The Entropy of Six-Vertex Model with Variety of Different Boundary - PowerPoint PPT Presentation

The Entropy of Six-Vertex Model with Variety of Different Boundary Conditions Thiago Silva Tavares collaboration with G.A.P. Ribeiro and V.E. Korepin State University of S˜ ao Carlos, Brazil 11/06/2015 Tavares (UFSCar) Florence 2015 11/06/2015 1 / 37

Outline First Part: Free and Toroidal Boundary Conditions 1 Introduction Free-Boundary decomposition Homogeneous Toroidal Boundaries Mixed Toroidal Boundaries: First Row Second Part: Fixed Boundary Conditions 2 The Domain-Wall descendants N´ eel Boundary Condition Merge type Boundaries Domain-Wall - Ferroelectric Fusion N´ eel - Ferroelectric Fusion Final Remarks and Open Questions 3 Tavares (UFSCar) Florence 2015 11/06/2015 2 / 37

First Part: Free and Toroidal Boundary Conditions Introduction A non-trivial problem in combinatorics Six-vertex model was proposed as a 2D realization of the counting problem of ice residual entropy Solved by Lieb under periodic (toroidal) boundary condition: � 4 S = 3 � 2 ln . 3 Why Periodic Boundary Conditions? Should we always expect intensive properties to be independent of boundary conditions? The first Counter-examples! Are they exceptions to the rule? Tavares (UFSCar) Florence 2015 11/06/2015 3 / 37

First Part: Free and Toroidal Boundary Conditions Introduction A non-trivial problem in combinatorics Six-vertex model was proposed as a 2D realization of the counting problem of ice residual entropy Solved by Lieb under periodic (toroidal) boundary condition: � 4 S = 3 � 2 ln . 3 Why Periodic Boundary Conditions? Should we always expect intensive properties to be independent of boundary conditions? The first Counter-examples! Are they exceptions to the rule? Tavares (UFSCar) Florence 2015 11/06/2015 3 / 37

First Part: Free and Toroidal Boundary Conditions Introduction Brascamp et al prove that, for rectangular lattices with even number of sites, the free-energy of free boundary conditions and periodic boundary conditions are the same(1973). Batchelor et al prove that toroidal boundary conditions with antiperiodic closing on the horizontal and periodic closing on the vertical still gives the same free-energy as PBC(1995). Nevertheless, the number of lines must be even otherwise partition function is zero. Korepin and Zinn-Justin prove that Domain-Wall boundary conditions 2 ln( 3 3 gives a different free-energy. The residual entropy is S = 1 2 4 ). What is really happening? Are those kinds of boundary really exceptions? Tavares (UFSCar) Florence 2015 11/06/2015 4 / 37

First Part: Free and Toroidal Boundary Conditions Free-Boundary decomposition Arrows can be either equal or opposite at closing! Horizontal: 0 ⇒ T (0) = A + D , 1 ⇒ T (1) = B + C Vertical: 0 ⇒ G (0) = Id , 1 ⇒ G (1) = σ x     L N L � � G ( φ k ) 1 1 T ( θ j ) ( λ j ) ( A ( λ ) + D ( λ ) + B ( λ ) + C ( λ )) N � � �  = Tr V � Z free = Tr V  , (1)   Vk 1 1 k φ k ,θ j =0 , 1 k =1 j =1 k =1 Θ 2 � 0 Θ 2 � 0 Θ 2 � 0 Θ 2 � 0 Θ 1 � 0 Θ 1 � 0 Θ 1 � 0 Θ 1 � 0 Φ 1 � 0 Φ 2 � 0 Φ 1 � 0 Φ 2 � 1 Φ 1 � 1 Φ 2 � 0 Φ 1 � 1 Φ 2 � 1 Θ 2 � 1 Θ 2 � 1 Θ 2 � 1 Θ 2 � 1 Θ 1 � 0 Θ 1 � 0 Θ 1 � 0 Θ 1 � 0 Φ 1 � 0 Φ 2 � 0 Φ 1 � 0 Φ 2 � 1 Φ 1 � 1 Φ 2 � 0 Φ 1 � 1 Φ 2 � 1 Θ 2 � 0 Θ 2 � 0 Θ 2 � 0 Θ 2 � 0 Θ 1 � 1 Θ 1 � 1 Θ 1 � 1 Θ 1 � 1 Φ 1 � 0 Φ 2 � 0 Φ 1 � 0 Φ 2 � 1 Φ 1 � 1 Φ 2 � 0 Φ 1 � 1 Φ 2 � 1 Θ 2 � 1 Θ 2 � 1 Θ 2 � 1 Θ 2 � 1 Θ 1 � 1 Θ 1 � 1 Θ 1 � 1 Θ 1 � 1 Φ 1 � 0 Φ 2 � 0 Φ 1 � 0 Φ 2 � 1 Φ 1 � 1 Φ 2 � 0 Φ 1 � 1 Φ 2 � 1 Tavares (UFSCar) Florence 2015 11/06/2015 5 / 37

First Part: Free and Toroidal Boundary Conditions Free-Boundary decomposition Each component of the previous sum can be viewed as a particular toroidal boundary condition, which mix periodic and anti-periodic closings. One may organize these contributions in a matrix M N , L whose elements are the partitions Z j , k such j − 1 = θ 1 2 0 + θ 2 2 1 + · · · + θ N 2 N − 1 , k − 1 = φ 1 2 0 + φ 2 2 1 + · · · + φ L 2 L − 1   Z 1 , 1 Z 1 , 2 · · · Z 1 , 2 L 2 N 2 L · · · Z 2 , 1 Z 2 , 2 Z 2 , 2 L   � � M N , L =  , Z free = Z j , k . (2) . . .  ...  . . .   . . .  j =1 k =1 Z 2 N , 1 Z 2 N , 2 · · · Z 2 N , 2 L Tavares (UFSCar) Florence 2015 11/06/2015 6 / 37

First Part: Free and Toroidal Boundary Conditions Free-Boundary decomposition a = b = c = 1  18 0 0 8   44 0 0 20 0 20 20 0  0 10 10 0 0 26 24 0 26 0 0 16     M 2 , 2 = M 2 , 3 =     0 10 10 0 0 26 24 0 26 0 0 16     8 0 0 8 26 0 0 20 0 20 20 0 44 0 0 26 148 0 0 84 0 84 84 0     0 26 26 0 0 94 84 0 94 0 0 72     0 24 24 0 0 84 80 0 84 0 0 72         20 0 0 20 84 0 0 74 0 72 74 0     M 3 , 2 =   M 3 , 3 =   0 26 26 0 0 94 84 0 94 0 0 72         20 0 0 20 84 0 0 72 0 76 72 0         20 0 0 20 84 0 0 74 0 72 74 0     0 16 16 0 0 72 72 0 72 0 0 68 selection rule Mod [Φ − Θ , 2] = 0 Z N × L = Z L × N j , k k , j Z 1 , 1 = Ω P , P is the largest element for ∆ = 1 2 Ω PP ≤ Ω free ≤ 2 L + N − 1 Ω PP ⇒ S PP = S free . (3) Tavares (UFSCar) Florence 2015 11/06/2015 7 / 37

First Part: Free and Toroidal Boundary Conditions Free-Boundary decomposition ∆ � = 1 2 We have more generally that the largest element is Z PP for ∆ ≥ − 1 and Largest contribution for ∆ < − 1 L even, N even Z PP L even, N odd Z PA L odd, N even Z AP L odd, N odd Z AA This scenario was verified for L , N up to six. F free = F max (4) Tavares (UFSCar) Florence 2015 11/06/2015 8 / 37

First Part: Free and Toroidal Boundary Conditions Homogeneous Toroidal Boundaries The homogenous toroidal boundaries are those where there is no change from periodic to anti-periodic along the horizontal or the vertical direction. They are: �� T (0) � N � Z 11 = Z PP = Tr V (5) �� T (1) � N � Z 2 N 1 = Z AP = Tr V (6) � T (0) � N � Π x � Z 12 L = Z PA = Tr V (7) � T (1) � N � Π x � Z 2 N 2 L = Z AA = Tr V (8) Both T (0) and T (1) can be diagonalized, and due to the discrete symmetries: � T (0) ( λ ) , Π x � � T (0) ( λ ) , Π z � = = 0 , (9) � T (1) ( λ ) , Π x � � T (1) ( λ ) , Π z � = + = 0 , (10) Π x Π z = ( − 1) L Π z Π x , (11) where Π x = � L m =1 σ x is the reflection operator and Π z = � L m =1 σ z is the parity operator, we can see that all four free-energies above can be obtained. Tavares (UFSCar) Florence 2015 11/06/2015 9 / 37

First Part: Free and Toroidal Boundary Conditions Homogeneous Toroidal Boundaries The homogenous toroidal boundaries are those where there is no change from periodic to anti-periodic along the horizontal or the vertical direction. They are: 2 L �� T (0) � N � � N � Λ (0) � Z 11 = Z PP = Tr V = (5) j j =1 2 L − 1 �� T (1) � N � = (1 + ( − 1) N ) � N � Λ (1) � Z 2 N 1 = Z AP = Tr V (6) j j =1 2 L ( − 1) px � T (0) � N � � N Π x � � Λ 0 � j Z 12 L = Z PA = Tr V = (7) j j =1 2 L − 1 ( − 1) px � T (1) � N � � N Π x � � Λ (1) = (1 + ( − 1) N + L ) � j Z 2 N 2 L = Z AA = Tr V (8) j j =1 Both T (0) and T (1) can be diagonalized, and due to the discrete symmetries: � T (0) ( λ ) , Π x � � T (0) ( λ ) , Π z � = = 0 , (9) � T (1) ( λ ) , Π x � � T (1) ( λ ) , Π z � = + = 0 , (10) Π x Π z = ( − 1) L Π z Π x , (11) where Π x = � L m =1 σ x is the reflection operator and Π z = � L m =1 σ z is the parity operator, we can see that all four free-energies above can be obtained. Tavares (UFSCar) Florence 2015 11/06/2015 9 / 37

First Part: Free and Toroidal Boundary Conditions Homogeneous Toroidal Boundaries Batchelor et. al. ⇒ F AP = F PP for ∆ < − 1, therefore 1 1 L ln Λ (1) L ln Λ (0) lim max = lim ∆ < − 1 , (12) max L →∞ L →∞ hence we have F PP = F AP = F PA = F AA , (13) whenever they are allowed by selection rule. Therefore F free = F PP ∀ ∆ (14) Note that there is no restriction over the parity of lattice size Tavares (UFSCar) Florence 2015 11/06/2015 10 / 37

First Part: Free and Toroidal Boundary Conditions Mixed Toroidal Boundaries: First Row The First Row of M N , L is given by: 2 L L � � � N � G ( φ m ) Λ (0) f { φ m } � V m ( T (0) ) N � Z 1 , j = Tr V = , (15) g L , g m =1 g =1 L g (0) � � f { φ m } � G ( φ m ) � g (0) � � = (16) � � L , g V m � m =1 Since T (0) commutes with S z , we can choose eigenvectors to live in a definite sector of S z . Therefore we have to have Φ even, otherwise f { φ m } will be zero. L , g Perron-Frobenius theorem ⇒ f { φ m } is non-negative for maximal L , g eigenvectors of each sector. How could f { φ m } change the free-energy? It should decay as fast as L , g e − δ LN . But this impossible since it only depends on L ! Tavares (UFSCar) Florence 2015 11/06/2015 11 / 37