The scalar meson puzzle from a linear sigma model perspective

Transcription

1 Montpellier, December 009 The scalar meson puzzle from a linear sigma model perspective Renata Jora (Grup de Fisica Teorica and IFAE, Universitat Autonoma de Barcelona) Collaborators: Amir Fariborz(SUNY Institute of Technology), Joseph Schechter(Syracuse University). 1

2 OUTLINE Introduction Scalars A generalized linear sigma model Conclusions

3 Introduction As it is well known QCD coupling constant is large at small energies thus making usual perturbation theory unapplicable. In order to describe this regime we need other methods. One of the tools for investigating theories otherwise untractable is the use of effective Lagrangians like chiral perturbation one or linear sigma models. Advantage of the linear sigma model: the scalars are present from the beginning( tree level) in the theory. Another advantage of linear sigma models: can be generalized to include four quark states thus being able to shed light on the scalar meson puzzle. But first let us see what is the problem with the scalar meson spectrum. 3

4 Scalar mesons Scalar meson Mass Width Isospin I f 0 (600) I = 0 κ(800) 841 ± ± 90 I = 1 f 0 (980) 980 ± to 100 I = 0 a 0 (980) ± to 100 I = 1 f 0 ( to to 500 I = 0 K0 (1430) 1414 ± 6 90 ± 1 I = 1 a 0 (1450) 1474 ± ± 13 I = 1 f 0 (1500) 1507 ± ± 7 I = 0 f 0 (1710) 1718 ± ± 8 I = 0 Table 1: List of the scalar mesons as taken from PDG. 4

5 The scalar meson puzzle For the light vector meson nonet the masses of the particles increase with the strange quark content: ρ ω uū d d uū + d d 770 MeV 780 MeV K u s 880 MeV φ s s 100 MeV (1) The same mass order is approximately valid for the pseudoscalars. However for a conceivably q q low lying scalar nonet this ordering is not respected. σ(600) s s κ(800) u s a 0 (980) 500 MeV 800 MeV uū d d 980 MeV f 0 (980) uū + d d 980 MeV () 5

6 Four quark states Four quark states have been introduced by Jaffe(PRD1977) in the context of the MIT bag model. A possible solution to the scalar meson puzzle; it is conjectured that the light scalar mesons (f 0 (500), f 0 (980), a 0 (980), κ(900) are an admixture of two quark and four quark states. (Fariborz, Jora, Schechter, PRD05). There are various color blind combination that can be formed as: a) molecule type made out of two quark-antiquark states. b) bound state of a diquark and an anti-diquark(here there are two possibilites depending on the choice of color representation). 6

7 Various four quark realizations ( as considered for a linear sigma model but with larger validity) The schematic chiral quark-antiquark field M b a = (q ba ) γ γ 5 q aa, (3) The two quark-two antiquark field One possibility the light scalars are molecules made out of two pseudoscalar mesons. M ()b a = ɛ acd ɛ bef ( M ) c( e M ) d. (4) f Another possibility bound states of a dual quark and dual antiquark. Dual antiquark a 3 representation of flavor SU(3), a 3 of color and a spin singlet L ge = ɛ gab ɛ EAB q T aac γ 5 R ge = ɛ gab ɛ EAB q T aac 1 1 γ 5 q bb, q bb, (5) M (3)f g = ( L ga) R fa. (6) 7

8 Dual antiquark a 6 representation of color, spin 1 L g µν,ab = Lg µν,ba = ɛgab q T aac 1 σ µν 1 + γ 5 R g µν,ab = Rg µν,ba = ɛgab q T aac 1 σ µν 1 γ 5 M (4)f g = q bb, q bb, (7) ( ) R L g f µν,ab µν,ab, (8) Explicit quark realization of the four quark fields [ ] [ ] = ɛ gab ɛ fde q LaA q RdA q LbB q ReB. (9) M ()f g [ ] [ ] g = ɛ gab ɛ fde ( q LaA q RdA q LbB q ReB [ ] [ ] q LaA q RdB q LbB q ReA ). (10) M (3)f [ ] [ ] g = 4ɛ gab ɛ fde ( q LaA q RdA q LbB q ReB [ ] [ ] + q LaA q RdB q LbB q ReA ). (11) M (4)f Using Fierz identities we obtain: M ()b a = M (3)b a M (4)b a 8 (1) 8

9 The most general linear sigma model The Lagrangian L = 1 T r( µm µ M ) 1 T r( µm µ M ) V 0 (M, M ) V SB (13) Here M and M correspond to two quark and four quark fields, respectively. M = S + iφ M = S + iφ (14) where S, S are the scalars and Φ and Φ are the pseudoscalars. S, S, Φ and Φ are all hermitian matrices. The Lagrangian has an SU(3) L SU(3) R symmetry broken explicitly by the symmetry breaking term V SB. 9

10 Examples of symmetric terms in the Lagrangian V 0 = c T r(mm + c 3 (det(m) + h.c.) +c a 4T r(mm MM + c b 4(T r(mm ) +d T r(m M ) + d 3 (det M + h.c.) +d a 4T r(m M M M ) + d b 4(T r(m M )) +e (T r(mm + h.c.) +e b 3(ɛ abc ɛ def M a d M b e M c f +e b 3(ɛ abc ɛ def Md a M e b M f c + h.c.) + h.c.) e a 4T r(mm M M + e b 4T r(mm M M ) +e c 4[T r(mm MM ) + h.c.] + e d 4[T r(mm MM ) + h.c.] +e e 4[T r(m M M M + h.c.] +e f 4 T r(mm )T r(m M +e g 4 T r(mm T r(m M ) + e h 4[(T r(mm )) + h.c.] +e i 4[T r(mm T r(mm ) + h.c.] +e j 4 [T r(m M )T r(m M ) + h.c.] (15) 10

11 Example of symmetry breaking terms in the Lagrangian V SB = k 1 [T r(am) + h.c.] + k [T r(am ) + h.c] +k 3 [T r(amm M) + h.c.] + k 4 [T r(amm M ) + h.c.] k 5 [T r(amm M ) + h.c.] + k 6 [T r(amm M) + h.c.] +k 7 [T r(am M M ) + h.c.] + k 8 [T r(am M M) + h.c.] +k 9 [T r(am M M) + h.c.] + k 10 [T r(am M M ) + h.c.] +k 11 [T r(am) + h.c.]t r(mm ) +k 1 [T r(am) + h.c]t r(m M ) +k 13 [T r(am)t r(mm ) + h.c.] +k 14 [T r(am)t r(m M ) + h.c.] +k 15 [T r(am ) + h.c.]t R(MM ) +k 16 [T r(am ) + h.c]t r(m M ) +k 17 [T r(am )T r(mm ) + h.c.] +k 18 [T r(am )T r(m M ) + h.c.] +k 19 A b aɛ bcd ɛ aef M c e M d f + h.c. +k 0 A b aɛ bcd ɛ aef M e c M f d + h.c. +k 1 A b aɛ bcd ɛ aef M c e M d f + h.c. (16) 11

12 Methods In the context of a linear sigma model with SU(3) L SU(3) R two possible approaches exist: to consider the most general renormalizable SU(3) L SU(3) R potential V 0 and a particular choice of the symmetry breaking term V SB. In this case an incomplete picture of the masses and mixings is obtained using the generating equations(the symmetries of the model). to consider a particular(a limited number of terms) potential V 0 and a particular symmetry breaking term V SB References: Phys Rev D [68,034001], [76, ], [76, ], [77, ], [79, ]. 1

13 By using the general approach we can: find the masses and the mixings for the π π, K K and κ κ systems prove the current algebra results for the limit of massless quarks and also for an approximation of the massive case for which tan θ π = β α key ingredient We need a specific potential in order to find more information about mixing and masses and about the corrections to the current algebra for pion-pion scattering. The specific potential has been studied for various cases of the symmetry breaking term: a)v SB = 0 b)v SB = T r(as) with A 1 = A = A 3 SU(3) invariant limit c)v SB = T r(as) with A 1 = A SU() invariant limit 13

14 Specific potential First we list the terms in the potential: V 0 = c Tr(MM ) + c a 4 Tr(MM MM ) + d Tr(M M ) + e a 3(ɛ abc ɛ def M a d M b e M c f + h.c.) + c 3 [γ 1 ln( detm detm ) + (1 γ 1) T r(mm ) T r(m M ) ]. (17) All the terms except the last two have been chosen to also possess the U(1) A invariance. Those terms are clearly non-renormalizable and violate U(1) A invariance such that the resulting Lagrangian can exactly mock up the U(1) A anomaly of QCD. We impose the criterion that effective vertices describing the smallest numbers of quarks plus antiquarks be retained. This quantity, representing the total number of fermion lines at each effective vertex can be written as, N = n + 4n, (18) Here n is the number of times M or M appears in each term while n is the number of times M or M appears in each term. We consider N max = 8. 14

15 The minimum equations and the counting of parameters The minmum equations read: V0 S + VSB S = 0, V0 S = 0. (19) There are 1 parameters in the model which get reduced to 8 by the minimum equations: The main inputs are given by: m[a 0 (980)] = MeV m[a 0 (1450)] = MeV m[π(1300)] = MeV m π = 137 MeV F π = 131 MeV (0) Also as input we take the ratio A 3 A 1 over an appropriate range. which is varied 15

16 Mass matrices One obtains respectively matrices for the pions, the kaons, the a s and the kappa s. As an example: ( M π ) = 4 ea 3 β 3 c + 4 c a 4 α1 4 e a 3 α 3 4 e a 3 α 3 d (1) ( X a ) = 4 ea 3 β 3 c + 1 c 4 α1 4 e a 3 α 3 4 e 3 α 3 d () One obtains 4 4 matrices for the I=0 scalars and pseudoscalars. For this matrix the basis states are consecutively: f a = S1 1 + S f b = S 3 3 s s, f c = S S n n, ns n s, f d = S 3 3 nn n n. (3) 16

17 The I=0 pseudoscalars The 4 4 mass matrix for the I = 0 pseudoscalars is quite complicated.there are two additional parameters that contribute c 3 and γ 1. These are coming from the instanton type of terms in the potential. The parameters c 3 and γ 1 are determined from the following equations: Tr ( Mη ) = Tr ( Mη ) exp det ( Mη ) = det ( Mη ) (4) We identify the lighest two η s predicted by our model (i.e. η 1 and η ) with η(547) and η (958) with experimental masses: exp m exp. [η(547)] = ± 0.04 MeV, m exp. [η (958)] = ± 0.4 MeV. (5) 17

18 I=0 pseudoscalars For the two heaviest states there are four experimental candidates. m exp. [η(195)] = 194 ± 4 MeV, m exp. [η(1405)] = ±.5 MeV, m exp. [η(1475)] = 1476 ± 4 MeV, m exp. [η(1760)] = 1756 ± 9 MeV. (6) We consider all six possible choices for identifying η 3 and η 4 with two of the indicated candidates. This fixes A 3 /A 1 = 30 and m π (1300) = 1.5 GeV 18

19 Predicted masses for kaons and kappa s K / with A 3 / A 1 = 30 K with A 3 / A 1 = 30 K / with A 3 / A 1 = 0 K with A 3 / A 1 = κ / with A 3 / A 1 = 30 κ with A 3 / A 1 = 30 κ / with A 3 / A 1 = 0 κ with A 3 / A 1 = Figure 1: Predicted masses for kaons and kappa s vs m[π(1300)] (GeV). We obtain: K(515), K (1195), κ(1067) and κ (164). 19

20 Predicted masses for I=0 scalars I=0 scalars with A 3 / A 1 = I=0 scalars with A 3 / A 1 = Figure : Predicted masses for isoscalars vs m[π(1300)] (GeV). We obtain: f 1 (74), f (1085), f 3 (1493) and f 4 (1784). 0

21 I=0 pseudoscalars; for the most realistic scenario The first mass is around 550 MeV (consistent with identifying it with η(547)), the second mass is in the range of MeV (consistent with identifying it with η(985)), the third mass is in the range of MeV (consistent with identifying it with η(195)) and the fourth mass is around MeV (consistent with identifying it with η(1760)) I=0 pseudoscalars with A 3 / A 1 = Figure 3: Predicted η masses vs m[π(1300)] (GeV). 1

22 Four quark percentages a 0 κ π K A 3 / A 1 = a 0 κ π K A 3 / A 1 = Figure 4: Predicted percentage of four quark contents vs m[π(1300)] (GeV).

23 Four quark percentages for the isoscalars There are four isoscalars as mentioned: The lowest I=0 scalar f 1 = σ for A 3 /A 1 = 30 and m[π(1300)] = 115 MeV has percentages in each of the f a, f b, f c, f d bases: (7) so σ is about 40 percent -quark and 60 percent 4-quark The next heaviest scalar f has 95 percent 4-quark and 5 percent -quark. The two heaviest scalars have both mostly -quark percentages: thus f 3 is 63 percent -quark and f 4 is 93 percent -quark. 3

24 Four quark percentages for I=0 pseudoscalars For all six scenarios considered a consistent picture emerges. As opposed to the scalars the lowest states(η(547)and η(985)) are mostly two quark structures while the heaviest states(η(195) and η(1760)) are mostly four quark structures η 1 with A 3 / A 1 = Figure 5: Predicted quark content for η 1 identified with η(547) vs m[π(1300)] (GeV). 4

25 Partial conclusions How do the results for an SU() symmetry breaking terms compare to the other two cases we considered, the limit of massless quark and an SU(3) invariant symmetry breaking term?. From the point of view of masses and predictions the SU() case is far more accurate and in better agreement with the experimental data, as expected. However the same qualitative picture emerges: the lowest lying pseudoscalars are mainly two quark fields as opposed to the lowest lying scalars that are mainly four quark structures. The reverse is true for the heaviest states. This can explain the inverted mass spectrum for the light scalars. Thus the main features of the model do not really depend on the symmetry breaking term. The two simplifying cases mentioned are very useful for analyzing other aspects like pion pion scattering amplitude and the low energy theorems. 5

26 PION-PION SCATTERING First we proved that the current algebra results hold for a general potential in the limit of zero quark masses, even for this case when we have a mixing between two quark and four quark states(thus the contribution from four scalars; they also hold for an SU(3) invariant symmetry breaking term if we make the approximation tan θ π = β α, so that can write the following expansion for A(s, t, u): A(s, t, u) = g g i + m i i m π [ (s m π) F π [ 1 + s m π m i m π + (s m π) i + ( s ] m π m i ) + m π ] g i (m i m π) 3 + (8). 6

27 SCATTERING LENGTHS The partial wave scattering lengths calculated in this model are: m π a 0 0 = m π a 0 = 1 3π 1 3π [ [ 5g + i g + i gi gi m i ( 3 m i 4m π ] + m i ) ],. (9) Note that the isospin label, I and the angular momentum label, J appear as a I J. The current algebra results give: m π a 0 0 = 7m π 16πF π, m π a 0 = m π 16πFπ. (30) 7

28 The results of our numerical calculation are shown in Figure a 0 0 a m [π(1300)] (GeV) Figure 6: Top curve: I = J = 0 scattering length, m π a 0 0 vs. m[π(1300)]. Bottom curve: I =, J = 0 scattering length, m π a 0 vs. m[π(1300)]. The error bars reflect the uncertainty of m[a 0 (1450)]. 8

29 SCATTERING LENGTHS The partial wave scattering lengths coming form the experiment: NA48/ collaboration: m π +(a 0 0 a 0) = 0.64 ± (31) m π +a 0 0 = 0.56 ± (3) E865 Collaboration: m π +a 0 0 = 0.16 ± (33) DIRAC Collaboration: m π +a 0 0 = (34) The value of a 0 0 calculated in the model gives good agreement with the experiment for m[π(1300)] > 115 MeV. The current algebra prediction is too low (0.15). The result for the a 0 also gives good agreement with the experiment. 9

30 DISCUSSION OF THE SCATTERING AMPLITUDE Four-pion S 0 (1) S 0 () S 8 (1) S 8 () Total m [π(1300)] (GeV) Figure 7: Individual contributions in the equation for A(s, t, u) (which contains only the first two terms in the expansion) at threshold. The main contribution before the cancelation doesn t come from the lightest scalar. 30

31 DISCUSSION OF THE SCATTERING AMPLITUDE (1) S 0 () S 0 (1) S 8 () S m [π(1300)] (GeV) Figure 8: Individual contributions to the fourth term in the expansion for A(s, t, u) at threshold. The main contribution after cancelation comes from the lightest scalar. 31

32 Conclusions Using a generalized linear sigma model which contains both -quark and 4-quark fields we were able to shed light on the scalar meson puzzle. We obtained that the lightest scalar mesons have a large four quark percentage as opposed to the heaviest one. This can explain their inverted mass spectrum. The lightest pseudoscalars are mostly two quark structures while the heaviest have alarge four quark component. These features are present for the three cases studied: in the massless quark limit for an SU(3) invariant symmetry breaking term and for an SU() invariant one. We also proved that the current algebra results hold, even in this case when we have a mixing of two chiral nonets, in the limit of massless quark or as a first order approximation for an SU(3) symmetry breaking term. The detailed treatment of the pion-pion scattering shows that the model can give controlled corrections to the current algebra scattering formula. 3