Matching the circular Wilson loop with dual open string solution at - PDF document

Matching the circular Wilson loop with dual open string solution at 1 -loop in strong coupling M. Kruczenski and A. Tirziu arXiv:0803.0315 [hep-th] • Compute the 1 -loop correction to the effective action for the string solution ending on a straight line at the boundary. • Compute the 1 -loop correction to the effective action for the string solution in AdS 5 × S 5 dual to the circular Wilson loop. • More generically, the method we use can be applied whenever the two dimensional spectral problem factorizes, to regularize and define the fluctuation determinants in terms of solutions of one-dimensional differential equations.

A. Tirziu • It can be applied to non-homogeneous solutions both for open and closed strings and to various boundary conditions. • Circular Wilson loop, 1 -loop partition function result matches, up to a factor of two, the expectation from the exact gauge theory computation. The discrepancy can be attributed to an overall constant in the string partition function coming from the measure, which we have not fixed.

A. Tirziu Wilson loop solutions Checking AdS/CFT: • Matching the anomalous dimension of certain operators in the gauge theory to the energy of a corresponding closed string in AdS 5 × S 5 . • Comparing expectation value of Wilson loops with the string partition function of a dual string solution that at the boundary of AdS ends on the loop. < W > = Z W = 1 � � � x µ + Φ i | ˙ x | θ i ) ds N TrP exp ( iA µ ˙ • Duality should be true to all orders in 1 /N expansion and all orders in gauge theory coupling g 2 , and on string theory side to full quantum string and all orders in string coupling. λ = g 2 N, 4 πg s = g 2

A. Tirziu Planar level N = ∞ , free string g s = 0 and test the correspondence as a function of λ . Gauge theory weakly coupled at small λ . String theory is weakly coupled at large λ . Hard to check correspondence in general. Wilson Loops constructed in N. Drukker, D. J. Gross and H. Ooguri, [arXiv:hep-th/9904191] These are loops at the boundary of AdS 5 . • Single straight line - globally supersymmetric BPS object. • Parallel lines separated by a length L . Related to the computation of quark-antiquark potential. • Cusp Wilson loop –discussed recently in connection with scattering amplitudes of gluons in L. F. Alday and J. M. Maldacena, “Gluon scattering amplitudes at strong coupling,” [arXiv:0705.0303 [hep-th]] • Circular Wilson loop – not invariant under all conformal transformations On string theory side, string solutions minimize the area bounded by these loops at the boundary of AdS 5 . 1 -loop string corrections to effective action.

A. Tirziu For straight string we should get zero for the 1 -loop effective action. For circular string we expect from gauge theory √ λ + 3 4 ln λ + 1 2 ln π 2 + 3 1 √ Γ = − + ... 8 λ It was shown in J. K. Erickson, G. W. Semenoff and that the expectation K. Zarembo, [arXiv:hep-th/0003055] value of circular Wilson loop computed exactly at planar level and all orders in λ . The proposed result is √ < W > = 2 √ I 1 ( λ ) λ Computation can be expressed in terms of a Gaussian matrix model. Gauge theory computation extended to all orders in 1 /N expansion in N. Drukker and D. J. Gross, This was fully checked recently [arXiv:hep-th/0010274] by direct gauge theory computation in V. Pestun, arXiv:0712.2824 [hep-th] • Goal: check the gauge theory expression at strong coupling against the string theory beyond classical level in string theory. We need to compute 1 -loop corrections to the effective action.

A. Tirziu Straight string solution AdS 5 metric ds 2 = 1 z 2 ( dx 2 0 + dx 2 1 + dx 2 2 + dx 2 3 + dz 2 ) solution is x 0 = τ, z = σ 0 ≤ z < ∞ For τ we take a large interval 0 ≤ τ < 2 πT , T large. Induced metric is AdS 2 2 = 1 σ 2 ( dτ 2 + dσ 2 ) ds 2 with 2 d curvature R (2) = − 2 .

A. Tirziu The classical action is √ λT S = ǫ Action actually is singular so we introduced a cutoff ǫ at small z . Linear divergency is proportional to the length of the Wilson loop • Need to regularize it to get physical result Consider Lagrange transform that introduces a boundary term making area finite. Action is proportional to the volume part of the Euler number √ χ v = 1 � d 2 σ √ gR (2) = − T S = − λχ v , 4 π ǫ M Natural topological way to regularize is to add a term proportional to boundary part of Euler number χ b = 1 � dsκ g = T 2 π ǫ ∂M making Euler number finite and integer. Regularized action is √ S = − λχ, χ = χ v + χ b = 0

A. Tirziu One loop correction to the effective action GS string in AdS 5 × S 5 . Bosonic part √ λ � d 2 σ √ gg ij G µν ( x ) ∂ i x µ ∂ j x ν S = 4 π and quadratic fermionic part √ λ � d 2 σL 2 F S F = 2 π L 2 F = − i ( √ gg ij δ IJ − ǫ ij s IJ )¯ θ I ρ i D j θ J where ρ i = Γ A e A i D i θ I = δ IJ ∇ i − 1 ∇ i = ∂ i + 1 2 iǫ IJ ρ i θ J , 4Ω AB Γ AB i

A. Tirziu Consider fluctuations near a particular solution N. Drukker, D. J. Gross and A. A. Tseytlin, [arXiv:hep- th/0001204] • shown that the 1 -loop effective action is finite for any string solution and any background metric We fix background metric the induced metric g ij = h ij = 1 σ 2 δ ij Transverse bosonic string fluctuation action √ � � λ � dτdσ 1 σ 2 ∂ i ζ A ∂ j ζ A + 2( ζ 1 ) 2 + 2( ζ 2 ) 2 + 2( ζ 3 ) 2 S = σ 2 4 π three bosonic fluctuations with mass squared = 2, and five with mass squared =0 Longitudinal fluctuation Lagrangian the same as that for conformal ghosts. Spectral problem needed to solve is 1 ) + 2 = −∇ 2 + 2 L = σ 2 ( − ∂ 2 0 − ∂ 2 Lf = Λ f,

A. Tirziu • Boundary conditions. Since we want to compare the results between straight and circular string solutions we choose periodic boundary condition in τ . In σ we choose Dirichlet boundary conditions. n g n ( σ ) e imτ with m = n With f ( τ, σ ) = � T then determinant of the operator is � � � σ 2 ( − ∂ 2 1 + m 2 ) + 2 det L = det m T large, at the end replace sum by integral Fermionic quadratic Lagrangian for straight string solution is L 2 F = − 2 i √ g ¯ θD F θ D F = − σ Γ 0 ∂ 0 + σ Γ 4 ∂ 1 − 1 2Γ 4 + i Γ 0 Γ 4 Can choose representation for Gamma matrices: Γ 0 = iσ 2 × I 8 , Γ 4 = σ 1 × I 8 , then Γ 0 , Γ 4 play the role of worldsheet Dirac matrices.

A. Tirziu Squaring Dirac operator we obtain the spectral problem L F θ = Λ θ L F = −∇ i ∇ i + R (2) 1 + m 2 ) + 3 + 1 = σ 2 ( − ∂ 2 4 + Γ 04 mσ 4 We have eight fermions with mass squared = 1 Putting together bosons and fermions we obtain 1 -loop partition function det 8 / 2 ( −∇ 2 + R (2) + 1) 4 Z = det 3 / 2 ( −∇ 2 + 2) det 5 / 2 ( −∇ 2 ) Computation of functional determinants difficult in general. When they reduce to one-dimensional operators can use a nice method to compute ratio of determinants.

A. Tirziu Computation of ratio of determinants Method found long ago in I.M. Gelfand and A.M. Yaglom, J. Math. Phys., 1 : 48-69,1960 For two operators defined in an interval x ∈ [ a, b ] and with Dirichlet boundary conditions L = − P 0 ( σ ) d 2 dσ 2 + P 1 ( σ ) d dσ + P 2 ( σ ) L = − P 0 ( σ ) d 2 P 1 ( σ ) d ˆ dσ 2 + ˆ dσ + ˆ P 2 ( σ ) the ratio of determinants can be computed as R b a dσP 1 ( σ ) P − 1 = e − 1 ( σ ) det L ψ ( b ) 2 0 R b det ˆ ˆ P 1 ( σ ) P − 1 e − 1 a dσ ˆ L ψ ( b ) ( σ ) 2 0 ˆ where ψ and ψ are solutions of the initial value problems L ˆ ψ ( a ) = ˆ ψ ′ ( a ) = ˆ ˆ ψ ′ ( a ) = 1 Lψ = 0 , ψ = 0 , ψ ( a ) = 0 ,

A. Tirziu For P 1 ( σ ) = ˆ P 1 ( σ ) = 0 this reduces to det L = ψ ( b ) det ˆ ˆ L ψ ( b ) • can be generalized to any boundary conditions. • can have different boundary conditions at the two ends • both operators must have the same boundary conditions • needs modification if operators have zero modes Return to the determinants of interest and compute ratio. 1 -loop effective action is Γ 1 = 1 � 2 ln P m m 1 + m 2 + 2 det 3 [ − ∂ 2 σ 2 ] det 5 [ − ∂ 2 1 + m 2 ] P m = 1 + m 2 + 1 + m 2 + det 4 [ − ∂ 2 σ ] det 4 [ − ∂ 2 4 σ 2 + m 3 4 σ 2 − m 3 σ ] (1) Singularity at σ = 0 , introduce a cutoff ǫ . Singularity is reflected in the initial value solutions.

A. Tirziu • We take finite interval σ ∈ [ ǫ, R ] , and at the end take the limit R → ∞ . It is crucial that the final result is independent of R . After the limit R → ∞ at the very end of the computation we take ǫ → 0 . Initial value solutions: transversal bosonic operator � m 2 + 2 � − g ′′ + g ′ ( ǫ ) = 1 g = 0 , g ( ǫ ) = 0 , σ 2 solution 1 g ( σ ) = m 3 ǫσ [ m ( σ − ǫ ) cosh m ( σ − ǫ ) − (1 − ǫm 2 σ ) sinh m ( σ − ǫ )] For fermions we need � � 1 + m 2 + 3 4 σ 2 + m − ∂ 2 θ = 0 σ 1 � � (2 mσ − 1) e m ( σ − ǫ ) − (2 mǫ − 1) e − m ( σ − ǫ ) θ ( σ ) = 4 m 2 √ ǫσ Solutions blow up when boundary is at ǫ = 0 . We expect to have 1 /ǫ divergency also at 1 -loop order.