# Hybrid Charmonium and theρ−π Puzzle

Authors: Leonard S. Kisslinger; Diana Parno; Seamus Riordan
Journal: Advances in High Energy Physics
 (2008)

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## 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 theΨ′(2S) state, about 600 MeV above the J/Ψ(1S) state. Since our solution is not consistent with a pure hybrid at this energy, we conclude that the Ψ′(2S) state is probably an admixed cc¯ and hybrid cc¯g state. From this conjecture, we find a possible
explanation of the famous ρ−π puzzle.


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## Body

## 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 Ψ′(2S) state, and to find a possible explanation of
the long-standing ρ−π puzzle.Theρ−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:
(1.1)R=B(Ψ′(2S)→h)B(J/Ψ(1S)→h)=B(Ψ′(2S)→e+e−)B(J/Ψ(1S)→e+e−)≃0.13,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 [6]. This is the ρ−π 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 Ψ′(2S) decay. Our present work suggests that one can
obtain such a cancellation by including valence gluonic structure.In Section2, we show that the energy of the lowest
hybrid charmonium state with the quantum numbers JPC=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 Section 3, we show
that this can provide a solution to the ρ−π puzzle. In Section 4, we give our conclusions
and compare our results to lattice gauge calculations.


## 2. Hybrid1−− 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 numbersJPC=1−− 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(2.1)Πμν(x)=〈T[JHμ(x)JHμ(0)]〉,with the current JHμ(x) creating the hybrid state being studied. The
QCD sum rule is obtained by evaluating Πμν in two ways. First, after a Fourier transform
to momentum space, a dispersion relation gives the left-hand side (LHS) of the
sum rule:
(2.2)Π(q)lhsμν=ImΠμν(MA)π(MA2−q2)+∫so∞dsImΠμν(s)π(s−q2),where MA is the mass of the state A (assuming zero width) and so 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−MA2) term in ImΠ) and the higher-lying states produced by JHμ shown as the continuum, is illustrated in
Figure 1.Figure 1
QCD sum rule
study of a stateA with mass MA (no width).Next,Πμν(q) is evaluated by an operator product expansion
(OPE), giving the right-hand side (RHS) of the sum rule:
(2.3)Π(q)rhsμν=∑kck(q)〈0|𝒪k|0〉,with increasing k corresponding to increasing dimension of 𝒪k.After a Borel transform,ℬ,
defined in Appendix B, in which the q variable is replaced by the Borel mass, MB,
the final QCD sum rule has the form
(2.4)1πe−MHHM2/MB2+ℬ∫so∞dsIm[Π(s)]π(s−q2)=ℬ∑kck(q)〈0|𝒪k|0〉,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 s≥MB2.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 so,
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 MB (with the approximation of a fit to the
continuum, there should be a minimum or maximum in the value of MHHM versus MB,
and the value of MHHM at this extremum should be approximately the
value of MB); (3) there should be a gap between the
solution for MHHM2 and so,
which reduces the contribution of the continuum to be smaller than the pole
term due to the factor of e−s/MB2 after the Borel transform, as we will explain
in what follows. If the value of so 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 as1−+,
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 [3, 4]
predicted 1−+ 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 so 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 1−+ was found to be about 4.4 GeV [13], 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 will discuss in what follows.


### 2.3. Heavy-Hybrid Meson Correlator

For a hybrid meson with quantum numbers1−− we use the standard current [3, 4]:
(2.5)JHμ=Ψ¯ΓνGμνΨ,with Γν=Cγν,
where C is the charge conjugation operator, γν is the usual Dirac matrix, and Ψ is the heavy-quark field. Carrying out a
four-dimensional Fourier transform, the correlator in momentum space is
(2.6)Πμν(p)=∫d4p1(2π)4Tr[SabΓαSbaΓβ](p−p1)Tr[GμαGνβ](p1),where Sab is the quark propagator, with colors a and b.
The color properties of Gμα,
the gluon color field, with μ,α Dirac indices, are given in Appendix A. Note
that the traces are both fermion and color traces. Details of Tr[SabΓαSbaΓβ](p) and Tr[GμαGνβ](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.Figure 2
Lowest-order term in sum rule.Using the standard quark and gluon propagators (see
AppendixA for some details), we find the following formula for this
lowest-dimensional process:
(2.7)Π1μν(p)=−6∫d4p1(2π)4{gμν[p1⋅(p−p1)]2p12−p1⋅(p−p1)p12[pμp1ν+pν(p−p1)μ−2p1μ(p−p1)ν]}×{23MQ2(p−p1)2+[94(p−p1)2−416MQ2−23MQ41(p−p1)2]I0(p−p1)}, with
(2.8)I0(p)=∫01dαα(1−α)p2−MQ2.Extracting the scalar correlatorΠS,
defined by Πμν(p)=(pμpν/p2−gμν)ΠV(p)+(pμpν/p2)ΠS(p),
and carrying out the p1 integrals, one finds for Π1S(p),
the scalar term of Π1μν(p),
(2.9)Π1S(p)=−3(4π)2∫01dαα(1−α)×{∫01dβp2βp2(1−β)−MQ2/(α−α2)×[−43MQ6α−α2−15512MQ6(α−α2)2+31912p2MQ4α−α2−43MQ4p2−413MQ2p4]+p42∫01dβ−1p2(1−β)−MQ2/(α−α2)×[553MQ2(p2−MQ2α−α2)(−3β+β2−β33)−83MQ4(4β−3β2+β33)]}+termswiththreeandfourintegrals.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 Figure3.Figure 3
Gluon condensate term in sum rule.For this
process, the correlatorΠ2μν(p) has the same form as (2.3), except that the
gluon trace used to obtain Π1μν(p) is replaced with the trace over the gluon
condensate [1–4], which gives
(2.10)Tr[GμαGνβ](p1)=(2π)4δ4(p1)1296〈G2〉(gμνgαβ−gμβgνα).The fermionic factor is the same
as for Figure 2. From this, one finds
(2.11)Π2μν(p)=−3i2〈G2〉(4π)2{gμν(−MQ2−p24)(p2−4MQ2)I0+pμpν(112p2−533MQ2+43MQ4p2)I0+43MQ2p2},giving for the scalar part of Figure 3(2.12)Π2S(p)=−3i2〈G2〉(4π)2(114p4−596MQ2p2+143MQ4)I0.


### 2.5. Borel Transform of the Correlator

To ensure convergence of the OPE of the correlator,
one performs a Borel transformℬ [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)≃Π1S(p)+Π2S(p),
with higher-order terms being very small. We shall see that even Π2 is essentially negligible within the accuracy
of the method, and the convergence after the Borel transform is established.Using the equations in AppendixB, from (B.1), (B.2), and (B.3), where the quantities B1 through B8 are given, with ℬΠ1S(p)≡Π˜1S(MB),
we find
(2.13)Π˜1S(MB)=−14(4π)2[−16MQ6B1−155MQ6B2+319B3−16MQ4B4−164B5−110(B6−MQ2B7)−16MQ4B8]=−12(4π)2MQ4∫0∞dδe−2(MQ2/MB2)(1+δ)×{[−310δ1+δ+638δ−656(1+δ)]K3(2MQ2MB2(1+δ))+[−1892δ1+δ+606δ−1968(1+δ)−32(4δ−3δ21+δ+δ33(1+δ)2)]K2(2MQ2MB2(1+δ))+[−4778δ1+δ+9442δ−4920(1+δ)−128(4δ−3δ21+δ+δ33(1+δ)2)]K1(2MQ2MB2(1+δ))+[−1356δ1+δ+6284δ−3280(1+δ)−96(4δ−3δ21+δ+δ33(1+δ)2)]K0(2MQ2MB2(1+δ))}+multipleintegralterms.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 ofΠ2S(p),
with ℬΠ2S(p)≡Π˜2S(MB), we find
(2.14)Π˜2S(MB)=−i32(4π)2MQ4e−2(MQ2/MB2)[11K2(2MQ2MB2)+143K1(2MQ2MB2)+18K0(2MQ2MB2)].


### 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:Πlhs = dispersion relation, Πrhs = OPE. Neglecting the width of the hybrid
meson, the dispersion relation has the form of a pole and a continuum: ∫0∞ds(ImΠ(s)/(s−p2)).
The dispersion relation is evaluated in Euclidean space, p2→−Q2,
and the continuum is assumed to start at s=so.
After the Borel transform, the form we use for the LHS is
(2.15)Π˜lhs(MB2)=Fe−MHH2/MB2+(L1MB2+L2MB4)e−so/MB2,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 MB.
In theory, the exact solution should be independent of MB.For convenience in carrying out the sum rule, we have
fit the RHS of the correlator to a polynomial in the Borel mass,Π˜1(MB)=a1MB2+a2MB4+a3MB6+a4MB8 and Π˜2(MB)=b1MB2+b2MB4+b3MB6+b4MB8 with a1=122.29,a2=−143.88,a3=45.79,a4=−0.346,b1=616.0,b2=−279.1,b3=−226.56, and b4=144.515,
with units GeV6.We find that for all values ofMB relevant to the sum rule, Π˜2 is just about 1% of Π˜1,
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 ∂1/MB2Π˜.
To do this, we use the relations
(2.16)∂1/MB2e−A/MB2Kν(−AMB2)|ν>0=−Ae−A/MB2Kν(−AMB2)+A2e−A/MB2[Kν−1(−AMB2)−Kν+1(−AMB2)],∂1/MB2e−A/MB2K0(−AMB2)=−Ae−A/MB2[K0(−AMB2)−K1(−AMB2)].Taking the ratio of∂1/MB2Π˜lhs(MB)=∂1/MB2Π˜rhs(MB) to the equation Π˜lhs(MB)=Π˜rhs(MB),
one obtains the sum rule for the mass of the heavy charmonium hybrid meson:
(2.17)MHH2={e−so/MB2[so(L1MB2+L2MB4)+L1MB4+2L2MB6]+∂1/MB2Π˜1S}×{e−so/MB2[(L1MB2+L2MB4)−Π˜1S}−1.The result of the QCD sum rule fit is shown in Figure4, with so=60.0GeV2, L1=−99.0GeV4,
and L2=6.06GeV2.Figure 4
QCD sum rule for heavy-hybrid charm meson.The parametersL1 and L2 have magnitudes expected for the fit to the
continunmm. Except for the large value of so,
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 MB in the region of stability. The mass predicted for the hybrid is
(2.18)MHHM≃3.66GeV,within a 10% accuracy of the sum
rule method, while the experimental mass of the Ψ′(2S) state [15] is
(2.19)M(Ψ′(2S))=3.68GeV.The large value of so,
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.


## 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(2.1)Πμν(x)=〈T[JHμ(x)JHμ(0)]〉,with the current JHμ(x) creating the hybrid state being studied. The
QCD sum rule is obtained by evaluating Πμν in two ways. First, after a Fourier transform
to momentum space, a dispersion relation gives the left-hand side (LHS) of the
sum rule:
(2.2)Π(q)lhsμν=ImΠμν(MA)π(MA2−q2)+∫so∞dsImΠμν(s)π(s−q2),where MA is the mass of the state A (assuming zero width) and so 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−MA2) term in ImΠ) and the higher-lying states produced by JHμ shown as the continuum, is illustrated in
Figure 1.Figure 1
QCD sum rule
study of a stateA with mass MA (no width).Next,Πμν(q) is evaluated by an operator product expansion
(OPE), giving the right-hand side (RHS) of the sum rule:
(2.3)Π(q)rhsμν=∑kck(q)〈0|𝒪k|0〉,with increasing k corresponding to increasing dimension of 𝒪k.After a Borel transform,ℬ,
defined in Appendix B, in which the q variable is replaced by the Borel mass, MB,
the final QCD sum rule has the form
(2.4)1πe−MHHM2/MB2+ℬ∫so∞dsIm[Π(s)]π(s−q2)=ℬ∑kck(q)〈0|𝒪k|0〉,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 s≥MB2.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 so,
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 MB (with the approximation of a fit to the
continuum, there should be a minimum or maximum in the value of MHHM versus MB,
and the value of MHHM at this extremum should be approximately the
value of MB); (3) there should be a gap between the
solution for MHHM2 and so,
which reduces the contribution of the continuum to be smaller than the pole
term due to the factor of e−s/MB2 after the Borel transform, as we will explain
in what follows. If the value of so 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 as1−+,
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 [3, 4]
predicted 1−+ 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 so 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 1−+ was found to be about 4.4 GeV [13], 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 will discuss in what follows.


## 2.3. Heavy-Hybrid Meson Correlator

For a hybrid meson with quantum numbers1−− we use the standard current [3, 4]:
(2.5)JHμ=Ψ¯ΓνGμνΨ,with Γν=Cγν,
where C is the charge conjugation operator, γν is the usual Dirac matrix, and Ψ is the heavy-quark field. Carrying out a
four-dimensional Fourier transform, the correlator in momentum space is
(2.6)Πμν(p)=∫d4p1(2π)4Tr[SabΓαSbaΓβ](p−p1)Tr[GμαGνβ](p1),where Sab is the quark propagator, with colors a and b.
The color properties of Gμα,
the gluon color field, with μ,α Dirac indices, are given in Appendix A. Note
that the traces are both fermion and color traces. Details of Tr[SabΓαSbaΓβ](p) and Tr[GμαGνβ](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.Figure 2
Lowest-order term in sum rule.Using the standard quark and gluon propagators (see
AppendixA for some details), we find the following formula for this
lowest-dimensional process:
(2.7)Π1μν(p)=−6∫d4p1(2π)4{gμν[p1⋅(p−p1)]2p12−p1⋅(p−p1)p12[pμp1ν+pν(p−p1)μ−2p1μ(p−p1)ν]}×{23MQ2(p−p1)2+[94(p−p1)2−416MQ2−23MQ41(p−p1)2]I0(p−p1)}, with
(2.8)I0(p)=∫01dαα(1−α)p2−MQ2.Extracting the scalar correlatorΠS,
defined by Πμν(p)=(pμpν/p2−gμν)ΠV(p)+(pμpν/p2)ΠS(p),
and carrying out the p1 integrals, one finds for Π1S(p),
the scalar term of Π1μν(p),
(2.9)Π1S(p)=−3(4π)2∫01dαα(1−α)×{∫01dβp2βp2(1−β)−MQ2/(α−α2)×[−43MQ6α−α2−15512MQ6(α−α2)2+31912p2MQ4α−α2−43MQ4p2−413MQ2p4]+p42∫01dβ−1p2(1−β)−MQ2/(α−α2)×[553MQ2(p2−MQ2α−α2)(−3β+β2−β33)−83MQ4(4β−3β2+β33)]}+termswiththreeandfourintegrals.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 Figure3.Figure 3
Gluon condensate term in sum rule.For this
process, the correlatorΠ2μν(p) has the same form as (2.3), except that the
gluon trace used to obtain Π1μν(p) is replaced with the trace over the gluon
condensate [1–4], which gives
(2.10)Tr[GμαGνβ](p1)=(2π)4δ4(p1)1296〈G2〉(gμνgαβ−gμβgνα).The fermionic factor is the same
as for Figure 2. From this, one finds
(2.11)Π2μν(p)=−3i2〈G2〉(4π)2{gμν(−MQ2−p24)(p2−4MQ2)I0+pμpν(112p2−533MQ2+43MQ4p2)I0+43MQ2p2},giving for the scalar part of Figure 3(2.12)Π2S(p)=−3i2〈G2〉(4π)2(114p4−596MQ2p2+143MQ4)I0.


## 2.5. Borel Transform of the Correlator

To ensure convergence of the OPE of the correlator,
one performs a Borel transformℬ [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)≃Π1S(p)+Π2S(p),
with higher-order terms being very small. We shall see that even Π2 is essentially negligible within the accuracy
of the method, and the convergence after the Borel transform is established.Using the equations in AppendixB, from (B.1), (B.2), and (B.3), where the quantities B1 through B8 are given, with ℬΠ1S(p)≡Π˜1S(MB),
we find
(2.13)Π˜1S(MB)=−14(4π)2[−16MQ6B1−155MQ6B2+319B3−16MQ4B4−164B5−110(B6−MQ2B7)−16MQ4B8]=−12(4π)2MQ4∫0∞dδe−2(MQ2/MB2)(1+δ)×{[−310δ1+δ+638δ−656(1+δ)]K3(2MQ2MB2(1+δ))+[−1892δ1+δ+606δ−1968(1+δ)−32(4δ−3δ21+δ+δ33(1+δ)2)]K2(2MQ2MB2(1+δ))+[−4778δ1+δ+9442δ−4920(1+δ)−128(4δ−3δ21+δ+δ33(1+δ)2)]K1(2MQ2MB2(1+δ))+[−1356δ1+δ+6284δ−3280(1+δ)−96(4δ−3δ21+δ+δ33(1+δ)2)]K0(2MQ2MB2(1+δ))}+multipleintegralterms.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 ofΠ2S(p),
with ℬΠ2S(p)≡Π˜2S(MB), we find
(2.14)Π˜2S(MB)=−i32(4π)2MQ4e−2(MQ2/MB2)[11K2(2MQ2MB2)+143K1(2MQ2MB2)+18K0(2MQ2MB2)].


## 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:Πlhs = dispersion relation, Πrhs = OPE. Neglecting the width of the hybrid
meson, the dispersion relation has the form of a pole and a continuum: ∫0∞ds(ImΠ(s)/(s−p2)).
The dispersion relation is evaluated in Euclidean space, p2→−Q2,
and the continuum is assumed to start at s=so.
After the Borel transform, the form we use for the LHS is
(2.15)Π˜lhs(MB2)=Fe−MHH2/MB2+(L1MB2+L2MB4)e−so/MB2,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 MB.
In theory, the exact solution should be independent of MB.For convenience in carrying out the sum rule, we have
fit the RHS of the correlator to a polynomial in the Borel mass,Π˜1(MB)=a1MB2+a2MB4+a3MB6+a4MB8 and Π˜2(MB)=b1MB2+b2MB4+b3MB6+b4MB8 with a1=122.29,a2=−143.88,a3=45.79,a4=−0.346,b1=616.0,b2=−279.1,b3=−226.56, and b4=144.515,
with units GeV6.We find that for all values ofMB relevant to the sum rule, Π˜2 is just about 1% of Π˜1,
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 ∂1/MB2Π˜.
To do this, we use the relations
(2.16)∂1/MB2e−A/MB2Kν(−AMB2)|ν>0=−Ae−A/MB2Kν(−AMB2)+A2e−A/MB2[Kν−1(−AMB2)−Kν+1(−AMB2)],∂1/MB2e−A/MB2K0(−AMB2)=−Ae−A/MB2[K0(−AMB2)−K1(−AMB2)].Taking the ratio of∂1/MB2Π˜lhs(MB)=∂1/MB2Π˜rhs(MB) to the equation Π˜lhs(MB)=Π˜rhs(MB),
one obtains the sum rule for the mass of the heavy charmonium hybrid meson:
(2.17)MHH2={e−so/MB2[so(L1MB2+L2MB4)+L1MB4+2L2MB6]+∂1/MB2Π˜1S}×{e−so/MB2[(L1MB2+L2MB4)−Π˜1S}−1.The result of the QCD sum rule fit is shown in Figure4, with so=60.0GeV2, L1=−99.0GeV4,
and L2=6.06GeV2.Figure 4
QCD sum rule for heavy-hybrid charm meson.The parametersL1 and L2 have magnitudes expected for the fit to the
continunmm. Except for the large value of so,
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 MB in the region of stability. The mass predicted for the hybrid is
(2.18)MHHM≃3.66GeV,within a 10% accuracy of the sum
rule method, while the experimental mass of the Ψ′(2S) state [15] is
(2.19)M(Ψ′(2S))=3.68GeV.The large value of so,
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.


## 3. Hybrid-Normal Charmonium and a Possible Solution to theρ−π Puzzle

In our treatment of hadronic decays of theΨ′(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 that0++ 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
(3.1)JG(x)=αsG2,while a scalar meson current has the form
(3.2)Jm(x)=12(u¯(x)u(x)−d¯(x)d(x)).We use for our 0++ current [17] (with Mo needed for correct dimensions)
(3.3)J0++=βMoJm+(1−|β|)JG.The QCD sum rule calculation makes use of the correlatorΠ(x)=〈T[J0++J0++]〉.
The cross term between Jm and JG is evaluated by using the scalar
glueball-meson coupling theorem [19, 20]:
(3.4)∫dx〈T[JG(x)Jm(0)]〉≃−329〈q¯q〉,with 〈q¯q〉≡thequarkcondensate, which is illustrated in Figure 5.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→ the f0 (1500);80% scalar meson at 1350 MeV→ the f0 (1370);Light Scalar Glueball 400–600 MeV→ the Sigma/Glueball.The Sigma/Glueball model follows from the existence of
theσ,
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 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 cc¯g admixture for charmonium systems.


### 3.2. Hybrid Mixing Model and theρ−π Puzzle

The solution to theρ−π puzzle by the mixing of hybrid and normal
meson components of the Ψ(2S) state can be understood from Figures 6 and 7, which illustrate the decay of a cc¯ and a cc¯g state into two hadrons. The cc¯ decay involves the matrix element 〈πρ|O|Ψ′(cc¯,2S)〉.Figure 6
Lowest-order PQCD diagram for acc¯ decay into two hadrons.Figure 7
Decay of acc¯g state into two hadrons.The corresponding hybrid decay involves the matrix
element〈πρ|O′|Ψ′(cc¯g,2S)〉,
with the diagram shown in Figure 7.Assuming that the 2 s state is acc¯-cc¯g admixture,
(3.5)|Ψ′(2S)〉=b|Ψ′(cc¯,2S)〉+1−b2|Ψ′(cc¯g,2S)〉, and recognizing that the O and O′ matrix elements are approximately equal, we
see that the solution to the ρ−π puzzle requires
(3.6)b+1−b2≃0.1.
The solution of the ρ−π puzzle would be given if
(3.7)b≃−.7. 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.


## 3.1. The Sigma/Glueball Model

In energy regions where there are both scalar mesons
and scalar glueballs, it is expected that0++ 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
(3.1)JG(x)=αsG2,while a scalar meson current has the form
(3.2)Jm(x)=12(u¯(x)u(x)−d¯(x)d(x)).We use for our 0++ current [17] (with Mo needed for correct dimensions)
(3.3)J0++=βMoJm+(1−|β|)JG.The QCD sum rule calculation makes use of the correlatorΠ(x)=〈T[J0++J0++]〉.
The cross term between Jm and JG is evaluated by using the scalar
glueball-meson coupling theorem [19, 20]:
(3.4)∫dx〈T[JG(x)Jm(0)]〉≃−329〈q¯q〉,with 〈q¯q〉≡thequarkcondensate, which is illustrated in Figure 5.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→ the f0 (1500);80% scalar meson at 1350 MeV→ the f0 (1370);Light Scalar Glueball 400–600 MeV→ the Sigma/Glueball.The Sigma/Glueball model follows from the existence of
theσ,
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 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 cc¯g admixture for charmonium systems.


## 3.2. Hybrid Mixing Model and theρ−π Puzzle

The solution to theρ−π puzzle by the mixing of hybrid and normal
meson components of the Ψ(2S) state can be understood from Figures 6 and 7, which illustrate the decay of a cc¯ and a cc¯g state into two hadrons. The cc¯ decay involves the matrix element 〈πρ|O|Ψ′(cc¯,2S)〉.Figure 6
Lowest-order PQCD diagram for acc¯ decay into two hadrons.Figure 7
Decay of acc¯g state into two hadrons.The corresponding hybrid decay involves the matrix
element〈πρ|O′|Ψ′(cc¯g,2S)〉,
with the diagram shown in Figure 7.Assuming that the 2 s state is acc¯-cc¯g admixture,
(3.5)|Ψ′(2S)〉=b|Ψ′(cc¯,2S)〉+1−b2|Ψ′(cc¯g,2S)〉, and recognizing that the O and O′ matrix elements are approximately equal, we
see that the solution to the ρ−π puzzle requires
(3.6)b+1−b2≃0.1.
The solution of the ρ−π puzzle would be given if
(3.7)b≃−.7. 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.


## 4. Conclusion

Using the method of QCD sum rules, we have shown that
theΨ′(2S) state cannot be a pure charmonium hybrid. We
have found that the energy of the lowest JPC=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-quark1−+ 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 1−+ hybrid at 1.9 to 2.1 GeV. The lowest energy 1−+ charmonium hybrid found in lattice
calculations is at 4.4 GeV [13],
about 800 MeV above our 1−− hybrid charmonium solution. This is consistent
with our large value of so.
For the 1−− state, we have shown that one must use a mixed cc¯ and cc¯g 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
theϒ(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
[23]. Investigation of
this system is a topic of future research.



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*Source: 982341-2009-02-03.xml*