Determining s by using the gradient flow in the quenched theory - PowerPoint PPT Presentation

Determining α s by using the gradient flow in the quenched theory Eliana Lambrou in collaboration with Szabolcs Borsanyi and Zoltan Fodor Bergische Universit¨ at Wuppertal 34th International Symposium on Lattice Field Theory Southampton, 25th July 2016

Introduction - Motivation

Current state of α s determinations • Many attempts to estimate α s and Λ parameter in literature • Summary and combined value by Flag Working Group [arXiv:1607.00299] • Criteria: • Renormalization scale: all points must have α eff < 0 . 2 • Perturbative behaviour: should be verified over a range of a factor 4 change in α n 1 eff (or α eff = 0 . 01 is reached) • Continuum extrapolation: at α eff = 0 . 3 have three lattice spacing with µ a < 0 . 5 for full O (a) improvement. • Finite-size effects: scale is determined in large enough volumes • Topology sampling 1

Current State of α s determination - quenched case Cont. Extrap. e Pert. Behav. l a c s . n e R Collaboration r 0 Λ MS Method ◦ ⋆ ⋆ CP-PACS 04 Schr¨ odinger Functional ◦ ⋆ ⋆ ALPHA 98 0.602(48) Schr¨ odinger Functional ◦ ◦ ⋆ L¨ uscher 93 0.590(60) Schr¨ odinger Functional ◦ ◦ 0 . 637( +32 Brambilla 10 ⋆ − 30 ) Heavy quark Potential ◦ ⋆ UKQCD 92 � 0.686(54) Heavy quark Potential ◦ ⋆ Bali 92 � 0.661(27) Heavy quark Potential ⋆ ⋆ ⋆ FlowQCD 15 0.618(11) Lattice spacing scale ◦ ⋆ ⋆ QCDSF/UKQCD 05 0.614(2)(5) Lattice spacing scale ⋆ SESAM 99 � � Lattice spacing scale ⋆ Wingate 95 � � Lattice spacing scale Davies 94 ⋆ � � Lattice spacing scale ◦ El-Khadra 92 ⋆ � 0.560(24) Lattice spacing scale ⋆ ⋆ Sternbeck 10 � 0 . 62(1) QCD vertices Ilgenfritz 10 ⋆ ⋆ � QCD vertices ◦ 0 . 59(1)( +2 Boucaud 08 ⋆ � − 1 ) QCD vertices ⋆ Boucaud 05 � � 0.62(7) QCD vertices In this talk: α s in the quenched case using the gradient flow 2

Gradient Flow - Setting the scale • Gradient Flow has many applications (scale setting, operator relation, topology etc . . . ) L¨ uscher (2010) • Simplest gauge invariant quantity: action density E ( t , x ) = 1 4 G a µν G a µν • Its expectation value � E ( t , x ) � serves as a non-perturbative definition of a reference scale • t 0 first introduced as a reference scale L¨ uscher (2010) t 2 � E ( t ) � � t = t 0 = 0 . 3 � • w 0 can also be used as a reference scale BMW Collaboration (2012) t d dt t 2 � E ( t ) � � 0 = 0 . 3 � t = w 2 3

Gradient Flow - Perturbative relation √ 8 t = 0.2 fm √ 8 t = 0.5 fm t 2 〈 E 〉 0.5 0.4 0.3 0.2 0.1 t 0 0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 t/r 02 Courtesy L¨ uscher (2010) Perturbative relation for its expectation value for QCD ( N A = 8) in MS scheme up to NNLO t 2 � E ( t ) � = 3 α s � 1 + α s k 1 + α 2 s k 2 + O ( α 3 � s ) 4 π k 1 = 1 . 09778674 L¨ uscher (2010) k 2 = − 0 . 9822456 Harlander and Neumann (2016) 4

Brute-Force determination of α s

Procedure Simulation details Lattices used β N a (in r 0 ) # cfgs • Use fine-lattices at T = 0 5.3570 48 0.05651 529 5.3669 48 0.05583 4680 • Keep the physical volume 5.4500 56 0.05011 215 constant LT c ≃ 2 5.4700 56 0.04911 222 5.5000 56 0.04690 197 • Periodic Boundary Conditions 5.5830 64 0.04178 161 5.6000 64 0.04115 200 • Tree-level Symanzik action, 5.8000 80 0.03229 400 Wilson flow, Clover-leaf 5.9500 96 0.02699 234 6.0500 112 0.02395 100 definition of observable 6.1500 128 0.02138 50 6.3600 160 0.01648 103 • Q = 0 configurations selected • w Q =0 / w 0 = 0 . 992(4) 0 • Use w 1 to set the scale: t d dt t 2 � E ( t ) � � 1 = 0 . 03 � t = w 2 • w 1 / r 0 = 0 . 115(2) 5

Improvement • Discretization correction terms at tree-level Fodor et al. (2014) t 2 � E ( t ) � = 3 α s C ( a 2 / t ) + O ( α s ) � � 4 π where ∞ a 2 m C ( a 2 / t ) = 1 + � C 2 m t m m =1 Coefficients known up to O ( a 8 ) • Finite-Volume correction Fodor et al. (2012) t 2 � E � = 3 α s 1 + δ ( t / L 2 ) � � 4 π where δ = 1 − 64 t 2 π + 8e − L 2 / 8 t + 24e − L 4 / 4 t + . . . 3 L 2 6

Improvement Improved vs Unimproved flow 0.06 0.05 0.04 t 2 <E> 0.03 0.02 β =6.05 β =6.05 impoved β =6.15 0.01 β =6.15 impoved β =6.36 β =6.36 impoved 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4  8 √  t / w 1 7

Improvement 0.05 0.045 0.04 t 2 E 0.035 sqrt(8t)/w 1 =1.0, unimproved 0.03 sqrt(8t)/w 1 =1.0, improved sqrt(8t)/w 1 =0.6, unimproved sqrt(8t)/w 1 =0.6, improved 0.025 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 (a/w 1 ) 2 8

Λ parameter 1. t 2 � E ( t ) � ⇒ α s from perturbative relation 2. Use 4-loop β -function in the MS -scheme to run α s at a high scale Ritbergen, Vermaseren, Larin (1997) t 2 � E ( t ) � = 3 α s 1 + α s k 1 + α 2 s k 2 + O ( α 3 � � s ) 4 π Λ -parameter from flow 0.7 0.695 0.69 0.685 0.68 r 0 Λ 0.675 0.67 0.665 0.66 0.655 0.05 0.1 0.15 0.2 0.25 0.3 √  8  t / r 0 9

Λ parameter t 2 � E ( t ) � = 3 α s 1 + α s k 1 + α 2 s k 2 + α 3 s k 3 + O ( α 4 � � s ) 4 π Λ -parameter from flow including k 3 k 3 =-2.0 0.74 k 3 =0.0 k 3 =2.0 0.72 0.7 r 0 Λ 0.68 0.66 0.64 0.05 0.1 0.15 0.2 0.25 0.3 √  8  t / r 0 10

Λ parameter We want to eliminate k 3 contribution t dt 2 � E � A ( t ) ≡ ( t 2 � E � ) 2 + C � � dt � 3 β 1 � �� (4 π ) 2 + 3 β 0 C 9 18 k 1 (4 π ) 3 + 6 k 1 β 0 � � = α 2 + α 3 (4 π ) 2 + C s s (4 π ) 2 (4 π ) 2 � 3 β 2 � �� 9( k 2 1 + 2 k 2 ) (4 π ) 4 + 6 k 1 β 1 (4 π ) 3 + 9 k 2 β 0 + α 4 + C s (4 π ) 2 (4 π ) 2 � 3 β 3 � �� 9(2 k 1 k 2 + 2 k 3 ) (4 π ) 5 + 6 k 1 β 2 (4 π ) 4 + 9 k 2 β 1 (4 π ) 3 + 12 k 3 β 0 + α 5 + C s (4 π ) 2 (4 π ) 2 By requiring combination of k 3 -terms to be zero ⇒ C = − 0 . 13636364 11

Λ parameter We follow the same procedure as previously but now α s determined via A ( t ) function Λ -parameter from A(t) 0.72 0.7 0.68 r 0 Λ 0.66 0.64 α =0.1708 α =0.2757 0.62 0.6 0.05 0.1 0.15 0.2 0.25 0.3  8 √  t / r 0 r 0 Λ=0.664(14) 12

Step-scaling

Step-Scaling procedure • Lattice sizes 14,16,20,24,28,32,40,48 • Choose c-value (0.1,0.12) ⇒ t = ( cN ) 2 • For each β of pairs ( N , 2 N ) find the difference 1 1 � � D ( t 2 � E � � µ ) = 2 µ − � � � t 2 � E � t 2 � E � � � µ • Find the function D in the continuum 1 1 µ = 4 π 1 1 � � � �� α s ( µ ) + (2 k 2 2 µ − α s (2 µ ) − 1 − k 2 ) � α (2 µ ) − α ( µ ) � � t 2 � E � t 2 � E � 3 � � = 4 π � 2 β 0 π ln2 + 2 β 1 π 2 ln24 π � � 3 t 2 � E � µ + . . . � 3 � • Keep w 1 / L fixed and use 14,16,20,24 to do step scaling 13

Step-scaling function D for c = 0 . 1 Step scaling function 6.8 6.6 6.4 6.2 6 D 5.8 5.6 5.4 Step-scaling function pair 14-28 5.2 pair 16-32 5 pair 20-40 pair 24-48 4.8 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 t 2 <E> 14

Λ parameter from T = 0 and step-scaling Λ -parameter using step scaling c=0.10 from w 1 /L=0.10 0.69 c=0.10 from w 1 /L=0.20 c=0.12 from w 1 /L=0.10 c=0.12 from w 1 /L=0.20 0.68 0.67 r 0 Λ 0.66 0.65 0.64 0.63 1e-05 0.0001 0.001 0.01 0.1 1 10  8 √  t / r 0 • c=0.10 from w 1 / L = 0 . 10 ⇒ t 2 � E � = 0 . 0107 • c=0.10 from w 1 / L = 0 . 20 ⇒ t 2 � E � = 0 . 0102 ⇒ For all α s < 0 . 01 • c=0.12 from w 1 / L = 0 . 10 ⇒ t 2 � E � = 0 . 0109 • c=0.12 from w 1 / L = 0 . 20 ⇒ t 2 � E � = 0 . 0103 15

Conclusions • We present a way of determining the Λ-parameter (and α s ) using the gradient flow • By brute-force elements we find a good plateau for the Λ-parameter that meets the criteria of a good α s determination ⇒ r 0 Λ = 0.664(14) • Step-scaling results are also in a good agreement with those from T = 0 simulations • Still work in progress so... be patient for final results soon 16

Thank you for your attention! 16

w 1 to w 0 and r 0 relations • w 1 / w Q =0 = 0.340(7) 0 • w 0 / r 0 = 0.341(2) Sommer et al. (2014) fixed volume: L=2/T c fixed volume: L=2/T c 1.005 0.35 Q=0 /w 0 w 0 plain improved 1 0.345 [Q=0] /w [Q=0] 0.995 0.34 0 0.99 0.335 w 1 0.985 0.33 0.98 0.325 0.975 0.32 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 (a/w 0 ) 2 (a/w 1 ) 2

Continuum function D -Sketch of analytic derivation up to NLO From d α s / d ln µ we can find d (1 /α s ) = − 1 d ln µ = 2 β 0 d α s + 2 β 1 π 2 α s + 2 β 2 π 3 α 2 s α 2 d ln µ π s Then we can find the difference � ln2 µ 1 1 d (1 /α s ) d ln ( µ ′ /µ ) d ln ( µ ′ /µ ) α s (2 µ ) − α s ( µ ) = ln µ � ln2 � ln2 = 2 β 0 π ln µ + 2 β 1 α s ( µ ′ ) dln ( µ ′ /µ ) + 2 β 2 α 2 s ( µ ′ ) dln ( µ ′ /µ ) π 2 π 3 0 0 Using similar procedure we find α s (2 µ ) − α s ( µ ) Then we fit using 5 . 0831+15 . 71 x + ax 2 + bx 3 + cx 4 + dx 2 1 N 2 + ex 3 1 N 2 + fx 4 1 N 2 + g (5 . 0831+15 . 71 x ) 1 N 2