Charmonium spectral functions from 2+1 flavour lattice QCD asztor 12 - PowerPoint PPT Presentation

Charmonium spectral functions from 2+1 flavour lattice QCD asztor 12 Attila P´ apasztor@bodri.elte.hu In collaboriation with anyi 3 urr 34 Z. Fodor 134 C. Hoelbling 3 S. Bors´ S. D¨ S. D. Katz 12 S. Krieg 34 S. Mages 5 adi 1 afer 5 D. N´ ogr´ A. Sch¨ oth 3 as 1 o 34 B. C. T´ N. Trombit´ K. K. Szab´ 1E¨ otv¨ os University 2 MTA-ELTE Lattice Gauge Theory Group 3 University of Wuppertal 4 J¨ ulich Supercomputing Center 5 University of Regensburg BGZ Triangle Meeting, 2014, Balatonf¨ ured

Motivation Motivation • We will consider correlators of charmonium currents in the pseudoscalar(PS) and vector(V) channels, i.e. ¯ c γ 5 c and � 3 i =1 ¯ c γ i c • These correspond to the η c and J / Ψ mesons • J / Ψ suppression is regarded as an important signal of QGP formation • Charmonium states are regarded as a possible thermometer for the QGP • This study is with 2+1 dynamical quarks, previous studies: mostly quenched, one in 2 flavour The talk is based on: JHEP 1404 (2014) 132

Preliminaries Spectral function ∼ im part of real-time retarded correlator A ( ω ) = (2 π ) 2 � e − E n / T − e − E m / T � |� n | J H (0) | m �| 2 δ ( p − k n + k m ) � Z m , n Quasipaticle ∼ peaks in the SF, melting/dissociation ∼ peaks not well separated anymore, they become part of the continuum Relation to the Euclidean time correlator � ∞ p ) K ( ω, τ ) where K ( ω, τ ) = cosh( ω ( τ − 1 / 2T )) G ( τ,� p ) = d ω A ( ω,� sinh( ω/ 2T ) 0 Left hand side is calculable with lattice techniques. This is very hard to invert.

The Maximum Entropy Method The method in a nutshell Q = α S − 1 2 χ 2 � A ( ω ) � �� � S = d ω A ( ω ) − m ( ω ) − A ( ω ) log m ( ω ) χ 2 = � ( G fit − G data ) C − 1 ij ( G fit − G data ) i i j j i , j � G i = A ( ω ) K ( ω, τ i ) d ω m ( ω ) is a function, summarizing our prior knowledge of the solution. Then we average over α . The conditional probability P [ α | data , m ] is given by Bayes’ theorem.

The Maximum Entropy Method Conclusions from mock data analysis • The peak positions OK, the shapes not so much • With not too noise data points, O(10) points are OK • Features that remain unchanged by varying the prior are restricted by the data • Resultion becomes worse at bigger ω • Peaks close in position can be merged into one broader peak. Take home message: the stable thing is just the position of the first (ground state)

Simulation details Lattice details Gauge action = Symanzik tree-level improved gauge action Fermion action = 2+1 dynamical Wilson fermions with 6 step stout smearing ( ρ = 0 . 11) and tree-level clover improvement Same configurations as in JHEP 1208 (2012) 126 a = 0 . 057(1)fm m π = 545MeV N s = 64 N t = 28 ... 12 T = 123 ... 288MeV We measured the charm meson correlators on these lattices.

Outline of MEM procedure Stability test at the lowest temp • Drop data points, emulating the number of data points available at the lowest temperature ( N t = 28) • Do the same analysis as with the higher temperature correlators. If the ground state peak cannot be reconstructed, the given number of data points is not reliable • RESULT: N t =12 NOT OK, N t =14,16,18,20 OK Error analysis • Systematic error analysis: vary ∆ ω , N ω , the shape of the prior function: m 0 , m 0 ω 2 , 1 / ( m 0 + ω ) , m 0 ω and m 0 =0.01, 0.1, 1.0, 10.0. • Statistical error analysis: given set of parameters, 20 jackknife samples

Sensitivity on prior function This is the PS channel, but V looks similar PSEUDOSCALAR CHANNEL, N t =20, T/T c ≈ 0.9 0.5 m( ω )=0.01/a 2 0.45 m( ω )=0.1 ω 2 0.4 m( ω )=1.0/a 2 /(10.0+ ω a) 0.35 0.3 A( ω )/ ω 2 0.25 0.2 0.15 0.1 0.05 0 0.5 1 1.5 2 2.5 3 3.5 ω a

Temperature dependence PSEUDOSCALAR CHANNEL, m( ω ) = 0.1 ω 2 0.25 N t =14, T/T c ≈ 1.30 N t =16, T/T c ≈ 1.14 N t =18, T/T c ≈ 1.01 0.2 N t =20, T/T c ≈ 0.91 0.15 A( ω )/ ω 2 0.1 0.05 0 0 0.5 1 1.5 2 2.5 3 3.5 a ω

Temperature dependence Vector channel VECTOR CHANNEL, m( ω ) = 0.1 ω 2 0.25 N t =14, T/T c ≈ 1.30 N t =16, T/T c ≈ 1.14 N t =18, T/T c ≈ 1.01 0.2 N t =20, T/T c ≈ 0.91 0.15 A( ω )/ ω 2 0.1 0.05 0 0 0.5 1 1.5 2 2.5 3 3.5 a ω

Results - MEM Pseudoscalar channel - position of 1st peak 1.15 Position of first peak in lattice units 1.1 1.05 1 0.95 0.9 0.85 0.8 0.75 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 aT=1/N t

Results - MEM Vector channel - position of 1st peak 1.25 Position of first peak in lattice units 1.2 1.15 1.1 1.05 1 0.95 0.9 0.85 0.8 0.75 0.7 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 aT=1/N t

Charm diffusion coefficient Kubo-formula 3 D = 1 A ii ( ω, T ) � 6 χ lim , ω ω → 0 i =1 If D > 0 we have ρ/ω > 0 for small ω implying a transport peak Narrow transport peak In the case of a narrow transport peak, we can use the ansatz: A transport ( ω, T ) = f ( T ) ωδ ( ω − 0 + ) This does not mean, that the diffusion coefficient is infinite. But in case of a narrow transport peak, the Euclidean correlator � G ( τ, T ) = K ( ω, τ ) A ( ω, T ) is not sensitive to the full shape of the peak, only the area. The contrubtion of the transport peak will be a temperature dependent constant (zero mode).

Some indication of a transport peak N t = 16 not conclusive 0.009 m( ω )=0.01 a 2 m( ω )=a -2 /(10+ ω a) m( ω )=0.1 ω 2 m( ω )=1.0 ω /a 0.006 m( ω )=0.01 ω /a A( ω )/( ω T) 0.003 0 0 0.025 0.05 ω a

A different method Definition of G / G rec Datta, Karsch, Petreczky, Wetzorke 2004 G ( t , T ) G ( t , T ) G rec ( t , T ) = � A ( ω, T ref ) K ( ω, t , T ) d ω Midpoint subtracted version G − G ( t , T ) − G ( N t / 2 , T ) = G rec ( t , T ) − G rec ( N t / 2 , T ) = G − rec G ( t , T ) − G ( N t / 2 , T ) � A ( ω, T ref ) [ K ( ω, t , T ) − K ( ω, N t / 2 , T )] d ω This removes the zero mode.

Results: G / G rec Pseudoscalar channel 1.06 1.04 1.02 G/G rec 1 0.98 N t =12 N t =14 0.96 N t =16 N t =18 0.94 N t =20 2 4 6 8 10 t/a

Results: G / G rec Vector channel 1.15 1.1 1.05 G/G rec 1 N t =12 N t =14 0.95 N t =16 N t =18 N t =20 0.9 2 4 6 8 10 t/a

Results: G − / G − rec Pseudoscalar channel, midpoint subtracted version 1.1 1.05 rec G - /G - 1 N t =12 0.95 N t =14 N t =16 N t =18 N t =20 0.9 2 4 6 8 10 t/a

Results: G − / G − rec Vector channel, midpoint subtracted version 1.1 1.05 rec G - /G - 1 N t =12 0.95 N t =14 N t =16 N t =18 N t =20 0.9 2 4 6 8 10 t/a

Results: G − G rec Vector channel 1e-05 5e-06 G-G rec 0 -5e-06 N t =16 N t =14 N t =12 -1e-05 2 4 6 8 10 t/a

Conclusions MEM analysis • There seems to be no change in the SF in the PS channel up to 1 . 4 T c • There seems to be some change in SF in the V channel • Indications of a transport peak in the V channel G / G rec analyis • No change in the PS channel • In the V channel, results are consistent with a temperature independent ω > 0 part and a temperature dependent zero mode (narrow transport peak), described by the ansatz A ( ω ) = f ( T ) ωδ ( ω − 0 + ) + A ( ω, T = 0).

Backup - implementation details 1 MEM continued... It can be shown, that the maximum of Q is in an N data dimensional subspace: � N data � � A ( ω ) = m ( ω ) exp s i f i ( ω ) i =1 Two choice for basis functions: Bryan (Eur. Biophys J. 18, 165 (1990)) or Jakov´ ac et al (Phys.Rev. D75 014506 (2007). We use the latter. In this case the maximization of Q is equivalent to the minimization of � ω max N data N data U = α � � G data s i C ij s j + d ω A ( ω ) − s i . i 2 0 i , j =1 i = 1 Comment: Have to use arbitrary precision arithmetics with both methods.

Backup - implementation details 2 Problem: stopping criterion -0.25 -0.3 -0.35 -0.4 -0.45 U -0.5 -0.55 -0.6 -0.65 -0.7 2000 4000 6000 8000 10000 12000 14000 16000 # of iteration steps

Backup - implementation details 3 Solution: going back to the N ω dimensions 6.3 6.25 6.2 6.15 -Q 6.1 6.05 6 5.95 5000 10000 15000 20000 25000 30000 35000 40000 # of iteration steps

Backup - charm mass tuning From Davies et al PRL 104, 132003 (2010) m c / m s = 11 . 85. Because of additive renormalization, it is impossible to use this directly. We use ( m c − m s ) / ( m s − m ud ) where the additive renormalization constant cancels. We know that for ud and s the masses used in the simulation correspond to a mass ratio of 1.5 (Durr et al. Phys. Lett. B701 (2011) 265), from this we get ( m c − m s ) / ( m s − m ud ) = 32 . 55 We check if the meson masses are indeed in the right ballpark: J P m i ma ma / m D ∗ s a m exp / m D ∗ s 0 − m s , m c D s 0.54(1) 0.95(2) 0.932 0 − m c , m c η c 0.8192(7) 1.437(4) 1.411 1 − m s , m c D ∗ 0.570(1) 1 1 s 1 − m c , m c J / Ψ 0.8388(8) 1.472(2) 1.466 3 / 2 + 3 m s Ω 0.478(8) 0.84(2) 0.791