On the replica approach for statistical mechanics of random system - PowerPoint PPT Presentation

On the replica approach for statistical mechanics of random system Giorgio Parisi • Spin glasses in the infinite range limit (the Sherrington Kirkpatrick model). • The computation of the free energy in the SK model (mean field approximation). • The heuristic approach base on replicas. • A first step toward a di ff erent approach. • Some interesting conjectures.

The simplest Ising spin glass (SK) has the following random Hamiltonian: � H J [ ⃗ σ ] = J i,k σ ( i ) σ ( k ) i,k =1 ,N σ i = ± 1, i = 1 , N . The J are random (e.g. Gaussian distributed with zero average). E [ J 2 i,k = N − 1 i,k ] ≡ J 2 E [ J i,k ] ≡ J i,k = 0 , Equivalently H J [ ⃗ σ 1 ] is a random Gaussian function: � H J [ ⃗ σ 1 ] = 0; H J [ ⃗ σ 1 ] H J [ ⃗ σ 2 ] = N ( ⃗ σ 1 , ⃗ σ 2 ) ≡ N σ 1 ( i ) σ 2 ( i ) i The probability distribution of the spectrum of J is known: the Dyson semicircle law.

Partition function: � � 2 N terms � � Z J = exp( − β H J [ ⃗ σ ]) H J [ ⃗ σ ] = J i,k σ ( i ) σ ( k ) i,k =1 ,N σ ⃗ Free energy: β Nf N J ( β ) = − log( Z J ) We want to compute ˜ N →∞ f N f ( β ) = lim J ( β ) The overline represent the average over the J ’s.

Which is the result for the value of the free energy? We start with a function q ( x ) defined in the interval 0 ≤ x ≤ 1. The function g q ( x, h ) is defined in the strip 0 ≤ x ≤ 1, −∞ < h < ∞ . Boundary condition: g q (1 , h ) = log(cosh( β h )); the function g q ( x, h ) satisfies the following q ( x ) dependent antiparabolic equation: � � 2 � ∂ 2 g q ( x, h ) � ∂ g q ( x, h ) ∂ g q ( x, h ) = − dq + x ∂ h 2 ∂ x dx ∂ h � 1 F [ q ] = 1 1 − q ( x ) 2 � � dx − g q (0 , 0) 2 β 0 ˜ f ( β ) = max q ( x ) F [ q ]

This formula was found using replica approach. We need to compute − β Nf ( β , N ) ≡ ln( Z )( β , N ) while it is simple to compute for integer n : � � Z ( β , N ) n log f ( n ; β , N ) = − n → 0 f ( n ; β , N ) = f . lim β Nn Nicola d’Oresme trick (1353). Starting form the definition of A n , A n − 1 √ A 1 / 2 = A, ln( A ) = lim n n → 0

After some algebra and Gaussian integrations we find the exact formula � exp( − β nNf ( n ; β , N )) = dQ exp( − β NnF [ Q ]) ⎛ �⎞ �� nF [ Q ] = − 1 2 β Tr Q 2 + β − 1 log ⎝� exp Q a,b σ a σ b ⎠ { σ } a,b The matrix Q is symmetric, zero on the diagonal ( Q a.a = 0). We have n variables σ a that takes the value ± 1. The integral is done on n ( n − 1) variables. 2

Symmetries The symmetry group is S n . If π is a permutation ( Q π ) a,b = Q π ( a ) , π ( b ) F [ Q π ] = F [ Q ] The proof is trivial: ⎛ �⎞ �� nF [ Q ] = − 1 2 β Tr Q 2 + β − 1 log ⎝� exp Q a,b σ a σ b ⎠ a,b { σ } �� � �� � �� � � � Q π exp = exp Q a,b σ π ( a ) σ π ( b ) = exp Q a,b σ a,b σ a σ b a,b a,b a,b { σ } { σ } { σ }

Point of maximum for N → ∞ . � exp( − β nNf ( n, β , N )) = dQ exp( − β NnF [ Q ]) N →∞ f ( n, β , N ) = F [ Q ∗ ] = min f ( n, β ) ≡ lim Q F [ Q ] We can compute f ( n ) on the integers: the maximum is at Q ∗ a,b = q , ∀ a, b , i.e. the only matrix left invariant by the whole permutation group. We compute everything for integer n and we perform an analytic continuation at n = 0. This gives a wrong result at high β . The function f ( n ) must have a singularity at 0 < n < 1: e.g. � 5 f ( n ) = 0 for n > 1 � n − 1 for n < 1 2; f ( n ) = 2 2

Putting the finger on the origin of troubles. If F [ Q ] has minimum at Q ∗ we must have ∂ 2 F [ Q ] ∂ F [ Q ] = 0 H ≥ 0 H a,b ; cd ≡ ∂ Q a,b ∂ Q a,b ∂ Q c,d The non-negativity of the Hessian ( H ≥ 0) is supposed to be equivalent (also for non-integer n ) to: The spectrum of H is non-negative. At high β the analytic continuation of the spectrum of H acquire a negative part for 0 < n < n ∗ ( β ): the de Almeida Touless instability.

A bold approach: we do the maximum point approximation at n = 0! This lead to a strange mathematics: one introduces a is n × n matrix Q and the symmetry group is S n : eventually n → 0. If π is a permutation ( Q π ) a,b = Q π ( a ) , π ( b ) F [ Q π ] = F [ Q ] F [ Q ] has a minimum at Q ∗ . We call S ∗ the subgroup of S n that leaves Q ∗ invariant i.e. the stabilizer subgroup (the little group). S n ⊃ S ∗ If S ∗ ̸ = S n the (replica) simmetry group is ”spontaneously broken”.

Explicit construction non-symmetric Q ∗ Q ∗ a,b = q 1 if I ( a/m ) = I ( b/m ) Q ∗ a,b = q 0 if I ( a/m ) ̸ = I ( b/m ). Q a,a = 0. i.e. n obiects are divided into n/m classes of m elements. The little group corresponding to Q ∗ is S ∗ = S n/m ⊗ ( S m ) n/m . When n → 0, S n/m → S 0 so that S 0 ⊃ S 0 . S 0 is an infinite group!

The new estimate is (all functions depend also on β ) f RSB = max q 1 ,q 0 ,m f ( q 1 , q 0 , m ) 1 The minimum is at m ∗ with 0 < m ∗ < 1. The old result f RSB is given by 0 f RSB = max f ( q, q, m ) f ( q, q, m ) does not depend on m 0 q In a recursive way ( S 0 ⊃ S 0 ) we can define f RSB ( β ) in such a way that k f RSB ≤ f RSB ≤ f RSB . . . ≤ f RSB . . . ≤ f RSB k →∞ f RSB ≡ lim 0 1 2 k k ∞ In 1979 it was conjectured that f RSB = f . In 2002 Guerra proved that ∞ f RSB ≤ f ∀ k . Less than one year later Talagrand twisted Guerra proof k to prove that f ≤ sup k f RSB . Hence k f RSB ≤ f ≤ f RSB f RSB = f → ∞ ∞ ∞

Can the replica derivation sligtly modified in such a way that it makes sense? It should not involve sets whose cardinality is a non-integer real number !!! Here I present a computation of ˜ f ≡ max q,m f RSB ( q, 0 , m ) that 1 correspond in replicas to Q ∗ a,b = q if I ( a/m ) = I ( b/m ) Q ∗ a,b = 0 if I ( a/m ) ̸ = I ( b/m ). for non-integer m (following Campellone, G.P. and Virasoro, inspired from Derrida). I will start with writing some identities. I will exchange limits with integral, I will treat non convergent asymptotis series as convergent, but these are minor sins! The cardinality of sets will always be an integer number.

� ∞ dt � � ln Z N = exp( − t ) − exp( − tZ N ) . t 0 Let us define 1 � k !( − t ) k Z k exp( − φ ( t, N )) ≡ exp( − tZ N ) = N . k =0 , ∞ Z k � N = dQ exp( − NF ( k, Q )), the integral is done over k × k matrices. At the end of the game we have to evaluate exp( − φ ( t, N )) for very large t . We need a very good control of Z k N : Z k N ≈ exp( − NF ( k, Q ∗ )) F ( k, Q ∗ ) ≡ min Q F ( k, Q ) gives the wrong replica symmetric result!

A di ff erent approximation could be to sum over all the critical points: � ∂ F ( k, Q ) � � Z k exp( − NF ( k, Q ∗ j )) = 0 N ≈ � ∂ Q a,b � Q ∗ j j All critical poins are beyond my command. I will make an arbitrary selection: We partition the set of k elements into l sets of size m i , where � l i =1 m i = k . (Here l is the total number of blocks of the matrix.) The o ff diagonal elements, i.e. Q ab , have a constant value q i if a and b belong to the same set. Q ab is zero if a and b do not belong to the same set. If we we take all the m i = m and q i = q we recover the replica computation for q 1 = q and q 0 = 0. Here we stick to integer m i !

In this way each stationary point of this kind depends is characterized (apart from permutations) the size of the blocks m i and by the values of q i that are a function of the m i . After some algebra we get ∞ ( − t ) m � − φ ( t, N ) = exp( mNf ( m )) . m ! m =1 When N → ∞ we need to evaluate the previous formula when t goes to to ∞ at constant y = ln t/N : t is very large, i.e. O (exp( yN )). We can transform the previous sum into an integral in the complex plane − φ ( t, N ) = 1 � dm exp( N ( my + mf ( m ))) . 2 i Γ [1 + m ] sin[ π ] C where y = ln t/N and C is an appropriate integration path in the complex plane and crosses the real line for 0 < x < 1.

We deform the path C to a new path going from − i ∞ to + i ∞ crossing the real line for 0 < m < 1. We look for a saddle point in the complex plane. The equation for the saddle point (i.e. m sp ) is f ( m sp ( y )) + m sp ( y ) f ′ ( m sp ( y )) + y = 0 . Let us assume, for simplicity that 0 < m sp < 1. In this case we have at the leading order φ ( t, N ) ≈ exp ( Nm sp ( y )( y + f ( m sp ( y ))) . where f ′ ( m sp ( y )) = 0 , After computing the integral on t and after some simple estimates we recover the result of the replica approach.

Crucial points! ∞ ( − t ) m exp( mNf ( m )) = 1 dm exp( N ( my + mf ( m ))) � � − φ ( t, N ) = . m ! 2 i Γ [1 + m ] sin[ π ] C m =1 • We have a function f ( m ) that we can write in an explicit way and we continue it from integer to non-integer values. • We use this analytic continuation to write the result as a complex contour integral. • We estimate the integral with the saddle point method. • In a simple case, we can do the computation and we obtain the replica result.