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Advances in High Energy Physics
VolumeΒ 2008, Article IDΒ 982341, 16 pages
http://dx.doi.org/10.1155/2008/982341
Research Article

Hybrid Charmonium and the πœŒβˆ’πœ‹ Puzzle

Department of Physics, Carnegie Mellon University, Pittsburgh, PA 15213, USA

Received 15 August 2008; Revised 25 October 2008; Accepted 19 November 2008

Academic Editor: SandipΒ Pakvasa

Copyright Β© 2008 Leonard S. Kisslinger et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Using the method of QCD sum rules, we estimate the energy of the lowest hybrid charmonium state, and find it to be at the energy of the Ξ¨ξ…ž(2𝑆) state, about 600 MeV above the 𝐽/Ξ¨(1𝑆) state. Since our solution is not consistent with a pure hybrid at this energy, we conclude that the Ξ¨ξ…ž(2𝑆) state is probably an admixed 𝑐𝑐 and hybrid 𝑐𝑐𝑔 state. From this conjecture, we find a possible explanation of the famous πœŒβˆ’πœ‹ puzzle.

1. Introduction

A hybrid meson is a state composed of a quark and antiquark color octet, along with a valence gluon, giving a color zero particle. Hybrids are of great interest in studying the nature of QCD. The nonperturbative method of QCD sum rules [1, 2] has long been used to predict the energies of light quark hybrid mesons [3, 4] and hybrid baryons (see [5]). In the present work, we use this method to estimate the energy of the lowest charmonium hybrid. A major motivation for the present work is to understand the nature of the Ξ¨ξ…ž(2𝑆) state, and to find a possible explanation of the long-standing πœŒβˆ’πœ‹ puzzle.

The πœŒβˆ’πœ‹ puzzle concerns the branching ratios for hadronic decays of the Ξ¨ξ…ž(2𝑆) state compared to the 𝐽/Ξ¨(1𝑆) state. By taking ratios of hadronic decays to gamma decays for these heavy-quark states, the wave functions at the origin cancel, and by using the lowest-order diagrams, one obtains the ratios of branching rates for two charmonium states: 𝑅=𝐡(Ξ¨ξ…ž(2𝑆)β†’β„Ž)=𝐡(𝐽/Ξ¨(1𝑆)β†’β„Ž)𝐡(Ξ¨ξ…ž(2𝑆)→𝑒+π‘’βˆ’)𝐡(𝐽/Ξ¨(1𝑆)→𝑒+π‘’βˆ’)≃0.13,(1.1)the so-called 13% rule. For the Ξ¨ξ…ž(2𝑆) state compared to the 𝐽/Ξ¨(1𝑆) state, however, the hadronic (e.g., πœŒβˆ’πœ‹) decay ratio is more than an order of magnitude smaller than predicted [6]. This is the πœŒβˆ’πœ‹ puzzle. There have been many, many theoretical attempts to explain this puzzle: Chen and Braaten [7] review earlier work by Hou and Soni, Brodsky, Lepage and Tuan, Karl and Roberts, Chaichian and Tornqvist, Pinsky, Brodsky and Karliner, and Li, Bugg and Zou. All seem to agree that this is an unsolved puzzle. More recently, there has been an attempt to locate the source of the problem [8], with the suggestion that there is a cancellation of two processes in the Ξ¨ξ…ž(2𝑆) decay. Our present work suggests that one can obtain such a cancellation by including valence gluonic structure.

In Section 2, we show that the energy of the lowest hybrid charmonium state with the quantum numbers 𝐽𝑃𝐢=1βˆ’βˆ’ is that of the Ξ¨ξ…ž(2𝑆), but our solution is not consistent with a pure hybrid. Since the 𝑐𝑐(2𝑆) state is also expected to have that energy (about 600 MeV above that of the 𝐽/Ξ¨(1𝑆) state), we predict that the Ξ¨ξ…ž(2𝑆) state is an admixture of 𝑐𝑐(2𝑆) and hybrid components. In Section 3, we show that this can provide a solution to the πœŒβˆ’πœ‹ puzzle. In Section 4, we give our conclusions and compare our results to lattice gauge calculations.

2. Hybrid 1βˆ’βˆ’ Charmonium Using QCD Sum Rules

We now will use the method of QCD sum rules to attempt to find the lowest hybrid charmonium state, assuming that such a pure hybrid charmonium meson with quantum numbers 𝐽𝑃𝐢=1βˆ’βˆ’ exists. First, let us review the method and the criteria for determining if one has obtained a satisfactory and accurate solution.

2.1. Method of QCD Sum Rules for a Hybrid Charmonium Meson

The starting point of the method of QCD sum rules is the correlator, which for a hybrid meson is Ξ πœ‡πœˆ(π‘₯)=βŸ¨π‘‡[π½πœ‡π»(π‘₯)π½πœ‡π»(0)]⟩,(2.1)with the current π½πœ‡π»(π‘₯) creating the hybrid state being studied. The QCD sum rule is obtained by evaluating Ξ πœ‡πœˆ in two ways. First, after a Fourier transform to momentum space, a dispersion relation gives the left-hand side (LHS) of the sum rule: Ξ (π‘ž)πœ‡πœˆlhs=ImΞ πœ‡πœˆ(𝑀𝐴)πœ‹(𝑀2π΄βˆ’π‘ž2)+ξ€œβˆžπ‘ π‘œπ‘‘π‘ ImΞ πœ‡πœˆ(𝑠)πœ‹(π‘ βˆ’π‘ž2),(2.2)where 𝑀𝐴 is the mass of the state 𝐴 (assuming zero width) and π‘ π‘œ is the start of the continuumβ€”a parameter to be determined. The imaginary part of Ξ (𝑠), with the term for the state we are seeking shown as a pole (corresponding to a 𝛿(π‘ βˆ’π‘€2𝐴) term in ImΞ ) and the higher-lying states produced by π½πœ‡π» shown as the continuum, is illustrated in Figure 1.

982341.fig.001
Figure 1: QCD sum rule study of a state 𝐴 with mass 𝑀𝐴 (no width).

Next, Ξ πœ‡πœˆ(π‘ž) is evaluated by an operator product expansion (OPE), giving the right-hand side (RHS) of the sum rule: Ξ (π‘ž)πœ‡πœˆrhs=ξ“π‘˜π‘π‘˜(π‘ž)⟨0|π’ͺπ‘˜|0⟩,(2.3)with increasing π‘˜ corresponding to increasing dimension of π’ͺπ‘˜.

After a Borel transform, ℬ, defined in Appendix B, in which the π‘ž variable is replaced by the Borel mass, 𝑀𝐡, the final QCD sum rule has the form 1πœ‹π‘’βˆ’π‘€2HHM/𝑀2π΅ξ€œ+β„¬βˆžπ‘ π‘œπ‘‘π‘ Im[Ξ (𝑠)]πœ‹(π‘ βˆ’π‘ž2)=β„¬π‘˜π‘π‘˜(π‘ž)⟨0|π’ͺπ‘˜|0⟩,(2.4)as we will show in what follows. Note that the Borel transform produces an exponential decrease with increasing values of 𝑠, as shown by the pole term in (2.4). This reduces the contribution of the continuum, where 𝑠β‰₯𝑀2𝐡.

This sum rule is used to estimate the heavy-hybrid mass, 𝑀HHM. One of the main sources of error is the treatment of the continuum. In addition to the parameter π‘ π‘œ, one must parameterize the effective shape of the continuum. The criteria for a satisfactory solution are: (1) the contribution of the continuum should not be as large as the pole term in the LHS; (2) with an exact sum rule, the value of 𝑀HHM is independent of the value of 𝑀𝐡 (with the approximation of a fit to the continuum, there should be a minimum or maximum in the value of 𝑀HHM versus 𝑀𝐡, and the value of 𝑀HHM at this extremum should be approximately the value of 𝑀𝐡); (3) there should be a gap between the solution for 𝑀2HHM and π‘ π‘œ, which reduces the contribution of the continuum to be smaller than the pole term due to the factor of π‘’βˆ’π‘ /𝑀2𝐡 after the Borel transform, as we will explain in what follows. If the value of π‘ π‘œ is much larger than the expected excited hybrid states, however, the solution is not physical.

2.2. Previous Results for Hybrids Using QCD Sum Rules and Lattice QCD Methods

Since mesons with certain quantum numbers, such as 1βˆ’+, cannot have a standard π‘žπ‘ž meson composition, there is a strong motivation for both experimental and theoretical searches for hybrid mesons, which can have such states. These states are called exotic or hermaphrodite mesons. Shortly after the introduction of QCD sum rules, they were used to attempt to find the masses and widths of exotics. The most accurate calculations [3, 4] predicted 1βˆ’+ hybrid light-quark mesons in the 1.3–1.7 GeV region, where such a state has been found [9]. The solutions satisfy the criteria for a good solution, as explained above. For example, for a solution that has a hybrid mass of 1.3 GeV, the value of π‘ π‘œ was 1.7 GeV, which is a reasonable separation of the lowest from the higher hybrid states. On the other hand, for these light meson hybrids, many terms in the OPE are needed, which significantly increases the uncertainty. This explains why previous calculations [3, 4] have found a rather wide range of values for the light-quark exotic hybrid.

For heavy-quark hybrids, the higher-order terms in the OPE are quite small, so the QCD sum rule method is more accurate. Since we are not trying to predict exotic hybrids, however, there is the serious complication of nonhybrid meson-hybrid meson mixing, which will be discussed in detail in what follows. This is an important aspect of our present work.

There have been many lattice QCD calculations of glueballs and light-quark hybrids [10]. The most recent calculations of light-quark hybrids find the lightest exotics to be about 2 GeV [11], quite a bit higher in energy than the QCD sum rule calculations. This probably is due to the fact that lattice QCD calculations for light quarks have some inconsistencies at the present time, while they are much more accurate for heavy-quark systems [12], as are QCD sum rules. Exotic charmonium states have been calculated using lattice QCD, and the 1βˆ’+ was found to be about 4.4 GeV [13], with the expectation that the 1βˆ’βˆ’ hybrid charmonium state is at a similar energy. For the calculation of nonexotic hybrid mesons, such as 1βˆ’βˆ’ hybrids, there are other difficulties for both methods, as we will discuss in what follows.

2.3. Heavy-Hybrid Meson Correlator

For a hybrid meson with quantum numbers 1βˆ’βˆ’ we use the standard current [3, 4]: π½πœ‡π»=Ξ¨Ξ“πœˆπΊπœ‡πœˆΞ¨,(2.5)with Ξ“πœˆ=πΆπ›Ύπœˆ, where 𝐢 is the charge conjugation operator, π›Ύπœˆ is the usual Dirac matrix, and Ξ¨ is the heavy-quark field. Carrying out a four-dimensional Fourier transform, the correlator in momentum space is Ξ πœ‡πœˆξ€œπ‘‘(𝑝)=4𝑝1(2πœ‹)4Tr[π‘†π‘Žπ‘Ξ“π›Όπ‘†π‘π‘ŽΞ“π›½](π‘βˆ’π‘1)Tr[πΊπœ‡π›ΌπΊπœˆπ›½](𝑝1),(2.6)where π‘†π‘Žπ‘ is the quark propagator, with colors π‘Ž and 𝑏. The color properties of πΊπœ‡π›Ό, the gluon color field, with πœ‡,𝛼 Dirac indices, are given in Appendix A. Note that the traces are both fermion and color traces. Details of Tr[π‘†π‘Žπ‘Ξ“π›Όπ‘†π‘π‘ŽΞ“π›½](𝑝) and Tr[πΊπœ‡π›ΌπΊπœˆπ›½](𝑝1) are given in Appendix A.

It is important to recognize that the correlator used in the QCD sum rule method is similar to the correlator used in the lattice gauge approach, with the same objective of finding the mass of a heavy-hybrid meson.

As described in the preceding subsection, the correlator is evaluated in the method of QCD sum rules via an OPE and a dispersion relation.

2.4. OPE of the Scalar Correlator in Momentum Space

The QCD sum rule method uses an OPE in dimension (or inverse momentum) (2.3). For the hybrid meson, the lowest-order diagram [14] is shown in Figure 2.

982341.fig.002
Figure 2: Lowest-order term in sum rule.

Using the standard quark and gluon propagators (see Appendix A for some details), we find the following formula for this lowest-dimensional process: Ξ 1πœ‡πœˆξ€œπ‘‘(𝑝)=βˆ’64𝑝1(2πœ‹)4{π‘”πœ‡πœˆ[𝑝1β‹…(π‘βˆ’π‘1)]2𝑝21βˆ’π‘1β‹…(π‘βˆ’π‘1)𝑝21[π‘πœ‡π‘πœˆ1+π‘πœˆ(π‘βˆ’π‘1)πœ‡βˆ’2π‘πœ‡1(π‘βˆ’π‘1)𝜈2]}Γ—{3𝑀2𝑄(π‘βˆ’π‘1)29+[4(π‘βˆ’π‘1)2βˆ’416𝑀2π‘„βˆ’23𝑀4𝑄1(π‘βˆ’π‘1)2]𝐼0(π‘βˆ’π‘1)},(2.7) with 𝐼0ξ€œ(𝑝)=10𝑑𝛼𝛼(1βˆ’π›Ό)𝑝2βˆ’π‘€2𝑄.(2.8)

Extracting the scalar correlator Π𝑆, defined by Ξ πœ‡πœˆ(𝑝)=(π‘πœ‡π‘πœˆ/𝑝2βˆ’π‘”πœ‡πœˆ)Π𝑉(𝑝)+(π‘πœ‡π‘πœˆ/𝑝2)Π𝑆(𝑝), and carrying out the 𝑝1 integrals, one finds for Π𝑆1(𝑝), the scalar term of Ξ 1πœ‡πœˆ(𝑝), Π𝑆13(𝑝)=βˆ’(4πœ‹)2ξ€œ10π‘‘π›Όξ€œπ›Ό(1βˆ’π›Ό)Γ—{10𝑑𝛽𝑝2𝛽𝑝2(1βˆ’π›½)βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2)4Γ—[βˆ’3𝑀6π‘„π›Όβˆ’π›Ό2βˆ’155𝑀126𝑄(π›Όβˆ’π›Ό2)2+319𝑝122𝑀4π‘„π›Όβˆ’π›Ό2βˆ’43𝑀4𝑄𝑝2βˆ’413𝑀2𝑄𝑝4]+𝑝42ξ€œ10π‘‘π›½βˆ’1𝑝2(1βˆ’π›½)βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2)Γ—[553𝑀2𝑄(𝑝2βˆ’π‘€2π‘„π›Όβˆ’π›Ό2)(βˆ’3𝛽+𝛽2βˆ’π›½338)βˆ’3𝑀4𝑄(4π›½βˆ’3𝛽2+𝛽33+)]}termswiththreeandfourintegrals.(2.9)We find that the terms with three and four integrals are very small, and we do not include them in our calculation.

The second term in the OPE for the heavy-quark hybrid correlator includes the gluon condensate, illustrated in Figure 3.

982341.fig.003
Figure 3: Gluon condensate term in sum rule.

For this process, the correlator Ξ 2πœ‡πœˆ(𝑝) has the same form as (2.3), except that the gluon trace used to obtain Ξ 1πœ‡πœˆ(𝑝) is replaced with the trace over the gluon condensate [1–4], which gives Tr[πΊπœ‡π›ΌπΊπœˆπ›½](𝑝1)=(2πœ‹)4𝛿4(𝑝1)1296⟨𝐺2⟩(π‘”πœ‡πœˆπ‘”π›Όπ›½βˆ’π‘”πœ‡π›½π‘”πœˆπ›Ό).(2.10)The fermionic factor is the same as for Figure 2. From this, one finds Ξ 2πœ‡πœˆ(𝑝)=βˆ’3𝑖2⟨𝐺2⟩(4πœ‹)2{π‘”πœ‡πœˆ(βˆ’π‘€2π‘„βˆ’π‘24)(𝑝2βˆ’4𝑀2𝑄)𝐼0+π‘πœ‡π‘πœˆ(112𝑝2βˆ’533𝑀2𝑄+43𝑀4𝑄𝑝2)𝐼0+43𝑀2𝑄𝑝2},(2.11)giving for the scalar part of Figure 3Π𝑆2(𝑝)=βˆ’3𝑖2⟨𝐺2⟩(4πœ‹)2(114𝑝4βˆ’596𝑀2𝑄𝑝2+143𝑀4𝑄)𝐼0.(2.12)

2.5. Borel Transform of the Correlator

To ensure convergence of the OPE of the correlator, one performs a Borel transform ℬ [1, 2], defined in Appendix B. This is discussed in detail in the early papers on QCD sum rules [1–4]. For our problem, we have assumed that Π𝑆(𝑝)≃Π𝑆1(𝑝)+Π𝑆2(𝑝), with higher-order terms being very small. We shall see that even Ξ 2 is essentially negligible within the accuracy of the method, and the convergence after the Borel transform is established.

Using the equations in Appendix B, from (B.1), (B.2), and (B.3), where the quantities 𝐡1 through 𝐡8 are given, with ℬΠ𝑆1Π(𝑝)≑𝑆1(𝑀𝐡), we find Π𝑆1(𝑀𝐡)1=βˆ’4(4πœ‹)2[βˆ’16𝑀6𝑄𝐡1βˆ’155𝑀6𝑄𝐡2+319𝐡3βˆ’16𝑀4𝑄𝐡4βˆ’164𝐡5βˆ’110(𝐡6βˆ’π‘€2𝑄𝐡7)βˆ’16𝑀4𝑄𝐡8]1=βˆ’2(4πœ‹)2𝑀4π‘„ξ€œβˆž0π‘‘π›Ώπ‘’βˆ’2(𝑀2𝑄/𝑀2𝐡)(1+𝛿)𝛿×{[βˆ’3101+𝛿+638π›Ώβˆ’656(1+𝛿)]𝐾3𝑀(22𝑄𝑀2𝐡𝛿(1+𝛿))+[βˆ’1892𝛿1+𝛿+606π›Ώβˆ’1968(1+𝛿)βˆ’32(4π›Ώβˆ’32+𝛿1+𝛿33(1+𝛿)2)]𝐾2𝑀(22𝑄𝑀2𝐡𝛿(1+𝛿))+[βˆ’4778𝛿1+𝛿+9442π›Ώβˆ’4920(1+𝛿)βˆ’128(4π›Ώβˆ’32+𝛿1+𝛿33(1+𝛿)2)]𝐾1𝑀(22𝑄𝑀2𝐡𝛿(1+𝛿))+[βˆ’1356𝛿1+𝛿+6284π›Ώβˆ’3280(1+𝛿)βˆ’96(4π›Ώβˆ’32+𝛿1+𝛿33(1+𝛿)2)]𝐾0𝑀(22𝑄𝑀2𝐡+(1+𝛿))}multipleintegralterms.(2.13)The multiple integral terms in (2.11) are small and are dropped, and 𝛿 is a variable of integration.

In a similar way, taking the Borel transform of Π𝑆2(𝑝), with ℬΠ𝑆2Π(𝑝)≑𝑆2(𝑀𝐡), we find Π𝑆2(𝑀𝐡3)=βˆ’π‘–2(4πœ‹)2𝑀4π‘„π‘’βˆ’2(𝑀2𝑄/𝑀2𝐡)[11𝐾2𝑀(22𝑄𝑀2𝐡)+143𝐾1𝑀(22𝑄𝑀2𝐡)+18𝐾0𝑀(22𝑄𝑀2𝐡)].(2.14)

2.6. QCD Sum Rule for Hybrid Charm Meson

The method of QCD sum rules uses a dispersion relation for the correlator, which it equates to the correlator's operator product. Following the usual convention, we call the dispersion relation the LHS and the OPE the RHS: Ξ lhs = dispersion relation, Ξ rhs = OPE. Neglecting the width of the hybrid meson, the dispersion relation has the form of a pole and a continuum: ∫∞0𝑑𝑠(ImΞ (𝑠)/(π‘ βˆ’π‘2)). The dispersion relation is evaluated in Euclidean space, 𝑝2β†’βˆ’π‘„2, and the continuum is assumed to start at 𝑠=π‘ π‘œ. After the Borel transform, the form we use for the LHS is Πlhs(𝑀2𝐡)=πΉπ‘’βˆ’π‘€2HH/𝑀2𝐡+(𝐿1𝑀2𝐡+𝐿2𝑀4𝐡)π‘’βˆ’π‘ π‘œ/𝑀2𝐡,(2.15)with 𝐹 the numerator of the pole, and 𝐿1 and 𝐿2 constants used to fit the form of the continuum. We should use a standard method to eliminate 𝐹, and fit the sum rule requirement that the solution should not be sensitive to 𝑀𝐡. In theory, the exact solution should be independent of 𝑀𝐡.

For convenience in carrying out the sum rule, we have fit the RHS of the correlator to a polynomial in the Borel mass, Π1(𝑀𝐡)=π‘Ž1𝑀2𝐡+π‘Ž2𝑀4𝐡+π‘Ž3𝑀6𝐡+π‘Ž4𝑀8𝐡 and Π2(𝑀𝐡)=𝑏1𝑀2𝐡+𝑏2𝑀4𝐡+𝑏3𝑀6𝐡+𝑏4𝑀8𝐡 with π‘Ž1=122.29,π‘Ž2=βˆ’143.88,π‘Ž3=45.79,π‘Ž4=βˆ’0.346,𝑏1=616.0,𝑏2=βˆ’279.1,𝑏3=βˆ’226.56, and 𝑏4=144.515, with units GeV6.

We find that for all values of 𝑀𝐡 relevant to the sum rule, Π2 is just about 1% of Π1, and drop it, as the method is only valid to a few percent. The sum rule is obtained by taking the ratio of Π to πœ•1/𝑀2𝐡Π. To do this, we use the relations πœ•1/𝑀2π΅π‘’βˆ’π΄/𝑀2𝐡𝐾𝜈𝐴(βˆ’π‘€2𝐡)|𝜈>0=βˆ’π΄π‘’βˆ’π΄/𝑀2𝐡𝐾𝜈𝐴(βˆ’π‘€2𝐡)+𝐴2π‘’βˆ’π΄/𝑀2𝐡[πΎπœˆβˆ’1𝐴(βˆ’π‘€2𝐡)βˆ’πΎπœˆ+1𝐴(βˆ’π‘€2π΅πœ•)],1/𝑀2π΅π‘’βˆ’π΄/𝑀2𝐡𝐾0𝐴(βˆ’π‘€2𝐡)=βˆ’π΄π‘’βˆ’π΄/𝑀2𝐡[𝐾0𝐴(βˆ’π‘€2𝐡)βˆ’πΎ1𝐴(βˆ’π‘€2𝐡)].(2.16)

Taking the ratio of πœ•1/𝑀2𝐡Πlhs(𝑀𝐡)=πœ•1/𝑀2𝐡Πrhs(𝑀𝐡) to the equation Πlhs(𝑀𝐡Π)=rhs(𝑀𝐡), one obtains the sum rule for the mass of the heavy charmonium hybrid meson: 𝑀2HH={π‘’βˆ’π‘ π‘œ/𝑀2𝐡[π‘ π‘œ(𝐿1𝑀2𝐡+𝐿2𝑀4𝐡)+𝐿1𝑀4𝐡+2𝐿2𝑀6𝐡]+πœ•1/𝑀2𝐡Π𝑆1}Γ—{π‘’βˆ’π‘ π‘œ/𝑀2𝐡[(𝐿1𝑀2𝐡+𝐿2𝑀4𝐡Π)βˆ’π‘†1}βˆ’1.(2.17)

The result of the QCD sum rule fit is shown in Figure 4, with π‘ π‘œ=60.0GeV2, 𝐿1=βˆ’99.0GeV4, and 𝐿2=6.06GeV2.

982341.fig.004
Figure 4: QCD sum rule for heavy-hybrid charm meson.

The parameters 𝐿1 and 𝐿2 have magnitudes expected for the fit to the continunmm. Except for the large value of π‘ π‘œ, the solution satisfies the criteria for a successful QCD sum rule, as Figure 4 shows. The solution for the mass is a few percent of the value of 𝑀𝐡 in the region of stability. The mass predicted for the hybrid is 𝑀HHM≃3.66GeV,(2.18)within a 10% accuracy of the sum rule method, while the experimental mass of the Ξ¨ξ…ž(2𝑆) state [15] is 𝑀(Ξ¨ξ…ž(2𝑆))=3.68GeV.(2.19)The large value of π‘ π‘œ, however, predicts that the excited hybrids are at a very high energy, as found in lattice gauge calculations, and that the Ξ¨ξ…ž(2𝑆) cannot be a pure hybrid. Therefore, we expect that the physical Ξ¨ξ…ž(2𝑆) is an admixture with a charm meson and a hybrid charm component. A second, orthogonal mixed state will be in the continuum. As we now show, this can provide a solution to the πœŒβˆ’πœ‹ puzzle.

3. Hybrid-Normal Charmonium and a Possible Solution to the πœŒβˆ’πœ‹ Puzzle

In our treatment of hadronic decays of the Ξ¨ξ…ž(2𝑆) state, we use the Sigma/Glueball model, which was motivated by the BES analysis of glueball decay [16] and the study of scalar mesons and scalar glueballs using QCD sum rules [17, 18]. We briefly review this model in what follows.

3.1. The Sigma/Glueball Model

In energy regions where there are both scalar mesons and scalar glueballs, it is expected that 0++ states will be an admixture of mesons and glueballs. For this reason, when using QCD sum rules to find such states, one must use currents that are a linear combination of glueball and meson currents. A scalar glueball current can have the form 𝐽𝐺(π‘₯)=𝛼𝑠𝐺2,(3.1)while a scalar meson current has the form π½π‘š1(π‘₯)=2(𝑒(π‘₯)𝑒(π‘₯)βˆ’π‘‘(π‘₯)𝑑(π‘₯)).(3.2)We use for our 0++ current [17] (with π‘€π‘œ needed for correct dimensions) 𝐽0++=π›½π‘€π‘œπ½π‘š+(1βˆ’|𝛽|)𝐽𝐺.(3.3)

The QCD sum rule calculation makes use of the correlator Ξ (π‘₯)=βŸ¨π‘‡[𝐽0++𝐽0++]⟩. The cross term between π½π‘š and 𝐽𝐺 is evaluated by using the scalar glueball-meson coupling theorem [19, 20]: ξ€œπ‘‘π‘₯βŸ¨π‘‡[𝐽𝐺(π‘₯)π½π‘š(0)]βŸ©β‰ƒβˆ’329βŸ¨π‘žπ‘žβŸ©,(3.4)with βŸ¨π‘žπ‘žβŸ©β‰‘thequarkcondensate, which is illustrated in Figure 5.

982341.fig.005
Figure 5: Scalar glueball-meson coupling theorem.

The results of the QCD sum rule calculations [17, 18] are that there are three solutions:  80% scalar glueball at 1500 MeV β†’ the 𝑓0 (1500); 80% scalar meson at 1350 MeV β†’ the 𝑓0 (1370); Light Scalar Glueball 400–600 MeV β†’ the Sigma/Glueball.

The Sigma/Glueball model follows from the existence of the 𝜎, a scalar πœ‹βˆ’πœ‹ resonance with a broad width and the same mass as the scalar glueball found in the sum rule calculations, and makes use of the glueball-meson coupling shown in Figure 5. It has been used for the study of the Roper resonance decay into a nucleon and a 𝜎 [5, 21], and the prediction of 𝜎 production in proton-proton high-energy collisions [22], and other applications. The relevance for our present work is that just as it is possible to determine the mixing parameter for a state consisting of a scalar meson and a scalar glueball, it should be possible to determine the 𝑐𝑐 and hybrid 𝑐𝑐𝑔 admixture for charmonium systems.

3.2. Hybrid Mixing Model and the πœŒβˆ’πœ‹ Puzzle

The solution to the πœŒβˆ’πœ‹ puzzle by the mixing of hybrid and normal meson components of the Ξ¨(2𝑆) state can be understood from Figures 6 and 7, which illustrate the decay of a 𝑐𝑐 and a 𝑐𝑐𝑔 state into two hadrons. The 𝑐𝑐 decay involves the matrix element βŸ¨πœ‹πœŒ|𝑂|Ξ¨ξ…ž(𝑐𝑐,2𝑆)⟩.

982341.fig.006
Figure 6: Lowest-order PQCD diagram for a 𝑐𝑐 decay into two hadrons.
982341.fig.007
Figure 7: Decay of a 𝑐𝑐𝑔 state into two hadrons.

The corresponding hybrid decay involves the matrix element βŸ¨πœ‹πœŒ|π‘‚ξ…ž|Ξ¨ξ…ž(𝑐𝑐𝑔,2𝑆)⟩, with the diagram shown in Figure 7.

Assuming that the 2 s state is a 𝑐𝑐-𝑐𝑐𝑔 admixture, |Ξ¨ξ…ž(2𝑆)⟩=𝑏|Ξ¨ξ…ž(π‘βˆšπ‘,2𝑆)⟩+1βˆ’π‘2|Ξ¨ξ…ž(𝑐𝑐𝑔,2𝑆)⟩,(3.5) and recognizing that the 𝑂 and π‘‚ξ…ž matrix elements are approximately equal, we see that the solution to the πœŒβˆ’πœ‹ puzzle requires βˆšπ‘+1βˆ’π‘2≃0.1.(3.6) The solution of the πœŒβˆ’πœ‹ puzzle would be given if π‘β‰ƒβˆ’.7.(3.7) In other words, if π‘β‰ƒβˆ’.7, we have found a solution of the πœŒβˆ’πœ‹ puzzle. The evaluation of 𝑏 is rather complicated, and the testing of this conjecture will be carried out in future work.

4. Conclusion

Using the method of QCD sum rules, we have shown that the Ξ¨ξ…ž(2𝑆) state cannot be a pure charmonium hybrid. We have found that the energy of the lowest 𝐽𝑃𝐢=1βˆ’βˆ’ hybrid charmonium state is approximately the same as the Ξ¨ξ…ž(2𝑆) state, about 600 MeV above the 𝐽/Ξ¨(1𝑆) state, but that the QCD sum rule solution is not consistent with a pure hybrid. The standard model prediction for 𝑐𝑐(2𝑆) is at approximately the same energy. Therefore, we expect that the physical Ξ¨ξ…ž(2𝑆) state is an admixture of a 𝑐𝑐(2𝑆) and a 𝑐𝑐(8)𝑔(8)(2𝑆). Using this picture, we find a possible solution to the famous πœŒβˆ’πœ‹ puzzle.

There have been many lattice calculations of exotic hybrid mesons. There is experimental evidence for an exotic light-quark 1βˆ’+ meson (see [9] for a discussion) at 1.4 to 1.6 GeV, which is consistent with QCD sum rule calculations [3, 4], while lattice calculations find the lowest 1βˆ’+ hybrid at 1.9 to 2.1 GeV. The lowest energy 1βˆ’+ charmonium hybrid found in lattice calculations is at 4.4 GeV [13], about 800 MeV above our 1βˆ’βˆ’ hybrid charmonium solution. This is consistent with our large value of π‘ π‘œ. For the 1βˆ’βˆ’ state, we have shown that one must use a mixed 𝑐𝑐 and 𝑐𝑐𝑔 current, which we shall use in future work. This use of a mixed current to define the correlator has not been done in lattice QCD caluclations, but with the QCD sum rule method, it can be done in a rather straight-forward calculation.

It is interesting that the energy difference between the Ξ₯(2𝑆) and the Ξ₯(1𝑆) is also approximately 600 MeV. If this is the energy of a Ξ₯(𝑛𝑆) hybrid, this could provide a solution to the puzzling 𝜎 decays of Ξ₯(𝑛𝑆) states that have recently been observed [23]. Investigation of this system is a topic of future research.

Appendices

A. Heavy-Quark Hybrid Correlator in Momentum Space

The current to create a heavy-quark hybrid meson with 𝐽𝑃𝐢=1βˆ’βˆ’ is π½πœ‡π»=Ξ¨Ξ“πœˆπΊπœ‡πœˆΞ¨,(A.1)where Ξ¨ is the heavy-quark field, Ξ“πœˆ=πΆπ›Ύπœˆ, π›Ύπœˆ is the usual Dirac matrix, 𝐢 is the charge conjugation operator, and the gluon color field is πΊπœ‡πœˆ=8ξ“π‘Ž=1πœ†π‘Ž2πΊπ‘Žπœ‡πœˆ,(A.2)with πœ†π‘Ž the SU(3) generator (Tr[πœ†π‘Žπœ†π‘]=2π›Ώπ‘Žπ‘). From this, one finds for the correlator of a heavy-quark hybrid meson Ξ πœ‡πœˆξ€œπ‘‘(𝑝)=4π‘₯𝑒𝑖𝑝⋅π‘₯βŸ¨π‘‡[π½πœ‡π»(π‘₯)π½πœ‡π»=ξ€œπ‘‘(0)]⟩4𝑝1(2πœ‹)4Tr[π‘†π‘Žπ‘Ξ“π›Όπ‘†π‘π‘ŽΞ“π›½](π‘βˆ’π‘1)Tr[πΊπœ‡π›ΌπΊπœˆπ›½](𝑝1),(A.3)where π‘†π‘Žπ‘ is the quark propagator, a standard Dirac propagator for a fermion with mass 𝑀𝑄, with colors π‘Ž and 𝑏. Using 𝐢=𝑖𝛾2𝛾0, Tr[𝛾𝛼𝛾𝛽]=4𝑔𝛼𝛽, and Tr[π›Ύπœ†π›Ύπ›Όπ›Ύπ›Ώπ›Ύπ›½]=4(π‘”πœ†π›Όπ‘”π›Ώπ›½βˆ’π‘”πœ†π›Ώπ‘”π›Όπ›½+π‘”πœ†π›½π‘”π›Ώπ›Ό): Tr[π‘†π‘Žπ‘Ξ“π›Όπ‘†π‘π‘ŽΞ“π›½ξ€œπ‘‘](𝑝)=(βˆ’24𝑖)π·π‘˜(2πœ‹)𝐷[βˆ’π‘€2𝑄𝑔𝛼𝛽+(π‘πœ†π‘˜π›Ώβˆ’π‘˜πœ†π‘˜π›Ώ)(π‘”πœ†π›Όπ‘”π›Ώπ›½βˆ’π‘”πœ†π›Ώπ‘”π›Όπ›½+π‘”πœ†π›½π‘”π›Ώπ›Ό1)](π‘˜2βˆ’π‘€2𝑄)[(π‘βˆ’π‘˜)2βˆ’π‘€2𝑄].(A.4)To use dimensional regularization, we define the quantity 𝐷=4βˆ’πœ–, and let πœ–β†’0 to complete the integrals. One can then show ξ€œπ‘‘π·π‘˜(2πœ‹)𝐷1(π‘˜2βˆ’π‘€2𝑄)[(π‘βˆ’π‘˜)2βˆ’π‘€2𝑄]=𝐼0(8πœ‹)2,(A.5)with 𝐼0=ξ€œ10𝑑𝛼(π›Όβˆ’π›Ό2)𝑝2βˆ’π‘€2𝑄.(A.6)

The first term in the OPE, shown in Figure 2, has a standard gluon propagator. For the gluon trace [3, 4] one finds Tr[πΊπœ‡π›ΌπΊπœˆπ›½](𝑝1)=βˆ’4πœ‹2𝑖(π‘”π›Όπ›½π‘πœ‡1π‘πœˆ1𝑝21+π‘”πœ‡πœˆπ‘π›Ό1𝑝𝛽1𝑝21βˆ’π‘”πœ‡π›½π‘π›Ό1π‘πœˆ1𝑝21βˆ’π‘”π›Όπœˆπ‘πœ‡1𝑝𝛽1𝑝21).(A.7)

The correlator for the process shown in Figure 2 is found to be Ξ 1πœ‡πœˆξ€œπ‘‘(𝑝)=βˆ’64𝑝1(2πœ‹)4{π‘”πœ‡πœˆ[𝑝1β‹…(π‘βˆ’π‘1)]2𝑝21βˆ’π‘1β‹…(π‘βˆ’π‘1)𝑝21[π‘πœ‡π‘πœˆ1+π‘πœˆ(π‘βˆ’π‘1)πœ‡βˆ’2π‘πœ‡1(π‘βˆ’π‘1)𝜈2]}Γ—{3𝑀2𝑄(π‘βˆ’π‘1)29+[4(π‘βˆ’π‘1)2βˆ’416𝑀2π‘„βˆ’23𝑀4𝑄1(π‘βˆ’π‘1)2]𝐼0(π‘βˆ’π‘1)}.(A.8)

The next term in the OPE for this heavy-quark system, where quark condensates are negligible, is the gluon condensate term, shown in Figure 3. The trace over the quark propagators is the same as in (A.4). The gluon field trace for this term is ⟨Tr[πΊπœ‡π›ΌπΊπœˆπ›½](𝑝1)⟩=(2πœ‹)4𝛿4(𝑝1)1296⟨𝐺2⟩(π‘”πœ‡πœˆπ‘”π›Όπ›½βˆ’π‘”πœ‡π›½π‘”πœˆπ›Ό).(A.9)

From this, one finds Ξ 2πœ‡πœˆ, given in (2.11).

B. Borel Transforms

A key method that enables one to use the operator expansion to get accurate sum rules is the use of the Borel transform [1, 2], ℬ, defined by ℬ=limπ‘ž2,π‘›β†’βˆž1(π‘›βˆ’1)!(π‘ž2)𝑛𝑑(βˆ’π‘‘π‘ž2)𝑛|π‘ž2/𝑛=𝑀2𝐡.(B.1)

Two key equations which we need are (with 𝐾𝜈 the modified Bessel functions) 1ℬ(π‘š2βˆ’π‘2)π‘˜=π‘’βˆ’π‘š2/𝑀2𝐡(π‘˜βˆ’1)!(𝑀2𝐡)π‘˜βˆ’1,ξ€œβˆž0π‘₯πœˆβˆ’1π‘’π‘Ž/π‘₯βˆ’π‘π‘₯π‘Ž=2(𝑏)𝜈/2𝐾𝜈√(2π‘Žπ‘).(B.2)

Transforms used in the body of the paper are ℬ𝐼0=ξ€œ10𝑑𝛼(π›Όβˆ’π›Ό2)ℬ(𝑝2βˆ’π‘€2𝑄(π›Όβˆ’π›Ό2))βˆ’1=2π‘’βˆ’2𝑀2𝑄/𝑀2𝐡𝐾0(2𝑀2𝑄𝑀2𝐡𝐡),1ξ€œ=ℬ10𝑑𝛼(π›Όβˆ’π›Ό2)2𝑝2ξ€œ10𝑑𝛽𝛽(1βˆ’π›½)[𝑝2βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2)(1βˆ’π›½)]=2𝑀2π‘„ξ€œβˆž0𝛿𝑑𝛿𝑒1+π›Ώβˆ’2(𝑀2𝑄/𝑀2𝐡)(1+𝛿)[2𝐾2𝑀(22𝑄𝑀2𝐡(1+𝛿))+8𝐾1𝑀(22𝑄𝑀2𝐡(1+𝛿))+6𝐾0𝑀(22𝑄𝑀2𝐡𝐡(1+𝛿))],2ξ€œ=ℬ10𝑑𝛼(π›Όβˆ’π›Ό2)3𝑝2ξ€œ10𝑑𝛽𝛽(1βˆ’π›½)[𝑝2βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2)(1βˆ’π›½)]=2𝑀2π‘„ξ€œβˆž0𝛿𝑑𝛿𝑒1+π›Ώβˆ’2(𝑀2𝑄/𝑀2𝐡)(1+𝛿)[2𝐾3𝑀(22𝑄𝑀2𝐡(1+𝛿))+12𝐾2𝑀(22𝑄𝑀2𝐡(1+𝛿))+30𝐾1𝑀(22𝑄𝑀2𝐡(1+𝛿))+20𝐾0𝑀(22𝑄𝑀2𝐡𝐡(1+𝛿))],3ξ€œ=ℬ10𝑑𝛼(π›Όβˆ’π›Ό2)3𝑝4ξ€œ10𝑑𝛽𝛽(1βˆ’π›½)[𝑝2βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2)(1βˆ’π›½)]=2𝑀4π‘„ξ€œβˆž0π‘‘π›Ώπ›Ώπ‘’βˆ’2(𝑀2𝑄/𝑀2𝐡)(1+𝛿)[2𝐾3𝑀(22𝑄𝑀2𝐡(1+𝛿))+12𝐾2𝑀(22𝑄𝑀2𝐡(1+𝛿))+30𝐾1𝑀(22𝑄𝑀2𝐡(1+𝛿))+20𝐾0𝑀(22𝑄𝑀2𝐡𝐡(1+𝛿))],4ξ€œ=ℬ10𝑑𝛼(π›Όβˆ’π›Ό2)𝑝4ξ€œ10𝑑𝛽𝛽(1βˆ’π›½)[𝑝2βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2)(1βˆ’π›½)]=2𝑀4π‘„ξ€œβˆž0π‘‘π›Ώπ›Ώπ‘’βˆ’2(𝑀2𝑄/𝑀2𝐡)(1+𝛿)[2𝐾2𝑀(22𝑄𝑀2𝐡𝐡(1+𝛿))],5ξ€œ=ℬ10𝑑𝛼(π›Όβˆ’π›Ό2)𝑝6ξ€œ10𝑑𝛽𝛽(1βˆ’π›½)[𝑝2βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2)(1βˆ’π›½)]=2𝑀6π‘„ξ€œβˆž0𝑑𝛿𝛿(1+𝛿)π‘’βˆ’2(𝑀2𝑄/𝑀2𝐡)(1+𝛿)[2𝐾3𝑀(22𝑄𝑀2𝐡(1+𝛿))+12𝐾2𝑀(22𝑄𝑀2𝐡(1+𝛿))+30𝐾1𝑀(22𝑄𝑀2𝐡(1+𝛿))+20𝐾0𝑀(22𝑄𝑀2𝐡(1+𝛿))],(1)𝐡6ξ€œ=ℬ10𝑑𝛼(π›Όβˆ’π›Ό2)𝑝6ξ€œ10𝑑𝛽(βˆ’3𝛽+𝛽2βˆ’π›½3/3)(1βˆ’π›½)[𝑝2βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2)(1βˆ’π›½)]=2𝑀6π‘„ξ€œβˆž0𝛿𝑑𝛿(βˆ’3𝛿+2βˆ’π›Ώ1+𝛿33(1+𝛿)2)π‘’βˆ’2(𝑀2𝑄/𝑀2𝐡)(1+𝛿)Γ—[2𝐾3𝑀(22𝑄𝑀2𝐡(1+𝛿))+12𝐾2𝑀(22𝑄𝑀2𝐡(1+𝛿))+30𝐾1𝑀(22𝑄𝑀2𝐡(1+𝛿))+20𝐾0𝑀(22𝑄𝑀2𝐡𝐡(1+𝛿))],7ξ€œ=ℬ10𝑑𝛼(π›Όβˆ’π›Ό2)2𝑝4ξ€œ10𝑑𝛽(βˆ’3𝛽+𝛽2βˆ’π›½3/3)(1βˆ’π›½)[𝑝2βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2=1)(1βˆ’π›½)]𝑀2𝑄𝐡6,𝐡8ξ€œ=ℬ10𝑑𝛼(π›Όβˆ’π›Ό2)𝑝4ξ€œ10𝑑𝛽(4π›½βˆ’3𝛽2+𝛽3/3)(1βˆ’π›½)[𝑝2βˆ’π‘€2𝑄/(π›Όβˆ’π›Ό2)(1βˆ’π›½)]=2𝑀4π‘„ξ€œβˆž0𝑑𝛿(4π›Ώβˆ’3𝛿2+𝛿1+𝛿33(1+𝛿)2)π‘’βˆ’2(𝑀2𝑄/𝑀2𝐡)(1+𝛿)Γ—[2𝐾2𝑀(22𝑄𝑀2𝐡(1+𝛿))+8𝐾1𝑀(22𝑄𝑀2𝐡(1+𝛿))+6𝐾0𝑀(22𝑄𝑀2𝐡(1+𝛿))].(B.3)

Acknowledgments

This work was supported in part by the NSF/INT Grant no. 0529828. The authors thank Professors Pengnian Shen, Wei-xing Ma, and other IHEP, Beijing colleagues for helpful discussions. They thank Professor Y. Chen for discussions of lattice QCD in comparison to QCD sum rules for hybrid states.

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