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Light Tetraquarks and the Chiral Phase Transition

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Light Tetraquarks and the Chiral Phase Transition Light tetraquarks and the chiral phase transition Vladimir Skokov, RIKEN/BNL Research Center & RDP: 1606.04111 XYZ, Pc states: strong evidence for tetraquark & pentaquark states, composed of both light and heavy quarks Why do we need heavy quarks to see tetraquark states? Jaffe ’79…Schechter…Close…Tornqvist…Maiani…Giacosa….Peleaz ’15: “the” σ meson is “a” tetraquark (diquark anti-diquark). But situation murky…. Punchline: tetraquarks must be included to understand the phase diagram of QCD (versus quark mass, plane of temperature T and baryon chemical potential μ) Tetraquarks: for three (not two) flavors of very light quarks, tetraquarks may generate a second chiral phase transition For QCD, in the plane of T & μ plane: direct connection between tetraquarks and color superconductivity in effective modesl, need tetraquarks to determine the Critical EndPoint Tetraquarks for two light flavors: meh, fuh gedda ‘boud it Chiral symmetry for two flavors Classically, chiral symmetry Gcl = SU(2)L x SU(2)R x U(1)A = O(4) x O(2) Use φ, complex 4-component vector φ =(σ + i ⌘ , ~a + i ~⇡ ) Linear σ model for exact χ symmetry: 0 cl 1 2 1 2 2 = (@ φ) + m φ⇤ φ + λ (φ⇤ φ) + ... L 2 µ 2 · · Quantum mechanically, axial anomaly reduces U(1)A -> Z(2)A : φ → (−) φ Simplest term which is only Z(2)A invariant: A =+m2 (Im φ)2 + ... L A With Gcl , η degenerate with π . With axial anomaly, η splits from massless π Directly induced by instantons + … Diquarks and tetraquarks for two flavors Jaffe ’79: most attractive channel for quark-quark scattering is antisym. in both flavor and color. Color: 3 x 3 = 3 (antisym) + 6 (sym) Two flavors: 2 x 2 = 1 (antisym) + 3 (sym) For two flavors diquark is a color triplet, flavor singlet, A ABC ab aB T 1 bC χ = ✏ ✏ (q ) − q L L C L (A, B, C = color; a, b = flavor) Also χR. χL and χR singlets under Z(2)A. One complex valued tetraquark field: A A ⇣ =(χR)⇤ χL Sigma models and tetraquarks for two flavors The tetraquark field ζ is a singlet under flavor and Z(2)A. Split complex ζ into its real and imaginary parts, ζr and ζi. QCD is even under parity, so only even powers of ζi can appears, forget ζi. But any powers of ζr can! A = h ⇣ + m2 ⇣2 + ⇣3 + λ ⇣4 V⇣r r r r r r r r r Hence ⟨ζr ⟩ is always nonzero! Couplings to φ start with U(1)A inv.: ⇣ = ⇣φ⇤ φ + ... V · The tetraquark ζr is just a massive field with a v.e.v. : boring! Chiral transition for two flavors 2 m 2 4 Assume anomaly is large up to Tχ . = φ + λφ Vφ 2 T = 0: m2 < 0, 2 ⟨φ⟩ ≠ 0 => T= Tχ: m = 0, ⟨φ⟩ = 0 ⇓ Chiral transition second order, m2 = 0 => infinite correlation length O(4) universality class. T ≫ fπ: m2 > 0, ⟨φ⟩ = 0 => Tetraquark field just doesn’t matter: massive field with cubic couplings Tetraquarks for three light flavors: must be included Chiral symmetry, three flavors Quark Lagrangian 1 γ qk = q Dq= q Dq + q Dq ,q = ± 5 q L 6 L 6 L R 6 R L,R 2 Classical Gcl = SU(3)L x SU(3)R x U(1)A : i↵/2 +i↵/2 q e− U q ,q e U q L ! L L R ! R R Order parameter for χ symmetry breaking (a, b… = flavor; A,B… = color) ab bA aA +i↵ Φ = q q Φ e U Φ U † L R ! R L Axial anomaly reduces U(1)A to Z(3)A . σ models for χ symmetry U(1)A invariant terms: 2 2 2 = m tr Φ†Φ + λ tr Φ†Φ + λ tr Φ†Φ VΦ O(18) Drop last term, λO(18) ~ λ/50. 0 = tr H Φ† + Φ χ sym. broken by background field: VH − To split the η’ from the π, K, & η, A = (det Φ +c.c.) add Z(3)A invariant term V Example: SU(3) symmetric case, H = h 1. JP = 0− : octet π = K = η & singlet η’. JP = 0+: octet a0 = κ = σ8 & singlet σ h ≠ 0 so π’s massive . Satisfy (like ’t Hooft ’86): m2 m2 = m2 m2 ⌘0 − ⇡ a0 − σ The anomaly moves η’ up from the π, but also moves singlet σ down from the a0! Light σ is a problem for σ models Tetraquarks for three flavors Three flavors: 3 x 3 = 3 + 6 . aA abc ABC bB T 1 cC χ = ✏ ✏ (q ) − q Diquark field flavor anti-triplet, 3 L L C L LR tetraquark field ζ transforms identically to ab aA bA Φ under Gχ = SU(3)L x SU(3)R ⇣ =(χR )⇤ χL Under U(1)A, Φ has charge +1, ζ charge -2. Since ζ & Φ in same representation of Gχ, direct mixing term. Z(3)A invariant: A 2 ⇣Φ,2 = m tr ⇣†Φ + Φ†⇣ Black, Fariborz, Schechter ph/9808415 + ….; V ’t Hooft, Isidori, Maiani, Polosa 0801.2288 + …. e An extra dozen couplings. abc a0b0c0 aa0 bb0 cc0 e.g., U(1)A inv. cubic term: ⇣1Φ,3 = ✏ ✏ ⇣ Φ Φ +c.c. V 1 ⇣ ⌘ “Mirror” model, T = 0 Spectrum : Φ = ⇡,K,⌘, ⌘0; a0, , σ8, σ0 ; ⇣ = ⇡, K,⌘, ⌘0; a0, , σ8, σ0. General model has 20 couplings: e e e e e e e e Fariborz, Jora, & Schechter: ph/0506170; 0707.0843; 0801.2552. Pelaez, 1510.00653 Instead study “mirror” model, where Φ and ζ start with identical couplings 2 + 2 = m tr Φ†Φ (det Φ +c.c.)+ λ tr Φ Φ VΦ − 2 + 2 = m tr ⇣†⇣ (det ⇣ +c.c.)+ λ tr ⇣ ⇣ V⇣ − A 2 Assume only ζΦ coupling is mass term: = m tr ⇣†Φ + Φ†⇣ V⇣Φ,2 e Simple, because mass only mixes: ⇡ ⇡ ,K K... $ $ e e Spectrum of the mirror model In the chiral limit, the mass eigenstates: (need to assume m 2 < 0 ) 2 2 ⇡, ⇡ =0, 2 m ; ⌘0, ⌘ 0 =3, 3 e 2 m − − a , a = m2 + +6λφ2 m2 ; σ , σ = m2 2 +6λφ2 m2 0 0 e e ± e 0 − e ± All states are mixtures of Φ and ζ. Of course 8 Goldstone bosons. e e e e Satisfy generalized ’t Hooft relation m2 + m2 m2 m2 = m2 + m2 m2 m2 ⌘0 ⌘0 − ⇡ − ⇡ a0 a0 − σ − σ e e e e Since every multiplet is doubled, this can easily be satisfied (unlike if just one). Even with same couplings, all masses are split by the mixing term. At nonzero T, the thermal masses of the Φ and ζ cannot be equal! Chiral transition for three flavors, no tetraquark 2 m 2 3 4 Cubic terms always generate φ = φ + λφ first order transitions. V 2 − T = 0: m2 < 0, ⟨φ⟩ ≠ 0 => T= Tχ: cannot flatten the potential => 2 T ≫ fπ: m > 0, At Tχ, two degenerate minima, ⟨φ⟩ = 0 => with barrier between them. Transition is first order. With tetraquarks, maybe two chiral transitions In chiral limit, may have have two chiral 45 phase transitions. => 40 35 At first, both jump, remain nonzero. ←⟨φ⟩ At second, both jump to zero. 30 25 ⟨ζ⟩→ 2 2 2 20 mφ(T )=3T + m 15 2 2 2 m⇣ (T )=5T + m 10 ⇣ 2 2 m = (100) 5 φ − T→ 0 Tχ Tχ 45 Tχ Tχ e g 40 " " 35 e 30 ←⟨φ⟩ 25 ⟨ζ⟩→ <= Also possible to have single chiral 20 phase transition, tetraquark crossover 15 ⇣ 10 2 2 5 φ m = (120) 0 − Tχ e ”Columbia” phase diagram for light quarks Lattice: chiral transition crossover in QCD If two chiral phase transitions for three massless flavors, persists for nonzero mass Implies new phase diagram in the plane of mu = md versus ms: critical line→ ms X = QCD, crossover tricritical point→ ←critical line C = crossover critical line→ I I = one II chiral transition mu,d II = two chiral transitions Tetraquarks in the plane of T and μ Diquark fields are identical to the order parameters for color superconductivity. Tetraquark condensate = gauge invariant square of CS condensate. Suggests: T χ ↓chiral crossover line ↓ color superconducting line T↑ ~ χ ❋Critical EndPoint? ↑tetraquark ←1st order chiral line crossover µB μ→ Line for chiral crossover might end, meet line for first order chiral transition at Critical EndPoint (CEP). Massless σ at CEP. Rajagopal, Shuryak & Stephanov, ’99 In effective models, to find the CEP, must include tetraquarks: need the right σ! Need tetraquarks to find the critical endpoint Kovacs, Szep, & Wolf, 1601.05291: Use linear σ model + Polyakov loop model And add vector mesons (good!). But do not include tetraquarks. Find a critical endpoint (CEP), but T so low, CEP would have been seen exp’y! T↑ ❋←CEP μbaryon→ Using the lattice to fix the σ model? Briceno, Dudek, Edwards, & Wilson (Hadron Spectrum Coll.), 1607.05900 First study of the σ meson on the lattice with light, dynamical quarks mπ = 391 MeV: σ bound state resonance just below ππ threshold mπ = 236 MeV: σ broad resonance, well above ππ threshold Perhaps: fix the parameters of σ model plus tetraquarks from the lattice Very difficult, but of direct relevance for the CEP. 0 half 300 500 700 900 width↑ -100 -200 -300 Four flavors, three colors: hexaquarks Diquark 2-index antisymmetric (ab)A abcd ABC cB T 1 dC χ = ✏ ✏ (q ) − q tensor: L L C L (ab),(cd) (ab)A † (cd)A So LR tetraquark is same: ⇣ = χR χL ⇣ ⌘ Tetraquark couples to usual Φ through cubic, quadratic terms, so what. a abcd ABC bA cB T 1 dC Instead, consider triquark field: χ = ✏ ✏ q (q ) − q L L L C L Triquark is a color singlet, fundamental rep.
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