MIT OpenCourseWarehttp://ocw.mit.edu 8.821 String Theory Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� 8.821 F2008 Lecture 14: Wave equation in AdS, Green’s function Lecturer: McGreevy November 11, 2008 Topics for this lecture • Find φ[φ0](x) by Green functions in x-space (efficient) • Compute �OO�, counter terms • Redo in p-space (general) References • Witten, hep-th/9802150 • GKP, hep-th/9802109 Solving Wave Equation I (Witten’s method) Let’s study the wave equation in AdS in some detail. This first method uses a trick by Witten which is efficient but slightly obscure. If we know “bulk-to-boundary” Green’s function K regular in the bulk, such that (−� + m 2)Kp(z, x) = 0 (1) Kp(z, x) → �Δ− δ�D(x − p), z → � (2) where p is some point on the boundary, then the field in the bulk dD x� φRen Δ− φRenφ[φ0](z, x) = 0 (x�)Kx� (z, x) z 0 (x)→ solves (1). 1� � � � Euclidean AdS Recall the metric on AdS with curvature scale L in the upper half plane coordinates: ds2 = L2 dz2 + dx2 2zNow here comes some fancy tricks, thanks to Ed: Trick (1): Pick p = “point at ∞”. This implies that the Green’s function K∞(z, x) is x-independent. The wave equation at k = 0: 0 = −z D+1∂zz−D+1∂t + m 2K∞(z) can easily be solved. The solution is power law (recall that in the general-k wave equation, it was the terms proportional to k2 that ruined the power-law behavior away from the boundary) K∞(z) = c+z Δ+ + c−z Δ− We can eliminate one of the constants: c = 0, whose justification will come with the result. − Trick (2): Use AdS isometries to map p = ∞ to finite x. Let xA = (xµ, z), take xA (x�)A = BxA/(x xB). The inversion of this mapping is: → zI : xµ → 2x+µ x2 z z z2+x2→ Claim: I A) is an isometry of AdS (also Minkowski version, see pset 4) B) is not connected to 1 in SO(D,2) C) maps p = ∞ to x = 0, i.e., I : K∞(z, x) → K∞(z�, x�) = K0(z, x) = c+zΔ+ /(z2 + x2)Δ+ . Some notes: (i) That this solves the wave equation (1) as neccessary can be checked directly. (ii) The Green’s function is Δ+zKx� (z, x) = c+ (z2 + (x − x�)2)Δ+ ≡ K(z, x; x�) (iii) The limit of the Green’s function as z 0 , i.e. the boundary is → czΔ+ 0, if x = x�K(z, x; x�) → cz−Δ+ → if x �= x�→ ∞, 2� � � � � �� � ��� � � ��� ��� � � � � � ��� � � �� (recall that Δ+ > 0 for any D, m). More specifically, the Green’s function approaches a delta function: K(z, x; x�) → const · �Δ− δD(x − x�). Clearly it has support only near x = x�, but to check this claim we need to show that it has finite measure: c�Δ+−Δ−dDx �−Δ− K0(�, x) = dD x (�2 + xc�D�2Δ+−D 2)Δ+ 1 dD x¯= (1 + ¯x2)Δ+�2Δ+ D π 2 Γ(Δ+ − D )2= c .Γ(Δ+) We will choose the constant c to set this last expression equal to one. Hence, φ[φ0](z, x) = dD x� Kx� (z, x)φRen 0 (x�) Δ+ dD x�cxφRen 0 (x�) ; = (z2 + (x − x�)2)Δ+ this solves (1) and approaches �Δ− φRen 0 (x) as z �.→ The action is related to expectation values of operators on the boundary: S � [φ ]0φ R φ0O� √γ φ n ∂φ · = − ln�e− η = − 2 ∂AdS = − ηdD x√g g zzφ(z, x)∂zφ(z, x)2 z=� = − η 2 dD x1dD x2 φRen(x1)φRen(x2)F�(x1, x2)0 0 where the “flux factor” is . dD xK(z, x; x1)z∂zK(z, x; x2)F�(x1, x2) ≡ D z=�z The boundary behavior of K is: KΔ+ (z, x; x�) c = �Δ− δD(x − x�) + O(�2) + �Δ+ + O(�2)(x − x�)2Δ−z=� D 2 Γ(Δ+ − D 2the first terms sets: c−1 )/Γ(Δ+), the second term is subleading in z.= π K(z, x; x�) 1= Δ+�Δ− δ(x − x�) + Δ+cz Δ+ + . . . z∂z(x − x�)2Δ+z=� Ok, now for the 2-point correlation function on the boundary: G2(x1, x2) ≡ �O(x1)O(x2)�c δ δ = δφ0(x1) δφ0(x2) −S 3 φ[φ0] = ηF�(x1, x2).� � � � � � � � � � � � � � � We must be careful when evaluating the cases x1 = x2 and x1 = x2, which we do in turn. �Firstly, if x1 =� x2: η � � Δ+czΔ+ G2(x1 =� x2) = 2 dD xz−D z Δ− δD(x − x1) + O(z 2) (ignore by x1 =� x2) + (x1 − x2)2Δ+ + O(z 2) z=� η 1 =2 cΔ+�−D+Δ−+Δ+ (x1 − x2)2Δ+ + O(�2) ηcΔ+ = .2(x1 − x2)2Δ+ Good. This is the correct form for a two point function of a conformal primary of dimension Δ+ in a CFT; this is a check on the prescription. Secondly, if x1 = x2: G2(x1, x2) = η Δ−�2Δ−−DδD(x1 − x2) + cΔ+ + Δ+c 2�2Δ+−D dD x 1 (x1 − x2)2Δ+ (x − x1)2Δ+ (x − x2)2Δ+ As � 0, the first term is divergent, the second term is finite, and the third term vanishes. The →first term is called a “divergent contact term”. It is scheme-dependent and useless. Remedy: Holographic Renormalization. Add to Sgeometry the contact term −�2Δ−−D (φRenΔS = Sc.t. = η 2 dD x −Δ0 (x))2 bdy = −Δ− η 2 ∂AdS,z=� √γ φ2(z, x). Note that this doesn’t affect the equations of motion. Nor does it affect G2(x1 =� x2). Solving Wave Equation II (k-space) Since the previous approach isn’t always available (for example if there is a black hole in the spacetime), let’s redo the calculation in k-space. Return to wave equation 0 = z D+1∂z z−D+1∂z − m 2L2 − z 2k2fk(z) with k2 = −ω2 + k2 > 0. The solution is D D fk(z) = AK z 2 Kν (kz) + AI z 2 Iν (kz), with ν = (D/2)2 + m2L2 = Δ+ − D/2. Assume k ∈ R (real time issues later). As z Kν ∼ e−kz and Iν ∼ ekz . The latter is not okay, so AI = 0. → ∞: 4� At boundary: � � nν (b0 + b1n2 + b2n4 + . . .), ν /∈ R Kν (n) ∼ n−ν a0 + a1n 2 + a2n 4 + . . . + nν ln n (b0 + b1n2 + b2n4 + . . .), ν ∈ R Hence D fk(z) = AK zD/2Kν (kz) ∼ z 2 ±ν = z Δ± , as z → 0.
View Full Document