Anomalies and discrete chiral symmetries

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1 Anomalies and discrete chiral symmetries 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 Talk 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 MC 1995: Why is chiral symmetry so hard on the lattice Why rehash old ideas? consequences have recently raised bitter 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 m 0 quarks: 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 m u = 0 cannot solve the strong CP problem 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 assume regulator allows defining composite fields 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 in a concave regions 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 Tr γ 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 = γ 5 Dγ 5 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 states go? continuum: lost at infinity above the cutoff Wilson: real eigenvalues in doubler region overlap: modes on opposite side of unitarity circle Dγ 5 = ˆγ 5 D Tr ˆγ 5 = 2ν This phenomenon involves both short and long distances zero modes compensated by modes lost at the cutoff 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 Back to effective field language At least two minima in the σ, η plane (σ, η ) = (0, ±v) η? v v σ? Question: do we know anything else about the potential in the σ, η plane? Yes! there are actually N f equivalent minima Michael Creutz BNL & U. Mainz 14

15 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 15

16 (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 valid symmetry Michael Creutz BNL & U. Mainz 16

17 η 1 σ Mass term mψψ tilts effective potential picks one vacuum (v 0 ) 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 17

18 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 18

19 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 19

20 ) 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 20

21 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! Theta dependence invisible to perturbation theory Michael Creutz BNL & U. Mainz 21

22 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 22

23 N f = 1: No chiral symmetry at all! η unique vacuum ψψ σ 0 from t Hooft vertex N =1 f 0 σ not a spontaneous symmetry breaking No singularity at m = 0 m = 0 not protected: renormalon ambiguity For small mass no first order transition at Θ = π larger masses? N f = 0: pure gauge theory Θ = π behavior unknown Michael Creutz BNL & U. Mainz 23

24 Michael Creutz BNL & U. Mainz 24

25 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 25

26 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 26

27 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 27

28 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! η 1 Extra minima from Z 4 expected to drive η mass down 2 0 σ testable but difficult 3 Michael Creutz BNL & U. Mainz 28

29 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 29

30 More on the m u issue General two flavor mass term and the strong CP problem Renormalizable: look at Lorentz invariant fermion bilinears m 1 ψψ + m 2 ψτ 3 ψ + im 3 ψγ 5 ψ + im 4 ψγ 5 τ 3 ψ m 1 average quark mass, isoscalar m 2 quark mass difference, isovector m 3 CP violating m 4 twisted mass (useful with Wilson fermions) Not independent Fermion kinetic term symmetric under ψ e iφγ 5τ 3 ψ mixes m 1 with m 4 and m 2 with m 3 Michael Creutz BNL & U. Mainz 30

31 Can set any single m i = 0 and another m j > 0 Choose convention m 4 = 0 and m 1 > 0 m 1 ψψ + m 2 ψτ 3 ψ + im 3 ψγ 5 ψ 2 flavor QCD depends on three independent quark mass parameters m 2 π ± m 1 m 2 π ± m 2 π 0 m 2 2 (+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 31

32 m 1 and m 2 transform differently under isospin m 2 /m 1 renormalization can depend on scheme m p, m π± and m π0 constant m p, m π± and m 2 /m 1 constant not equivalent non-perturbatively ( renormalon ambiguity ) m u = m 1 + m 2 + im 3 = 0 m 3 = 0 not even wrong Michael Creutz BNL & U. Mainz 32

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