Hybrid Charmonium and the $\rho-\pi$ Puzzle

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 $\Psi'(2S)$ state, about 600 Mev above the $J/\Psi(1S)$ state. Since our solution is not consistent with a pure hybrid at this energy, we conclude that the $\Psi'(2S)$ state is probably an admixed $c \bar{c}$ and hybrid $c \bar{c}g$ state. From this conjecture we find a possible explanation of the famous $\rho-\pi$ puzzle.


I. 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] has long been used to predict the energies of light quark hybrid mesons [2] and hybrid baryons (see Ref. [3]). 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 Ψ ′ (2S) state, and to find a possible explanation of the long-standing ρ − π puzzle.
The ρ − π puzzle concerns the branching ratios for hadronic decays of the Ψ ′ (2S) state compared to the J/Ψ(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 the so-called 13 % rule. For the Ψ ′ (2S) state compared to the J/Ψ(1S) state, however, the hadronic (e.g.ρ − π) decay ratio is more than an order of magnitude smaller than predicted [4]. This is the ρ − π puzzle. There have been many, many theoretical attempts to explain this puzzle: Chen and Braatan [5] 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 [6], with the suggestion that there is a cancellation of two processes in the Ψ ′ (2S) decay. Our present work suggests that one can obtain such a cancellation by including valence gluonic structure.
In Sec. II we show that the energy of the lowest hybrid charmonium state with the quantum numbers J P C = 1 −− is that of the Ψ ′ (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/Ψ(1S) state), we predict that the Ψ ′ (2S) state is an admixture of cc(2S) and hybrid components. In Sec. III we show that this can provide a solution to the ρ − π puzzle. In Sec. IV we give our conclusions and compare our results to lattice gauge calculations.
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 J P C = 1 −− exists. First, let us review the method and the criteria for determining if one has obtained a satisfactory and accurate solution.

A. 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 with the current J µ H (x) creating the hybrid state being studied. The QCD sum rule is obtained by eval-uating Π µν in two ways. First, after a Fourier transform to momentum space, a dispersion relation gives the left-hand side (lhs) of the sum rule: where M A is the mass of the state A (assuming zero width) and s o is the start of the continuum-a parameter to be determined. The imaginary part of Π(s), with the term for the state we are seeking shown as a pole (corresponding to a δ(s − M 2 A ) term in ImΠ) and the higher-lying states produced by J µ H shown as the continuum, is illustrated in the figure: Next Π µν (q) is evaluated by an operator product expansion (O.P.E.), giving the right-hand side (rhs) of the sum rule with increasing k corresponding to increasing dimension of O k . After a Borel transform, B, defined in Appendix B, in which the q variable is replaced by the Borel mass, M B , the final QCD sum rule has the form 1 π e −M 2 as we shall show below. Note that the Borel transform produces an exponential decrease with increasing values of s, as shown by the pole term in Eq (5). This reduces the contribution of the continuum, where s ≥ M 2 B . This sum rule is used to estimate the heavy hybrid mass, M HHM . One of the main sources of error is the treatment of the continuum. In addition to the parameter s o , 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. 3) There should be a gap between the solution for M 2 HHM and s o , which reduces the contribution of the continuum to be smaller than the pole term due to the factor of e −s/M 2 B after the Borel transform, as we shall explain below. If the value of s o is much larger than the expected excited hybrid states, however, the solution is not physical.

B. Previous Results for Hybrids Using QCD Sum Rules and Lattice QCD Methods
Since mesons with certain quantum numbers, such as 1 −+ , cannot have a standard qq 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 [2] predicted 1 −+ hybrid light-quark mesons in the 1.3-1.7 GeV region, where such a state has been found [7]. 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 s o 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 O.P.E. are needed, which significantly increases the uncertainty. This explains why previous calculations [2] have found a rather wide range of values for the light-quark exotic hybrid.
For heavy-quark hybrids the higher-order terms in the O.P.E. 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 below. This is an important aspect of our present work.
There have been many lattice QCD calculations of glueballs and light-quark hybrids [8]. The most recent calculations of light-quark hybrids find the lightest exotics to be about 2 GeV [9], 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 [10], as are QCD sum rules. Exotic charmonium states have been calculated using lattice QCD, and the 1 −+ was found to be about 4.4 GeV [11], with the expectation that the 1 −− hybrid charmonium state is at a similar energy. For the calculation of nonexotic hybrid mesons, such as 1 −− hybrids, there are other difficulties for both methods, as we shall discuss below.

C. Heavy Hybrid Meson Correlator
For a hybrid meson with quantum numbers 1 −− we use the standard current [2] with Γ ν = Cγ ν , where C is the charge conjugation operator, γ ν is the usual Dirac matrix, and Ψ is the heavy quark field. Carrying out a fourdimensional Fourier transform, the correlator in momentum space is where S ab is the quark propagator, with colors a and b. The color properties of G µα , the gluon color field, with µ, α Dirac indices, are given in Appendix A. Note that the traces are both fermion and color traces.
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 O.P.E. and a dispersion relation. The QCD sum rule method uses an operator product expansion (O.P.E.) in dimension (or inverse mo-mentum), Eq(4). For the hybrid meson the lowestorder diagram [12] is shown in Fig. 2:

FIG. 2: Lowest-order term in sum rule
Using the standard quark and gluon propagators (see Appendix A for some details) we find for this lowest-dimensional process Extracting the scalar correlator Π S , defined by Π µν (p) = (p µ p ν /p 2 − g µν )Π V (p) + (p µ p ν /p 2 )Π S (p), and carrying out the p 1 integrals, one finds for Π S 1 (p), the scalar term of Π µν 1 (p), +terms with three and four integrals .
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 O.P.E for the heavy-quark hybrid correlator includes the gluon condensate, illustrated in Fig. 3:

FIG. 3: Gluon condensate term in sum rule
For this process, the correlator Π µν 2 (p) has the same form as Eq. (4), except that the gluon trace used to obtain Π µν 1 (p) is replaced with the trace over the gluon condensate [1,2], which gives The fermionic factor is the same as for Fig. 2. From this one finds giving for the scalar part of Fig. 3 Π

E. Borel Transform of the Correlator
To ensure convergence of the O.P.E. of the correlator, one performs a Borel transform B [1], defined in Appendix B. This is discussed in detail in the early papers on QCD sum rules [1,2]. For our problem we have assumed that Π S (p) ≃ Π S 1 (p) + Π S 2 (p), with higher-order terms being very small. We shall see that even Π 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 Eqs. (37,37,38), where the quantities B 1 through B 8 are given, with BΠ S 1 (p) ≡Π S 1 (M B ), we find (1 + δ)) + multiple integral terms .
The multiple integral terms in Eq.(12) are small and are dropped, and δ is a variable of integration.
In a similar way, taking the Borel transform of Π S 2 (p), with BΠ S 2 (p) ≡Π S 2 (M B ), we find

F. 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 lefthand side and the operator product expansion the right-hand side: Π lhs = dispersion relation, Π rhs = operator product expansion. Neglecting the width of the hybrid meson, the dispersion relation has the form of a pole and a continuum: The dispersion relation is evaluated in Euclidean space, p 2 → −Q 2 , and the continuum is assumed to start at s = s o . After the Borel transform, the form we use for the lhs is with F the numerator of the pole, and L 1 and L 2 constants used to fit the form of the continuum. We shall use a standard method to eliminate F , and fit the sum rule requirement that the solution should not be sensitive to M B . In theory, the exact solution should be independent of M B . For convenience in carrying out the sum rule, we have fit the rhs of the correlator to a polynomial in the Borel mass, We find that, for all values of M B relevant to the sum rule,Π 2 is just about 1 % ofΠ 1 , and drop it, as the method is only valid to a few percent. The sum rule is obtained by taking the ratio ofΠ to To do this we use the relations Taking the ratio of , one obtains the sum rule for the mass of the heavy charmonium hybrid meson: The parameters L 1 and L 2 have magnitudes expected for the fit to the continunmm. Except for the large value of s o , the solution satisfies the criteria for a successful QCD sum rule: as Fig. 4 shows, the solution for the mass is within a few per cent of the value of M B in the region of stability. The mass predicted for the hybrid is within a 10% accuracy of the sum rule method, while the experimental mass of the Ψ ′ (2S) state [19] is The large value of s o , however, predicts that the excited hybrids are at a very high energy, as found in lattice gauge calculations, and that the Ψ ′ (2S) cannot be a pure hybrid. Therefore, we expect that the physical Ψ ′ (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 ρ − π puzzle.

III. HYBRID-NORMAL CHARMONIUM AND A POSSIBLE SOLUTION TO THE ρ − π PUZZLE
In our treatment of hadronic decays of the Ψ ′ (2S) state, we use the Sigma/Glueball model, which was motivated by the BES analysis of glueball decay [13] and the study of scalar mesons and scalar glueballs using QCD sum rules [14,15]. We briefly review this model below.

A. 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 while a scalar meson current has the form We use for our 0 ++ current [14] (with M o needed for correct dimensions) The QCD sum rule calculation makes use of the correlator Π(x) = T [J 0 ++ J 0 ++ ] . The cross term between J m and J G is evaluated by using the scalar glueball-meson coupling theorem [16]: with <qq >≡ the quark condensate , which is illustrated in Fig. 5: The results of the QCD sum rule calculations [14,15] are that there are three solutions: The Sigma/Glueball model follows from the existence of the σ, a scalar π − π 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 Fig. 5. It has been used for the study of the Roper resonance decay into a nucleon and a σ [3,17], and the prediction of σ production in proton-proton high energy collisions [18], 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, and for Υ systems.

B. Hybrid Mixing Model and the ρ − π Puzzle
The solution to the ρ − π puzzle by the mixing of hybrid and normal meson components of the Ψ(2S) state can be understood from Fig. 6 and Fig. 7, which illustrate the decay of a cc and a ccg state into two hadrons. The cc decay involves the matrix element < πρ|O|Ψ ′ (cc, 2S) >: The corresponding hybrid decay involves the matrix element < πρ|O ′ |Ψ ′ (ccg, 2S) >, with the diagram shown in Fig. 7.
Assuming that the 2s state is a cc-ccg admixture In other words, if b ≃ −.7, we have found a solution of the ρ − π puzzle. The evaluation of b is rather complicated, and the testing of this conjecture will be carried out in future work.

IV. CONCLUSION
Using the method of QCD sum rules, we have shown that the Ψ ′ (2S) state cannot be a pure charmonium hybrid. We have found that the energy of the lowest J P C = 1 −− hybrid charmonium state is approximately the same as the Ψ ′ (2S) state, about 600 MeV above the J/Ψ(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 Ψ ′ (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 ρ − π puzzle.
There have been many lattice calculations of exotic hybrid mesons. There is experimental evidence for an exotic light-quark 1 −+ meson (see Ref [7] for a discussion) at 1.4 to 1.6 GeV, which is consistent with QCD sum rule calculations [2], while lattice calculations find the lowest 1 −+ hybrid at 1.9 to 2.1 GeV. The lowest energy 1 −+ charmonium hybrid found in lattice calculations is at 4.4 GeV (see Ref [11]), about 800 MeV above our 1 −− hybrid charmonium solution. This is consistent with our large value of s o . For the 1 −− 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 Υ(2S) and the Υ(1S) is also approximately 600 MeV. If this is the energy of a Υ(nS) hybrid, this could provide a solution to the puzzling σ decays of Υ(nS) states that have recently been observed [20]. Investigation of this system is a topic of future research.

A. Heavy-Quark Hybrid Correlator in Momentum Space
The current to create a heavy-quark hybrid meson with J P C = 1 −− is where Ψ is the heavy quark field, Γ ν = Cγ ν , γ ν is the usual Dirac matrix, C is the charge conjugation operator, and the gluon color field is with λ a the SU(3) generator (T r[λ a λ b ] = 2δ ab ). From this one finds for the correlator of a heavyquark hybrid meson where S ab is the quark propagator, a standard Dirac propagator for a fermion with mass M Q , with colors a and b. Using C = iγ 2 γ 0 , T r[γ α γ β ] = 4g αβ , and T r[γ λ γ α γ δ γ β ] = 4(g λα g δβ − g λδ g αβ + g λβ g δα ): T r[S ab Γ α S ba Γ β ](p) = (−24i)