European Physical Journal C: Particles and Fields (May 2018)

A non-perturbative exploration of the high energy regime in $$N_{\mathrm{f}}=3$$ Nf=3 QCD

  • Mattia Dalla Brida,
  • Patrick Fritzsch,
  • Tomasz Korzec,
  • Alberto Ramos,
  • Stefan Sint,
  • Rainer Sommer

DOI
https://doi.org/10.1140/epjc/s10052-018-5838-5
Journal volume & issue
Vol. 78, no. 5
pp. 1 – 25

Abstract

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Abstract Using continuum extrapolated lattice data we trace a family of running couplings in three-flavour QCD over a large range of scales from about 4 to 128 GeV. The scale is set by the finite space time volume so that recursive finite size techniques can be applied, and Schrödinger functional (SF) boundary conditions enable direct simulations in the chiral limit. Compared to earlier studies we have improved on both statistical and systematic errors. Using the SF coupling to implicitly define a reference scale $$1/L_0\approx 4$$ 1/L0≈4 GeV through $$\bar{g}^2(L_0) =2.012$$ g¯2(L0)=2.012 , we quote $$L_0 \Lambda ^{N_\mathrm{f}=3}_{{\overline{\mathrm{MS}}}} =0.0791(21)$$ L0ΛMS¯Nf=3=0.0791(21) . This error is dominated by statistics; in particular, the remnant perturbative uncertainty is negligible and very well controlled, by connecting to infinite renormalization scale from different scales $$2^n/L_0$$ 2n/L0 for $$n=0,1,\ldots ,5$$ n=0,1,…,5 . An intermediate step in this connection may involve any member of a one-parameter family of SF couplings. This provides an excellent opportunity for tests of perturbation theory some of which have been published in a letter (ALPHA collaboration, M. Dalla Brida et al. in Phys Rev Lett 117(18):182001, 2016). The results indicate that for our target precision of 3 per cent in $$L_0 \Lambda ^{N_\mathrm{f}=3}_{{\overline{\mathrm{MS}}}}$$ L0ΛMS¯Nf=3 , a reliable estimate of the truncation error requires non-perturbative data for a sufficiently large range of values of $$\alpha _s=\bar{g}^2/(4\pi )$$ αs=g¯2/(4π) . In the present work we reach this precision by studying scales that vary by a factor $$2^5= 32$$ 25=32 , reaching down to $$\alpha _s\approx 0.1$$ αs≈0.1 . We here provide the details of our analysis and an extended discussion.