Craig A. Tracy Joint work with Harold Widom

The Asymmetric Simple 

Exclusion Process: 

Integrable Structure 

and 

Limit Theorems 

Craig A. Tracy 

Joint work with Harold Widom 

Sunday, March 29, 2009 

1

The asymmetric simple exclusion 

process (ASEP): Introduced in 

1970 by Frank Spitzer in 

“Interaction of Markov Processes.” 

The “default stochastic model for 

transport phenomena”. The `Ìsing 

model of nonequilibrium 

phenomena’’. 

ASEP is a model for interacting 

particles on a lattice. 

Frank Spitzer 


2

ASEP on Integer Lattice 

q p 

 

p≠q 


3


q p 

 

p≠q 

● 

Each particle has an alarm clock -- 

exponential distribution with parameter one 


3


q p 

 

p≠q 

● 



● 

When alarm rings particle jumps to right with 

probability p and to the left with probability q 


3


q p 

 

p≠q 

● 



● 

When alarm rings particle jumps to right with 

probability p and to the left with probability q 

● 

Jumps are suppressed if neighbor is occupied 


3

Initial Conditions 


4


Step Initial Condition, q>p 


4



Flat Initial Condition 


4



Flat Initial Condition 

Random: Product Bernoulli measure 


4

Growth Processes & ASEP 

● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 

● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 


5

Growth Processes & ASEP 

● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 

● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 


5

KPZ Equation & Growth Processes 

❘ 

❘ 

↵ 

Kardar, Parisi & Zhang 

( 

∂h 

∂t = ν ∂2 h ∂h 

∂x 2 + λ ∂x 

) 2 

+ w 

diffusion growth noise 

u(x, t) = ∂h 

∂x 

Noisy Burgers eqn 


6

Remarks 


7

• Physicists expect KPZ equation to describe a 

large class of stochastically growing interfaces: 

KPZ Universality 

Remarks 


7




Remarks 

• KPZ eqn mathematically difficult to handle so 

make discrete approximation in space 


7




Remarks 



• ASEP is one discrete version of KPZ; thus 

expect `ùniversal behavior’’ in limit theorems 


7




Remarks 



• ASEP is one discrete version of KPZ; thus 

expect `ùniversal behavior’’ in limit theorems 

• TASEP is ASEP with jumps only to left or jumps 

only to right. TASEP is a simpler model 

(determinantal process) 


7

Total Current I(x,t) 

Step Initial Condition 

Take q>p net drift to left 

I(x,t) = # of particles ≤ x at time t, x ≤ 0 

Let η(x,t)=1 if particle at x at time t 

otherwise 0 

so I(x,t)=∑ y≤x η(y,t) 


8

Current & Position of m th particle 


9


Step initial condition 

x m (t)=position of m th particle from 

left at time t, x m (0)=m 


9





Event: 

{I(x,t)=m}={x m (t)≤x, x m+1 (t)>x} 


9





Event: 

{I(x,t)=m}={x m (t)≤x, x m+1 (t)>x} 

From this and exclusion property: 

Prob(I(x,t)≤m) = 1-Prob(x m+1 (t)≤x) 


9

Integrable Structure of ASEP 

We solve the Kolmogorov 

forward equation (“master 

equation”) for the transition 

probability Y→X: 

P Y (X; t) 

Main idea comes from the 

Bethe Ansatz (1931) 

Hans Bethe in 1967 


10

N=1 ASEP 

probability q 

probability p 

● ● ● ● ● ● ● ● ● ● 

Let P Y (X;t)=probability Y→X at time t. 

Master equation: 

dP 

dt 

= pP(x − 1; t)+qP(x + 1; t) − P (x; t) 

∫ 

P y (x; t) = 

ξ x−y−1 e tε(ξ) dξ 

C 

ε(ξ) = p ξ + q ξ − 1 11 

Sunday, March 29, 2009

N=2 ASEP 

● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 

Master equation takes simple form for this configuration 

● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 

Master equation reflects exclusion for this configuration 


12

N=2 ASEP 

● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 

Master equation takes simple form for this configuration 

● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 

Master equation reflects exclusion for this configuration 

Impose boundary conditions for first equation so 

that if satisfied the second equation is 

automatically satisfied --- Bethe’s Idea 


12

Important Point 

New boundary conditions arise for N=3, 4,... 

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ 

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ 

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ 

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ 

Last configuration requires new BC -- 

automatically satisfied by 2-particle BC 


13

Bethe Ansatz Solution of Master 

Equation 


14


Equation 

For any ξ 1 ,...,ξ N ∈ C\{0} and any permutation 

σ a solution is 

∏ 

j 

ξ x j 

σ(j) etε(ξ j) 


14


Equation 

For any ξ 1 ,...,ξ N ∈ C\{0} and any permutation 

σ a solution is 

∏ 

j 

ξ x j 

σ(j) etε(ξ j) 

Can take linear combination or integral of 

a linear combination & have a solution: 

∫ ∑ 

∏ 

e tε(ξj) d N ξ 

σ∈S N 

F σ (ξ) ∏ j 

ξ x j 

σ(j) 

j 


14

Remarks 


15

Remarks 

• Want BC to be satisfied. Bethe’s idea: This 

can be applied pointwise to the integrand 


15

Remarks 



• Gives condition on coefficients with result 

(this part same as the Yang & Yang analysis 

of XXZ spin Hamiltonian). If 

A σ = sgn(σ) 

∏ 

i

Remarks 



• Gives condition on coefficients with result 

(this part same as the Yang & Yang analysis 

of XXZ spin Hamiltonian). If 

A σ = sgn(σ) 

∏ 

i

Remarks 


16

Remarks 

• To recap: We have found a solution to the 

master equation satisfying the BCs. 


16

Remarks 



• Must show the solution at X={x 1 ,...x N } & t=0 

reduces to δ X,Y where Y={y 1 ,...,y N } is the initial 

configuration of N particles. 


16

Remarks 






• Term for σ=id satisfies initial condition. Must 

show remaining N!-1 terms give zero at t=0. 

This is the new part of the problem. 


16

Remarks 






• Term for σ=id satisfies initial condition. Must 

show remaining N!-1 terms give zero at t=0. 

This is the new part of the problem. 

• Have not yet specified the contours. 


16

Theorem (TW): If p ≠ 0 and r is small enough then 

P Y (X; t) = ∑ 

σ∈S N 

∫C N r 

A σ (ξ) ∏ i 

ξ x i 

σ(i) 

∏ 

i 

( 

ξ −y i−1 

i e ε(ξ i) t ) d N ξ. 

where 

and satisfies 

A σ = sgn σ 

∏ 

i

Marginal Distributions 

P(x m (t)≤x) 


18


P(x m (t)≤x) 

Case m=1: 

Fix x 1 =x, sum P Y (X;t) over allowed x 2 , x 3 , x 4 ,... 

Can do this since contours are small: |ξ i |< 1 


18


P(x m (t)≤x) 

Case m=1: 

Fix x 1 =x, sum P Y (X;t) over allowed x 2 , x 3 , x 4 ,... 

Can do this since contours are small: |ξ i |< 1 

Result is an expression involving N! terms. Use 

first miraculous identity to reduce sum to one 

term! 

Here’s the identity: 


18

First Identity 

∑ 

sgn σ 

⎛ 

⎝ ∏ i

• Using this identity we get for m=1 an 

expression for P(x 1 (t)≤x) as a single N- 

dimensional integral with a product 

integrand. This expression is for finite-N 

ASEP 

I(x, Y, ξ) = ∏ i

Prob(x 1 (t)=x) = 

∫ 

p N(N−1)/2 ∫C r 

· · · 

C r 

I(x, Y, ξ) dξ 1 · · · dξ N 

(p ≠ 0) 


21

Prob(x 1 (t)=x) = 

∫ 

p N(N−1)/2 ∫C r 

· · · 

C r 

I(x, Y, ξ) dξ 1 · · · dξ N 

(p ≠ 0) 

• Sum of N! integrals reduced to one integral 


21

Prob(x 1 (t)=x) = 

∫ 

p N(N−1)/2 ∫C r 

· · · 

C r 

I(x, Y, ξ) dξ 1 · · · dξ N 

(p ≠ 0) 


• Form is not so useful to take N→∞ 


21

Prob(x 1 (t)=x) = 

∫ 

p N(N−1)/2 ∫C r 

· · · 

C r 

I(x, Y, ξ) dξ 1 · · · dξ N 

(p ≠ 0) 



• We now expand contour outwards -- only 

residues that contribute come from ξ i =1. 


21

Prob(x 1 (t)=x) = 

∫ 

p N(N−1)/2 ∫C r 

· · · 

C r 

I(x, Y, ξ) dξ 1 · · · dξ N 

(p ≠ 0) 



• We now expand contour outwards -- only 

residues that contribute come from ξ i =1. 

• Can take N→∞ in resulting expression to 

obtain 


21

σ(S) : = ∑ i∈S 

i 

∑ 

P(x 1 (t) =x) = 

S 

∫ 

p σ(S)−|S| 

q σ(S)−|S|(|S|+1)/2 × 

∫ 

C R 

· · · 

C R 

I(x, Y S , ξ) d |S| ξ 

The sum is over all nonempty subsets of Z 

When p=0 only one term is nonzero, S={1}. 


22

Remarks 


23

Remarks 

• To go beyond the left-most particle, m=1, 

there are new complications 


23

Remarks 



• These come from the fact that we must 

sum over all x j > x m and all x i < x m . 

Some contours must be small (former) 

and some must be large (latter) to 

obtain convergence of geometric series 


23

Remarks 



• These come from the fact that we must 

sum over all x j > x m and all x i < x m . 

Some contours must be small (former) 

and some must be large (latter) to 

obtain convergence of geometric series 

• This involves finding a new identity 


23

Second Identity 

S ranges over subsets of {1, 2, . . . , N} 

∑ 

|S|=m 

∏ 

p + qξ i ξ j − ξ i 

ξ j − ξ i 

i∈S, j∈S c 

· (1 − ∏ 

j∈S c ξ j ) 

= q m [ N 

m 

] 

(1 − 

N∏ 

ξ j ). 

j=1 

[N] = pN − q N 

p − q 

[ ] N [N]! 

= 

m 

, [N]! = [N][N − 1] · · · [1], 

[m]! [N − m]! , (q − binomial coefficient), 24 


Final series result for case Y = Z + 

P (x m (t) ≤ x) = (−1) ∑ m k≥m 

∫ ∫ 

× · · · 

C R 

× ∏ i 

1 

k! 

[ ] k − 1 

k − m 

τ 

p (k−m)(k−m+1)/2 q km+(k−m)(k+m−1)/2 

∏ ξ j − ξ i 

∏ 1 

C R 

p + qξ i ξ j − ξ i (1 − ξ 

i≠j 

i i )(qξ i − p) 

( ) ξi x e ε(ξ i)t 

dξ 1 · · · dξ k 

25 



P (x m (t) ≤ x) = (−1) ∑ m k≥m 

∫ ∫ 

× · · · 

C R 

× ∏ i 

1 

k! 

C R 

∏ 

[ ] k − 1 

k − m 

τ 

i≠j 

p (k−m)(k−m+1)/2 q km+(k−m)(k+m−1)/2 

ξ j − ξ i 

∏ 


( 

ξ x i e ε(ξ i)t ) dξ 1 · · · dξ k 

i 

1 

(1 − ξ i )(qξ i − p) 

● For p=0 only k=m term is nonzero 


25


P (x m (t) ≤ x) = (−1) ∑ m k≥m 

∫ ∫ 

× · · · 

C R 

× ∏ i 

1 

k! 

C R 

∏ 

[ ] k − 1 

k − m 

τ 

i≠j 

p (k−m)(k−m+1)/2 q km+(k−m)(k+m−1)/2 

ξ j − ξ i 

∏ 


( 


i 

1 

(1 − ξ i )(qξ i − p) 


• Recognize double product as a determinant 

whose entries are a kernel, i.e. K(ξ i ,ξ j ) 


25


P (x m (t) ≤ x) = (−1) ∑ m k≥m 

∫ ∫ 

× · · · 

C R 

× ∏ i 

1 

k! 

C R 

∏ 

[ ] k − 1 

k − m 

τ 

i≠j 

p (k−m)(k−m+1)/2 q km+(k−m)(k+m−1)/2 

ξ j − ξ i 

∏ 


( 


i 

1 

(1 − ξ i )(qξ i − p) 


• Recognize double product as a determinant 

whose entries are a kernel, i.e. K(ξ i ,ξ j ) 

• Result can then be expressed as a contour 

integral whose integrand is a Fredholm 

determinant 


25

Fredholm determinant 

• Let K(x,y) be a kernel function 

• Fredholm expansion of det(I-λK): 

(−1) n 

n! 

∫ 

· · · 

∫ 

det (K(ξ i , ξ j ) 1≤i,j≤n 

dξ 1 · · · dξ n = 

∫ 

C 

det (I − λK) 

dλ 

λ n+1 

•Can then do sum over k (q-Binomial theorem): 


26

Final expression for m th particle distribution fn. 


Set γ = q − p>0, τ = q/p and 

define an integral operator K on 

circle C R : 

Then 

K(ξ, ξ ′ )=q ξ′x e tε(ξ′ )/γ 

p + qξξ ′ − ξ 

P (x m (t/γ) ≤ x) = 

∫ 

det(I − λK) 

∏ m−1 

k=0 (1 − λτ k−1 ) 

The contour in the λ-plane encloses 

all of the singularities of 

the integrand. 

dλ 

λ 

↑ k-1→k 


27

Asymptotic analysis 

We now transform the operator 

K so that we can perform a 

steepest descent analysis. 

Recall that the generic behavior 

for the coalescence of two saddle 

points leads to the Airy function 

Ai(x) 

George Airy 


28

ξ −→ 1 − τη 

1 − η , τ = p q < 1, 

K(ξ, ξ ′ ) −→ K 2 (η, η ′ )= ϕ(η′ ) 

η ′ − τη 

ϕ(η) = 

( 1 − τη 

1 − η 

) x 

e [ 1 

1−η − 1 

1−τη] 

t 

Introduce: K 1 (η, η ′ ) = ϕ(τη) 

η ′ − τη 


29

Two Preliminary Results 


30


Proposition: 

Let Γ be any closed curve going around η=1 

once counterclockwise with η=1/τ on the 

outside. Then the Fredholm determinant of 

K(ξ,ξ’) acting on C R has the same Fredholm 

determinant as K 1 (η,η’)-K 2 (η,η’) acting on Γ. 


30


Proposition: 

Let Γ be any closed curve going around η=1 

once counterclockwise with η=1/τ on the 

outside. Then the Fredholm determinant of 

K(ξ,ξ’) acting on C R has the same Fredholm 

determinant as K 1 (η,η’)-K 2 (η,η’) acting on Γ. 

Proposition: 

Suppose the contour Γ is star-shaped with 

respect to η=0. Then the Fredholm 

determinant of K 1 acting on Γ is equal to 

∞∏ 


k=0 

(1 − λτ k ) 

30

Denote by R the resolvent kernel of K 1 

Factor determinant: 

det(I-λ K)=det(I-λ K 1 ) det(I+K 2 (I+R)) 

Set λ=τ -m μ so formula for distr. fn becomes 

∫ 

∞ 

∏ 

k=0 

(1 − µτ k ) det ( I + τ −m µK 2 (I + R) ) dµ 

µ 

μ runs over a circle of radius > τ 


31

By a perturbative expansion of R, followed 

by a deformation of operators, we show 

det (I + λK 2 (I + R)) = det (I + µJ) 

∫ 

J(η, η ′ ϕ∞ (ζ) ζ m f(µ, ζ/η ′ ) 

) = 

ϕ ∞ (η ′ ) (η ′ ) m+1 ζ − η 

ϕ ∞ (η) = (1 − η) −x e ηt 

1−η 

f(µ, z) = 

∞∑ 

k=−∞ 

τ k 

1 − τ k µ zk 

dζ 

The kernel J(η,η’), which acts on a circle centered 

at 0 with radius less than τ, is analyzed by the 

steepest descent method. 

Note: m now appears inside the kernel! 


32

Main Result 

We set 

σ = m t ,c 1 = −1+2 √ σ,c 2 = σ −1/6 (1− √ σ) 2/3 , γ = q−p 

Theorem (TW). When 0 ≤ p

Total Current Fluctuations 

I(x,t) = # of particles ≤ x at time t, x ≤0 

Theorem. 

( I([−vt], t/γ) − a1 

lim P t 

t→∞ a 2 t 1/3 

≤ s 

) 

=1− F 2 (−s) 

where 

0 ≤ v

Remarks 


35

Remarks 

• Theorems generalize Kurt Johansson’s results 

for TASEP to ASEP. 


35

Remarks 



• Coefficients a 1 & c 1 known since 1980s, Liggett, 

Rost. 


35

Remarks 




Rost. 

• These theorems establish KPZ Universality at 

the fluctuation level for ASEP. 


35

Remarks 




Rost. 

• These theorems establish KPZ Universality at 

the fluctuation level for ASEP. 

• Balázs & Seppäläinen; Quastel & Valkó prove 

t ⅓ fluctuations for ASEP with Bernoulli product 

initial condition -- general probabilistic 

methods 


35


36

● Would like a conceptual understanding of why identities 

& cancellations appear in ASEP proofs. 


36



● Extend ASEP results to other initial conditions, e.g. 

flat initial conditions. Do we see F 1 as in TASEP? 


36





● Can we apply Bethe Ansatz methods to other growth 

models? 


36





● Can we apply Bethe Ansatz methods to other growth 

models? 

● Ultimately we want universality theorems not to rely 

upon integrable stucture of ASEP. For ⅓ exponent 

progress by Balázs, Seppäläinen, Quastel & Valkó. 


36

Thanks to Anne Schilling & Doron Zeilberger 

for advice with the combinatorial identities 


37

Craig A. Tracy Joint work with Harold Widom

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