Functional Fit Approach (FFA) for Density of States method: SU(3) spin system and SU(3) gauge theory with static quarks Mario Giuliani Christof Gattringer, Pascal Törek Karl Franzens Universität Graz Mario Giuliani (Universität Graz) Southampton, 26th July 2015 1 / 22
Sign Problem: a brief review Different approaches to solve the sign problem: – Reweighting – Expansion methods – Stochastic differential equations – Mapping to dual variables – Et cetera... Density of states approach: – Method used FFA Functional Fit Approach ( ARXIV: 1503.04947, 1607.07340 ) – See also LLR Linear Logarithmic Relaxation by K. Langfeld, B. Lucini and A. Rago ( ARXIV:1204.3243, 1509.08391 ) Mario Giuliani (Universität Graz) Southampton, 26th July 2015 2 / 22
Density of States Method In QFT we want to compute for our theory: � �O� = 1 � D [ ψ ] e − S [ ψ ] D [ ψ ] O [ ψ ] e − S [ ψ ] Z = Z In the density of states approach we divide the action into two parts: S [ ψ ] = S ρ [ ψ ] + c X [ ψ ] * S ρ [ ψ ] and X [ ψ ] are real functionals of the fields ψ * S ρ [ ψ ] is the part of the action that we include in the weighted density ρ * Here c is purely imaginary: c = i ξ Mario Giuliani (Universität Graz) Southampton, 26th July 2015 3 / 22
Density of States Method The weighted density is defined as: � D [ ψ ] e − S ρ [ ψ ] δ ( X [ ψ ] − x ) ρ ( x ) = Using ρ ( x ) we can write Z and �O� s: � x max � x max �O� = 1 dx ρ ( x ) e − i ξ x dx ρ ( x ) e − i ξ x O [ x ] Z = Z x min x min → ψ ′ such that we can write: Usually there is a symmetry ψ − � x max Z = 2 dx ρ ( x ) cos ( ξ x ) 0 ... and for the observables O : � x max �O� = 2 dx ρ ( x ) { cos ( ξ x ) O even ( x ) − i sin ( ξ x ) O odd ( x ) } Z 0 Where O even = O ( x )+ O ( − x ) and O odd = O ( x ) −O ( − x ) 2 2 Mario Giuliani (Universität Graz) Southampton, 26th July 2015 4 / 22
First example: SU(3)-spin model Arxiv:1607.07340 SU ( 3 ) spin model is a 3D effective theory for heavy dense QCD Relevant d.o.f. is the Polyakov loop P ( n ) ∈ SU ( 3 ) (static quark source at n ) The model has a real and positive dual representation ⇒ reference data We have an action: 3 � TrP ( n ) TrP ( n + ν ) † + c . c . � [ e µ TrP ( n )+ e − µ TrP ( n ) † ] � � � S [ P ] = − τ − κ n ν = 1 n The action depends only on the trace ⇒ simple parametrization e i θ 1 ( n ) 0 0 e i θ 2 ( n ) P ( n ) = 0 0 e − i ( θ 1 ( n )+ θ 2 ( n )) 0 0 Mario Giuliani (Universität Graz) Southampton, 26th July 2015 5 / 22
Definition of density of states We define the weighted density of states with S ρ [ P ] = Re [ S [ P ]] and Im [ S [ P ]] = 2 κ sinh ( µ ) X [ P ] : � D [ P ] e − S ρ [ P ] δ ( x − X [ P ]) ρ ( x ) = x ∈ [ − x max , x max ] Symmetry P ( n ) → P ( n ) ∗ implies ρ ( − x ) = ρ ( x ) This simplifies the partition function: x max x max � � Z = dx ρ ( x ) cos ( 2 κ sinh ( µ ) x ) = 2 dx ρ ( x ) cos ( 2 κ sinh ( µ ) x ) − x max 0 x max � = 2 � � � � � �O X dx ρ ( x ) O E ( x ) cos ( 2 κ sinh ( µ ) x ) + i O O ( x ) sin ( 2 κ sinh ( µ ) x ) Z 0 Mario Giuliani (Universität Graz) Southampton, 26th July 2015 6 / 22
Parametrization of the density ρ ( x ) Ansatz for the density: ρ ( x ) = e − L ( x ) , normalization ρ ( 0 ) = 1 ⇒ L ( 0 ) = 0 We divide the interval [ 0 , x max ] into N intervals n = 0 , 1 , . . . , N − 1. L ( x ) is continuous and linear on each of the intervals, with a slope k n : 5 k N − 1 4 k N − 2 3 k 6 k 5 k 4 2 k 3 k 1 k 2 1 k 0 0 ∆ N − 2 ∆ N − 1 ∆ 0 ∆ 1 ∆ 2 ∆ 3 ∆ 4 ∆ 5 ∆ 6 l max Mario Giuliani (Universität Graz) Southampton, 26th July 2015 7 / 22
Determination of the slopes k n How do we find the slopes k n ? Restricted expectation values which depend on a parameter λ ∈ R : 1 � D [ P ] e − S ρ [ P ]+ λ X [ P ] O � � � � ��O�� n ( λ ) = X [ P ] θ n X [ P ] Z n ( λ ) � D [ P ] e − S ρ [ P ]+ λ X [ P ] θ n � � Z n ( λ ) = X [ P ] � 1 for x ∈ [ x n , x n + 1 ] � � θ n x = 0 otherwise Update with a restricted Monte Carlo Vary the parameter λ to fully explore the density Mario Giuliani (Universität Graz) Southampton, 26th July 2015 8 / 22
Functional Fit Approach FFA Expressed in terms of the density: x max x n + 1 x n + 1 � � � dx ρ ( x ) e λ x = c dx ρ ( x ) e λ x θ n � � dx e ( − k n + λ ) x Z n ( λ ) = = x − x max x n x n = c e ( λ x − k n ) x n + 1 − e ( λ x − k n ) x n λ − k n For computing the slopes we use as observable X [ P ] : x n + 1 1 � dx ρ ( x ) e λ x x = ∂ � � �� X [ P ] �� n ( λ ) = Z n ( λ ) ∂λ ln Z n ( λ ) x n Mario Giuliani (Universität Graz) Southampton, 26th July 2015 9 / 22
Functional Fit Approach FFA Explicit expression for restricted expectation values: n − 1 1 � � − 1 � � � �� X [ P ] �� n ( λ ) − ∆ j 2 = h ( λ − k n )∆ n ∆ n j = 0 1 − e − r − 1 1 r − 1 h ( r ) = 2 Strategy to find k n : Evaluate �� X [ P ] �� n ( λ ) for different values of λ 1 Fit these Monte Carlo data h (( λ − k n )∆ n ) 2 k n are obtained from simple one parameter fits 3 The quality of the fit provide a self-consistent check of our simulation Mario Giuliani (Universität Graz) Southampton, 26th July 2015 10 / 22
Fit of slopes ⇒ density ρ ( l ) Example: 8 3 , κ = 0 . 005 , µ = 0 . 0 τ = 0 . 075 1 2 - 0 0.5 ) 0 j 20 ∆ ln( ρ (x)) n-1 ∑ j=0 0.4 40 )- -2000 λ 0.3 ( 60 n 〉 〉 F[P] 80 0.2 -4000 〈 100 τ =0.025 〈 ( 0.1 n 1 120 τ =0.050 ∆ -6000 0 140 τ =0.075 τ =0.100 − 160 0.1 -8000 τ =0.125 180 − 0.2 τ =0.150 − 0.3 -10000 − 0.4 -12000 − 0.5 − − 4 2 0 2 4 λ 0 200 400 600 800 1000 1200 1400 x ρ ( x ) = e − L ( x ) L ( x ) k n Mario Giuliani (Universität Graz) Southampton, 26th July 2015 11 / 22
Observables 1 Particle number density n : x max n = 1 1 ∂ sinh ( µ ) ln Z = 1 2 ∂ � dx ρ ( x ) sin ( 2 κ sinh ( µ ) x ) x V 2 κ V Z 0 2 ... and the corresponding susceptibility χ n : χ n = 1 ∂ ∂ sinh ( µ ) n 2 κ � 2 x max x max � 2 = 1 � 2 � � � dx ρ ( x ) cos ( 2 κ sinh ( µ ) x ) x 2 + dx ρ ( x ) sin ( 2 κ sinh ( µ ) x V Z Z 0 0 Mario Giuliani (Universität Graz) Southampton, 26th July 2015 12 / 22
Comparison with dual approach With large statistic and small intervals we are able to explore results up to µ = 4: Particle number density n Lattice 8 3 , τ = 0 . 130 and κ = 0 . 005: 0.07 n Density of states Dual formulation 0.06 0.05 0.04 0.03 0.02 0.01 0.00 µ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 We find a good agreement for chemical potential up to µ ≈ 4 . 0 Mario Giuliani (Universität Graz) Southampton, 26th July 2015 13 / 22
Comparison with dual approach Susceptibility χ n Lattice 8 3 , τ = 0 . 130 and κ = 0 . 005: 3.5 χ n 3.0 2.5 2.0 1.5 1.0 Density of states Dual formulation 0.5 0.0 µ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 The density performs fine up to µ ≈ 4 Mario Giuliani (Universität Graz) Southampton, 26th July 2015 14 / 22
Comparison with dual approach We can also go to bigger lattice size with the same parameters: Particle number density n Lattice 12 3 , τ = 0 . 130 and κ = 0 . 005: 0.07 n Density of states Dual formulation 0.06 0.05 0.04 0.03 0.02 0.01 0.00 µ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 We find a good agreement for chemical potential up to µ ≈ 4 . 0 Mario Giuliani (Universität Graz) Southampton, 26th July 2015 15 / 22
Comparison with dual approach Susceptibility χ n Lattice 12 3 , τ = 0 . 130 and κ = 0 . 005: 3.5 χ n 3.0 2.5 2.0 1.5 1.0 Density of states Dual formulation 0.5 0.0 µ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 The density performs fine up to µ = 4 Mario Giuliani (Universität Graz) Southampton, 26th July 2015 16 / 22
We find smaller error bars for larger µ 2 π The oscillating factor is bigger, but we still have: ∆ n ≪ 2 κ sinh µ This can be explained looking at the different densities: ln( ρ (x)) 0 -50 µ =0.1 -100 µ =0.5 µ =1.0 µ =1.5 -150 µ =2.0 µ =2.5 µ =3.0 -200 µ =3.5 µ =4.0 -250 0 100 200 300 400 500 600 x The changed shape of the density above µ ≈ 2 . 25 weakens the piling up of the errors on the singles k n Mario Giuliani (Universität Graz) Southampton, 26th July 2015 17 / 22
Second example: SU(3) static quarks Further step towards a real QCD system SU ( 3 ) static quarks is a 4D effective theory for heavy dense QCD A SU ( 3 ) gauge theory plus the static quarks represented by Polyakov loops We have the following action: S [ U ] = − β � � � n ) † � � � e µ N T � n ) + e − µ N t � TrU µν ( n ) − κ P ( � P ( � Re 3 n µ<ν n � n � Where the Polyakov loops are: N T − 1 n ) = 1 � P ( � 3 Tr U 4 ( � n , n 4 ) n 4 = 0 Mario Giuliani (Universität Graz) Southampton, 26th July 2015 18 / 22
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