Relaxation dynamics of the asymmetric simple exclusion process with Langmuir kinetics on a ring

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by  Jun Sato

Relaxation dynamics of the asymmetric simple exclusion process with Langmuir
kinetics on a ring
Jun Sato and Katsuhiro Nishinari
Research Center for Advanced Science and Technology, University of Tokyo,
4-6-1 Komaba, Meguro-ku, Tokyo 153-8904, Japan
(Dated: April12, 2016)
6 WeconsidertheasymmetricsimpleexclusionprocesswithLangmuirkineticsonaperiodiclattice.
1 Weanalyticallyobtaintheexacttimeevolutionofcorrelationfunctionswitharbitrarylengthstarting
0
from the initial state with no particle in the system. The exact stationary state of this model has
2
been known for the totally asymmetric case. We propose a basis transformation which simplifies

r theproofofthestationarityofthisstateandenablesthegeneralization tothepartiallyasymmetric
p case. Moreover, we construct low-energy excitations and obtain theexact relaxation time.
A
PACSnumbers: 05.10.Gg
1
1

Introduction. Nonequilibrium statistical mechan-
]
h ics has been intensively studied in recent years. How
c and why a nonequilibrium state reaches thermal equilib-
e
rium is a long-standing problem. Despite much effort in
m
this field, complete understanding of relaxation dynam-
-
t ics starting from a specific initial state is far from being
a
fulfilled. As for the static property of a nonequilibrium
t FIG. 1: ASEP with Langmuir kinetics
s steadystate,however,exactsolutionsareknownforsome
.
t simple models.
a
m Amongthem,theasymmetricsimpleexclusionprocess expression of the time evolution of the correlation func-
- (ASEP) is one of the most fundamental and nontrivial tions in the ASEP with Langmuir kinetics (ASEP-LK)
d model describing the nonequilibrium transport phenom- model on a periodic lattice.
n ena[1–6],whichhaveawiderangeofapplicationssuchas TheASEP-LKmodeldescribesanattachmentandde-
o
c biological transport [1], and pedestrian and traffic flows tachment of particles as well as the exclusive hopping
[ [5, 6]. The ASEP is a continuous time Markov process process[17]. Themostnotablepointinthis modelis the
describing the asymmetric diffusion of particles with an nonconservation of the number of particles even in the
3
v exclusion principle on a one-dimensional lattice. In a closed ring. Steady state properties of this model with
1 time interval dt, a particle hops to the right (left) site openboundaryconditionshavebeen studiedindetailby
5 with the probability pdt (qdt) if it is vacant. use of mean field approximation and Monte Carlo simu-
6 The most outstanding advantage of this model is the lations, which revealsinteresting phenomena suchas the
2
amenability to exact analytical methods such as matrix coexistenceofhigh-andlow-densityphasesseparatedby
0
. productansatz[7]andBetheansatz[8]. Thesteadystate the shock wave in the density profile [18, 19]. The exact
1 of the ASEP is exactly constructed by the matrix prod- stationarystate is constructedinthe case of TASEP-LK
0
uct ansatz, which offers exact derivations of interesting withaperiodic lattice[20],inwhichparticlehoppingoc-
6
1 phenomena such as boundary induced phase transitions curs only in one direction (q =0).
: [7, 9, 10]. In this paper we introduce a basis transformation
v
As for the dynamics, however, knowledge about the whichsimplifiestheproofofthestationarityofthisstate
i
X full spectrum of the Markov matrix is needed for the and enables generalization to the partially asymmetric
r complete understanding of the relaxation process. As case (q 6= 0). We construct low-energy excitations and
a
a first step toward this goal, spectral gaps of the first obtain the exact relaxation time. Furthermore, we an-
excited states are examined by use of the Bethe ansatz alytically obtain the exact time evolution of correlation
method, which provides asymptotic information about functions with arbitrary length starting from the initial
dynamics such as relaxationtime. The finite size scaling state with no particle in the system.
of it reveals that the model is governed by the Kardar- Model. WeconsidertheASEPwithLangmuirkinetics
Parisi-Zhanguniversality class [8, 11–15]. described in Fig. 1. A particle is attached on a site with
Furthermore, full relaxation dynamics in the totally rate ω if it is vacant and detached from the site with
a
asymmetric simple exclusion process (TASEP) on a ring rate ω if it is occupied.
d
isexactlyexaminedbynumericalimplementationsofthe Hereafter we impose a periodic boundary condition
Bethe ansatz equations and the determinant formula for with length L. We associate a Boolean variable τ to
n
formfactors [16]. In this paper,we presentananalytical every site n to represent whether a particle is present
2
(τ = 1) or not (τ = 0). Let |τ = 0i and |τ = 1i where we define ω := ω + ω . It is noted that this
n n n n a d
denote the standard basis vectors in this order for the transformationdiagonalizes the Langmuir kinetics term.
vector space C2. We consider the L-fold tensor product Oneofthemostremarkablepointsinthistransformation
of this basis |τ1τ2···τLi := |τ1i⊗|τ2i⊗···⊗|τLi, the isthatthestationarystateinthisbasisbecomesasimple
dimension of which is 2L. Then we can write a state of pure vector |00···0i. In the following we prove
the system at time t in a vector form with each element
M˜|00···0i=0. (7)
being a probability distribution
It is obvious that
|P(t)i= P(τ1,··· ,τL|t)|τ1,··· ,τLi. (1)
τ1,X···,τL h˜ |0i =−ω 0 0 1 =0. (8)
n n 0 1 0
Thetimeevolutionofthisstateisdescribedbythemaster (cid:18) (cid:19)n(cid:18) (cid:19)n
equation
Moreover,we have
d
|P(t)i=M|P(t)i, (2) L α(p−q) L
dt M˜n,n+1|00···0i= (|ni−|n+1i)
1+α
where the Markov matrix M is given by n=1 n=1
X X
=0, (9)
L
M= (Mn,n+1+hn), where |ni denotes the state with a single particle being
n=1 atthesiten. Thelastequalityisduetothetranslational
X
0 0 0 0 invariance of the system. This finishes the proof of (7).
Mn,n+1 =00 −qq −pp 00 , hn =(cid:18)−ωωaa −ωωdd(cid:19)n wrTithteuns atshe stationary state |Si in the original basis is
0 0 0 0
 n,n+1 |Si=U|00···0i
  (3)
1 1 1
The subscripts at the bottom right of the matrix rep- = ⊗ ⊗···⊗
α α α
resent the vector space on which the matrix is acting. (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
L
The Langmuir kinetics term hn can be regarded as an = αN |τ1,··· ,τLi . (10)
off-diagonal magnetic field, which induces a nonconser- !
vation of number of particles in the system. NX=0 τ1+··X·+τL=N
The stationary state |Si belongs to the eigenvector of InthesubspacewithafixedparticlenumberN,theprob-
the Markov matrix with the eigenvalue zero: ability distributions of L states are all proportionalto
N
αN. This means that P(τ1,··· ,τL) is determined from
M|Si=0. (4) (cid:0) (cid:1)
particle number N = τ1+···+τL only, independent of
The first step is to find a stationary state |Si explicitly. the explicit particle configuration (τ1,··· ,τL).
This result coincides with that given in [20] for the
In orderto do so,we introduce a basis transformationin
totally asymmetric case q = 0. We note that our con-
the following.
struction of the stationary state does not depend on the
Stationary state. Let U be the L-fold tensor product
value of q.
of a 2×2 matrix defined by
Relaxation time. Next we construct the first excited
1 1 ⊗L ωa stateandevaluatetherelaxationtime. Inthelanguageof
U = , α= , (5)
α −1 ω quantum spin chains, the stationary state in the trans-
(cid:18) (cid:19) d
formed basis |00···0i is considered to be a completely
and consider the basis transformation
ferromagneticstate. Inthiscasethelowlyingexcitations
L aregivenbyspinwavestates. Accordingtothis analogy,
M˜ =U−1MU = M˜n,n+1+h˜n , wecanfindlowlyingexcitationsoftheASEP-LKmodel.
nX=1(cid:16) (cid:17) Let us define S± = σx ± iσy, where σx,y,z denote
M˜n,n+1 = 1 standard Pauli matrices. These correspond to the “spin
1+α flip operators.” We further define total flip operators
0 0 0 0 S± = L S± and the state
T n=1 n
−α(p−q) −(q+pα) (p+qα) p−q
  , P L
α(p−q) (q+pα) −(p+qα) −(p−q)
|S−i:=S−|00···0i= |ni, (11)
 0 0 0 0  T T
 n,n+1 n=1
  X
0 0
h˜ =−ω , (6) whichisconsideredtobeaspinwaveexcitationwiththe
n 0 1
(cid:18) (cid:19)n wavenumberk =0. Thisturnsouttobethefirstexcited
3
state in the transformed basis. In the following we show ω Eigenvalues
that 0.2 0 0.2 0.4 0.6 0.8 z1 z1∗ 1.4 1.4 z2 z2∗ z3 z3∗ x1 x2 3.4
∗ ∗
M˜|S−i=−ω|S−i. (12) 2 0 2 z4 z4 4 x3 5 5 x4 x5 6 z5 z5 7 x6 8
T T
TABLE I: All the eigenvalues for L=4 with ω =0.2 and 2.
We first note that M˜n,n+1 is written in the form Numerical values in the table are given by
z1=1.27024+0.369916i, z2 =1.52976+0.369916i,
0 0 0 0 z3=1.86397+0.287025i, x1 =2.31872, x2=2.55334,
−a −b c d z4=3.02539+0.243086i, x3 =4.10419, x4=5.06102,
M˜n,n+1 = a b −c −d , (13) x5=5.86836, z5 =6.97461+0.243086i, x6=7.96644.
 0 0 0 0 
 n,n+1
  hasthesameformas(13),wherea′ =−(p−q)(1−α)/(1+
where a = α(p−q)/(1+α), b = (q+pα)/(1+α), c =
α), b′ = −c′ = −(p − q)/(1 + α), and d′ = 0. Since
(p+qα)/(1+α),andd=(p−q)/(1+α). Onecaneasily
obtain the derivation of Lℓ=1M˜ℓ,ℓ+1|ST−i = 0 is independent
of the explicit values of a,b,c and d, we can inductively
L show that Lℓ=1MP˜ℓ,ℓ+1|(ST−)mi = 0 with respect to m.
M˜ℓ,ℓ+1|ni=a(|n,n+1i−|n−1,ni) Together aPgain with Lℓ=1h˜ℓ|(ST−)mi = −mω|(ST−)mi,
Xℓ=1 we arrive at the eigenvalue equation (16).
P
+b(|n−1i−|ni)+c(|n+1i−|ni), (14) Direct diagonalizations for small systems show that,
up to (L+1) th largest eigenvalues, they are given by
where |n1,n2i denotes the state with two particles be- {0,−ω,−2ω,··· ,−Lω} for sufficiently small ω. If we
ing at the sites n1 and n2: |n1,n2i := Sn−1Sn−2|00···0i. increasethevalueofω,othercomplexeigenvaluesλwith
This yields Lℓ=1M˜ℓ,ℓ+1|ST−i=0 from the translational the real part −ω > Reλ > −Lω enter into this group.
symmetry. Together with the fact that L h˜ |S−i = However, we observe that the second largest eigenvalue
P ℓ=1 ℓ T
−ω|S−i, we can conclude the eigenvalue equation (12). is always −ω. Some examples are listed in Table I.
T P
Fromdirect diagonalizationsof the Markovmatrix for Full dynamics. Finallyletusconsiderthetimeevolu-
small systems, we observe that this eigenvalue −ω is al- tion of physical quantities. Let A be a physical quantity
waysthe secondlargestrealpartamong allthe eigenval- which takes a value A = A(τ) in a particle configura-
ues. SomeexamplesarelistedinTableI. Soweconclude tion τ := (τ1,··· ,τL). We think of A as an operator on
that the relaxationtime T for the ASEP with Langmuir a state vector |τi acting diagonally as A|τi = A(τ)|τi.
kinetics is given by Thenthe expectationvalue ofthe physicalquantityAin
a state |Pi= P(τ)|τi is written as
τ
1
T =ω−1 = . (15)
ω +ω P P(τ)A(τ)
a d hAi= τ =hT|A|Pi/hT|Pi, (19)
P(τ)
Note that this result is independent of system size L. P τ
Unlike the case with usual ASEP, the energy gapalways where|Ti:= |Pτi. Sincethestatevector|Piobeysthe
τ
has a finite value, which causes an exponentially fast re- master equation (2), the time evolution of the physical
laxation to the stationary state. quantity A stParting from the initial state |Ii reads
Higher excitations. Next we construct further low
lying excitations. In the following we show that hA(t)i=hT|AeMt|Ii/hT|Ii. (20)
M˜|(S−)mi=−mω|(S−)mi (m=0,1,2,··· ,L), (16) Here we haveused hT|eMt =hT|,whichfollows fromthe
T T
property of the probability transition matrix hT|M =0.
where Hereafter we take the no-particle state |00···0i as an
initial state |Ii. Note that hT|Ii = 1. By the basis
1
|(S−)mi:= (S−)m|00···0i transformationthe time evolutionofAcanbewrittenas
T m! T
= |n1,n2,··· ,nmi. (17) hA(t)i=h00···0|A˜eM˜t|Si (21)
1≤n1<X···<nm≤L where we have used hT˜|=hT|U =(1+α)Lh00···0| and
First we note that the following commutator |I˜i=U−1|Ii=(1+α)−L|Si.
Wewishtoexpandtheabovequantitybytheeigenvec-
0 0 0 0
tors of Markov matrix M. However, they are not neces-
−a′ −b′ c′ d′
[M˜n,n+1,ST−]= a′ b′ −c′ −d′ (18) sarily orthogonal to each other since the Markov matrix
M is not normal: M†M6=MM†. However,we numer-
 0 0 0 0 n,n+1 icallyconfirmthatthevectors|(S−)miareorthogonalto
  T
4
anyothereigenvectors,andtherestoftheeigenspacecan
alwaysbe orthogonalizedby the Gram-Schmidtmethod.
These facts lead to the following expansion:
L h00···0|A˜eM˜t|(S−)mih(S−)m|Si
hA(t)i= T T
h(S−)m|(S−)mi
m=0 T T
X
h00···0|A˜eM˜t|λ˜ihλ˜|Si
+ . (22)
hλ˜|λ˜i
Xλ˜
Noting that the stationary state |Si can be expanded as
|Si= L αm|(S−)mi, we easily obtain
m=0 T
hλ˜|Si=P0 unless |λ˜i=|(S−)mi (m=0,1,2,··· ,L), FIG. 2: Time evolution of the particle density ρ(t) starting
T from the empty state is plotted against the scaled time t/T,
(23)
where T is the relaxation time (15). Parameters are chosen
where |λ˜i are the eigenvectors of M˜. Consequently we to be (p,q) = (0.7,0.3), (ωa,ωd) = (0.1,0.4). The thick red
lineistheplotofEq. (25). Thethingreenlineistheresultof
arriveattheformulaforthetimeevolutionofthephysical a one-time Monte Carlo simulation. The dashed blue line is
quantity A starting from the empty state |00···0i as a 100-times average of Monte Carlo simulations. We set the
follows: lattice length to L = 100 in the simulation. The plot starts
from zero and saturates to thevalueα/(1+α) known as the
L Langmuir isotherm.
hA(t)i= e−mωtαmh00···0|A˜|(S−)mi. (24)
T
m=0
X
increases and saturates to the value α/(1+α)2. On the
Density, current, and correlation functions. First
otherhand,inthecaseω >ω ,thecurrentismaximized
let us apply the above formula to the particle density a d
to the value 1/4at t=T ln 2α , then begins to decrease
ρ(t) := hτ1(t)i, which represents the probability for a α−1
and approaches the value α/(1+α)2. This decrease of
particle to exist at the site 1. From the translational
the current is due to the congestion of particles induced
symmetry,multiplication ofL to this quantity yields the
by the rapid increase of density by the large attachment
total number of particles in the system. Let A be A =
(1−σz)/2. By the basis transformation we have A˜ = rate ωa. The steady state current J(∞)=α/(1+α)2 is
1
maximized at the case α=1, namely ω =ω .
1 α −1 a d
. Then we can easily obtain the time More generally, we can easily prove by use of
1+α −α 1 !1 Eq. (24) that the time evolution of any cor-
evolution of the particle density ρ(t)=hA(t)i as relation function hx1x2···xn(t)i is decomposed into
α hx1(t)ihx2(t)i···hxn(t)i, where xℓ is τℓ or 1−τℓ. This
ρ(t)= 1−e−ωt , (25) decomposition results in
1+α
wCharicloh siismsuhlaotwionnsi.n FNiogt.e t2h,a(cid:0)ttotgheitsheisr(cid:1)awnitehxatchteaMnaolnytte- hx1x2···xn(t)i= (1+αmα)n(1−e−ωt)m(1+αe−ωt)n−m,
ical result without any approximations. In the long (27)
time limit t → ∞, it approaches the value α/(1 + α)
known as the Langmuir isotherm [21]. Next let us where m is the number of τℓ’s in {x1,x2,··· ,xn}.
Conclusion. In this paper we examine the dynamical
consider the current J(t) := hτ1(1 − τ2)(t)i and set
A = (1 −σz)(1 + σz)/4. By the basis transformation property of the ASEP with Langmuir kinetics on a peri-
1 2
odic lattice. We introduce a basis transformation which
1 α −1 1 1
we have A˜ = . Substitut- enables us to construct the ground state and low lying
(1+α)2 −α 1 !1 α α!2 excitations in a transparent way. As a result we obtain
ing this into the Eq. (24), we obtain the time evolution the exact stationary state and relaxation time. Further-
of the current, J(t)=hA(t)i, as more, we analytically obtain the exact time evolution of
α the density and current starting from the initial state
J(t)= (1+α)2(1−e−ωt)(1+αe−ωt), (26) with no particle in the system. We also show that corre-
lationfunctionswitharbitrarylengthcanbedecomposed
which is shownin Fig. 3, together with the Monte Carlo into the product of one-point functions. Relaxation dy-
simulations. NotethatJ(t)=ρ(t)(1−ρ(t)),whichmeans namics from the other initial states and the transport
that the mean field approximation gives the exact dy- property with an open boundary are the next problems
namics. In the case ω < ω , the current monotonically to be solved.
a d
5
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TheauthorsthankS.IchikiandK.Sakaifortheiruse-
fuldiscussions. ThisworkwassupportedbyJSPSKAK-
ENHI Grant No. 25287026.