- Email: [email protected]
RHIC experiments
Nuclear Physics A 844 (2010) 171c–175c www.elsevier.com/locate/nuclphysa
Heavy quark hybrids, decay puzzles, and BES/CLEO/RHIC experiments Leonard S. Kisslingera∗† a
Department of Physics, Carnegie Mellon University, Pittsburgh, PA 15213, USA We estimate the energy of the lowest charmonium and upsilon states with hybrid admixtures using the method of QCD Sum Rules. Our results show that the Ψ (2S) and Υ(3S) states both have about a 50% admixture of hybrid and meson components. From this we find explanations of both the famous ρ − π puzzle for charmonium, and the unusual pattern of σ decays that have been found in Υ decays. Moreover, this picture can be used for predictions of heavy quark production with the octet model for RHIC. 1. INTRODUCTION There is a great interest in studying states with active glue, such as hybrid mesons, a color singlet composed of a quark-antiquark in a color octet and a gluon, in order to better understand nonperturbative QCD. In the present work we use QCD Sum Rules for J P C = 1−− vector states to find the lowest mixed meson-hybrid meson states for both charmonium and upsilon systems, since we found in earlier work [1] that there was no pure hybrid with these quantum numbers. In addition to the importance of finding states with active glue, we are motivated by several experimental considerations. First, the ratio of hadronic decays of the charmonium Ψ (2S) compared to the J/Ψ(1S) state is more than an order of magnitude smaller than predicted by perturbative QCD (PQCD), the so-called ρ − π puzzle. Second, the Υ(nS) states have an unusual pattern of decays into two pions, which also cannot be consistent with PQCD [2] using standard q q¯ models. Third, our theory of heavy quark states provides a basis for the color octet model predictions of RHIC heavy quark production. We first review the heavy quark puzzles and the octet model and then the method of QCD Sum Rules. Then we discuss our results for for mixed meson-hybrid meson charmonium and upsilon states, and show how this provides a solution to the heavy quark puzzles, and can be used to calculate an important part of the octet model for RHIC physics. 2. HEAVY QUARK PUZZLES AND RHIC EXPERIMENTS The lightest charmonium 1−− states are the J/ψ(1S) and ψ (2S), with masses 3.097 and 3.686 GeV, respectively. The lightest upsilon 1−− states are Υ(1S), Υ(2S), Υ(3S), Υ(4S). ∗ †
This work was supported in part by the NSF/INT grant number 0529828. Session in honor of the 85th birthday of Ernest M. Henley.
0375-9474/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.nuclphysa.2010.05.030
172c
L.S. Kisslinger / Nuclear Physics A 844 (2010) 171c–175c
The mass of the Υ(1S) is 9.46 GeV, and the separation energy of the Υ(2S) and the Υ(1S) states is nearly the same as that between the ψ (2S) and J/ψ(1S) states, which we shall see is somewhat misleading. 2.1. The ρ − π puzzle The ρ − π puzzle for c¯ c 1−− states is based on the two processes for PQCD and electromagnetic decay of such states. By taking ratios the wave functions at the origin cancel, and this predicts the ratio of branching ratios for c¯ c decays into hadrons (h) R =
B(Ψ (2S)→h) B(J/Ψ(1S)→h)
+ −
B(Ψ (2S)→e e ) = B(J/Ψ(1S)→e + e− ) 0.12,
(1)
the famous 12% rule. However,the Ψ (2S) to J/Ψ ratios for ρ−π and other hadron decays are more than an order of magnitude smaller than predicted by Eq. (1). Many theorists have tried and failed to explain this puzzle. 2.2. The σ decays of Υ(nS) states puzzle The puzzle of sigma decays of b¯b 1−− (Υ(nS)) states is given by the following. The sigma is a low-energy broad two-pion scalar resonance. Experiments on Υ(nS) states find [2] Υ(2S) → Υ(1S) + 2π has a large branching ratio, but no σ Υ(3s) → Υ(1S) + 2π has a large branching ratio to σ Δn = 2, emit σ Δn = 2, no σ emitted. This is the Vogel Δn = 2 puzzle, which cannot be understood using perturbative QCD, as expected for heavy bottomium states. 2.3. The Octet model for RHIC and hybrids The major goal of modern RHIC (Relativistic Heavy Ion Collision) experiments is to produce and study the quark-gluon plasma (QGP) which existed in the early universe before the QCD phase trasition. One important signal of this QGP is the production of heavy quark (charmonium and upsilon) states via q q¯ interactions in the early universe. ¯ in which an octet q q¯ produces an octet The most natural mechanism is q q¯ → g → QQ, ¯ which is just PQCD, followed by the nonperturbative (NPQCD) process in which QQ, ¯ becomes a singlet QQ ¯ with the emission of a gluon (or other color octet) the octet QQ [3]. As we shall see, our determination of mixed heavy quark and heavy quark hybrid mesons will provide a mechanism for predicting the NPQCD matrix elements. 3. MIXED HEAVY QUARK HYBRID HEAVY QUARK 1−− STATES AND QCD SUM RULES Before we discuss our calculation of mixed heavy hybrids, we briefly review the method of QCD Sum Rules
L.S. Kisslinger / Nuclear Physics A 844 (2010) 171c–175c
173c
3.1. Review of QCD sum rules The correlator for use in the QCD sum rule method for a hybrid or mixed hybrid baryon state, |H > ΠH (x) =< 0|T [JH (x)JH (0)]|0 >,
(2)
with the current JH creating a state |H > (JH | >= |H > + a continuum of states with similar properties).The QCD sum rule method is based on equating a dispersion relation, called the left hand side, to an operator product expansion (OPE) of the correlator, called the right hand side. Π(q)H lhs = H Π(q)rhs =
ImΠH (MH ) 2 −q 2 ) π(MH
H
(s) + s∞ ds ImΠ π(s−q 2 ) 0 k ck (q)0|Ok |0,
(3)
where MH is the mass of the state (assuming zero width) and s0 is the start of the continuum (a parameter to be determined), ck (q) are the Wilson coefficients and 0|Ok |0 are gauge invariant operators constructed from quark and gluon fields, with increasing k corresponding to increasing dimension of Ok . After a Borel transform, B, in which the q variable is replaced by the Borel mass, MB , the final QCD sum rule has the form BΠH (q)(LHS) = BΠH (q)(RHS).
(4)
For a succesful solution the value of MH (MB ) should have a minimum near the value of MB , should not depend very much on MB , and the value of s0 should be approximately the magnitude of the square of the mass of the next higher excited state. 3.2. Mixed charmonium-hybrid charmonium and mixed upsilon states Here we give a brief overview of the QCD Sum Rule calculation. For details see Ref. [4]. Recognizing that there is strong mixing between a heavy quark meson and a hybrid heavy quark meson with the same quantum numbers (as shown below), and that the fact that a pure hybrid charmonium solution is not a physical state [1], we now attempt to find the lowest J P C = 1−− charmonium state with a sizable admixture of a charmonium meson and a hybrid charmonium meson. An appropriate mixed vector (J P C = 1−− ) charmonium, hybrid charmonium current to use in QCD Sum Rules is √ μ ; (5) J μ = bJHμ + 1 − b2 JHH with q¯ca γ μ qca JHμ = μ ¯ JHH = ΨΓν Gμν Ψ,
(6)
μ where JHμ is the standard current for a 1−− charmonium state, and JHH is the heavy charmonium hybrid current, where Ψ is the heavy quark field, Γν = Cγν , γν is the usual Dirac matrix, C is the charge conjugation operator, and the gluon color field is Gμν = 8a=1 λ2a Gμν a with λa the SU(3) generator (T r[λa λb ] = 2δab ). The correlator is μ ν Πμν H−HH (x) =< 0|T [J (x)J (0)]|0 > .
(7)
174c
L.S. Kisslinger / Nuclear Physics A 844 (2010) 171c–175c
Using the QCD Sum Rule method briefly described above, after a Borel transform we obtain a sum rule by equating the dispersion relation to the OPE. By taking the ratio of the derivative of the sum rule with respect to 1/MB2 to the sum rule, we obtain an expression for the mass of the heavy mixed hybrid as a function of the Borel mass, MB 2 MH−HH =
{[so (K0 = K1MB2 + K2MB4 + K3MB6 ) −
so 2
+K1Mb4 + 2K2MB6 + 3K3MB8 ]e MB +∂1/MB2 sH−HH } × {(K0 + K1MB2 + K2MB4 +K3MB6 )e
− so2 M B
−
(8)
S
−1 H−HH } .
The range of b for which a satisfactory solution is obtained is b=−0.7 ± 0.1, with the result for b=-.7 shown in the figure. See Ref. [4] for ΠSH−HH 13.66 2 MC−HC 13.65
13.64 13.50
13.55
13.60
13.65
13.70
13.75
MB2
Figure 1. Mixed charmonium-hybrid charmonium mass = 3.69GeV.
We find the mass of the lowest-energy mixed charmonium-hybrid charmonium to be about the energy of the Ψ (2S) state, 3.69 GeV, with so =20 GeV2 , b= -0.7 ⇒ 50-50 per cent charmonium-hybrid charmonium. It satisfies the criteria for about a fifteen per cent accuracy. The values of the other parameters are K0 = −15.9, K1 = 0.224, K2 = −0.00015, K3 = 0.00009. The only solutions satisfying the sum rule criteria are those with the value of b about −.7 ± .1, so that we find the state to be about a 50-50 per cent meson-hybrid meson. As we shall see, this gives a solution to the ρ − π puzzle. For the upsilon case we find an essentially identical solution. The surprising thing, however, is that it is the Υ(3S), rather than the Υ(2S) state, that is the mixed hybrid upsilon state. As we shall now see, from this we have obtained a solution to the Vogel Δn = 2 puzzle. 4. MIXED MESON-HYBRID MESON, HEAVY QUARK DECAY PUZZLES, AND OCTET MODEL In this section we briefly explain how our mixed heavy hybrid picture gives the solution to the puzzles and can be used for RHIC physics. 4.1. The ρ − π puzzle Note that the matrix element < πρ|O|ψ (c¯ c, 2S) > for ρ − π decay of |c¯ c(2S) > is given by the PQCD diagram with three gluons emitted, while for the hybrid component
L.S. Kisslinger / Nuclear Physics A 844 (2010) 171c–175c
175c
there are two gluons emitted, and the third is the hybrid constituent. Since we find that |Ψ (2S) > −0.7|c¯ c(2S) > +0.7|c¯ cg(2S) >, so the charmonium and hybrid charmonium approximately cancel, we obtain for all 2 hadron decays, including ρ + π decay, R=
B(Ψ (2S) → ρ + π) << 0.12, B(J/Ψ(1S) → ρ + π)
(9)
which is our proposed solution to the ρ − π puzzle. 4.2. σ decays of Υ(nS) states puzzle The solution to the Vogel Δn = 2 puzzle is based on the aplication of the glueball/sigma model, based on the study of scalar mesons and scalar glueballs [5], which was motivated by the BES analysis of glueball decay, and our solution for the lowest mixed state to be the Υ(3S) state. Using the gluon-meson theorm [6], there is a strong decay of the hybrid component of the Υ(3S) to the Υ(1S) with sigma emission. Since the 2S and 4S have no hybrid component, this provides a solution to the Vogel Δn = 2 puzzle. 4.3. Octet model for RHIC The differential cross section for the production of a charmonium state in a A-A collision in the color octet model is [7] dσ ¯ [pp → ψ(c¯ c)] = fq/A fq/A dσ [qq → C C(8) → ψ(c¯ c)] dpT dt (10) ψ dσ ¯ [qq → C C(8) → ψ(c¯ c)] = [perturbative QCD]× < 0|O8 Q |0 >, dt
ψ
In brief, we can determine the NPQCD matrix elements < 0|O8 Q |0 >, coupling octet to singlet states, using our hybrid solutions and processes. 5. CONCLUSION In summary, we find that the ψ (2S) is approximately 50% charmonium and 50% hybrid charmonium; and the Υ(3S) is approximately 50% bottomium and 50% hybrid bottomium. This solves the ρ − π problem for charmonium decays, and the Vogel Δn = 2 puzzle for sigma decays of upsilon states. It also serves as the basis for determining matrix elements for heavy quark state production in the octet model. REFERENCES 1. L. S. Kisslinger, D. Parno and S. Riordan, arXiv:0805.1943; Adv. in High Energy Phys. (2009). 2. H. Vogel, hep-ex/060601, Proceedings of 4th Flavor Physics and CP Violation Conference (FPCP’06) (2006), www.slac.stanford/econf/C060409. 3. G. C. Nayak, M. X. Liu, and F. Cooper, Phys. Rev. D 68 (2003) 034003; F. Cooper, M. X. Liu, and G. C. Nyak, Phys. Rev. Lett. 93 (2004) 171801. 4. L. S. Kisslinger, Phys Rev. D 79 (2009) 114026. 5. L. S. Kisslinger, J. Gardner and C. Vanderstraeten, Phys. Lett. B410 (1997) 1. 6. V. A. Novikov, M. A. Shifman, A. I. Vainstein, V. I. Zakharov, Nucl. Phys. B165 (1980) 67; Nucl. Phys. B191 (1981) 301. 7. P. Cho and A. K. Leibovich, Phys. Rev. D53 (1996) 150; 6203.
Heavy quark hybrids, decay puzzles, and BES/CLEO/RHIC experiments Leonard S. Kisslingera∗† a
Department of Physics, Carnegie Mellon University, Pittsburgh, PA 15213, USA We estimate the energy of the lowest charmonium and upsilon states with hybrid admixtures using the method of QCD Sum Rules. Our results show that the Ψ (2S) and Υ(3S) states both have about a 50% admixture of hybrid and meson components. From this we find explanations of both the famous ρ − π puzzle for charmonium, and the unusual pattern of σ decays that have been found in Υ decays. Moreover, this picture can be used for predictions of heavy quark production with the octet model for RHIC. 1. INTRODUCTION There is a great interest in studying states with active glue, such as hybrid mesons, a color singlet composed of a quark-antiquark in a color octet and a gluon, in order to better understand nonperturbative QCD. In the present work we use QCD Sum Rules for J P C = 1−− vector states to find the lowest mixed meson-hybrid meson states for both charmonium and upsilon systems, since we found in earlier work [1] that there was no pure hybrid with these quantum numbers. In addition to the importance of finding states with active glue, we are motivated by several experimental considerations. First, the ratio of hadronic decays of the charmonium Ψ (2S) compared to the J/Ψ(1S) state is more than an order of magnitude smaller than predicted by perturbative QCD (PQCD), the so-called ρ − π puzzle. Second, the Υ(nS) states have an unusual pattern of decays into two pions, which also cannot be consistent with PQCD [2] using standard q q¯ models. Third, our theory of heavy quark states provides a basis for the color octet model predictions of RHIC heavy quark production. We first review the heavy quark puzzles and the octet model and then the method of QCD Sum Rules. Then we discuss our results for for mixed meson-hybrid meson charmonium and upsilon states, and show how this provides a solution to the heavy quark puzzles, and can be used to calculate an important part of the octet model for RHIC physics. 2. HEAVY QUARK PUZZLES AND RHIC EXPERIMENTS The lightest charmonium 1−− states are the J/ψ(1S) and ψ (2S), with masses 3.097 and 3.686 GeV, respectively. The lightest upsilon 1−− states are Υ(1S), Υ(2S), Υ(3S), Υ(4S). ∗ †
This work was supported in part by the NSF/INT grant number 0529828. Session in honor of the 85th birthday of Ernest M. Henley.
0375-9474/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.nuclphysa.2010.05.030
172c
L.S. Kisslinger / Nuclear Physics A 844 (2010) 171c–175c
The mass of the Υ(1S) is 9.46 GeV, and the separation energy of the Υ(2S) and the Υ(1S) states is nearly the same as that between the ψ (2S) and J/ψ(1S) states, which we shall see is somewhat misleading. 2.1. The ρ − π puzzle The ρ − π puzzle for c¯ c 1−− states is based on the two processes for PQCD and electromagnetic decay of such states. By taking ratios the wave functions at the origin cancel, and this predicts the ratio of branching ratios for c¯ c decays into hadrons (h) R =
B(Ψ (2S)→h) B(J/Ψ(1S)→h)
+ −
B(Ψ (2S)→e e ) = B(J/Ψ(1S)→e + e− ) 0.12,
(1)
the famous 12% rule. However,the Ψ (2S) to J/Ψ ratios for ρ−π and other hadron decays are more than an order of magnitude smaller than predicted by Eq. (1). Many theorists have tried and failed to explain this puzzle. 2.2. The σ decays of Υ(nS) states puzzle The puzzle of sigma decays of b¯b 1−− (Υ(nS)) states is given by the following. The sigma is a low-energy broad two-pion scalar resonance. Experiments on Υ(nS) states find [2] Υ(2S) → Υ(1S) + 2π has a large branching ratio, but no σ Υ(3s) → Υ(1S) + 2π has a large branching ratio to σ Δn = 2, emit σ Δn = 2, no σ emitted. This is the Vogel Δn = 2 puzzle, which cannot be understood using perturbative QCD, as expected for heavy bottomium states. 2.3. The Octet model for RHIC and hybrids The major goal of modern RHIC (Relativistic Heavy Ion Collision) experiments is to produce and study the quark-gluon plasma (QGP) which existed in the early universe before the QCD phase trasition. One important signal of this QGP is the production of heavy quark (charmonium and upsilon) states via q q¯ interactions in the early universe. ¯ in which an octet q q¯ produces an octet The most natural mechanism is q q¯ → g → QQ, ¯ which is just PQCD, followed by the nonperturbative (NPQCD) process in which QQ, ¯ becomes a singlet QQ ¯ with the emission of a gluon (or other color octet) the octet QQ [3]. As we shall see, our determination of mixed heavy quark and heavy quark hybrid mesons will provide a mechanism for predicting the NPQCD matrix elements. 3. MIXED HEAVY QUARK HYBRID HEAVY QUARK 1−− STATES AND QCD SUM RULES Before we discuss our calculation of mixed heavy hybrids, we briefly review the method of QCD Sum Rules
L.S. Kisslinger / Nuclear Physics A 844 (2010) 171c–175c
173c
3.1. Review of QCD sum rules The correlator for use in the QCD sum rule method for a hybrid or mixed hybrid baryon state, |H > ΠH (x) =< 0|T [JH (x)JH (0)]|0 >,
(2)
with the current JH creating a state |H > (JH | >= |H > + a continuum of states with similar properties).The QCD sum rule method is based on equating a dispersion relation, called the left hand side, to an operator product expansion (OPE) of the correlator, called the right hand side. Π(q)H lhs = H Π(q)rhs =
ImΠH (MH ) 2 −q 2 ) π(MH
H
(s) + s∞ ds ImΠ π(s−q 2 ) 0 k ck (q)0|Ok |0,
(3)
where MH is the mass of the state (assuming zero width) and s0 is the start of the continuum (a parameter to be determined), ck (q) are the Wilson coefficients and 0|Ok |0 are gauge invariant operators constructed from quark and gluon fields, with increasing k corresponding to increasing dimension of Ok . After a Borel transform, B, in which the q variable is replaced by the Borel mass, MB , the final QCD sum rule has the form BΠH (q)(LHS) = BΠH (q)(RHS).
(4)
For a succesful solution the value of MH (MB ) should have a minimum near the value of MB , should not depend very much on MB , and the value of s0 should be approximately the magnitude of the square of the mass of the next higher excited state. 3.2. Mixed charmonium-hybrid charmonium and mixed upsilon states Here we give a brief overview of the QCD Sum Rule calculation. For details see Ref. [4]. Recognizing that there is strong mixing between a heavy quark meson and a hybrid heavy quark meson with the same quantum numbers (as shown below), and that the fact that a pure hybrid charmonium solution is not a physical state [1], we now attempt to find the lowest J P C = 1−− charmonium state with a sizable admixture of a charmonium meson and a hybrid charmonium meson. An appropriate mixed vector (J P C = 1−− ) charmonium, hybrid charmonium current to use in QCD Sum Rules is √ μ ; (5) J μ = bJHμ + 1 − b2 JHH with q¯ca γ μ qca JHμ = μ ¯ JHH = ΨΓν Gμν Ψ,
(6)
μ where JHμ is the standard current for a 1−− charmonium state, and JHH is the heavy charmonium hybrid current, where Ψ is the heavy quark field, Γν = Cγν , γν is the usual Dirac matrix, C is the charge conjugation operator, and the gluon color field is Gμν = 8a=1 λ2a Gμν a with λa the SU(3) generator (T r[λa λb ] = 2δab ). The correlator is μ ν Πμν H−HH (x) =< 0|T [J (x)J (0)]|0 > .
(7)
174c
L.S. Kisslinger / Nuclear Physics A 844 (2010) 171c–175c
Using the QCD Sum Rule method briefly described above, after a Borel transform we obtain a sum rule by equating the dispersion relation to the OPE. By taking the ratio of the derivative of the sum rule with respect to 1/MB2 to the sum rule, we obtain an expression for the mass of the heavy mixed hybrid as a function of the Borel mass, MB 2 MH−HH =
{[so (K0 = K1MB2 + K2MB4 + K3MB6 ) −
so 2
+K1Mb4 + 2K2MB6 + 3K3MB8 ]e MB +∂1/MB2 sH−HH } × {(K0 + K1MB2 + K2MB4 +K3MB6 )e
− so2 M B
−
(8)
S
−1 H−HH } .
The range of b for which a satisfactory solution is obtained is b=−0.7 ± 0.1, with the result for b=-.7 shown in the figure. See Ref. [4] for ΠSH−HH 13.66 2 MC−HC 13.65
13.64 13.50
13.55
13.60
13.65
13.70
13.75
MB2
Figure 1. Mixed charmonium-hybrid charmonium mass = 3.69GeV.
We find the mass of the lowest-energy mixed charmonium-hybrid charmonium to be about the energy of the Ψ (2S) state, 3.69 GeV, with so =20 GeV2 , b= -0.7 ⇒ 50-50 per cent charmonium-hybrid charmonium. It satisfies the criteria for about a fifteen per cent accuracy. The values of the other parameters are K0 = −15.9, K1 = 0.224, K2 = −0.00015, K3 = 0.00009. The only solutions satisfying the sum rule criteria are those with the value of b about −.7 ± .1, so that we find the state to be about a 50-50 per cent meson-hybrid meson. As we shall see, this gives a solution to the ρ − π puzzle. For the upsilon case we find an essentially identical solution. The surprising thing, however, is that it is the Υ(3S), rather than the Υ(2S) state, that is the mixed hybrid upsilon state. As we shall now see, from this we have obtained a solution to the Vogel Δn = 2 puzzle. 4. MIXED MESON-HYBRID MESON, HEAVY QUARK DECAY PUZZLES, AND OCTET MODEL In this section we briefly explain how our mixed heavy hybrid picture gives the solution to the puzzles and can be used for RHIC physics. 4.1. The ρ − π puzzle Note that the matrix element < πρ|O|ψ (c¯ c, 2S) > for ρ − π decay of |c¯ c(2S) > is given by the PQCD diagram with three gluons emitted, while for the hybrid component
L.S. Kisslinger / Nuclear Physics A 844 (2010) 171c–175c
175c
there are two gluons emitted, and the third is the hybrid constituent. Since we find that |Ψ (2S) > −0.7|c¯ c(2S) > +0.7|c¯ cg(2S) >, so the charmonium and hybrid charmonium approximately cancel, we obtain for all 2 hadron decays, including ρ + π decay, R=
B(Ψ (2S) → ρ + π) << 0.12, B(J/Ψ(1S) → ρ + π)
(9)
which is our proposed solution to the ρ − π puzzle. 4.2. σ decays of Υ(nS) states puzzle The solution to the Vogel Δn = 2 puzzle is based on the aplication of the glueball/sigma model, based on the study of scalar mesons and scalar glueballs [5], which was motivated by the BES analysis of glueball decay, and our solution for the lowest mixed state to be the Υ(3S) state. Using the gluon-meson theorm [6], there is a strong decay of the hybrid component of the Υ(3S) to the Υ(1S) with sigma emission. Since the 2S and 4S have no hybrid component, this provides a solution to the Vogel Δn = 2 puzzle. 4.3. Octet model for RHIC The differential cross section for the production of a charmonium state in a A-A collision in the color octet model is [7] dσ ¯ [pp → ψ(c¯ c)] = fq/A fq/A dσ [qq → C C(8) → ψ(c¯ c)] dpT dt (10) ψ dσ ¯ [qq → C C(8) → ψ(c¯ c)] = [perturbative QCD]× < 0|O8 Q |0 >, dt
ψ
In brief, we can determine the NPQCD matrix elements < 0|O8 Q |0 >, coupling octet to singlet states, using our hybrid solutions and processes. 5. CONCLUSION In summary, we find that the ψ (2S) is approximately 50% charmonium and 50% hybrid charmonium; and the Υ(3S) is approximately 50% bottomium and 50% hybrid bottomium. This solves the ρ − π problem for charmonium decays, and the Vogel Δn = 2 puzzle for sigma decays of upsilon states. It also serves as the basis for determining matrix elements for heavy quark state production in the octet model. REFERENCES 1. L. S. Kisslinger, D. Parno and S. Riordan, arXiv:0805.1943; Adv. in High Energy Phys. (2009). 2. H. Vogel, hep-ex/060601, Proceedings of 4th Flavor Physics and CP Violation Conference (FPCP’06) (2006), www.slac.stanford/econf/C060409. 3. G. C. Nayak, M. X. Liu, and F. Cooper, Phys. Rev. D 68 (2003) 034003; F. Cooper, M. X. Liu, and G. C. Nyak, Phys. Rev. Lett. 93 (2004) 171801. 4. L. S. Kisslinger, Phys Rev. D 79 (2009) 114026. 5. L. S. Kisslinger, J. Gardner and C. Vanderstraeten, Phys. Lett. B410 (1997) 1. 6. V. A. Novikov, M. A. Shifman, A. I. Vainstein, V. I. Zakharov, Nucl. Phys. B165 (1980) 67; Nucl. Phys. B191 (1981) 301. 7. P. Cho and A. K. Leibovich, Phys. Rev. D53 (1996) 150; 6203.