QCD and a Holographic Model of Hadrons

Transcription

1 QCD and a Holographic Model of Hadrons M. Stephanov U. of Illinois at Chicago AdS/QCD p.1/18

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. Bottom-up approach. A holographic model: J. Erlich, E. Katz, D. Son and M.S., hep-ph/ (L. Da Rold, A. Pomarol, hep-ph/ ) ABC of AdS/CFT (holography) This talk: A simple model: chiral symmetry breaking quark-hadron duality, sum rules OPE etc. AdS/QCD p.2/18

3 AdS/CFT correspondence: formulation Begin with S 4 [G, q] = 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,...): Z 4 [φ 0 (x)] = D[G, q] exp { } is 4 + i φ 0 O. x 4 x 1, x 2, x 3, x 4 5d bulk metric: ds 2 = z 2 ( dz 2 + η µν dx µ dx ν). η µν = diag(+1, 1, 1, 1). z φ(z, x) φ 0 (x) (Note: x m λx m ). Polchinski-Strassler slice Z 5 [φ 0 (x)] = z = z m IR D[φ]e is 5[φ] φ(x,ɛ)=φ 0 (x) z = ɛ UV Z 4 = Z 5 (Generating functional) [4d sources φ 0 (x)] = (Effective action) [fields φ 0 (x)]. AdS/QCD p.3/18

4 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., Z 4 [V ] = D[G, q] exp { } is 4 + i V 0 J. x 4 We shall look at 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, ɛ) = V 0 (x). Then take 2 variational derivatives wrt V 0 (x) and V 0 (y) to find J(x)J(y). AdS/QCD p.4/18

5 Example: current (contd) V 5 = 0 gauge; linearize; Fourier x µ q µ : 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=ɛ. Let V (q, z) be a solution with V (q, z) = 1, then we need (V a µ ) = V a µ0(q)v (q, z). Π(Q 2 ) = 1 2g z V (q, z). Q 2 z z=ɛ Thus V (q, z) = (Qz)K 1 (Qz) = 1 + Q2 z 2 Π(Q 2 ) = 1 2g ln(qz) + O(z 2 ). ln Q 2 + contact terms AdS/QCD p.5/18

6 5d coupling and N c In QCD In AdS 5 : Π(Q) = N c 24π 2 ln Q Π(Q 2 ) = 1 2g 2 5 ln Q Thus g 2 5 = 12π2 N c Large N c small coupling AdS/QCD p.6/18

7 Example: current (summary notes) δs 5,eff δ(v 0 ) = 0 corresponds to µ J µ = 0. log Q 2 scale invariance of the 5d theory (metric). Dictionary : 4d 5d W 4 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 AdS/QCD p.7/18

8 Dimension of operator and 5d mass L 5 = 1 ( g g mn m φ n φ m 2 5 φ 2) 2 z 0 (qz 1) solution of LE equations for L 5 : φ z φ with ( φ 4) φ m 2 5 = 0. m 2 5 = 0 : φ const = φ 0 OK; m : φ z φ const = φ 0. [φ] = 0 [φ 0 ] = + φ ([x] = 1) Thus [O] = 4 φ O and m 2 5 = ( φ 4) φ = 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. AdS/QCD p.8/18

9 Spontaneous symmetry breaking S 5 = 1 d 5 x g g mn m φ n φ with b.c. at z = 0: φ z φ = φ 0. The extremum: φ sol = φ 0 z φ + A z O ( φ + O = 4). Vary the source: φ 0 φ 0 + δφ 0 : δs 5 = d 4 x z 3 δφ z φ +... = ( O φ ) z=0 d 4 x δφ 0 A Compare to W 4 : δw 4 = d 4 x δφ 0 O Therefore A 0 as φ 0 0: A = 1 2 O 4 O spontaneous symmetry breaking AdS/QCD p.9/18

10 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 { DX X 2 1 } (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, D z X = 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. AdS/QCD p.10/18

11 Hadrons and QCD sum rule z ( 1 z zv ) + q2 z V = 0. Normalizable modes: ψ ρ (ɛ) = 0, z ψ ρ (z m ) = 0, (dz/z) ψρ (z) 2 = 1. Π V ( q 2 ) = 1 2g z V (q, z) Q 2 z = 1 g 2 5 ρ [ψ ρ (ɛ)/ɛ] 2 (q 2 m 2 ρ)m 2 ρ. F ρ = 1 g 5 ψ ρ(ɛ) ɛ. QCD sum rule: Π V (Q 2 ) = 1 2g 2 5 ln Q AdS/QCD p.11/18

12 Chiral symmetry breaking and GOR relation A = (A L A R )/2, X = X 0 exp(i2π a t a ), 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; δa : z ( z 1 z ϕ ) + z 3 g 2 5v 2 (π ϕ) = 0; δa z : q 2 z ϕ + z 2 g 2 5v 2 z π = 0. Π A ( q 2 ) q 0 f2 π. Analogous to V : A(q, ɛ) = 1. q2 fπ 2 = 1 z A(0, z) g5 2. z z=ɛ u A(0, u)/u u 3 v 2 π(z) For m q 0: ϕ(z) = A(0, z) 1, π(z) = z, u π(z) = m 2 π z 0 du u3 v(u) 2 1 g 2 5 u ua(0, u) = m 2 πf 2 π 1 2m q σ for z m q /σ. Thus m 2 πf 2 π = 2m q σ + O(m 2 q). (Still holds for σ 3, or deformed AdS.) AdS/QCD p.12/18

13 Arbitrary σ (v(z) = m q z m + σ z σ ): GOR (contd) 1 = m 2 πf 2 π 0 dz z 3 (m q z m + σ z σ) 2 = m2 πf 2 π 1 ( σ m )m q σ = m2 πf 2 π 2m q qq ( σ m )σ = 2 qq AdS/QCD p.13/18

14 Meson wavefunctions and couplings 3 π φ 2 1 ρ ρ a Couplings: -1 g ρππ = g 5 dz v2 (z) z 3 z/z m (π ϕ)(z) π(z) ψ ρ (z), AdS/QCD p.14/18

15 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 1720± MeV N c g 5 = 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. AdS/QCD p.15/18

16 Holographic model vs 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 Σ 3 Σ 1 SU(2) R Σ 2 Σ K 1 Σ K Σ K+1 Folded: 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 AdS/QCD p.16/18

17 OPE and higher order terms Π V (Q 2 ) = 1 2g 2 5 ln Q 2 ; Π A (Q 2 ) = Π V (Q 2 ) + # m ) qσ (m Q + O 2q, σ2 ; 4 Q 6 (In open moose: Π A Π V exp( #Q).) In QCD: Π V (Q 2 ) =... m qσ Q 4 ; L 5 = g Tr Π A(Q 2 ) =... + m qσ Q 4 ; [ ] γ XFXF (X F L XF R ) + γ X 2 F 2(XX FL 2 + X XFR) 2 Π A Π V = 4 3 m qσ (4γ XFXF + 1) 1 Q 4 = 2m q qq 1 Q 4 γ XFXF = 1 8 Similarly: Π V =... + # α sg 2 Q 4. Need source h 0 for operator α s G 2 massless scalar field h in 5d. Classical solution: h = h 0 + A h z 4. A h = (1/4) α s G 2. Coupling L 5 = γh(f 2 L + F 2 R) gives Π V =... #γa h /Q 4. AdS/QCD p.17/18

18 Outlook h(z, x) glueball spectrum (Polchinski-Strassler, Boschi-Filho-Braga) strange mesons chiral anomaly (WZW) Baryons (Teramond-Brodsky) finite density? running of α s (log violations of scaling) AdS/QCD p.18/18