LECTURE III Asymmetric Simple Exclusion Process: Bethe Ansatz Approach to the Kolmogorov Equations. Craig A. Tracy UC Davis

Transcription

1 LECTURE III Asymmetric Simple Exclusion Process: Bethe Ansatz Approach to the Kolmogorov Equations Craig A. Tracy UC Davis Bielefeld, August 2013

2 ASEP on Integer Lattice q p suppressed both suppressed

3 What is Bethe Ansatz? Wiktionary: Ansatz is German word that means a mathematical expression to describe a certain phenomenon, but without a proved justification for its use. From Wolfram MathWorld: an assumed form for a mathematical statement that is not based on any underlying theory or principle. Google Bethe Ansatz gives about 167,000 results. Google Hans Bethe gives about 278,000 results. Google Bethe gives about 3,030,000 results.

4 Figure: Hans Bethe, Nobel Prize in Physics, 1967, for his contributions to the theory of nuclear reactions, especially his discoveries concerning the energy production in stars.

5 H. Bethe, 1931: On the Theory of Metals, I. Eigenvalues and Eigenfunctions of a Linear Chain of Atoms [English translation]. In this paper Bethe gave a method to diagonalize a certain quantum spin system (called the Heisenberg spin chain or XXX-spin chain). This is the only paper Bethe ever wrote on Bethe Ansatz. Method was developed in the 1960s by Lieb, McGuire, C.N. Yang, C.P. Yang, Sutherland,.... In 1966 Yang & Yang extended the Bethe Ansatz to study the spectral theory of the XXZ quantum spin chain. The Hamiltonian is defined on a Hilbert space j C2 j H = ( ) σj x σj+1 x + σ y j σy j+1 + σz j σj+1 z j where σj α are Pauli spin matrices acting in slot j and the identity elsewhere. (Bethe considered = 1.) M. T. Bachelor, The Bethe Ansatz After 75 Years, Physics Today, January 2007.

6 Nice story, but what has all this to do with ASEP?

7 Nice story, but what has all this to do with ASEP? The generator of the Markov process ASEP is a similarity (not unitary!) transformation of the XXZ quantum spin system. This observation goes back at least to Gwa & Spohn (1992).

8 Nice story, but what has all this to do with ASEP? The generator of the Markov process ASEP is a similarity (not unitary!) transformation of the XXZ quantum spin system. This observation goes back at least to Gwa & Spohn (1992). This suggests we use the ideas of Bethe Ansatz and apply them to ASEP.

9 Nice story, but what has all this to do with ASEP? The generator of the Markov process ASEP is a similarity (not unitary!) transformation of the XXZ quantum spin system. This observation goes back at least to Gwa & Spohn (1992). This suggests we use the ideas of Bethe Ansatz and apply them to ASEP. Here s the rub: Bethe Ansatz was initially designed for the spectral theory of the operator. We want the transition probability. Not so easy to go from one to the other in practice.

10 Nice story, but what has all this to do with ASEP? The generator of the Markov process ASEP is a similarity (not unitary!) transformation of the XXZ quantum spin system. This observation goes back at least to Gwa & Spohn (1992). This suggests we use the ideas of Bethe Ansatz and apply them to ASEP. Here s the rub: Bethe Ansatz was initially designed for the spectral theory of the operator. We want the transition probability. Not so easy to go from one to the other in practice. After this introduction, let s get to work!

11 Start with N-particle ASEP A state X = (x 1,..., x N ) is specified by giving the location x 1 < x 2 < < x N, x i Z of the N particles on the lattice Z. Want P Y (X ; t) = probability of state X at time t given that we are in state Y at t = 0 P Y (X ; t) satisfies a differential equation, called the Kolmogorov forward equation or the master equation.

12 N = 1: The differential equation is dp (x; t) = pp(x 1; t) + qp(x + 1; t) P(x; t), x Z. dt Want solution that satisfies the initial condition P(x; 0) = δ x,y DE is linear and separable in x and t

13 N = 1: The differential equation is dp (x; t) = pp(x 1; t) + qp(x + 1; t) P(x; t), x Z. dt Want solution that satisfies the initial condition P(x; 0) = δ x,y DE is linear and separable in x and t Let ξ C be arbitrary. Easy to see there are solutions of the form u(x; t) = f (ξ)ξ x e tε(ξ), ε(ξ) = p ξ + qξ 1. where f is any function of ξ.

14 N = 1: The differential equation is dp (x; t) = pp(x 1; t) + qp(x + 1; t) P(x; t), x Z. dt Want solution that satisfies the initial condition P(x; 0) = δ x,y DE is linear and separable in x and t Let ξ C be arbitrary. Easy to see there are solutions of the form u(x; t) = f (ξ)ξ x e tε(ξ), ε(ξ) = p ξ + qξ 1. where f is any function of ξ. DE is linear take linear superposition f (ξ)ξ x e tε(ξ) dξ

15 Set f (ξ) = ξ y 1 /(2πi) and choose contour of integration to be a circle centered at the origin of radius r P y (x; t) = 1 ξ x y 1 e tε(ξ) dξ 2πi C r

16 Set f (ξ) = ξ y 1 /(2πi) and choose contour of integration to be a circle centered at the origin of radius r P y (x; t) = 1 ξ x y 1 e tε(ξ) dξ 2πi C r Satisfies initial condition by residue theorem. P y (x; 0) = δ x,y

17 Set f (ξ) = ξ y 1 /(2πi) and choose contour of integration to be a circle centered at the origin of radius r P y (x; t) = 1 ξ x y 1 e tε(ξ) dξ 2πi C r Satisfies initial condition by residue theorem. P y (x; 0) = δ x,y This solves N = 1 ASEP. Solution is in Feller though not derived in the manner here.

18 N = 2 ASEP: X = (x 1, x 2 ) If x 2 > x 1 + 1: dp dt (x 1, x 2 ) = pp(x 1 1, x 2 ) + qp(x 1 + 1, x 2 ) + pp(x 1, x 2 1) + qp(x 1, x 2 + 1) 2P(x 1, x 2 ) (1) If x 2 = x 1 + 1: dp dt (x 1, x 2 ) = pp(x 1 1, x 2 ) + qp(x 1, x 2 + 1) P(x 1, x 2 ) (2)

19 N = 2 ASEP: X = (x 1, x 2 ) If x 2 > x 1 + 1: dp dt (x 1, x 2 ) = pp(x 1 1, x 2 ) + qp(x 1 + 1, x 2 ) + pp(x 1, x 2 1) + qp(x 1, x 2 + 1) 2P(x 1, x 2 ) (1) If x 2 = x 1 + 1: dp dt (x 1, x 2 ) = pp(x 1 1, x 2 ) + qp(x 1, x 2 + 1) P(x 1, x 2 ) (2) First equation is just two N = 1 problems. Second equation takes into account the exclusion. DE is now no longer constant coefficient.

20 N = 2 ASEP: X = (x 1, x 2 ) If x 2 > x 1 + 1: dp dt (x 1, x 2 ) = pp(x 1 1, x 2 ) + qp(x 1 + 1, x 2 ) + pp(x 1, x 2 1) + qp(x 1, x 2 + 1) 2P(x 1, x 2 ) (1) If x 2 = x 1 + 1: dp dt (x 1, x 2 ) = pp(x 1 1, x 2 ) + qp(x 1, x 2 + 1) P(x 1, x 2 ) (2) First equation is just two N = 1 problems. Second equation takes into account the exclusion. DE is now no longer constant coefficient. Bethe s first idea: Incorporate hard equation (2) into a boundary condition so that we have only to solve the easy equation (1).

21 We now consider the easy equation on all of X = (x 1, x 2 ) Z 2 : du dt (x 1, x 2 ) = pu(x 1 1, x 2 ) + qu(x 1 + 1, x 2 ) + pu(x 1, x 2 1) + qu(x 1, x 2 + 1) 2u(x 1, x 2 ) (3) Require that the solution to (3) satisfy the boundary condition pu(x 1, x 1 ) + qu(x 1 + 1, x 1 + 1) u(x 1, x 1 + 1) = 0, x 1 Z (4)

22 We now consider the easy equation on all of X = (x 1, x 2 ) Z 2 : du dt (x 1, x 2 ) = pu(x 1 1, x 2 ) + qu(x 1 + 1, x 2 ) + pu(x 1, x 2 1) + qu(x 1, x 2 + 1) 2u(x 1, x 2 ) (3) Require that the solution to (3) satisfy the boundary condition pu(x 1, x 1 ) + qu(x 1 + 1, x 1 + 1) u(x 1, x 1 + 1) = 0, x 1 Z (4) Observe that if u(x 1, x 2 ) satisfies (4) then for x 2 = x it satisfies the hard equation (2).

23 We now consider the easy equation on all of X = (x 1, x 2 ) Z 2 : du dt (x 1, x 2 ) = pu(x 1 1, x 2 ) + qu(x 1 + 1, x 2 ) + pu(x 1, x 2 1) + qu(x 1, x 2 + 1) 2u(x 1, x 2 ) (3) Require that the solution to (3) satisfy the boundary condition pu(x 1, x 1 ) + qu(x 1 + 1, x 1 + 1) u(x 1, x 1 + 1) = 0, x 1 Z (4) Observe that if u(x 1, x 2 ) satisfies (4) then for x 2 = x it satisfies the hard equation (2). Thus want solution to (3) that satisfies boundary condition (4) and initial condition u(x 1, x 2 ; 0) = δ x1,y 1 δ x2,y 2.

24 We now consider the easy equation on all of X = (x 1, x 2 ) Z 2 : du dt (x 1, x 2 ) = pu(x 1 1, x 2 ) + qu(x 1 + 1, x 2 ) + pu(x 1, x 2 1) + qu(x 1, x 2 + 1) 2u(x 1, x 2 ) (3) Require that the solution to (3) satisfy the boundary condition pu(x 1, x 1 ) + qu(x 1 + 1, x 1 + 1) u(x 1, x 1 + 1) = 0, x 1 Z (4) Observe that if u(x 1, x 2 ) satisfies (4) then for x 2 = x it satisfies the hard equation (2). Thus want solution to (3) that satisfies boundary condition (4) and initial condition u(x 1, x 2 ; 0) = δ x1,y 1 δ x2,y 2. Note that since (3) holds in all Z 2 it is constant coefficient DE. How to find the solution that satisfies the boundary condition? Bethe s second idea.

25 Let ξ 1, ξ 2 C. Then A 12 (ξ) ξ x 1 1 ξx 2 2 et(ε(ξ 1)+ε(ξ 2 )) is a solution to (3)

26 Let ξ 1, ξ 2 C. Then is a solution to (3) A 12 (ξ) ξ x 1 1 ξx 2 2 et(ε(ξ 1)+ε(ξ 2 )) But we can permute ξ 1 ξ 2 and still get a solution. Thus is a solution. {A 12 (ξ) ξ x 1 1 ξx A 21(ξ) ξ x 1 2 ξx 2 1 } et(ε(ξ 1)+ε(ξ 2 ))

27 Let ξ 1, ξ 2 C. Then is a solution to (3) A 12 (ξ) ξ x 1 1 ξx 2 2 et(ε(ξ 1)+ε(ξ 2 )) But we can permute ξ 1 ξ 2 and still get a solution. Thus is a solution. {A 12 (ξ) ξ x 1 1 ξx A 21(ξ) ξ x 1 2 ξx 2 1 } et(ε(ξ 1)+ε(ξ 2 )) Require this solution satisfy the boundary condition. Simple computation shows if we choose where then boundary condition satisfied. A 21 = S(ξ 2, ξ 1 )A 12 S(ξ, ξ ) = p + qξξ ξ p + qξξ ξ

28 C Choose A 12 = ξ y ξ y 2 1 2, then we have the solution { } ξ x 1 y ξ x 2 y S(ξ 2, ξ 1 ) ξ x 1 y ξ x 2 y e t(ε(ξ 1)+ε(ξ 2 )) dξ 1 dξ 2 C Here we ve incorporated a factor of (2πi) 1 with each integration.

29 C Choose A 12 = ξ y ξ y 2 1 2, then we have the solution { } ξ x 1 y ξ x 2 y S(ξ 2, ξ 1 ) ξ x 1 y ξ x 2 y e t(ε(ξ 1)+ε(ξ 2 )) dξ 1 dξ 2 C Here we ve incorporated a factor of (2πi) 1 with each integration. Does this solution satisfy the initial condition? If contour C r is chosen to be a circle of radius r centered at the origin, then the first term satisfies the initial condition. This means the second term must vanish at t = 0.

30 C Choose A 12 = ξ y ξ y 2 1 2, then we have the solution { } ξ x 1 y ξ x 2 y S(ξ 2, ξ 1 ) ξ x 1 y ξ x 2 y e t(ε(ξ 1)+ε(ξ 2 )) dξ 1 dξ 2 C Here we ve incorporated a factor of (2πi) 1 with each integration. Does this solution satisfy the initial condition? If contour C r is chosen to be a circle of radius r centered at the origin, then the first term satisfies the initial condition. This means the second term must vanish at t = 0. Actually, we need second term to vanish only in the physical region x 1 < x 2. In this region it does vanish if we choose r sufficiently small so that the poles coming from the zeros of the denominator of S lie outside of C r.

31 C Choose A 12 = ξ y ξ y 2 1 2, then we have the solution { } ξ x 1 y ξ x 2 y S(ξ 2, ξ 1 ) ξ x 1 y ξ x 2 y e t(ε(ξ 1)+ε(ξ 2 )) dξ 1 dξ 2 C Here we ve incorporated a factor of (2πi) 1 with each integration. Does this solution satisfy the initial condition? If contour C r is chosen to be a circle of radius r centered at the origin, then the first term satisfies the initial condition. This means the second term must vanish at t = 0. Actually, we need second term to vanish only in the physical region x 1 < x 2. In this region it does vanish if we choose r sufficiently small so that the poles coming from the zeros of the denominator of S lie outside of C r. Thus have solved for the transition probability for N = 2 ASEP.

32 C Choose A 12 = ξ y ξ y 2 1 2, then we have the solution { } ξ x 1 y ξ x 2 y S(ξ 2, ξ 1 ) ξ x 1 y ξ x 2 y e t(ε(ξ 1)+ε(ξ 2 )) dξ 1 dξ 2 C Here we ve incorporated a factor of (2πi) 1 with each integration. Does this solution satisfy the initial condition? If contour C r is chosen to be a circle of radius r centered at the origin, then the first term satisfies the initial condition. This means the second term must vanish at t = 0. Actually, we need second term to vanish only in the physical region x 1 < x 2. In this region it does vanish if we choose r sufficiently small so that the poles coming from the zeros of the denominator of S lie outside of C r. Thus have solved for the transition probability for N = 2 ASEP. S is the Yang-Yang S-matrix in ASEP variables.

33 P Y (X ; t) for N-particle ASEP X Z N X ± i = {x 1,..., x i 1, x i ± 1, x i+1,..., x N } The free equation on Z N R is du N dt (X ) = ( pu(x i ; t) + qu(x + i ; t) u(x ; t) ) i=1

34 P Y (X ; t) for N-particle ASEP X Z N X ± i = {x 1,..., x i 1, x i ± 1, x i+1,..., x N } The free equation on Z N R is du N dt (X ) = ( pu(x i ; t) + qu(x + i ; t) u(x ; t) ) i=1 The boundary conditions are pu(x 1,..., x i, x i,..., x N ; t) + qu(x 1,..., x i + 1, x i + 1,..., x N ) = u(x 1,..., x i, x i + 1,..., x N, i = 1, 2,..., N 1 This boundary condition comes when particle at x i is neighbor to particle at x i+1 = x i + 1

35 P Y (X ; t) for N-particle ASEP X Z N X ± i = {x 1,..., x i 1, x i ± 1, x i+1,..., x N } The free equation on Z N R is du N dt (X ) = ( pu(x i ; t) + qu(x + i ; t) u(x ; t) ) i=1 The boundary conditions are pu(x 1,..., x i, x i,..., x N ; t) + qu(x 1,..., x i + 1, x i + 1,..., x N ) = u(x 1,..., x i, x i + 1,..., x N, i = 1, 2,..., N 1 This boundary condition comes when particle at x i is neighbor to particle at x i+1 = x i + 1 Check that no new boundary conditions are needed, e.g. when 3 or more particles are all adjacent.

36 P Y (X ; t) for N-particle ASEP X Z N X ± i = {x 1,..., x i 1, x i ± 1, x i+1,..., x N } The free equation on Z N R is du N dt (X ) = ( pu(x i ; t) + qu(x + i ; t) u(x ; t) ) i=1 The boundary conditions are pu(x 1,..., x i, x i,..., x N ; t) + qu(x 1,..., x i + 1, x i + 1,..., x N ) = u(x 1,..., x i, x i + 1,..., x N, i = 1, 2,..., N 1 This boundary condition comes when particle at x i is neighbor to particle at x i+1 = x i + 1 Check that no new boundary conditions are needed, e.g. when 3 or more particles are all adjacent. Require initial condition u(x ; 0) = δ X,Y in physical region.

37 Look for solutions of the form (Bethe s second idea) u(x ; t) = A σ (ξ) ξ x i y σ(i) 1 σ(i) e t P i ε(ξ i ) dξ 1 dξ N C r C r σ S N i S N is the permutation group.

38 Look for solutions of the form (Bethe s second idea) u(x ; t) = A σ (ξ) ξ x i y σ(i) 1 σ(i) e t P i ε(ξ i ) dξ 1 dξ N C r C r σ S N i S N is the permutation group. Find boundary conditions are satisfied if the A σ satisfy A σ (ξ) = {S(ξ β, ξ α ) : {β, α} is an inversion in σ} The inversions in σ = (3, 1, 4, 2) are {3, 1}, {3, 2}, {4, 2}. Thus A id = 1.

39 Look for solutions of the form (Bethe s second idea) u(x ; t) = A σ (ξ) ξ x i y σ(i) 1 σ(i) e t P i ε(ξ i ) dξ 1 dξ N C r C r σ S N i S N is the permutation group. Find boundary conditions are satisfied if the A σ satisfy A σ (ξ) = {S(ξ β, ξ α ) : {β, α} is an inversion in σ} The inversions in σ = (3, 1, 4, 2) are {3, 1}, {3, 2}, {4, 2}. Thus A id = 1. Final step: Show u(x ; t) satisfies the initial condition. As before, the term corresponding to the identity permutation gives δ X,Y. We must show the sum of the N! 1 other terms sum to zero in the physical region! This turns out to be quite involved. It will be the case if r is chosen so that all singularities coming from the A σ lie outside the contour C r (we assume p 0). Our original article had an error. See the erratum.

40 Where do we go from here? P Y (X ; t) is a complicated object: A sum of N! terms each term a N-dimensional integral.

41 Where do we go from here? P Y (X ; t) is a complicated object: A sum of N! terms each term a N-dimensional integral. How will we ever let N?

42 Where do we go from here? P Y (X ; t) is a complicated object: A sum of N! terms each term a N-dimensional integral. How will we ever let N? Take q > p (drift towards the left) and consider the marginal distribution P(x 1 (t) = x) = P Y ({x, x 2,..., x N }; t) x<x 2 < <x N

43 Where do we go from here? P Y (X ; t) is a complicated object: A sum of N! terms each term a N-dimensional integral. How will we ever let N? Take q > p (drift towards the left) and consider the marginal distribution P(x 1 (t) = x) = P Y ({x, x 2,..., x N }; t) x<x 2 < <x N For the T(totally)ASEP case, q = 1, p = 0 this is particularly simple since the first particle is not influenced by the other particles. How does this change for ASEP?

44 Where do we go from here? P Y (X ; t) is a complicated object: A sum of N! terms each term a N-dimensional integral. How will we ever let N? Take q > p (drift towards the left) and consider the marginal distribution P(x 1 (t) = x) = P Y ({x, x 2,..., x N }; t) x<x 2 < <x N For the T(totally)ASEP case, q = 1, p = 0 this is particularly simple since the first particle is not influenced by the other particles. How does this change for ASEP? Will next show that P(x 1 (t) = x) can be expressed as a single N-dimensional integral. This will require one of our combinatorial identities.

45 Where do we go from here? P Y (X ; t) is a complicated object: A sum of N! terms each term a N-dimensional integral. How will we ever let N? Take q > p (drift towards the left) and consider the marginal distribution P(x 1 (t) = x) = P Y ({x, x 2,..., x N }; t) x<x 2 < <x N For the T(totally)ASEP case, q = 1, p = 0 this is particularly simple since the first particle is not influenced by the other particles. How does this change for ASEP? Will next show that P(x 1 (t) = x) can be expressed as a single N-dimensional integral. This will require one of our combinatorial identities. Indeed, we will encounter several miraculous identities. An open problem is to explain in more structural terms why such identities arise and to find new identities. More of this in Lecture III.

46 Summary Let χ N = { (x 1,..., x N ) Z N : x 1 < < x N } Semigroup e tl : P Y (X ; t) = C r e tl e Y = C r X χ N P Y (X ; t)e X, Y χ N, σ S N A σ (ξ) N i=1 ξ x i y σ(i) 1 σ(i) S N is the permutation group, ε(ξ) = p/ξ + qξ 1, and e t P i ε(ξ i ) dξ 1 dξ N A σ (ξ) = {S(ξ β, ξ α ) : {β, α} is an inversion in σ} S(ξ, ξ ) = p + qξξ ξ p + qξξ ξ Choose r 1 so that all poles from A σ lie outside of C r. Each dξ carries a factor (2πi) 1.

47 Alternative form for A σ Set then f (ξ, ξ ) = p + qξξ ξ A σ = sgn(σ) i<j f (ξ σ(i), ξ σ(j) ) i<j f (ξ i, ξ j ) Take p 0 P(x 1 (t) = x) : = = Cr N Marginal Distribution for x 1 (t) x<x 2 < <x N P Y ({x, x 2,..., x N }; t) ξ σ(2) ξσ(3) 2 A σ (ξ) ξn 1 σ(n) (1 ξ σ S σ(2) ξ σ(n) )(1 ξ σ(3) ξ σ(n) ) (1 ξ σ(n) ) N ξ x y i 1 i e t P i ε(ξ i ) dξ 1 dξ N i

48 First Combinatorial Identity ( i<j f (ξ σ(i), ξ σ(j) )) ξ σ(2) ξσ(3) 2 ξn 1 σ(n) sgn(σ) (1 ξ σ S σ(1) ξ σ(n) ) (1 ξ σ(n 1) ξ σ(n) )(1 ξ σ(n) ) N = p N(N 1)/2 i<j (ξ j ξ i ) j (1 ξ j) Thus we have, p 0, P(x 1 (t) = x) = p N(N 1)/2 ξ j ξ i 1 ξ 1 ξ N ( f (ξ i, ξ j ) i (1 ξ i) C N r i<j A single N-dimensional integral! i ξ x y i 1 i e tε(ξ i ) ) dξ 1 dξ N

49 Back Story to Proof of Identity I Led to conjecture identity from special cases discovered using Mathematica.

50 Back Story to Proof of Identity I I Led to conjecture identity from special cases discovered using Mathematica. But how to prove identity?

51 Back Story to Proof of Identity I I Led to conjecture identity from special cases discovered using Mathematica. But how to prove identity?

52 So we called Doron... He suggested that Problem VII.47 of Pólya & Szegö, an identity of Issai Schur, had a similar look about it and might be proved in a similar way. Doron was right.

53 So we called Doron... He suggested that Problem VII.47 of Pólya & Szegö, an identity of Issai Schur, had a similar look about it and might be proved in a similar way. Doron was right. Idea of proof: Use induction on N

54 So we called Doron... He suggested that Problem VII.47 of Pólya & Szegö, an identity of Issai Schur, had a similar look about it and might be proved in a similar way. Doron was right. Idea of proof: Use induction on N Call left-hand side ϕ N (ξ 1,..., ξ N ) and sum over all permutations such that σ(1) = k. This gives an expression involving ϕ N 1.

55 So we called Doron... He suggested that Problem VII.47 of Pólya & Szegö, an identity of Issai Schur, had a similar look about it and might be proved in a similar way. Doron was right. Idea of proof: Use induction on N Call left-hand side ϕ N (ξ 1,..., ξ N ) and sum over all permutations such that σ(1) = k. This gives an expression involving ϕ N 1. Use induction hypothesis to get a simpler identity to prove.

56 So we called Doron... He suggested that Problem VII.47 of Pólya & Szegö, an identity of Issai Schur, had a similar look about it and might be proved in a similar way. Doron was right. Idea of proof: Use induction on N Call left-hand side ϕ N (ξ 1,..., ξ N ) and sum over all permutations such that σ(1) = k. This gives an expression involving ϕ N 1. Use induction hypothesis to get a simpler identity to prove. Construct a function f = f (z) so (1) integral over a large circle gives zero and (2) sum of residues of poles enclosed by the contour give the identity.

57 So we called Doron... He suggested that Problem VII.47 of Pólya & Szegö, an identity of Issai Schur, had a similar look about it and might be proved in a similar way. Doron was right. Idea of proof: Use induction on N Call left-hand side ϕ N (ξ 1,..., ξ N ) and sum over all permutations such that σ(1) = k. This gives an expression involving ϕ N 1. Use induction hypothesis to get a simpler identity to prove. Construct a function f = f (z) so (1) integral over a large circle gives zero and (2) sum of residues of poles enclosed by the contour give the identity. Nice simplification but how do we take N?

58 Large Contour Expansion Expand the contours C r to large contours C R, R 1 in order to let N.

59 Large Contour Expansion Expand the contours C r to large contours C R, R 1 in order to let N. In deforming the contours to large contours we encounter two types of poles:

60 Large Contour Expansion Expand the contours C r to large contours C R, R 1 in order to let N. In deforming the contours to large contours we encounter two types of poles: Poles coming from zeros of denominators at ξ k = 1. Poles coming from zeros of f (ξi, ξ j ). Remarkably the residues from the poles of the second type are zero. Can then let N, Y = {y1, y 2,..., }, y 1 < y 2 < +

61 Large Contour Expansion Expand the contours C r to large contours C R, R 1 in order to let N. In deforming the contours to large contours we encounter two types of poles: Poles coming from zeros of denominators at ξ k = 1. Poles coming from zeros of f (ξi, ξ j ). Remarkably the residues from the poles of the second type are zero. Can then let N, Y = {y1, y 2,..., }, y 1 < y 2 < + P Y (x 1 (t) = x) = p σ(s) S q σ(s) S ( S +1)/2 I (x, Y S, ξ) d S ξ S C S R where all the poles of the integrand lie inside C R. The sum runs over all nonempty, finite subsets S of Z +. Here σ(s) = i S i I (x, Y, ξ) = ξ j ξ i 1 ξ 1 ξ N ( ) ξ x y i 1 i e tε(ξ i ) f (ξ i, ξ j ) (1 ξ 1 ) (1 ξ N ) i<j i

62 Position of the mth particle Want marginal distribution P Y (x m (t) = x) where x m (t) is the position of the mth particle from the left at time t.

63 Want marginal distribution Position of the mth particle P Y (x m (t) = x) where x m (t) is the position of the mth particle from the left at time t. For step initial condition, Y = Z +, the distribution of the current I(x, t) = # of particles x at time t is related to the position of the mth particle by P Z + (I(x, t) m) = 1 P Z + (x m+1 (t) x) and the current fluctuations can be related to the height fluctuations.

64 Want marginal distribution Position of the mth particle P Y (x m (t) = x) where x m (t) is the position of the mth particle from the left at time t. For step initial condition, Y = Z +, the distribution of the current I(x, t) = # of particles x at time t is related to the position of the mth particle by P Z + (I(x, t) m) = 1 P Z + (x m+1 (t) x) and the current fluctuations can be related to the height fluctuations. P Y (x m (t) = x) = x 1 < <x m 1 <x<x m+1 < <x N P Y (X ; t) Problem with doing this sum need combination of small contours and large contours.

65 Want marginal distribution Position of the mth particle P Y (x m (t) = x) where x m (t) is the position of the mth particle from the left at time t. For step initial condition, Y = Z +, the distribution of the current I(x, t) = # of particles x at time t is related to the position of the mth particle by P Z + (I(x, t) m) = 1 P Z + (x m+1 (t) x) and the current fluctuations can be related to the height fluctuations. P Y (x m (t) = x) = x 1 < <x m 1 <x<x m+1 < <x N P Y (X ; t) Problem with doing this sum need combination of small contours and large contours. This requires new combinatorial identities

66 Identity: Combinatorial Identity #2 [N] = pn q N, [N]! = [N][N 1] [1], p q S =m i S j S c f (ξ i, ξ j ) ξ j ξ i 1 j S c ξ j [ ] N = m = q m [ N 1 m [N]! [m]![n m]! ] 1 The sum runs over all subsets of {1,..., N} with cardinality m and S c denotes the complement of S in {1,..., N}. N j=1 ξ j

67 where Case of Step Initial Condition Y = Z + P Z +(x m (t) x) = ( 1) m+1 q m(m 1)/2 [ 1 k 1 k! k m k m J k (x, ξ) = i j J k (x, ξ) dξ 1 dξ k CR k ξ j ξ i f (ξ i, ξ j ) ] p (k m)(k m+1)/2 q k(k+1)/2 1 i (1 ξ i)(qξ i p) i ξ x 1 i e tε(ξ i ) Have a somewhat more complicated formula for arbitrary initial Y.

68 where Case of Step Initial Condition Y = Z + P Z +(x m (t) x) = ( 1) m+1 q m(m 1)/2 [ 1 k 1 k! k m k m J k (x, ξ) = i j J k (x, ξ) dξ 1 dξ k CR k ξ j ξ i f (ξ i, ξ j ) ] p (k m)(k m+1)/2 q k(k+1)/2 1 i (1 ξ i)(qξ i p) i ξ x 1 i e tε(ξ i ) Have a somewhat more complicated formula for arbitrary initial Y. But how does one analyze this for large t?

69 In the integrand for J k ( ) ξ j ξ i f (ξ i, ξ j ) = det 1 w(ξ i ) f (ξ i, ξ j ) i j i

70 In the integrand for J k ( ) ξ j ξ i f (ξ i, ξ j ) = det 1 w(ξ i ) f (ξ i, ξ j ) i j Thus recognize integral as a coefficient in the Fredholm expansion of det(i λk) where K is an integral operator acting on L 2 (C R ) K(ξ, ξ ) = ξx e tε(ξ) f (ξ, ξ ) i

71 In the integrand for J k ( ) ξ j ξ i f (ξ i, ξ j ) = det 1 w(ξ i ) f (ξ i, ξ j ) i j Thus recognize integral as a coefficient in the Fredholm expansion of det(i λk) where K is an integral operator acting on L 2 (C R ) K(ξ, ξ ) = ξx e tε(ξ) f (ξ, ξ ) Can do sum over k to get P Z + (x m (t) x) = where C det(i λk) (λ; τ) m i dλ λ τ = p q, (λ, τ) m = (1 λ)(1 λτ) (1 λτ m 1 ) and C is a circle centered at the origin containing all the singularities of the integrand.

72 Deformation of Fredholm Determinant: Final Simplification of P Z + (x m (t) x) Though we have reduced the problem to a single contour integral involving a Fredholm determinant, this determinant is difficult to analyze asymptotically. Develop a deformation theory K J so that Fredholm determinants remain equal. This final representation is amenable to asymptotic analysis much along the lines as encountered in random matrix theory and determinantal processes.

73 Limit Theorems Theorem (TW) Let m = [σt], γ = q p fixed, then ( lim P Z t + x m (t/γ) c 1 (σ)t + c 2 (σ) s t 1/3) = F 2 (s) uniformly for σ in compact subsets of (0, 1) where c 1 (σ) = σ, c 2 (σ) = σ 1/6 (1 σ) 2/3.

74 Limit Theorems Theorem (TW) Let m = [σt], γ = q p fixed, then ( lim P Z t + x m (t/γ) c 1 (σ)t + c 2 (σ) s t 1/3) = F 2 (s) uniformly for σ in compact subsets of (0, 1) where c 1 (σ) = σ, c 2 (σ) = σ 1/6 (1 σ) 2/3. Theorem (ACQ, SS) Let then Z ε (T, X ) = p(t, X )e Fε(T,X ), p = heat kernel F T (s) = lim P(F ε (T, X ) + T s) = KPZ crossover distribution ε 0 4! Remark: Explicit formulas for F T (s).

75 Corollary(ACQ, SS) lim T F T ( ) 2 1/3 T 1/3 s = F 2 (s) The KPZ Equation is in the KPZ Universality Class!

76 Corollary(ACQ, SS) lim T F T ( ) 2 1/3 T 1/3 s = F 2 (s) The KPZ Equation is in the KPZ Universality Class! Summary of KPZ Universality Scaling exponent 1 3 does not depend upon initial configuration Droplet initial conditions: Long time one-point fluctuations described by F 2. Flat initial conditions: Long time one-point fluctuations described by F 1. Not (yet) a rigorous proof of this for KPZ equation.

77 Thank you for your attention