Lattice QCD+QED. Towards a Quantitative Understanding of the Stability of Matter. G. Schierholz. Deutsches Elektronen-Synchrotron DESY

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1 Lattice QCD+QED Towards a Quantitative Understanding of the Stability of Matter G. Schierholz Deutsches Elektronen-Synchrotron DESY

2 The Challenge (Mn Mp)QED [MeV] Helium Stars Exp No Fusion α EM No Fusion Helium Stars (M n M p ) QCD [MeV] m u /m d Having an analytic expression for the nucleon mass as a function of quark masses and α EM, we can visualize the allowed region

3 If m u /m d 0.8 even, protons would decay spontaneously to neutrons

4 The neutron proton mass difference is one of the most consequential quantities of physics. It is extremely fine tuned for the stability of matter as we know it and the existence of our Universe. This calls for a calculation from first principles Lattice Gauge Theory is the method of choice. Lattice calculations are now reaching a level of precision, where it is possible to address isospin breaking effects These effects have two sources, the mass difference of u and d quarks, and electromagnetic interactions Both effects are of the same order of magnitude and cannot be separated unambiguously due to the nonperturbative nature of the strong interactions, which makes a direct calculation from QCD + QED necessary

5 Further issues A primary concern is to understand the pattern of flavor and isospin symmetry breaking in QCD + QED In particular, there is the prospect of making precise predictions for appropriate isospin violating processes Lattice calculations of hadronic processes are approaching O(1%) precision. At this level electromagnetic corrections must be included in the calculation Simulations of QCD + QED with electrically charged quarks allow to probe the effect of dynamical quarks on the confining properties of the QCD vacuum

6 Outline Lattice QCD + QED Vacuum Structure Flavor Physics and Spectroscopy Isospin Splittings Conclusions

7 Lattice QCD + QED BMW arxiv: QCDSF arxiv: arxiv: arxiv: Octet baryons Vacuum structure Octet mesons Octet baryons Light quark masses Underpinned by group theory

8 With J. Charvetto (Adelaide), R. Horsley (Edinburgh), Y. Nakamura (Kobe), H. Perlt (Leipzig), D. Pleiter (Jülich), D. Leinweber (Adelaide), P.E.L. Rakow (Liverpool), A. Schiller (Leipzig), H. Stüben (Hamburg), R. Young (Adelaide), J. Zanotti (Adelaide)

9 Action S = S G + S QED + S u F + Sd F + Ss F S G = 6 g 2 x,µ<ν S QED = 1 2e 2 S q F = x 1 { } 3 Tr c 0 [1 U µν (x)] + c 1 [1 R µν (x)] x,µ<ν (A µ (x) + A ν (x + µ) A µ (x + ν) A ν (x)) 2 noncompact { [q(x) γ µ 1 e ie qaµ(x)ũ µ (x)q(x + ˆµ) 2 µ q(x) γ µ + 1 e ieqaµ(x ˆµ) Ũ µ (x ˆµ)q(x ˆµ)] κ q q(x)q(x) 1 4 c SW µν } q(x)σ µν F µν (x)q(x) Lattice spacing a implicit e u = 2 3, e d = e s = 1 3

10 Lattices V κ u κ d κ s } Symmetric point µ D u = µd d = µd s 1/κ u + 1/κ d + 1/κ s = constant Couplings β 6 g 2 = 5.50, α EM e2 4π = 0.10 corresponding to a = 0.068(2)fermi. Later on we extrapolate the results to α EM = 1/137

11 Efficiency Q ρa(τ) # trajectory τ

12 Gauge fixing The actions S QED and S q F are invariant under U(1) gauge transformations A µ (x) A µ (x) + µ α(x), q(x) e ie qα(x) q(x) This is not the case for propagators of charged particles, which demands gauge fixing. We choose Landau gauge µ A µ (x) = 0, which, however, does not eliminate all gauge degrees of freedom, but allows for shifts µ α(x) with 2 α(x) = 0. Maintaining (anti-)periodicity of the quark fields, this redundancy can be eliminated by adding multiples of 2π/e q L µ to A µ (x) such that π e q L µ < B µ π e q L µ, B µ = 1 V A µ (x) x The constant background field can be factored out from the link matrices and absorbed into the quark momenta p q(x) e ie qbx q(x), q(x) q(x)e ie qbx which amounts to a shift p p + e q B. This leaves us with photon propagators that are devoid of zero modes

13 In the presence of a constant background field B µ the correlator of a single hadron H then becomes 0 H(t) H(0) 0 Z H 2 e which amounts to a shift M H MH 2 + e2 H B 2 M 2 H + ( p+e H B ) 2 t To extract masses we remove the influence of the background field by subtracting the associated kinetic energy from the ensemble averaged lattice energy Bin1 Bin2 Bin EnAv B am π Bin3 Bin1 Bin Trajectory Number B 2

14 Vacuum Structure Density of QCD (aqua) and QED (yellow) actions Electromagnetic field strength repelled by chromoelectric one

15 Density of positive (red) and negative (purple) charge compared with QCD action density Density of positive (red) and negative (purple) charge compared with QED action density (Charvetto)

16 Chiral Magnetic Effect Excess of right-handed quarks due to chiral anomaly = p s p J, s B We find evidence for J B to be correlated with position of instanton Instanton

17 Flavor Physics and Spectroscopy K 0 (d s) Y K + (u s) n(udd) Y p(uud).. π (dū) π 0 π + (u d) η 8 1. I 3 Σ (dds) Σ 0 1 Λ 8 1. Σ + (uus) I K (sū) 1 K0 (s d).... Ξ (dss) 1 Ξ 0 (uss) < 1%

18 Strategy arxiv: QCD interactions are flavor blind. The only difference between flavors comes from the quark mass matrix In lattice calculations one can vary the quark masses freely, which helps to illuminate the pattern of flavor symmetry breaking One has the best theoretical understanding when all quark masses are equal, because one can use the full power of flavor SU(3) We interpolate between the symmetric point µ u = µ d = µ s and the physical point by keeping the sum of the quark masses (µ u + µ d + µ s )/3 m fixed at its physical value, which is particularly instructive The symmetry of the electromagnetic current is similar to the symmetry of the quark mass matrix The simplifications from keeping m = constant in the mass expansion are analog to the simplifications from the identity e u +e d +e s = 0. We thus can read off the QED corrections from the mass expansion changing masses to charges

19 Benefit Unitary M 2 (aab) = M α 1(2δm a + δm b ) + α 2 (δm a δm b ) + β EM 0 (e 2 u + e2 d + e2 s ) + βem 1 (2e 2 a + e2 b ) + βem 2 (e a e b ) 2 + β EM 3 (e 2 a e2 b ) δm q = m q m = 1/2κ sea q 1/ κ q Partially quenched M 2 (aab) = M α 1(2δµ a + δµ b ) + α 2 (δµ a δµ b ) + β EM 0 (e 2 u + e2 d + e2 s ) + βem 1 (2e 2 a + e2 b ) + βem 2 (e a e b ) 2 + β EM 3 (e 2 a e2 b ) δµ q = µ q m = 1/2κ val q 1/ κ q

20 QCD Gell-Mann Okubo quadratic , 27 M 2 (a b) = M α(δµ a + δµ b ) + M 2 (aab) = M α 1(2δµ a + δµ b ) + α 2 (δµ a δµ b ) + δµ q = µ q m δµ u + δµ d + δµ s = π K η 1.4 N Λ Σ Ξ M PS 2 /X2 π M N 2 /X2 N (δµ u +δµ d )/ (δµ u +δµ d )/2 X 2 π =(M2 K 0 +M 2 K + +2M 2 π 0 M 2 π + )/3 X 2 N =(M2 n +M2 p +M2 Σ +M 2 Σ + +M 2 Ξ +M 2 Ξ 0 )/6

21 3 M PS 2 (a b)/x π π K η δµ q = µ q m arxiv: (δµ a +δµ b )/2

22 Comparison with ChPT M 2 0 = χ [ 1 16 χ f 2 0 (3L 4 + L 5 6L 6 2L 8 ) + χ 24π 2 f 2 0 ln χ Λ 2 χ ] α = Q 0 [ 1 16 χ f 2 0 (3L 4 + 2L 5 6L 6 4L 8 ) + χ 8π 2 f 2 0 ln χ Λ 2 χ ] χ = 2Q 0 Z S m Z NS m m arxiv:

23 Spectrum 80% of mass M [MeV] π K η s ρ K* ϕ s N Λ Σ Ξ Σ* Ξ* Ω Dual superconductor

24 QCD + QED M 2 (a b) = M α(δµ a + δµ b ) + β EM 1 (e a e b ) 2 Dashen scheme + γ EM 1 (e a e b ) 2 (δµ a + δµ b ) + γ EM 2 (e 2 a e2 b )(δµ a δµ b ) M 2 π 0 = M α(δµ u + δµ d ) M 2 K 0 = M α(δµ d + δµ s ) }{{} Determine physical quark masses

25 Dashen scheme Define symmetric point by M PS (uū) = M PS (d d) = M PS (s s) = M PS (d s) = M PS (s d) Renormalize (rescale) quark masses δµ q δµ q = δµ q (1 + Ke 2 q ) such that a µq µ u = µ d = µ s at symmetric point Dashen scheme bare 0 1/9 4/9 e 2 q

26 Neutral mesons η s π 0 K 0 η M PS 2 (aā)/m K 0 π M PS 2 /X2 π δµ s δµ u δµ a (δµ u +δµ d )/

27 1 8 M 2 (aab) = M α 1(2δµ a + δµ b ) + α 2 (δµ a δµ b ) + β EM 1 (2e 2 a + e2 b ) + βem 2 (e a e b ) 2 + β 3 (e 2 a e2 b ) e N = 1 e N = a 2 M N 2 (aab) a 2 M N 2 (aab) δµ a +δµ b 2δµ a +δµ b Σ,Ξ p,σ +

28 Extrapolation Extrapolate linearly from α EM = 0.1 to α EM = 1/ α1 [MeV 1 ] α EM Recover QCD results

29 Isospin Splittings Quark Masses M 2 π 0 = M α(δµ u + δµ d ) = α(µ u + µ d ) M 2 K 0 = M α(δµ d + δµ s ) = α(µ d + µ s ) } µ u +µ d +µ s = constant m q = Z MS m (2GeV) ZMS D µ q m u = 2.49(14)MeV m d = 4.80(27)MeV m s = 94.5(52) MeV RBC-UKQCD RM123 MILC QCDSF m u m d = 0.52(2), m s m d = 19.7(9) m u /m d

30 µ D q = ( 1 + α EM e 2 q 2.20(9) )µ q numerically Changing schemes µ MS q (2GeV) = µ MS q (2GeV) = ( ) 1 + α EM e 2 q O(α EMg 2 ) µ q one loop (1 α EM e 2 q 1.0(5) ) = Z MS D (2GeV)µD q µ D q µ D q

31 Meson Octet MPS[MeV] π + π 0 K + K a/l M π + = α ( EM 2L c ) + 2πα EM M π +L 3L 3 ( 1 + 4π ) M π +L c 1 r 2 π ++ Davoudi & Savage

32 Baryon Octet 10 MN [MeV] Ξ Ξ 0 p n -10 Σ + Σ a/l M p = α ( EM 2L c ) + 2πα EM M p L 3L 3 ( 1 + 4π ) M π +L c 1 r 2 p + Davoudi & Savage

33 Splittings M QCD + QED QED Experiment M π + M π (20) 4.59 M K 0 M K (10) 1.66(6) 3.93 M n M p 1.35(18)(8) 2.20(28)(10) 1.30 M Σ M Σ (73)(8) 0.63(8)(6) 8.08 M Ξ M Ξ (55)(45) 1.26(16)(13) M [MeV] π K N Σ Ξ M n M p + M Σ + M Σ + M Ξ M Ξ 0 = 0 10 Coleman Glashow

34 Changing schemes [M D QED ]2 = M 2 (α EM,{µ D q }) M2 (0,{µ D q }) [M MS QED ]2 = M 2 (α EM,{µ D q }) M2 (0,{µ MS q }) Taking the difference gives [M MS QED ]2 [M D QED ]2 = M 2 (0,{µ D q }) M2 (0,{µ MS q }) In particular ( 4 [M MS QED ]2 p [MD QED ]2 n = α EM1.0(5) (α 1 + 2α 2 ) 9 δµd u 1 9 δµd d )

35 QCD vs QED arxiv: Exp (Mn Mp)QED [MeV] (M n M p ) QCD [MeV] Dashen scheme MS

36 Analytic Solution? Renormalization Group Tells us how the bare parameters of the theory must behave to keep the physics constant as the cut-off is varied ln µ u /µ d ln µ = 0 e2 8π 2 + e2 12π 2 ( 4 µ2 u µ µ2 d 2 µ 2 ) µ = 1/a QCD QED γ EM q = 3e2 q 8π 2 [ 1 µ2 q µ 2 ln ( 1 + µ2 µ 2 q )] Solution µ u 2 1 e µ d 2 µ 8π 2 = 1 2 e2 16π ln µ = m u = 1 2 m d 2 m q = Z MS µ q

37 Conclusions Flavor and isospin symmetry breaking of hadron masses follow a very simple pattern, made visible by systematic lattice simulations of QCD + QED So far we have investigated isospin breaking of pseudoscalar meson and octet baryon masses. That allowed us to look simultaneously at both sources of isospin breaking, the quark mass differences and electromagnetic interactions, which are of comparable importance The stability of matter, and the existence of the Universe as we know it, largely hinges on the ratio of up to down quark mass From a broader perspective, we can look forward to a better understanding of the QCD vacuum and the mechanism of confinement and chiral symmetry breaking With increased computer power it will be possible to improve on the precision of the calculation, which is still limited