Chiral Suppression and SU(3) Symmetry in Scalar Glueball Decays

CHIN.PHYS.LETT.

Vol.25,No.5(2008)1609

Chiral Suppression and SU(3)Symmetry

in Scalar Glueball Decays

?

JIN Hong-Ying(金洪英)1??,LIU Shao-Min(刘少敏)1,ZHANG Zhu-Feng(张珠峰)1,LI Xue-Qian(李学潜)2

1

Zhejiang Institute of Modern Physics,Zhejiang University,Hangzhou 310027

2

Department of Physics,Nankai University,Tianjin 300071

(Received 11February 2008)

We theoretically evaluate the decay rates of a scalar glueball up to the order O (αs ),where only two Feynman diagrams contribute and a special attention is paid on possible ?avour SU(3)symmetry breaking in the process.It is concluded that the SU(3)?avour symmetry may be respected in any case.However,due to chiral suppression in G 0→ˉq q ,Γ(G 0→ˉq q ˉq q )very likely is larger than Γ(G 0→ˉq q ).These results are supported by the experimental

data on the decays of χc 0(1P )(0++

).Based on this result,we propose a criterion to identify the scalar glueball.

PACS:12.38.Lg,12.39.Mk

To identify glueballs in experiments is really a tough job.Recently an interesting criterion has been proposed by Chanowitz.[1]Based on the perturbative theory,Chanowitz suggested that the amplitude of G 0→ˉq q is chiral-suppressed and proportional to the current quark mass (see Fig.1.In that scenario,the ?avour SU(3)symmetry would be strongly broken in two-body decays of scalar glueballs.If the perturba-tive QCD were overwhelmingly dominant,this allega-tion would be absolutely valid.However in the con-cerned cases the non-perturbative QCD e?ects play an important role,one must check whether such con-clusion is still reliable.Certainly,such works would be best to be carried out via lattice QCD (LQCD)or other model-independent methods.However,in the absence of LQCD result ,we would like to get some hints from the present non-perturbative methods.First,if this allegation is valid,mixing between scalar glueballs and regular ˉq q quarkonia should be chiral-suppressed.When the non-perturbative e?ect are accounted,the picture would be di?erent.In the constituent quark model,the quarks are dressed with gluons,so that the masses of u and d quarks are around 300MeV and the mass of s quark is about 400MeV.Then the di?erences of the constituent quark masses are not so drastic as that of the current

masses,so the production rates of ˉs s and ˉu u ,ˉdd

should not be so apart as claimed by the authors of Ref.[1].Certainly,chiral suppression also disappears.It is well understood that the U(1)chiral symmetry can be bro-ken by either QCD vacuum or light quark masses.On the other aspect,since QCD is blind to ?avours,its vacuum does not break the SU(3)?avour symmetry whereas the second one obviously does.The famous U(1)puzzle indicates that the QCD vacuum domi-nantly a?ects the low energy phenomenology,namely a left-handed fermion can convert into a right-handed fermion by exchanging instantons.Such instanton-induced e?ects can be estimated in terms of the In-stanton Liquid Model.In this Letter,we give such an estimate.

The processes G 0→ππ,KK can occur via the diagrams shown in Fig.2.Probably,the contribution from Fig.2is larger than that from Fig.1,since Fig.2does not su?er from a chiral suppression.If that is true,the SU(3)symmetry should be well respected in the decays of a scalar glueball.We make a check by comparing the contribution from Fig.2with that from Fig.1.

Fig.1.G 0→ˉq q de-

cay.

Fig.2.G 0→ππ,KK

decay.

In order to make a comparison with the results of Ref.[1],following the method given by the authors of Ref.[1],we assume that the two gluons are on the mass shell and restrict our calculations only up to the leading order of αs .Indeed,a full order-by-order com-parison is impossible at present.

We obtain the ratio of the decay widths where the Γ1includes only the perturbative contribution and Γ2refers to the instanton e?ects on the widths.They are,respectively,

Γ1(G 0→ˉs s )Γ2(G 0→ˉq q )=916π2m 2s m 2

M 4G βs βq

·ln 2((1+βs )/(1?βs ))n 2ρ6I 2(1+β2q

),(1)where q can be u,d,s ,βx =

1?4m 2x /M 2G ,and M G

?Supported by the National Natural Science Foundation of China under Grant No 10775117.

??

Email:jhy@https://www.360docs.net/doc/776326239.html,

c

2008Chinese Physical Society and IOP Publishing Ltd

1610JIN Hong-Ying et al.Vol.25

is the mass of the scalar glueball.I is a parameter and de?ned as

I = θΛ0dθsin θ 10

dα

·

K 1((M G /2)ρ 1+β2q +2βq cos θ/

α(1?α))α(1?α)(M G /2) 1+β2q +2βq

cos θ·(3+βq cos θ)+ πθΛdθsin θ 1

0dα

·

K 1(Λρ/

α(1?α)(M G /2) 1+β2q +2βq

cos θ·(3+βq cos θ).

(2)

The upper bound of the integral θΛis set as

θΛ=cos ?1

12βq

(4Λ2/M 2G ?1?β2

q ),(3)and Λis a scale cut-o?which removes the infrared divergence in the McDonald function K 1.

The other parameters employed in the phenomeno-logical analysis are [3,4]ρ=1/0.6GeV ,n =8×10

?4

GeV 4

,

(4)

m =86MeV ,

m s =150MeV ,

M G =1.65GeV .

(5)

The cut-o?Λis set within the range of

200MeV ～500MeV.In fact,the ‘real’cut-o?which a?ects the physical transition amplitudes is Λ/ α(1?α),so that Λshould be su?ciently large,for example,setting Λ=500MeV.Our numerical result indicates that Γ1(G 0→ˉs s )/Γ2(G 0→ˉq q )is in the region of 0.04～0.43.If we choose the glueball mass as M G =5GeV,Γ1(G 0→ˉs s )/Γ2(G 0→ˉq q )is in the region of 0.09～1.0.Γ1is proportional to m 2s while Γ2induced by the instanton e?ect only mildly depends on the quark mass via parameter βq .It is noted that if a scalar glueball decays via the diagram in Fig.1,the instanton e?ect is comparable with Γ1as the strange quark mass takes reasonable values.That means very likely that there is still chiral suppression by the current mass of strange quarks but SU(3)sym-metry is not broken.However,it is still not the end of the story.Based on the e?ective Lagrangian given by Ref.[1],the authors of Ref.[2]showed that the decays G 0→ππ,KK are most possibly through the ?gure Fig.2rather than Fig.1,so there is not chiral suppression at all.Although Ref.[2]is also criticized by Chanowitz,[5]an important point indicated by the authors of Ref.[2]is that in an e?ective Lagrangian for the decay of scalar glueball there are not only the term F μνF μνG ,but also terms such as m ˉq qG ,ˉq q ˉq qG ···,where the mass parameter m somehow includes non-perturbative QCD e?ects.Therefore any conclusion which is drawn based on only the ?rst term is not

convincing.When considering contributions from the terms m ˉq qG ,ˉq q ˉq qG ,etc.,one needs more information about the coe?cients of these interactions,however so far they have not been fully determined due to lack of su?cient experimental data at present.Thus,a better choice may be starting from the full QCD.By the perturbative QCD,scalar glueballs can decay into ππand KK at the leading order in αs ,and the corre-sponding Feynman diagram is shown in Fig.2.Very likely,the contribution of Fig.2is larger than that of Fig.1just as shown in Ref.[2].In the following,we estimate the contribution of Fig.2.

Now we consider two decay modes of a 0++scalar glueball,namely the process G 0→ˉs s which is realized via Fig.1and G 0→ππvia Fig.2.Any exact calcu-lation of such processes is really tough.What we will do in following is only an estimate,in which we need use some ansatzes that have been widely used in other places.We hope uncertainties from these ansatzes are under control.

For the calculation,we need to know the wavefunc-tion of the glueball ?rst.It is reasonable to employ the Bethe–Salpeter wavefunction for the scalar glueball:

0|A a μ(x )A b

ν(y )|G (p ) =e

?i p

2·(x +y ) 0|A a μ((x ?y )/2)·A b ν(?(x ?y )/2)|G (p ) ,

(6) 0|A a μ(z )A b ν(?z )|G (p ) =δab (ψ1g μν+ψ2p μp ν/M 2

G

+ψ3z μp ν+ψ4z νp μ+ψ5z μz νM 2G ),

(7)

where ψi ,i =1,5are scalar functions of z and p .Since z =(x ?y )/2is small,[6]we can ignore the last three terms in Eq.(7)and obtain

0|A a μ(z )A b ν(?z )|G (p ) .=δab (ψ1g μν+ψ2p μp ν/M 2

G ).

(8)

We further assume that ψ1and ψ2in Eq.(8)can be expressed as A?G (z,p )and B?G (z,p ),thus the equation is somehow simpli?ed and becomes

0|A a μ(z )A b

ν(?z )|G (p )

=δab (Ag μν+Bp μp ν/M 2G )?G (z,p ),

(9)

where the function ?G (z,p )is normalized as

?G (0,p )=1,and A =ψ1(0,p ),B =ψ2(0,p ).The numerical results con?rm that this approximation is not too far from reality as it ?rst looks.Indeed,when the axial gauge A 0=0in the rest frame of the glueball is employed,ψ1(z,p )=?ψ2(z,p ).

Any four-momentum q μcan be divided into two parts:longitudinal and transverse.In our case,we project q μonto the glueball momentum p ,that is,

q μ=q μ +q μ⊥=q ·p M 2G p μ+q μ⊥=ξp μ

+q μ⊥.(10)With the help of Eq.(10),we can rewrite our glue-ball wavefunction as

?G (z,p )=

d 4q (2π)4

e ?iq ·z

f G (q 2

,q ·p )

No.5JIN Hong-Ying et al.1611

=

dξ

d3q⊥

(2π)3

e?iξz·p e?iz·q⊥f G(ξ,q2⊥).

(11)

In the rest frame of the glueball,q⊥is the three-dimensional relative momentum of the two‘con-stituent gluons’.The wavefunction of the two S-wave constituents with q⊥could be recast into the form[7,8]

f G(ξ,q2⊥)=f1(ξ)e?|q⊥|2/2β,(12) whereβis a con?nement parameter and should not be too far fromΛQCD;and f1(ξ)could be considered as the light-cone wavefunction.For a0++glueball, f1(ξ)=f1(?ξ)and f1(ξ)→0when|ξ|→1,[6]so the essential structure of f1(ξ)is～1?ξ2.In this work, we consider two possible forms for f1(ξ):～(1?ξ2) and～ξ2(1?ξ2)and will not consider the forms which include higher orders ofξ.

Collecting all these relations together,we obtain

?1G(z,p)=3(2π)3/2

4β3

e?iξz·p e?iz·q⊥

·e?|q⊥|2/2β2(1?ξ2)dξ

d3q⊥

(2π)3

,

(13)

?2G(z,p)=15(2π)3/2

4β3

e?iξz·p e?iz·q⊥

·e?|q⊥|2/2β2ξ2(1?ξ2)dξ

d3q⊥

(2π)3

.

(14)

With the explicit glueball wavefunction,we can start our analysis.Our?rst goal is to calculate the am-plitude corresponding to the process shown in Fig.1, i.e.a scalar glueball decays into an sˉs pair.

Using Eq.(13)or(14),we obtain the?nal expres-sion of the amplitude for G0→ˉs s:

M(G0→ˉs s)=128αsπ

3

·

m s[2A+B]

M2

G

(1+β2s)+4m2s

ˉsλ(p1)sλ(p2).

(15)

Our next task is to calculate the amplitude for a scalar glueball decaying into two pions(see Fig.2). Using Eq.(13)or(14)and the leading twist pion wave-function?π(ξ,μ),which is de?ned in the form[6]

0|ˉd(z)γμγ5u(?z)|π(q)

=ifπqμ

1

?1

dξe iξz·p?π(ξ,μ),z2=0,

(16)

whereμis the renormalization scale,we have

M(G0→π+π?)=8αsπ2M G f2π

(2π)3/2

β3

·[A(1+β2π)+B]·X i,

(17) where

X1=

3

4

1

?1

dξ1

1

?1

dξ2?π(ξ1,μ)?π(ξ2,μ)

·e?(ξ1+ξ2)2M2Gβ2π/8β2[1?(ξ2?ξ1)2/4],(18) X2=

15

16

1

?1

dξ1

1

?1

dξ2?π(ξ1,μ)?π(ξ2,μ)

·e?(ξ1+ξ2)2M2Gβ2π/8β2(ξ2?ξ1)2[1?(ξ2?ξ1)2/4].

(19)

Thus we obtain the ratio of the two decay widths as

Γ(G0→ˉs s)

Γ(G0→ππ)

=

128

9π

M2

G

?4m2s

M2

G

?4mπ

·

m2sβ2sβ6

f4π[M2

G

(1+β2s)+4m2s]2

·

[2A+B]2

[A(1+β2π)+B]2X i2

.(20)

Sinceβx≈1,this ratio does not depend on the pa-rameters A and B.

Several parameters that will be used in our numer-ical analysis are[9]

mπ=138MeV,fπ=130MeV,

β=200MeV?1000MeV.(21) Moreover,we set the renormalization scale atμ= 500MeV and use the following pion wavefunction:[6]?π(ξ,500MeV)=

15

4

ξ2(1?ξ2)(22) in our later calculations.

Table1.Value ofΓ(G0→ˉs s)/Γ(G0→ππ)(Γ1/Γi

2

)with various con?nement parameterβ.

β(GeV)0.20.30.40.50.60.70.80.9 1.0

Γ1/Γ2

10.00030.0020.0090.0250.0560.110.200.340.57

Γ1/Γ2

2

0.00010.000760.00350.0120.0330.0790.170.330.62

Table1presents the numerical results of our cal-culations.The values in Table1show that when β<500MeV,the width of G0→ˉs s is much smaller than that of G0→ππ.It is understood as Fig.2is non-factorizable and would be infrared divergent as β→0.Therefore,the theoretically estimated ratios shown in Table1are not reliable ifβtakes a small value;in other words,due to the infrared behaviour of Fig.2,the value ofβcannot be smaller than500MeV by our observation.On another aspect,the param-eterβrepresents the con?nement of the constituents in hadrons(glueball),and it must be somehow related

1612JIN Hong-Ying et al.Vol.25

toΛQCD which is about250～300MeV,and should not be too large.We believe that its reasonable value should be smaller than1GeV.Therefore one only can trust thatβis set in the region of500MeV<β< 1000MeV.It is quite consistent with the result ob-tained in Ref.[7].Moreover,the ratio is not sensitive to the form of the light-cone wavefunction in this re-gion.

By the values given in Table1,even if we cannot determine the concrete values of the ratio,we may conclude that the width of G0→ˉs s should not be larger than the width of G0→ππ.Obviously,the scalar glueball may decay mainly via the mechanism Fig.2rather than Fig.1.

Finally we brie?y discuss our observation and make a summary.

We have shown that the SU(3)?avour-symmetry most possibly is respected in Fig.1although we can-not exclude the contribution from Fig.1which is likely suppressed by the current mass of strange quarks. And our estimate shows that for decays of a scalar glueball,Fig.2very likely is more important than Fig.1which breaks the SU(3)?avour-symmetry,thus the SU(3)?avour-symmetry should be retained in all decay channels of scalar glueballs.This conclusion can be directly tested by the experimental data on the decays ofχc0(1P)(0++).Thisˉc c charmonium ?rst transits into two gluons and then the two gluons convert into light hadrons.Except a small di?erence in the phase space of the?nal states which is negligi-ble in our case as long as the mass of the glueball is much heavier than pion and kaon,the SU(3)?avour symmetry predictsΓ(χc0(1P)→KK)/Γ(χc0(1P)→ππ)=4/3.The experimentally measured ratio is Γ(χc0(1P)→KK)/Γ(χc0(1P)→ππ)=1.5±0.2[9] and it con?rms the SU(3)?avour symmetry in the de-cay.We note that the authors of Ref.[2]calculated Γ(χc0(1P)→KK)/Γ(χc0(1P)→ππ)～(f K/fπ)4, and predicted that the SU(3)symmetry would be slightly broken by a factor(f K/fπ)4.This allegation may be questionable,because they neglected a pos-sible source of the SU(3)symmetry breaking which is hidden in the light-cone wavefunctions ofπand K.[6]Certainly,if one includes all mass-dependent ef-fects,the light-cone wavefunctions are indeed depen-dent on the quark masses,[10,11]but such mass e?ects are smaller than a factor(f K/fπ)4=2.[12]

The ratio ofΓ(G0→ˉs s)/Γ(G0→ππ)<1indi-catesΓ(G0→ˉq qˉq q)>Γ(G0→ˉq q).By the QCD the-ory,it cannot happen for decays of a normalˉq q meson since theˉq qˉq q channel is O(αs)suppressed compared with the channelˉq q.This observation probably can be applied to identify scalar glueballs which have a com-pletely di?erent decay pattern from theˉq q mesons. Since a pair of hard quark and anti-quark can form a meson only when they are moving in parallel,there is a large possibility that the four particlesˉq qˉq q even-tually form three or even more hadrons.To our pic-ture,one can conjecture that compared with scalarˉq q mesons,the probability of a scalar glueball decaying into multi-hadron?nal states should be remarkably increased.The data on decays of low-lyingˉc c char-monia indeed exhibit such characters.Forχc0(1P)of 0++,the widths of three-body and four-body decays are almost four times larger than that for two-body decays.In the case of the0?+ηc(1S),the widths of three-body decays are twice that of two-body decays. In contrast with the above case,for J/ψ(1S)of1??, the widths of two-body decays are several times larger than that of multi-body decays.Although some ob-servable multi-body decays are secondary products of the two-body decays,such contrast is still obvious.

Let us look at three possible candidates for scalar glueballs:f0(1370),f0(1500)and f0(1710).The ex-perimental data show f0(1710)dominantly decays into two hadrons,[9]so very likely it is only aˉq q meson. Whereas,both f0(1370)and f0(1500)have a large fraction into4π,we expect that the further experi-ments should conduct more accurate measurements to determine the fractions of such scalars decaying into 4π,by which we may draw a de?nite conclusion about their quark structure,namely if they are indeed glue-balls.As a matter of fact,nowadays,people tend to believe that pure glueballs may not exist,but hide in a mixture with regular quarkonia,on other aspect, our work indicates that by accurate experimental mea-surements,one can determine the fraction of glueball components in the mixture,if it is su?ciently large, we may really name the mixture as a glueball. References

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