Review of Scalar Mesons

1. Review of Scalar Mesons D.V. Bugg, Queen Mary, London • A brief history • Sigma and Kappa, f0(980) and a0(980) • DOs and DONTs • How resonances synchronise on thresholds • 4. Q-Q states and Glueballs • 5. Where do we go from here?

2. A brief history 1. Treiman and Goldberger • Gell-Mann and Levy, s model • Goldstone • Adler zeros • Weinberg and current algebra; soft pions • Chiral Perturbation Theory

3. pp -> p-p+ 2 t = m p-p -> p-p p+p ->p+p f = N(s)/D(s), where N(s) describes exchanges from u and t channels; D(s) describes rescattering from the right-hand cut and therefore the phase. D(s) should be common to all processes for a single resonance

4. 2 2 A The Adler zero needs to be built into the formalism, since it is a direct consequence of Chiral Symmetry Breaking, We want to describe the S-wave, which needs to be projected out. In the S-wave, the zero appears at s = M - m /2 f(elastic) = N(s)/D(s) = K/(1 – iKr) K = b(s – s ) in the simplest possible form -> b(s – s ) exp[-(s – M )/B] . . . . Bing Song Zou -> (b1 + b2 s)(s – s )exp[-(s – M )/B] ….DVB for BES data The CRUCIAL point is that the Adler zero appears in the numerator, making the pp amplitude small near threshold. But logically it MUST also appear in the denominator in the term iKr. For the kappa, Bing Song’s form is adequate For sigma and Kappa in production reactions, N(s) = constant Im D = -N A 2 A 2 A Oset, Oller, Pelaez et al have made extensive fits to data from 1993 to the present including the effects of Chiral Symmetry Breaking and the Adler zero. They give formulae which resemble closely those used here – the crucial feature being the Adler zero.

5. BES J/y -> wpp (39K signal) s b1(1235) s

6. +16 +9 Caprini et al: 441 - i (272 ) MeV BES pole=500 – i 264 MeV Combined fit: 472 – i 271 MeV -8 -12.5 Essential difference: Caprini et al fit a quadratic in s, The BES data near the KK threshold demand a cubic Caprini et al BES data see J. Phys. G34 (2007) 151, hep-ph/0608081 remark: the coupling constant of the r is proportional to its width; if this changes, so does the s. The same remark applies to the K*(890) and k.

7. Extended Unitarity: arXiv 0801.1908 Central production data from the ISR require constructive interference of f0(980) with sigma on the lower side of f0(980). Phasedifference between sigma and f0(980) = 36 +- 5 deg, while EU predicts 90 deg. For ELASTIC scattering, T(res) a g . Amplitudes are constrained to the unitary circle, so phases of s and fo add. Im T(el) = T(el) T(el)* -> T = T(s) + exp[2i d(s)] T(fo). In a production process, T(prodn) a Gg , allowing T(prod) a [T(s) + b T(f0)], with b = 0.6. I propose whatever linear combination is produced rescatters as itself, giving a different unitarity relation: Im T(prod) = T(prod) T(prod)* . Then T = T(s) + exp(2iY) b T(f0), where 2Y = 2d(s) - d(fo) + sin [b sin d(fo)]. This prescription fits the phase angle between sigma and f0 perfectly. CONCLUSION: Aitchison chose the wrong unitarity relation. Elastic scattering is a SPECIAL case when b -> 1. 2 p p -1

8. In the K-matrix approach, it is common practice to use complex coupling constants G for each resonance: f = G1*K1 + G2*K2 ---------------------- . (1 – i K(el) rho) However this conflicts with EU, which says that the phase should be given purely by the denominator. So if EU were correct, G1 and G2 should be real. But then you cannot fit the data. By allowing them to be complex, you fit data correctly, but strictly speaking the denominator should be replaced by the phase relation from the previous page. CONCLUSION: data have been fitted (almost) correctly, but under a misunderstanding about the correct formula. The only place where the new unitarity relation has much effect is if two broad resonances overlap strongly with b close to 1 (very rare).

9. 2 j j j 2 Kp A My way of fitting the Kappa D(s) = Do - s – i S g (s) rho (s) g (s) = G (s – s ) exp [ -g(s – A)] The Adler zero is needed in BOTH k and Ko(1430) amplitudes The coupling of both to Kh’is important. A simultaneous fit is made to LASS, BES and E791 data k only D ->Kpp Ko(1430)

10. T(k -> Kp) = Lexp(-aq ) 2 +30 - 55 RESULT: M = 750 - i (342 +- 60) MeV Buttiker, Moussallam et al use the Roy equations like Caprini et al and find M=658 +- 13 MeV, G = 557 +- 24 MeV. No doubt this pole exists. However they fit only up to 1200 MeV/c and omit the low mass tail of K0(1430) and the Kh’ threshold. Because of this their errors are too small and their value for the mass too small. Zheng, Zhou et al (Peiking Univ) also found that if the Adler zero is included, the k pole is unavoidable and well determined by Lass data. ____________________

11. pp Adler zero physical region Re s Unitarity distorts the contours on the physical s axis Pole: s =0.15 – i0.26 How to make sense of the poles Im s Alternative way of saying the same thing: The Cauchy-Riemann equations require d Re f = d Im f d Re f = d Im f d Re s d Im s d Im s d Re s For the k, the pole is almost beneath the Kp threshold

12. The effect of thresholds Threshold cusps 0802.0934 f0(980) as an example 2 FF = exp(-3k ) 2 D(s)=M - s – P(s) Im P = g r(s)F (s) Q(s – thr) Re P = 1 P ds’ Im P(s’) 2 2 p s’ - s 2 2 Near threshold, Re P = 2g [4m (K) – s] Behaves like a scaled version of M - s At threshold, Re P is positive definite IMPORTANT: The Flatte prescription i[4M2(K)/s – 1] below threshold is WRONG ______________ p s 2 1/2

13. From numerical tests with f0(980), perturbing its parameters, a bare cusp does not generate a resonance. However, it can add to meson exchanges, which vary slowly with s, and the coherent sum generates a resonance locked to the threshold. As mass and coupling constants vary, the pole position changes little and the peak in pp hardly moves. Likewise, if a threshold adds coherently to a resonance created by confinement, it can attract the resonance over at least +-100 MeV. Examples: f2(1565), Ko(1430), L (2940), P11(1710), P13(1720). The X(3872) may be the c-cbar 3P1 radial excitation captured by the D(1865)D(2007) threshold. The a0(980) likewise appears to be captured by the KK threshold. It moves away from the hp threshold because of the Adler zero in that channel – see van Beveren, Bugg, Kleefeld, and Rupp, Phys. Lett. 92 (2006) 265, where the whole nonet of s, k, a0(980) and f0(980) is reproduced with a single coupling constant L and SU3 factors. The movement of the poles with L is tabulated. The a0(980) disappears if L is reduces by 40%; s, k and a0(980) become bound states below their thresholds if L increases by a factor ~2.5 C

14. Near the opening threshold, a resonance has a long tail like the deuteron. This tail reduces the zero-point energy (quite large for a state confined in a sphere of 0.8 fm or less). This further stabilises narrow states at thresholds. A formula of Tornqvist shows that f0(980) is ~60% KK; the a0(980) is ~35% KK. A conseqence is that f0(980) has an abnormally strong branching ratio to KK. Comment on Kloe data on f -> g(p p ) and g(hp ): The line-shape does not determine coupling to KK at all well; it is strongly correlated with (M - s). However, the absolute rate for f decays depends on g^2(phi->KK)g^2(KK)g^2(pp or hp). It is important to USE this absolute rate to determine g^2(KK). This stabilises the fit enormously and eliminates correlations between parameters, see hep-ex/0603023. 0 0 0 2 ___________

15. 0 Glueballs The f0(1370) has been reconfirmed in pp -> 3p data from Crystal Barrel (700K events): hep-ex/0706.1341. It is also seen as a clear peak in pp –> hhp ; no confusion is possible there with f2(1270), whose decay rate to hh is far too small to account for the peak. Ochs objects that it is not seen clearly in Cern-Munich data. However, above the KK threshold, inelasticity h and phase shift d are VERY strongly correlated, with the result that resonant loops cannot be traced accurately – hence the failure to detect f0(1500) there until Crystal Barrel found it, and also to detect r(1450) and f2(1565). I DO fit the Cern-Munich data accurately allowing some mixing between s, f0(1370) and f0(1500). The f0(1370) also appears as a clear peak in BES data on J/y -> fpp. 1310

16. The f0(1370/1310), f0(1500) and f0(1710) require a glueball mixing with n-nbar and s-sbar. The f0(1710) is very unlikely to be dominantly glueball because it only decays to KK. [BES 2 data identify clearly a separate f0(1790) in pi-pi and 4pi; this is likely to be the radial excitation of f0(1370/1310)]. The largest glueball component is in f0(1500). BES 2 see an X(1812) -> fw. Itfalls too fast above 1812 MeV for consistency with phase space, and must therefore be a decay channel of f0(1790). A glueball has composition (uu + dd + ss)(uu + dd + ss). 2(uu + dd)ss -> 4fw or 4K*K* or 4KK. No K*K* is observed at 1790 MeV, probably because decays to KK have more phase space. _ _ _ _ _ _ _ _ _

17. The f0(2100) observed by Crystal Barrel is seen strongly in hh and only weakly in pp. Both it and f0(1500) are also seen strongly in E865 data on pbar-p -> hh. The f0(2100) is remarkable for being the strongest state observed in all Crystal Barrel data in flight. It is unlikely to be s-sbar, since other states like f2(1525) and f0(1710) are produced very weakly if at all. The mass ratios of f0(1500), f2(1980), f0(2100) and h(2190) agree remarkably well predicted ratios from Lattice Gauge calculations of Morningstar and Peardon, see Phys. Lett. B486 (2000) 49.

18. A Conjecture Oset, Oller et al find they can generate many states from meson exchanges, (including Adler zeros). We know that s, k, ao(980) and fo(980) can be generated by meson exchanges (including Alder zeros). We also know that most resonances are generated by q-qbar forces (coloured gluons). But the r has the right decay width to generate the s, and K*(890) has the right width to generate the k. I suggest this is not accidental. Chiral Symmetry Breaking is a phase transition and so is Confinement. I suggest there is an intimate relation between them: H11 V Y = E Y V H22 The Variational Principle ensures the minimum E is the Eigenstate. Effectively the Confinement potential is created partly by coloured gluons at short range and partly by meson exchanges at long range. There is a close analogy to the covalent bond in chemistry

19. Conclusions • It is important to resolve the question whether there is there any evidence against Chiral Symmetry Breaking in the Kp system? That seems unlikely since that evidence is strong for pp, pN and even hp. • 2. Mini-summary of how to fit the k : • (i) include the Adler zero in D(s) with Bing Song’s form for BOTH k and Ko(1430). • (ii) Include the Kh’ channel with my new formula for Re(Kh’->Kp) below its threshold. • (iii) do NOT include non-resonant interfering background. • (iv) Fit ALL available sets of data simultaneously. • Meson exchanges (including Adler zeros) account for s, k, ao(980) and fo(980). Why look further? • Cusps are generated in the real part of the amplitude at an opening threshold. The KK threshold captures ao(980) and fo(980). Cusps can also capture q-qbar resonances, e.g. f2(1565) at the ww threshold. • If (q-qbar)(q-qbar) states exist, they will `melt’ near thresholds, because they unavoidably have long-range mesonic tails.