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

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

Determining 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,


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SLIDE 1

Determining αs 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

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SLIDE 2

Introduction - Motivation

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SLIDE 3

Current state of αs determinations

  • Many attempts to estimate αs and Λ parameter in literature
  • Summary and combined value by Flag Working Group [arXiv:1607.00299]
  • Criteria:
  • Renormalization scale: all points must have αeff < 0.2
  • Perturbative behaviour: should be verified over a range of a factor 4

change in αn1

eff (or αeff = 0.01 is reached)

  • Continuum extrapolation: at αeff = 0.3 have three lattice spacing

with µa < 0.5 for full O(a) improvement.

  • Finite-size effects: scale is determined in large enough volumes
  • Topology sampling

1

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SLIDE 4

Current State of αs determination - quenched case

Collaboration

R e n . s c a l e

  • Pert. Behav.
  • Cont. Extrap.

r0ΛMS Method CP-PACS 04 ⋆ ⋆

  • Schr¨
  • dinger Functional

ALPHA 98 ⋆ ⋆

  • 0.602(48)

Schr¨

  • dinger Functional

L¨ uscher 93 ⋆

  • 0.590(60)

Schr¨

  • dinger Functional

Brambilla 10

  • 0.637(+32

−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

Boucaud 08

  • 0.59(1)(+2

−1)

QCD vertices Boucaud 05

  • 0.62(7)

QCD vertices

In this talk: αs in the quenched case using the gradient flow

2

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SLIDE 5

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 4G a

µνG a µν

  • Its expectation value E(t, x) serves as a non-perturbative

definition of a reference scale

  • t0 first introduced as a reference scale L¨

uscher (2010)

t2E(t)

  • t=t0 = 0.3
  • w0 can also be used as a reference scale BMW Collaboration (2012)

t d dt t2E(t)

  • t=w 2

0 = 0.3

3

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SLIDE 6

Gradient Flow - Perturbative relation

0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

t/r02

0.1 0.2 0.3 0.4 0.5

t2〈E〉 t0

√ 8t = 0.2 fm √ 8t = 0.5 fm

Courtesy L¨ uscher (2010)

Perturbative relation for its expectation value for QCD (NA = 8) in MS scheme up to NNLO t2E(t) = 3αs 4π

  • 1 + αsk1 + α2

sk2 + O(α3 s)

  • k1 = 1.09778674 L¨

uscher (2010)

k2 = −0.9822456 Harlander and Neumann (2016)

4

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SLIDE 7

Brute-Force determination of αs

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SLIDE 8

Procedure

Simulation details

  • Use fine-lattices at T = 0
  • Keep the physical volume

constant LTc ≃ 2

  • Periodic Boundary Conditions
  • Tree-level Symanzik action,

Wilson flow, Clover-leaf definition of observable

  • Q = 0 configurations selected

Lattices used

β N a (in r0) # cfgs 5.3570 48 0.05651 529 5.3669 48 0.05583 4680 5.4500 56 0.05011 215 5.4700 56 0.04911 222 5.5000 56 0.04690 197 5.5830 64 0.04178 161 5.6000 64 0.04115 200 5.8000 80 0.03229 400 5.9500 96 0.02699 234 6.0500 112 0.02395 100 6.1500 128 0.02138 50 6.3600 160 0.01648 103

  • w Q=0

/w0 = 0.992(4)

  • Use w1 to set the scale: t d

dt t2E(t)

  • t=w 2

1 = 0.03

  • w1/r0 = 0.115(2)

5

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SLIDE 9

Improvement

  • Discretization correction terms at tree-level Fodor et al. (2014)

t2E(t) = 3αs 4π

  • C(a2/t) + O(αs)
  • where

C(a2/t) = 1 +

  • m=1

C2m a2m tm Coefficients known up to O(a8)

  • Finite-Volume correction Fodor et al. (2012)

t2E = 3αs 4π

  • 1 + δ(t/L2)
  • where

δ = 1 − 64t2π 3L2 + 8e−L2/8t + 24e−L4/4t + . . .

6

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SLIDE 10

Improvement

0.01 0.02 0.03 0.04 0.05 0.06 0.2 0.4 0.6 0.8 1 1.2 1.4 t2<E> √ 8 t / w1 Improved vs Unimproved flow β=6.05 β=6.05 impoved β=6.15 β=6.15 impoved β=6.36 β=6.36 impoved

7

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SLIDE 11

Improvement

0.025 0.03 0.035 0.04 0.045 0.05 0.02 0.04 0.06 0.08 0.1 0.12 0.14 t2E (a/w1)2 sqrt(8t)/w1=1.0, unimproved sqrt(8t)/w1=1.0, improved sqrt(8t)/w1=0.6, unimproved sqrt(8t)/w1=0.6, improved

8

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Λ parameter

  • 1. t2E(t) ⇒ αs from perturbative relation
  • 2. Use 4-loop β-function in the MS-scheme to run αs at a high scale

Ritbergen, Vermaseren, Larin (1997)

t2E(t) = 3αs 4π

  • 1 + αsk1 + α2

sk2 + O(α3 s)

  • 0.655

0.66 0.665 0.67 0.675 0.68 0.685 0.69 0.695 0.7 0.05 0.1 0.15 0.2 0.25 0.3 r0Λ √ 8 t / r0 Λ-parameter from flow

9

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SLIDE 13

Λ parameter

t2E(t) = 3αs 4π

  • 1 + αsk1 + α2

sk2 + α3 sk3 + O(α4 s)

  • 0.64

0.66 0.68 0.7 0.72 0.74 0.05 0.1 0.15 0.2 0.25 0.3 r0Λ √ 8 t / r0 Λ-parameter from flow including k3 k3=-2.0 k3=0.0 k3=2.0

10

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SLIDE 14

Λ parameter

We want to eliminate k3 contribution A(t) ≡ (t2E)2 + C

  • t dt2E

dt

  • = α2

s

  • 9

(4π)2 + 3β0C (4π)2

  • + α3

s

  • 18k1

(4π)2 + C 3β1 (4π)3 + 6k1β0 (4π)2

  • + α4

s

  • 9(k2

1 + 2k2)

(4π)2 + C 3β2 (4π)4 + 6k1β1 (4π)3 + 9k2β0 (4π)2

  • + α5

s

  • 9(2k1k2 + 2k3)

(4π)2 + C 3β3 (4π)5 + 6k1β2 (4π)4 + 9k2β1 (4π)3 + 12k3β0 (4π)2

  • By requiring combination of k3-terms to be zero ⇒ C = −0.13636364

11

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SLIDE 15

Λ parameter

We follow the same procedure as previously but now αs determined via A(t) function

0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.05 0.1 0.15 0.2 0.25 0.3 r0Λ √ 8 t / r0 Λ-parameter from A(t) α=0.1708 α=0.2757

r0Λ=0.664(14)

12

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SLIDE 16

Step-scaling

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SLIDE 17

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, 2N) find the difference

D(t2E

  • µ) =

1 t2E

  • 2µ −

1 t2E

  • µ
  • Find the function D in the continuum

1 t2E

  • 2µ −

1 t2E

  • µ = 4π

3

  • 1

αs(2µ) − 1 αs(µ) + (2k2

1 − k2)

  • α(2µ) − α(µ)
  • = 4π

3 2β0 π ln2 + 2β1 π2 ln24π 3 t2E

  • µ + . . .
  • Keep w1/L fixed and use 14,16,20,24 to do step scaling

13

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SLIDE 18

Step-scaling function D for c = 0.1

4.8 5 5.2 5.4 5.6 5.8 6 6.2 6.4 6.6 6.8 0.02 0.04 0.06 0.08 0.1 0.12 0.14 D t2<E> Step scaling function Step-scaling function pair 14-28 pair 16-32 pair 20-40 pair 24-48 14

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SLIDE 19

Λ parameter from T = 0 and step-scaling

0.63 0.64 0.65 0.66 0.67 0.68 0.69 1e-05 0.0001 0.001 0.01 0.1 1 10 r0Λ √ 8 t / r0 Λ-parameter using step scaling c=0.10 from w1/L=0.10 c=0.10 from w1/L=0.20 c=0.12 from w1/L=0.10 c=0.12 from w1/L=0.20

  • c=0.10 from w1/L = 0.10 ⇒ t2E = 0.0107
  • c=0.10 from w1/L = 0.20 ⇒ t2E = 0.0102
  • c=0.12 from w1/L = 0.10 ⇒ t2E = 0.0109
  • c=0.12 from w1/L = 0.20 ⇒ t2E = 0.0103

⇒ For all αs < 0.01 15

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SLIDE 20

Conclusions

  • We present a way of determining the Λ-parameter (and αs) using

the gradient flow

  • By brute-force elements we find a good plateau for the Λ-parameter

that meets the criteria of a good αs determination ⇒ r0Λ = 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

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SLIDE 21

Thank you for your attention!

16

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SLIDE 22

w1 to w0 and r0 relations

  • w1/w Q=0

= 0.340(7)

  • w0/r0 = 0.341(2)

Sommer et al. (2014)

0.975 0.98 0.985 0.99 0.995 1 1.005 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 (a/w0)2 fixed volume: L=2/Tc w0

Q=0/w0

0.32 0.325 0.33 0.335 0.34 0.345 0.35 0.01 0.02 0.03 0.04 0.05 0.06 0.07 w1

[Q=0]/w[Q=0]

(a/w1)2 fixed volume: L=2/Tc plain improved

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SLIDE 23

Continuum function D-Sketch of analytic derivation up to NLO

From dαs/dlnµ we can find d(1/αs) dlnµ = − 1 α2

s

dαs dlnµ = 2β0 π + 2β1 π2 αs + 2β2 π3 α2

s

Then we can find the difference 1 αs(2µ) − 1 αs(µ) = ln2µ

lnµ

d(1/αs) dln(µ′/µ)dln(µ′/µ) = 2β0 π lnµ + 2β1 π2 ln2 αs(µ′)dln(µ′/µ) + 2β2 π3 ln2 α2

s(µ′)dln(µ′/µ)

Using similar procedure we find αs(2µ) − αs(µ) Then we fit using

5.0831+15.71x +ax2 +bx3 +cx4 +dx2 1 N2 +ex3 1 N2 +fx4 1 N2 +g(5.0831+15.71x) 1 N2