matrix (2.2) is the identity matrix and, otherwise, it is the zero matrix.
Recalling that x1 ≥ y1, x2 ≥ y2, x1 < x2, and y1 < y2, we can show that (2.12) satisfies this
initial condition. Hence,
PY (x1, x2; t) = −
∫
C
−
∫
C
(
ξx1−y1−11 ξx2−y2−12 I4 + ξx1−y2−12 ξx2−y1−11 S21
)
eεtdξ1dξ2.
2.2 Three-particle system
Let U(x1, x2, x3; t) be an 8 × 8 matrix whose elements are functions defined on Z3 × [0,∞).
Suppose that U(x1, x2, x3; t) is a solution of
d
dt
U(x1, x2, x3; t) = U(x1 − 1, x2, x3; t) + U(x1, x2 − 1, x3; t)
+ U(x1, x2 − 1, x3 − 1; t)− 3U(x1, x2, x3; t) (2.13)
and is subject to
U(x, x, x3; t) = (B⊗ I2)U(x, x+ 1, x3; t) for all x, x3 ∈ Z,
U(x1, x, x; t) = (I2 ⊗B)U(x1, x, x+ 1; t) for all x1, x ∈ Z, (2.14)
where ⊗ means the tensor product of matrices. Then, as in N = 2, it is possible to show that
for each (x1, x2, x3) ∈W3, U(x1, x2, x3; t) satisfies the differential equation for PY (x1, x2, x3; t).
As the Bethe Ansatz solution of (2.13), we put
U(x1, x2, x3; t) =
∑
σ∈S3
Aσξ
x1
σ(1)ξ
x2
σ(2)ξ
x3
σ(3)e
εt, (2.15)
where
ε =
1
ξ1
+
1
ξ2
+
1
ξ3
− 3
8 E. Lee
and ξi, (i = 1, 2, 3) are nonzero complex numbers with 0 < |ξi| < 1, and Aσ are 8 × 8 matrices
of complex numbers independent of x1, x2, x3. Here, the sum is over all permutations σ in the
symmetric group S3. Substituting (2.15) into (2.14), we obtain
A213 = −(I8 −B⊗ I2ξ1)
−1(I8 −B⊗ I2ξ2)A123,
A312 = −(I8 −B⊗ I2ξ1)
−1(I8 −B⊗ I2ξ3)A132,
A321 = −(I8 −B⊗ I2ξ2)
−1(I8 −B⊗ I2ξ3)A231, (2.16)
and
A132 = −(I8 − I2 ⊗Bξ2)
−1(I8 − I2 ⊗Bξ3)A123,
A231 = −(I8 − I2 ⊗Bξ1)
−1(I8 − I2 ⊗Bξ3)A213,
A321 = −(I8 − I2 ⊗Bξ1)
−1(I8 − I2 ⊗Bξ2)A312. (2.17)
Using the properties of the tensor product of (invertible) linear operators
(A⊗B)(C⊗D) = AC⊗BD, (A⊗B)−1 = A−1 ⊗B−1,
and recalling the notation of Sβα in (2.11), we have
I2 ⊗ Sβα = −I2 ⊗ (I4 −Bξα)−1(I4 −Bξβ) = −(I8 − I2 ⊗Bξα)−1(I8 − I2 ⊗Bξβ), (2.18)
Sβα ⊗ I2 = −(I4 −Bξα)−1(I4 −Bξβ)⊗ I2 = −(I8 −B⊗ I2ξα)−1(I8 −B⊗ I2ξβ). (2.19)
Let α = σ(i), β = σ(i+1) and let Ti ∈ Sn be a simple transposition for any symmetric group Sn
such that (Tiσ)(i) = σ(i+ 1) and (Tiσ)(i+ 1) = σ(i), and (Tiσ)(k) = σ(k) for k 6= i, i+ 1. Then,
(2.16) and (2.17) can be simply written as
AT1σ = (Sβα ⊗ I2)Aσ, AT2σ = (I2 ⊗ Sβα)Aσ,
respectively. It can be shown that (2.18) and (2.19) satisfy the following relations by simple
computations
(i) (Sγβ ⊗ I2)(I2 ⊗ Sγα)(Sβα ⊗ I2) = (I2 ⊗ Sβα)(Sγα ⊗ I2)(I2 ⊗ Sγβ), (2.20)
(ii) (Sβα ⊗ I2)(Sαβ ⊗ I2) = I8 = (I2 ⊗ Sβα)(I2 ⊗ Sαβ). (2.21)
Hence, we have the following expressions for Aσ:
A213 = (S21 ⊗ I2)A123,
A132 = (I2 ⊗ S32)A123,
A312 = (S31 ⊗ I2)(I2 ⊗ S32)A123,
A231 = (I2 ⊗ S31)(S21 ⊗ I2)A123,
A321 = (S32 ⊗ I2)(I2 ⊗ S31)(S21 ⊗ I2)A123.
If we put A123 = I8
3∏
i=1
ξ−yi−1i and integrate (2.15) over counterclockwise circles C centered at
the origin with radii less than 1, then we obtain
PY (x1, x2, x3; t) = −
∫
C
−
∫
C
−
∫
C
∑
σ∈S3
Aσ
3∏
i=1
ξxiσ(i)e
εtdξ1dξ2dξ3.
On the TASEP with Second Class Particles 9
2.3 N -particle system
2.3.1 Bethe Ansatz solution
Let U(x1, . . . , xN ; t) be a 2N × 2N matrix whose elements are functions defined on ZN × [0,∞).
Suppose that U(x1, . . . , xN ; t) is a solution of
d
dt
U(x1, . . . , xN ; t) =
N∑
i=1
U(x1, . . . , xi−1, xi − 1, xi+1, . . . , xN ; t)−NU(x1, . . . , xN ; t)
and is subject to
U(x1, . . . , xi−1, xi, xi, xi+2, . . . , xN ; t) = I
⊗(i−1)
2 ⊗B⊗ I
⊗(N−i−1)
2
×U(x1, . . . , xi−1, xi, xi + 1, xi+2, . . . , xN ; t) for all i = 1, . . . , N − 1. (2.22)
Then, it is possible to show that for each (x1, . . . , xN ) ∈ WN , U(x1, . . . , xN ; t) satisfies the
differential equation for PY (x1, . . . , xN ; t). For the initial positions of particles Y = (y1, . . . , yN )
∈WN , we put the Bethe Ansatz solution
U(x1, . . . , xN ; t) =
∑
σ∈SN
Aσ
N∏
i=1
(
ξ
xi−yσ(i)−1
σ(i) eε(ξi)t
)
, (2.23)
where
ε(ξi) =
1
ξi
− 1
and ξi, i = 1, . . . , N , are complex numbers with 0 < |ξi| < 1, and Aσ’s are 2N × 2N matrices of
complex numbers.
2.3.2 Matrices Aσ
Let Ti, i = 1, . . . , N − 1, be simple transpositions in SN . It is well known that T1, . . . , TN−1
generate SN and satisfy the following relations [5, Chapter 4]: for all i, j = 1, . . . , N − 1,
TiTj = TjTi if |i− j| ≥ 2,
TiTjTi = TjTiTj if |i− j| = 1,
T 2
i = 1. (2.24)
Substituting (2.23) into (2.22), we see that (2.23) satisfies (2.22) provided that
ATiσ = −
(
I2N − I
⊗(i−1)
2 ⊗B⊗ I
⊗(N−i−1)
2 ξβ
)−1(
I2N − I
⊗(i−1)
2 ⊗B⊗ I
⊗(N−i−1)
2 ξα
)
Aσ
=
(
I2(i−1) ⊗ Sβα ⊗ I2(N−i−1)
)
Aσ, (2.25)
where α = σ(i) and β = σ(i + 1) for all σ and all i = 1, . . . , N − 1. Since T1, . . . , TN−1
generate SN , for each given σ ∈ SN there exists a finite sequence (ai)
n
i=1 of integers 1, . . . , N −1
such that
σ = Tan · · ·Ta1 . (2.26)
(Here, the representation (2.26) is not unique because of (2.24).) Suppose that Tai in (2.26)
interchanges α and β, that is,
Tai(· · ·αβ · · · ) = (· · ·βα · · · ).
For each Tai we define a 2N × 2N matrix denoted by Tai = Tai(α, β) by the tensor product of
N − 2 identity matrices and Sβα as follows.
10 E. Lee
Definition 2.2. Let Sβα be given by (2.11). We define
Tl = Tl(α, β) := I
⊗(l−1)
2 ⊗ Sβα ⊗ I
⊗(N−l−1)
2
for l = 1, . . . , N − 1.
The matrices T1, . . . ,TN−1 satisfy the matrix version of the relations (2.24),
(i) Ti(α, β)Tj(γ, δ) = Tj(γ, δ)Ti(α, β) if |i− j| ≥ 2, (2.27)
(ii) Ti(β, γ)Tj(α, γ)Ti(α, β) = Tj(α, β)Ti(α, γ)Tj(β, γ) if |i− j| = 1, (2.28)
(iii) Ti(β, α)Ti(α, β) = I2N . (2.29)
Here, (ii) and (iii) generalize (2.20) and (2.21), respectively.
Definition 2.3. For given σ = Tan · · ·Ta1 ∈ SN we define
Aσ := Tan · · ·Ta1 . (2.30)
Here, σ may have a different representation σ = Tbm · · ·Tb1 but both Tan · · ·Ta1 and Tbm · · ·Tb1
define the same matrix because of (2.27), (2.28) and (2.29). Thus, (2.30) is well-defined.
Lemma 2.4. If Aσ is given by (2.30), then (2.25) holds. Hence (2.23) satisfies (2.22) for all
(x1, . . . , xN ) ∈ ZN .
Proof. Let σ = Tan · · ·Ta1 and Aσ = Tan · · ·Ta1 . Suppose that σ(i) = α and σ(i + 1) = β.
Since Tiσ = TiTan · · ·Ta1 and
Ti = I2(i−1) ⊗ Sβα ⊗ I2(N−i−1) ,
we immediately obtain
ATiσ = TiTan · · ·Ta1 = TiAσ =
(
I2(i−1) ⊗ Sβα ⊗ I2(N−i−1)
)
Aσ. �
Finally, if we integrate (2.23) over circles centered at the origin with radii less than 1, we
obtain
PY (X; t) = −
∫
C
· · · −
∫
C
∑
σ∈SN
Aσ
N∏
i=1
(
ξ
xi−yσ(i)−1
σ(i) eε(ξi)t
)
dξ1 · · · dξN . (2.31)
The proof of
PY (X; 0) =
{
I2N if X = Y,
0 if X 6= Y
is given in [17].
3 Proofs of the main results
3.1 Proof of Theorem 1.2
To prove Theorem 1.2, we need to sum P(Y,21...1)(X, 21 . . . 1; t) over all allowed configurations,
that is, we need to compute
P(Y,21...1)(Et) =
∑
x=x1<···<xN
P(Y,21...1)(X, 21 . . . 1; t). (3.1)
On the TASEP with Second Class Particles 11
Putting x1 = x, x2 = x+ i1, . . . , xN = x+ i1 + · · ·+ iN−1, (3.1) becomes
∞∑
i1,...,iN−1=1
−
∫
C
· · · −
∫
C
∑
σ∈SN
[
Aσ
](
ξσ(2) · · · ξσ(N)
)i1(ξσ(3) · · · ξσ(N)
)i2 · · · (ξσ(N)
)iN
×
N∏
i=1
(
ξx−yi−1i eε(ξi)t
)
dξ1 · · · dξN
= −
∫
C
· · · −
∫
C
∑
σ∈SN
[
Aσ
] ξσ(2)ξ
2
σ(3) · · · ξ
N−1
σN
(1− ξσ(2) · · · ξσ(N))(1− ξσ(3) · · · ξσ(N)) · · · (1− ξσ(N))
×
N∏
i=1
(
ξx−yi−1i eε(ξi)t
)
dξ1 · · · dξN ,
where [Aσ] is the (2N−1 + 1, 2N−1 + 1)th element of Aσ. (We write [A]i,j for the (i, j)th element
of matrix A but we will simply write [A] for i, j = 2N−1 + 1.) In order to compute
∑
σ∈SN
[Aσ]
ξσ(2)ξ
2
σ(3) · · · ξ
N−1
σN
(1− ξσ(2) · · · ξσ(N))(1− ξσ(3) · · · ξσ(N)) · · · (1− ξσ(N))
, (3.2)
first, we will find an expression for [Aσ].
Lemma 3.1. Let Tl = Tl(α, β) be a 2N × 2N matrix in Definition 2.2 and let Aσ be given
by (2.30).
(a) If l 6= 1, then [Tl(α, β)] = Sβα, and if l = 1, then [Tl(α, β)] = −1.
(b) Let (a1, . . . , a2N ) be the (2N−1 + 1)th row vector of Tl. Then, ak = 0 for all k 6= 2N−1 + 1.
(c) Tl is an upper-triangular matrix.
(d) The diagonal terms of Aσ are given by
[Aσ]l,l =
{
[Tan ]l,l · · · [Ta2 ]l,l[Ta1 ]l,l if σ 6= 1,
1 if σ = 1.
Proof. All these properties are easily verified by observing the forms of Tl. �
Remark 3.2. Unlike Lemma 3.1(d), to the best of the author’s knowledge, it is nontrivial to find
the non-diagonal terms of Aσ without directly performing matrices multiplication Tan · · ·Ta1 .
In general, this makes it inconvenient to find an explicit form of P(Y,ν)(X,π; t) for π 6= ν.
Now, we give an expression for [Aσ].
Lemma 3.3. If N ≥ 2,
[Aσ] = sgn(σ)
N−2∏
i=0
(
1− ξ2+i
1− ξσ(2+i)
)i
. (3.3)
Proof. If σ = 1, the statement is trivial. Suppose that σ 6= 1. We prove the statement by
induction on N . When N = 2, it is easy to verify the statement if we observe (2.10). Let
σ′ ∈ SN and suppose that (3.3) holds for N , that is,
[Aσ′ ] = sgn(σ′)
(
1− ξ3
1− ξσ′(3)
)
· · ·
(
1− ξN−1
1− ξσ′(N−1)
)N−3( 1− ξN
1− ξσ′(N)
)N−2
12 E. Lee
holds. For any σ ∈ SN+1, there are σ′ ∈ SN and an integer K = {1, . . . , N + 1} such that
σ(i) = σ′(i) for 1 ≤ i ≤ K − 1, σ(K) = N + 1 and σ(i) = σ′(i − 1) for K + 1 ≤ i ≤ N + 1.
Suppose that σ′ = Tan · · ·Ta1 for some finite sequence a1, . . . , an taking values in {1, . . . , N−1}.
If we view Tan , . . . , Ta1 as simple transpositions in SN+1, then
σ = TKTK+1 · · ·TNTan · · ·Ta1
(
1 2 · · ·N (N + 1)
)
and
[Aσ]2N+1,2N+1 = [TK ]2N+1,2N+1 · · · [TN ]2N+1,2N+1[Tan · · ·Ta1 ]2N+1,2N+1
= [TK ]2N+1,2N+1 · · · [TN ]2N+1,2N+1[Aσ′ ]
by Lemma 3.1(d). Using the induction hypothesis and Lemma 3.1(a), we obtain
[Aσ]2N+1,2N+1 =
(
− 1− ξN+1
1− ξσ′(K)
)(
− 1− ξN+1
1− ξσ′(K+1)
)
· · ·
(
− 1− ξN+1
1− ξσ′(N)
)
× sgn(σ′)
(
1− ξ3
1− ξσ′(3)
)
· · ·
(
1− ξN−1
1− ξσ′(N−1)
)N−3( 1− ξN
1− ξσ′(N)
)N−2
. (3.4)
If we recall that σ(i) = σ′(i) for 1 ≤ i ≤ K − 1, σ(K) = N + 1 and σ(i) = σ′(i − 1) for
K + 1 ≤ i ≤ N + 1 and note that (−1)N−K+1 sgn(σ′) = sgn(σ), then we see that (3.4) is equal
to
sgn(σ)
(
1− ξ3
1− ξσ(3)
)(
1− ξ4
1− ξσ(4)
)2
· · ·
(
1− ξN+1
1− ξσ(N+1)
)N−1
. �
Now, let us come back to the expression (3.2). The sum in (3.2) with [Aσ] in (3.3) will be
simplified by the algebraic identity
∑
σ∈SN
[Aσ]
ξσ(2)ξ
2
σ(3) · · · ξ
N−1
σN
(1− ξσ(2) · · · ξσ(N))(1− ξσ(3) · · · ξσ(N)) · · · (1− ξσ(N−1)ξσ(N))(1− ξσ(N))
= (1− ξ1)
∏
1≤i<j≤N
ξj − ξi
1− ξi
N∏
i=1
1
1− ξi
, N ≥ 2. (3.5)
Once this identity is proved, the proof of Theorem 1.2 is completed.
Remark 3.4. Although this paper does not deal with the ASEP with second class particles,
it is believed that the generalization of the identity (3.5) to the ASEP with second class par-
ticles is possible. This generalization was confirmed for small systems by the author. Since
P(Y,ν)(X, ν; t) in the ASEP with second class particles is not determinantal, it seems that the
determinantal structure of P(Y,ν)(X, ν; t) in the TASEP with second class particles is not essen-
tial in deriving (3.5). But, we do not have a result corresponding to Lemma 3.3 for the ASEP
with second class particles.
If we observe that
1− ξ1
N−2∏
i=0
(1− ξ2+i)i
1∏
i<j
(1− ξi)
=
N∏
i=1
1
(1− ξi)N−2
,
then we see that (3.5) is equivalent to∑
σ∈SN
sgn(σ)
1
(1− ξσ(3))(1− ξσ(4))2 · · · (1− ξσ(N))N−2
On the TASEP with Second Class Particles 13
×
ξσ(2)ξ
2
σ(3)ξ
3
σ(4) · · · ξ
N−1
σ(N)
(1− ξσ(2) · · · ξσ(N))(1− ξσ(3) · · · ξσ(N)) · · · (1− ξσ(N−1)ξσ(N))(1− ξσ(N))
=
N∏
i=1
1
(1− ξi)N−1
∏
1≤i<j≤N
(ξj − ξi), N ≥ 2. (3.6)
Also, if we substitute 1/ξN−i+1 for ξi in (3.6), we obtain another equivalent identity∑
σ∈SN
sgn(σ)
1
(ξσ(1) − 1)N−2(ξσ(2) − 1)N−3 · · · (ξσ(N−2) − 1)
×
ξσ(N−2)ξ
2
σ(N−3) · · · ξ
N−2
σ(1)
(ξσ(1) · · · ξσ(N−1) − 1)(ξσ(1) · · · ξσ(N−2) − 1) · · · (ξσ(1)ξσ(2) − 1)(ξσ(1) − 1)
=
N∏
i=1
1
(ξi − 1)N−1
∏
1≤i<j≤N
(ξj − ξi). (3.7)
We will prove the identity (3.7). In order to prove (3.7), we will use an identity for the TASEP.
In (3.2) with p = 1 in [15], Aσ can be written as
Aσ = sgn(σ)
(
1− ξ2
1− ξσ(2)
)(
1− ξ3
1− ξσ(3)
)2
· · ·
(
1− ξN
1− ξσ(N)
)N−1
.
Using (1.6) with p = 1 in [15], we have
∑
σ∈SN
Aσ
ξσ(2)ξ
2
σ(3)ξ
3
σ(4) · · · ξ
N−1
σ(N)
(1− ξσ(2) · · · ξσ(N))(1− ξσ(3) · · · ξσ(N)) · · · (1− ξσ(N−1)ξσ(N))(1− ξσ(N))
= (1− ξ1 · · · ξN )
N∏
i=1
1
1− ξi
∏
1≤i<j≤N
ξj − ξi
1− ξi
, N ≥ 2,
equivalently,∑
σ∈SN
sgn(σ)
1
(1− ξσ(2))(1− ξσ(3))2 · · · (1− ξσ(N))N−1
×
ξσ(2)ξ
2
σ(3)ξ
3
σ(4) · · · ξ
N−1
σ(N)
(1− ξσ(2) · · · ξσ(N))(1− ξσ(3) · · · ξσ(N)) · · · (1− ξσ(N−1)ξσ(N))(1− ξσ(N))
= (1− ξ1 · · · ξN )
N∏
i=1
1
(1− ξi)N
∏
1≤i<j≤N
(ξj − ξi), N ≥ 2. (3.8)
If we substitute 1/ξN−i+1 for ξi in (3.8), we obtain an equivalent version of (3.8),∑
σ∈SN
sgn(σ)
1
(ξσ(1) − 1)N−1(ξσ(2) − 1)N−2 · · · (ξσ(N−1) − 1)
×
ξσ(N−1)ξ
2
σ(N−2) · · · ξ
N−1
σ(1)
(ξσ(1) · · · ξσ(N−1) − 1)(ξσ(1) · · · ξσ(N−2) − 1) · · · (ξσ(1)ξσ(2) − 1)(ξσ(1) − 1)
= (ξ1 · · · ξN − 1)
N∏
i=1
1
(ξi − 1)N
∏
1≤i<j≤N
(ξj − ξi). (3.9)
This is the identity we will use in order to prove (3.7). Also, we need the following result to
prove (3.7).
14 E. Lee
Lemma 3.5 (Vandermonde determinants).
N∑
α=1
(−1)N+α(ξα − 1)N−1
∏
1≤i<j≤N
i,j 6=α
(ξj − ξi) =
∏
1≤i<j≤N
(ξj − ξi). (3.10)
Proof. The left hand side of (3.10) is a cofactor expansion of
det
1 1 · · · 1
ξ1 ξ2 · · · ξN
...
...
...
ξN−21 ξN−22 · · · ξN−2N
(ξ1 − 1)N−1 (ξ2 − 1)N−1 · · · (ξN − 1)N−1
.
Note that
(ξα − 1)N−1 =
N∑
k=1
aN−kξ
N−k
α
for some constants aN−1, . . . , a0. (Here, aN−1 = 1.) If we add a multiple of the first row by the
constant −a0 to the Nth row, the determinant does not change and is equal to
det
1 1 · · · 1
ξ1 ξ2 · · · ξN
...
...
...
ξN−21 ξN−22 · · · ξN−2N
N−1∑
k=1
aN−kξ
N−k
1
N−1∑
k=1
aN−kξ
N−k
2 · · ·
N−1∑
k=1
aN−kξ
N−k
N
.
We successively perform these row operations to obtain
det
1 1 · · · 1
ξ1 ξ2 · · · ξN
...
...
...
ξN−21 ξN−22 · · · ξN−2N
ξN−11 ξN−22 · · · ξN−1N
,
which is the Vadermonde determinant. �
Now, we prove (3.7).
Proof of identity (3.7). When N = 2, it is easy to show (3.7). We will show (3.7) for N ≥ 3.
The left hand side of (3.7) is an antisymmetric function of ξ1, . . . , ξN because the sum is an
anti-symmetrized sum, so it is divisible by the Vandermonde determinant. Hence, the left hand
side of (3.7) can be written as
G(ξ1, . . . , ξN )×
∏
1≤i<j≤N
(ξj − ξi)
where G(ξ1, . . . , ξN ) is a symmetric function of ξ1, . . . , ξN . So, we want to show that
G(ξ1, . . . , ξN ) =
N∏
i=1
1
(ξi − 1)N−1
.
On the TASEP with Second Class Particles 15
Fix α ∈ {1, . . . , N}. Let σ′ be a bijective mapping from {1, . . . , N−1} onto {1, . . . , N}\{α} and
let S′N−1 be the set of σ′. Let σα ∈ SN be a permutation such that σα(N) = α and σα(i) = σ′(i)
for i = 1, . . . , N − 1. Then, the left hand side of (3.7) is equal to
N∑
α=1
∑
σα∈SN
sgn(σα)
1
(ξσα(1) − 1)N−2(ξσα(2) − 1)N−3 · · · (ξσα(N−2) − 1)
×
ξσα(N−2)ξ
2
σα(N−3) · · · ξ
N−2
σα(1)
(ξσα(1) · · · ξσα(N−1) − 1)(ξσα(1) · · · ξσα(N−2) − 1) · · · (ξσα(1) − 1)
. (3.11)
The sum over σα ∈ SN for a fixed α in (3.11) is equal to
∑
σ′∈S′N−1
(−1)N−α sgn(σ′)
1
(ξσ′(1) − 1)N−2(ξσ′(2) − 1)N−3 · · · (ξσ′(N−2) − 1)
×
ξσ′(N−2)ξ
2
σ′(N−3) · · · ξ
N−2
σ′(1)
(ξσ′(1) · · · ξσ′(N−1) − 1)(ξσ′(1) · · · ξσ′(N−2) − 1) · · · (ξσ′(1) − 1)
, (3.12)
because σα(i) = σ′(i) for i = 1, . . . , N − 1 and sgn(σα) = (−1)N−α sgn(σ′). Therefore, if we
apply (3.9) for N − 1 to (3.12) and sum over α, we obtain
N∑
α=1
(−1)N−α
N−1∏
i=1
i 6=α
1
(ξi − 1)N−1
∏
1≤i<j≤N−1
i,j 6=α
(ξj − ξi) =
N∏
i=1
1
(ξi − 1)N−1
∏
1≤i<j≤N
(ξj − ξi)
by using Lemma 3.5. �
3.2 Proof of Theorem 1.4 and Corollary 1.5
Lemma 3.6. Let ki be any integer such that 0 ≤ ki ≤ i − 2 for i = 2, . . . , N , and l be any
nonnegative integer. Let us consider
det
ξN−1+l1 ξN−2+k21 ξN−3+k31 ξN−4+k41 · · · ξ
1+kN−1
1 ξkN1
ξN−1+l2 ξN−2+k22 ξN−3+k32 ξN−4+k42 · · · ξ
1+kN−1
2 ξkN2
...
...
...
...
...
...
...
ξN−1+lN ξN−2+k2N ξN−3+k3N ξN−4+k4N · · · ξ
1+kN−1
N ξkNN
. (3.13)
(a) If there is a nonzero ki for some i = 3, . . . , N , the determinant is zero.
(b) If ki = 0 for all i, the determinant is
hl(ξ1, . . . , ξN )
∏
1≤i<j≤N
(ξi − ξj).
Proof. (a) Suppose that kα 6= 0 for some α ≥ 3 but the determinant is nonzero. Since the
power of the terms in the second column is (N − 2) and the determinant should be nonzero, the
power of the terms in the third column must be (N − 3). If we repeat this process, we see that
the power of the terms in the ith column should be (N − i), i = 2, . . . , α−1, for the determinant
to be nonzero. If kα 6= 0, then the power of the terms in the αth column is one of N − α + kα
where kα = 1, 2, . . . , α− 2, and hence the determinant is zero, which is a contradiction.
16 E. Lee
(b) If ki = 0 for all i, then we have
det
ξN−1+l1 ξN−21 ξN−31 · · · 1
ξN−1+l2 ξN−22 ξN−32 · · · 1
...
...
...
...
...
ξN−1+lN ξN−2N ξN−3N · · · 1
= det
[
ξ
λj+N−j
i
]N
i,j=1
= hl
∏
1≤i<j≤N
(ξi − ξj)
where λ = (λ1, λ2, . . . , λN ) = (l, 0, . . . , 0) is a partition and hl = hl(ξ1, . . . , ξN ) is the complete
symmetric polynomial of degree l [10, Chapter I.3]. �
Now, we apply the initial condition (1.2) to (1.1). Then, we have, after some manipulations,
P(Y,21...1)(Et) = −
∫
C
· · · −
∫
C
∏
1≤i<j≤N
(ξj − ξi)
N∏
i=1
1
(1− ξi)N−1
N∏
i=1
(
ξx−N−1−li eε(ξi)t
)
×
(
N−2∏
i=0
(1− ξ2+i)i
)
ξN+l−1
1 ξN−22 ξN−33 · · · ξN−1dξ1 · · · dξN
=
1
N !
−
∫
C
· · · −
∫
C
∏
i<j
(ξj − ξi)
N∏
i=1
1
(1− ξi)N−1
N∏
i=1
(
ξx−N−1−li eε(ξi)t
)
×
∑
σ∈SN
sgn(σ)
(
N−2∏
i=0
(1− ξσ(2+i))i
)
ξN+l−1
σ(1) ξN−2σ(2) ξ
N−3
σ(3) · · · ξσ(N−1)
dξ1 · · · dξN .
If we expand
N−2∏
i=0
(1− ξσ(2+i))i, then we see that each term is in the form of
C · ξk3σ(3)ξ
k4
σ(4) · · · ξ
kN
σ(N)
where ki are some integers such that 0 ≤ ki ≤ i− 2 and C is some constant, and∑
σ∈SN
sgn(σ)ξN−1+lσ(1) ξN−2σ(2) ξ
N−3+k3
σ(3) · · · ξ1+kN−1
σ(N−1) ξ
kN
σ(N)
is the determinant (3.13). Hence, by Lemma 3.6, we obtain
P(Y,21...1)(Et) =
(−1)N(N−1)/2
N !
−
∫
C
· · · −
∫
C
hl(ξ1, . . . , ξN )
∏
1≤i<j≤N
(ξj − ξi)2
N∏
i=1
1
(ξi − 1)N−1
×
N∏
i=1
(
ξx−N−l−1i eε(ξi)t
)
dξ1 · · · dξN . (3.14)
Finally, we immediately obtain (1.4) if l = 0 in (3.14).
Acknowledgements
This work was supported by the social policy grant from Nazarbayev University. The author is
grateful to the anonymous referees for valuables comments and suggestions.
On the TASEP with Second Class Particles 17
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1 Introduction
1.1 Statement of the main results
1.2 Previous results
2 Preliminary
2.1 Two-particle system
2.2 Three-particle system
2.3 N-particle system
2.3.1 Bethe Ansatz solution
2.3.2 Matrices A
3 Proofs of the main results
3.1 Proof of Theorem 1.2
3.2 Proof of Theorem 1.4 and Corollary 1.5
References
|
| id | nasplib_isofts_kiev_ua-123456789-209458 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2025-12-02T00:39:27Z |
| publishDate | 2018 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Lee, E. 2025-11-21T19:13:37Z 2018 On the TASEP with Second Class Particles / E. Lee // Symmetry, Integrability and Geometry: Methods and Applications. — 2018. — Т. 14. — Бібліогр.: 18 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 60J25; 60K35; 82B23 arXiv: 1705.10544 https://nasplib.isofts.kiev.ua/handle/123456789/209458 https://doi.org/10.3842/SIGMA.2018.006 In this paper, we study some conditional probabilities for the totally asymmetric simple exclusion processes (TASEP) with second-class particles. To be more specific, we consider a finite system with one first-class particle and N−1 second-class particles, and we assume that the first-class particle is initially at the leftmost position. In this case, we find the probability that the first class particle is at x and it is still the leftmost particle at time t. In particular, we show that this probability is expressed by the determinant of an N×N matrix of contour integrals if the initial positions of particles satisfy the step initial condition. The resulting formula is very similar to a known formula in the (usual) TASEP with the step initial condition, which was used for asymptotics by Nagao and Sasamoto [Nuclear Phys. B 699 (2004), 487-502]. This work was supported by the social policy grant from Nazarbayev University. The author is grateful to the anonymous referees for valuable comments and suggestions. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications On the TASEP with Second Class Particles Article published earlier |
| spellingShingle | On the TASEP with Second Class Particles Lee, E. |
| title | On the TASEP with Second Class Particles |
| title_full | On the TASEP with Second Class Particles |
| title_fullStr | On the TASEP with Second Class Particles |
| title_full_unstemmed | On the TASEP with Second Class Particles |
| title_short | On the TASEP with Second Class Particles |
| title_sort | on the tasep with second class particles |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/209458 |
| work_keys_str_mv | AT leee onthetasepwithsecondclassparticles |