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ISSN 1063-7788, Physics of Atomic Nuclei, 2006, Vol. 69, No. 2, pp. 328–331. c© Pleiades Publishing, Inc., 2006.Original Russian Text c© M.A. Durnev, 2006, published in Yadernaya Fizika, 2006, Vol. 69, No. 2, pp. 350–353.
ELEMENTARY PARTICLES AND FIELDSTheory
Molecular Structure of Exotic Mesons
M. A. Durnev*
St. Petersburg State University, Universitetskaya nab. 7–9, St. Petersburg, 199034 RussiaReceived November 23, 2004; in final form, May 16, 2005
Abstract—The 0−− exotic meson state is considered as the P-wave molecular state ρ′(1465) of η(550)mesons. The mass and decay width of the exotic 0−− meson are calculated on the basis of the N/Ddispersion method.
PACS numbers : 11.55.Fv, 12.39.Ki, 12.39.Mk, 12.40.YxDOI: 10.1134/S1063778806020177
1. INTRODUCTION
The hypothesis that exotic mesons have a molecu-lar structure is inspired by QCD. The possibility thatlight scalar mesons are q2q2 rather than ordinary qqstates was studied by Alford and Jaffe [1]. Disregard-ing quark loops and quark annihilation, they calcu-lated scattering amplitudes for a pseudoscalar mesonwithin lattice QCD, and the results of those calcu-lations showed evidence that a JPC = 0++ stablefour-quark bound state of nonexotic flavor quantumnumbers exists for rather heavy quarks.
The existence of various color-singlet combina-tions encumbers the investigation of mixtures of twoquarks with two antiquarks. The first combinationknown as a four-quark state features no color-singletsubcomponent. In the second combination, there aretwo color-singlet quark–antiquark subcomponents,either being a meson. This mixture of two mesons isknown as a molecule. In the limit of large numberNc of colors, there are no four-quark states withinQCD; only meson–meson molecules exist [2]. Themolecular picture becomes more precise if the dis-tance between the mesons in the molecule signifi-cantly exceeds their size. This is possible if the bind-ing energy of the two mesons is much lower thanthe quark–antiquark energy within each individualmeson (the latter is close to the meson mass if thequarks involved are light). A theoretical calculationof this low binding energy is complicated by the factthat the binding energy is the result of delicate com-pensations of mutual attractions and repulsions, eachcomponent taken individually corresponding to anenergy that substantially exceeds the binding energysince such components are controlled by the QCDscale ΛQCD, which is much greater than the binding
*E-mail: [email protected]
energy. In [1, 3, 4], the binding energy was calcu-lated within SU(2) lattice QCD by using the adia-batic approximation to derive the potential of inter-action between two light–heavy mesons. An anal-ysis of meson–meson systems shows that the po-tential in question is attractive at short and inter-mediate distances. The binding energy was also cal-culated on the basis of QCD sum rules [5], modelsinvolving the exchange of quark–antiquark pairs [6,7], the four-quark potential model [8, 9], the color-tube model [10], the relativistic Faddeev–Yakubovskymodel [11], and meson-exchange molecular mod-els [12, 13].
Following the hypothesis of the molecular struc-ture of exotic mesons, we assume in this study thatone can construct many meson–meson bound statesand their mixtures that are pure in the quantum num-bers JPC . The binding energy of molecular states islow; therefore, the sum of the masses of mesons thatform the molecule being considered is expected tobe close to the mass of the exotic state. Accordingto [14], ρ
′(1465) (1−−) η(550) (0−+) of total mass
mρ′ (1465) + mη(550) ≈ 2015 MeV is one of the ap-propriate combinations. In this molecule, the con-stituent mesons have a relative angular momentumequal to unity (P wave). Summation of the quantumnumbers of ρ
′(1465) and η(550) yields the following:
1−− + 0−+ = 1+−. Upon taking into account the 1−
P-wave state, we arrive at the set of states 0−−, 1−−,and 2−−, the 0−− state being the lowest exotic state.According to [14], the interaction between ρ
′(1465)
and η(550) is mediated by the exchange of the vectormeson ρ(770). The calculation within the relativisticFaddeev–Yakubovsky model yields M = 1970 MeVfor the mass of the 0−− state [15].
In this study, we calculate the mass and the decaywidth within the N/D dispersion method [16, 17],
328
MOLECULAR STRUCTURE OF EXOTIC MESONS 329
ρ
ηρ
′
P
-wave
ρ ρ
η
ηη
ρ
′
ρ
′
ρ
′
Fig. 1. Meson molecule.
η
ρ ′
0 + + +
η
ρ
′
0
...
η
ρ
′
0
η
ρ
′
0=
Fig. 2. Summation of diagrams within the N/D method.
which is used to find the poles of the amplitude thatcorrespond to bound states; in turn, the pole posi-tions determine the masses and widths of these boundstates.
2. RESULTS OF THE CALCULATION
We consider the exchange of the ρ meson. In theapproximation used here, this particle has a ratherhigh mass (like a massive gluon) [18]. Therefore,we consider the approximation where the interactionpoints in Fig. 1 merge into one interaction point, as isshown in Fig. 2.
The molecular amplitude is
A(s) =N(s)D(s)
, D(s) = 1 − g20B(s), (1)
where N(s) is the N function, which depends onenergy only slightly; g0 is a dimensionless constant;and
B(s) =
Λ∫
(m1+m2)2
ds′ ρ(s
′)
s′ − s. (2)
Here, m1 ≡ mρ′ ; m2 ≡ mη; Λ is the cutoff parameter;
and ρ(s) is the two-body P-wave phase space, whichis given by
ρ(s) =q2s
(m1 + m2)2
×√
(s − (m1 + m2)2)(s − (m1 − m2)2)s
,
where
q2s =
14s
(s − (m1 + m2)2)(s − (m1 − m2)2).
The poles of the amplitude correspond to bound statesand determine meson masses. We obtain the follow-ing equation for s:
1 − g20B(s) = 0. (3)
Upon explicitly calculating the function B andanalytically continuing it in s to the region (m1 +m2)2 < s < Λ, we arrive at an ultimate expression forB(s) (see Appendix).
By substituting the calculated function B(s)into (3), we find the roots s0 of Eq. (3). Equation (3)can be solved numerically, but it is more convenientto use a graphical method to find all roots s0 inthe complex-plane region under investigation. Theimaginary and the real part of the expression 1 −g20B(Res, Ims) correspond individually to a certain
M(g0) and Γ(g0) at fixed Λ
g0 M , GeV Γ, GeV
Λ = 10 GeV2
1.47 2.20 0.257
1.28 2.30 0.389
1.13 2.40 0.511
1.01 2.50 0.620
0.92 2.60 0.719
Λ = 5 GeV2
2.29 2.20 0.259
1.91 2.30 0.382
1.63 2.40 0.528
1.42 2.50 0.687
1.25 2.60 0.846
PHYSICS OF ATOMIC NUCLEI Vol. 69 No. 2 2006
330 DURNEV
Im(
g
02
B
(
s
) – 1) = 0, Re(
g
02
B
(
s
) – 1) = 0
–0.5
–1.0
–1.5
–2.04 5 6 7
Re
s
(
a
)
Im
s
Im(
g
02
B
( s ) – 1) = 0
0
–0.5
–1.0
–1.5
–2.04 5 6 7
Re
s
(
b
)
Im
s
Re(
g
02
B
(
s
) – 1) = 0
0
–1.549
–1.551
–1.553
6.248 6.250 6.252Re
s
(
d
)Im
s
–0.5
–1.0
–1.5
–2.04 5 6 7
Re
s
(
c
)
Im
s
0
Fig. 3. Graphical method.
three-dimensional surface above the s plane. A sec-tion of each such surface by the s plane (that is,z = 0) gives a certain set of planar curves in the z = 0plane. The intersection of these sets of planar curvesgives the set of roots s0. Upon plotting the imaginary(Fig. 3a) and the real (Fig. 3b) part calculated bya computer individually, one finds their intersection(Fig. 3c) and visually seeks the roots in the complex-plane region under consideration. The accuracy of thesearch may be improved by enlarging the scale of thegraph (Fig. 3d). Considering that s0 = M2 − iMΓ,we can obtain the mass M and the decay width Γ as
M2 = Res0, MΓ = −Ims0.
The calculations show that there is only one roots0 in the complex-plane region of interest; at a givenvalue of Λ, the mass and the width of the state corre-sponding to this root depend only on g0 (see table).
In our case, we consider low-energy physics andconstrain the energy region from above by
√Λ ≈
3 GeV.We constrain the region of the expected state to
the mass interval between 2.2 and 2.6 GeV. As aresult, we arrive at a state of mass M > (m1 + m2).
The P-wave state does not decay owing to the po-tential barrier [proportional to L(L + 1)/2] that pre-vents this.
3. CONCLUSIONAlthough the proposed method for calculating the
mass and width of the exotic 0−− state involves arather small number of parameters, it gives quite ac-curate results. The method for constructing the 0−−
molecular state makes it possible to determine correctmasses and widths. Based on mesons rather than onquarks, it is free from the confinement problem.
The approximate model considered here relies onthe exchange of a ρ meson, which is a rather heavyparticle, this simplifying the problem significantlywithout loss of physical meaning. Therefore, we usedthe approximation where the interaction points mergeinto one point. In this way, we take into accountonly the short-range part of the interaction. At lowvalues of the cutoff parameter (Λ ≤ 10 GeV2), thereis a stable solution; that is, the mass and the widthdepend only slightly on Λ (see table). This is nottrue at high values of Λ, however, possibly becauseof our approximation, which neglects the long-rangepart of the interaction. Therefore, our result is onlyqualitative in that case.
ACKNOWLEDGMENTSI am grateful to S.M. Gerasyuta for stimulating
discussions and for the formulation of an interestingproblem.
PHYSICS OF ATOMIC NUCLEI Vol. 69 No. 2 2006
MOLECULAR STRUCTURE OF EXOTIC MESONS 331
APPENDIX
B(s) =(
αs
(m1 + m2)2+ β +
1sδ
)
×√
(s − (m1 + m2)2)(s − (m1 − m2)2)s
×
⎛⎝ln
−√
s−(m1+m2)2
s−(m1−m2)2 +√
Λ−(m1+m2)2
Λ−(m1−m2)2√s−(m1+m2)2
s−(m1−m2)2+
√Λ−(m1+m2)2
Λ−(m1−m2)2
+ iπ
⎞⎠
+ α
√(Λ − (m1 + m2)2)(Λ − (m1 − m2)2)
(m1 + m2)2
+δ
s
√(Λ − (m1 + m2)2)(Λ − (m1 − m2)2)
Λ
+(
β + αs − m2
1 − m22
(m1 + m2)2+
1sδ
)
× ln1 +
√Λ−(m1+m2)2
Λ−(m1−m2)2
1 −√
Λ−(m1+m2)2
Λ−(m1−m2)2
+(−m2
1 − m22
sβ +
(m2
1 + m22
m21 − m2
2
− m21 − m2
2
s
)δ
s
)
× ln1 + m1−m2
m1+m2
√Λ−(m1+m2)2
Λ−(m1−m2)2
1 − m1−m2m1+m2
√Λ−(m1+m2)2
Λ−(m1−m2)2
,
α, β, and δ are coefficients that are different for dif-ferent states; that is, they are state-dependent. In thecase considered here (JPC = 0−−), we have
α =14, β = − (m2
1 + m22)
2(m1 + m2)2, δ =
(m1 − m2)2
4.
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Translated by M. Kobrinsky
PHYSICS OF ATOMIC NUCLEI Vol. 69 No. 2 2006