aa r X i v : . [ m a t h . P R ] M a y Blocks in ASEP with step-Bernoulli initialcondition
Kyle Johnson
Department of MathematicsUniversity of CaliforniaDavis, CA 95616, USAemail: [email protected]
May 30, 2019
Abstract
This paper extends work by Tracy and Widom on blocks in theasymmetric simple exclusion process (ASEP) to the case of step-Bernoulli initial condition. We consider the probability that a particleat site x is the beginning of a block of L consecutive particles at time t in ASEP with step-Bernoulli initial condition. A Fredholm determi-nant representation for this probability is derived, and the asymptoticsare computed for the KPZ regime. The asymmetric simple exclusion process (ASEP) is a stochastic process onthe integer lattice Z where each particle waits an exponential time, thenjumps to the left with probability p and to the right with probability q = 1 − p ,unless the site to which it would jump is occupied in which case the particleremains where it is.In the case of step initial condition, where at time zero positive inte-ger sites are occupied and all other sites are unoccupied, a formula for thedistribution of the m th particle from the left [4] was the starting point for1he one-point probability distribution of the height function for the Kardar-Parisi-Zhang (KPZ) equation with narrow wedge initial conditions [1, 3].In [5], a formula for the probability that a block of L particles starts at site x at time t was derived; and in [6], the asymptotics (as x, m → ∞ in the KPZregime) of this probability were computed for step initial condition. In [7],a formula for the probability that the m th particle is in site x at time t wasderived for step-Bernoulli initial condition, where positive integer sites areoccupied with probability ρ and the other sites are unoccupied. Asymptoticsof this system (as x, m, t → ∞ ) were computed in the KPZ regime. Here,we combine the two cases above and consider the probability that the m thparticle is at the beginning of a block of length L at time t starting from step-Bernoulli initial condition, and then compute the asymptotics. Formally, thisis the probability P L,ρ ( x, m, t ) of the event x m ( t ) = x, x m +1 ( t ) = x + 1 , . . . , x m + L − ( t ) = x + L − ρ , where x m ( t )is the position of the m th particle at time t .We work under the assumption that q = 0 and we define τ = p/q , γ = q − p >
0. To state our first theorem, we must introduce some notation. Let ζ , ζ ′ ∈ C and define U ( ζ , ζ ′ ) = p + qζ ζ ′ − ζζ ′ − ζǫ ( ζ ) = pζ − + qζ − . Let C R be the circle centered at the origin with radius R oriented coun-terclockwise and let z ∈ C L . Let K x , K x,ρ and K L,x,ρ ( z ) be the integraloperators acting on C R with kernels K x ( ξ, ξ ′ ) = ξ x e ǫ ( ξ ) t p + qξξ ′ − ξ ,K x,ρ ( ξ, ξ ′ ) = qK x ( ξ, ξ ′ ) ρ ( ξ − τ ) ξ − ρ (1 − τ ) , We use the same notation for the operator as its kernel. When we say an operator acts on a subset of C , we mean that it acts on L functionson that set. K L,x,ρ ( ξ, ξ ′ ; z ) = K x,ρ ( ξ, ξ ′ ) L Y j =1 U ( z j , ξ )respectively. Here R is chosen to be large enough so that C R contains all thesingularities of the integrand.Let ( λ ; τ ) m = Q m − j =0 (1 − λτ j ) be the τ -Pochhammer symbol.All contour integrals are to be given a factor of 1 / πi . The empty productis taken to be 1. Theorem 1
Assume p, q = 0 . Then P L,ρ ( x, m, t ) =( − L − p L ( L +1) / τ (1 − m )( L − × Z Γ ,τ · · · Z Γ ,τ z L ( qz − p ) z L − ( qz − p ) · · · z L ( qz L − p ) × (cid:20)Z det( I − p − L λK L,x + L − ,ρ ( z ))( λ ; τ ) m dλλ L (cid:21) Y ≤ i
In the limit as t → ∞ , P L,ρ ( x, m, t/γ ) = c − σ ( L − / F ′ ( s ) t − / + o ( t − / ) when σ < ρ ,c − σ ( L − / dds ( F ( s ) ) t − / + o ( t − / ) when σ = ρ , ρ < ,c ′− σ ( L − / G ′ ( s ) t − / + o ( t − / ) when σ > ρ , ρ < . Where in the first two cases s = ( x − c t ) c − t − / and in the third case s = ( x − c ′ t ) c ′− t − / . These hold uniformly for s in a bounded set, anduniformly for σ in a compact subset of its domain. Notice that when σ < ρ and ρ = 1 we recover the result of [6]. Corollary
For all < ρ ≤ , given that the m th particle from the left is in position x attime t the conditional probability that it is the beginning of a block of length L converges to σ ( L − / as t → ∞ . This follows from the fact that the conditional probability is equal to. P L,ρ ( x, m, t/γ ) P ,ρ ( x, m, t/γ ) = σ ( L − / + o (1) .
4n Section 2, we prove Theorem 1. In Section 3 we derive an alternateformula for P L,ρ ( x, m, t ) that will be used in the proof of Theorem 2. InSection 4 we prove Theorem 2 in three cases. The first two cases of Theorem2 are highly similar to [6] and so the details are left to Appendices B and C.Appendix A contains a result of Harold Widom [9] which is used in Section4.3. We begin by introducing notation. Let f L ( ξ ) = p L ( L +1) / − LN Y i ξ Li Z Γ ξ · · · Z Γ ξ φ L ( z , . . . , z L ; ξ ) dz · · · dz L , and I L ( x, k, ξ ) = Y ≤ i U, V ⊂ Z , define σ ( U, V ) = { ( u, v ) : u ∈ U, v ∈ V } . Let P L,Y ( x, m, t ) by the probability that the mth particle from the left is atsite x at time t starting from the initial conditions where the sites Y areoccupied and all other sites are unoccupied. Theorem 2 of [5], after a littlealgebra, is P L,Y ( x, m, t ) = X k ≥ m + L − c m,k,L X S ⊂ Y | S | = k τ σ ( S,Y ) Z | ξ i | = R I L ( x, S, ξ ) d | S | ξ, (1)5here R is large enough for the poles of I L to be contained inside the contour, c m,k,L = q k ( k − / ( − m +1 τ m ( m − / τ − km (cid:20) k − Lm − (cid:21) τ , and (cid:20) nm (cid:21) τ = m Y i =1 − τ n − i − − τ i − is the τ -binomial coefficient.To find the probability for step-Bernoulli initial condition, we first takea weighted average over all initial configurations Y ⊂ { , ...N } . As in [7],the only factor in (1) that depends on Y is τ σ ( S,Y ) and the probability of theinitial condition Y is given by ρ | Y | (1 − ρ ) N −| Y | . From [7] we have that X S ⊂ Y ⊂ [1 ,N ] ρ | Y | (1 − ρ ) N −| Y | τ σ ( S,Y ) = τ k ( k +1) / ρ k Y i (1 − ρ + τ k − i +1 ρ ) t i (2)where we define t i = s i − s i − − S (rather than just the size of S ) is Y ξ − s i i = Y ξ − ( t + ··· + t i + i ) i = Y ξ − ii · Y ( ξ i ξ i +1 · · · ξ k ) − t i . Multiplying this by (2) and summing over all S ⊂ { , ...N } we obtain τ k ( k +1) / ρ k Y ξ − ii X P t i ≤ N − k Y i (cid:18) − ρ + τ k − i +1 ρξ i · · · ξ k (cid:19) t i . Letting N → ∞ then gives us τ k ( k +1) / ρ k Y i ξ i · · · ξ k − ρ − τ k − i +1 ρ . And so we may conclude that P L,ρ ( x, m, t ) = X k ≥ m + L − τ k ( k +1) / ρ k c m,k,L × Z | ξ i | = R I L ( x, k, ξ ) Y i ξ i · · · ξ k − ρ − τ k − i +1 ρ Y i dξ i f L is symmetric in the ξ i and so applying combinatorial iden-tity (9) from [7] gives us P L,ρ ( x, m, t ) = X k ≥ m + L − k ! q k ( k − / τ k ( k +1) / c m,k,L × Z | ξ i | = R Y i = j U ( ξ i , ξ j ) f L ( ξ ) × Y i ρξ i − ρ (1 − τ ) Y i ξ x − i e ǫ ( ξ i ) t − ξ i dξ i . Applying the alternative expression of f L from [5], the above integrandbecomes( − L p L ( L − / − kL Z Γ ,τ · · · Z Γ ,τ z L ( qz − p ) z L − ( qz − p ) · · · z L ( qz L − p ) Y i 11 + τ n αη , where, as in [7], α = − ρρ . Let f ( µ, z ) = X k ∈ Z τ k − τ k µ z k , and V ( ζ , η ′ ; w ) = wζ − τwη ′ − τ . Let J L,x,m,ρ ( w ) be the operator with kernel J L,x,m,ρ ( η, η ′ ; w ) = Z ϕ ∞ ( ζ ) ϕ ∞ ( η ) ζ m − L ( η ′ ) m − L +1 f ( µ, ζ /η ′ ) ζ − η L Y j =1 V ( ζ , η ′ ; w j ) dζ . (5)When ρ > / − ( τ n α ) − lie outside the unit circle and we8ake the ζ -contour to be a circle slightly larger than the unit circle. In thiscase the operator J L,x,m,ρ acts on a circle slightly smaller than the unit circle.When ρ ≤ / ζ -contour is the circle with diameter [ − α − + δ, δ ] and J L,x,m,ρ acts on the circle with diameter [ − α − + 2 δ, − δ ]. Lemma Let the contours Γ ,τ be defined as in Theorem 1. Then P L,ρ ( x, m, t ) = − τ − ( L − L +2) / Z Γ ,τ · · · Z Γ ,τ L Y j =1 ( w j − L − j w j ( w j − τ ) L − j +1 Y i 2, the singularities in this product are all greaterthan τ − so the argument from [6] goes over unchanged.The right side of Theorem 1 is an analytic function of ρ on (0 , ρ on (0 , 1) by varyingthe ζ , η , and w i -contours such that the singularities − ( τ n α ) − lie outside the ζ -contour, the η -contour remains inside the ζ -contour, and the w i -contourremains inside the η contour. The expressions agree for ρ ∈ (1 / , 1] and soagree for all ρ . Theorem 2 naturally breaks into three cases. In all cases, we assume m = σt .In the first two cases, we assume x = c t + sc t / and, as in [6], compute theasymptotics using the substitutions η → ξ + c − t − / η, η ′ → ξ + c − t − / η ′ , ζ → ξ + c − t − / ζ . (7) σ < ρ When σ < ρ the argument from [6] goes through unchanged since the polestructure of the integrand is the same and the value of ϕ ∞ ( ζ ) /ϕ ∞ ( η ′ ) tends9o 1 near the saddle point. Thus, we may conclude that, as t → ∞ , P L,ρ ( x, m, t/γ ) = c − σ ( L − / F ′ ( s ) t − / + o ( t − / ) . This is the first part of Theorem 2. For a more detailed argument see Ap-pendix B. σ = ρ In this case, as in [6], we still have that the operator µJ x,L,m,ρ has the sameFredholm determinant as the sum of J (0) + o (1)and J (1) L X j =1 w j w j ξ − τ c − t − / + o ( t − / )where o (1) and o ( t − / ) denote operators whose trace norms are o (1) and o ( t − / ) respectively. However, J (0) and J (1) have kernels J (0) ( η, η ′ ) = Z Γ ζ e ζ / sζ +( η ′ ) / − sη ′ ( ζ − η )( η ′ − ζ ) ηζ dζ and J (1) ( η, η ′ ) = − Z Γ ζ e ζ / sζ +( η ′ ) / − sη ′ ζ − η ηζ dζ . This is because, after the substitutions (7), L Y j =1 V ( ζ , η ′ ; w j ) → ζ − η ) L X j =1 w j w j ξ − τ c − t − / + E ( ζ , η ′ ; w ) , (8)where E ( ζ , η ′ ; w ) is a polynomial in ζ − η ′ with O ( t − / ) coefficients. If wedistribute the other factors in our kernel through this sum, we can consider J L,x,m,ρ as the sum of three operators. The operator corresponding to theadditive 1 in (8) is J (0) + o (1), by the argument in [7]. Although our kernelhas m − L where the kernel in [7] has m , this results only in an O (1 /t ) changein σ . The argument in [7] begins by setting σ = ρ + o ( t − / ) and so the10 (1 /t ) change makes no difference. Similarly, the operator corresponding tothe second term in (8) is J (1) L X j =1 w j w j ξ − τ c − t − / + o ( t − / ) . The operator corresponding to the error term E ( ζ , η ′ ; w ) has trace norm O ( t − / ).Since J (1) = dds J (0) and J (0) is independent of w , the argument from [6]can be followed word for word to conclude P L,ρ ( x, m, t/γ ) = c − σ ( L − / dds (cid:0) F ( s ) (cid:1) t − / + o ( t − / ) . σ > ρ Define ψ such that e ψ ( ζ ) = (1 − ζ ) − x − L +1 e ζ − ζ t ζ m − L . (9)In [7] the determinant of the J -kernel was computed by deforming the η -contour to the component of the level curve | e ψ ( η ) /e ψ ( − α − ) | = 1 − δ withthe saddle point outside and a small indentation to the left of η = 1, and the ζ -contour to the component of the level curve | e ψ ( ζ ) /e ψ ( − α − ) | = 1 − δ withthe saddle point inside and a small indentation to the right of ζ = 1. Thisresulted in the norm of the operator represented by the new ζ integral beingexponentially small.The extra factors found when considering blocks do not depend on t , and sothey only change the norm of the operator by O (1). Thus, if we change thecontours of J as above, we pick up the residue e ψ ( − α − ) e ψ ( η ) η − f ( µ, − ( αη ) − ) (cid:18) − η α − (cid:19) c ′ st / ∞ Y n =1 τ n αη − τ n L Y j =1 − α − w − τwη − τ from the pole at ζ = − α − . Therefore, with exponentially small error in t ,det( I + µJ ) equals 11+ µ Z e ψ ( − α − ) e ψ ( η ) η − f ( µ, − ( αη ) − ) (cid:18) − η α − (cid:19) c ′ st / ∞ Y n =1 τ n αη − τ n L Y j =1 − α − w − τwη − τ dη. We can ignore the first summand (which is 1) because, by appendix B of [5], Z Γ ,τ · · · Z Γ ,τ L Y j =1 ( w j − L − j w j ( w j − τ ) L − j +1 Y i 1) so ( ξ − − ρ ) = − 1. Combining this with(11) and (6) gives us that P L,ρ ( x, m, t/γ ) = t − / c ′− √ σ L − G ′ ( s ) + o ( t − / )as t → ∞ . Appendix A w -integrals The following proof is due to Harold Widom [9].All integrals in this appendix are over Γ ,τ . LetΦ L ( w , . . . , w L ) = Y j ≥ ( w j − L − j w j ( w j − τ ) L − j +1 Y i 1) log(1 − ζ )+ ζ − ζ t +( m − L ) log( ζ ) . Differentiating this expression provides the saddle point equation x + L − − ζ + t (1 − ζ ) + m − Lζ = 0 . The two saddle points coincide when m − L = ( x + L − t ) t . (16)If m − L = σt and x + L − c t then (16) gives us σ = ( c + 1) c = − √ σ (we take the positive root because c should increase with σ ). Thus, the saddle point is at ξ = − √ σ − √ σ . Let c = σ − / (1 − √ σ ) / and set x + L − c t + c st / . Let ψ be as in(9). Taking the Taylor expansion of ψ around the point ζ = ξ , we have ψ ( ζ ) = − c t ( ζ − ξ ) / c st / ( ζ − ξ ) + O ( t ( ζ − ξ ) ) + O ( t / ( ζ − ξ ) ) . (17)where c = σ − / (1 − √ σ ) / Define ψ ( ζ ) = ( x + L − 1) log(1 − ζ ) + ζ − ζ t +( m − L ) log( ζ ) and ψ ( ζ ) = ψ ( ζ ) − ψ ( ξ ). Lemma 5 of [4] gives us the exis-tence of contours Γ ζ and Γ η with the following properties:(i) The part of Γ η in a neighborhood N η of η = ξ is a pair of rays ξ in thedirections ± π/ ζ in a neighborhood N ζ of ζ = ξ is a pairof rays from ξ − t − / in the directions ± π/ δ > ψ ( ζ )) < − δ on Γ ζ \ N ζ and Re( ψ ( η )) > δ on Γ η .(iii) The circular η - and ζ –contours for J L,x,m,ρ can be simultaneously de-formed to Γ η and G ζ respectively, so that during the deformation the inte-grand in (5) remains analytic in all variables (this requires σ < ρ ).16y condition (iii) of the above lemma and Proposition 1 of [4], det J remainsthe same if J acts on Γ η and the integral (5) is over Γ ζ .Aside from the e ψ ( ζ ) /e ψ ( η ) factor, the J -kernel is uniformly O ( t / ). Thisis because ζ − η = t − / , and none of the remaining factors grow with t .Together with (ii), this gives us that when we restrict to ζ ∈ Γ ζ \ N ζ and η ∈ Γ η \ N η , J has exponentially small trace norm. For a < / 3, we mayfurther restrict η and ζ to rays of length t − a , since the kernel has trace norm O ( e δt − a ) outside of a t − a -neighborhood of ξ by (17).On these rays, we make the substitutions (7). Each V ( ζ , η ′ ; w ) becomes1 + ( ζ − η ′ ) wwξ − τ c − t − / [1 + O (min(1 , t − / | η ′ | ))], the product Q V ( ζ , η ′ ; w j )becomes (8), and so J can be considered as T + T + T , corresponding tothe terms 1, ( ζ − η ) P w j w j ξ − τ c − t − / , and E ( ζ , η ′ ; w ) in (8). T can be written as the product A B where, before substitution, A : L (Γ ζ ) → L (Γ η ) and B : L (Γ η ) → L (Γ ζ ) have kernels A ( η, ζ ) = e ψ ( ζ ) ζ − η ∞ Y n =0 τ n αη τ n αζ , B ( ζ , η ′ ) = µf ( µ, ζ /η ) ηe ψ ( η ) (18)respectively. After substitution, the factor 1 / ( ζ − η ) is unchanged. Theproduct in the kernel of A becomes ∞ Y n =0 c t / ( α − + τ n ξ ) + τ n ηc t / ( α − + τ n ξ ) + τ n ζ . (19)Each factor is 1 + o (1) since, for some ǫ > 0, for all n ≥ α − + τ n ξ > ǫ (this is where we need the assumption σ < ρ ).Notice that near z = 1, f ( µ, z ) = O (cid:18) | − z | (cid:19) and f ( µ, z ) = µ − − z + O (1) , so µf ( µ, ζ /η ) /η in the kernel of B becomes O (cid:18) | η − ζ | (cid:19) and 1 η − ζ + O ( t − / ) (20)after the substitutions. From (17), we have that, for some δ > e ψ ( ζ ) and e − ψ ( η ) are, respectively, O ( e − δ | ζ | ) and O ( e − δ | η | ) after rescaling. Thus we canbound the rescaled kernels by constants times e − δ | ζ | | ζ − η | , e − δ | η | | η − ζ | , a > / 4, so doesthe error term in (17). Thus the kernels have pointwise limits e − ζ / sζ ζ − η , e η / − sη η − ζ , respectively. Thus, T converges to the operator J (0) with kernel J (0) ( η, η ′ ) = Z Γ ζ e − ζ / sζ +( η ′ ) / − sη ′ ( ζ − η )( η ′ − ζ ) dζ . (21)The contour Γ ζ converges to the rays from − c to − c + ∞ e ± πi/ andthe contour Γ η converges to the rays from 0 to ∞ e ± πi/ .Since Re( ζ − η ′ ) < ζ ∈ Γ ζ and η ′ ∈ Γ η , it follows that e s ( ζ − η ′ ) η ′ − ζ = Z ∞ s e x ( ζ − η ′ ) dx. Thus (21) is equal to Z ∞ s Z Γ ζ e − ζ / η ′ ) / x ( ζ − η ′ ) ζ − η dζ dx. Thus, the limiting operator can be written as the product of an operatorfrom L ( s, ∞ ) to L (Γ η ) with kernel Z Γ ζ e − ζ / xζ ζ − η dζ and an operator from L (Γ η ) to L ( s, ∞ ) with kernel e − xη + η / . Changing the order of the operators preserves the Fredholm determinant,and Z Γ ζ Z Γ η e − ζ / η / yζ − xη ζ − η dηdζ = − K Airy ( x, y )18o we can concludedet( I + µT ) → det( I − K Airy χ ( s, ∞ ) ) = F ( s ) . (22)The operator T can be written as A B L X j =1 w j w j ξ − τ c − t − / where, before substitution, A : L (Γ ζ ) → L (Γ η ) and B : L (Γ η ) → L (Γ ζ )have kernels e ψ ( ζ ) ζ − η ∞ Y n =0 τ n αη τ n αζ , ( ζ − η ) µf ( µ, ζ /η ) ηe ψ ( η ) respectively. Notice A = A and B is bounded by e − δ | η | , which is Hilbert-Schmidt, so A B converges in trace norm to its pointwise limit J (1) , withkernel J (1) ( η, η ′ ) = − Z Γ ζ e − ζ / sζ +( η ′ ) / − sη ′ ζ − η dζ . Notice that J (1) = dds J (0) .For the operator T , consider E ( ζ , η ′ ; w ) = X ζ n h k ( η ′ )where h k ( η ′ ) = O ( t − / | η ′ | k +1 ) and the (finite) sum ranges over appropriatevalues of n, k . Thus we have T ( η, η ′ ) = X Z Γ ζ (cid:0) A ( η, ζ ) ζ n (cid:1)(cid:0) B ( ζ , η ′ ) h k ( η ′ ) (cid:1) dζ where A ( η, ζ ), B ( ζ , η ′ ) are as in (18). From the bounds given in (B), we seethat A ( η, ζ ) and B ( ζ , η ′ ) are decaying exponentially in ζ , η ′ , respectively,and so after multiplying the kernels by | ζ | n and | η ′ | k , respectively, they remainHilbert-Schmidt. Thus the trace norm of T is O ( t − / ).We have shown that T = J (0) + o (1) (where the o (1) term is independentof all w j ), T = J (1) P Lj =1 w j w j ξ − τ c − t − / + o ( t − / ), and T = O ( t − / ). This19ives usdet( I + µJ x,L,m,ρ ( w )) = det I + J (0) + o (1) + J (1) L X j =1 w j w j ξ − τ c − t − / + o ( t − / ) ! = det( I + J (0) + o (1)) det I + ( I + J (0) ) − J (1) L X j =1 w j w j ξ − τ c − t − / + o ( t − / ) ! = ( F ( s ) + o (1)) " I + J (0) ) − J (1) ) L X j =1 w j w j ξ − τ c − t − / + o ( t − / ) Recall J (1) = dds J (0) , so it follows thattr(( I + J (0) ) − J (1) ) = dds log det( I + J (0) ) = F ′ ( s ) F ( s ) , thereforedet( I + µJ x,L,m,ρ ( w )) = F ( s ) + o (1) + F ′ ( s ) L X j =1 w j w j ξ − τ c − t − / + o ( t − / ) . (23)The o (1) is independent of the w j , as before. All terms are independent of µ , so to evaluate the µ -integral in (6), it suffices to compute Z ( τ L µτ ) ∞ dµµ L = ( − L − τ ( L − L − / (1 − τ ) · · · (1 − τ L − ) . (24)From section IV of [5], we know Z Γ ,τ · · · Z Γ ,τ L Y j =1 ( w j − L − j w j ( w j − τ ) L − j +1 Y i The author thanks Harold Widom for providing the proof in Appendix A, andCraig Tracy for helping in all stages of this work. This work was supportedby the National Science Foundation through grants DMS-1207995 and DMS-1809311. 22 eferences [1] Amir, G., Corwin, I., Quastel, J.: Probability distribution of the freeenergy of the continuum directed random polymer in 1+1 dimensions ,Commun. Pure Appl. Math. , 466-537 (2011)[2] Ferrari, P.L., Spohn, H.: A determinantal formula for the GOE Tracy-Widom distribution. J.Phys. A: Math. Gen. , L557-L561 (2005)[3] Sasamoto T., Spohn, H., The crossover regime for the weakly asymmetricexclusion process , J. Stat. 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