Holographic QCD. M. Stephanov. U. of Illinois at Chicago. Holographic QCD p. 1/2

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1 Holographic QCD p. 1/2 Holographic QCD M. Stephanov U. of Illinois at Chicago

2 Holographic QCD p. 2/2 Motivation and plan Large N c : planar diagrams dominate resonances are infinitely narrow Effective theory in terms of resonances is weakly coupled? What is this effective theory? String theory explicit examples suggest it is a 5d effective theory. QCD/phenomenology hint: ρ, ρ, ρ,... resonant modes in a cavity (KK)? Bottom-up approach (AdS from QCD vs QCD from AdS). A holographic model: Erlich, Katz, Son and MS; Da Rold, Pomarol ABC of AdS/CFT (holography) This talk: A simple model generic features: chiral symmetry breaking quark-hadron duality, sum rules VMD, etc. Excited meson spectrum: m 2 n + S

3 Holographic QCD p. 3/2 Holographic correspondence: formulation Begin with S 4 [G, q] = R d 4 x L[G, q]. Generating functional for correlators of an operator O (examples of O: G a µνg µνa, qq, qγ µ t a q,...): 4 [φ 0 (x)] = D[G, q] exp j ff is 4 + i φ 0 O. x 4 5d bulk metric: ds 2 = z ` dz η µν dx µ dx ν. η µν = diag(+1, 1, 1, 1). (Note: x m λx m ). z = z m φ(x, z) z x 1, x 2, x 3, x 4 φ 0 (x) z = ǫ 0 AdS slice IR bulk UV boundary 5 [φ 0 (x)] = φ(x,ǫ)=φ 0 (x) D[φ]e is 5[φ] 4 = 5 (Generating functnl) [4d sources φ 0 (x)] = (Effective action) [fields φ 0 (x)].

4 Holographic QCD p. 4/2 Example: conserved current Let O be a current: J µa = qγ µ t a q. φ 0 : source for J µa is a vector potential V µa 0. I.e., We shall look at 4 [V ] = D[G, q] exp j ff is 4 + i V 0 J. x 4 d 4 xe iqx J µa (x)j νb (y) = δ ab (q µ q ν q 2 η µν )Π( q 2 ). In QCD, scale invariance in the UV means Π(Q 2 ) ln(q 2 ). 5d action for V a m? Let us take S 5 = 1 4g 2 5 d 5 x g V a mnv amn. At tree level, we need to minimize S 5 wrt V with the b.c. V (x, ǫ) = V 0 (x). Then take 2 variational derivatives wrt V 0 (x) and V 0 (y) to find J(x)J(y).

5 Holographic QCD p. 5/2 Example: current (contd) V 5 = 0 gauge; linearize; Fourier x µ q µ ; q µ V µ = 0: z 1 z zv «+ q2 z V = 0. z V (q, z) V 0 (q) 0 Action to quadratic order in V, on eqs. of motion (int. by parts): S 5 = 1 2g 2 5 d 4 x 1 z V a µ z V aµ z=ǫ. Let V (q, z) be a solution with V (q, ǫ) = 1. We need V (q, z) = V 0 (q)v (q, z). Π(Q 2 ) = 1 g z V (q, z). (Q Q 2 z z=ǫ 2 q 2 ) Thus V (q, z) = (Qz)K 1 (Qz) = 1 + Q2 z 2 ln(qz) + O(z 2 ). 2 Π(Q 2 ) = 1 ln Q + contact terms g5 2

6 Holographic QCD p. 6/2 5d coupling and N c In QCD Π(Q 2 ) = N c 24π 2 ln Q In AdS 5 : Π(Q 2 ) = 1 2g 2 5 ln Q Thus g 2 5 = 12π2 N c Large N c small coupling

7 Holographic QCD p. 7/2 Example: current (endnotes) The role of 5d gauge inv: z 1 z zv «= 0 δs 5,eff δ(v 0 ) = 0 corresponds to µ J µ = 0. log Q 2 in UV scale inv. of the 5d theory near z = 0. [x m ] = 1; [g mn ] = [z 2 ] = 2 : S 5 = 1 4g 2 5 d 5 x g {z } 0 V a mnv amn {z } 0 [g 5 ] = 0

8 Holographic QCD p. 8/2 Dictionary 4d 5d generating func. W 4 eff. action S 5,eff operator O(x) (φ 0 source) field φ(x, z) (φ 0 boundary value) scale invariance (log Q) scale invariance µ J µ = 0 gauge invariance large N c small g 5 large Q small z dimension of O mass of φ

9 Holographic QCD p. 9/2 Dimension of 4d operator and 5d mass L 5 = 1 g `gmn m φ n φ m 2 5 φ 2 2 z 0 (at fixed q, i.e., qz 1) extremum satisfies z (z 3 z φ) z 5 m 2 5φ = 0 φ z φ with ( φ 4) φ m 2 5 = 0. φ z φ const = φ 0 the source for O [φ] = 0 [φ 0 ] = + φ ([x] = 1) Thus [O] = 4 φ O and m 2 5 = O ( O 4) For example, T µ µ : m 2 5 = 0; ψψ : m 2 5 = 3; J µ : m 2 5 = ( 3)( 1) = 0 protected.

10 Spontaneous symmetry breaking S 5 = 1 2 d 5 x g g mn m φ n φ +... with b.c. at z = 0: φ z φ φ 0. The extremum is a linear combination: Vary the source: φ 0 φ 0 + δφ 0 : φ sol = φ 0 z φ + A z O ( φ + O = 4). (Freedman et al, Klebanov-Witten) δs 5 = d 4 x z 3 δφ z φ +... = ( O φ ) d z=0 4 x δφ 0 A Compare to W 4 : δw 4 = d 4 x δφ 0 O Therefore A = 1 2 O 4 O A 0 as φ 0 0: spontaneous symmetry breaking Holographic QCD p. 10/2

11 Holographic QCD p. 11/2 The model 4D: O(x) 5D: φ(x, z) p O (m 5 ) 2 q L γ µ t a q L A a Lµ q R γ µ t a q R A a Rµ q α R qβ L (2/z)X αβ S = zm 0 d 5 x g Tr n DX X 2 1 o (F 2 4g5 2 L + FR) 2 Symmetries: X LXR, F L LF L L, F R RF R R. Boundary conditions at z = z m : F zµ = 0. Chiral symmetry breaking: X 0 (z) = 1 2 Mz Σz3. Matching to QCD: Σ αβ = q α q β. We take M = m q 1 and Σ = σ1. Four free parameters: m q, σ, z m and g 5. Compared to three in QCD: m q, Λ QCD and N c.

12 Holographic QCD p. 12/2 Hadrons and QCD sum rule Now consider q 2 > 0: z 1 z zv «+ q2 z V = 0. Normalizable modes: ψ ρ (ǫ) = 0, z ψ ρ (z m ) = 0, Exist only for discrete values of q 2 = m 2 ρ > 0, where the V (q, ǫ) = 1 solution is thus not unique. Expanding V (q, z) = P ρ a ρ(q)ψ ρ (z), R (dz/z) ψρ (z) 2 = 1. Π V ( q 2 ) = 1 g z V (q, z) = 1 Q 2 z z=ǫ g5 2 X ρ [ψ ρ (ǫ)/ǫ] 2 (q 2 m 2 ρ)m 2 ρ. Decay constants: F ρ = 1 g 5 ψ ρ(ǫ) ǫ. QCD sum rule is automatic: the same function at q 2 Q 2 < 0 is Π V (Q 2 ) = 1 2g 2 5 ln Q

13 Chiral symmetry breaking and pion mass m 2 πf 2 π = 2m q σ + O(m 2 q) with A = (A L A R )/2, A µ = A µ + µ ϕ, v(z) = m q z + σz 3 δa : z `z 1 z A + z 1 q 2 A z 3 g 2 5v 2 A = 0; Π A ( q 2 ) q 0 with X = X 0 exp(i2π a t a ) f2 π q 2. As for V : A(q, ǫ) = 1. f2 π = 1 g 2 5 z A(0, z) z z=ǫ. δa : z `z 1 z ϕ + z 3 g 2 5v 2 (π ϕ) = 0; A(0, z) π(z) δa z : q 2 z ϕ + z 2 g 2 5v 2 z π = z 3 v 2 For q 0: ϕ(z) = A(0, z) 1, π(z) = 1. φ(0) = 0, but π(0) = 0 requires q 0 and UV m q /σ z IR z π(z ) = q 2 0 v(z) 1 2 g5 2z za(0, z) = q 2 fπ 2 dz z3 (Continues to hold for σ 3, or deformed AdS.) 1 2m q σ = 1 for z p m q /σ. Holographic QCD p. 13/2

14 Holographic QCD p. 14/2 Meson wavefunctions and VMD z/z m π φ ρ ρ a 1 VMD F ρ g ρππ m 2 ρ ( 1) n n 2.5 n Couplings: g ρππ = g 5 «φ 2 dz ψ ρ (z) g5 2 z + v2 (π φ) 2. z 3 {z } pion 2 no nodes π γ q π γ F ρ ρ g ρππ etc. Isospin sum rule for π formfactor: G(q 2 = 0) = X ρ F ρ g ρππ m 2 ρ = 1

15 Holographic QCD p. 15/2 Comparison with experiment Observable Measured Model Units m π 139.6± MeV m ρ 775.8± MeV m a1 1230± MeV f π 92.4± MeV F 1/2 ρ 345±8 329 MeV Fa 1/ ± MeV g ρππ 6.03± m ρ 1465± MeV * fitted N c g 5 = p 12π 2 /N c = 2π. m ρ = 2.405/z m z m = (323 MeV) 1. f π and m π σ = (327 MeV) 3 and m q = 2.29 MeV.

16 Holographic QCD p. 16/2 Linear confinement: m 2 n n Problem for high n: m 2 n n 2. High n mesons are large IR regime. Thus high n large z. I = d 5 x g e Φ L, A. Karch et al. Instead of hard cutoff at z m Φ z 2 at large z Mode equation z e B z v n + m 2 ne B v n = 0, 0 z m B Φ(z) A(z), with e A(z) = z 1 from ds 2 = e 2A (dx dx dz 2 ). Substitute v n = e B/2 ψ n ψ n + V (z)ψ n = m 2 nψ n, V (z) = 1 4 (B ) B. With Φ = z 2 : V = z z 2 2d harmonic oscillator (radial eqn., m = 1). m 2 n = 4(n + 1)

17 Holographic QCD p. 17/2 Higher spins Gauge (δφ M1...M S = (M1 ξ M2...M S )) + general coord. invariance: I = 1 2 d 5 x n o g e Φ N φ M1...M S N φ M 1...M S + M 2 (z)φ M1...M S φ M 1...M S +... Axial gauge φ z... = 0: δφ z... = z ξ... + (. ξ...)z = ξ... 2(S 1)A ξ... = 0, and there is residual gauge inv.: ξ(x, z)... = e 2(S 1)A(z) ξ... (x) under which a rescaled field φ... = e 2(S 1)A φ... just shifts: δ φ... = (. ξ...) This residual shift invariance requires I = 1 n o d 5 x e 5A e Φ e 4(S 1)A e 2A(1+S) N φµ1...µ 2 S N φµ1...µ S. which means B = Φ (2S 1)A and V = z 2 + S2 1/4 z 2 + 2(S 1) and m 2 = 4(n + S).

18 Holographic QCD p. 18/2 Higher spins (comments) m 2 = 4(n + S). (Classical) Nambu-Gotto string also predicts: dm2 dn = dm2 ds. Note: B = Φ (2S 1)A no z 2 should be in A (metric), or else m n ns Since φ... const at boundary is a solution [O... ] = 4 [ φ... ] Since [ φ... ] = [φ... ] 2(S 1) = 2 S we find [O... ] = 2 + S i.e., twist = 2.

19 Holographic QCD p. 19/2 Holographic model and open moose K = 0 : π K = 1 : π,ρ K = 2 : π,ρ,a 1 (ga µ ) 1 (ga µ ) 2 (ga µ ) K 1 (ga µ ) K SU(2) L SU(2) R Σ 2 Σ 3 Σ K 1 Σ K+1 Σ 1 Similar to Sakai-Sugimoto. Problem: Π LR = Π V V Π AA exp( Q) too small. Solution: fold Σ K A K/ A K 1 A K Σ K Σ K+1 X K/2... X 2 X 1 SU(2) L Σ 2 A K/2... A 2 A 1 Σ 1 SU(2) R

20 Holographic QCD p. 20/2 Instanton as a baryon In the chiral Lagrangian framework baryon is a Skyrmion. Also known fact vector mesons stabilize Skyrmion no need for Skyrme term (Adkins, Nappi) Static solution in 5d is a 4d instanton. Pion fields in the instanton have Skyrmion topology. A µ = τ a η a µνx ν x 2 (x 2 + ρ 2 ). Here A µ is a four-vector with coordinates µ running through 1, 2, 3, and 5 (i.e., x, y, z and u) and x 2 = x 2 + u 2. Note that the metric signature for these 4 coordinates is Euclidean (,,, ). A 0 = 0. Pion field π A 5 : τ x A 5 = x 2 (x 2 + ρ 2 ). at fixed u it is a hedgehog. Can also show that topological charge of the instanton is equal to the baryon charge of the pion-field Skyrmion.

21 Holographic QCD p. 21/2 Outlook Chiral symmetry and high resonances (restoration?) (Shifman,Vainshtein) Baryons Dirac fields (Teramond-Brodsky, Hong-Inami-Yee) Skyrmions/instantons (Hong-Rho-Yee-Yi, Hata-Sugimoto-Yamata) finite density (Domokos-Harvey) chiral anomaly (WW) (Hill-achos, Sakai-Sugimoto) strange mesons (Shock-Wu,Schvellingeret al), U(1) A and η (Katz-Schwartz) (scalar) glueball spectrum (Polchinski-Strassler, Boschi-Filho-Braga) higher order chiral Lagrangian (DaRold-Pomarol, Hirn-Sanz) running of α s log corrections to warp factor in UV (Kiritsis talk) power corrections to OPE higher order terms in 5d Structure functions thermodynamics, transport... (Herzog, Kajantie-Tahkokallio-Yee, Nakano-Teraguchi-Wen)