One-loop correction to the energy of a wavy line string in AdS_5
aa r X i v : . [ h e p - t h ] N ov One-loop correction to the energyof a wavy line string in
AdS
School of Physics M013, The University of Western Australia,35 Stirling Highway, Crawley W.A. 6009, Australia
A. A. Tseytlin ∗ The Blackett Laboratory, Imperial College, London SW7 2AZ, U.K.
Abstract
We consider a computation of one-loop
AdS × S superstring correction to the energyradiated by the end-point of a string which moves along a wavy line at the boundary of AdS with a small transverse acceleration (the corresponding classical solution was de-scribed by Mikhailov in hep-th/0305196). We also compute the one-loop effective actionfor an arbitrary small transverse string fluctuation background. It is related by an ana-lytic continuation to the Euclidean effective action describing one-loop correction to theexpectation value of a wavy Wilson line. We show that both the one-loop contributionto the energy and to the Wilson line are controlled by the subleading term in the strong-coupling expansion of the function B ( λ ) as suggested by Correa, Henn, Maldacena andSever in arXiv:1202.4455. ∗ Also at Lebedev Institute, Moscow
Introduction
In [1] it was suggested that there is universality in a class of near-BPS problems: expectationvalues of small-cusp Euclidean Wilson loop, Euclidean wavy Wilson line, energy radiated by amoving particle in Minkowski space. They all are controlled by the same function B ( λ ) B ( λ ) = √ λ π I ( √ λ ) I ( √ λ ) = √ λ π − π + 332 π √ λ + . . . . (1.1)In the generalized cusp case the corresponding one-loop strong-coupling corrections were alreadydiscussed in [2]. In this paper we attempt to further check this prediction at strong couplingin the case of radiation and wavy line, reproducing the first subleading coefficient in (1.1) by adirect one-loop superstring computation.We shall start in section 2 with a review of a classical string solution in Minkowski-signature AdS ending on a nearly-straight time-like line at the boundary. It is described in the staticgauge by the Mikhailov’s solution [3] (see also [4]). We compute the classical string action andthe energy on this solution. In the remaining sections we shall compute the one-loop correctionsto these classical expressions.In section 3 we shall first consider the contribution from the bosonic string fluctuationsusing static gauge. Similar computations (see [5, 6, 7]) involve several subtle points that wewill not directly address here (assuming cancelation of UV divergences and conformal anomalyshould happen in the full theory as it does in the conformal gauge). In particular, we will thusignore the one-loop path integral over the S string modes as it should contribute only to thedivergences and the conformal anomaly.In section 4 we will compute the one-loop fermionic contributions. We avoid detailed studyof cancellation of UV divergences by using a heuristic (but natural) argument of how to extractthe relevant finite contribution. Summing up the bosonic and the fermionic contributions wefind that the full one-loop correction to the radiated energy is indeed proportional to the next-to-leading coefficient in the expansion in (1.1).Finally, in section 5 we will consider the Euclidean wavy Wilson line solution [8, 9, 10].The one-loop correction to the expectation value of the wavy Wilson line is given by theEuclidean one-loop effective action which can be obtained by an analytic continuation from thecorresponding Minkowski-signature expression computed in Sections 3 and 4. As a result, wefind that this one-loop correction is also governed by the next-to-leading coefficient in (1.1).1 Classical solution for a string ending on a wavy line
We shall parametrize the worldsheet by ( τ, σ ) and
AdS by the Poincare coordinates( X µ , Z ) = ( X , X i , Z ) , µ = 0 , . . . , , i = 1 , , ,ds = − ( dX ) + ( dX i ) + dZ Z . (2.1)Except for the last section, in this paper both the worldsheet and AdS are assumed to haveMinkowski signature with τ and X being timelike. We will consider the static gauge X = τ , Z = σ . (2.2)The Nambu-Goto action is then given by S cl = − √ λ π Z dτ dσ √− g , (2.3)where the induced metric is g αβ = 1 Z η µν ∂ α X µ ∂ β X ν = g αβ + 1 σ ∂ α X i ∂ β X i , α, β = 1 , g αβ being the AdS metric g αβ = σ diag( − , X i ( τ, σ = 0) = 0 . We shall consider more general solution where a string end-pointmoves along small deviation of that straight line. To leading order in X i the string action (2.3)is given by S cl = √ λ π Z dτ dσ σ [( ∂ τ X i ) − ( ∂ σ X i ) ] (2.5)and the equation of motion is − ∂ τ X i + ∂ σ X i − σ ∂ σ X i = 0 . (2.6)Its solution with boundary conditions X i ( τ, σ = 0) ≡ x i ( τ ) = x i + ( τ ) + x i − ( τ ) (2.7)is given by [3] X i = x i + ( τ + σ ) − σ ˙ x i + ( τ + σ ) + x i − ( τ − σ ) + σ ˙ x i − ( τ − σ ) , (2.8)where x i + , x i − are arbitrary functions and by dot we denote the derivative over τ . The solutionis uniquely defined by the boundary curve (2.7) the third normal derivative ( ∂ σ X i ) (cid:12)(cid:12) σ =0 [3]. Inthis paper, for simplicity, we will setx i + ( τ ) = 0 , x i − ( τ ) = x i ( τ ) . (2.9)2e shall reserve the notation x i for x i − ( τ − σ ) using x i for the function x i − of τ only.Let us evaluate the action and the energy on this solution in terms of the boundary datax i ( τ ) [3, 4]. Using equations of motion and ignoring the total τ -derivative the classical actioncan be written as S cl = √ λ π Z dτσ ( X i ∂ σ X i ) (cid:12)(cid:12)(cid:12) ∞ σ = ǫ = − √ λ π Z dτσ X i ∂ σ X i (cid:12)(cid:12)(cid:12) σ = ǫ , (2.10)where we introduced the cut-off near the boundary and assumed that X i vanishes at infinity.Using the solution (2.8),(2.9) we obtain1 σ X i ∂ σ X i = − σ x i ( τ − σ )¨ x i ( τ − σ ) − ˙ x i ( τ − σ )¨ x i ( τ − σ ) . (2.11)Expanding x i ( τ − σ ) near σ = ǫ → σ X i ∂ σ X i = − ǫ x i ( τ )¨x i ( τ ) + x i ( τ )...x i ( τ ) . (2.12)Substituting in (2.10) and integrating by parts gives S cl = − √ λ π ǫ Z dτ ( v i ) + √ λ π Z dτ v i a i , (2.13)where v i = ˙x i ( τ ) , a i = ¨x i ( τ ) (2.14)are the velocity and the acceleration of the string’s end-point. Hence, the finite part of theaction is given by S cl,fin = √ λ π Z dτ v i a i , (2.15)Now let us evaluate the classical energy. We define the target space energy as E cl = − Z dσ ∂L∂∂ τ X , (2.16)where L is the string Lagrangian in (2.3). Eq. (2.16) may be interpreted as the energy radiatedby the end-point particle moving with acceleration. To compute (2.16) in static gauge (2.2)we have to restore the dependence on X in the induced metric (2.4) and set X = τ after wetake the derivative. Alternatively, since in the static gauge we have X = τ , the energy (2.16)coincides with the two-dimensional Hamiltonian i.e. E cl ( τ ) = √ λ π Z dσσ [( ∂ τ X i ) + ( ∂ σ X i ) ] . (2.17)To find E cl in terms of the boundary data we first differentiate (2.17) with respect to τ , ∂ τ E cl ( τ ) = √ λ π Z dσσ [ ∂ τ X i ∂ τ X i + ∂ σ X i ∂ τ ∂ σ X i ] , (2.18)3hen integrate by parts and then use equations of motion (2.6). As a result, ∂ τ E cl ( τ ) = − √ λ π σ ∂ τ X i ∂ σ X i (cid:12)(cid:12)(cid:12) σ = ǫ . (2.19)From the solution (2.8) we find that − σ ∂ τ X i ∂ σ X i = 1 σ ˙ x i ¨ x i + (¨ x i ) . (2.20)Expanding near the boundary σ = ǫ → − σ ∂ τ X i ∂ σ X i | σ = ǫ = 1 ǫ ˙x i ¨x i − ˙x i ...x i = 1 ǫ v i a i − ∂ τ ( v i a i ) + ( a i ) . (2.21)Ignoring the divergence and the total derivative we find that the energy radiated over someperiod Σ is given by E cl = √ λ π Z Σ0 dτ ( a i ) = 2 πB ( λ ) Z Σ0 dτ ( a i ) , (2.22)where to leading order in large tension expansion the coefficient B ( λ ) is thus given by B ( λ ) = √ λ π . (2.23)The above classical consideration can be extended to the general case of non-linear dependenceon X i [3, 11, 12] but it will not be discussed here. Let us now consider the one-loop corrections to (2.13), (2.22). We will compute the one-loop effective action for an arbitrary transverse background X i ( τ, σ ) that solves the linearizedequations of motion (2.6). Let us reserve the notation X µ for the classical background anddenote the full AdS string fields as ( Y , Y i , Y ), i.e. ds = − ( dY ) + ( dY i ) + ( dY ) ( Y ) , (3.1)where Y will be the radial direction. The S string coordinates will be denoted as χ a , a =1 , . . . ,
5. We will impose the quantum static gauge Y = X = τ , Y = Z = σ , (3.2) Like in a light-cone gauge, there is no non-trivial ghost determinant in a static gauge: this gauge is fixeddirectly in terms of string coordinates, but their variation under reparametrizations involve the gauge parameteronly algebraically ( δX = ξ a ∂ a X ). The resulting ghost determinant is ultralocal, i.e. contributes only δ (0) terms.The same remark will apply to the κ -symmetry gauge we will use. Y , Y will not fluctuate. We will also split Y i as Y i = X i + y i λ / , (3.3)where y i are the quantum fluctuations. Since the classical solution is non-zero only in AdS ,the S fields χ a have only fluctuating part (which we also rescale by λ − / ). To compute theeffective action to one-loop order we need to expand the Nambu-Goto action S b = − √ λ π Z dτ dσ √− G , G αβ = g αβ ( Y ) + 1 √ λ ∂ α χ a ∂ β χ a , (3.4)to quadratic order in the quantum fields y i , χ a (here g αβ is the induced metric depending onthe AdS fields Y m ).Expanding (3.4) in powers of y i and χ a we will get the classical action, then linear termswhich vanish since the background satisfies the equations of motion, and, finally, the quadraticterms on which we will concentrate. S The quadratic term in χ a is S S = − π Z dτ dσ p − g ( X ) g αβ ( X ) ∂ α χ a ∂ β χ a , (3.5)where g αβ ( X ) is given by (2.4).In general, we can find a coordinate system where g αβ ( X ) is conformally flat. Then p − g ( X ) g αβ ( X ) = η αβ (3.6)and (3.5) is independent of X i . To perform such change of variables in the path integral onehas to use a regularization covariant with respect to the induced metric g αβ ( X ). The integra-tion over χ a will produce quadratic and logarithmic divergences as well as a contribution toconformal anomaly (see, e.g., [13]). On the other hand, in critical string theory the divergencesand the conformal anomaly are expected to cancel in the total expression for the partitionfunction. Hence we may ignore a non-trivial 1-loop contribution from (3.5) as it should cancelwhen we add to it similar contributions from AdS modes and the fermions. If we vary (3.5)with respect to ∂∂ τ X to find the energy we obtain ∂L S ∂∂ τ X = − π p − g ( X ) ∂g αβ ( X ) ∂∂ τ X T αβ ( χ ) , (3.7) T αβ ( χ ) = ∂ α χ a ∂ β χ a − g αβ ( X ) g γδ ( X ) ∂ γ χ a ∂ δ χ a . (3.8) Note that this cancellation is non-perturbative in X i as it is based on a regularization covariant with respectto the full metric g αβ ( X ) which cannot be imposed when we expand g αβ ( X ) in powers of X . h T αβ i = 0 for the full stress-energy tensor of all the bosonic and fermionic contributions. Note that on physical groundsone should not of course expect the S fluctuations to contribute non-trivially to the energysince the string propagates only in AdS . To finish this subsection let us find the propagator of the fields χ a as we will need it in thenext section. Using (3.6) we find that that χ a have the action of massless fields in AdS S S = 14 π Z dτ dσ [( ∂ τ χ a ) − ( ∂ σ χ a ) ] . (3.9)Hence, the propagator G ( τ , σ ; τ , σ ) satisfies( − ∂ τ + ∂ σ ) G ( τ , σ ; τ , σ ) = δ ( τ − τ ) δ ( σ − σ ) . (3.10)Though this looks like the equation in flat space, G has to satisfy the AdS (half-plane)boundary conditions G | σ =0 = 0 , G | σ =0 = 0 . (3.11)The corresponding solution is then given by G = − π log ( σ − σ ) − ( τ − τ ) ( σ + σ ) − ( τ − τ ) . (3.12)Note that this is a function of a single variable ℓ which is (half) the geodesic distance in AdS G = − π log ℓℓ + 1 , (3.13) ℓ = 12 ( σ − σ ) − ( τ − τ ) σ σ . (3.14) AdS The
AdS part of the action is S AdS = − √ λ π Z dτ dσ p − g ( Y ) , (3.15)where Y = X = τ , Y = Z = σ and the transverse coordinates are given by (3.3). Then theinduced metric will split as g ( Y ) = g ( X ) + g ( X, y ) + g ( y ) , (3.16) g αβ ( X ) = g αβ + 1 σ ∂ α X i ∂ β X i , (3.17) g αβ ( X, y ) ≡ h αβ = 1 λ / σ [ ∂ α X i ∂ β y i + ∂ α y i ∂ β X i ] , (3.18) g αβ ( y ) = 1 √ λ σ ∂ α y i ∂ β y i . (3.19) This is obvious in conformal gauge but should be true also in static gauge assuming the conformal anomalycancellation takes place. g αβ is the metric of AdS and we used the static gauge conditions X = τ , Z = σ on thebackground fields and y = 0 , y = 0 on the quantum fields.Now we expand (3.15) to quadratic order in y i keeping the background X i arbitrary. Weget p − g ( Y ) = p − g ( X ) det / [1 + g − ( X ) · g ( y ) + g − ( X ) · h ] , (3.20)where in the second factor we denote by g ( y ) and h the matrices g αβ ( y ) and h αβ , by g − ( X )the matrix inverse to g αβ ( X ) and by · the matrix multiplication. We may then write (3.20) as p − g ( X ) exp (cid:2)
12 tr log(1 + g ( X ) − · g ( y ) + g − ( X ) · h ) (cid:3) , (3.21)where the trace is over the worldsheet indices ( α, β ). Now we can expand (3.21) to quadraticorder in y p − g ( Y ) = p − g ( X )[1+ 12 tr g − ( X ) · g ( y )+ 12 tr g − ( X ) · h −
14 tr( g ( X ) · h ) + 18 (tr g − ( X ) · h ) + . . . ] . (3.22)This is the most general expression up to terms of order y for an arbitrary background. Thefirst term p − g ( X ) is the classical action. The third term (linear in h ) is linear in y and henceit vanishes on the equations of motion for X i . Let us consider the contribution coming fromthe second term S (1) AdS = − π Z dτ dσ σ p − g ( X ) g αβ ( X ) ∂ α y i ∂ β y i , (3.23)where we used g αβ ( y ) in (3.17). This term is analogous to (3.5), so by our assumption discussedin the previous subsection its contribution should cancel against other similar contributions fromthe sphere and fermion terms. Hence, the non-trivial bosonic contribution should come fromthe last two terms in (3.22) quadratic in the metric h αβ in (3.18).The fourth term in (3.22) is S (2) AdS = √ λ π Z dτ dσ p − g ( X ) g αβ ( X ) g γδ ( X ) h βγ h δα . (3.24)Since h is linear in X i , h βγ h δα is already quadratic in X i so we can replace g ( X ) with the AdS metric g .Similarly, the last term in (3.22) is S (3) AdS = − √ λ π Z dτ dσ p − g ( X ) g αβ ( X ) g γδ ( X ) h αβ h γδ . (3.25)Here we can also replace g ( X ) with g . Both (3.24) and (3.25) contain only terms quadratic in X i . Hence, the one-loop contribution comes from Feynman graph in Figure 1.7igure 1: The Feynman graph describing the bosonic contribution to the one-loop effective action.The internal lines correspond to the quantum fields and the external lines are the background. Sucha diagram is proportional to the Green’s function at coinciding points.
The bosonic one-loop effective action is then given by e i Γ b = Z [ dy i ] e i ( S (0) AdS + S (2) AdS + S (3) AdS ) , (3.26)where S (0) AdS is the free action given by (see eq. (3.23)) S (0) AdS = 14 π Z dτ dσ σ [( ∂ τ y i ) − ( ∂ σ y i ) ] . (3.27)We will need to find Γ b to order X . The propagator h y i y j i comes from the free action (3.27)and is thus diagonal. Let us denote h y i ( τ , σ ) y j ( τ , σ ) i = 2 πG ( τ , σ ; τ , σ ) δ ij . (3.28)To find G we change variables y i = σz i . Then S (0) AdS = 14 π Z dτ dσ h ( ∂ τ z i ) − ( ∂ σ z i ) − z i ) σ i . (3.29)This is the action of a massive field in AdS with m = 2. The equation for the propagator isthen (cid:0) − ∂ τ + ∂ σ − σ (cid:1) G = δ ( τ − τ ) δ ( σ − σ ) . (3.30)The solution to this equation satisfying the AdS boundary conditions G | σ =0 = 0 , G | σ =0 = 0is given by G = 12 π h − − σ + σ − ( τ − τ ) σ σ log ( σ − σ ) − ( τ − τ ) ( σ + σ ) − ( τ − τ ) i . (3.31)Note that it can be written as a function of the geodesic distance parameter ℓ defined in (3.14) G = 12 π h − − ( ℓ + 12 ) log ℓℓ + 1 i , (3.32)8ow let us consider S (2) AdS and S (3) AdS . Using the explicit form of the matrix h in (3.18) weobtain to quadratic order in X i S (2) AdS = 18 π Z dτ dσ σ η αβ η γδ ( ∂ β X i ∂ γ y i + ∂ β y i ∂ γ X i ) ( ∂ δ X j ∂ α y j + ∂ δ y j ∂ α X j )= 14 π Z dτ dσ σ [( ∂ α X i ∂ α X j )( ∂ β y i ∂ β y j ) + ( ∂ α X i ∂ α y j )( ∂ β X j ∂ β y i )] , (3.33) S (2) AdS = − π Z dτ dσ σ ( ∂ α X i ∂ α y i )( ∂ β X j ∂ β y j ) , (3.34)where α, β are contracted using η αβ . To compute Γ b to order X we need to expand exp[ i ( S (2) AdS + S (3) AdS )] to order X , i.e. we need to compute the expectation value of the operators in (3.33),(3.34). Since h y i y j i ∼ δ ij we can simplify (3.33), (3.34) by taking only the diagonal contributionin y i y j . In other words, we can set j = i and sum over i , i.e. keep only( S (2) AdS + S (3) AdS ) trace = 14 π Z dτ dσ σ X i ( ∂ α X i ∂ α X i )( ∂ β y i ∂ β y i )= 14 π Z dτ dσ σ X i [( ∂ τ X i ) − ( ∂ σ X i ) ] [( ∂ τ y i ) − ( ∂ σ y i ) ] . (3.35)Replacing y i with z i = y i /σ we get( S (2) AdS + S (3) AdS ) trace = 14 π Z dτ dσ X i h ( ∂ τ X i ) − ( ∂ σ X i ) ih ( ∂ τ z i ) − ( ∂ σ z i ) − σ z i ∂ σ z i − ( z i ) σ i (3.36)It is natural to define the z -dependent operator so that it does not contain first derivatives. Forthis we integrate the third term in the second bracket by part ignoring the terms with higherderivatives on X as well as the boundary term, thus obtaining( S (2) AdS + S (3) AdS ) trace = 14 π Z dτ dσ X i [( ∂ τ X i ) − ( ∂ σ X i ) ] O i , (3.37) O i = ( ∂ τ z i ) − ( ∂ σ z i ) − z i ) σ . (3.38)The one-loop effective action is then given byΓ b = 14 π Z dτ dσ X i [( ∂ τ X i ) − ( ∂ σ X i ) ] hO i i . (3.39) hO i i = (cid:2) ∂ τ ∂ τ − ∂ σ ∂ σ − σ σ (cid:3) πG ( τ , σ ; τ , σ ) (cid:12)(cid:12)(cid:12) τ → τ → τ, σ → σ → σ , (3.40)(3.41)where G is given by (3.31). Substituting G into (3.40) we find hO i i = C + C log ( σ − σ ) − ( τ − τ ) ( σ + σ ) − ( τ − τ ) , (3.42)9here C = ( σ + σ ) ( σ + σ ) − σ + 5 σ σ + 2 σ )( τ − τ ) + 3( τ − τ ) σ σ [( σ + σ ) − ( τ − τ ) ] (cid:12)(cid:12)(cid:12) τ → τ → τ, σ → σ → σ ,C = ( σ − σ ) − τ − τ ) σ σ (cid:12)(cid:12)(cid:12) τ → τ → τ, σ → σ → σ . (3.43)Taking the limit we get C = 12 σ , C = 0 , (3.44)so that (for any i ) hO i i = 12 σ , (3.45)and the bosonic one-loop effective action (3.39) is thusΓ b = 18 π Z dτ dσ σ [( ∂ τ X i ) − ( ∂ σ X i ) ] . (3.46)Note that it has exactly the same form as the classical action (2.5) to order X . Also notethat (3.46) is valid for an arbitrary classical background. In particular, on the Mikhailov’ssolution (2.8) it is given by Γ b = 18 π Z dτ v i a i . (3.47)Now let us find the one-loop correction to the energy. We can do it by computing E b = − Z dσ h ∂L b ∂∂ τ X i . (3.48)For this we consider (3.24), (3.25) and restore the dependence on X in the induced metric g ( X ). That is, we replace in (3.24), (3.25) the metric g with the metric ˜ g (see eq. (2.4))˜ g ττ = − σ ( ∂ τ X ) , ˜ g τσ = 0 , ˜ g σσ = 1 σ (3.49)and set ∂ τ X = 1 after we take the derivative. Since all the dependence on ∂ τ X comes fromthe induced metric we get ∂S (2) AdS ∂∂ τ X = √ λ π Z dτ dσ p − ˜ g ∂ ˜ g µν ∂∂ τ X [ −
12 ˜ g µν ˜ g αβ ˜ g γδ h βγ h δα + 2˜ g αβ h αν h βµ ] , (3.50) ∂S (3) AdS ∂∂ τ X = − √ λ π Z dτ dσ p − ˜ g ∂ ˜ g µν ∂∂ τ X [ −
12 ˜ g µν ˜ g αβ ˜ g γδ h αβ h γδ + 2˜ g αβ h αβ h µν ] . (3.51)Using ∂ ˜ g ττ ∂∂ τ X = 2 σ , ∂ ˜ g τσ ∂∂ τ X = 0 , ∂ ˜ g σσ ∂∂ τ X = 0 (3.52)10nd substituting the explicit expression for h we obtain ∂S (2) AdS ∂∂ τ X = 14 π Z dτ dσ σ [ A + 2 A ττ ] , ∂S (3) AdS ∂∂ τ X = − π Z dτ dσ σ [ B + 2 B ττ ] , (3.53)where (the worldsheet indices are contracted with η αβ ) A = ( ∂ α X i ∂ α X j )( ∂ β y i ∂ β y j ) + ( ∂ α X i ∂ α y j )( ∂ β X j ∂ β y i ) ,B = ( ∂ α X i ∂ α y i ) ,A αβ = ( ∂ γ X i ∂ γ X j ) ∂ α y j ∂ β y i + ( ∂ γ y i ∂ γ y j ) ∂ α X j ∂ β X i ,B αβ = ( ∂ γ X i ∂ γ y i )( ∂ α X j ∂ β y j + ∂ α y j ∂ β X j ) . (3.54)Just like in the case of the effective action discussed above, in taking the vev in (3.53) only thediagonal terms in y i y i will contribute. Thus in adding together the two expressions in (3.53)we can set j = i and sum over i . We obtain( A − B ) trace = X i ( ∂ α X i ∂ α X i )( ∂ β y i ∂ β y i ) = X i [( ∂ τ X i ) − ( ∂ σ X i ) ] [( ∂ τ y i ) − ( ∂ σ y i ) ] , ( A ττ − B ττ ) trace = X i ( ∂ α X i ∂ α X i ) ∂ τ y i ∂ τ y i + X i ( ∂ α y i ∂ α y i ) ∂ τ X i ∂ τ X i = − X i [( ∂ τ X i ) − ( ∂ σ X i ) ] ∂ τ y i ∂ τ y i − X i [( ∂ τ y i ) − ( ∂ σ y i ) ] ∂ τ y i ∂ τ y i . (3.55)Note that the sum of ( A − B ) trace and ( A ττ − B ττ ) trace yields the Legendre transform of (3.35)with respect to both background and quantum fields. Explicitly, summing the above twoequations we obtain the two contributions to the energy. The first one is E b = 14 π Z dτ dσ σ X i [( ∂ τ X i ) + ( ∂ σ X i ) ] h ( ∂ τ y i ) − ( ∂ σ y i ) i . (3.56)This term has exactly the same y -dependence as in (3.35). Changing the variables y i = σz i and integrating by parts we obtain E b = 14 π Z dτ dσ X i [( ∂ τ X i ) + ( ∂ σ X i ) ] hO i i , (3.57)where the operator O i is given in (3.38). Using (3.45) we get E b = 18 π Z dτ dσ σ [( ∂ τ X i ) + ( ∂ σ X i ) ] . (3.58)However, eqs. (3.53)–(3.55) give rise to an additional unexpected correction of the form˜ E b = 14 π Z dτ dσ σ X i [( ∂ τ X i ) − ( ∂ σ X i ) ] h ∂ τ y i ∂ τ y i i , (3.59)11et us carefully examine this term on the solution (2.8), (2.9), i.e. X i = x i ( τ − σ ) + σ ˙ x i ( τ − σ ).First, on dimensional grounds, h ∂ τ y i ∂ τ y i i is independent of τ and σ . Then integrating by partsand using equations of motion we find˜ E b ∼ Z dσσ ∂ τ ( X i ∂ τ X i ) − Z dσ∂ σ (cid:0) X i ∂ σ X i σ (cid:1) , (3.60)where the proportionality coefficient depends on h ∂ τ y i ∂ τ y i i . Since the classical solution can bewritten as X i = x i − σx i ′ , x i ′ = ∂ σ x i , ∂ σ X i = − σx i ′′ , ∂ τ X i = − x ′ i + σx i ′′ . (3.61)we get Z dσ∂ σ (cid:0) X i ∂ σ X i σ (cid:1) = Z dσ∂ σ ( x i ′ x i ′′ − σ x i x i ′′ ) (3.62) Z dσσ ∂ τ ( X i ∂ τ X i ) = Z dσσ ∂ σ ( x i x i ′ ) − Z dσσ ∂ σ [( x i ′ ) + x i x i ′′ ] + Z dσ∂ σ ( x i ′ x i ′′ ) (3.63)where we used that ∂ τ acting on any function of τ − σ can be replaced with − ∂ σ . Now wecan integrate by parts the second term in (3.63). The result will cancel the first term up to aboundary contribution. Hence, we get Z dσσ ∂ τ ( X i ∂ τ X i ) = − Z dσ∂ σ (cid:2) ( x i ′ ) + x i x i ′′ σ (cid:3) + Z dσ∂ σ ( x i ′ x i ′′ ) . (3.64)Substituting (3.62) and (3.64) in (3.60) we obtain˜ E b ∼ Z dσ ∂ σ (cid:2) ( x i ′ ) σ (cid:3) = − σ ( x i ′ ) (cid:12)(cid:12)(cid:12) σ = ǫ → . (3.65)Expanding near σ = ǫ → x i ′ = − ˙ x i = − ˙x i − ǫ ¨x i + · · · = − v i − ǫa i + . . . , (3.66)so that ˜ E b ∼ ǫ ( v i ) + 2 v i a i . (3.67)These are precisely the first two terms in (2.21) which we have previously ignored in theclassical expression. Hence, the whole contribution ˜ E b can be ignored and the bosonic one-loopcorrection is given by (3.58). 12 The one-loop contribution from the fermionic sector
The quadratic fermionic action has the following form [5, 14] S f = 12 π Z dτ dσ (cid:2) − i √− GG αβ ¯ θρ α ∇ β θ + ǫ αβ ¯ θρ α Γ ∗ ρ β θ (cid:3) , (4.1)where we imposed the κ -symmetry gauge θ = θ = θ (which is ghost-free). In (4.1) θ is a32-component spinor satisfying the Majorana-Weyl condition, G αβ is the induced metric whichto one-loop order is simply the background metric g αβ ( X ). The matrices ρ α are ρ α = Γ A E AM ∂ α Y M , (4.2)where Γ A , A = 0 , . . . , ×
32 flat Dirac matrices satisfying { Γ A , Γ B } = 2 η AB , η AB = diag( − , , , , . (4.3) Y M are the coordinates in AdS , which can be replaced by their classical parts: Y = X = τ, Y = Z = σ, Y i = X i . E AM are the AdS vielbeins E AM = 1 Z δ AM = 1 σ δ AM . (4.4)The matrices ρ α are then given by ρ τ = 1 σ Γ + 1 σ Γ i ∂ τ X i , ρ σ = 1 σ Γ + 1 σ Γ i ∂ σ X i . (4.5)They satisfy { ρ α , ρ β } = 2 g αβ ( X ) . (4.6)For the string AdS background the covariant derivative ∇ α is given by ∇ α = ∂ α + 14 Ω ABM Γ A Γ B ∂ α X M , (4.7)i.e. ∇ τ = ∂ τ − σ Γ Γ − σ Γ i Γ ∂ τ X i , ∇ σ = ∂ σ − σ Γ i Γ ∂ σ X i . (4.8)Finally, the matrix Γ ∗ is given byΓ ∗ = i Γ Γ Γ Γ Γ , Γ ∗ = 1 , [Γ ∗ , Γ A ] = 0 . (4.9)We will use the basis where Γ is antisymmetric and antihermitian while Γ i , Γ are symmetricand hermitian. Then Γ ∗ is antisymmetric and hermitian.13igure 2: The Feynman graph describing the fermionic contribution to the one-loop effective action.The internal lines correspond to the fermionic loop and the external lines are the background.
Anticommuting Majorana spinors satisfy¯ θ Γ A A ...A n θ = ¯ θ Γ A A ...A n θ , n = 3 , , ¯ θ Γ A A ...A n θ = − ¯ θ Γ A A ...A n θ , n = 3 , . (4.10)In particular, it follows that ¯ θ Γ A A ...A n θ = 0 unless n = 3 ,
7. Note that the second term in (4.1)contains 7 gamma-matrices and, hence, is non-zero.According to our discussion in the previous section, it is natural to assume that the kineticterm in (4.1) combines with (3.5) and (3.23) to cancel the conformal anomaly so that the entirenon-trivial fermionic contribution should come from the second term in (4.1). Expanding thissecond term to order X gives2 σ ¯ θ [ i Γ Γ Γ ∗ − Γ Γ i Γ ∗ ∂ τ X i + Γ Γ i Γ ∗ ∂ σ X i + Γ ij Γ ∗ ∂ τ X i ∂ σ X j ] θ . (4.11)Note that it contains linear terms in X i and, hence to find order X terms we have to considerthe Feynman graph in Figure 2. In flat space due to translational invariance in all directions wecan go to the momentum space and single out the integral over the loop momentum. However,this cannot be done in AdS since there is no translational invariance in the radial direction.Therefore, we will use a different approach.Let is write the action (4.1) as S f = − iπ π Z dτ dσ p − g ( X ) ¯ θ D θ , (4.12)where D is the covariant (with respect to the induced metric g αβ ( X )) Dirac operator D = g αβ ρ α ∇ β + i p − g ( X ) ǫ αβ ρ α Γ ∗ ρ β . (4.13)Formally “squaring” this operator leads toˆ O ≡ p − g ( X ) D = p − g ( X ) g αβ ˆ ∇ α ˆ ∇ β − p − g ( X ) R (2) − p − g ( X ) , (4.14)14here ˆ ∇ contains extra connection term in addition to the 2d spinor connection. Here R (2) isthe curvature of the induced metric g αβ ( X ). The first term in (4.14) is a counterpart of (3.5)and (3.23). By our assumption the contribution from this term should cancel the similar bosoniccontributions. The second term is a total derivative and its contribution can also be ignored.We shall thus assume that, in the first two terms, we can ignore the X i -dependence so that wecan effectively replace the metric g ( X ) with the AdS metric g . The non-trivial contributionthen comes from the last term which is the square of the second term in (4.1). The operatorˆ O we will write in the formˆ O = ( η αβ ∇ α ∇ β − R − σ ) − ( p − g ( X ) − σ ) ≡ ˆ O + ˆ O int . (4.15)Here ˆ O is the free operatorˆ O = η αβ ∇ α ∇ β − R − σ = − ∂ τ + ∂ σ + 1 σ Γ Γ ∂ τ − σ . (4.16)The operator ˆ O int contains interaction terms depending on the backgroundˆ O int = p − g ( X ) − σ = − σ [( ∂ τ X i ) − ( ∂ σ X i ) ] + . . . . (4.17)Hence, to order X the operator ˆ O can be written asˆ O = ˆ O + 12 σ [( ∂ τ X i ) − ( ∂ σ X i ) ] = ˆ O (cid:16) G · σ [( ∂ τ X i ) − ( ∂ σ X i ) ] (cid:17) , (4.18)where G is the Green’s function of the operator ˆ O . The fermionic contribution to the one-loopeffective action is then given by Γ f = 14 Tr log ˆ O = 14 Tr log (cid:16) G · σ [( ∂ τ X i ) − ( ∂ σ X i ) ] (cid:17) = 18 Z dτ dσ σ [( ∂ τ X i ) − ( ∂ σ X i ) ] tr G (0) + . . . , (4.19)where G (0) is the Green’s function with coincident arguments and tr is the trace over the spinorindices.Unfortunately, it appears to be complicated to find the Green’s function of the opera-tor (4.16) directly. Though the operator √− g ˆ O is self-adjoint with respect to the measure In few simple cases of string backgrounds like parallel lines [5] or folded string [14] there is a non-trivialrotation of fermions that puts the derivative ˆ ∇ with induced connection into the form of 2d spinor covariantderivative. That need not be the case if the background is general enough (an example is circular string with 3spins in [15]). We thank S. Giombi for pointing out the relevance of this issue in the present case. We have the coefficient 1/4 because we integrate over the real spinors θ and then square the Dirac operator. dτ dσ √− g its explicit form in (4.16) is not symmetric (or hermitian) in the usual sense. Sowe will follow an indirect approach. The equation for G ( τ , σ ; τ , σ ) can be written as follows h ρ α ∇ α + i √− g ρ τ ρ σ Γ ∗ i G (1; 2) = 1 √− g δ ( τ − τ ) δ ( σ − σ ) . (4.20)Computing the square of the first-order operator in the left hand side of (4.20) and multiplyingit by √− g = σ − gives precisely (4.16). The result of (4.20) can be written in the form of aconvolution G (1; 2) = Z dτ dσ p − g ( τ , σ ) G f (1; 3) G f (3; 2) , (4.21)where G f is the fermionic Green’s function satisfying the first order equation (cid:2) ρ α ∇ α + i √− g ρ τ ρ σ Γ ∗ (cid:3) G f (1; 2) = 1 √− g δ ( τ − τ ) δ ( σ − σ ) . (4.22)Using the explicit expressions for ρ α , ∇ ,α we can write (4.22) in the form1 σ (cid:2) − Γ ∂ τ + Γ ∂ σ + iσ Γ Γ Γ ∗ (cid:3) G f (1; 2) = δ ( τ − τ ) δ ( σ − σ ) . (4.23)Let us define G f = √ σ σ G ′ f . Then G ′ f satisfies[ − Γ ∂ τ + Γ ∂ σ + iσ Γ Γ Γ ∗ ] G ′ f (1; 2) = r σ σ δ ( τ − τ ) δ ( σ − σ ) . (4.24)Due to the δ -function we can replace q σ σ with 1. Hence, we need to solve the equation DG ′ f (1; 2) = δ ( τ − τ ) δ ( σ − σ ) , D = − Γ ∂ τ + Γ ∂ σ + iσ Γ Γ Γ ∗ , (4.25)where D acts on the first argument. Let us square this operator, i.e. consider the equation D G s (1; 2) = (cid:2) − Γ ∂ τ + Γ ∂ σ + iσ Γ Γ Γ ∗ (cid:3) G s (1; 2) = δ ( τ − τ ) δ ( σ − σ ) . (4.26)If we know the solution to this equation G s then G ′ f (1; 2) = DG s (1; 2) , G f (1; 2) = √ σ σ DG s (1; 2) . (4.27)Squaring D gives explicitly (cid:2) − ∂ τ + ∂ σ − σ + 1 σ i Γ Γ ∗ (cid:3) G s (1; 2) = δ ( τ − τ ) δ ( σ − σ ) . (4.28)The matrix i Γ Γ ∗ is symmetric and squares to 1, so we can introduce the orthogonal projectors P ± = 12 (1 ± i Γ Γ ∗ ) , P ± = P ± , P + + P − = 1 , P + P − = P − P + = 0 . (4.29)16ow let us look for a solution in the form G s = P + G + + P − G − . (4.30)It then follows that G + and G − should satisfy the following equations( − ∂ τ + ∂ σ ) G + (1; 2) = δ ( τ − τ ) δ ( σ − σ ) , ( − ∂ τ + ∂ σ − σ ) G − (1; 2) = δ ( τ − τ ) δ ( σ − σ ) . (4.31)Thus G + is the Green’s function of a scalar field with m = 0 which is given in (3.12). Similarly, G − is the Green’s function of a scalar field with m = 2 which is given in (3.31). Hence, wefinally obtain G f (1; 2) = 12 π √ σ σ D [ P + G (1; 2) + P − G (1; 2)] , (4.32)where D always acts on the first argument. Substituting G f into (4.21) gives G (1 ,
2) = 14 π Z dτ dσ p − g (3) √ σ σ √ σ σ × D [ P + G (1; 3) + P − G (1; 3)] D [ P + G (3; 2) + P − G (3; 2)] . (4.33)We need to evaluate tr G (1; 2) in the limit τ → τ → τ, σ → σ → σ . The integral is of thesame difficulty as the loop integral in Figure 2 and we were not able to evaluate it explicitly.We suggest the following indirect way to extract the finite contribution. First, we will write D [ P + G (1; 3) + P − G (1; 3)] = − D [ P + G (1; 3) + P − G (1; 3)] . (4.34)where sub-indices on D indicate the arguments on which it acts. Then we integrate by part toform D ignoring all the additional terms. After integrating by parts we get D [ P + G (3; 2) + P − G (3; 2)] = 2 πδ ( τ − τ ) δ ( σ − σ ) . (4.35)Then tr G (0) = 12 π tr( P + G + P − G ) = 12 π · ·
12 ( G + G ) . (4.36)From the expressions for G , G in (3.13), (3.32) we find G + G = − ℓ + 1) log ℓℓ + 1 . (4.37) Since θ is a Weyl spinor we should also introduce the Weyl projector. However, for simplicity, we will ignoreit and use the fact that in the space of Weyl spinors the trace of the unit matrix gives 16 rather than 32. Note that despite being a fermionic Green’s function, G f is not explicitly antisymmetric. This is due to theboundary conditions in AdS , in particular, to the absence of the translational invariance along the σ -direction.Nevertheless, one can expect that in Feynman graphs one can use it as if it were antisymmetric.
17e see that there is a natural finite contribution -1 in (4.37). Thus the finite part of tr G (0) isgiven by tr G (0) = − π . (4.38)Substituting this into (4.19) givesΓ f = − π Z dτ dσ σ [( ∂ τ X i ) − ( ∂ σ X i ) ] . (4.39)This is the final expression for the one-loop correction to the order X effective action comingfrom the fermionic sector.To find the corresponding contribution to the energy we will start with the general expressionfor the one-loop effective actionΓ f = 14 Tr log ˆ O = 14 Tr log h − G · ( p − g ( X ) − σ ) i , (4.40)and vary it with respect to ∂ τ X . Taking the functional derivative gives − ∂ Γ f ∂∂ τ X = 14 Tr h − G · ( p − g ( X ) − σ ) · G · p − g ( X ) g ττ ( X ) ∂g ττ ( X ) ∂∂ τ X i . (4.41)Expanding (4.41) to order X produces two contributions. One comes from expanding p − g ( X )in the denominator. It is proportional to ( ∂ τ X i ) − ( ∂ σ X i ) . This kind of term was discussedat the end of the previous section in eqs. (3.59)–(3.67) and was shown to be irrelevant. Therelevant contribution comes from expanding the numerator p − g ( X ) g ττ ( X ) which gives − − [ ∂ τ X i ) + ( ∂ σ X i ) ]. Using ∂g ττ ( X ) ∂∂ τ X = − σ (4.42)we find (ignoring the X -independent term) − ∂ Γ f ∂∂ τ X = 14 Tr (cid:0) G · σ [ ∂ τ X i ) + ( ∂ σ X i ) ] (cid:1) . (4.43)The one-loop fermionic contribution to the energy is then given by E f = 18 Z dσ σ [ ∂ τ X i ) + ( ∂ σ X i ) ] tr G (0) = − π Z dσ σ [ ∂ τ X i ) + ( ∂ σ X i ) ] , (4.44)where we used eq. (4.38).Let us now combine together the bosonic and fermionic one-loop contributionsΓ − loop = Γ b + Γ f = − π Z dτ dσ σ [ ∂ τ X i ) − ( ∂ σ X i ) ] , (4.45) E − loop = E b + E f = − π Z dσ σ [ ∂ τ X i ) + ( ∂ σ X i ) ] . (4.46) Note that one-half of the geodesic distance ℓ is a usual regulator in the point-splitting regularization in fieldtheory in curved space-time, for a review see, e.g., [16]. − loop = − π Z dτ v i a i , (4.47) E − loop = − π Z dτ ( a i ) = 2 πB ( λ ) Z dτ ( a i ) , (4.48)where the one-loop contribution to B ( λ ) is given by (cf. (2.23)) B ( λ ) = − π . (4.49)This is the same as the second coefficient in (1.1). We thus provided a direct string perturbativecheck of the exact expression (1.1) proposed in [1]. Let us now perform an analytic continuation along τ and X and consider the Euclideanworldsheet and the Euclidean AdS space. We will also work in the static gauge (2.2). A wavyWilson line is a small deviation from a straight line X = τ, Z = σ in the transverse directions X i in Euclidean space [8, 9, 10]. To quadratic order in the transverse fields the Euclidean stringaction is S e = √ λ π Z dτ dσσ (cid:2) ( ∂ τ X i ) + ( ∂ σ X i ) (cid:3) , (5.1)and the equations of motion are ∂ τ X i + ∂ σ X i − σ ∂ σ X i = 0 . (5.2)A wavy line is a solution to these equations with the boundary condition X i ( τ, σ ) (cid:12)(cid:12)(cid:12) σ =0 = x i ( τ ) , (5.3)where x i ( τ ) is an arbitrary curve on the boundary. The precise form of the solution is givenby [10] X i ( τ, σ ) = Z dτ ′ π x i ( τ ′ ) 2 σ (( τ − τ ′ ) + σ ) . (5.4)Substituting it into the action gives S e = − √ λ π Z dτ dτ ′ [( ˙x( τ ) − ˙x( τ ′ )] ( τ − τ ′ ) . (5.5) It is straightforward to show that (5.5) is obtained starting with (5.1) (after dropping a total derivative).The minus sign in front of the Euclidean action in [10] is a misprint. h W wavy i = e − S e = 1 + √ λ π Z dτ dτ ′ [( ˙x( τ ) − ˙x( τ ′ )] ( τ − τ ′ ) + . . . = 1 + 12 B ( λ ) Z dτ dτ ′ [( ˙x( τ ) − ˙x( τ ′ )] ( τ − τ ′ ) + . . . , (5.6)where we ignored the higher order terms in x i and B ( λ ) = √ λ π . (5.7)To the one-loop order the expectation value of h W wavy i is h W wavy i = e − S e − Γ e, − loop , (5.8)where Γ e, − loop is the Euclidean one-loop effective action computed on the solution (5.4). Inthe previous sections we computed the one-loop correction to the Minkowskian effective action(4.46) for an arbitrary classical background X i . Its Euclidean analog can be found by a simpleanalytic continuation: Γ e, − loop = − π Z dτ dσ σ (cid:2) ∂ τ X i ) + ( ∂ σ X i ) (cid:3) . (5.9)Evaluating it on the solution (5.4) givesΓ e, − loop = 316 π Z dτ dτ ′ [( ˙x( τ ) − ˙x( τ ′ )] ( τ − τ ′ ) . (5.10)Hence, to quadratic order in x i we get h W wavy i = 1 + 12 B ( λ ) Z dτ dτ ′ [( ˙x( τ ) − ˙x( τ ′ )] ( τ − τ ′ ) + . . . , (5.11)where the one-loop correction to B ( λ ) is given by B ( λ ) = B + B + ... , B = − π , (5.12)i.e. is the same as in (4.49). Thus we checked that the one-loop correction to the expectationvalue of the wavy Wilson line is controlled by the same function B ( λ ) as the energy radiatedby a moving particle in agreement with the claim of [1]. Acknowledgements
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