Thermal correlators in integrable models

Transcript

1 Thermal correlators in integrable models G. Takács HAS Theoretical Physics Research Group Eötvös University, Budapest In collaboration with: B. Pozsgay (Institute for Theoretical Physics, Amsterdam) I. Szécsényi (Eötvös University) Talk given at the 8th Bologna Workshop on: CFT and Intergrable Models and their Applications to Quantum Field Theory, Statistical Mechanics and Condensed Matter Physics September 12-15, 2011

2 Outline of talk 1. Introduction Thermal correlators The problem 2. Ingredients of the solution Finite volume as regulator Multidimensional residue 3. The thermal two-point function 4. How to proceed further? Conclusions and outlook

3 What do we wish to compute and why? Goal: correlators in dimensional (integrable) QFTs at finite temperature O 1... O n T = Tr e RH O 1... O n Tr e RH R = 1 T

4 What do we wish to compute and why? Goal: correlators in dimensional (integrable) QFTs at finite temperature O 1... O n T = Tr e RH O 1... O n Tr e RH R = 1 T Why? A large number of condensed matter systems can be described by such models (impurities, (quasi) one-dimensional conductors, spin chains/ladders) What is a correlator good for? n = 1: order parameters n = 2: response functions (susceptibility etc)

5 One-point functions O(t, x) T = e REn n O(0, 0) n n n e REn

6 One-point functions O(t, x) T = N=0 k=1 e REn n O(0, 0) n n n e REn Solution is known: LeClair-Mussardo formula (1999) 1 N (ˆ ) dθk 1 O(t, x) T,µ = N! 2π 1 + e ɛ(θ F O k) 2N (θ 1,....θ N ) conn

7 One-point functions O(t, x) T = N=0 k=1 e REn n O(0, 0) n n n e REn Solution is known: LeClair-Mussardo formula (1999) 1 N (ˆ ) dθk 1 O(t, x) T,µ = N! 2π 1 + e ɛ(θ F O k) 2N (θ 1,....θ N ) conn ˆ dθ TBA: ɛ(θ) = mr cosh θ µ 2π ϕ(θ θ ) log(1 + e ɛ(θ ) ) ϕ(θ) = i log S(θ) θ F O 2N (θ 1,....θ N ) conn = θ 1,....θ N O(0, 0) θ 1,....θ N connected

8 One-point functions O(t, x) T = N=0 k=1 e REn n O(0, 0) n n n e REn Solution is known: LeClair-Mussardo formula (1999) 1 N (ˆ ) dθk 1 O(t, x) T,µ = N! 2π 1 + e ɛ(θ F O k) 2N (θ 1,....θ N ) conn ˆ dθ TBA: ɛ(θ) = mr cosh θ µ 2π ϕ(θ θ ) log(1 + e ɛ(θ ) ) ϕ(θ) = i log S(θ) θ F O 2N (θ 1,....θ N ) conn = θ 1,....θ N O(0, 0) θ 1,....θ N connected Pozsgay-Takács (2007): finite volume regularization, to O ( e 3mR) Pozsgay (2010): proof to all orders

9 Two-point functions R R t t x O 1 (t, x)o 2 (0, 0) T = 1 e REn n O 1 (0, 0) m e (R t)em e ix(pm Pn) m O 2 (0, 0) n Z n,m (t: Euclidean time). n : N particles, m : M particles Massive QFT: E n N m, E m M m double series in powers of e mr and e m(r t)

10 Two-point functions R R t t x O 1 (t, x)o 2 (0, 0) T = 1 e REn n O 1 (0, 0) m e (R t)em e ix(pm Pn) m O 2 (0, 0) n Z n,m (t: Euclidean time). n : N particles, m : M particles Massive QFT: E n N m, E m M m double series in powers of e mr and e m(r t) What do we have? 1. Large-distance asymptotics (Altshuler, Konik, Tsvelik 2004) 2. Partial expansion (Essler-Konik 2008,2009 only works for N = 1 or M = 1)

11 Two-point functions R R t t x O 1 (t, x)o 2 (0, 0) T = 1 e REn n O 1 (0, 0) m e (R t)em e ix(pm Pn) m O 2 (0, 0) n Z n,m (t: Euclidean time). n : N particles, m : M particles Massive QFT: E n N m, E m M m double series in powers of e mr and e m(r t) What do we have? 1. Large-distance asymptotics (Altshuler, Konik, Tsvelik 2004) 2. Partial expansion (Essler-Konik 2008,2009 only works for N = 1 or M = 1) Goal: to have a systematic expansion!

12 LeClair-Mussardo proposal (1999) O 1 (x, t)o 2 (0, 0) T = 1 N [ˆ ] dθj N! 2π f σ j (θ j )e σ j (tɛ j +ixk j ) N=0 σ i =±1 j=1 F O 1 N (θ 1 iπ σ 1,..., θ N iπ σ N ) F O 2 N (θ 1 iπ σ 1,..., θ N iπ σ N ) where σ j = (1 σ j )/2 {0, 1}, f σj (θ j ) = 1/(1 + e σ j ɛ(θ j ) ), ɛ j = ɛ(θ j )/R és k j = k(θ j ) ˆ k(θ) = m sinh θ + dθ δ(θ θ )ρ 1 (θ ) ˆ 2πρ 1 (θ)(1 + e ɛ(θ) ) = m cosh θ + dθ ϕ(θ θ )ρ 1 (θ ) FN O (θ 1,..., θ N ) = 0 O(0, 0) θ 1,..., θ N

13 LeClair-Mussardo proposal (1999) O 1 (x, t)o 2 (0, 0) T = 1 N [ˆ ] dθj N! 2π f σ j (θ j )e σ j (tɛ j +ixk j ) N=0 σ i =±1 j=1 F O 1 N (θ 1 iπ σ 1,..., θ N iπ σ N ) F O 2 N (θ 1 iπ σ 1,..., θ N iπ σ N ) where σ j = (1 σ j )/2 {0, 1}, f σj (θ j ) = 1/(1 + e σ j ɛ(θ j ) ), ɛ j = ɛ(θ j )/R és k j = k(θ j ) ˆ k(θ) = m sinh θ + dθ δ(θ θ )ρ 1 (θ ) ˆ 2πρ 1 (θ)(1 + e ɛ(θ) ) = m cosh θ + dθ ϕ(θ θ )ρ 1 (θ ) FN O (θ 1,..., θ N ) = 0 O(0, 0) θ 1,..., θ N Saleur (1999): doubts concerning validity...

14 Form factors Asymptotic states: { θ 1,..., θ n in : θ 1 > θ 2 > > θ n θ 1,..., θ n = θ 1,..., θ n out θ θ = 2πδ(θ θ) : θ 1 < θ 2 < < θ n Factorized scattering: θ 1,..., θ k, θ k+1,..., θ n = S(θ k θ k+1 ) θ 1,..., θ k+1, θ k,..., θ n Form factors: F O mn(θ 1,..., θ m θ 1,..., θ n ) = θ 1,..., θ m O(0, 0) θ 1,..., θ n

15 Form factors Asymptotic states: { θ 1,..., θ n in : θ 1 > θ 2 > > θ n θ 1,..., θ n = θ 1,..., θ n out θ θ = 2πδ(θ θ) : θ 1 < θ 2 < < θ n Factorized scattering: θ 1,..., θ k, θ k+1,..., θ n = S(θ k θ k+1 ) θ 1,..., θ k+1, θ k,..., θ n Form factors: F O mn(θ 1,..., θ m θ 1,..., θ n ) = θ 1,..., θ m O(0, 0) θ 1,..., θ n Crossing: Fmn(θ O 1,..., θ m θ 1,..., θ n) = Fm 1n+1(θ O 1,..., θ m 1 θ m + iπ, θ 1,..., θ n) + n k 1 2πδ(θ m θ k ) S(θ l θ k )Fm 1n 1(θ O 1,..., θ m 1 θ 1,..., θ k 1, θ k+1..., θ n) k=1 l=1 Elementary form factors: F O n (θ 1,..., θ n ) = 0 O(0, 0) θ 1,..., θ n

16 The problem: disconnected pieces O 1 (t, x)o 2 (0, 0) T = 1 e REn n O 1 (0, 0) m e (R t)em e ix(pm Pn) m O 2 (0, 0) n Z n,m

17 The problem: disconnected pieces O 1 (t, x)o 2 (0, 0) T = 1 e REn n O 1 (0, 0) m e (R t)em e ix(pm Pn) m O 2 (0, 0) n Z n,m Disconnected pieces: δ and δ 2 terms.

18 The problem: disconnected pieces O 1 (t, x)o 2 (0, 0) T = 1 e REn n O 1 (0, 0) m e (R t)em e ix(pm Pn) m O 2 (0, 0) n Z n,m Disconnected pieces: δ and δ 2 terms. Well-known treatment: put in box but here it is important to take into account particle interactions.

19 The problem: disconnected pieces O 1 (t, x)o 2 (0, 0) T = 1 e REn n O 1 (0, 0) m e (R t)em e ix(pm Pn) m O 2 (0, 0) n Z n,m Disconnected pieces: δ and δ 2 terms. Well-known treatment: put in box but here it is important to take into account particle interactions. Why? Because the relevant contribution is subleading in the volume dependence!

20 The problem: disconnected pieces O 1 (t, x)o 2 (0, 0) T = 1 e REn n O 1 (0, 0) m e (R t)em e ix(pm Pn) m O 2 (0, 0) n Z n,m Disconnected pieces: δ and δ 2 terms. Well-known treatment: put in box but here it is important to take into account particle interactions. Why? Because the relevant contribution is subleading in the volume dependence! One-point function: all terms are maximally disconnected! O(t, x) T = 1 e REn n O(0, 0) n Z n

21 Form factors in finite volume I Spectrum in finite volume: {I 1,..., I n } L I 1 I n Bethe-Yang equations: e iml sinh θ k S( θ k θ l ) = 1 l k Using phaseshift S(θ) = e iδ(θ) : Q k ( θ 1,..., θ n ) = ml sinh θ k + δ( θ k θ l ) = 2πI k l k, k = 1,..., n Density of states: E E 0 = n m cosh θ k + O k=1 (e µl) ρ(θ 1,..., θ n ) = det J (n), J (n) kl = Q k(θ 1,..., θ n ) θ l

22 Form factors in finite volume II General case: {I 1,..., I m} O(0, 0) {I 1,..., I n } L = F m+n( θ O m + iπ,..., θ 1 + iπ, θ 1,..., θ n ) ρ( θ 1,..., θ n )ρ( θ 1,..., θ m) +O(e µl )

23 Form factors in finite volume II General case: {I 1,..., I m} O(0, 0) {I 1,..., I n } L = F m+n( θ O m + iπ,..., θ 1 + iπ, θ 1,..., θ n ) ρ( θ 1,..., θ n )ρ( θ 1,..., θ m) Diagonal case: {I 1,..., I m} = {I 1,..., I n } F(A) L ρ({1,..., n} \ A) L {I 1... I n } O {I 1... I n } L = A {1,2,...n} where ρ({k 1,..., k r }) L = ρ( θ k1,..., θ kr ) and the symmetric evaluation F({k 1,..., k r }) L = F s 2r ( θ k1,..., θ kr ) +O(e µl ) ρ({1,..., n}) L +O(e µl ) F s 2l(θ 1,..., θ l ) = lim ɛ 0 F O 2l (θ l + iπ + ɛ,..., θ 1 + iπ + ɛ, θ 1,..., θ l ) (Pozsgay-Takács, 2007)

24 Finite volume as regulator All matrix elements are finite in finite volume: δ(θ θ ) ml cosh θδ II

25 Finite volume as regulator All matrix elements are finite in finite volume: but: care is needed δ(θ θ ) ml cosh θδ II δ(θ 1 θ 1)δ(θ 2 θ 2) ml cosh θ 1 δ I1 I 1 ml cosh θ 2δ I2 I 2 there are also + O(L) contributions!

26 Finite volume as regulator All matrix elements are finite in finite volume: but: care is needed δ(θ θ ) ml cosh θδ II δ(θ 1 θ 1)δ(θ 2 θ 2) ml cosh θ 1 δ I1 I 1 ml cosh θ 2δ I2 I 2 there are also + O(L) contributions! Solution: take correlator in finite volume O 1 (t, x)o 2 (0, 0) T,L = 1 e REn(L) e (R t)em(l) e ix(pm(l) Pn(L)) Z(R, L) n,m n O 1 (0, 0) m L m O 2 (0, 0) n L Z(R, L) = n e REn(L) L only at the end! (Pozsgay-Takács 2007: 1pt functions)

27 Reordering the series expansion I where C NM = I 1...I N O 1 (x, t)o 2 (0) T,L = 1 Z N,M C NM J 1...J M {I 1... I N } O 1 (0) {J 1... J M } L {J 1... J M } O 2 (0) {I 1... I N } L e i(p 1 P 2 )x e E 1(R t) e E 2t u and v keep track of powers of e mt and e m(r t) (at the end set u = v = 1) O 1 (x, t)o 2 (0) T,L = 1 Z Z = N u N v M C NM N,M (uv) N Z N

28 Reordering the series expansion II Z 1 = N (uv) N Z N Z 0 = Z 0 = 1 Z 1 = Z 1 Z 2 = Z 2 1 Z 2 From this: O 1 (x, t)o 2 (0) T,L = u N v N D NM D NM = l C N l,m l Z l

29 Reordering the series expansion II Z 1 = N (uv) N Z N Z 0 = Z 0 = 1 Z 1 = Z 1 Z 2 = Z 2 1 Z 2 From this: O 1 (x, t)o 2 (0) T,L = u N v N D NM D NM = l First few coefficients: D 1M = C 1M Z 1 C 0,M 1 C N l,m l Z l D 2M = C 2M Z 1 C 1,M 1 + (Z 2 1 Z 2 )C 0,M 2

30 Reordering the series expansion II Z 1 = N (uv) N Z N Z 0 = Z 0 = 1 Z 1 = Z 1 Z 2 = Z 2 1 Z 2 From this: O 1 (x, t)o 2 (0) T,L = u N v N D NM D NM = l First few coefficients: D 1M = C 1M Z 1 C 0,M 1 Infinite volume limit: C N l,m l Z l D 2M = C 2M Z 1 C 1,M 1 + (Z 2 1 Z 2 )C 0,M 2 O 1 (x, t)o 2 (0) T = lim O 1(x, t)o 2 (0) T,L = D NM L N,M D NM = lim D NM L

31 Partition function Z 1 = I e mr cosh θ = ˆ dθ 2π ml cosh θe mr cosh θ ( ml sinh θ = 2πI )

32 Partition function Z 1 = I e mr cosh θ = ˆ dθ 2π ml cosh θe mr cosh θ ( ml sinh θ = 2πI ) Two-particle contribution Z 2 = I 1<I 2 e mr(cosh θ1+cosh θ2) ml sinh θ 1 + δ(θ 1 θ 2 ) = 2πI 1, ml sinh θ 2 + δ(θ 2 θ 1 ) = 2πI 2 2 I 1 <I 2 = = I 1,I 2 I 1 =I 2 ˆ dθ1 2π ˆ dθ2 ρ2(θ1, θ2) 2π ˆ dθ1 ml cosh θ1... 2π ρ 2(θ 1, θ 2) = m 2 L 2 cosh θ 1 cosh θ 2 + ml(cosh θ 1 + cosh θ 2)ϕ(θ 1 θ 2) Z 2 = 1 ˆ dθ1 2 2π 1 2 ˆ dθ2 2π ρ2(θ1, θ2)e mr(cosh θ 1+cosh θ 2 ) ˆ dθ1 2π ml cosh θ1e 2mR cosh θ 1

33 Multi-dimensional residue theorem Performing sums over quantum numbers (at finite L!): Turn them into integrals!

34 Multi-dimensional residue theorem Performing sums over quantum numbers (at finite L!): Turn them into integrals! Theorem: Suppose g(z), f 1 (z),..., f n (z) are analytic functions, where z = (z 1,..., z n ). Let C C n be a monomial multi-contour (C = C 1 C n ). Assume that f k (z) = 0 k = 1,..., n has a solution z = (z 1,..., z n ) such that for all k z k is in the interior of C k and ( ) fk det 0 z l z=z Then C dz 1 2πi... dz n g(z) 2πi f 1 (z)... f n (z) = g(z ) ( ) det fk z=z z l

35 Multi-dimensional residue theorem Performing sums over quantum numbers (at finite L!): Turn them into integrals! Theorem: Suppose g(z), f 1 (z),..., f n (z) are analytic functions, where z = (z 1,..., z n ). Let C C n be a monomial multi-contour (C = C 1 C n ). Assume that f k (z) = 0 k = 1,..., n has a solution z = (z 1,..., z n ) such that for all k z k is in the interior of C k and ( ) fk det 0 z l z=z Then C dz 1 2πi... dz n g(z) 2πi f 1 (z)... f n (z) = g(z ) ( ) det fk z=z z l Remarks: Standard contour deformations work away from the det f = 0 variety. General multi-contour: sum of moniomial ones

36 Evaluating D 11 I D 11 = lim D 11 L D 11 = C 11 Z 1 C 00, C 00 = O 1 O 2 C 11 = I,J {I } O 1 (0) {J} L {J} O 2 (0) {I } L e i(p1 p2)x e E1(R t) e E2t

37 Evaluating D 11 I D 11 = lim D 11 L D 11 = C 11 Z 1 C 00, C 00 = O 1 O 2 C 11 = I,J {I } O 1 (0) {J} L {J} O 2 (0) {I } L e i(p1 p2)x e E1(R t) e E2t where ml sinh θ = 2πI, ml sinh θ = 2πJ E 1 = m cosh θ, p 1 = m sinh θ E 2 = m cosh θ, p 2 = m sinh θ {I } O 1 (0) {J} L = F O1 2 (θ + iπ, θ ) ρ1 (θ)ρ 1 (θ ) + δ IJ O 1 {J} O 2 (0) {I } L = F O2 2 (θ + iπ, θ) ρ1 (θ)ρ 1 (θ ) + δ IJ O 2

38 Evaluating D 11 II From this C 11 = F O1 2 (θ + iπ, θ )F O2 2 (θ + iπ, θ) ρ 1 (θ)ρ 1 (θ e i(p1 p2)x e E1(R t) e E2t ) I,J + O 1 J F O2 2 (θ + iπ, θ ) ρ 1 (θ e E2R + O 2 ) I F O1 2 (θ + iπ, θ) ρ 1 (θ) e E1R + I O 1 O 2 e E1R

39 Evaluating D 11 II From this C 11 = F O1 2 (θ + iπ, θ )F O2 2 (θ + iπ, θ) ρ 1 (θ)ρ 1 (θ e i(p1 p2)x e E1(R t) e E2t ) I,J + O 1 J F O2 2 (θ + iπ, θ ) ρ 1 (θ e E2R + O 2 ) I F O1 2 (θ + iπ, θ) ρ 1 (θ) e E1R + I O 1 O 2 e E1R Last term is O(L), but cancelled by Z 1 C 00! D 11 = C 11 Z 1 C 00, C 00 = O 1 O 2

40 Evaluating D 11 II From this C 11 = F O1 2 (θ + iπ, θ )F O2 2 (θ + iπ, θ) ρ 1 (θ)ρ 1 (θ e i(p1 p2)x e E1(R t) e E2t ) I,J + O 1 J F O2 2 (θ + iπ, θ ) ρ 1 (θ e E2R + O 2 ) I F O1 2 (θ + iπ, θ) ρ 1 (θ) e E1R + I O 1 O 2 e E1R Last term is O(L), but cancelled by Z 1 C 00! D 11 = C 11 Z 1 C 00, C 00 = O 1 O 2 End result: D 11 = ˆ ˆ dθ dθ 2π 2π F O 1 2 (θ + iπ, θ )F O 2 2 (θ + iπ, θ) e imx(sinh θ sinh θ ) m(r t) cosh θ mt cosh θ ˆ +( O 1 F O 2 2s + O 2 F O 1 dθ cosh θ 2s ) e mr 2π

41 Evaluating D 22 Using gives We need: C 22 = I 1>I 2 D 22 = C 22 Z 1 C 11 + (Z 2 1 Z 2 )C 00 K (R) t,x (θ 1, θ 2 ; θ 1, θ 2) D 11 = C 11 Z 1 C 00 D 22 = C 22 Z 1 D 11 Z 2 C 00 J 1>J 2 {I 1, I 2 } O 1 {J 1, J 2 } L {J 1, J 2 } O 2 {I 1, I 2 } L K (R) t,x (θ 1, θ 2 ; θ 1, θ 2) = e imx(sinh θ1+sinh θ2 sinh θ 1 sinh θ 2 ) e m(r t)(cosh θ1+cosh θ2) e mt(cosh θ 1 +cosh θ 2 ) Separating diagonal pieces: ({J 1, J 2 } = {I 1, I 2 } tagok) + I 1 >I 2 = I 1 >I 2 J 1 >J 2 I 1 >I 2 J 1 >J 2

42 Diagonal pieces I {I 1, I 2} O {I 1, I 2} L = F O 4s (θ 1, θ 2) + ρ 1(θ 1)F O 2c + ρ 1(θ 2)F O 2c + ρ 2(θ 1, θ 2) O ρ 2(θ 1, θ 2) Put into diagonal part: I 1 >I 2 {I 1, I 2} O 1 {I 1, I 2} L {I 1, I 2} O 2 {I 1, I 2} L e mr(cosh θ1+cosh θ2) F 4s F 4s terms: mr(cosh θ e 1 +cosh θ 2 ) F O 1 ρ I 1 >I 2(θ 1, θ 2) 2 4s (θ 1, θ 2)F O 2 4s (θ 1, θ 2) = 2 1 e mr(cosh θ 1+cosh θ 2 ) F O 1 2 ρ 2(θ 1, θ 2) 2 4s (θ 1, θ 2)F O 2 4s (θ 1, θ 2) = 1 2 I 1,I 2 ˆ dθ1 2π ˆ dθ1 2π I 1 =I 2 ˆ dθ2 2π O(L 2 ) + O(L 3 ): 0 for L. e mr(cosh θ 1+cosh θ 2 ) F O 1 4s (θ 1, θ 2)F O 2 4s (θ 1, θ 2) ρ 2(θ 1, θ 2) e 2mR(cosh θ 1) ml cosh θ 1 ρ 2(θ 1, θ 1) 2 F O 1 4s (θ 1, θ 1)F O 2 4s (θ 1, θ 1)

43 Diagonal pieces II Similarly for other pieces, the total is: C diag 22 = ˆ dθ1 2π ˆ dθ2 2π [ 1 2 e mr(cosh θ 1+cosh θ 2 ) F O 1 2c F O 2 2c (cosh θ 1 + cosh θ 2 ) 2 cosh θ 1 cosh θ e mr(cosh θ 1+cosh θ 2 ) ( m 2 L 2 cosh θ 1 cosh θ 2 + ml(cosh θ 1 + cosh θ 2 )ϕ(θ 1 θ 2 ) ) ( ) ] O 1 O 2 + ml cosh θ 1 e mr(cosh θ 1+cosh θ 2 ) O 1 F O 2 2c + O 2 F O 1 2c 1 2 ˆ Counter terms : dθ [ ( )] 2π e 2mR cosh θ ml cosh θ O 1 O O 1 F O 2 2c + O 2 F O 1 2c Z 2 C 00 = 1 ˆ dθ 2 2π ml cosh θe 2mR cosh θ O 1 O 2 1 ˆ ˆ dθ1 dθ2 [ m 2 L 2 cosh θ 1 cosh θ 2 2 2π 2π +ml(cosh θ 1 + cosh θ 2 )ϕ(θ 1 θ 2 ) ] e mr(cosh θ 1+cosh θ 2 ) O 1 O 2 ˆ ˆ dθ1 dθ2 Z 1 D 11 = Z 1 2π 2π F O 1 2 (θ 1 + iπ, θ 2 )F O 2 2 (θ 1, θ 2 + iπ) e imx(sinh θ 1 sinh θ 2 ) e m(r t) cosh θ 1 e mt cosh θ 2 ˆ dθ ( ) Z 1 F O 1 2c 2π O 2 + F O 2 2c O mr cosh θ 1 e

44 Diagonal part III Divergent parts drop out: D (diag) 22 = 1 ˆ dθ1 dθ 2 2 2π 2π e mr(cosh θ1+cosh θ2)( F O1 4s (θ 1, θ 2 ) O 2 ) F O2 4s (θ 1, θ 2 ) O 1 ˆ dθ1 But: Z 1 D 11 remains! Z 1 ˆ dθ1 2π ˆ dθ2 2π e mr(cosh θ1+cosh θ2) F O1 2π ˆ dθ ( 2π e 2mR cosh θ O 1 F O2 2c 2c F O2 2c + O 2 F O1 2c ˆ dθ2 2π F O 1 2 (θ 1 + iπ, θ 2 )F O 2 2 (θ 1, θ 2 + iπ) e imx(sinh θ 1 sinh θ 2 ) e m(r t) cosh θ 1 e mt cosh θ 2 and will regulate the nondiagonal part. ) (cosh θ 1 + cosh θ 2 ) 2 cosh θ 1 cosh θ 2

45 Nondiagonal part of D 22 I C (nondiag) 22 = I 1>I 2 J 1>J 2 {I 1, I 2 } O 1 {J 1, J 2 } L {J 1, J 2 } O 2 {I 1, I 2 } L K (R) t,x (θ 1, θ 2 ; θ 1, θ 2) K (R) t,x (θ 1, θ 2 ; θ 1, θ 2) = e imx(sinh θ1+sinh θ2 sinh θ 1 sinh θ 2 ) m(r t)(cosh θ1+cosh θ2) e e mt(cosh θ 1 +cosh θ 2 ) {I 1, I 2 } O 1 {J 1, J 2 } L = F O1 4 (θ 2 + iπ, θ 1 + iπ, θ 1, θ 2 ) ρ 2 (θ 1, θ 2 ) 1/2 ρ 2 (θ 1, θ 2 )1/2 {J 1, J 2 } O 2 {I 1, I 2 } L = F O2 4 (θ 2 + iπ, θ 1 + iπ, θ 1, θ 2 ) ρ 2 (θ 1, θ 2 ) 1/2 ρ 2 (θ 1, θ 2 )1/2 Q 1 (θ 1, θ 2 ) = ml sinh θ 1 + δ(θ 1 θ 2 ) = 2πI 1 Q 2 (θ 1, θ 2 ) = ml sinh θ 2 + δ(θ 2 θ 1 ) = 2πI 2 Q 1 (θ 1, θ 2) = ml sinh θ 1 + δ(θ 1 θ 2) = 2πJ 1 Q 2 (θ 1, θ 2) = ml sinh θ 2 + δ(θ 2 θ 1) = 2πJ 2

46 Nondiagonal part of D 22 II Using the residue theorem {I 1, I 2 } O 1 {J 1, J 2 } L {J 1, J 2 } O 2 {I 1, I 2 } L K (R) t,x (θ 1, θ 2 ; θ 1, θ 2) 1 dθ 1 dθ 2 = ρ 2 (θ 1, θ 2 ) C J1 J 2 2π 2π F O1 4 (θ 2 + iπ, θ 1 + iπ, θ 1, θ 2) 4 (θ 2 + iπ, θ 1 K (R) t,x (θ 1, θ 2 ; θ 1 + iπ, θ 1, θ 2 ), θ 2 ( ) e iq 1 (θ 1,θ 2 ) + 1 ) ( e iq 2 (θ 1,θ 2 ) + 1 ) F O2 where C J1 J 2 surrounds the solution of Q 1 (θ 1, θ 2) = ml sinh θ 1 + δ(θ 1 θ 2) = 2πJ 1 Q 2 (θ 1, θ 2) = ml sinh θ 2 + δ(θ 2 θ 1) = 2πJ 2 This is the contour we want to open : 1 2 J 1 J 2 C J1J 2 C 1 C 2 θ, θ, 1 2

47 Nondiagonal part of D 22 III But: new singularities encountered. 1. QF : partly from Q, partly from F θ 1 = θ 1 and Q 2 (θ 1, θ 2) = 2πJ 2 θ 1 = θ 2 and Q 2 (θ 1, θ 2) = 2πJ 2 θ 2 = θ 1 and Q 1 (θ 1, θ 2) = 2πJ 1 θ 2 = θ 2 and Q 1 (θ 1, θ 2) = 2πJ 1 2. FF : from F 3. QQ (vanishes for L ): θ 1 = θ 2 = θ 1 θ 1 = θ 2 = θ 2 θ 1 = θ 1 és θ 2 = θ 2 θ 1 = θ 2 és θ 2 = θ 1

48 Nondiagonal part of D 22 IV So: J 1 >J 2 dθ 1 C J1 2π J 2 dθ 2 2π = 1 dθ 1 2 C 2π dθ 2 (FF terms) (QF terms) 2π where C = C 1 C 2 (the open multi-contour). (J 1 = J 2 is not a singularity, because form factors vanish). Kinematical singularity axiom i Res θ=θ F O n+2(θ+iπ, θ, θ 1,..., θ n ) = ( 1 ) n S(θ θ k ) Fn O (θ 1,..., θ n ) k=1 can be used to compute QF and QQ residues. (Tedious and the result is lengthy, but straightforward).

49 Nondiagonal part of D 22 V Divergent part comes from second order poles in QF terms: D QF = dθ1 2π dθ 2 2π ˆ dθ 2 2π F O1 2 (θ 2 + iπ, θ 2)F O2 2 (θ 2 + iπ, θ 2 ) e imx(sinh θ2 sinh θ 2 ) e mr cosh θ1 e m(r t) cosh θ2 e mt cosh θ 2 ( (mx cosh θ 1 imt sinh θ 1 )(1 S(θ 2 θ 1 )S(θ 1 θ 2 )) ) +ϕ(θ 1 θ 2)S(θ 2 θ 1 )S(θ 1 θ 2 ) + ml cosh θ 1 ˆ dθ1 2π ˆ dθ 2 2π F O1 2 (θ 1 + iπ, θ 2)F O2 2 (θ 2 + iπ, θ 1 ) e imx(sinh θ1 sinh θ 2 ) e m(2r t) cosh θ1 e mt cosh θ 2 and is exactly cancelled by the remaining counter-term : ˆ ˆ ˆ dθ cosh θ dθ1 dθ2 ml cosh θe mr 2π 2π 2π F O1 2 (θ 1 + iπ, θ 2 )F O2 2 (θ 1, θ 2 + iπ) e imx(sinh θ1 sinh θ2) e m(r t) cosh θ1 mt cosh θ2 e

50 Conclusion 1. Systematic form factor expansion for thermal correlators Consistency: L limit is finite Explicitly constructed: D 0N, D M0, D 1N, D M1 and D Comparison to LeClair-Mussardo 2pt function formula: 2.1 LM expression not even well-defined 2.2 Higher corrections look different 3. Method was also applied to 3.1 T = 0 3-point function 3.2 second order form factor perturbation theory 4. Kormos-Pozsgay (2010): residue method works for one-point functions of bulk fields on an interval with boundaries.

51 Open issues 1. Is it possible to sum up the T -dependence in dressing factors (a la LeClair Mussardo) 1-point function: OK (but we knew that beforehand) 2-point function: need higher terms (e.g. D 2N ) to verify: in progress Can the method be simplified? (Lots of residue-crunching and redundant terms) 3. Applications: need extension of finite volume form factors to nondiagonal theories. Work in progress... result expected to be under the Christmas tree! E.g.: O(3) σ-model (Essler-Konik: M = 1, N = 2 term for spin-spin correlator).