Is the up-quark massless? Hartmut Wittig DESY

Transcription

1 Is the up-quark massless? Hartmut Wittig DESY Wuppertal, 5 November 2001

2 Quark mass ratios in Chiral Perturbation Theory Leutwyler s ellipse: ( mu m d ) Q 2 ( ms m d ) 2 = 1 25 m s m d 38 R M (Leutwyler, hep-ph/ ) m u m d Q 2 = m2 K m 2 π m 2 K m2 π (m 2 K 0 m 2 K + ) QCD m 2 K m 2 π = m s + m m u + m d {1 + M + O(m 2 s) } 1

3 Results at next-to-leading order: m u m d = ± 0.043, m s m d = 18.9 ± 0.8 Value of correction term M not known precisely... Cannot be estimated from phenomenology and chiral symmetry alone Large N c arguments and additional theoretical considerations suggest that M is small and positive (Leutwyler, hep-ph/ ): 0 < M < 0.13 However: magnitude and sign of M have not been derived from first principles If M were large and negative the up-quark could be massless solution to the strong CP problem 2

4 The strong CP problem and m u = 0 QCD action with a θ-term: θ g2 64π 2 d 4 x F F, 0 θ 2π Induces CP violation in the strong sector, but no experimental evidence Value of θ is shifted by the quark mass matrix: θ = θ + arg (det M ) explain why θ 0 If one of the quarks were massless one could perform a chiral U(1) rotation with 2N f (ϕ R ϕ L ) = arg (det M ) which cancels the effects from θ : θ = 0 3

5 Require an approach to determine M based on first principles M is related to low-energy constants in the effective chiral Lagrangian compute low-energy constants using lattice simulations of QCD combine Chiral Perturbation Theory (ChPT) with Lattice QCD Outline: I. Basic concepts of Chiral Perturbation Theory II. Strategy to compute low-energy constants III. Results IV. Summary 4

6 I. Basic concepts Effective chiral Lagrangian L eff L eff = L (2) eff + L(4) eff + L(6) eff +... Simultaneous expansion in powers of momentum p and quark masses m u, m d, m s Order p 2 : L (2) eff = F Tr ( µ U µ U ) + F 0 2 B 0 2 ( i T a ϕ a ) U = exp, M = m u 2F 0 Tr ( M(U + U ) ) m d F 0, B 0 : coupling ( low-energy ) constants m s Mathematical basis for soft pion theorems of the 1960 s Order p 4 :... many more interaction terms... coupling constants L 1, L 2,..., L 10 (, L 11, L 12 ) 5

7 Loop corrections: One-loop corrections to L (2) eff divergences, but generate logarithmic form of counterterms not included in lowest order L eff Chiral Perturbation Theory (ChPT) is non-renormalisable Structure of L eff determined by chiral symmetry: L eff parameterised in terms of empirical constants F 0, B 0, L i determine low-energy constants from phenomenology Examples: L 1, L 2, L 3 : π π scattering L 5 : F K = 1 + chiral log s + 4(m2 K m2 π) F π F 2 π L r 5 L r 5(m ρ ) = (1.4 ± 0.5)

8 Need additional theoretical input to fix complete set of low-energy constants: Quark masses, B 0 : m 2 π = 2B 0 m, m 2 K = B 0 (m s + m) m = 1 2 (m u + m d ) cannot determine B 0 or absolute magnitude of m, m s without explicit knowledge of quark mass dependence Hidden symmetry in L eff : L eff = L (2) eff + L(4) eff invariant under m u m u + λm d m s + cyclic perm. L 6,7 L 6,7 + λf B 0, L 8 L 8 λf B 0, λ : arbitrary parameter Chiral symmetry cannot distinguish between different sets M λ, L λ i Kaplan-Manohar ambiguity (1986) 7

9 Kaplan-Manohar ambiguity and m u = 0 Leutwyler s ellipse is mapped onto itself Expression for correction term M : M = 8(m2 K m2 π) F 2 π (2L 8 L 5 ) + chiral log s Experimental information on L 8 is afflicted with the KM ambiguity: GMO = 8(m2 K m2 π) F 2 π (L 5 12L λ 7 6L λ 8) + log s There is no direct experimental information on L 6, L 7, L 8 or M QCD is not afflicted with KM ambiguity Test hypothesis m u = 0 in lattice QCD: Direct determination (difficult!) Compute quantities which provide information on L 8 or M and are not subject to KM ambiguity define effective chiral Lagrangian by determining the low-energy constants on the lattice from first principles 8

10 Use lattice simulations to (1) Determine L i with high precision ChPT: analytic tool with numerically determined coupling constants test whether m u = 0 (2) Determine B 0 and the absolute normalisation of quark masses; combine with quark mass ratios obtained in ChPT m s, m d, m u Aside: problem (2) has been solved in the quenched approximation : M s / M = 24.4 ± 1.5, M = 1 2 (M u + M d ) (Leutwyler, Phys Lett B378 (1996) 313) M s + M = 140 ± 5 MeV (Garden, Heitger, Sommer, HW, Nucl Phys B571 (2000) 237) M s = 134 ± 5 MeV m MS s (2 GeV) = 97 ± 4 MeV (all errors except quenching) 9

11 Lattice QCD: Quark masses are used as input parameters: m m PS (m), F PS (m) Can map out the quark mass dependence of m PS and F PS without knowing the physical quark masses Phenomenology: m π = m PS ( m), m K = m PS ( m + m s ) partially quenched QCD: can vary sea and valence quark masses independently m val γ 5 γ 5 m sea Values of L i s depend only on N f Can determine their values even for unphysical quark masses 10

12 II. Strategy ALPHA Coll. (Heitger, Sommer, HW), Nucl Phys B588 (2000) 377 Modify and extend other proposals: Cohen, Kaplan & Nelson, JHEP 11 (1999) 027 Sharpe & Shoresh, Phys Rev D62 (2000) Introduce reference quark mass m ref and define R F (x) = F PS (m)/f PS (m ref ) ( ) /( ) FPS (m) FPS (m ref ) R M (x) = G PS (m) G PS (m ref ) ( ) /( ) 2m 2mref = m 2 PS (m) m 2 PS (m ref) F PS m PS = 0 A 0 (0) PS, G PS = 0 P (0) PS m = xm ref x : dimensionless mass parameter R F, R M are universal functions of x : Renormalisation factors drop out; straightforward extrapolation to the continuum limit Control over lattice artefacts: can isolate dynamical quark effects unambiguously Can compute R X (x) with high statistical precision 11

13 Expressions for R X in ChPT Consider QCD with N f = 3 degenerate flavours (S. Sharpe 1997) Notation: y = 2B 0m (4πF 0 ) 2, x = y/y ref, α i = 8(4π) 2 L i Numerator m sea = m val = m; denominator: m sea = m val = m ref R M (x) = [x ln x + (x 1) ln y ref] y ref (x 1)[(2α 8 α 5 ) + 3(2α 6 α 4 )] R F (x) = [x ln x + (x 1) ln y ref] + y ref (x 1) 1 2 (α 5 + 3α 4 ) Numerator m val = m, m sea = m ref ; denominator: m sea = m val = m ref sensitive to different combinations of α i s, e.g. R F (x) = [ (x + 1) ln ( 1 2 (x + 1)) + (x 1) ln y ref ] + y ref (x 1) 1 2 α 5 Can determine α 4, α 5, α 6 and α 8 12

14 Quenched approximation: associated with flavour- Additional parameters in L (2) eff singlet field R M (x) = α [ ( Φ x ln x + (x 1) ln y )] ref ( ) + δ ln x y ref (x 1) 2α (0) 8 α (0) 5 R F (x) = 1 y ref (x 1) 1 2 α(0) 5 Estimates for δ, α Φ : δ δ ± (CP-PACS, hep-lat/ ) (FNAL, hep-lat/ ) α Φ 0.6 (S. Sharpe, Lattice 96) Meson spectroscopy data by CP-PACS described well for δ = 0.12 ± 0.02, α Φ = 0 or δ = 0.05 ± 0.02, α Φ = 0.5 but not δ = 0.12 ± 0.02, α Φ =

15 Extracting the low-energy constants from R X Obtain α i from least-χ 2 fits to R X (x) over a suitably chosen interval in x. Alternatively, consider R X (x 1, x 2 ) R X (x 1 ) R X (x 2 ), x 1 < x 2 Obtain α i from simple algebraic expressions; vary x 1, x 2 to check stability Example: 5 = 2 Rquen F (x 1, x 2 ) (x 1 x 2 )y ref α (0) 14

16 III. Results Test viability of the method: statistical precision in continuum limit stability in extraction of α i s use quenched approximation Numerical details: 4 values of a: a fm L 1.5 fm O(a) improved Wilson fermions Reference point defined at (m PS r 0 ) 2 = 3 { mref m s y ref = Interpolate R X to common values of x = 0.75, 0.80,..., 1.40 Observe stable continuum extrapolations 15

17 R M (x), R F (x) in the continuum limit: Linearity of R F (x) matched by expression in quenched ChPT, c.f. 1 R F (x) = 1 + (x 1)y ref 2 α(0) 5 Several competing effects have to produce flat behaviour of R M (x) Restrict x-interval to 0.75 x

18 Obtain very stable estimates for α (0) 5 (independent of δ, α Φ ) α (0) 5 = ± (stat. error) Insert estimates for δ and α Φ into expression for R M ; determine (2α (0) 8 α (0) 5 ) (2α (0) 8 α (0) 5 ) = { 0.35(5) δ = 0.12, αφ = (5) δ = 0.05, α Φ =

19 Stability analysis / Error estimates Higher orders in chiral expansion may modify estimates for α i s. systematic study requires data at smaller values of x Estimate uncertainty due to higher orders; assume that O(x 2 ) ( O(x) ) 2, O(x) 0.20 at x = 1 alternatively, fit to R F (x) = 1 + (x 1)y ref 1 2 α(0) 5 + d F y 2 ref(x 2 1) 0.75 x 1.4 uncertainty in α (0) 5, (2α(0) 8 α (0) 5 ) amounts to ±

20 QCD with N f = 2 (A Irving, C McNeile, C Michael, K Sharkey, HW, Phys Lett B518 (2001) 243) Simulation with N f = 2 degenerate flavours of dynamical quarks by UKQCD Collaboration; identify with physical u and d quarks Numerical details: only one value of a: a 0.1 fm L 1.5 fm No control over continuum limit sea quarks: m PS /m V 0.58 (cf. m π /m ρ = 0.169) O(a) improved Wilson fermions Reference point defined at (m PS r 0 ) 2 = { mref 0.7m s y ref = Consider degenerate & non-degenerate combinations of valence quarks: m val 1 = m val 2 = xm ref, m sea = m ref, VV m val 1 = xm ref, m val 2 = m sea = m ref, VS 19

21 Results for VV and VS cases Consistency between VV and VS cases Uncertainties due lattice artefacts statistical error Uncertainties due to higher order orders ±0.2 20

22 Results & Comparison Renormalisation scale: µ = 4πF π = 1175 MeV Quenched (N f = 0) Theory: α (0) 5 = 0.99 ± 0.06 ± 0.2 α (0) 8 = { 0.67 ± 0.04 ± 0.2, δ = 0.12, αφ = ± 0.04 ± 0.2, δ = 0.05, α Φ = 0.5 Partially quenched (N f = 2) Theory α (2) 5 = 1.22 ± 0.13 ± 0.25 α (2) 8 = 0.79 ± 0.06 ± 0.21 Comparison with conventional ChPT: α (3) 5 = 0.5 ± 0.6, standard α (3) 8 = { 0.76 ± 0.4, standard 0.9 ± 0.4, m u = 0 21

23 The value of F K /F π Experimental result F K /F π = 1.22 ± 0.01 serves to determine α 5 can use lattice estimates to predict F K /F π check if results make sense phenomenologically Analyse lattice data for N f = 2 as if they had been generated with N f = 3 α (3) 5 = 0.98 ± 0.13 ± 0.24 α (3) 8 = 0.59 ± 0.06 ± 0.21 F K F π = 1 + chiral log s m 2 K m2 π (4πF π ) 2 = ± ± α(3) 5 suggests that quark mass dependence for m < m s is only weakly distorted by neglecting dynamical quarks. Compute mass correction M : M = chiral log s + m2 K m2 π (4πF π ) 2 (2α(3) 8 α (3) 5 ) = 0.04 ± 0.05 ±

24 Confront lattice results with the hypothesis m u = 0 : dashed line: expected behaviour if m u = 0 23

25 IV. Summary Question whether m u = 0 can be studied from first principles by combining Lattice QCD with Chiral Perturbation Theory: Resolve ambiguity caused by hidden symmetry in L eff Provide an absolute normalisation (e.g. m + m s ) m u, m d, m s Obtain accurate estimates for low-energy constants through ratios of matrix elements conceptionally clean; good control over continuum limit statistical precision of order ±0.05 uncertainties due to neglecting higher orders need to be investigated further first results for N f = 0, 2 do not support m u = 0 Future: require Monte Carlo data for N f = 3 flavours of dynamical quarks Algorithmic challenge: simulate odd N f efficiently for m PS /m V < Statistical precision may be relaxed to study whether m u = 0 24