Compact N -quark Hadron Mass Dependence on N 4 : A Classical Field Picture
aa r X i v : . [ phy s i c s . g e n - ph ] S e p Compact N -quark Hadron Mass Dependence on N :A Classical Field Picture Rui-Cheng LI ∗ School of Physical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
September 18, 2020
Abstract
We give a hypothesis on the mass spectrum of compact N -quark hadron states in a classical fieldpicture, which indicates that there would be a mass dependence on about N . We call our model“bag-tube oscillation model”, which can be seemed as a kind of combination of quark-bag model andflux-tube model. The large decay widths due to large masses might be the reason why the compact N -quark hadrons still disappear so far. PACS numbers
Key words multi-quark state, constituent quark, glueball, bag model, flux-tube model, hadronmass spectrum ∗ [email protected] Introduction
It is reasonable and necessary to go back to the classical field picture from the quantum field picture. Theclassical field picture has been applied into the renormalization in quantum field theory. For example, thephysical mass of an electron in the quantum electrodynamics (QED) was seemed as the combination of thebare mass of electron and the effective mass of electric field surrounding the electron (or, the self-energycorrection of electron). Similarly for a quark, besides of the bare mass and the electric field, the colorfield should also be combined into the physical mass. Thus, we need to evaluate the effective mass of thecolor field. This goal has been realized in the perturbative sector of the quantum chromodynamics (QCD),however, this task has not been finished and it might be complicate in the nonperturbative sector.On the other hand, instead of the physical mass of a quark, the constituent quark degree of freedom(d.o.f) was introduced into a hadron system, which is dominated by the nonperturbative sector of QCD.Then, some interesting question arise. For example, why is it so different between the pion defined as twoquark system and the proton defined as three quark system? Why is it so different between the massesof constituent quarks in a pion and the ones in a proton, and, does this depend on the evaluation to theeffective masse of the color field in the nonperturbative sector?So, if we can evaluate the effective mass of the color field, either quantitatively or qualitatively, thenwe might get some clues on the masses of constituent quarks, and we might also get some information onthe mass spectrum of hadrons with multiple quarks. In this work, we will evaluate the effective mass ofthe color field in a classical field picture.The remainder of this paper is organized as follows. In Section 2, we review the introduction of currentquarks and give a definition on the physical d.o.f by going back to classical field picture from quantumfield picture. In Section 3, we will try to give a hypothesis on the mass spectrum of compact multi-quarkhadrons and glueballs, and try to explain why the compact N -quark hadrons still disappear so far. InSection 4, we will try to give some new understanding to the constituent quarks. Finally, the conclusionsare given in Section 5. Besides of the mass renormalization, we will show, the physical dynamical quantum vacuum state will berelated to the classical field. The crucial reason is that, the high excited state | N ~ ω i with energy N ~ ω of a simple harmonic oscillator system is equivalent to the Fock state of an N -particle system (or, thedirect product of each single particle state ~ ω with energy ~ ω ); and, like the laser, by statistics on theprobability amplitude of single particle, the energy density distribution (not the normalized state vector)of an N -particle system is equivalent to the field strength of a classical field.For a simple harmonic oscillator system, the Hamiltonian is defined as [1] H = X k ω k (cid:18) a † k a k + 12 (cid:19) ; (1)and, the ground state (vacuum state) | i and the single particle state | i , the creation operator a † k andthe annihilation operator a k , are defined as | i = Y k | i k , h | i = 1 , (2) a k | i = 0 , a † k | i = | i , a k | i = | i . (3)The energy of vacuum state (the zero-point energy) is E = h | H | i = X k ω k = ∞ ; (4)and, to avoid the infinity, a renormalized Hamiltonian is defined as H → H R ≡ H − h | H | i = X k ω k a † k a k , (5)1ith a “single current quark” state | m i with the mass m as the eigenfunction of H R defined as H R | m i ≡ ω m | m i , ω m = √ k + m . (6)However, by recalling the bare mass, the energy of a free particle should be just infinity, and the infinityvacuum expectation value (VEV) should not be subtracted roughly; on the other hand, the eigen-equationof H R in (6) is not a real Schrodinger equation and can not include the full information of the system.Indeed, for a full Hamiltonian H , the vacuum state | i and the “single real quark” state | M i = a † | i with the mass M should be H | i = E | i , (7) H | M i = E | M i , (8)= ( H + H R ) | M i = ( E + ω ) | M i ; (9)then, for a state | m i defined in (6), there will be | m i = a | i + a | M i + a | M i + ...., (10) ⇒ ω m = ω, < ω m < + ∞ , (11)or, inversely, from (10) there will be | M i = b | m i + b | i + b | M i + ... ; (12)where the coefficients a i and b i might be functions of many parameters (e.g., the energy scale µ ). Thatmeans, the current quark state | m i should be a mixed state of the “single real quark” state | M i andother higher energy level states; and, in the perturbative sector, there should be a ≃ a ≃ | m i ≃ | M i .When the interaction H int is included, there will be the eigenvector of the total Hamiltonian H , | M i = α † | Ω i = c | m i + c | Ω i + c | M i + ..., with α = a, | Ω i 6 = | i , (13)where | Ω i is the new vacuum state, the coefficients c i could be functions of many parameters (e.g., theenergy scale µ ), and α † can be seemed as a Bogliubov transform of a † ; we will call | M i as the “singlephysical quark” state with the mass M . From (13), it means that, | M i should be a mixed state of thecurrent quark state | m i and other multi-particle states; and the mass relation will be M = | c | m + | c | m Ω + ¯ M (ren.) −−−→ | c | m + ¯ M . (14)In (14), m Ω ≃ ∞ is the effective mass of | Ω i state and it will be counteracted by the “infinite part” ofthe color field surrounding the quark (by omitting the electric field), denoted by E ( ∞ ) c ; besides, ¯ M is theeffective mass of a superposition state | ¯ M i ≡ c | M i + ... , and it will be corresponding to the “finite part”of the color field surrounding the quark, denoted by E ( ¯ M ) c . As mentioned above, the high excited state | N M i of an effective harmonic oscillator is equivalent to the Fock state of an N -particle system, and, likethe laser, the energy ¯ M (not the normalized state vector) of an N -particle system is equivalent to the fieldstrength of a classical field E ( ¯ M ) c . In a word, the mass of physical state | M i is the combination of themass of current quark state | m i and the mass of an effective classical field E ( ¯ M ) c , or, the mass correctionto a system of multiple quarks is from the potential energy (interaction ) H int , which is just the energy ofcolor field!We should point out that, the “single physical quark” state | M i (eigenstates of the full Hamiltonian H ) is indeed a “quasi-particle” state rather than a real “particle” state, since it cannot exist due to theconfine character of QCD. Instead, the “single current quark” state | m i is seemed as physical excitationquanta in QCD in the perturbation sense, while the state | M i is traditionally called “dressed quark” since2he mass ¯ M could be seemed as a non-perturbative self-energy correction on a current quark, or called“constituent quark” [2] which will be discussed in Sect.4.Our model is different with the decomposition picture in Ref. [3]. In Ref. [3], the field operators ofquarks and gluons are redefined to the combination of a background field part and a quantum field partand then a class of new Feynman rules of quarks and gluons were given, that means, the field theory isstill constructed in a quantized scheme. People have made lots of attempts to interpret the mass spectrum of compact N -quark hadrons in manymethods [4] [5] [6], such as the lattice quantum chromodynamics (LQCD) [7], QCD sum rules [8], and theconstituent quark models, et al. Most of the constituent quark models are nonrelativistic, that means, onlythe constituent quark d.o.fs existed. In these nonrelativistic potential models, due to the character of QCD,one leading part of the potential is always constructed to a confining form, such as: a volume-dependentform (e.g., the quark-bag model [9]), a harmonic oscillator form (e.g., the Isgur-Karl model [6] [10] orSkyrme model in large N c limit [11]), a linear form (e.g., the Cornell model [12]), et al; and the otherleading part of the potential is always constructed to a Coulomb form or a Yukawa form; in addition, as thenext-to-leading part, some hyperfine potential terms will be constructed as well to generate the complicatehadronic mass spectrum, such as the Regge trajectory terms (orbital and spin interactions) [13], the spin-flavor coupling terms (the G¨ u rsey-Radicati terms) [14], and so on. As converting to the language of fieldtheory, e.g., in the QCD framework, a linear form confinement potential will arise in the nonperturbativeregime of QCD, which can be derived from the lattice QCD (LQCD) calculation in the heavy quarklimit [15]; some logarithmical form potential [16] and the Coumlomb form potential due to one gluonexchange processes will arise in the perturbative regime of QCD [17]. Or, instead the Coulomb form, aYukawa form potential from one Goldstone boson exchange processes in chiral perturbative theory (ChPT)will arise [18] [19].To understand the nonperturbative effect of QCD, especially the linear potential from LQCD, somepseudo-particle d.o.fs are proposed, since gluon d.o.fs are not good ones in the nonperturbative regime.On one hand, motivated by the linear Regge trajectories, mesons are treated as strings; on the other hand,based on the QCD picture in strong-coupling regime, as what are shown in LQCD, it is found that a classof collective string-like flux tubes (or “flux links” on the lattice) generated from the gluon condensationscould be seemed as good natural d.o.fs (for the reason, e.g., in lattice QCD, for the color-electric field E , | E | has a definite nonzero eigenvalue). So, a type of so-called flux-tube model (or called string-like model,collective model or hypercentral model) is established [20] [21]. However, that does not mean flux tubesare really physical objects; instead, they are just pseudo-particle d.o.fs. Here we just list some details onthe flux-tube model which will be useful to introduce our own model, as below:(1) flux tubes are directed, with a quark or antiquark acting as a unit source (sink) of flux;(2) the configurations of mesons, baryons and 4-quark hadron states are shown in Fig. 1-(a), 1-(b)and 1-(c), respectively; for baryons in Fig. 1-(b), there is a junction of 3 flux tubes; for 4-quark hadronsor multi-quark hadrons in Fig. 1-(c), there exists mixing between different topological configurations; thedynamics of multi-quark states are highly not like mesons and baryons, and the system might not becompletely confined [20] [22];(3) dynamics: hadrons decay via flux tube breaking, due to quark pair creation;(4) for color-singlet hadrons, the residual force (van der Waals-type force) will be suppressed to ashort-range one, due to the confinement or color screening;(5) when all the flux tubes in a hadron are “frozen” (on the so-called adiabatic surface), the hadron isa purely quark-flavored state with definite quantum numbers, and with energy E = b P Nn =1 R n , where N is the number of flux tubes in the hadron; when there are excitation modes (phonon) with energy { ω n , n = 1 , , ..., n max } on the flux tubes, the hadron will become a hybrid, with energy E + P n ω n ;that means, the energy of a hadron includes a linear potential energy from the string-like tensional energybetween the static quarks as leading part, with a kind of harmonic vibrational energy as perturbation;(6) glueball states are purely consisted with flux tube loops without quarks; etc.3igure 1: In the “flux-tube model” [20], flux tube configurations for (a) mesons, (b) baryons and (c)4-quark hadrons, respectively; particularly, (a) a flux tube (or ”flux link” on the lattice) is a directedelement (or ”string”); (b) three units of flux all directed toward (away from) a “junction” can annihilate(be created) there.Figure 2: In our “bag-tube oscillation model”, the tube configurations of color fields for (a) mesons, (b)baryons and (c) 4-quark hadrons, respectively, with the superscripts a, b, c denoting the color indices.Now, we will propose our own model. Our model can be seemed as a kind of combination of quark-bagmodel [9] and flux tube model, and we call our model “ bag-tube model ” or “ bag-tube oscillationmodel ”. Here we only list some tentative definitions for our model, as below:(1) we define that, the space configuration of a hadron is in an oscillation between a bag-like one anda tube-like one, i.e., color fields induced by the color charges (quarks, antiquarks, or gluons) in a hadronare confined in a bag when charges are in short distances, while the color fields are confined in the tubevolumes when charges are in long distances; we want to stress that, each color charge does not only actas a unit source of color fields in all the tubes linked with itself, it will also influence the strength of colorfields in all the other tubes not linked with itself in the whole hadron, as shown in formula (16) in Sect.3.2;(2) we propose that, the flux tubes should exist between each two sources, because “a quark could notknow which one(s) it would choose to be teamed with” (introducing the topological mixing in Fig. 1-(c)is just a solution to this “problem”); in our picture, color charges are only responsible for producing colorfields, and, the confined space configuration and quantum numbers of a hadron should be fixed until (or,by) a projection onto the Hilbert space; so, the configurations of mesons, baryons and 4-quark hadronstates are shown in Fig. 2-(a), 2-(b) and 2-(c), respectively, e.g.: for baryons in Fig. 2-(b), the configurationof tubes is a “ △ -type” instead of the “Y-type” in Fig. 1-(b), and for 4-quark hadrons, the configurationof tubes is the one in Fig. 2-(c) instead of the one in Fig. 1-(c); we ignore the direction of flux tube at thefirst sight; the gray bags in Fig. 2-(a), (b) and (c) are the quark-bag configurations; all the multi-quarksystems could be confined, but not all of their lives would be long, see Sect. 3.5;(3) dynamics: in flux tube models reviewed above, hadrons decay via flux tube breaking due toquark pair creation, however, we should ask: where will a string break down (or, where is the mostsolid/firm/stable place of a sting), at the high energy density region or low energy density region (or, atthe weak force region or strong force region)? where is the high energy density region (or, where is thestrong force region), the region of quarks colletion or the region of quarks diffusion? in our model, we skipthe string picture; instead, in our picture, we define, a hadron can decay via emitting color-singlet quark4lusters (hadrons), which are generated by the recombination of color charges when a hadron turns to thebag configuration phase during the oscillation;(4) for color-singlet hadrons, the residual force (van der Waals-type force) will be suppressed to ashort-range one, due to the confinement or color screening;(5) we define that, the energy of a compact N -quark hadron has a leading part of E = ρ cN V N = ρ cN · nV , (15)where n is the number of tubes in the hadron, V is the equal constant volume value of all the tubes in ahadron, see (19) in Sect. 3.2, that means, the shape of the tubes can vary but its volume will not change;(6) in glueball states, it is also the gluonic color charges that act as sources of color fields, like in themulti-quark states, see Sect. 3.6; etc. N Like the Coulomb fields, for an effective classical field E ( ¯ M ) c induced by a source with color charge quantumnumber Q c , it is reasonable to measure the field energy density as ρ c ∼ | E ( ¯ M ) c | ∼ | Q c | ; (16)so, for an N -quark hadron state with definite flavor and color wavefunction hadron = ˆ O ir (cid:2) q c f q c f . . . q c N f N (cid:3) , (17)where the notation ˆ O ir denotes an operation to pick out one eigenstate in the irreducible representationof the direct product group of flavor-symmetry and color-symmetry, it is reasonable to measure the fieldenergy density as ρ cN ∼ | N · E ( ¯ M ) c | ∼ | N Q c | = N | Q c | = N ρ c , (18)where we take ρ c to denote the color field energy density stimulated by one “single physical quark”.To evaluate the “effective volume” of the color field configuration in a hadron, we have recalled thetube configuration [1] of a color field between two color sources, see Fig. 2, and it will be reasonable tomeasure the effective volume as V N ∼ C N V ∼ N ( N − V N ≫ −−−→ N V , N ≥ , (19)where V is taken to denote the volume of one single tube formed by 2 quarks, and C N ≡ N !( N − = N ( N − is the binomial coefficient. Here we should not confuse the color field with the electromagnetic field, forexample, the r.m.s electric charge radius of pion (about 0 . f m ) and proton (about 0 . f m ) [23] willgive a ratio of the electromagnetic volume 0 . / . ≃ / . / N -quark hadron will roughly have a dependenceon N in the increasement of N , as¯ M N ≡ ρ cN V N ∼ N · N ( N −
1) ¯ M (20) N ≫ −−−→ N ¯ M , N ≥ , ¯ M ≡ ρ c V . (21)Coincidentally, if we formally treat the Hamiltonian operator as ˆ H ∼ | E ih E | for an energy eigenstate | E i of a “single physical quark”, then, after mapping the state | E i to a classical field strength E , the energyof a “single physical quark” will be E = h E | ˆ H | E i ∼ h E | · | E i · h E | · | E i ∼ | E | , and the energy of an N -quark hadron will become E N ∼ N | E | .In combination with the current quark mass m , as shown in (14), the total mass of a compact N -quarkhadron will be M N ∼ ( m + m + . . . + m N ) + ¯ M N ∼ ( m + m + . . . + m N ) + N · N ( N − · ¯ M ; (22)5oreover, to include the information of Regge trajectory on the angular momentum L and the total spin J of a hadron, we modify the total mass above to be a hypothesis, as M JLN = ( m + m + . . . + m N ) + N · N ( N − · a JLN + b JLN , (23)where a JLN and b JLN are dimensionful constants for definite { N, J, L } configuration. The coefficient | c | in(14) has been absorbed into m i in (22,23). m i defined in a perturbation sense Now we will concentrate on the values of current quark masses m i in (23). What is m i ? From (6), m i isoriginally defined as the mass of free current quark, but, as said in the end of Sect. 2, the current quark isseemed as physical d.o.f in QCD in the perturbation sense, so m i can be also seemed as the pole mass ofthe current quark defined by the pole position in the full propagator in the perturbation sense in QCD [23].In the perturbation sense, there is a relation between the pole mass m f of current quark with flavor f andthe M S “running mass m f ( µ ) in the perturbation sense [23] [24], and m f ( µ ) is conventionally defined ata scale µ ≫ Λ χ , where Λ χ ∼ GeV is the non-perturbative scale of dynamical chiral symmetry breaking,for example, m s (2 GeV ) ∼ M eV , m c ( m c ) = 1270 M eV , m b ( m b ) = 4180 M eV .However, here the challenge is an inverse problem, that is, how can we determine the energy scale µ , orfurthermore, how can we determine m i ( µ ), a JLN ( µ ), b JLN ( µ ) and M JLN ( µ ) (since generally the variables m i , a JLN , b JLN and M JLN in (23) are all energy scale dependent)? Moreover, is it possible for µ ≃ Λ χ (located inthe non-perturbative region) rather than µ ≫ Λ χ (located in the perturbative region)?On the other hand, how to examine our “ N mass rule” hypothesis in (23)? One method is to checkthe reasonability of derived values of m i after accepting the “ N mass rule”, that is, the validity of the“ N mass rule” can be indirectly confirmed by the reasonability of derived values of m i . In detail, afterfixing the values of m u,d , a JLN and b JLN , then with Eq. (23), we will compute m s,c,b for hadrons, and we cancheck whether the m s,c,b values follow appropriate requirements or not. We will choose the values below(only for L = 0): m q = m u = m d = 10 M eV, ( q = u, d ) , (24) b J,L =0 N ≃ S ( S + 1) · C N M eV, ( S = J = 0 , , ,
32 ; N = 2 , , (25) a J =0 ,L =0 N =2 = a J = ,L =0 N =3 = a J =1 ,L =0 N =2 = a J = ,L =0 N =3 ≃ M eV ; (26)here Eq. (24) is motivated by the mass running effect, i.e., the value of m q in hadrons should be larger thanthe current d quark mass value 4 M eV at µ = 2 GeV [23]; Eq. (25) is motivated by the mass differencebetween π (140) and ρ (775) and the mass difference between p (940) and ∆(1232); Eq. (26) is derived byinserting (24,25) and the mass of π (140) into Eq. (23); and we should note that, here we only define valuesfor the L = 0 case, so we can ignore the √ L type terms of Regge Trajectories. The results of m s,c,b arelisted in Table 1, 2 3 respectively.Table 1: Values of current quark mass m s computed with m u = m d = 0.flavor mass (MeV) m s /M eV prediction mass (MeV) m s /M eV prediction q ¯ q π (140) ρ, ω (775) q ¯ s K (494) 364 K ∗ (892) 152 s ¯ s η (550) ? mixing? φ (1020) 145 qqq p, n (940) ∆(1232) qqs Σ(1200) , Λ(1116) 293 ,
209 Σ ∗ (1385) 173 qss Ξ(1320) 211 Ξ ∗ (1530) 164 sss −− −− −− Ω(1672) 160In Fig. 3, the results of the current quark masses m s,c,b extracted in a non-perturbative approachvia our model (23) are plotted as discrete points (rounded, triangled or squared, respectively), while theresults naively derived in the perturbative approach within the M S scheme [24] are plotted as curves. We6able 2: Values of current quark mass m c computed from Eq. (23) with m u = m d = 0, m s = m s (2 GeV ) ∼ M eV . Each hadron mass range (in the unit
M eV ) in the “prediction” columns denoted as “( , )”, iscorresponding to a m c range put by hand and denoted with a colon as “: ( , )”.flavor mass (MeV) m c /M eV prediction mass (MeV) m c /M eV prediction c ¯ q D (1870) 1740 D ∗ (2010) 1270 c ¯ s D s (1968) 1748 D ∗ s (2112) 1282 c ¯ c η c (1 S )(2980) 1430 J/ψ (3097) 1183 qqc Σ c (2455) , Λ c (2286) 1548 , ∗ c (2520) 1308 qsc Ξ c (2468), Ξ ′ c (2578) 1471 , ∗ c (2645) 1343 qcc Ξ cc (3621?) 1362? unknown Ξ ∗ cc (3621?) 1209? unknown ssc −− −− −− Ω c (2770) 1378 scc −− −− −− Ω cc (?) : (1270 , , ccc −− −− −− Ω ccc (?) : (1270 , , m b computed from Eq. (23) with m u = m d = 0, m s = m s (2 GeV ) ∼ M eV and m c = m c ( m c ) = 1270 M eV . Each hadron mass range (in the unit
M eV ) in the “prediction”columns denoted as “( , )”, is corresponding to a m b range put by hand and denoted with a colon as“: ( , )”.flavor mass (MeV) m b /M eV prediction mass (MeV) m b /M eV prediction b ¯ q B (5279) 5149 B ∗ (5325) 4585 b ¯ s B s (5366) 5146 B ∗ s (5413) 4583 b ¯ c B c (6275) 4885 B ∗ c (?) b ¯ b η b (1 S )(9300) 4590 Υ(1 S )(9460) 4365 qqb Σ b (5810) , Λ b (5620) 4903 , ∗ b (5830) 4618 qsb Ξ b (5797),Ξ ′ b (5935) 4800 , ∗ b (5955) 4653 qcb Ξ cb (?) : (4180 , , ∗ cb (?) : (4180 , , qbb Ξ bb (?) : (4180 , , ∗ bb (?) : (4180 , , ssb −− −− −− Ω b (6046) 4654 scb −− −− −− Ω cb (?) : (4180 , , sbb −− −− −− Ω bb (?) : (4180 , , ccb −− −− −− Ω ccb (?) : (4180 , , cbb −− −− −− Ω cbb (?) : (4180 , , bbb −− −− −− Ω bbb (?) : (4180 , , èèèèèèèè òòòòòòòòòòò òò òò ŸŸŸŸ Ÿ ŸŸŸŸŸŸ Ÿ ŸŸ Μ (cid:144) GeV123456 m f (cid:144) GeV
Figure 3: The solid, dashed, dotted lines are respectively for dependence of the running mass m f ofcurrent quark with flavor f = s, c, b on the energy scale µ , which are naively derived in the perturbativeapproach within the M S scheme [24]; the rounded, triangled, squared points are respectively for thecurrent quark masses of m s , m c , m b extracted in a non-perturbatively approach via our model (23), withthe corresponding µ values naively set to the mass of each corresponding hadron only just for convenience.For the m b line, we set α s ( m b ) = 0 .
223 [23] and N f = 5 in the range of m b ( m b ) < µ < GeV , with N f = 4 in 1 GeV < µ < m b ( m b ); for the m c line, we set N f = 4 in m c ( m c ) < µ < m b ( m b ), with N f = 3in 0 . GeV < µ < m c ( m c ); for the m s line, we set N f = 4 in m c ( m c ) < µ < GeV , with N f = 3 in0 . GeV < µ < m c ( m c ).should point out that, the energy scale µ of each discrete point plotted in Fig. 3 is just naively set tothe mass of each corresponding hadron for convenience, however, the typical momentum transfer betweentwo quarks in the hadrons should not be so large; it is shown that, at least, the range of of m s (or m c , m b ) represented by the discrete points can be qualitatively consistent with the curves. So, it would bereasonable to interpret the points in Fig. 3 as the corresponding results of running current quark massesin the non-perturbative region of the (full) QCD, that means, our model (23) is reasonable in some sense.Besides, we want to discuss a special invalid sector of our N model, that is, the singlet η (550) sector.Mesons can be embeded in the representations and of the SU (3) group of flavor symmetry, as : η ( ω ) = ( u ¯ u + d ¯ d − s ¯ s ) / √ , (27) : η ( ω ) = ( u ¯ u + d ¯ d + s ¯ s ) / √ , (28)while the physical states (mass eigenstates) are always the (non-ideal) mixing states, such as:the pseudoscalar ones (with the mixing angle θ P = − . ◦ ) η (550) = η cos θ P − η sin θ P ≃ . · u ¯ u + 0 . · d ¯ d − . · s ¯ s, (29) η ′ (958) = η sin θ P + η cos θ P ≃ . · u ¯ u + 0 . · d ¯ d + 0 . · s ¯ s, (30) ⇒ m η = 620 M eV, m η = 888 M eV, (31)and the vector ones (with the mixing angle θ V = 36 . ◦ ) φ (1020) = ω cos θ V − ω sin θ V ≃ − . · u ¯ u − . · d ¯ d − . · s ¯ s, (32) ω (782) = ω sin θ V + ω cos θ V ≃ . · u ¯ u + 0 . · d ¯ d − . · s ¯ s, (33) ⇒ m ω = 936 M eV, m ω = 866 M eV. (34)From (27) and (28), there would be m η > m η and m ω > m ω due to the larger proportion of s quarkin η ( ω ) than the one in η ( ω ), however, there exsit m η < m η . What does this imply? One of thereasonable possibility is that, there exist heavier partner particles with the same quantum numbers as ofthe singlet η ( q ¯ q ), e.g., the multi-quark states ( q ¯ qq ¯ q ) or the hybrid states ( q ¯ qg ¯ g ), which could enlarge the8ass of m η due to the mixing. So, the N -rule in our mass model (23) would not hold well for the lighterhadrons η ( ω ) and η ( ω ), neither for η (550)( φ (1020)) and η ′ (958)( ω (782)). So, in Table 1, we have notextracted the mass of current s quark for η (550). Besides, we treat ω as pure q ¯ q state due to the very fewproportion of s quark. In chiral perturbative theory (ChPT), the Goldstone bosons will become pseudo-Goldstone bosons withnonzero masses due to the chiral symmetry breaking Lagrangian terms from small nonzero current quarkmass m q ( m q ≪ Λ QCD ) [2], L m q = m q ¯ qq, q = u, d, s, (35) → L m χ = vT r ( m † χ Σ + m χ Σ † ) , m χ ≡ m q , (36)where v ≡ h Ω | ¯ q ( x ) q ( x ) | Ω i is the vacuum expectation value (VEV) of quark condensate, and Σ = exp[ i M /f π ] ∼ ¯ q jR ( x ) q kL ( x ) /v gives the local orientation of the quark condensate, with M a 3 × f π the decay constant of pions. For example, the pions will have a mass dependence as m π ∼ vf π m χ , (37)that is, if we identify m χ ≡ m q , then, m π is linear on the current quark mass m q and it will become zeroin the chiral limit. Thus, does that mean, m π is linearly dependent on the quark mass m q and m π is notdependent on the “volume energy” in (15) at all, or, our model is wrong? Solutions to this doubt can belisted below: (1) One choice is to modify our model. For a meson, it can be massless in the chiral limit, provided its volume is zero; this condition can besatisfied, because a massless particle will move in speed of light, so its volume will automatically be zero.For a baryon, it can automatically avoid the massless case even in the chiral limit, since there is alwaysnonzero spin term b JLN in our mass rule (23); that is consistent with the case in ChPT. That means, ourmodel can still hold in the chiral limit.Besides, like the relation in (37) which is constructed to describe the restoration of chiral symmetryfrom breaking phase, here we also need an assumption on the continuous transition from nonzero to zerofor the volume of pion, V π , or the common parameter V in our model (by recalling V π = V ), as V = V π ∼ m χ ≡ m q , (only for m χ ≪ Λ QCD ) . (38)However, we should stress that, m χ ≡ m q is introduced as just a parameter in (38) rather than the real massof quark, and, once the parameter m χ is fixed, the common volume parameter V is independent on theflavors in a hadron. Moreover, like what people have done in ChPT, that is, only for the m χ ≡ m q ≪ Λ QCD case, the chiral symmetry breaking effects can expressed in an expansion on m q (or m q Λ QCD ) in (36) or ina mass relation in (37), or in our volume relation in (38); otherwise, for large m q , the chiral symmetry isexplicitly broken and there could not perform an perturbative expansion on m q Λ QCD based on a symmetrytheory any more, or to say, there even could not exist a ChPT any more.Although the real quark masses are actually not zero and we can avoid the m π = 0 case in our model,nevertheless, the continuous transition from nonzero to zero for the masses of Goldstone bosons is veryimportant to avoid criticisms for a theory on approximate symmetry. (2) The other choice is to modify ChPT. We would ask, does the Goldstone theorem still hold even for composite particles generated from thequark condensate due to a strong interaction? Or, are the mesons really massless in the chiral limit? Isit possible that, mesons are always massive, no matter whether quarks are massive or not? Could wereinterpret the results in ChPT? For these questions, we are motivated by the three details below:(i) In (35) and (36), although they are both the explicitly chiral symmetry breaking terms, m χ in (36)is not necessarily and certainly to be m q in (35)! Otherwise, if m χ is the quark mass m q , one shouldanswer, what is the value of corresponding energy scale µ to define this m q ?9ii) Even in chiral symmetry reserved case, there can still generate meson mass terms, e.g., from thefour quark coupling terms, L q ∼ λv ¯ f L f L ¯ f R f R , λ ≡ m q . (39)In (39), if the dimensional parameter λ is defined to m q , i.e., λ ≡ m q , then the meson can get a mass m π ∼ vλf π ; so, is that just the underlying reason why the relation (37) holds so well?(iii) To investigate the property of a meson in ChPT, since mesons are generated from quark condensate,one should consider both the VEV v (including spontaneously vacuum symmetry breaking information)and mass parameter m q or m χ (including the explicit chiral symmetry breaking information) at the sametime rather than separately.Here we want to give a new interpretation to some results in the ChPT. For example, at the quarklevel, if we take the assumptions for v and f π as vf π ≡ h Ω | f π ¯ q ( x ) q ( x ) | Ω i ≡ h ϕ ( x ) f π i ∼ ̺ g + ̺ q ( m u + m d ) 1 f π , (40) f π m π ∼ f π h | ¯ qγ γ q | π i ∼ f π · Φ ¯ qq ( x = 0) · M ¯ qq →| i ∼ V , (41)then we can get the mass relation m π ∼ ( ̺ g + ̺ q ) V, (42)where the ̺ g V part can match with our hypothesis on effective energy ¯ M N = ρ cN V N for gluon fields in(20). Here in (40) we understand v ≡ h ϕ ( x ) i as the energy density of a higgs-type field ϕ ( x ) (with anextra factor f π as a normalization factor), which should include both the energy density of quark fields ̺ q and the energy density of gluon fields ̺ g (with the factor m u + m d ) as a normalization factor); and, in(41) we treat the pion transition matrix element M π →| i ≡ h | ¯ qγ γ q | π i being proportional to Φ ¯ qq ( x = 0)( i.e., the value of wavefunction Φ ¯ qq ( x ) at x = 0 point in the coordinate space), thus being inverse tothe characteristic size L (or the volume V = L ) of pion by imposing a Gaussian type wavefunction ina constituent quark model scheme. That will mean, the relation (37) in ChPT can be embedding in ourmodel (42) by introducing assumptions in (40,41). N -quark Hadrons Disappear? By inserting m u,d , b J,LN and a J,LN defined in (24,25,26) into Eq. (23), we can get the masses of a compact4-quark hadron with quantum numbers J = 0, L = 0 M J =0 ,L =0 N =4 ( qq ¯ q ¯ q ) ≃ · · (4 − · M eV = 2880
M eV, (43) M J =0 ,L =0 N =4 ( { qq ¯ q ¯ c, qq ¯ c ¯ c, qc ¯ c ¯ c, cc ¯ c ¯ c } ) ≃ M eV + 1270 · { , , , } M eV (44)= { , , , } M eV, (45) M J =0 ,L =0 N =4 ( { qq ¯ q ¯ b, qq ¯ b ¯ b, qb ¯ b ¯ b, bb ¯ b ¯ b } ) ≃ M eV + 4200 · { , , , } M eV (46)= { , , , } M eV, (47)and the masses of a compact 5-quark hadron with quantum numbers J = , L = 0 as M J = ,L =0 N =5 ( qqqq ¯ q ) ≃ · · (5 − · M eV = 7500
M eV, (48) M J = ,L =0 N =5 ( { qqqq ¯ c, qqqc ¯ c, qqcc ¯ c, qccc ¯ c, cccc ¯ c } ) ≃ M eV + 1270 · { , , , , } M eV (49)= { , , , , } M eV, (50) M J = ,L =0 N =5 ( { qqqq ¯ b, qqqb ¯ b, qqbb ¯ b, qbbb ¯ b, bbbb ¯ b } ) ≃ M eV + 4200 · { , , , , } M eV (51)= { , , , , } M eV ; (52)and we can see the masses are rather larger than the results in constituent quark models [4].10y comparing with the electro-weak decays of the constituent quarks and the q ¯ q annihilation decays inthe compact N -quark hadrons, because the mass of one compact 2 N -quark meson | q N ¯ q N i (or, 3 N -quarkbaryon | q N i ) will be much larger than N compact 2-quark mesons | q ¯ q i (or, N compact 3-quark baryons | qqq i ), the width of the strong decay processes | q N ¯ q N i → N | q ¯ q i (or, | q N i → N | qqq i ) will be very large.This can be easily understood by noting that, both the | q N ¯ q N i (or, | q N i ) and the N | q ¯ q i (or, N | qqq i ) arethe eigenstates of the full Hamiltonian and the decay probability is proportional to the energy differencebetween the higher energy level and the lower energy level in the quantum mechanics perturbative theroy.Therefore, on one hand, it is difficult to discover compact N -quark hadrons due to the too large decaywidths; on the other hand, it is difficult to produce compact N -quark hadrons on the colliders due tothe smaller phase space of motion for the larger N value; these are just the reasons why there are rarelydefinite signals of compact N -quark hadrons so far.Unlike the compact 2 N -quark mesons, all of which will be at last decay via the q ¯ q annihilations, if thereare other unknown mechanisms (e.g., in the high temperature, high density and high pressure environment)forbidding the decays of | q N i → N | qqq i , there will exist neutral stable compact 3 N -quark baryons (as allthe charged ones have decayed to the neutral ones via the weak interaction) deposited in the core of starsand the neutral stable ones will not be so easy to detected and discovered. With the so-called Cho-Duan-Ge decomposition [25], we can gauge independently decompose the 8 gluonsin SU (3) QCD to 2 color-neutral binding gluons (also called “neuron” or“neuton”) and 6 (or three complex)colored valence gluons (also called “chromon” or “coloron”), then the 6 chromons can condense to glueballs(also called the “chromoballs”). According to the mass spectrum hypothesis above, i.e., Eq. (23), M JLN = ( m + m + . . . + m N ) + N · N ( N − · a JLN + b JLN , (23)with the same b JLN and a JLN values in (25) and (26), we can fit a nonzero “current gluon” mass in thecondensate occurring case as m g = 120 M eV, (53)by treating the unidentified particle X (360) [23] [26] (with the quantum numbers I G ( J P C ) not identifiedyet) as a 2-gluon glueball state | g ¯ g i with the quantum numbers I G ( J P C ) = 0 + (0 ++ ), by recalling that thequantum numbers of gluon are I ( J P ) = 0(1 − ). Moreover, by setting the number of color charge of gluonsto be the same as quarks, the 2-gluon and 3-gluon glueball states would have the masses I G ( J P C ) = 0 + (0 ++ ) , M J =0 ,L =0 g ¯ g ≃ M eV, (54) I G ( J P C ) = 0 + (1 + − ) , M J =1 ,L =0 g ¯ g ≃ M eV, (55) I G ( J P C ) = 0 + (2 ++ ) , M J =2 ,L =0 g ¯ g ≃ M eV, (56)and I ( J P ) = 0(1 − ) , M J =1 ,L =0 ggg ≃ M eV, (57) I ( J P ) = 0(2 − ) , M J =2 ,L =0 ggg ≃ M eV, (58) I ( J P ) = 0(3 − ) , M J =3 ,L =0 ggg ≃ M eV. (59)Since the dynamically reversible transition processes gg ↔ q ¯ q are going on all the time, it is mostlikely that the 2-gluon glueball states | g ¯ g i will be mixed with the quarkonium states | q ¯ q i and they wouldnot be distinctly discovered. Nevertheless, the glueball states with total spin J = 2 , L = 0) might be more pure, and they would be more expectable to search. More glueballstates with other I G ( J P C ) quantum numbers (e.g., see Ref. [4] [25]) are also allowed in our model.
What is an N -quark hadron? For a hadron state with definite quantum numbers, we can write out theFock expansion as | hadron, P i = h | P ; p p i + h | P ; p p p i + .... + ..., (60)11here p i is momentum of the i -th constituent, such as constituent quarks and the colored valence gluonsbut not the free current quarks and the free gluons; and, the coefficients h i might be functions of manydifferent parameters, such as the energy scale µ . Generally, each Fock state is allowed, and we do notknow which one is the most dominant. For example, the parton picture can be seemed as the i = ∞ Fockstate; although the i = 2 Fock states are always seemed as the leading-order ones in a meson state, thereare no absolutely sufficient reasons to ignore the | ¯ qqg ¯ g i states but only for the simplicity. That means, ahadron state is “defined to be” an N -quark state in the Fock expansion language.Moreover, for each Fock state, besides of the quantum numbers (such as: the spins, the flavors, thecolors, etc), we do not know the dynamical details of the constituents, i.e., the masses and the interactions.That means, a constituent quark is indeed “defined to be” a physical quasi-particle d.o.f after we definethe hadron as an N -quark system, rather than directly deduced from the first principle of QCD.After we treat constituent quarks as physical quasi-particle d.o.f, i.e., the “single physical quark” states | M i = α † | Ω i , as said in (13) and the last two paragraphs in Sect. 2, we can get the masses of constituentquarks. If we just define the ordinary mesons as 2-quark states and the ordinary baryons as 3-quark states, | meson, P i = | P ; p p i , | baryon, P i = | P ; p p p i , (61)then the masses of constituent quarks would be about M meson and M baryon , respectively. Why are theconstituent quark model so successful? It is just because that the total mass of the constituent quarksis the dominant part of the hadron mass, while the residual effective potential energy can be seemed asperturbation. Following this approach, one can solve the wavefunction h | ¯ q ( x ) q ( y ) | hadron i after somesuitable potential model for the interaction of constituent quarks is constructed. In this paper we try to give a hypothesis on the mass spectrum of compact N -quark hadrons and glueballsin a classical field picture, which indicates that there would be a mass dependence on about N . We callour model “bag-tube oscillation model”, which can be seemed as a kind of combination of quark-bag modeland flux-tube model. The large decay widths due to large masses might be the reason why the compact N -quark hadrons still disappear so far. I am very grateful to Dr. Jia-Jun WU at University of Chinese Academy of Sciences (UCAS) for usefulcomments, and I am also very grateful to Prof. Xin-Heng GUO at Beijing Normal University and Dr.Xing-Hua WU at Yulin Normal University for guidance on field theories before.