Lectures on Quantum sine-gordon Models

Transcription

1 Lectures on Quantum sine-gordon Models Juan Mateos Guilarte 1, 1 Departamento de Física Fundamental (Universidad de Salamanca) IUFFyM (Universidad de Salamanca) Universidade Federal de Matto Grosso Cuiabá, Brazil, 010

2 Outline 1 3

3 Surfaces of constant negative curvature Let u, v be Tchebyshev coordinates in hyperbolic geometry (skip umbilical points): ds = du + cosφ(u, v)dudv + dv u = constant and v = constant are asymptotic lines (normal curvature null). φ: angle between the two asymptotic lines. Metric tensor components and Christoffel symbols: g 11 = g = 1, g 11 = g 1 = sin φ g 1 = g 1 = cosφ, g 1 = g 1 = cosφ Γ 1 11 = cotφ φ, Γ 1 u Γ 11 = cosecφ φ u = cosecφ φ v, Γ 1 = cotφ φ v sin φ, Γ 1 1 = Γ1 1 = 0, Γ 1 = Γ 1 = 0

4 Codazzi-Mainardi Scalar curvature, Gauss curvature and principal curvatures S = K = κ 1 κ = g ik R j ijk = cosecφ φ u v with Gaussian curvature K = 1 are tantamount to solutions of the sine-gordon φ = sinφ(u, v) (1) u v The sine-gordon (1) is the Codazzi- Mainardi of compatibility between the first and second fundamental forms I = Edu + Fdudv + Gdv = du + cosφ(u, v)dudv + dv II = Ldu + Mdudv + Ndv = sinφ(u, v)dudv ( ) = M Γ 1 11 Γ1 1, M u M ( v = M Γ Γ 1 )

5 Pseudospherical congruences Σ 1, Σ : in R 3. Let l : Σ 1 Σ be a diffeomorphism between the two. l is a pseudospherical congruence with constant θ if: 1 The line joining s 1 Σ 1 with s = l(s 1 ) Σ is tangent to both Σ 1 and Σ The angle between the normal of Σ 1 at s 1 and the normal of Σ at s = l(s 1 ) is θ 3 d(s 1, l(s 1 )) = sinθ, s 1 Σ 1 Theorem. If l : Σ 1 Σ is a pseudospherical congruence of constant θ: 1 Σ 1 and Σ are pseudospherical. The Tchebyshev coordinates u, v in Σ 1 are mapped into the Tchebyshev coordinates u, v in Σ. 3 Let φ 1 (u, v) the solution of the sge corresponding to Σ 1. The solution φ of the first-order PDE system φ u = φ ( ) 1 u + asin φ + φ 1 φ, v = φ 1 v + ( ) a sin φ φ 1, () where a = tan θ, is the solution of the sge corresponding to Σ. 4 The system () is solvable if φ 1 solves the sge.

6 BT check and one-soliton solution BT check: Derive the left in () and use the right in (): ( ) φ u v φ 1 φ + φ 1 = a cos u v v (φ + φ 1 ) ( ) ( ) φ φ 1 φ + φ 1 = sin cos = sinφ sinφ 1 One-soliton solution. Start from: φ 1 = 0. Integration of BT ( ) dφ sin φ = log tan φ = au + f (v) + c u = v + g(v) + cv 4 a One step BT transformation φ (u, v) = 4arctan[Cexp(au + v )], C = cu cv = constant a

7 Bianchi permutability theorem Bianchi miracle: after the first step, the procedure can be carried out algebraically. Consider the successive BT: 1 u (φ φ 1 ) = a 1 sin φ + φ 1 1, u (φ 3 φ 1 ) = a sin φ 3 + φ 1 1 u (φ 4 φ ) = a sin φ 4 + φ 1, u (φ 4 φ 3 ) = a 1 sin φ 4 + φ 3 Add and subtract these s in such a way that the derivatives disappear ( ) ( )] ( ) ( )] φ + φ 1 φ4 + φ 3 φ3 + φ 1 φ4 + φ a 1 [sin sin = a [sin sin tan φ 4 φ 1 4 Two step transformation = a 1 + a tan φ φ 3 a 1 a 4

8 Start from φ (u, v) = 4 arctan Two- φ 4 (u, v) = 4 arctan Two-soliton and breather solutions [ )] [ )] C 1 exp (a 1 u + va1, φ 3 (u, v) = 4 arctan C exp (a u + va [ a1 + a [ tan arctan (C 1 e v +a a 1 u ) 1 arctan (C e v +a a u )]] a 1 a Breather solutions. Choose C 1 = C = 1, a 1 = sin θ i cos θ, and a = sin θ + i cos θ [ sin θ sin( u v ] cos θ) φ 4 (u, v) = 4 arctan cos θ cosh( u+v sin θ) Three- ( v +a φ 6 (u, v) = 4 arctan C e a u ) + + [ [ [ [ ( a1 + a 3 a1 + a v +a 4 arctan tan arctan tan arctan C 1 e a 1 u ) ( v +a 1 arctan C e a u )]] a 1 a 3 a 1 a + + [ [ ( a + a v 3 +a arctan tan arctan C e a u ) ( v +a arctan C 3 e a 3 u )]]]] 3 a 3 a

9 Pseudospherical One-soliton : a = 1, tractroid, and a 1, Dini Two-soliton. Kuen and breather ( 1 cos φ Singularities: det cos φ 1 ) = 0 C. L. Terng and K. Uhlenbeck: Geometry of solitons, Notices of AMS, 47(000) 17-5