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206A: STAT Gibbs Measures Invited Speaker: Andrea Montanari Lecture 5: Random Energy Model Lecture date: September 12 Scribe: Sebastien Roch This is a guest lecture by Andrea Montanari (ENS Paris and Stanford) on the Random Energy Model (REM) in physics. 1 Disordered Models In this lecture, we consider so-called disordered models in statistical physics. These are particle systems where the energy function is random. Therefore, we have two levels of randomness. We use the notation E, P to denote averaging with respect to the energy and we use the notation to denote averaging with respect to the (random) Boltzmann distribution. The state space is {0, 1}N which we sometimes denote equivalently {1, . . . , 2N }. We denote the (random) energy function E : {0, 1}N R. Thus, for [0, + ], the Boltzmann distribution is 1 p(x) = exp ( E(x)) , Z( ) where the so-called partition function is Z( ) = x exp ( E(x)) . We are interested in the typical properties of p under P. Example: Random k-SAT. Let F : {0, 1}N {0, 1} be a Boolean function in k-CNF form. For example, F (x) = (x1 x2 x3 ) ( 2 x3 x4 ). Let x EF (x) = #{clauses violated by x}, (1) which we think of as an energy function. Imagine that we pick F uniformly at random over all k-SAT formulas with N variables and M = N clauses. In particular, with the energy in (1), if = + , the Boltzmann distribution is uniform over all satisfying assignments in F . The rst two moments of E under P are easily computed: E[E(x)] = N , 2k 5: Random Energy Model-1 and Cov[E(x), E(y)] = N (P[x & y violate an arbitrary clause] P[x violates an arbitrary clause]P[y violates an arbitrary clause]). Note that the correlation between two assignments x and y depends only on their Hamming distance. Note also that both assignments can violate a clause simultaneously only if they agree on all variables involved. Therefore, Cov[E(x), E(y)] = N 1 1 (1 )k 2k 2k 2 N f ( ), where the Hamming distance between x and y is N . For large k, the function f is nonnegligible from 0 up to O(k 1 ). 2 Random Energy Model The REM was introduced by Derrida [1]. It is de ned as follows. Each energy level is an independent Gaussian with mean 0 and variance N/2, i.e. {E(x)}x {0,1}N are i.i.d. N (0, N/2) and N Cov[E(x), E(y)] = 1{x = y}. 2 We follow Derrida s treatment of the REM [1]. The main quantities we want to compute are the so-called free energy 1 FN ( ) = log ZN ( ), the internal energy UN ( ) = E(x) = [ FN ( )], and the canonical entropy SN ( ) = H(p ) = 2 where H is the Shannon entropy. FN ( ), 3 Thermodynamic Properties Using the equivalent state space {1, . . . , 2N }, we compute the number of energy levels in the interval [N , N ( + )] A( , + ) = #{i : Ei [N , N ( + )]}, 5: Random Energy Model-2 which up to subexponential factors is, in expectation, + E[A( , + )] = 2N where Let N N x2 . e dx = eN maxx [ , + ] sQ (x) , = sQ (x) = log 2 x2 . log 2. Note that s is positive on ( , ). There are two cases of interest: 1. When [ , + ] [ , ] = , E[A( , + )] is exponentially large and the random variable A( , + ) is concentrated. 2. When [ , + ] [ , ] = , E[A( , + )] is exponentially small and the random variable A( , + ) is almost surely 0. Next, we compute the partition function. Let s(x) = We have ZN ( ) = i=1 sQ (x), x [ , 0, o.w. + ], 2N . e Ei = . eN (s(x) x) dx = eN maxx [s(x) x] . To summarize, we can prove the following. Proposition Let 1 ( ) = max[s(x) x]. x Then we have 1 log ZN ( ) = ( ). N + N lim P[| log ZN N | N ] e N 2 /2 Also, . 4 The Phase Transition There is a simple graphical way to compute ( ): nd the point on the curve of s(x) with slope . See Figure 5.2 in [2]. Therefore, it is easy to show that the free energy density f ( ) = ( ) 1 FN ( ) = = N + N lim log 2 , c , 4 log 2, > c , 5: Random Energy Model-3 where c = 2 log 2. Also, the entropy density is s( ) = and the energy density is u( ) = 1 UN ( ) = N + N lim c , , 2 log 2, > c . 1 SN ( ) = N + N lim + log 2, c , 4 0, > c , 2 There is a phase transition at the point c which is seen by the discontinuity in the second derivative of the free energy density. It is a so-called second order phase transition. The typical behavior in the two phases are: 1. At high temperature, i.e. c , there are exponentially many states with energy density /2 and the Boltzmann distribution is roughly uniform over them. 2. At low temperature, i.e. > c , the Boltzmann distribution is concentrated on a subexponential number of states of lowest energy density log 2. 5 Low Temperature Regime In this Section, we consider some more properties of the low temperature regime. 5.1 Ruelle s Reformulation A di erent characterization of the REM in terms of a point process was given by Ruelle [3]. Let 1 2 be a Poisson point process on R+ with intensity m 1 m for 0 m < 1. Consider the random variables i pi = . j j Then the values { i }i behave like the large N limit of the Bolztmann distribution of the p REM. To see this, consider the regime > c . As we pointed out before, the Boltzmann distribution is concentrated on a small number of states with energy density roughly log 2. Thus, consider the following rescaling of the energy Ei = N log 2 + zi . 5: Random Energy Model-4 By plugging this expression for Ei into the density of a N (0, N/2), it is easy to see that the zi s have a density roughly proportional to e c z . Now, consider again the original probabilities of the REM i pi = , j j where i e c zi . Then one can show that in the large N limit, the i s form a Poisson point process with intensity proportional to 1 c / . This corresponds to the Ruelle reformulation with m = c / . Note that this Poisson process has an accumulation point at 0. Also, notice that the larger is (i.e. the lower the temperature is), the fatter the tail of the process is, indicating that the Boltzmann distribution is dominated by a few large values. 5.2 Condensation To quantify further the condensation phenomenon at low temperature, consider the variable Qx,y = 1, if x = y, 0, o.w., where x and y are two states. The inverse of the quantity P[Qx,y = 1] = E i p2 , i measures the number of states on which the Boltzmann distribution is concentrated. Note that 2 Ei Z(2 ) ie E p2 = E =E . i Z( )2 Z( )2 i From our previous calculations for Z, one can derive N + lim P[Qx,y = 1] = 0, 1 c , c , > c . References [1] B. Derrida, Random-Energy Model: Limit of a Family of Disordered Models, PRL, 45(2):79 82, 1980. [2] M. Mezard and A. Montanari, Constraint Satisfaction Networks in Physics and Computation, Oxford University Press, In Press. [3] D. Ruelle, A mathematical reformulation of Derrida s REM and GREM, CMP, 108(2):225 239, 1987. 5: Random Energy Model-5

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