Large N c behaviour of an effective lattice theory for heavy dense - PowerPoint PPT Presentation

CRC - TR Large N c behaviour of an effective lattice theory for heavy dense QCD The 37 th International Symposium on Lattice Field Theory Owe Philipsen, Jonas Scheunert Goethe-Universit¨ at Frankfurt am Main Institut f¨ ur Theoretische Physik

Introduction

Motivation At finite baryon density, lattice QCD has a sign problem, which prohibits direct simulation. Approximate methods: Taylor expansion, reweighting, imaginary potential. → Fail for µ/ T � 1. Need alternative methods to probe cold and dense QCD. 1 / 13

3d Effective Theory Centre symmetric 3d effective actions for thermal SU(N) Yang-Mills from strong coupling series Langelage, J.; Lottini, S. & Philipsen, O. JHEP, 2011, 02, 057 Onset Transition to Cold Nuclear Matter from Lattice QCD with Heavy Quarks Fromm, M.; Langelage, J.; Lottini, S.; Neuman, M. & Philipsen, O. Phys. Rev. Lett., 2013, 110, 122001 Equation of state for cold and dense heavy QCD Glesaaen, J.; Neuman, M. & Philipsen, O. JHEP, 2016, 03, 100 2 / 13

3d Effective Theory Definition of effective theory ( U ∈ SU ( N c )): � Ψ e − S G [ U ] − S ( W ) [ U , Ψ , ¯ D U D Ψ D ¯ Ψ] Z = f � � D U 0 e − S eff SU (3) D L e − S eff [ L ] =: = �� � Ψ e − S G [ U ] − S ( W ) [ U , Ψ , ¯ D U i D Ψ D ¯ Ψ] ⇒ S eff [ U 0 ] = − log f Analytic determination using combined strong coupling (small β = 2 N c g 2 ) 1 and hopping expansion (small κ = 2 am +8 ). S eff has a mild sign problem and weak couplings → pertubative treatment also possible. 3 / 13

Systematics of the hopping expansion

Effective Theory at Strong Coupling At strong coupling, link integration factorizes [Rossi, Wolff 84] � ¯ D Ψ D ¯ Ψ(1+ T [ U 0 ])Ψ exp( − S eff [ U 0 ]) = Ψ e 3 � d U i ( n ) e κ tr � J i ( n ) U i ( n )+ U † i ( n ) K i ( n ) � � � × i =1 n ∈ Λ J i ( n ) ab = ¯ Ψ( n ) f α, b (1 − γ i ) αβ Ψ( n + e i ) f β, a K i ( n ) ab = ¯ Ψ( n ) f α, b (1 − γ i ) αβ Ψ( n + e i ) f β, b 4 / 13

Single Site integral Known for U(N) [Bars 80], for SU ( N ) use � � d U det( U ) q f ( U ) � d U f ( U ) = q ∈ Z SU ( N ) U ( N ) to obtain ∞ 1 − δ k , 0 � � �� det( J ) k + det( K ) k � d U e κ tr ( JU + U † K ) = � 2 k =0 SU ( N c ) a r ( κ ) b r , k ( κ ) χ r ( JK ) � × . d r r ∈ GL ( N c ) irreps Summands are of order O ( κ kN c +2 � Nc l =1 λ l ) ⇒ spatial Baryon hoppings surpressed for large N c . Due to Grassmann constraint kN c + 2 � N c l =1 ≤ 4 N f N c . 5 / 13

Grassmann integration After spatial link integration � ¯ e − S eff = D Ψ D ¯ Ψ(1+ T )Ψ Ψ e 3 � 1 + P (Ψ( n ) , Ψ( n + e i ) , ¯ Ψ( n ) , ¯ � � � Ψ( n + e i )) × i =1 n ∈ Λ After expanding product: Grassmann integration using Wick’s theorem. Note: (1 + T ) − 1 ( x , y ) ∼ δ x , y ⇒ Integration factorizes for Ψ’s with different spatial coordinates ⇒ Can be organized as an expansion of clusters of connected graphs on Λ s using the moment cumulant formalism. [Ruelle 69, M¨ unster 81] 6 / 13

Free Energy to NLO Perturbative treatment of effective theory for arbitrary N c and N f (degenerate) quark flavours: Free energy κ 2 N τ � z 11 � − f = log( z 0 ) − 6 N f , N c z 0 with the SU ( N c ) integrals � d W det(1 + h 1 W ) 2 N f , z 0 = SU ( N c ) � h 1 W � � d W det(1 + h 1 W ) 2 N f tr z 11 = , 1 + h 1 W SU ( N c ) µ − m where h 1 = (2 κ e a µ ) N τ = e T , m = − log(2 κ ). 7 / 13

Integration of temporal links The integrands only depend on the eigenvalues of the group element W . ⇒ Use eigenvalues for parametrization (reduced Haar measure). � d W det(1 + h 1 W ) 2 N f SU ( N c ) 1 �� � d ϕ i (1 + h 1 e i ϕ k ) 2 N f e i ( l − k + q ) ϕ i � = det (2 π ) N c 1 ≤ k , l ≤ N c q ∈ Z 2 N f �� �� 2 N f h pN c � = det . [Nishida 03] 1 i − j + p 1 ≤ i , j ≤ N c q =0 Specifically for N f = 1 z 0 = 1 + ( N c + 1) h N c 1 + h 2 N c . 1 8 / 13

Large N c limit

Nuclear liquid gas transition for general N c 0.994 0.998 1.002 1.006 1.01 9 / 13

Asymptotic Analysis For N f = 1 the κ 2 correction to the pressure: a 4 p 1 = − 6 κ 2 ( 1 2 N c ( N c + 1) h N c 1 + N c h 2 N c ) 2 1 ) 2 . N c (1 + h N c 1 (1 + N c ) + h 2 N c 1 h 1 < 1, for N c → ∞ expand about h N c = 0 1 a 4 p 1 = − 3 2 κ 2 N c ( N c + 1) 2 h 2 N c + O ( h 3 N c ) 1 1 ∼ − 3 2 κ 2 N 3 c h 2 N c . 1 h 1 > 1, expand about 1 / h N c = 0 1 a 4 p 1 = − 6 κ 2 N c + O (1 / h N c 1 ) ∼ − 6 κ 2 N c . 10 / 13

Asymptotic Analysis - results N f = 2, h 1 < 1: 1 1 + κ 4 3 N τ c h 2 N c c h 2 N c N 3 c h N c 1 − κ 2 48 a 4 N 7 800 a 4 N 8 + O ( κ 6 ) p ∼ 1 1 6 a 4 N τ 1 + κ 4 (9 N τ + 1) N τ 1 − κ 2 N τ 6 a 3 N 3 c h N c 24 a 3 N 7 c h 2 N c N 8 c h 2 N c + O ( κ 6 ) n B ∼ 1 1 1200 a 3 h 1 > 1: p ∼ 4 log( h 1 ) N c − κ 2 12 a 4 N c + κ 4 198 a 4 N c + O ( κ 6 ) N τ a 4 N 4 N 5 n B ∼ 4 a 3 − κ 2 N τ − κ 4 (59 N τ − 19) N τ c c + O ( κ 6 ) a 3 h N c 20 a 3 h N c 1 1 11 / 13

Conjecture large N c phase diagram ’t Hooft limit: N c → ∞ , hold λ = g 2 N c fixed [’t Hooft] [McLerran, Pisarski 09]: 12 / 13

Gauge corrections Include Gauge corrections using character expansion, to leading order: � 1 + 2 u F − u N t � 2 − f = log( z 0 ( h 1 , corr )) + κ 2 N t ( − 6 N f ) z 11 ( h 1 , corr ) F 1 − u F z 0 ( h 1 , corr ) N � � 1 + 2 u F − u N t �� F h 1 , corr = exp N t . 1 − u F In ’t Hooft limit [Gross, Witten 1979] u F = 1 λ ⇒ qualitative results unchanged. Open questions: Higher order corrections? Interchange strong strong-coupling and large N c limit? N c -dependence of a ? 13 / 13