Isospin chemical potential in holographic QCD

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1 Isospin chemical potential in holographic QCD Marija Zamaklar University of Durham based on work with Ofer Aharony (Weizmann) Cobi Sonnenschein (Tel Aviv) Kasper Peeters (Utrecht) and in progress Galileo Galilei Institute, May 6th 2008

2 Introduction AdS/CFT integrability high precision tests purest N = 4 a perfect generator of a huge # of integrable struct. non-ads/non-cft (direct) applications to realistic gauge theories zero temperature and chemical potentials (T = 0, µ = 0) glueball spectra masses of hadrons (mesons) hadron form factors Csaki et al. Karch & Katz... Polchinski & Strassler finite-temperature and µ = 0 theories Son, Starinets,... (viscosity of quark-gluon plasma) Finite chemical potential THIS TALK!

3 Setup: (I) Pure Glue pure QCD i.e. no matter instead, consider 4+1 dim max. susy YM compactify on circle impose anti-periodic bdy. cond. for fermions do not know geometry in IR, reduces to pure QCD, scalars and fermions decouple dual to near-horizon geometry of non-extremal D4-brane, doubly Wick rotated

4 The geometry [Witten, Sakai & Sugimoto,... ] ( u ) 3/2 [ηµν ds 2 = dx µ dx ν + f(u)dθ 2] + R ( ) R 3/2 [ du 2 ] u f(u) + u2 dω 4 world-volume our 3+1 world uλ 3 f(u) = 1 u θ is a compact Kaluza-Klein circle u: radial direction bounded from below u u Λ

5 Several remarks Solution characterised by two parameters: R D4 : RD4 3 = πg sls 3 N c R : R= 2π ( R 3 ) 1/2 D4 M Λ = 2π 3 R u Λ Relation to gauge-theory parameters: size of S 1 on D4 (i.e. M KK ) set by R λ g 2 YM N c = R3 D4 α R non-extremality of D-brane: angle θ identified with period R to avoid conical singularity Regime of validity: sugra OK if R 2 R 3 D4 /R α λ 1 (max curvature at the wall) valid as long as e φ = g s (u/r D4 ) 3/4 < 1 (min coupling at the wall) Problem : M KK M glueball M meson M Λ 1/R cannot decouple KK modes!

6 Overview u = energy scale θ u focus here u = u Λ IR wall ( ) 3/2 RD4 u Λ u 2 Λ = R 3/2 D4 u1/2 Λ

7 Setup : (II) Introducing matter Sakai-Sugimoto model Add D8 flavour (probe) branes to D4 stack strings between flavour & colour branes in fund. rep. of flavour & colour group Solve for the shape of the D8 D4 : θ D8 : }{{} flat }{{} curve u(θ) Solution to the 1st order equation gives embedding u(θ) coord. sys. adapted to D8 θ,ω 4 D8 θ D8 wall direction

8 Symmetry encoded in geometry Asymptotically exhibits full chiral symmetry SU(N f ) L SU(N f ) R Bending of the brane encodes spontaneous symmetry breaking in gauge theory in a geometrical way SU(N f) L SU(N f) isospin SU(N f) R spectrum of fluctuations contains (π ±,π 0 ) Goldstone bosons Brane geometry also reproduces chiral symmetry restoration above T > T c SU(N f) L SU(N f) R

9 Low spin mesons Spectrum is known only in the limits: Low-spin mesons: fluctuations on and of the flavour brane Fluctuations governed by Dirac-Born-Infeld action of the flavour brane S = V S 4 d 5 xe φ det (g µν + 2πα F µν ) + S Wess-Zumino = V S 4 d 4 xdz g F µν F ρλ g µρ g νλ +... Expand world-volume fields in modes meson spectrum & action

10 Effective action for light mesons Decompose the gauge fields F µν = n F uµ = n G (n) µν (x)ψ (n) (u), B (n) µ (x) uψ (n) (u), Fourier transform & factor out polarisation vectors, d 4 k B (m) µ B (n) µ [u 1/2 γ 1/2 (ω 2 )] k 2 )ψ (n) u (u 5/2 γ 1/2 u ψ (n) = 0. }{{} a Sturm-Liouville problem mass spectrum of mesons

11 High spin mesons Spectrum is known only in the limit: Sigma model (semiclass) high-spin glueballs (closed) & mesons (open) q q meson: u f1 u f2 region I projected m q u f1 u Λ u Λ region II N.B. High spin mass M high λm Λ vs. low spin mass M low M Λ M KK

12 Part II: Turning on an isospin chemical potential Chiral Langrangian

13 Isospin vs Baryon chemical potential Why isospin chemical potential is easier in holographic models than baryon chemical potential: large N c baryons much heavier than at finite N c mesons closer to the real-world baryons complicated solitons, mesons elementary fields so far only singular solitons known potentially comparable with the lattice (no sign problem) Bad feature: Artificial, no pure isospin systems exist in nature (weak decays) neutron stars

14 Chiral Lagrangian At small µ I happens chiral Lagrangian (with m q = 0) to get a feeling what L chiral = f2 π 4 Tr(D νud ν U ), U U(N f ). U e i fπ πa(x)t a Invariant under separate U g 1 L U, T a U(N f ) generators U Ug R The vacuum U = I preserves the vector-like U(N f ) symmetry, U g L Ug 1 R g L = g R. In U = I want to turn on a vector chemical potential µ L = µ R. Other global transformations move us around on the moduli space of vacua, M = U(N f) U(N f ) U(N f )

15 Chiral Lagrangian and µ 0 As usual, chemical potentials via D ν U = ν U 1 2 δ ν,0(µ L U Uµ R ) = ν U 1 2 δ ν,0([µ V,U] {µ A,U}) (µ L = µ V µ A, µ R = µ V + µ A). V χ = f2 ( ) π 4 Tr ([µ V,U] {µ A,U})([µ V,U ] + {µ A,U }) From V χ minima: (1) µ V = 0, µ A -any V χ -const. ρ A f 2 πµ A (2) µ A = 0, µ V = µ I σ 3 /2 U max = e iα (cos(β)i + i sin(β)σ 3 ) and U min = e iα (cos(β)σ 1 + sin(β)σ 2 ) in the U min : ρ V f 2 π µ I ρ A,I = 0. (3) µ V = µ I σ 3 /2, µ A = µ A,I σ 3 /2: { µ 2 A,I < µ2 V U min as in (2) µ 2 A,I > µ2 V U min opposite

16 Vectorial isospin potential Effects of µ V in U = U min effects of µ A in U = I vacuum

17 Aside: non-zero pion mass The chiral Lagrangian gives us the behaviour of the pions for small µ I, Son, Splittorf, Stephanov However, Sakai-Sugimoto has m π = 0, so we will at small µ I see

18 Beyond Chiral Langrangian Chiral Langrangian, valid up to the first massive vector meson, µ I m ρ Other operators are relevant, e.g. Skyrme term L Skyrme = 1 32e 2 Tr ( [U 1 µ U,U 1 ν U ] 2 ). This leads to a dispersion relation for pions ω 2 + k 2 + µ 2 I k2 µ 2 I e 2 f 2 π = 0. This suggests massive pions eventually become unstable. But, does not explain what the ρ does. Sakai-Sugimoto has pions and fixed couplings to other mesons. Study π s and ρ in this model as function of µ I.

19 Part III: Holographic isospin chemical potential

20 Beyond Chiral Langrangian µ I = 0 Cigar-shaped subspace with D8 s embedded, u = (1 + z 2 ) 1/3 No chemical potential no background field, trivial A µ = 0 vacuum. Meson massess from linearised DBI action around trivial vacuum. A µ (x µ,z) = U 1 (x) µ U(x)ψ + (z) + n 1 A z = 0 B (n) µ (x)ψ n (z), Can go beyond χ-perturbation theory: have χ-langrangian interacting with infinite tower of massive modes.

21 Beyond Chiral Langrangian µ I = 0 Effective action we use come from the truncated string effective action S = T [ ] d 4 xdu u 1/2 γ 1/2 Tr(F µν F µν ) + u 5/2 γ 1/2 Tr(F µu F µ u) +... where ignored DBI corrections to the YM, ((l 2 s F)n ) and beyond O(l 3 s F) For eg., just for pion this gives F zµ = U 1 µ U φ (0) (z) + B-stuff F µν = [U 1 µ U,U 1 ν U]ψ + (z) ( ψ + (z) 1 ) + B-stuff. which gives chiral Lagrangian plus Skyrme term, ( f S = d 4 2 x Tr π 4 (U 1 µ U) [ U 1 32e 2 µ U,U 1 ν U ] ) 2 + π B f 2 π λn cm 2 KK, e2 1 λn c,.

22 Sakai-Sugimoto and chiral symmetry In Sakai-Sugimoto, global symmetry is realised as large gauge transformation of bulk field, A µ ga µ g 1 + ig µ g 1 lim z g(z,xµ ) = g L SU(N f ) L, lim z + g(z,xµ ) = g R SU(N f ) R.

23 Sakai-Sugimoto and chiral symmetry And changes holonomy U = P exp (i dz A z ) g L g 1 changes the pion expectation value, since U = exp (iπ a (x)σ a /f π ). R. So if start with trivial vacuum A µ = A z = 0, the vectorial transformation g L = g R preserves vaccum, does not change U If g L g R, does not preseve vacuum i.e. changes holonomy χ-symmetry breaking

24 Turning on µ I 0 For SS model, bulk field A µ (x,u) A ν (x,u) B ν (x) (1 + O( 1u ) ) + ρ ν (x)u (1 3/2 + O( 1u ) ). here B µ (x) source term for gauge theory current J ν (x) ( R d 4 xb µj ν (x)) ρ ν (x) vev of J µ To add vectorial/axial chemical potential, solve for the even/odd bulk field with b.c. : A µ (x,z ) = µ L δ µ,0 A µ (x,z + ) = µ R δ µ,0

25 Isotropic & homogenious solution First ansatz, assume that condensate is x-independent A 0 (z),a i = 0 Isospin chemical potential background satisfies 5d YM equation (in A u = 0 gauge), [ ] z (1 + z 2 ) z A (3) V: A (3) 0 = µ V, 0 = 0 A: A (3) 0 = µ A arctan z. N.B Soln to YM action, neglect DBI corrections, i.e. valid for µ I λ/l Spectrum around vectorial soln (V) tachyonic i.e. free energy is unaffected, but fluctuations are affected! roll down to π (1) 0, then rotate back to trivial vacuum Properties of new vacuum: Effectively work with axial solution (A) f π unmodified, two massive and one massless pion

26 Instability of isotropic solution Soln found is unique isotropic soln: pions condensed. What about ρ et al? Are there any other ground states which dominate for higher µ I? Analyse general stability of soln For µ I λ 5 /L 2 can still use just nonabelian YM expand YM around axial solution Ā0 in U = I vacuum A 0 = Ā0(u) + δa (a) 0 (ω, k,u)σ a e iωt+i k x, A i = δa (a) i (ω, k,u)σ a e iωt+i k x, A u = 0

27 Transverse vectors and scalars The transverse vectors (δa 0 = 0, i δa i = 0) develop an instability: at k = 0 the dispersion relation is 1.75 ω Μ The scalars (fluations transverse to the brane) are unstable too, but only for much larger µ, ω Μ The main question: what about the pions & longitudinal vectors?

28 Pions and longitudinal vectors Both pions and longitudinal vectors are governed by A i ik i A T and A 0. Equations diagonal for δa (1) i = ±iδa (2) i, δa (1) 0 = ±iδa (2) 0. The difference is the boundary conditions The pion is pure large gauge, so impose F 0i = 0, A T (z + ) = π 2 + c 3 z +... A 0 (z + ) = (ω + πµ) π 2 + ( c3 k 2 Vectors asymptote to zero at z ±, A T (z + ) = 1 z +... A 0 (z + ) = ) 1 ω + πµ πµ z +... k2 1 ω + πµ z +... N.B µ I = 0 recover Lorentz inv. rels. (δa 0 = ωδa T pion and δa 0 = k 2 /ωa T, long. vec.)

29 Pions and longitudinal vectors cont. Similarly, imposing appropriate b.c. at z = fixes ω(µ,k). So the spectrum is π s, for small µ I mass up (as from χ L) modes change nature no-crossing for k 0 k = 0 special crossing of ρ and π ρ condenses

30 Vector instability The value of µ crit the same as for transverse ρ all components of ρ vector for k = 0 condense at µ crit 1.7m ρ Ω 2 1 transverse: Μ ρ meson instability.

31 Finding a new ground state What is the new ground state? Ansatz (inspired by linear analysis): with b.c. A (1) 3 (z) = ±ia(2) 3 (z), A(1) A (3) µ = δ µ,0a (3) 0 (z) A u = 0, A (3) i (z) = A (2) i (z) = 0 (i = 1, 2), 0 (z = ± ) = ±µ I/2, A (1) 3 (z = ± ) = 0 Solution of the nonlinear equations [ ] u u 5/2 γ 1/2 u A (3) [ ] u u 5/2 γ 1/2 u A (1) 0 3 = 4(A (1) 3 )2 A (3) 0 u 1/2 γ 1/2, = 4(A (3) 0 )2 A (1) 3 u 1/2 γ 1/2. Have two solutions { ( µ < µcrit : A (1) 3 = 0 A (3) 0 = µi π arctan µ > µ crit : A (1) 3 0 A (3) 0 0. ) z u Λ

32 The new ground state A numerical solution yields: µ crit 1.7m ρ, ρ µ µ crit. ρ-meson condensate forms: breaking rotational SO(3) SO(2) breaking the residual flavour U(1) (in addition, the pion condensate remains present)

33 Summary and todo Can we include the pion mass (using tachyon)? How does this depend on L (constituent quark masses)? Are there further instabilities at even higher µ? Corrections due to DBI and Chern-Simons? Behaviour in deconfined phase, as function of temperature?