Six-point gluon scattering amplitudes from -symmetric integrable model

Transcription

1 YITP Workshop 2010 Six-point gluon scattering amplitudes from -symmetric integrable model Yasuyuki Hatsuda (YITP) Based on arxiv: [hep-th] in collaboration with K. Ito (TITECH), K. Sakai (Keio Univ.), and Y. Satoh (Univ. of Tsukuba)

2 AdS/CFT Correspondence Concrete example of gauge/string duality Maldacena 97 In the t Hooft limit: N with λ held fixed, planar SYM/free string contribution is dominant The AdS/CFT correspondence is a strong/weak type duality Type IIB Strings N =4SU(N) vs. on AdS 5 S 5 Super Yang-Mills λ g 2 YM N =4πg sn = R4 α 02 Gauge theory side: weak coupling analysis String theory side: strong coupling analysis difficult to compare

3 Integrability in AdS/CFT Planar N =4SYM and AdS 5 S 5 string theories have integrable structures Minahan, Zarembo 02 Bena, Polchinski, Roiban 03 Integrability is a power tool to analyze the spectrum of both theories Integrability also plays an important role in studying scattering amplitudes Thermodynamic Bethe ansatz (TBA) appears Alday, Gaiotto, Maldacena 09 Motivation: With the help of integrability, we would like to find a formulation that connects weak and strong coupling analyses

4 Alday-Maldacena Program How to compute gluon scattering amplitudes at strong coupling by using AdS/CFT Alday, Maldacena 07 There is a duality between gluon amplitudes and expectation values of null polygonal Wilson loops The expectation value of Wilson loop can be computated by the area of minimal surface of open string AdS bulk T-dual AdS/CFT Gluons Wilson loop Boundary

5 Strategy Start with classical strings in AdS Alday, Gaiotto, Maldacena 09 Solve equation of motion with null polygonal boundary Solve a set of integral equations (TBA equations) Substitute a solution into action Compute free energy Area of minimal surface It is hard to construct solutions with polygonal boundaries Our goal is to know the area of minimal surface, not to construct solutions Alday, Gaiotto and Maldacena proposed a set of integral equations which determines the minimal area of the hexagonal Wilson loop in AdS(5)

6 String sigma-model in AdS(5) Pohlmeyer reduction EoMs + Virasoro constraints Hitchin equations Stokes phenomenon for solutions of Hitchin equations Consider a solution in each Stokes sector Define new functions from such solutions These Y s satisfy some functional relations (Y-system) We can rewrite Y-system as a set of integral equations Such equations are of the form of Thermodynamic Bethe ansatz (TBA) equations We studied the TBA equations for six-point case in detail

7 Y-system and TBA equations Y-system: TBA equations: ²(θ) =2 Z cosh θ + K 2 log(1 + e ² ) ²(θ) =2 2 Z cosh θ +2K 1 log(1 + e ² ) K 1 (θ) = f g = Y + 1 Y 1 =1+Y 2 Y + 2 Y 2 =(1+μY 1 )(1 + μ 1 Y 1 ) 1 2π cosh θ, K 2(θ) = Z ²(θ) logy 1 (θ), dθ 0 f(θ θ 0 )g(θ 0 ) ²(θ) logy 2 (θ) + K 1 log(1 + μe ² )(1 + μ 1 e ² ) 2coshθ π cosh2θ f ± f θ ± πi 4 + K 2 log(1 + μe ² )(1 + μ 1 e ² ) Y 1 Y 2 Y 3 = Y 1 Alday, Gaiotto, Maldacena 09

8 Minimal Area Although the area of the minimal surface is divergent, we can regularize it in a well understood way BDS conjecture A = A div + A BDS R remainder function R = R 1 Z 2 A free R 1 = 1 3X Li 2 (1 U k ) 4 k=1 cross ratios Bern, Dixon, Smirnov 05 x ij x i x j A free = 1 2π Z µ dθ U 1 = x2 14 x2 36 x 2 13 x2 46,U 2 = x2 25 x2 14,U 3 = x2 36 x2 25 x 2 24 x2 x x Z coshθ log(1 + μe ²(θ) )(1 + μ 1 e ²(θ) ) +2 2 Z coshθ log(1 + e ²(θ) )

9 Goal Our goal is to know the remainder function as a function of the cross ratios Three cross ratios are related to the Y-function Uk =1+Y2 µ (2k 1)πi Thus we can relate the cross ratios to three parameters in TBA systems in principle (U 1,U 2,U 3 ) ( Z, ϕ, μ) 4 iϕ (k =1, 2, 3) The TBA equations are easily solved numerically In some special cases, we can obtain analytical results

10 Exact Result at Massless Limit TBA equations can be solved in the massless limit In this limit, Y-functions are independent of Functional relations algebraic equations Y 2 =1+Y 1 2, Y2 2 =(1+μY 1)(1 + μ 1 Y 1 ) µ µ φ 2φ Y 1 =2cos, Y 2 =1+2cos 3 3 The free energy is given by A free = 1 π (L μ(y 1 )+L μ 1(Y 1 )+L 1 (Y 2 )) = π 6 φ2 3π L λ (x) 1 µ µ logx log 1+ λ µ 2Li 2 λ 2 x x Z 0 Alday, Gaiotto, Maldacena 09 θ μ = e iφ

11 In this limit, three cross ratios are all equal U 1 = U 2 = U 3 =4cos 2 µ φ 3 We obtain the exact expression of the remainder function R(U, U, U) = π 6 + φ2 µ φ U =4cos 2 3 3π 3 4 Li 2(1 U)

12 Analysis near Massless Limit We can also obtain analytical expression near by using the CFT technique Z 0 Recall that wide classes of 2d massive integrable models can be regarded as mass deformations of CFTs Z S = S CFT + λ d 2 x ε(x) YH, Ito, Sakai, Satoh, arxiv: ε = ε = 1 3 Zamolodchikov 87 parafermion CFT in our case (n = 6) The coupling constant is exactly related to the mass of TBA system Fateev 94 γ(x) Γ(x) Γ(1 x) λ =( ) Z 4/3

13 Free Energy near CFT point Partition function Z = h1i = h exp λ The free energy is perturbatively expanded as A free = A (CFT) free Z 2 + Z n Y j=2 Due to -symmetry ε ε, the terms with odd n vanish The first non-trivial correction is n = 2 case X Z n=1 i À d 2 x ε(x) 0 ( λ) n (2π) 4 3 n+2 n! evaluate by CFT action d 2 z j z j 4/3 hv (0)ε(1)ε(z 2 ) ε(z n )V ( )i 0,connected

14 For n = 2, we can evaluate the correlation function exactly hv (0)ε(1)ε(z 2 )V ( )i 0,connected The first correction of the free energy: = 1 z z2 2φ 3π This result is in good agreement with the numerical result!

15 Remainder Function near CFT point To compute the remainder function, we need to know the behavior of R 1 = 1 4 Recall that R 1 3X k=1 Li 2 (1 U k ) U k =1+Y 2 µ (2k 1)πi 4 iϕ (k =1, 2, 3) We assume that the Y-function is expanded as Y 2 (θ) = X n=0 Ỹ (n) 2 (θ, φ) Z 4 3 n

16 The first and second coefficients take the following forms Ỹ (0) 2 (θ, φ) =1+2cos Ỹ (1) 2 (θ, φ) =y (1) (φ)cosh µ 2φ The perturbative expansion of R 1 = X n=0 R (n) 1 (ϕ, φ) Z 4 3 n 3 µ 4θ 3 R 1 R (0) 1 (ϕ, φ) = 3 4 Li 2(1 4β 2 ) R (1) 1 (ϕ, φ) =0 R (2) 1 (ϕ, φ) = 3(4β2 1+log(4β 2 )) 64β 2 (4β 2 1) 2 y (1) (φ) 2 µ φ β =cos 3

17 At present, we could not fix the function y (1) (φ), but it can be evaluated numerically It should be fixed analytically In summary, the remainder function is expanded as R = R (0) + R (2) Z O( Z 4 ) R (0) = π 6 + φ2 3π 3 4 Li 2(1 4β 2 ) µ 1 R (2) = Cγ 3 + φ µ 1 γ 3π 3 φ 3π φ =0, ϕ = π (4β2 1+log(4β 2 )) y (1) (φ) 2 64β 2 (4β 2 1) 2 µ φ β =cos 3

18 Comment on Large Mass Limit Large mass case: Z À1 The TBA equations can be solved approximately ²(θ) = 2 Z cosh θ + (exponetial corrections) ²(θ) =2 2 Z cosh θ + (exponetial corrections) Free energy: A free 2 Z π Similarly we can evaluate the remainder function modified Bessel fonction of the second kind h (μ + μ 1 )K 1 (2 Z )+ 2K 1 (2 i 2 Z )

19 Summary Gluon scattering amplitude at strong coupling can be computed by the area of minimal surface with a null polygonal boundary The problem to determine the area of such minimal surface is mapped to a set of integral equations (TBA equations) We analyzed the TBA equations for six-point amplitudes in detail We obtained the analytical expression of the area up to an unknown function It is interesting to fix the analytic form of this unknown function Analysis of TBA equations for general n-point amplitudes Do TBA equations also appear if we consider -corrections?