Anomalies, gauge field topology, and the lattice

Transcription

1 Anomalies, gauge field topology, and the lattice Michael Creutz BNL & U. Mainz Three sources of chiral symmetry breaking in QCD spontaneous breaking ψψ 0 explains lightness of pions implicit breaking of U(1) by the anomaly explains why η is not so light explicit breaking from quark masses pions are not exactly massless Rich physics from the interplay of these three effects Michael Creutz BNL & U. Mainz 1

2 Based on very old ideas Dashen, 1971: possible spontaneous strong CP violation before QCD! t Hooft, 1976: ties between anomaly and gauge field topology Fujikawa, 1979: fermion measure and the anomaly Witten, 1980: connections with effective Lagrangians many extensions shortly after I won t use the large N aspects here MC 1995: Why is chiral symmetry so hard on the lattice Z n symmetry and transition at Θ = π Why rehash old ideas? consequences have raised bitter recent controversies Michael Creutz BNL & U. Mainz 2

3 Axial anomaly in N f flavor massless QCD leaves behind a residual Z Nf flavor-singlet chiral symmetry tied to gauge field topology and the QCD theta parameter Consequences degenerate quarks with m 0: first order transition at Θ = π sign of mass relevant for odd N f : perturbation theory incomplete N f = 1: no symmetry for mass protection nontrivial N f dependence: invalidates rooting controversial Michael Creutz BNL & U. Mainz 3

4 Consider QCD with N f light quarks and assume the field theory exists and confines spontaneous chiral symmetry breaking ψψ 0 SU(N f ) SU(N f ) chiral perturbation theory makes sense anomaly gives η a mass N f small enough to avoid any conformal phase Use continuum language imagine some non-perturbative regulator in place (lattice?) momentum space cutoff much larger than Λ QCD lattice spacing a much smaller than 1/Λ QCD Michael Creutz BNL & U. Mainz 4

5 Construct effective potential for meson fields represents vacuum energy density for a given field expectation formally via a Legendre transformation For simplicity initially consider degenerate quarks with small mass m N f even interesting subtleties for odd N f Michael Creutz BNL & U. Mainz 5

6 Work with composite fields σ ψψ π α iψλ α γ 5 ψ λ α : Gell-Mann matrices for SU(N f ) η iψγ 5 ψ Spontaneous symmetry breaking at m = 0 ( σ) (σ) has a double well structure vacuum has σ = v 0 minimum of (σ) = ±v σ Ignore convexity issues phase separation occurs when a field is in a concave region Michael Creutz BNL & U. Mainz 6

7 Nonsinglet pseudoscalars are Goldstone bosons symmetry under flavored rotations σ cos(φ)σ + sin(φ)πα π α cos(φ)π α (N f = 2) sin(φ)σ ψ e iφγ 5λ α ψ potential has Nf 2 1 flat directions one for each generator of SU(N f ) π σ Michael Creutz BNL & U. Mainz 7

8 Small mass selects vacuum mσ σ +v π = 0 Goldstones acquire mass m π σ Michael Creutz BNL & U. Mainz 8

9 Anomaly gives the η a mass even if m q = 0 m η = O(Λ QCD ) (σ, η ) not symmetric under ψ e iφγ 5 ψ σ σ cos(φ) + η sin(φ) η σ sin(φ) + η cos(φ) Expand the effective potential near the vacuum state σ v and η 0 (σ, η ) = m 2 σ(σ v) 2 + m 2 η η 2 + O((σ v) 3, η 4 ) both masses of order Λ QCD Michael Creutz BNL & U. Mainz 9

10 In quark language Classical symmetry ψ e iφγ5/2 ψ ψ ψe iφγ 5/2 mixes σ and η σ σ cos(φ) + η sin(φ) η σ sin(φ) + η cos(φ) This symmetry is anomalous any valid regulator must break chiral symmetry remnant of the breaking survives in the continuum Michael Creutz BNL & U. Mainz 10

11 ariable change alters fermion measure dψ e iφγ 5/2 dψ = e iφtrγ 5/2 dψ But doesn t Trγ 5 = 0??? Fujikawa: Not in the regulated theory!!! i.e. lim Λ Tr ( γ 5 e D2 /Λ 2) 0 Dirac action ψ(d + m)ψ D = D [D, γ 5 ] + = 0 Use eigenstates of D to define Trγ 5 D ψ i = λ i ψ i Trγ 5 = i ψ i γ 5 ψ i Michael Creutz BNL & U. Mainz 11

12 Index theorem with gauge winding ν, D has ν zero modes D ψ i = 0 modes are chiral: γ 5 ψ i = ± ψ i ν = n + n Non-zero eigenstates in chiral pairs D ψ = λ ψ Dγ 5 ψ = λγ 5 ψ = λ γ 5 ψ Space spanned by ψ and γ 5 ψ gives no contribution to Trγ 5 ψ γ 5 ψ = 0 when λ 0 only the zero modes count! Trγ 5 = i ψ i γ 5 ψ i = ν Michael Creutz BNL & U. Mainz 12

13 Where did the opposite chirality go? continuum: lost at infinity opposite chirality states above the cutoff overlap: modes on opposite side of unitarity circle Dγ 5 = ˆγ 5 D Tr ˆγ 5 = 2ν Wilson: real eigenvalues in doubler region This phenomenon involves both short and long distances zero modes compensated by modes lost at the cutoff cannot ignore instanton physics at short distances Cannot uniquely separate perturbative and non-perturbative effects small instantons can fall through the lattice scheme and scale dependent Michael Creutz BNL & U. Mainz 13

14 Under the transformation ψ e iφγ 5/2 ψ ψ ψe iφγ 5/2 Regulated fermion measure changes by e iφtrγ 5 = e iφν changes weighting of gauge configurations with winding non-zero φ introduces a sign problem for Monte Carlo Note: here φ is the conventional Θ/N f each flavor contributes equally Trγ 5 = N f ν Michael Creutz BNL & U. Mainz 14

15 Expanding in smooth gauge fields ( ) ν = Tr γ 5 e D2 /Λ 2 = Tr e 2 /Λ 2 g 2 Λ F 4 µν Fµν + O(Λ 5 ) trace over space, spinor, and color indices e 2 /Λ 2 heat kernel regulator for space Tr x (e 2 /Λ 2 f(x)) = Λ4 16π 2 d 4 xf(x) ν = g2 16π 2 d 4 xtr c F µν Fµν If F = F then S = d 4 x 1 2 F 2 = 8π2 g 2 ν classical instanton action Michael Creutz BNL & U. Mainz 15

16 Back to effective field language At least two minima in the σ, η plane η? v v σ? Question: do we know anything about the potential elsewhere in the σ, η plane? Yes! there are actually N f equivalent minima Michael Creutz BNL & U. Mainz 16

17 Define ψ L = 1+γ 5 2 ψ Singlet rotation ψ L e iφ ψ L not a good symmetry for generic φ Flavored rotation ψ L g L ψ L = e iφ αλ α ψ L is a symmetry for g L SU(N f ) For special discrete values of φ these rotations can cross g = e 2πi/N f Z Nf SU(N f ) A valid discrete singlet symmetry: σ σ cos(2π/n f ) + η sin(2π/n f ) η σ sin(2π/n f ) + η cos(2π/n f ) Michael Creutz BNL & U. Mainz 17

18 (σ, η ) has a Z Nf symmetry N f equivalent minima in the (σ, η ) plane N f = 4: η 1 σ At the chiral lagrangian level Z N is a subgroup of both SU(N) and U(1) At the quark level measure gets a contribution from each flavor ( t Hooft vertex) ψ L e 2πi/N f ψ L is a symmetry Michael Creutz BNL & U. Mainz 18

19 η 1 σ Mass term mψψ tilts effective potential picks one vacuum as the lowest in n th minimum m 2 π m cos(2πn/n f ) highest minima are unstable in the π α direction multiple meta-stable minima when N f > 4 Michael Creutz BNL & U. Mainz 19

20 Anomalous rotation of the mass term mψψ m cos(φ)ψψ + im sin(φ)ψγ 5 ψ twists tilt away from the σ direction An inequivalent theory! η 1 m φ σ as φ increases, vacuum jumps from one minimum to the next Michael Creutz BNL & U. Mainz 20

21 Here each flavor has been given the same phase Conventional notation uses Θ = N f φ Z Nf symmetry implies 2π periodicity in Θ Degenerate light quarks first order transition at Θ = π η Θ = π 1 σ Michael Creutz BNL & U. Mainz 21

22 ) Discrete symmetry in mass parameter space m m exp( 2πiγ5 N f for N f = 4: mψψ and imψγ 5 ψ mass terms give equivalent theories true if and only if N f is a multiple of 4 η 1 σ Michael Creutz BNL & U. Mainz 22

23 Odd number of flavors, N f = 2N is not in SU(2N + 1) 1 η N =3 f m > 0 and m < 0 not equivalent! m < 0 represents Θ = π 0 σ an inequivalent theory spontaneous CP violation: η 0 2 Inequivalent theories can have identical perturbative expansions! Michael Creutz BNL & U. Mainz 23

24 Center of SU(N f ) is a subgroup of U(1) 10,000 random SU(3) and SU(4) matrices: 3 SU(3) 4 SU(4) 2 3 Im Tr g Im Tr g Re Tr g Re Tr g region for SU(3) bounded by exp(iφλ 8 ) all SU(N) points enclosed by the U(1) circle e iφ boundary reached at center elements Michael Creutz BNL & U. Mainz 24

25 N f = 1: No chiral symmetry at all! unique vacuum ψψ σ 0 from t Hooft vertex not a spontaneous symmetry breaking η N =1 f σ No singularity at m = 0 m = 0 not protected 0 For small mass no first order transition at Θ = π larger masses? Michael Creutz BNL & U. Mainz 25

26 Instanton effects scheme dependent ( renormalon ambiguity) (Θ, m) singular coordinates when N f = 1 (Re m, Im m) better rough gauge configurations: ambiguous winding number overlap operator not unique: depends on domain wall height m = 0 for a non-degenerate quark is an ambiguous concept N f = 0: pure gauge theory Θ = π behavior unknown Michael Creutz BNL & U. Mainz 26

27 When is rooting OK? Starting with four flavors can we adjust N f using D + m D + m D + m D + m 1 4 = D + m? A vacuous question outside the context of a regulator! Rooting before removing regulator OK if regulator breaks anomalous symmetries on each factor i.e. four copies of the overlap operator: Dγ 5 = ˆγ 5 D winding ν from Trˆγ 5 = 2ν Michael Creutz BNL & U. Mainz 27

28 Forcing Z Nf symmetry with regulator in place D + me iπγ D + me iπγ D + me 3iπγ D + me 3iπγ 5 4 maintains m me iπγ 5/2 symmetry permutes flavors Four one-flavor theories with different values of Θ theta cancels out for the full four flavor theory Rooting averages four inequivalent theories: NOT OK ( D + M 1 D + M 2 ) 1/2 D + M1 M 2 Michael Creutz BNL & U. Mainz 28

29 Staggered fermions regulator maintains one exact chiral symmetry D s + m = D s + me iγ 5φ OK since actually a flavored symmetry separate into four effective tastes D i two tastes of each chirality Ds + me iγ 5φ D 1 +me iφγ D 2 +me iφγ D 3 +me iφγ D 4 +me iφγ 5 plus taste mixing (not the crucial issue) Michael Creutz BNL & U. Mainz 29

30 Rooting to get to one flavor NOT OK rooting does not remove the Z 4 tastes are not equivalent rooting averages inequivalent theories Rooted staggered fermions are not QCD! Extra minima from Z Nf expected to drive η mass down testable but difficult Michael Creutz BNL & U. Mainz 30

31 Summary QCD with N f massless flavors has a discrete Z Nf chiral symmetry flavor singlet Associated with a first order transition at Θ = π when m 0 Sign of mass significant for N f odd not seen in perturbation theory No symmetry for N f = 1 m = 0 unprotected Structure inconsistent with rooted staggered quarks References: Ann. Phys. 324 (2009), 1573: arxiv: arxiv: Michael Creutz BNL & U. Mainz 31

32 Extra slides: generalized quark masses Michael Creutz BNL & U. Mainz 32

33 General masses can give intricate phase diagrams: three flavors fixed m s vary m u, m d transition along m u = m dm s m d +m s ( ) Σ md Σ ( ) CP m s mu ms m s m s CP Σ ( ) Σ ( ) Michael Creutz BNL & U. Mainz 33

34 Generalized mass term Renormalizable: look at fermion bilinears Lorentz invariant Two flavors: m s ψψ + m v ψτ 3 ψ + im 3 ψγ 5 ψ + im 4 ψγ 5 τ 3 ψ m s average quark mass, isoscalar m v quark mass difference, isovector m 3 CP violating m 4 twisted mass (useful with Wilson fermions) These are not independent Fermion kinetic term symmetric under ψ e iφγ 5τ 3 ψ mixes m s with m 4 and m v with m 3 Michael Creutz BNL & U. Mainz 34

35 Can set any single m i = 0 and another m j > 0 Choose convention m 4 = 0 and m s > 0 m s ψψ + m v ψτ 3 ψ + im 3 ψγ 5 ψ 2 flavor QCD depends on three independent quark mass parameters m 2 π ± m s m 2 π ± m 2 π 0 m 2 v (+EM effects) neutron electric dipole moment m 3 Strong CP problem: why is m 3 experimentally extremely small? unification with CP violating weak interactions? Michael Creutz BNL & U. Mainz 35

36 Symmetries make each term multiplicatively renormalizable CP symmetry protects m 3 Isospin protects m v Flavored chiral symmetry protects m s m s and m v transform differently under isospin m v /m s renormalization can depend on scheme m p, m π± and m π0 constant m p, m π± and m v /m s constant not equivalent non-perturbatively ( renormalon ambiguity ) Michael Creutz BNL & U. Mainz 36

37 Renormalization group interpretation a dm da = mγ(g) = m(γ 0g 2 + γ 1 g ) + non-perturbative non-perturbative part vanishes faster in g than any power Bare quark masses run to zero logarithmically with the cutoff m = m r g γ 0/β 0 (1 + O(g 2 )) a 0 0 Renormalized quark mass m r = lim a 0 mg γ 0/β 0 integration constant of RG equation Numerical value of m r depends on scheme t Hooft vertex contributes a non-perturbative part m N f 1 not proporional to m for N f = 1 m η 1 a e 1/2β 0g 2 g β 1/β 2 0 similar form can appear in RG equation Michael Creutz BNL & U. Mainz 37

38 Example new scheme: ã = a g = g m = m m r g γ 0/β 0 e 1/2β 0 g2 g β 1 /β 2 0 Λa on RG trajectory the last factor approaches unity Non-perturbative redefinition of parameters makes m r lim a 0 m g γ 0/β 0 = m r m r = 0 m u = 0 is not a solution to the strong CP problem With degenerate quarks m π = 0 defines m = 0 Michael Creutz BNL & U. Mainz 38