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夸克在Bethe–Salpeter方程中的Y（3940）质量.pdf
Mass of Y(3940) in Bethe–Salpeter equation for quarks
Abstract
1 Introduction
2 Revised general form of the BS wave functions
3 The extended Bethe–Salpeter equation
4 Form factors of heavy vector mesons
5 Numerical result
6 Conclusion
References
Eur. Phys. J. C (2015) 75:98
DOI 10.1140/epjc/s10052-015-3315-y
Regular Article - Theoretical Physics
Mass of Y(3940) in Bethe–Salpeter equation for quarks
Xiaozhao Chen1,a, Xiaofu Lü2,3,4
1 Department of Foundational courses, Shandong University of Science and Technology, Taian 271019, China
2 Department of Physics, Sichuan University, Chengdu 610064, China
3 Institute of Theoretical Physics, The Chinese Academy of Sciences, Beijing 100080, China
4 CCAST (World Laboratory), P.O. Box 8730, Beijing 100080, China
Received: 2 December 2014 / Accepted: 10 February 2015 / Published online: 28 February 2015
© The Author(s) 2015. This article is published with open access at Springerlink.com
Abstract The general form of the Bethe–Salpeter wave
functions for the bound states composed of two vector fields
of arbitrary spin and definite parity is corrected. Using
the revised general formalism, we investigate the observed
Y(3940) state, which is considered as a molecule state con-
∗0 ¯D
∗0. Though the attractive potential between
sisting of D
∗0 including one light meson (σ , π, ω, ρ) exchange
D
is considered, we find that in our approach the contribution
from one-π exchange is equal to zero and consider SU(3)
symmetry breaking. The obtained mass of Y(3940) is con-
sistent with the experimental value.
∗0 and ¯D
1 Introduction
∗0 ¯D
The exotic state Y(3940) was discovered by the Belle collab-
oration [1] and then confirmed by the BABAR collaboration
[2]. The investigation of the structure of Y is of great signifi-
cance, while the conventional c¯c charmonium interpretation
for this state is disfavored [3]. Then possible alternative inter-
pretations have been proposed, such as hadronic molecule
and tetraquark states. Following the experimental results, it is
suggested in Ref. [4] that the Y(3940) is a hadronic molecule
∗0. However, in previous work [4], the numer-
state of D
ical result of the binding energy for the molecule state sensi-
tively depends on the value of typical cutoff in the effective
interaction potential between two heavy vector mesons, and
these two heavy mesons are considered as pointlike objects.
Furthermore, the spin–parity quantum numbers J P of the
Y(3940) are not unambiguously determined in experiment,
except for C = +. The molecule state hypothesis implies
, but Ref.
that the quantum numbers of Y(3940) are 0
[4] cannot deduce the definite quantum numbers in theory.
Though the general form of the Bethe–Salpeter (BS) wave
functions for the bound states consisting of two vector fields
+
+
or 2
a e-mail: chen.xzhao.hn@gmail.com
∗0 ¯D
of arbitrary spin and definite parity has been given in Ref.
[5], we find that the derivation of this formalism has a seri-
ous defect. In this work, the general formalism is firstly cor-
rected. Then we assume that the Y(3940) state is a molecule
∗0 and
state composed of two heavy vector mesons D
the revised general formalism is applied to the investigation
of this two-body system. To construct the interaction kernel
between two heavy mesons derived from one light meson (σ ,
π, ω, ρ) exchange, we consider that the heavy meson is not
a pointlike particle but a bound state composed of u-quark
and c-quark and then investigate the light meson interaction
with the u-quark in the heavy meson. Through the form fac-
tor we can obtain the light meson interaction with heavy
meson and the potential between two heavy mesons with-
out an extra parameter [6,7]. Obviously, this potential in our
approach contains more inspiration of quantum chromody-
namics (QCD). Finally, numerically solving the relativistic
Schrödinger-like equation with this potential, we can obtain
the mass of the molecule state and then deduce the definite
quantum numbers of the Y(3940) system.
In this work, one-π exchange is considered in the interac-
tion kernel between two heavy mesons. When investigating
this pseudoscalar meson interaction with the u-quark in the
heavy meson, we find that the coupling LI = igπ ¯uγ5uπ
should have no contribution to this interaction and represent it
as the derivative coupling Lagrangian LI = i fπ ¯uγμγ5u∂μπ.
In this approach we find that one-π exchange has no contri-
bution to the potential between two heavy vector mesons.
Besides, it should be noted that the flavor-SU(3) singlet and
octet states of vector mesons mix to form the physical ω and φ
mesons, so the exchange mesons between two heavy mesons
should not be the physical mesons but rather the singlet and
octet states. Then in the interaction kernel between two heavy
mesons SU(3) symmetry breaking should be considered.
This paper has the following structure. In Sect. 2 the
revised general form of BS wave functions for the bound
states composed of two vector fields with arbitrary spin and
123
98 Page 2 of 11
Eur. Phys. J. C (2015) 75 :98
+
∗0 ¯D
∗0 with J P = 0
definite parity is given. In Sect. 3 we show the BS wave func-
and
tions for the molecule states of D
+
. After constructing the interaction kernel between two
2
heavy vector mesons, we obtain the Schrödinger type equa-
tions in instantaneous approximation. In Sect. 4 we show how
. Then
to calculate the form factors of the heavy meson D
the interaction potential and the mass of Y are calculated.
Sections 5 and 6 give our numerical result and conclusion.
∗
2 Revised general form of the BS wave functions
If a bound state of spin j and parity ηP is composed of two
vector fields with masses M1 and M2, respectively, its BS
wave function is a 4 × 4 matrix
χ j
P(λτ )
(x
1
2
, x
) = 0|T Aλ(x
1
)Aτ (x
)|P
, j,
2
χ j
P(λτ )
which can be written as
) = ei P
·X
+ η2x
1
(x
, x
= η1x
1
2
),
χ j
P(λτ )
(x
− x
= x
2 and η1,2 are two pos-
where X
itive quantities such that η1,2 = M1,2/(M1 + M2). Then one
has the BS wave function in the momentum representation
2, x
1
χ j
λτ (P
, p
) =
−i p
·x
e
d4x
χ j
P(λτ )
(x
),
(3)
is the momentum of the bound state, p
= η2 p
where P
ative momentum of two vector fields and we have P
+ p
p
2
1
two vector fields, respectively. This is shown in Fig. 1.
is the rel-
=
2 are the momenta of
The polarization tensor ημ1μ2···μ j describing the spin of
1 and p
− η1 p
2; p
, p
1
the bound state can be separated,
χ j
λτ (P
, p
) = ημ1μ2···μ j
χμ1μ2···μ j λτ (P
, p
),
(4)
and the polarization tensor is totally symmetric, transverse,
and traceless:
ημ1μ2··· = ημ2μ1···, P
μ1
ημ1μ2··· = 0, ημ1
μ1μ2··· = 0.
(5)
(1)
(2)
Because of Eq. (5), χμ1···μ j λτ is totally symmetric with
respect to the indices μ1, . . . , μ j . It is necessary to note that
the polarization tensor ημ1μ2···μ j of the vector–vector bound
state should contain all contributions from the spins of two
vector fields and then χμ1···μ j λτ should be independent of the
polarization vectors of two vector fields. Therefore, from the
BS wave function (1) and Lorentz covariance, we have
λ) f2
λ) f6
λ) f7
λ) f8
λ) f9
[gλτ f1 + (P
τ + P
··· p
χμ1···μ j λτ = p
λ p
τ p
μ j
μ1
τ f4 + p
τ f5]
λ) f3 + P
+ (P
τ − P
λ p
λ p
τ p
λ P
··· p
··· p
τ + p
+ ( p
{μ2
{μ2
μ j gμ1}τ p
μ j gμ1}λ p
··· p
+ ( p
··· p
τ − p
{μ2
{μ2
μ j gμ1}τ p
μ j gμ1}λ p
··· p
τ + p
··· p
+ ( p
{μ2
{μ2
μ j gμ1}τ P
μ j gμ1}λ P
··· p
+ ( p
τ − p
··· p
{μ2
{μ2
μ j gμ1}λ P
μ j gμ1}τ P
+ p
··· p
λτ ξ ζ p
ξ P
ζ f10
μ j
μ1
··· p
+ p
{μ2
μ1}λτ ξ p
ξ f11
μ j
··· p
+ p
{μ2
μ1}λτ ξ P
ξ f12
μ j
+ ( p
··· p
{μ2
μ1}λξ ζ p
ξ P
ζ p
μ j
··· p
+ p
{μ2
μ1}τ ξ ζ p
λ) f13
ζ p
ξ P
μ j
··· p
+ ( p
{μ2
μ1}λξ ζ p
ζ p
ξ P
μ j
− p
··· p
{μ2
μ1}τ ξ ζ p
ξ P
ζ p
λ) f14
μ j
··· p
+ ( p
{μ2
μ1}λξ ζ p
ζ P
ξ P
μ j
τ
+ p
··· p
{μ2
μ1}τ ξ ζ p
λ) f15
ζ P
ξ P
μ j
+ ( p
··· p
{μ2
μ1}λξ ζ p
ξ P
ζ P
τ
μ j
··· p
− p
{μ2
μ1}τ ξ ζ p
ξ P
ζ P
λ) f16,
μ j
τ
τ
(6)
where{μ1, . . . , μ j} represents symmetrization of the indices
·
μ1, . . . , μ j . There are only 16 scalar functions fi (P
2)(i = 1, . . . , 16) in (6). In Ref. [5] the derivation of
, p
p
Eqs. (6, 7) has some errors, and they have been revised as (6)
in this paper. From the massive vector field commutators for
arbitrary times x
[Aλ(x
[Aλ(x
10 and x
)] = 0 for M1 = M2,
δλτ − ∂λ∂τ
)] = i
M 2
1
for M1 = M2,
(x
), Aτ (x
), Aτ (x
− x
1
1
2
2
20
1
2
)
Fig. 1 Bethe–Salpeter wave function for the bound state composed of
two vector fields
123
where the right side of second equation is a c-number func-
tion; we may write
0|T Aλ(x
1
)Aτ (x
2
)|P
, j = 0|T Aτ (x
2
)Aλ(x
)|P
, j.
1
(7)
Eur. Phys. J. C (2015) 75 :98
So we see that in the momentum representation the BS wave
function of two equal or different vector fields is invariant
under the substitutions p
,− p
→ p
).
2 and p
) = χ j
1, i.e.,
→ p
, p
χ j
λτ (P
τ λ(P
(8)
1
2
This invariance is similar to crossing symmetry, which
implies that the scalar functions in Eq. (6) have the following
properties: for j = 2n, n = 0, 1, 2, 3 . . .,
· p
, p
fi (P
2) = + fi (−P
· p
, p
2)
i = 1, 3, 4, 5, 6, 9, 10, 12, 14, 15,
· p
, p
2) = − fi (−P
· p
, p
2)
fi (P
i = 2, 7, 8, 11, 13, 16,
and, for j = 2n + 1, n = 0, 1, 2, 3 . . .,
2)
2) = + fi (−P
· p
· p
, p
, p
fi (P
i = 2, 7, 8, 11, 13, 16,
· p
, p
fi (P
2) = − fi (−P
· p
, p
2)
i = 1, 3, 4, 5, 6, 9, 10, 12, 14, 15.
(9a)
(9b)
(9c)
(9d)
η2 f1+(η2 P
· p
− p
2)( f2+ f3)+(η2 P
2 − P
· p
) f4=0,
(11a)
Page 3 of 11 98
and, for j = 0,
)( f2 + f3)
· p
2 + P
f1 + (η1 P
+(η1 P
2) f5 + j! f6 + j! f7 = 0,
· p
+ p
+ p
2)( f2 − f3)
η1 f1 + (η1 P
· p
· p
) f4 + j! f8 + j! f9 = 0,
2 + P
+(η1 P
2)( f2 + f3)
− p
· p
η2 f1 + (η2 P
) f4 − j! f8 + j! f9 = 0,
· p
2 − P
+(η2 P
+ p
2)( f6 − f7)
· p
(η1 P
+(η1 P
)( f8 − f9) = 0,
· p
2 + P
2)( f6 + f7)
− p
· p
(η2 P
)( f8 + f9) = 0,
2 − P
· p
+(η2 P
· p
η1 f11 − f12 + (η1 P
2)( f13 − f14)
+(η1 P
)( f15 − f16) = 0,
· p
2 + P
· p
−η2 f11 − f12 + (η2 P
2)( f13
+ f14) + (η2 P
2 − P
· p
For the sake of simplicity, we introduce φ1,2 and ψ1 to replace
f4,5 and f10, respectively, for j = 0,
f4 = −η1η2φ1 + [η1η2(η1η2 P
2 − η1 P
· p
2) − η2
2 − 2η2 P
· p
(η2
2 P
+ p
· p
2)]φ2,
2 − η1 P
· p
+ η2 P
+ p
· p
2)
· p
+ p
2)]φ2,
− p
)( f15 + f16) = 0.
+ η2 P
· p
− η2
(η2
1 P
f5 = φ1 − [(η1η2 P
2 + 2η1 P
2 − 2η2 P
− p
2 + 2η1 P
− p
+ p
+ p
· p
(11b)
2)
2)
1
2
+ (η2
1 P
+ (η2
2 P
f10 = ψ1,
2
1
+ p
2)
· p
+ η2 P
· p
− p
2+2η1 P
− η2
(η2
1 P
− p
2)φ3 − η2(η1 P
and introduce φ1,2,3,4 and ψ1,2,3,4,5 to replace f4,5,6,7 and
f10,13,14,15,16, respectively, for j = 0,
f4 = −η1η2φ1+[η1η2(η1η2 P
2 − η1 P
· p
2) − η2
2 − 2η2 P
· p
(η2
2 P
2)]φ2+η1(η2 P
+ p
· p
· p
+ p
2)φ4,
f5 = φ1 − [(η1η2 P
· p
· p
2 − η1 P
− p
+η2 P
2)
· p
2 − 2η2 P
2)+(η2
2+2η1 P
· p
+ p
+ (η2
1 P
2 P
)φ3 − (η1 P
+ p
2)]φ2 − (η2 P
· p
2 − P
· p
2+ P
· p
2 − P
+ p
· p
2+2η1 P
f6 = [(η2
)φ3
2)(η2 P
1 P
)φ4]/(2 j!),
· p
2+ P
+ (η2
2 − 2η2 P
· p
+ p
2)(η1 P
2 P
f7 = [(η2
2+2η1 P
· p
+ p
2 − P
· p
2)(η2 P
)φ3
1 P
)φ4]/(2 j!),
2+ P
· p
− (η2
· p
+ p
2 − 2η2 P
2)(η1 P
2 P
f16 = ψ5,
f14 = ψ3,
f10 = ψ1,
f13 = ψ2,
f15 = ψ4,
)φ4,
123
Then we will reduce the general form without any assumption
and approximation.
For the interacting massive vector field Aμ(x), the true
A Aμ = −2g jμ,
equation of motion is
∂νfνμ − M 2
where the field strength tensor fνμ is antisymmetric, MA is
the vector field mass, and g is the coupling constant. Because
jμ is a conserved current, one can obtain that Aμ with inter-
actions should satisfy the subsidiary condition ∂μ Aμ(x) = 0
[8]. Using this subsidiary condition for the massive vector
field and the equal-time commutation relation, we get
∂1λT Aλ(x
1
)Aτ (x
2
) = ∂2τ T Aλ(x
1
)Aτ (x
) = 0.
2
1
2
The proof has been given by Ref. [5]. The BS wave function
in Eq. (1) obeys this relation:
, j
∂1λ0|T Aλ(x
)Aτ (x
, j = 0,
)|P
= ∂2τ0|T Aλ(x
2
and in the momentum representation
) = 0.
1λχ j
)|P
)Aτ (x
) = p
2τ χ j
, p
, p
λτ (P
λτ (P
(10)
1
p
Substituting Eqs. (4) and (6) into (10), we obtain a set of
independent equations, for j = 0,
f1+(η1 P
η1 f1+(η1 P
· p
2+ P
)( f2+ f3)+(η1 P
2)( f2 − f3)+(η1 P
+ p
· p
· p
2+ P
+ p
· p
2) f5 = 0,
) f4 = 0,
98 Page 4 of 11
Eur. Phys. J. C (2015) 75 :98
· p
· p
, p
, p
2) and ψi (P
2) are independent
where φi (P
scalar functions. Solving the set of equations (11), we see
that f1,2,3,8,9,11,12 are the functions of φi and ψi : for j = 0,
f1 = (η1η2 P
+ (η2
1 P
+ η2 P
· p
+ η2(η2
1 P
+ η2 P
· p
+ p
2)(η2
2 P
2)/2 − η1(η2
2 − 2η2 P
2 P
+ p
2)]φ2,
+η2 P
· p
− p
2)φ1
2−2η2 P
· p
2 − η1 P
· p
f2 = (η1 − η2)φ1/2 + [(η2 − η1)(η1η2 P
− p
2 + 2η1 P
2 − η1 P
2+2η1 P
+ p
· p
+ p
· p
2−η1 P
2)φ2/2,
· p
· p
− p
2)φ2,
· p
2)
· p
· p
+η2 P
+ p
2 − η1 P
2+2η1 P
− p
· p
2)φ1
· p
2−2η2 P
+ p
2)(η2
2 P
f2 = (η1 − η2)φ1/2+[(η2 − η1)(η1η2 P
· p
2 − η1 P
+ p
· p
2 − 2η2 P
2)
2)φ2,
f3 = −φ1/2−(η1η2 P
and, for j = 0,
f1 = (η1η2 P
+ (η2
1 P
− p
+ η2 P
· p
+ η2(η2
2+2η1 P
1 P
+ [(η2 − η1)(η1η2 P
− η1(η2
2 − 2η2 P
2 P
+ [(η2 − η1)(η1η2 P
+ η2(η2
2+2η1 P
1 P
f3 = −φ1/2 − (η1η2 P
2)/2 − η1(η2
2 P
2)]φ2
+ p
· p
2 − η1 P
· p
2)]φ3
· p
+ p
· p
2 − η1 P
· p
+ p
2)]φ4,
· p
2 − η1 P
+η2 P
+η2 P
· p
− p
2)/2
· p
− p
2)/2
+η2 P
· p
− p
2)(φ2+φ3+φ4)/2,
2+2η1 P
f8 = [−(η2
· p
1 P
· p
2 − 2η2 P
− (η2
2 P
· p
f9 = [−(η2
2+2η1 P
1 P
· p
+ (η2
2 − 2η2 P
2 P
− η1 P
f11 = (η2 P
· p
· p
2 − η1 P
+ (η2 P
2 − 2P
· p
f12 = (2η1η2 P
+η2 p
2+η2 P
+ (2η1η2 P
+ p
+ p
+ p
+ p
− 2 p
· p
2 − η1 p
· p
− η1 P
· p
2)(η2 P
· p
2)(η1 P
· p
2)(η2 P
· p
2)(η1 P
2)ψ2+(P
)ψ4+ P
2)ψ2 − p
· p
− p
+ p
− p
+ p
· p
2ψ5,
2ψ3
)ψ4 − (P
2)φ3
2)φ4]/(2 j!),
2)φ3
2)φ4]/(2 j!),
)ψ3
· p
)ψ5.
Then the BS wave function of the bound state becomes
χ j=0
χ j=0
λτ φ1 + T 2
[ p
μ1
, p
, p
λτ φ2)
μ j
(P
(P
λτ
λτ
) = T 1
) = ημ1···μ j
+ T 3
+ T 5
λτ φ3 + T 4
λτ ψ2 + T 6
ξ P
λτ φ2 + λτ ξ ζ p
ζ ψ1,
λτ φ1 + T 2
··· p
(T 1
λτ φ4 + p
··· p
μ1
λτ ψ4 + T 8
λτ ψ3 + T 7
μ j
ξ P
ζ ψ1
(13)
λτ ξ ζ p
λτ ψ5],
(12)
where
λτ = (η1η2 P
T 1
− (η1η2 P
2 − η1 P
λ P
− p
· p
+ η2 P
· p
τ − η1 p
τ + η2 P
τ − p
λ P
λ p
2)gλτ
λ p
τ ),
123
+ p
2)gλτ
λτ = (η2
T 2
1 P
2+2η1 P
2−2η2 P
· p
+ p
· p
2)(η2
2 P
+ (η1η2 P
2 − η1 P
· p
+ η2 P
· p
− p
2)
τ + η2 p
τ − η1 P
τ − p
× (η1η2 P
λ P
λ p
λ p
λ P
τ )
− (η2
· p
2 − 2η2 P
+ p
τ + η1 P
2)(η2
2 P
λ p
1 P
λ P
+ η1 p
τ + p
λ P
λ p
τ )
+ p
· p
2 + 2η1 P
− (η2
2)(η2
1 P
2 P
τ + p
− η2 P
τ − η2 p
λ p
λ p
λ P
τ ),
λ P
τ
τ
2)
+ p
· p
2 + 2η1 P
··· p
λτ = 1
{μ2
μ j gμ1}λ(η2
j! p
T 3
1 P
× [(η2
+ p
2 − 2η2 P
· p
+ p
2)(η1 P
)τ
2 P
− p
· p
2−η1 P
· p
+η2 P
− (η1η2 P
2)(η2 P
− p
+ p
· p
2 − 2η2 P
[(η2
··· p
2)
2 P
μ j
μ1
× (η2
τ + η1 P
τ + η1 p
τ + p
λ P
λ p
λ p
1 P
λ P
τ )
− (η1η2 P
+ η2 P
2 − η1 P
· p
· p
− p
2)
τ − p
τ − η1 P
τ + η2 p
τ )],
× (η1η2 P
λ p
λ P
λ P
λ p
− p
)τ]
2)
{μ2
+ p
2)
· p
− p
λτ = (η2 P
T 5
λτ = 1
··· p
2 − 2η2 P
μ j gμ1}τ (η2
j! p
T 4
2 P
× [(η1η2 P
2 − η1 P
· p
· p
+ η2 P
+ p
× (η1 P
)λ
− p
)λ]
· p
2 + 2η1 P
− (η2
+ p
2)(η2 P
1 P
− p
+ p
· p
2 + 2η1 P
[(η2
··· p
2)
1 P
μ j
μ1
× (η2
τ + p
τ − η2 P
τ − η2 p
λ P
2 P
λ P
λ p
λ p
τ )
− (η1η2 P
− p
· p
+ η2 P
· p
2 − η1 P
2)
τ + η2 p
τ − η1 P
τ − p
τ )],
× (η1η2 P
λ p
λ P
λ p
λ P
· p
· p
− 2 p
··· p
− η1 P
{μ2
2) p
μ j
+η2 p
· p
+ (2η1η2 P
··· p
2−η1 p
{μ2
2) p
μ j
+ p
··· p
τ + p
{μ2
{μ2
μ1}λξ ζ p
ζ p
ξ P
μ j
··· p
μ1}τ ξ ζ p
ξ P
λ,
μ j
· p
{μ2
) p
··· p
{μ2
2 p
··· p
{μ2
μ j
··· p
{μ2
μ j
2 − η1 P
λτ = (η2 P
T 7
··· p
μ j
μ1}λξ ζ p
μ1}τ ξ ζ p
··· p
· p
2 − 2P
{μ2
) p
μ1}λτ ξ p
μ j
· p
− η1 P
· p
2 + η2 P
) p
{μ2
··· p
ξ + p
{μ2
μ1}λξ ζ p
ξ P
μ j
μ1}τ ξ ζ p
ξ P
ζ P
λ,
μ j
μ1}λτ ξ P
ξ P
ζ p
ξ P
ζ p
λτ = (P
T 6
− p
+ p
− p
+ (2η1η2 P
··· p
μ j
+ p
{μ2
μ1}λτ ξ P
··· p
μ1}λτ ξ p
λ,
ζ P
τ
ζ p
μ j
τ
ξ
ξ
ξ
μ1}λτ ξ p
μ1}λτ ξ P
ξ
ξ
Eur. Phys. J. C (2015) 75 :98
ξ
λτ = P
··· p
2 p
{μ2
μ1}λτ ξ p
T 8
μ j
− (P
· p
··· p
{μ2
) p
μ j
··· p
+ p
{μ2
μ1}λξ ζ p
μ j
− p
··· p
{μ2
μ1}τ ξ ζ p
μ j
μ1}λτ ξ P
ξ P
ξ P
ζ P
τ
ζ P
λ.
ξ
This derivation makes use of the fact that P
, and
ημ1···μ j are linearly independent. In Ref. [5] Eqs. (18–19)
are wrong, they are revised as (12) and (13) in this paper.
, p
i
Now, under space reflection
→ x
one has
= −x
= x
i
0
0
i
x
x
,
,
PAλ(x
Pλξ =
),
−1 = Pλξ Aξ (x
)P
⎛
−1
⎜⎜⎝
0
0
0
−1
0
0
0
⎞
⎟⎟⎠ ,
0
0
−1
0
0
0
0
1
, j,
, j = ηP|P
= (−P
, P
) and P
0
P|0 = |0, P|P
= (−x, x
with x
(14)
).
0
We obtain the properties of the BS wave function under
space reflection from Eqs. (1) and (14):
χ j
P(λτ )
(x
1
2
, x
) = ηPPλξ Pτ ζ χ j
P(ξ ζ )
1
(x
2
)
, x
and in the momentum representation
, p
) = ηPPλξ Pτ ζ χ j
χ j
λτ (P
, p
ξ ζ (P
).
(15)
(16)
From (4), (6), (12), (13), and (16), it is easy to derive, for
ηP = (−1) j ,
χ j=0
, p
χ j=0
, p
(P
(P
λτ
λτ
λτ φ1 + T 2
[ p
μ1
) = T 1
) = ημ1···μ j
+ T 3
λτ φ3 + T 4
λτ φ2,
··· p
μ j
λτ φ4],
λτ φ1 + T 2
(T 1
λτ φ2)
and, for ηP = (−1) j+1,
χ j=0
) = λτ ξ ζ p
ξ P
χ j=0
) = ημ1···μ j
+ T 5
, p
, p
(P
(P
( p
λτ
λτ
λτ ψ2 + T 6
ζ ψ1,
μ1
μ j
··· p
λτ ψ3 + T 7
λτ ξ ζ p
ζ ψ1
λτ ψ4 + T 8
ξ P
λτ ψ5).
(17)
(18)
(19)
(20)
The general form of the BS wave functions for the bound
states composed of two massive vector fields of arbitrary
spin and definite parity is obtained. No matter how high the
spin of the bound state is, its BS wave function should satisfy
Eqs. (17), (18), (19) or (20). From (17) and (18), we conclude
that the BS wave function of a bound state composed of two
Page 5 of 11 98
+
massive vector fields with spin j = 0 and parity (−1) j has
only four independent components and that of a bound state
with J P = 0
has only two independent components. From
(19) and (20), we conclude that the BS wave function of
a bound state with spin j and parity (−1) j+1 has only five
independent components except for j = 0 and one for j = 0.
Up to now, all the above analyses are model independent.
In the next section we will apply the general formalism to
investigate molecule states composed of two vector mesons.
3 The extended Bethe–Salpeter equation
+
or 2
, p
∗0 and ¯D
Assuming that the Y(3940) is a S-wave molecule state con-
∗0, one can
sisting of two heavy vector mesons D
have J P = 0
+
for this system. [4] From Eqs. (17)
and (18), we can obtain the BS wave function describing this
bound state, for J P = 0
+
,
λτF1 + T 2
) = T 1
+
λτF2,
χ 0
λτ (P
or, for J P = 2
+
,
+
χ 2
λτ (P
+T 3
The BS wave function of this bound state satisfies the equa-
tion
, p
λτG3 + T 4
) = ημ1μ2
λτG4].
λτG1 + T 2
(T 1
λτG2)
μ1 p
[ p
μ2
(21)
(22)
χλτ (P
=
)
, p
id4q
(2π )4
F λα( p
1
)Vαθ,βκ ( p
, q
; P
)χθ κ (P
, q
)Fβτ ( p
2
),
)
2
2
1
1
+M2
1λ p
1α
M2
1
1
1
+M2
)= (δλα+ p
(23)
where Vαθ,βκ is the interaction kernel, we have the propaga-
−i ,
tors for the spin 1 fields F λα( p
) = (δβτ + p
2β p
2τ
−i , and the bound state
Fβτ ( p
M2
= (0, 0, 0, i M) in the rest frame.
2
momentum is set as P
In Ref. [5], we have considered that the effective interaction
between two heavy mesons is derived from one light meson
(σ , ω, ρ) exchange and obtained the result that the molecule
∗0 lies above the threshold. In this work, one-π
state D
exchange is also considered and one-ω exchange is recon-
sidered, shown as in Fig. 2.
∗0 ¯D
2
2
)
p
p
1
2
∗
Now, we construct the kernel between two heavy vector
is
mesons from one-π exchange. The charmed meson D
composed of a heavy quark c and a light antiquark ¯u. Owing
to the large mass of the c-quark, the Lagrangian representing
⎛
⎞
the interaction of π-meson triplet with quarks should be
⎝ u
⎠ ,
⎛
√
2π+
√
⎝ π 0
2π− −π 0
0
0
LI =igπ
¯u
⎞
⎠
0
0
0
¯d
d
c
γ5
¯c
(24)
123
98 Page 6 of 11
Eur. Phys. J. C (2015) 75 :98
2
V ( p
=
)|J
2
)
2
+
ν |V (q
1
E2( p
2
)E2(q
¯h( p)(k2)νabc ε∗
a ( p
2
)εb (q
2
)( p
2
+q
2
)c ,
2
)
1
1
MH
10
20
p2 + M 2
= (q, iq
= (p, i p
= (−q, iq
MH (E H ( p)+MH )
), p
), k = p
1
(31)
= (−p, i p
), q
),
where p
= q
− p
− q
20
2
10
2 is the momentum
q
; h(k2) and ¯h(k2) are scalar
of the light meson and k = p−q
2
2
1
, i ε·p
(ε·p)p
functions, the four-vector ε( p) = (ε+
)
is the polarization vector of heavy vector meson with momen-
tum p, E H ( p) =
H , (ε, 0) is the polarization vec-
tor in the heavy meson rest frame. In our approach, con-
sidering that the exchange meson is off the mass shell, we
calculate the meson–meson interaction when k2 = −m2 and
the heavy meson form factors h(k2) and ¯h(k2) are neces-
sarily required. In Sect. 4, we will show that for the form
−(k2) = 0. Then the effective interac-
factors h
tion from one-π exchange should have the form given by Eq.
(27). Cutting the external lines containing the normalizations
and polarization vectors ε∗
), we
obtain the interaction kernel from one-π exchange
V π
αθ,βκ ( p
−(k2) = ¯h
), ε∗
; P
), εκ (q
), εθ (q
, q
β ( p
α( p
)ckμ
1
1
2
2
) = f 2
π h( p)
(k2)μαθc( p
1
×
1
k2 + m2
+ q
×( p
2
2
¯h( p)
1
)ckν .
1
+ q
1
(k2)νβκc
π
(32)
Then we reconsider one-ω exchange between two heavy
vector mesons. In hadronic physics, the physical ω and φ
mesons are linear combinations of the SU(3) octet V8 =
√
√
6 and singlet V1 = (u ¯u + d ¯d + s¯s)/
(u ¯u + d ¯d − 2s¯s)/
3
(33)
as
φ = −V8cosθ + V1sinθ, ω = V8sinθ + V1cosθ,
where the mixing angle θ = 38.58
◦
was obtained by KLOE
[10]. Because of the SU(3) symmetry, we consider that
the exchange mesons are not the physical mesons and the
exchanged mesons should be the octet V8 and singlet V1
states. From Eq. (33), we can obtain the relations of the octet–
quark coupling constant g8 and the singlet–quark coupling
constant g1
gφ = −g8cosθ + g1sinθ, gω = g8sinθ + g1cosθ,
where gω and gφ are the corresponding meson–quark cou-
φ = 13.0 [11]. Since the
pling constants, g2
SU(3) is broken, the masses of the singlet V1 and octet V8
states are approximatively identified with the two physical
masses of ω and φ mesons, respectively. The interaction ker-
nel derived from one scalar meson exchange and one vector
meson exchange has been given in Ref. [5], and the kernel
from one light meson (π, σ , ρ, V1, and V8) exchange becomes
ω = 2.42 and g2
(34)
Fig. 2 The light meson exchange between two heavy mesons
¯c
¯u
or
¯d
LI = i fπ
⎛
√
γμγ5
2∂μπ+
√
⎝ ∂μπ 0
×
2∂μπ− −∂μπ 0
0
0
⎞
⎠ ,
⎞
⎠
⎛
⎝ u
d
c
0
0
0
(25)
where gπ and fπ are the π-meson–quark coupling constants,
gπ = 340
97 [9]. Because the contribution of Fig. 2 is from the
term i ¯uγ5π 0u or i ¯uγμγ5∂μπ 0u, we set fπ = gπ
and mu
− =
is u-quark mass. Then the effective quark current is J
μ = i ¯uγμγ5u and the S-matrix element between
i ¯uγ5u or J
+
two heavy mesons is
Vπ = g2
)|J
−|V (q
−|V (q
V ( p
πV ( p
)|J
),
)
2mu
1
1
2
1
2
k2 + m2
π
(26)
or
Vπ = f 2
π
1
k2 + m2
)kμ
μ |V (q
)|J
πV ( p
+
1
1
),
)|J
× kνV ( p
ν |V (q
+
(27)
2
2
where V|J
−|V and V|J
μ |V represent the vertices of the
+
pseudoscalar meson interaction with the heavy vector meson,
respectively. The matrix elements of these quark currents can
be expressed as
−|V (q
V ( p
× h
1
2E1(q
1
1
)εd (q
) =
(28)
1b
1
1
),
)
)
c
( p
2E1( p
ε∗
1
1
E2( p
2
2b ε∗
c ( p
2
2aq
)E2(q
2
)εd (q
2
)
2
1
2
1aq
) =
−|V (q
)|J
−(k2)abcd p
V ( p
)|J
2
× ¯h
−(k2)abcd p
)|J
V ( p
μ |V (q
+
1
=
1
E1( p
1
)E1(q
)
1
1
2
)
123
),
(29)
h( p)(k2)μabcε∗
a
( p
1
1
)( p
1
+ q
1
)c,
)εb(q
(30)
Eur. Phys. J. C (2015) 75 :98
Page 7 of 11 98
Vαθ,βκ ( p
= h( p)
×
1
)
; P
, q
(k2)μαθc( p
¯h( p)
f 2
k2 + m2
π
g2
σ
(k2)
1
π
+ h(s)
1
k2 + m2
σ
1
)ckμ
+ q
1
(k2)νβκc ( p
2
+ q
2
)ckν
¯h(s)
1
(k2)δαθ δβκ
ρ
φ
1
1
ω
g2
ρ
g2
1
g2
8
+
+
k2 + m2
(k2)¯h(v)
× {h(v)
(k2)¯h(v)
− h(v)
+ ( p
+ q
1
1
+ ( p
+ q
2
2
+ q
2κ + δακ p
1αδθβ p
+
k2 + m2
k2 + m2
+ q
) · ( p
+ q
(k2)( p
1
1
2
(k2)δαθ[q
+ q
2β ( p
)κ
1
1
(k2)¯h(v)
2κ] − h(v)
(k2)[q
1θ]δβκ + h(v)
1α( p
2
(k2)¯h(v)
(k2)[q
1αq
2
2β + δαβ p
2κ]},
1θ p
2
)β p
1θ q
)α p
2
2
1
2
1
)δαθ δβκ
)θ
+ q
2
2β δθ κ
(35)
+
where k = (k, 0).
Firstly, we assume that the Y(3940) is a molecule state with
J P = 0
. Substituting its BS wave function given by Eq.
(21) and the kernel (35) into the BS equation (23), we find that
the integral of one term on the right-hand side of (21) has a
contribution to the one of itself and the other term. Moreover,
the cross terms contain the factors of 1/M 2
2 , which
are small for the masses of the heavy mesons are large. It is
difficult to strictly solve the BS equation, and in this paper
we use a simple approach to solve it as follows. Ignoring the
cross terms, one can obtain two individual equations:
1 and 1/M 2
),
),
)
, q
, q
, p
, p
; P
; P
· p
2) =
· p
2) =
(36)
)
F 1
λτ (P
F 2
λτ (P
×F 1
F λα( p
· q
, q
· p
, p
λτ (P
2) = T 2
λτF2(P
)Vαθ,βκ ( p
1
2)Fβτ ( p
2
)Vαθ,βκ ( p
1
2)Fβτ ( p
2
· p
id4q
(2π )4
θ κ (P
id4q
F λα( p
(2π )4
×F 2
· q
, q
(37)
θ κ (P
2) = T 1
·
where F 1
λτF1(P
λτ (P
· p
2). Solving these two equa-
, p
, p
p
tions, respectively, one can obtain two series of eigenval-
ues and eigenfunctions. Because the cross terms are small,
we can take the ground state BS wave function to be a lin-
, p
2) and
ear combination of two eigenstates F 10
· p
2) corresponding to the lowest energy in
, p
F 20
λτ (P
·
Eqs. (36) and (37). Then in the basis provided by F 10
λτ (P
2) = T 1
2) =
· p
, p
, p
λτF10(P
p
· p
, p
+
λτF20(P
T 2
λτ is consid-
ered as
λτ (P
2), the BS wave function χ 0
2) and F 20
2) and F 2
· p
· p
, p
, p
λτ (P
+
λτ (P
χ 0
, p
) = c1F 10
λτ (P
· p
, p
2)+c2F 20
λτ (P
· p
, p
2). (38)
Substituting (38) into the BS equation (23) and comparing
the tensor structures in both sides, we obtain an eigenvalue
equation,
c1F10(P
×
2) =
, p
− i
¯h(s)
1
+ M 2
− i
· p
h(s)
1
(k2)
2
1
2
2
p
p
2
1
σ
(k2)
{h(v)
1
(k2)¯h(v)
1
(k2)
1
1
+ M 2
g2
k2+m2
σ
+
id4q
(2π )4
g2
ρ
ρ
φ
1
2
2
1
ω
g2
8
g2
1
+
+
k2 + m2
k2 + m2
k2 + m2
(k2)¯h(v)
× ( p
+ q
) · ( p
+ q
) + 2h(v)
1
2
2
1
(k2)¯h(v)
· q
)] + 2h(v)
)−(q
· p
× [(q
2
1
2
1
)]}
· q
· q
− ( p
2)
, q
c1F10(P
2
1
g2
ρ
+
× {2h(v)
+ 2h(v)
× c2F20(P
id4q
(2π )4
(k2)¯h(v)
(k2)¯h(v)
· q
+
k2 + m2
k2 + m2
(k2)[q
− q
2
2
2
( p
1 q
1
2
(k2)[q
− q
2
2
2
1 q
( p
1
2
2
, q
2)
+
2
g2
1
ω
1
2
2
1
,
ρ
(k2)
(k2)[(q
1
· q
2
)
g2
8
φ
k2 + m2
· q
)]
2
)]}
· q
1
(39a)
1
M 2
1 p
2
2
id4q
(2π )4
c2F20(P
=
2)
· p
, p
1
+ M 2
1
g2
ρ
ρ
φ
1
1
2
2
)
)
ω
p
2
2
g2
8
2
1
1
+ M 2
− i
− i
p
+
+
×
g2
k2 + m2
k2 + m2
1
k2 + m2
(k2)¯h(v)
× {h(v)
· p
(k2)[( p
· q
)(q
2
2
1
1
2
1
(k2)¯h(v)
− ( p
)] + h(v)
· p
· q
(k2)
)(q
2
1
2
1
· ( p
+ q
· (q
− p
× [ p
)q
1
1
2
2
2
1
− (M 2
· ( p
· q
+ q
+ ( p
)]}
))q
2
2
2
1
1
× c1F10(P
2) +
· q
, q
id4q
(2π )4
¯h(s)
×
2
2
1
M 2
1 p
[M 2
(k2)q
h(s)
1
g2
σ
1
1
g2
ρ
g2
8
2
1
· q
(k2)
+ ( p
k2 + m2
k2 + m2
) − q
] +
(k2)¯h(v)
{h(v)
+
(k2)( p
k2 + m2
1
· q
+ ( p
[M 2
+ q
× ·( p
2
)q
1
2
2
2
(k2)¯h(v)
− q
] + h(v)
2
(k2)
1
× [M 2
· q
· q
) − M 2
2
1 q
(q
)( p
1
2
2
2
2
1
+ q
) − q
· q
· p
2
2
2
( p
1 q
( p
)
2
1
2
1
2
1
(k2)¯h(v)
× ( p
· q
)] + h(v)
(k2)
2
2
1
+
1
1
)
1
2
1
1
1
1
φ
ρ
σ
2
1
2
2
g2
1
k2 + m2
+ q
)
ω
2
· q
1
)
( p
123
98 Page 8 of 11
2
2
· ( p
[ p
× q
2
2
1
+ ( p
− (M 2
1
× c2F20(P
+ q
· q
))q
1
· q
, q
2)
1
· (q
)q
1
· ( p
2
1
1
− p
+ q
2
)
1
)]}
,
(39b)
2
1
λ P
λ P
λ P
− i
2) =
, p
τ + η2 P
λ p
1
+ M 2
2
2
, q
2),
where the eigenvalues are different from the eigenvalues in
(36) and (37). From this equation, we can obtain the eigenval-
ues and eigenfunctions which contain the contribution from
the cross terms.
τ − η1 p
τ −
Comparing the terms (η1η2 P
τ ) in the left and right sides of Eq. (36), we obtain
λ p
p
· p
F1(P
×
1
− i
+ M 2
2
p
p
1
; P
· q
)F1(P
, q
id4q
(40)
(2π )4 V1( p
; P
) contains all coefficients of the term
, q
where V1( p
τ + η2 P
τ − η1 p
τ ) in the right side
(η1η2 P
λ p
of (36). In this paper, we set k = (k, 0). Then the fourth
components of momenta of two heavy mesons have no
) = E1(q
= q
=
change: p
) = E2(q
10
20
). To simplify the potential, we replace
E2( p
) → E1 =
2
the heavy meson energies E1( p
+ M 2
) → E2 =
(M 2 − M 2
1
)/(2M), E2( p
+ M 2
(M 2− M 2
2
1
)/(2M). The potential depends on the three-
) ⇒ V (p, q, M). Integrat-
vector momentum V ( p
0 and multiplying by
ing both sides of Eq. (40) over p
(M + ω1 + ω2)(M 2 − (ω1 − ω2)2), we obtain the the
Schrödinger type equation
)
(p
), p
) = E1(q
) = E2(q
1
2
= E1( p
− p2
τ − p
= q
; P
, q
λ P
λ p
10
20
2
1
1
+
1
2
2
0
1
b2
(M)
1
2μR
=
2μR
d3k
(2π )3 V 0
1
+
, k)0
(p
1
and the potential between D
/MH expansion
order of the p
+
, k),
(p
∗0 and ¯D
(41)
∗0 up to the second
ρ
8
1
1
+
g2
σ
V 0
1
(k2)
(k2)
(k2)
+ g2
k2+m2
¯h(s)
k2 + m2
1
2E2
σ
g2
+ g2
ρ
k2+m2
k2+m2
−1 − 4p2 + 5k2
, k) = h(s)
1
(p
2E1
+ h(v)
× ¯h(v)
(p) =
where 0
0 F1(P
dp
E2) = [M 4 − (M 2
− M 2
1
2
(M1+ M2)2][M 2− (M1− M2)2]/(4M 2), ω1 =
and ω2 =
4E1 E2
· p
2), μR = E1 E2/(E1+
, p
)2]/(4M 3), b2(M) = [M 2 −
p2 + M 2
τ +
2 . Comparing the terms (η2
1 P
p2 + M 2
λ P
(k2)
(42)
+
ω
1
1
1
φ
,
123
Eur. Phys. J. C (2015) 75 :98
λ P
τ + p
λ p
τ ) in both sides of Eq. (37), we
τ + η1 p
2
1
− i
1
+ M 2
− i
2
p
2
· q
, q
2).
F2(P
(43)
· p
2), we obtain the
, p
F2(P
η1 P
λ p
obtain
2
F2(P
2
×
p
, p
2) =
· p
1
+ M 2
2
p
1
; P
2
)q
2
2
Setting 0
0 p
2
2
Schrödinger type equation
)
(p
, q
id4q
(2π )4 V2( p
(p) =
− p2
d p
+
+
0
2
2μR
b2
(M)
2
2μR
=
d3k
(2π )3 V 0
2
+
, k)0
(p
2
+
and the potential between D
/MH expansion
order of the p
, k)
(p
∗0 and ¯D
(44)
∗0 up to the second
+
V 0
2
(k2)
, k) = h(s)
1
(p
2E1
+ h(v)
× ¯h(v)
1
1
(k2)
(k2)
g2
σ
(k2)
¯h(s)
k2 + m2
1
2E2
σ
g2
+ g2
ρ
k2+m2
k2 + m2
−1− 2p2 + 2k2
1 − k2
M 2
1
+ g2
k2+m2
− 2p2 + 2k2
ω
1
8
φ
ρ
.
4M 2
1
4E1 E2
(45)
In instantaneous approximation the eigenfunctions in Eqs.
(36) and (37) can be calculated and the eigenvalue equation
⎛
(39) becomes
− λ
⎝ b2
⎞
⎠
(M)
10
2μR
= 0,
(46)
H12
b2
(M)
20
2μR
− λ
c
1
c
2
H21
where we have the matrix elements
H12 = H21 =
×
d3 p
0
+
10
)∗
+
g2
ρ
(p
k2 + m2
ρ
k2
E1 E2
(k2)
× ¯h(v)
1
d3k
(2π )3 h(v)
g2
1
+
1
k2 + m2
ω
(k2)
g2
8
k2 + m2
φ
+
0
20
, k),
(p
(47)
+
10 and 0
(M)/(2μR) and b2
20
and b2
(M)/(2μR) are the eigenvalues
10
corresponding to lowest energy in Eqs. (41) and (44), respec-
+
tively; 0
20 are the corresponding eigenfunctions.
In (42), (45), and (47) the contribution from one-π exchange
to the potential between two heavy vector mesons has van-
ished, but we still give the heavy meson form factors h( p)(k2)
and ¯h( p)(k2) in Sect. 4. Then applying the method above, we
can investigate the alternative J P = 2
Y state.
assignment for the
+
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