Review of Scalar Mesons

D.V. Bugg, Queen Mary, London

1. A brief history

2. Sigma and Kappa, f0(980) and a0(980) DOs and DONTs

1. How resonances synchronise on thresholds

4. Q-Q states and Glueballs

5. Where do we go from here?

A brief history

1. Treiman and Goldberger

2. Gell-Mann and Levy, model

3. Goldstone

4. Adler zeros

5. Weinberg and current algebra; soft pions

6. Chiral Perturbation Theory

pp ->

p -> p +p ->+p

t = 2

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

The Adler zero needs to be built into the formalism, since it is adirect 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 -

f(elastic) = N(s)/D(s) = K/(1 – iKK = 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 amplitude small near threshold.But logically it MUST also appear in the denominator in the termiK

For the kappa, Bing Song’s form is adequate

For sigma and Kappa in production reactions, N(s) = constant

2 2A

A

A 2

2A

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.

Im D = -N

BES J/signal)

b1(1235)

BES data

Caprini et al

Caprini et al: 441 - i (272 ) MeV

BES pole=500 – i 264 MeV

Combined fit: 472 – i 271 MeV

+16 +9-8 -12.5

Essential difference: Caprini et al fit a quadratic in s,

The BES data near the KK threshold demand a cubic

see J. Phys. G34 (2007) 151,

hep-ph/0608081

remark: the coupling constant of the is proportional to

its width; if this changes, so does the .

The same remark applies to the K*(890) and

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) deg, while EU predicts 90 deg.

For ELASTIC scattering, T(res) g . Amplitudes are constrained to the

unitary circle, so phases of and fo add. Im T(el) = T(el) T(el)* ->

T = T + exp[2i fo).

In a production process, T(prodn) Gg , allowing

T(prod) [T( + T(f0)], with = 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() + exp(2iT(f0), where

2fo) + sin [sin (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 -> 1.

-1

2

In the K-matrix approach, it is common practice touse 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 phaseshould be given purely by the denominator. So if EU were correct, G1 and G2 should be real. But then you cannot fit thedata. By allowing them to be complex, you fit data correctly, but strictly speaking the denominator should be replaced by the phaserelation 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 mucheffect is if two broad resonances overlap strongly withclose to 1 (very rare).

My way of fitting the Kappa

D(s) = Do - s – i g (s) rho (s)

g (s) = G (s – s ) exp [ -s – A)]

The Adler zero is needed in BOTH and Ko(1430) amplitudes

The coupling of both to K’is important.

A simultaneous fit is made to LASS, BES and E791 data

only

Ko(1430)

2

A

j j j

D ->K

T( -> Kexp(-q )2

RESULT: M = 750 - i (342 +- 60) MeV+30

- 55

Buttiker, Moussallam et al use the Roy equations like Caprini et al

and find M=658 +- 13 MeV, = 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 K’ 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 pole is unavoidable and well determined by Lass data.____________________

How to make sense of the poles

Re s

physical region

Adler zero

Im s

Pole: s =0.15 – i0.26

Unitarity distorts the contours

on the physical s axis

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 the pole is almost beneath the Kthreshold

D(s)=M - s – (s)

Im g (s)F (s) (s – thr)

Re = 1 P ds’ Im (s’)

2

2 2

s’ - s

Near threshold, Re = 2g [4m (K) – s]

Behaves like a scaled version of M - s

At threshold, Re is positive definite

IMPORTANT: The Flatte prescription i[4M2(K)/s – 1] below threshold is WRONG

2______________s

2

2

f0(980) as an example

FF = exp(-3k )2

Threshold cusps

1/2

0802.0934

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 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), (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 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 a0(980) and

f0(980) is reproduced with a single coupling constant and SU3

factors. The movement of the poles with is tabulated.

The a0(980) disappears if is reduces by 40%;

and a0(980) become bound states below their thresholds

if increases by a factor ~2.5

C

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 and ):

The line-shape does not determine coupling to KK at all well;

it is strongly correlated with (M - s). However, the absolute rate

for decays depends on g^2(phi->KK)g^2(KK)g^2(or ).

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

___________

Glueballs

The f0(1370) has been reconfirmed in pp -> 3 data from Crystal

Barrel (700K events): hep-ex/0706.1341. It is also seen as a clear

peak in pp –> ; no confusion is possible there with f2(1270),

whose decay rate to 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 and phase shift

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 (1450) and

f2(1565). I DO fit the Cern-Munich data accurately allowing

some mixing between f0(1370) and f0(1500). The f0(1370) also

appears as a clear peak in BES data on J/

0

1310

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) -> 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 -> 4or 4K*K* or 4KK.

No K*K* is observed at 1790 MeV, probably because decays to

KK have more phase space.

_ _ _ _ _ _

_ _ _

The f0(2100) observed by Crystal Barrel is seen strongly in

and only weakly in Both it and f0(1500) are also seen

strongly in E865 data on pbar-p -> 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 (2190) agree

remarkably well predicted ratios from Lattice Gauge calculations of

Morningstar and Peardon, see Phys. Lett. B486 (2000) 49.

A Conjecture

Oset, Oller et al find they can generate many states from meson exchanges,

(including Adler zeros).

We know that ao(980) and fo(980) can be generated by mesonexchanges (including Alder zeros).

We also know that most resonances are generated by q-qbarforces (coloured gluons). But the has the right decay width togenerate the and K*(890) has the right width to generate the I suggest this is not accidental. Chiral Symmetry Breaking is a phasetransition and so is Confinement. I suggest there is an intimate relationbetween them: H11 V E

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

Conclusions

1. It is important to resolve the question whether there is there any evidence against Chiral Symmetry Breaking in the Ksystem? That seems unlikely since that evidence is strong for N and even .

2. Mini-summary of how to fit the :

(i) include the Adler zero in D(s) with Bing Song’s form for BOTH and Ko(1430).

(ii) Include the K’ channel with my new formula for Re(K’->K) below its threshold.

(iii) do NOT include non-resonant interfering background.

(iv) Fit ALL available sets of data simultaneously.

3. Meson exchanges (including Adler zeros) account for ao(980) and fo(980). Why look further?

4. 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 threshold.

5. If (q-qbar)(q-qbar) states exist, they will `melt’ near thresholds, because they unavoidably have long-range mesonic tails.