Confinement in Polyakov gauge

Transcription

1 Confinement in Polyakov gauge Florian Marhauser arxiv:

2 QCD Phase Diagram chiral vs. deconfinement phase transition finite density critical point...

3 Confinement Order Parameter ( β ) φ( x) = L( x) = 1 N c Tr c P exp ig dτ A ( x, τ) puts a static quark into the theory Ψ τ = iga Ψ Ψ( x, τ) =P exp ( ig τ dτ A ) Ψ( x, )

4 Confinement Order Parameter ( β ) φ( x) = L( x) = 1 N c Tr c P exp ig dτ A ( x, τ) puts a static quark into the theory Ψ τ = iga Ψ Ψ( x, τ) =P exp ( ig τ dτ A ) Ψ( x, ) quark propagating in time j µ = δ µ β dτδ(x x(τ))

5 Polyakov Loop φ Tr c P e ig R d 4 xa µ (x) j µ (x) = exp ( βf q ) δ µ β dτδ(x x(τ)) Energy of an infinitely heavy quark

6 Polyakov Loop φ Tr c P e ig R d 4 xa µ (x) j µ (x) = exp ( βf q ) δ µ β dτδ(x x(τ)) Energy of an infinitely heavy quark Confinement F q φ = Deconfinement F q < φ Perturbation Theory Expansion around F q <

7 Polyakov Loop φ Tr c P e ig R d 4 xa µ (x) j µ (x) = exp ( βf q ) δ µ β dτδ(x x(τ)) Energy of an infinitely heavy quark Confinement F q φ = Deconfinement F q < φ Perturbation Theory Expansion around F q < Gauge group Universality class PT order SU(2) Ising 2nd SU(3) Potts 1st

8 Order Parameter reformulation order parameter: φ = L[A (x)] with the Jensen inequality L[ A ] L[A ] and properties of the Polyakov loop, we can show T < T c : L[ A ]= 1 2 gβ A ( x) = π 2, T > T c : L[ A ] 1 2 gβ A ( x) < π 2

9 Order Parameter reformulation order parameter: φ = L[A (x)] with the Jensen inequality L[ A ] L[A ] and properties of the Polyakov loop, we can show T < T c : L[ A ]= 1 2 gβ A ( x) = π 2, T > T c : L[ A ] 1 2 gβ A ( x) < π 2 L[ A (x) ] also serves as an order parameter

10 Gauge φ( x) = L( x) = 1 N c Tr c P exp ( ig β dτ A ( x, τ) ) Polyakov gauge in SU(2): A (x) =A c ( x)σ c Pauli matrix time-independent, removes path ordering L( x) = 1 [e Tr igβac( x)σc] [ ] g c = cos N c 2 βac ( x) compute in polyakov gauge via L[ A ] V eff ( A )

11 Perturbative Treatment (1-loop) Potential is periodic with period 2π [ g ] L( x) = cos 2 βac ( x) V Weiss.8 Weiss Potential V Weiss non-pert. Potential Π 2 Π 3 Π 2 2 Π g Β A c Π 2 Π 3 Π 2 2 Π g Β A c deconfining minimum away from edge gβ 2 Ac = {, π} L =1 smoothly 2 nd order PT

12 Non-Perturbative Treatment FRG Method: scale dependent effective action Γk include quantum fluctuations by lowering scale action Γ Γ k S cl scale k Λ

13 Non-Perturbative Treatment FRG Method: scale dependent effective action Γk include quantum fluctuations by lowering scale action Γ Γ k S cl scale k Λ flow equation for effective action [ ( Wetterich s equation ) ] k k Γ k = t Γ k = 1 2 Tr 1 t R k + R k Γ (2) k

14 Parametrisation [ g ] L( x) = cos 2 βac ( x) focus on A c effective potential sufficient to generate confinement } Γ k = dτ d 3 x { 1 2 Ac 2 A c + V k (A c ) +Γ ψ

15 Parametrisation [ g ] L( x) = cos 2 βac ( x) focus on A c effective potential sufficient to generate confinement } Γ k = dτ d 3 x { 1 2 Ac 2 A c + V k (A c ) +Γ ψ spatial gluons integrated out V k (A c )=V,k (A c )+ V k (A c ) analytical result

16 ^ ^ Flow of spatial gluons.8.6 ^ k = ^ k = 1.5 ^ k = 2 ^ k = 3 ^ k = 4 V,k.4.2 π / 2 π 3/2 π 2 π ϕ

17 Coupling for small momenta α s pβ 1 α s (p 2 ).1 T = MeV T = 15 MeV T = 3 MeV T = 6 MeV p [GeV] in flow identify k = p

18 Flow of V eff V Α 7 Α 6.1 Α 5.5 Α 4 Α α = kβ.5 Π 2 Π 3 Π 2 2 Π φ ϕ = gβa c.1 expectation value A c = V (Ac ) from A c A c = A c

19 ^ Effective Potential L[ A c ] = cos [ ] gβ 2 Ac phase transition T c = MeV V eff -.1 T c / σ = T = 5 MeV T = 4 MeV T = 35 MeV T = 3 MeV T = 25 MeV for comparison T c,lattice σ =.71 π /2 π 3/2 π 2 π ϕ

20 Phase Diagram L.4 α s (p 2 ).1 T = MeV T = 15 MeV T = 3 MeV T = 6 MeV p [GeV] zero temperature coupling finite temperature coupling T / T c

21 Phase Diagram polyakov gauge vs. landau gauge Landau gauge: Braun, Gies, Pawlowski, arxiv: L.4.2 Landau gauge Polyakov gauge T / T c

22 Critical Exponent ν 1.2 x 1-3 m 2 2 V (A c ) (A c )2 ϕ=ϕmin 8 x 1-4 m 2 4 x x 1-4 fit ν =.65 data (T-T c ) / T c

23 Outlook - SU(3) FM, J.Pawlowski, work in progress 2 instead of 1 variable no conceptual problems numerically demanding need 2D grid various methods for solving however, no full solution yet

24 Outlook - SU(3) FM, J.Pawlowski, work in progress 2 instead of 1 variable no conceptual problems numerically demanding need 2D grid various methods for solving however, no full solution yet

25 Outlook - SU(3) FM, J.Pawlowski, work in progress 2 instead of 1 variable no conceptual problems numerically demanding need 2D grid various methods for solving however, no full solution yet Yes we can!

26 Outlook - SU(3) phase transition - preliminary FM, J.Pawlowski, work in progress V(gβA3, ) st order phase transition between 2-3 MeV

27 Outlook - SU(3) key ingredient I: adequate grid FM, J.Pawlowski, work in progress Weyl chamber A8 A8 A3 A3 full information in one Weyl chamber key ingredient II: advantageous rewriting of the equations

28 Thank you for your attention

29 Backup slides

30 order parameter reformulation center symmetric phase L = 1 2 gβ A = arccos L = arccos L center trafo write L ZL δl ZδL L = L + δl 1 δl + O ( δl 2 ) 1 L 2 it follows δl 2n+1 = Z δl 2n+1 = all even powers vanish because arccos is an odd funtion 1 2 gβ A = arccos L = π 2 center broken phase L[ A ] L[A ] L > L > 1 2 gβa < π 2

31 gauge fixing time-independence and rotation into Cartan tr σ 3 A =, tr (σ 1 ± iσ 2 )A =, residual gauge fixing 1 dx tr σ 3 A 1 =, 2 dx dx 1 tr σ 3 A 2 =, 3 dx dx 1 dx 2 tr σ 3 A 3 = Faddeev-Popov determinant [ ( ) ] FP [A] = (2T ) 2 sin 2 ga 3 ( x) 2T x cancels longitudinal gauge fields in the static approximation

32 integrating out no backreaction of temporal on spatial gauge fields Flow schematically given by t Γ k = 1 2 Tr ( Γ (2) k 1 + R A ) [ t R k + Tr t ln(s (2) ] YM + R A) ii. k (k) 4T { ( V,k = (2π) 2 dpp 2 ln 1 2 cos(ϕ)e βk (k) ) } +e 2βk (k) ln(1 2 cos(ϕ)e βp + e 2βp ) + V W

33 ^ ^ matching scales T k 2 f(k/t ) = T n= 1 ω 2 n + k k / k ^ k

34 Deconfinement vs. χsb lattice results Karsch: T conf = T χsb Tr Fodor: T conf >T χsb (3)(3) Δ l,s.6.4 p4fat3: N τ = (2)(4).2 T [MeV] L ren Tr (3)(4) N τ =4 6 8 T [MeV] T[MeV]