Statistical Mechanics and Combinatorics : Lecture I

Transcription

1 Statistical Mechanics and Combinatorics : Lecture I 1 Overview We are considering interactions between three fields of mathematics: combinatorics, statistics and probability In short, combinatorics is the study on counting Statistics is the aspect on weighted counting and probability is also the aspect on weighted counting but with the condition weights = 1 Example 11 Consider paths from origin to (m, n) on integer lattice View of combinatorics: the number of such lattice paths is ( N m) where N = m + n View of stat mech: Let a, b be the weight for stepping right and up respectively This provides a weight for each path Let Z be the sum of all weighted path with length N Then Z = weight(path) = a R b U where R, U are the numbers of right and up steps of each path with length N respectively View of probability: Let a, b, R, U, N be the same notation as the previous case Let p = a, q = b be the probability for right and up steps respectively The sum Z of a+b a+b all weighted path with length N becomes Z = Pr(path) = p R q U 1

2 Remark 12 There is a underlying probability measure on paths Given any path γ 0, its probability P r(γ 0 ) is given by P r(γ 0 ) = weight(γ 0) weight(γ) pathγ Example 13 Consider paths from origin to (1, 1) and let a i, b i be the weights for i = 1, 2 as the following figure: Take the path γ 0 = a 1 b 2 Then 2 Boltzmann Measure Pr(γ 0 ) = a 1 b 2 a 1 b 2 + b 1 a 2 Given N identical particles, each falls into one of k states For each state, we associate a number E i for i = 1, k which is called energy of every state For a closed system, the total energy E and population numbers n i for energy state E i must satisfy the equations: { E = k E in i N = k N i The number W of possible ways to place particles for a given partition {N 1, N 2,, N k } is W = N! N 1!N 2! N k! By the law of large numbers, when N 1 the weight of the optimal partitioning dominates Thus we want to determine the specific partitioning which provides a maximal W Since the logarithm is a monotonically increasing function, we may consider for the maximum of ln W = (N ln N N) k (N i ln N i N i ) + err Apply the variational method with Lagrange multipliers We set f = ln W + α(n k N i ) + β(e k E i N i ) 2

3 with the undetermined multipliers α, β Since Therefore, It follows that 0 = f N i = ln N i (α + βe i ) N i = e α+βe i N i N = e βei k j=1 e βe j Let the particle function Z = k j=1 e βe j In physics, the quantity β = 1 where T denotes T temperature (usually this is 1/kT where this k stands for Boltzmann s constant) The free energy F is given by F = T ln Z Consequently, we can define the Boltzmann measure as the probability Pr(state i) of each particle in state i Now consider the average energy Ē Pr(state i) = e βe i Z Ē = E N = k E i N i N = k E i e βe i Z = β Z Z = ln Z β Lemma 21 ln Z is a convex function for β Proof We want to show that 2 ln Z is monotone that is ln Z > 0 β β 2 2 k β ln Z = 2 ( E i ) 2 e βe i Z = E 2 E 2 = var(e) > 0 ( k E ie βe i ) 2 Z 2 Definition 22 1 ln Z is called the free energy β When T, Pr(i) 1 When T 0, the probability is supported on the minimum k energy state(s) 3 Statistical models A statistical model is a pair (Ω, P), where Ω is the configuration space such that particles usually interact locally and P is a probability measure on Ω 3

4 Dimer model A dimer cover or perfect matching of a graph G = (V, E) is a subset of edges such that each vertex is the endpoint of exactly one edge Ω is the set of dimer covers If each edge has an associated energy the energy of a cover is the sum of energies of its edges Six-vertex model The configuration space Ω is a set of orientations of a subgraphs of Z 2 with two incoming and two outgoing edges at each vertex There are six energies associated to the six possible configurations at a vertex The energy of a configuration is the sum of the energies at each vertex Gradient model Let G = (V, E) be a graph Let the configuration space Ω be R V Let Φ, U : R R be potential functions A Hamiltonian is defined as Φ(f(v)) + U(f(v i ) f(v j )) v V e=v i v j E Gaussian free field As a special case of the gradient model, let G = (V, E) be any graph, usually a lattice in d-dimensional Euclidean space Define a Hamiltonian by H(f) = (f(v i ) f(v j )) 2 v i,v j E which is the Hooke s law with spring energy Then, the function f with probability density is e βh(f) which is a Gaussian density Ising model Consider the integer lattice Z d Let G = (V, E) be a graph on the lattice with the set of integer vertices V and the set of edges E For each vertex i, we assign a variable σ i {1, 1}, which provides a spin configuration Let Ω G = {1, 1} V be the configuration space and take σ Ω G The energy function or Hamiltonian H is given by H(σ) = σ i σ j i,j Ω G : i j =1 4

5 The sum is over all pairs of sites in G which are nearest neighbors The probability measure is given by Pr(σ) = e βh(σ) σ Ω G e βh(σ) Hard disk model Suppose there are k unit disks in a region Λ The energy function E is defined by { if two disks overlapping E = 0 else Let Ω be the configuration space of k disjoint disks in Λ The underlying probability measure is Lebesgue measure on Ω Λ k R k The partition function Z is the volume of Ω(k) Example 31 (1-dimensional hard disk model) Let Λ be the interval [0, N] configuration space Ω N,3 is the set of three disjoint unit intervals lying in Λ, ie The Ω N,3 = {(a 1, a 2, a 3 ) 0 a 1, a 2, a 3, a 1 + a 2 + a 3 N 3} a 1 a 2 0 N a 3 The volume of the configuration space Ω N,3 is given by formula vol(ω N,3 ) = (N 3)3 3! Generally, the configuration is defined as Ω N,k = {k disjoint unit intervals in Λ} Then the volume of Ω N,k becomes vol(ω N,k ) = (N k)k k! 5

6 Let s combine all the cases together by giving energy E 0 per particle Define z = e βe 0 as fugacity Then Z N (z) = = N z k vol(ω N,k ) k=0 N (N k) k k=0 k! z k 1 Exercise: If N, show that the free energy per unit length F (z) = lim N N log Z N(z) is given by 1 F (z) = lim N N log Z N(Z) = z z ! z3 43 4! z4 + Note that F (z) is the Lambert function also called product logarithm denoted by P L(z), and satisfies P L(z)e P L(z) = z The derivative of P L satisfies P L (z) = P L(z) z(1 + P L(z)) for z / {0, 1/e} So F (z) z z = P L(z) 1 + P L(z) 1 as z 6