Square-Triangle-Rhombus Random Tiling

Transcription

1 Square-Triangle-Rhombus Random Tiling Maria Tsarenko in collaboration with Jan de Gier Department of Mathematics and Statistics The University of Melbourne ANZAMP Meeting, Lorne December 4, 202

2 Crystals Crystals are invariant under discrete translations and rotations. Atomic structure can be modelled with periodic lattices. Translational symmetry and 2-, 3-, 4-, 6-fold rotational symmetries. Silicon (Si) hexagonal lattice Sodium Chloride (NaCl) square lattice 2 / 27

3 Quasi-periodicity in Nature Both exhibit 5-fold rotational symmetry but no translational: Atomic model of Al-Pd-Mn quasicrystal surface Mosaic on the wall of Darb-e Imam Shrine in Isfahan, Iran, circ / 27

4 Quasi-periodic Tilings possess no translational symmetry, only forbidden rotational symmetry Penrose tiling with rhombi 5-fold symmetry Square-Triangle tiling 2-fold symmetry 4 / 27

5 Quasi-periodic and Random Tilings Quasi-periodic tiling projection of a cut of a n-dimensional cube by a plane with an irrational slope. Random tiling projection of a cut of a n-dimensional cube by a surface. n= 9: Quasi-periodic tiling Random tiling 5 / 27

6 Random tilings are entropic models for quasicrystals. They do not require strict matchings and still exhibit forbidden rotational symmetries in a statistical sense. 6 / 27

7 Random tilings are entropic models for quasicrystals. They do not require strict matchings and still exhibit forbidden rotational symmetries in a statistical sense. Square-Triangle tiling was recently found to be related to puzzles of Knutson Tao Woodward, which compute Littlewood Richardson coefficients / 27

8 Square-Triangle-Rhombus Tiling Model Square-Triangle tiling Rhombus tiling Exact results for the square-triangle tiling model were obtained by Widom and Kalugin. Solved using the algebraic Bethe Ansatz by de Gier and Nienhuis. In unpublished work de Gier and Nienhuis proposed an extension of square-triangle tiling with a new thin rhombus tile: 8 / 27

9 Formulating the problem New model consists of filling a region of the plane with squares, triangles and thin rhombi in restricted orientations as shown: weight( ) = u, weight( )=e µ, weight(all other tiles) =. Want to compute entropy σ as a function of u and µ Z A = weight(config), config σ(u, µ) = lim A log Z A log u n µ n 9 / 27

10 Correspondence between the 0-Vertex Model and the Square-Triangle-Rhombus Tiling Vertex Model S-T-R tiling Cruicially, this lattice model is integrable. Can use the transfer matrix method and the method of algebraic Bethe Ansatz to compute σ. 0 / 27

11 Deformed M N lattice with p.b.c. and associated propagation of two types of particles: 2 s and 3 s. r s 0 s + s t + t / 27

12 The eigenvalue Λ of the transfer matrix is given by: M Λ(u) = u N i= + e M2µ u u i M i= u i u M 2 + e (M M2)µ v j u. j= Two types of particles / Bethe Ansatz root types: M of u i and M 2 of v j. M 2 j= u v j + 2 / 27

13 The eigenvalue Λ of the transfer matrix is given by: M Λ(u) = u N i= + e M2µ u u i M i= u i u M 2 + e (M M2)µ v j u. j= Two types of particles / Bethe Ansatz root types: M of u i and M 2 of v j. M 2 j= u v j + The Bethe Ansatz equations is a coupled system of non-linear equations: M 2 ui N e µ + ( ) M = 0, u i v j j= M e µ + ( ) M2 = 0. u i v j i= 3 / 27

14 Taking the logarithm of both sides, the BA equations can be rewritten as M 2 N ln(u i ) + ln(u v j ) (M )iπ + µm 2 = 2πiI i, I i Z, j= M i= Define F (u) and F 2 (v): ln(u i v) (M 2 )iπ + µm = 2πiĨj, Ĩ j Z F (u) = ln(u) + N F 2 (v) = N M i= M 2 j= ln(u v j ) + m 2 µ, m 2 = M 2 N, ln(u i v) + m µ, m = M N Then the BA equations and Λ can be rewritten in terms of F and F 2 : Re [F (u i )] = Re [F 2 (v j )] = 0, Λ(u) = u N e Nmµ ( e NF(u) + e NF(u) NF2(u) + e NF2(u)). 4 / 27

15 Differentiating F and F 2 and taking the thermodynamic limit (N ) the root density functions f and f 2 are given by a system of integral equations: f (u) = u f 2 (v) 2πi v u dv f 2 (v) = 2πi U V f (u) u v du. f is analytic on C\V {0} and f 2 is analytic on C\U. V U Γ V Γ U We are interested in the analytic continuation of f across the cut V, moving along the path Γ V, and, similarly, in the analytic continuation of f 2 across the cut U, moving along the path Γ U. 5 / 27

16 For a general linear combination, G(z) = a f (z) + a 2 f 2 (z), If endpoints of V and U coincide, then closed path Γ necessarily crosses both V and U. The analytic continuations of G(z) along Γ = Γ U Γ V is given by the monodromy matrix: ( ) ( ) ( ) a 0 a Γ :. a 2 and we observe that the monodromy group is isomorphic to Z 6 : (Γ U Γ V ) 6 = I a 2 6 / 27

17 b V U b.0 Figure : Bethe Ansatz roots (for N = 52, M = 60, M 2 = 39, and γ = 0.07). The roots {v j } and {u i } lie on a curves V and U respectively, in the complex plane. 7 / 27

18 f and f 2 are made single-valued functions with the change of variables: t = ( zb ) /6 + t 6 zb, z = b + bb t 6, z = b t = 0 z = b t In the original variable z, f and f 2 correspond to different sheets of the Riemann surface of the same function G(z). b is parametrised as follows: b = i b e iγ, γ ( π/2, π/2). 8 / 27

19 G (z)dz V F2 Images of the Bethe Ansatz roots in the t-plane. F2 F + F2 The t-plane is divided into 2 sectors. F F U Dots placed at points z = 0 and z =, with z = 0 lying on the lines corresponding to the images of the V -cut. 9 / 27

20 Reconstructing G(z)dz from its singularities We find G(z) = z C = 2 3 cos γ G(z)dz = 2 k= r k t t k dt. ( ) t + t C(t + t 5 ) C (t + t 5 ), 3 ( ) e iγ + e 2iγ/3 (im e iπ/6 ( + m 2 )). The images of U and V in the t-plane meet at t = 0 and t =, which implies that the G(z)dz vanishes there and that C = 0. This fixes m and m 2 to be: m = 2 cos π + γ, 3 3 m2 = 2 cos γ / 27

21 The eigenvalue Λ can be calculated by considering the following integral, e iθ t + t I (θ, γ) = Re 0 z dz 3 dt dt. F i (u) are given by I (θ, γ) for appropriate intervals of θ, where cos 3θ u(θ, γ) = b(γ) sin(γ 3θ). 2 / 27

22 Recall that Λ was given in terms of F i s: N log Λ(u) = m µ + log(u) min Re {F (u), F (u) + F 2 (u), F 2 (u)} We find the largest eigenvalue N log Λ max(u) = m µ + log(u) I (θ, γ), ( π with θ 6, π ) and γ 2 ( 0, π ) 2 Using the Legendre transformation, we were able to compute the entropy in terms of tile tensities: n r and n s+ σ(n s +, n r ) = µ(m n s + ) + ( n r ) log(u) I (θ, γ). 22 / 27

23 Bulk entropy σ Can solve for n r and n s+ in terms of u and µ(γ), and for all other tile densities in terms of n r, n s+ and BA root densities m, m 2. σ 2 n r 3 4 n t Bulk entropy σ as a function of the total triangle area fraction 3 4 n t 23 / 27

24 Previously known results without rhombi, ie. u = 0 Bulk entropy σ as a function of the total triangle area fraction 3 4 n t σ fold phase fold phase 6-fold phase 0.02 n t = n t+ n s+ = n s = n s0 3 4 n t 0 24 / 27

25 Line of maximum entropy and symmetry n s = n s0 n t = n t+ 2 n r r s 0 s + s t + t 3 4 n t Densities on the line of maximum entropy in terms of γ are: n r = 4 [ π sin γ ] [ γ ] sin 3 3 n t = 2 4 [ ] [ cos (π + 2γ) + 4 cos n s+ = [ ] [ ]) π + γ (4 cos 2 cos (π + 4γ) ] (π + 4γ) n s0 = m 2 = 2 3 cos γ / 27

26 σ max Expression for the line of maximum entropy in terms of γ: σ max line (γ) = [ ] ( log(432) + 2 log ( tan [ 2 = cos (π + 4γ) (π 2γ)] tan[γ] tan [ 2 (π + 2γ)]) 6 3 ( [ π + log tan 4 + γ ] [ ]) [ ]) π + γ tan (π + 2γ) sec [ ] ( ( π + γ [ π cos log tan γ ] [ ]) tan (π + 2γ) ( [ ] [ ]) ( [ ] )) π + γ +2 log tan (π 2γ) tan (π + 2γ) sin ( [ π log tan 6 + γ ] [ γ ]) ( tan, where γ 0, π ) / 27

27 Conclusion We solved the model on a two-dimensional surface, where the densities and all other quantities are parametrised in terms of u and µ. We computed an explicit expression for the line of maximum entropy, which is also the line of maximum symmetry. Further directions: solving for general U and V ; the extension of these methods to other integrable tiling models: octagonal and decagonal triangle-rectangle tiling models; possible connections to combinatorics. 27 / 27