Thermodynamics of hadrons using the Gaussian functional method in - PowerPoint PPT Presentation

Thermodynamics of hadrons using the Gaussian functional method in the linear sigma model Shotaro Imai 1 and Hua-Xin Chen 2 , Hiroshi Toki 3 , Li-Sheng Geng 2 Kyoto Univ. 1 , Beigang Univ. 2 , RCNP/Osaka Univ. 3 Oct. 27 2013 Chiral 13 @ Beihang Univ. arXiv:1309.0591 [nucl-th] 1 / 15

Introduction Chiral symmetry breaking ▶ Mass generation for (massless or light) fermion ▶ Nambu-Goldstone boson ▶ Restoration at high temperature (phase transition) Non-perturbative interaction among mesons The interaction term of linear sigma model ( λ ∼ O (10) ) L int = λ 4 ( σ 2 + π 2 ) 2 Chiral symmetry with the fluctuations of mesons around their mean field values at finite temperature ▶ The Cornwall-Jackiw-Tomboulis (CJT) formalism J. M. Cornwall, R. Jackiw and E. Tomboulis PRD 10 (1974) . . . ▶ The optimized perturbation theory S. Chiku and T. Hatsuda PRD 58 (1998) . . . ▶ The Gaussian Functional Method: corresponding to Hartree-Fock approx. + RPA T. Barnes and G. Ghandour PRD 22 (1980) . . . 2 / 15

The Gaussian Functional Method (view of my talk) Gaussian Functional Method Barnes and Ghandour PRD 22 (1980), Nakamura and Domitrasinovic PTP 106 (2001) 1. Schr¨ odinger picture in field theory with the Gaussian ground state functional ansatz 2. The minimization condition: determination of the variational parameters ▶ The resulting (dressed) mass of Nambu-Goldstone (NG) boson is not zero due to the non-perturbative effect 3. Considering the bound state of mesons (4 quarks state): Bethe-Salpeter equation ▶ Emergence of the NG bosons → Physical mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ▶ Fixing the parameters with the sigma meson mass (500 MeV) ▶ Dressed mass vs. Physical mass 4. The phase transition at finite temperature 3 / 15

The Gaussian Functional Method I O(2) Linear sigma model ( φ = ( ϕ 0 , ϕ 1 , ϕ 2 , ϕ 3 ) = ( σ, π ) ) L =1 2( ∂ µ φ ) 2 + 1 2 µ 0 φ 2 − λ 0 4 ( φ 2 ) 2 + εσ The Hamiltonian δ 2 ∫ ( − 1 δϕ i ( x ) ϕ i ( y ) + 1 H [ φ ] = d y δ ( y − x ) 2 ∇ x ϕ i ( x ) ∇ y ϕ i ( y ) 2 ) − 1 2 µ 0 φ 2 + λ 0 4 ( φ 2 ) 2 + εσ The Gaussian ground state functional ) ( − 1 ∫ ] G − 1 Ψ[ φ ] = N exp d x d y [ ϕ i ( x ) − ⟨ ϕ i ( x ) ⟩ ij ( x , y )[ ϕ j ( y ) − ⟨ ϕ j ( y ) ⟩ ] 4 fluctuation V.E.V d 3 k G ij ( x , y ) =1 ∫ 1 e i k · ( x − y ) 2 δ ij (2 π ) 3 √ k 2 + M 2 i dressed mass Energy with the variational parameters M i and ⟨ ϕ i ⟩ ∫ E ( M i , ⟨ ϕ i ⟩ ) = D φ Ψ ∗ [ φ ] H [ φ ]Ψ[ φ ] 4 / 15

The Gaussian Functional Method II The minimization condition: Determination of parameters ( ⟨ ϕ i ⟩ , M i ) ( ∂ E ( M i , ⟨ ϕ i ⟩ ) ) = 0 , for i = 0 . . . 3 ∂ ⟨ ϕ i ⟩ , M i min ⇔ (One- and two-point) Schwinger-Dyson equations = 1 + 1 + 2 6 The mean field values: One-point SD equation ⟨ ϕ 0 ⟩ = v, ⟨ ϕ i ⟩ = 0 for i = 1 , 2 , 3 0 = − ε v 2 + 3 I 0 ( M σ ) + 3 I 0 ( M π ) µ 2 [ ] v + λ 0 d 4 k ∫ 1 I 0 ( M i ) = i k 2 − M 2 (2 π ) 4 i + iϵ 5 / 15

The Gaussian Functional Method III Two-point SD equation: (Determination of masses M i ) + 1 + 1 = 2 2 Dressed mass ( M 0 = M σ , M i = M π ) 3 v 2 + 3 I 0 ( M σ ) + 3 I 0 ( M π ) M 2 σ = − µ 2 [ ] 0 + λ 0 = ε v + 2 λ 0 v 2 v 2 + I 0 ( M σ ) + 5 I 0 ( M π ) M 2 π = − µ 2 [ ] 0 + λ 0 = ε v + 2 λ 0 [ I 0 ( M π ) − I 0 ( M σ )] non-perturbative effect The pion mass M π ̸ = 0 even in the chiral limit ε → 0 → DO NOT satisfy The Nambu-Goldstone theorem 6 / 15

The Nambu-Goldstone Theorem The dressed mass cannot satisfy the Nambu-Goldstone theorem Relation between M σ and M π with some cutoff Λ 2000 2000 2000 2000 1500 1500 1500 1500 M [MeV] M [MeV] 1000 1000 1000 1000 500 500 500 500 0 0 0 0 0 200 400 600 800 1000 0 200 400 600 800 1000 M [MeV] M [MeV] (b) ε ̸ = 0 (a) ε = 0 ▶ The dressed masses depend on the cutoff ▶ They cannot satisfy the NG theorem independently the cutoff ▶ There are no finite sigma mass and zero pion mass in the chiral limit ε = 0 ▶ The physical mass of sigma (600 MeV) and pion (140 MeV) cannot exist in ε ̸ = 0 We cannot identify these masses as physcal mass (NG boson) 7 / 15

The Bethe-Salpeter Equation I Physical masses appear as pole of the Bethe-Salpeter (four-point SD) euqarion (bound state of mesons) σ − π channel → Physical pion mass m π ( s = p 2 ) d 4 k ∫ 1 G σπ → σπ ( p 2 ) = i [ k 2 − M 2 σ + iϵ ] [( k − p ) 2 − M 2 (2 π ) 4 π + iϵ ] v 2 [ ( )] V σπ → σπ ( s ) =2 λ 0 1 + 2 λ 0 s − M 2 π T σπ → σπ ( s ) = V σπ → σπ ( s ) + V σπ → σπ ( s ) G σπ → σπ ( s ) T σπ → σπ ( s ) V σπ → σπ ( s ) = 1 − V σπ → σπ ( s ) G σπ → σπ ( s ) 8 / 15

The Bethe-Salpeter Equation II The coupled channel σ − σ and π − π → Physical sigma mass m σ  [ ] [ ]  1 + 3 2 λ 0 v 2 1 + 3 2 λ 0 v 2  3 ( V σσ → σσ ) V σσ → ππ s − M 2 s − M 2 V = = 2 λ 0 σ σ 1 [ ] [ ]  1 + 3 2 λ 0 v 2 5 + 3 2 λ 0 v 2 V ππ → σσ 3 V ππ → ππ 1 s − M 2 3 s − M 2 σ σ ( T σσ → σσ ) ( G σσ → σσ ) T σσ → ππ 0 T = , G = 1 T ππ → σσ 3 T ππ → ππ 0 3 G ππ → ππ T = V + 1 2 V GT =(1 − 1 2 V G ) − 1 V Physical mass m σ , m π vs. Dressed mass M π (NG boson) 1000 1000 800 800 [MeV] [MeV] 600 600 m m 400 400 m m 200 200 m m 0 0 -200 -200 0 200 400 600 800 1000 0 200 400 600 800 1000 M [MeV] M [MeV] (b) ε ̸ = 0 (a) ε = 0 9 / 15

Parameters π 0 = 93 × 142 2 MeV 3 ) We accept the parameters below ( ε = f π m 2 chiral limit breaking case λ 0 =83 . 6 λ 0 =75 . 5 µ 0 =1680 MeV µ 0 =1610 MeV Λ =800 MeV Λ =800 MeV ε =1 . 86 × 10 6 MeV 3 ε =0 MeV 3 ⇓ ⇓ M σ =1200 MeV M σ =1150 MeV M π =580 MeV M π =564 MeV m σ =500 MeV m σ =500 MeV v = f π =93 MeV v = f π = 93 MeV m π =0 MeV m π =138 MeV We fit the parameters to reproduce the pion decay constant f π and the pion mass m π 10 / 15

Finite Temperature I Finite temperature with the Matsubara formalism E ( v, M σ , M π ) → E ( v ( T ) , M σ ( T ) , M π ( T ); T ) The behavior of the free energy as a function of the mean field value v (fixing M σ , M π ) In the case of the chiral limit 10 100 8 80 4 6 v [MeV] 60 T = 1 9 4 4 40 9 T = 3 0 0 T = 1 9 0 2 20 T = 1 9 5 T = 0 0 0 -2 -20 0 50 100 150 200 0 100 200 300 v [MeV] T [MeV] (a) Free energy (b) Mean field values v The free energy is suddenly change at 195 MeV and the first order phase transition 11 / 15

Finite Temperature II In the case of the explicit chiral symmetry breaking E χSB = εv 10 100 8 80 4 6 v [MeV] 60 T = 1 9 8 .2 4 40 T = 3 0 0 9 T = 1 9 5 2 20 T = 1 9 9 T = 0 0 0 -2 -20 0 50 100 150 200 0 100 200 300 v [MeV] T [MeV] (a) Free energy (b) Mean field values v Similar behavior even in the case of explicit symmetry breaking: Suddenly change at 195 MeV and the first order phase transition ▶ The whole energy at transition temperature E ∼ − 10 8 MeV 4 ▶ The chiral symmetry breaking term E χSB ∼ − 10 7 MeV 4 Since the contribution of E χSB is 10 times smaller, the free energy suddenly change → In the MFA, they are comparable ∼ − 10 8 MeV 4 12 / 15

Finite Temperature III Solutions of the BS equation at finite temperature G ( s ) → G ( s, T ) 1000 1000 800 800 [MeV] [MeV] 600 m 600 m m m 400 400 m 200 200 m m m 0 0 -200 -200 0 100 200 300 0 100 200 300 T [MeV] T [MeV] (b) ε ̸ = 0 (a) ε = 0 ▶ Mesons bound state picture holds only in the symmetry broken phase ▶ In the symmetric phase, they are unbound and their masses π + 4 λ 0 v 2 G σπ → σπ ( m 2 π ) coincide the dressed mass m 2 π = M 2 1 − 2 λ 0 G σπ → σπ ( m 2 π ) The second term vanishes due to the symmetry restoration ( v → 0) 13 / 15

Summary 1. We treat the non-perturbative effect using Gaussian ground state functional ansatz 2. Determination of the variational parameters (One- and Two-point Schwinger-Dyson Equation) ▶ There are NO NG bosons: unphysical particle 3. Mesons bound state in the Bethe-Salpeter Equation ▶ Emergence of the NG bosons ▶ 4 quarks picture of mesons 4. The behavior of the chiral symmetry at finite temperature ▶ The phase transition ▶ The meson-meson bound state Future work ▶ Investigation of 3 flavor case ▶ Application to 2 color system at finite density 14 / 15

The end Thank you for your kind attention 15 / 15

Back up slides 16 / 15

Parameter dependence Chiral limit( ε = 0 ) 3000 3000 2000 2000 1500 1500 2000 2000 M [MeV] M 1000 1000 0 1000 1000 M 500 500 0 0 0 0 0 50 100 150 200 0 50 100 150 200 0 0 Explicit breaking( ε ̸ = 0 ) 3000 3000 2000 2000 1500 1500 2000 2000 M [MeV] M 1000 1000 0 1000 1000 M 500 500 0 0 0 0 0 50 100 150 200 0 50 100 150 200 0 0 (a) µ 0 vs. λ 0 (b) M σ , M π vs. λ 0 17 / 15

Dressed mass at finite temperature Dressed mass at finite temperature ε M 2 v ( T ) + 2 λ 0 v 2 ( T ) σ ( T ) = ε M 2 π ( T ) = v ( T ) + 2 λ 0 [ I 0 ( M π ( T )) − I 0 ( M σ ( T ))] 1500 1500 [MeV] M [MeV] M 1000 1000 500 500 M M M M M M 0 0 0 100 200 300 0 100 200 300 T [MeV] T [MeV] (b) ε ̸ = 0 (a) ε = 0 18 / 15