Second Class Current in QCD Sum Rules

26 Pages·0.23 MB·English

by  H. Shiomi

UTHEP-329, January 1996
Second Class Current in QCD Sum Rules
H. Shiomi
∗
Institute of Physics, University of Tsukuba, Tsukuba, Ibaraki 305, Japan
(February 1, 2008)
6
9
9
1

n
a
J Abstract

4
2
1 Induced tensor charge of the nucleon gT, which originates from G-parity
v violation, is evaluated from QCD sum rules. We find that gT/gA with
9 g being the axial charge is 0.0152 0.0053 which is proportional to
A
2 − ±
u-d quark mass difference. This result is small compared to preliminary
3
1 analysis of the experiment, but is consistent with the estimate in the MIT
0
bag model.
6
9
/
h PACS numbers: 12.38.Lg, 14.20.Dh
p
Key words: Induced tensor charge of the nucleon, QCD sum rules
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p
e
h
:
v
i
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Typeset using REVTEX
e-mail address: [email protected]
∗
1
I. INTRODUCTION
Deriving the coupling constants from QCD [1] is one of the most important themes
in hadron physics. Also, getting precise values of the coupling constants from the first
principle will enable us to make more quantitative predictions for nuclear systems.
For extracting the hadronic quantities from QCD, QCD Sum Rules (QSR) discussed
belowareknowntobeoneofthetwopowerfultools. (TheotherislatticeQCDsimulations
[2].) QSR was first proposed in a paper by Shifman Vainsthein, and Zakharov in 1978
[3], in which the main idea and the application to meson systems such as meson masses,
decay rates and ρ ω mixing, are shown. Later, many applications have been made [4,5].
−
The extension to the baryon systems was put forward by Ioffe [6]. QSR with external
field was also proposed by Ioffe and Smilga [7], which will be discussed later.
In this paper we will evaluate induced tensor charge of the nucleon (g ) with the help
T
of axial charge (g ) and nucleon sum rules.
A
g and g are defined by the nucleon matrix element of the axial current,
T A
P(p ) A5 N(p )
h 2 | µ| 1 i
γ g iγ σ qν
5 P 5 µν
= u¯ (p )(γ γ g + + g )u (p ), q = p p , (1)
p 2 µ 5 A T n 1 1 2
M +M M +M −
p n p n
where M (M ) is proton(neutron) mass, A5 = u¯γ γ d represents axial current and u (u )
p n µ µ 5 p n
reveals proton (neutron) wave function. g is called induced psudoscalar constant, which
P
will not be examined in this paper. The deviation of g from unity is a reflection of the
A
underlying composite structure of the nucleon and there have been many studies on g ,
A
e.g., QCD sum rules [8,9], quark models [10] , the bag model [11] and the skyrme model
[12]. The difference of g and g is classified by G-parity which is the charge conjugation
A T
C combined with the rotation of 180 around the y-axis in the isospin space
◦
G = CeiIyπ, (2)
where I is the rotational matrix around the y-axis in the isospin space.
y
Under the G-parity,
Gp¯γ γ nG 1 = p¯γ γ n, Gp¯σ γ nG 1 = p¯σ γ n, GA5G 1 = A5. (3)
µ 5 − − µ 5 µν 5 − µµ 5 µ − − µ
The first current in eq.(3) with the same sign as that of A5 under G- parity is called the
µ
first class current, while the second current in eq.(3) with the opposite sign as that of A5
µ
is referred to as the second class current [14].
There are two sources of G-parity violation in the standard model. One is from QED
(electric charges of u and d quarks are different) and another is from the mass term in the
QCD Hamiltonian (masses of u and d quarks are different). We will exclusively examine
the latter effect in this thesis.
The mass term in the QCD Hamiltonian is written as
1 1
¯ ¯
H = (m +m )(u¯u+dd)+ (m m )(u¯u dd). (4)
mass u d u d
2 2 − −
2
where the light quark masses are determined from analyses of the hadron mass splittings
in QCD sum rules [13];
m (µ = 1GeV) = (5.1 0.9)MeV, m (µ = 1GeV) = (9.0 1.6)MeV.
u d
± ±
Under G-parity, the first term in eq.(4) does not change the sign, but the second term
in eq.(4) changes sign, which means that g is represented as g (m m )/M since
T T u d N
∼ −
g is dimensionless. This implies that g will be much smaller than g because of the
T T A
small u d quark mass difference g 4MeV/1000MeV 0.004 where we have used the
T
− ∼ ∼
quark mass difference m m 5MeV.
d u
− ∼
This roughestimate of g using theG-parity violation, however, may not beconsistent
T
with the analysises of the experimental data given by measuring the beta-ray angular
distribution in aligned 12B and 12N [15]. In Ref. [15], results of the analyses using the
experimental data are quoted as
g /g = 0.14 0.10 in 1985, (5)
T A
±
g /g = 0.21 0.14 in 1992. (6)
T A
− ±
This shows that g is of order 10 % compared to g , which is order of magnitude larger
T A
than the naive expectation. Although the experimental error bars are large and even the
sign of g /g is not certain yet, the data poses a theoretical challenge togive more reliable
T A
estimate of g /g .
T A
It is in order here to show other examples of the G-parity violation which is more
firmly established than g [16]:
T
1)Proton-neutron mass difference.
Experimental mass difference is M M = 1.29 MeV, in which the contribution of
p n
− −
the u d quark mass difference after subtracting the theoritical electromagnetic effect
−
(0.76 0.3 MeV) is 2.05 MeV. This last number has been successfully reproduced in
± −
QSR calculations [17].
2)ρ0 ω mixing.
−
The ρ ω mixing is defined by the covariant matix element ρ0 H ω at the ρ0 ω
GPB
− h | | i −
mass shell with H being the second term in eq.(4). The recent measurement of the
GPB
e+e π+π shows an unambigous determination of the ρ0 ω mixing with negative
− −
→ −
sign, which ought to be dominated by the quark mass difference since the electromagnetic
effect by ρ γ ω is positive and small [18,19].
→ →
3)π± π0 mass differnce (mπ± mπ0=4.6 MeV).
− −
This is a typical example of the electromagnetic G-parity violation. Theoretical estimate
gives (mπ± mπ0)em = 4.6 0.1 MeV,while the effect of thequark mass difference appears
− ±
only in second order of the quark masses and is extremely small [20].
The ingredients of this paper are twofolds: i) to get QSR with an external field for g
T
and g in section II-V, and ii) to predict the value of g relative to g in section VI.
A T A
As for i), we will adopt a method proposed by loffe and Smilga [7] and independently
by Balitsky and Yung [21], in which two point functions with an external electromagnetic
field strength F is studied up to linear in F . The method has been applied for the
µν µν
magnetic moment of the nucleon and the results agree with the experimental data with
a good accuracy. A method on two point functions with an axial-vector field Z was also
µ
3
developed by Belyaev and Kogan [8], and later improved by Pasupathy et al. [9]. They
have considered terms proportional to Z for evaluating the axial charge g . The latter
µ A
method with Z replaced by the vector potential A is, however, not suitable for studing
µ µ
magnetic moments since the explicit momentum transfer must be retained.
Adopting QSR with the external field induces new parameters which are absent in
ordinary QSR. These parameters reflects the response of QCD vacuum to the external
field. For instance, 0 q¯σ q 0 , which is identical to zero in the vacuum, acquires the
µν E
h | | i
non-zerovalueduetothepresenceoftheexternalfield. Toevaluatethesenewcondensates,
QSR with the assumption of the vector dominance can be used [22].
There is another new feature of QSR with the external field compared to the ordinary
one. The phenomenological side of the correlation function with external field takes the
following double pole form near the nucleon resonance
0 η N N JE N N η¯0 (p2 M2) 2, (7)
h | | ih | | ih | | i − N −
where JE is a current coupled to the external field. Besides the double pole part which we
are interested in, single poles, which expresses transition from the ground state to excited
states, appears. Furthremore, the single pole term is not suppressed compared to the
double pole term after applying the Borel transform. This bears no resemblance to the
contribution of continuum which is exponentially suppressed by Borel transform. Hence
we must take into account both the double pole and the single pole in phenomenological
side on the same footing, which requires a procedure to subtract the single poles.
As for ii), one must remember that g originates from the G-parity violation induced
T
by the u-d quark mass difference and the electromagnetism. The experimental value of
g is still uncertain as we have mentioned above. Thereby we shall try to determine g
T T
from QSR with main emphasis on the effect of the u-d quark mass difference. Within our
knowledge, no serious evaluation of g has been done so far except for a rough estimate
T
using the MIT bag model [23,24]. We will therefore reexamine the bag model calculation
also and compare it with our QSR result.
The paper is organized as follows. Section II-IV are devoted to derive g and g sum
T A
rules in QSR with the external field. In section V, we estimate the quark and induced
condensates by using QSR.Insection VI,we analyse theg sum ruleandget its numerical
T
number. In section VII, discussions and summary are made.
II. WEAK INTERACTION IN HADRONIC SIDE AND QCD SIDE
We start with the two point function with an external field;
Π (p) = i d4xeipx 0 Tη (x)η¯ (0) 0 (8)
E · p n E
h | | i
Z
= F Π (p), (9)
µν µν
where ’E’ denotes external field of weak boson W+, F (x) = ∂ W+(x) ∂ W+(x), and
µν µ ν − ν µ
η (η ) corresponds to the proton (neutron) interpolating field defined as [6]
p n
η (x) = ǫ (ua(x)Cγ ub(x))γ γ dc(x), η (x) = η (u d) (10)
p abc µ 5 µ n p
↔
4
The hadronic side of sum rules can be saturated by a process where neutron turns into
proton by absorbing the W+ boson, matrix element of which is shown in eq.(1). Hence
we define an effective Lagrangian in which nucleon current couples to W+ field as
g g g
Lhad = j5W+ = p¯ g γ γ W+ + T γ σ ∂ W+ n, (11)
int −2√2 µ µ −2√2 A µ 5 µ Mp +Mn 5 µν ν µ !
where p (n) represents proton (neutron) field and g is associated with Fermi constant as
GF = g2 with M being W+ boson mass in the Glashow-Weinberg-Salam model [25].
√2 8M2 W
W
It is in order here to mention g . g is associated with g via PCAC and can be
p P A
directly measured by the muon capture [24]. However, our external field does not pick up
the contribution of g because we are using the field strength F instead of the vector
P µν
potential W+ as an external field.
µ
On the other hand, in the quark level, the interaction of quarks and the external field
is written as
g g
Lquark = − j5W+ = − u¯γ γ d W+, (12)
int 2√2 µ 2√2 µ 5 µ
where u and d are up and down quark respectively. The common factor g/2√2 in eq.(11)
and in eq.(12) is obtained by comparing the V-A theory [26] with the Glashow-Weinberg-
Salam model.
III. HADRONIC SIDE FOR TWO POINT FUNCTION WITH EXTERNAL
FIELD
In this section we examine the hadronic contribution to QSR with the external field.
Firstly, we consider Fig.1, which shows that a neutron with momentum p absorbs W+
1
boson and turns into proton with momentum p . Using eq.(11), we may write down Fig.1
2
as
1 iγ σ 1
5 µν
Fig.1 = g γ γ + g q (13)
A µ 5 T ν
pˆ2 Mp Mp +Mn ! pˆ1 Mn
− −
1
=
(p2 M2)(p2 M2)
2 − p 1 − n
qˆ iγ σ qˆ
5 µν
+pˆ+M g γ γ + g q +pˆ+M , (14)
p A µ 5 T ν n
× −2 ! Mp +Mn ! 2 !
where p (p ) is proton (neutron) momentum mentioned above, p = p1+p2 and q = p p .
2 1 2 1− 2
In the limit of soft external momentum q 0, we keep only the terms proportional to
µ
→
q in eq.(14) to extract F . Then eq.(14) is reduced to
µ µν
iγ q
5 ν
[P pˆσ +P σ pˆ+P σ +P (γ p γ p )pˆ] q Γ (p), (15)
(p2 M2)(p2 M2) 1 µν 2 µν 3 µν 4 µ ν − ν µ ≡ ν µν
− n − p
where
5
g M
A n
P = g , (16)
1 T
− 2 − M +M
n p
g M
A p
P = + g , (17)
2 T
− 2 M +M
n p
g M M p2
A n p
P = (M M )+( − )g , (18)
3 n p T
− 2 − M +M
p n
2ig
T
P = . (19)
4
M +M
p n
By using eq.(11), we evaluate eq.(8) in first order of the external field:
d4xeipx 0 T(η (x)η¯ (0)Lhad) 0
− · h | p n int | i
Z
g 1
= λ λ d4y d4lei(p l)yW+(y)
−2√2 p n (2π)4 − ·
Z Z
1 iγ σ 1
5 µν
g γ γ + g (l p)
×pˆ−Mp A µ 5 Mp +Mn T − ν! ˆl−Mn
g 1
= λ λ d4y d4qe iqyqνΓ (p)W+(y)
−2√2 p n(2π)4 − · µν
Z Z
ig
= λ λ Γ F (0)
p n µν µν
−4√2
gγ λ λ
5 n p
= F (0)
µν
4√2(p2 M2)(p2 M2)
− n − p
[P pˆσ +P σ pˆ+P σ +P (γ p γ p )pˆ], (20)
1 µν 2 µν 3 µν 4 µ ν ν µ
× −
where P ,P ,P and P are defined in eq.(16)-eq.(19), λ and λ are defined as 0 η N =
1 2 3 4 n p
h | | i
λ u(p) with u(p) being the nucleon Dirac spinor. Apart from the terms above, we must
N
take into account two other contributions. One is the single pole caused by a transition
of nucleon to resonance states as follows,
1 1
ΠE(p) λNλN∗ HNN∗ , (21)
∼ pˆ MN pˆ MN∗
− −
where N(N∗) is the nucleon (excited states, e.g., N(1440)), and HNN∗ is a transition
matrix from the nucleon to the excited states. As we have mentioned, the single pole is
not suppressed compared to the double pole after the Borel transform. Since we are not
interested in the single poles, we will subtract them, using a procedure shown later. The
other hadronic contribution is a continuum starting at a threshold S , which contains
0
only the excited states. The continuum can be exponentially suppressed by applying the
Borel transform.
IV. OPE FOR TWO POINT FUNCTION WITH EXTERNAL FIELD
In ordinary QSR with no external fields, only Lorenz invariant operators survive in
OPE. On the other hand, the external field induces new condensates. Relevant conden-
sates up to dimension 6 in our case read
6
g
0 u¯γ σ d 0 = (m m )F χ(0), (22)
5 µν E u d µν
h | | i − 2√2
g
¯
g 0 u¯γ G d 0 = dd u¯u F κ(0), (23)
s 5 µν E 0 µν
h | | i h − i 2√2
g
¯ ¯
g ǫ 0 dG u 0 , = dd u¯u iF ξ(0). (24)
s µνρω ρω E 0 µν
h | | i h − i 2√2
The above condensates are non-vanishing because the QCD vacuum is distorted by the
external field and the Lorenz invariance is broken. Note also that taking m = m
u d
makes all the above terms vanish. Using the fixed point gauge, we can rewrite the
i 0 T(η (x)η¯ (0)) 0 as follows:
p n E
h | | i
i 0 T(η (x)η¯ (0)) 0 = 4iǫ ǫ 0 γ γ iSce(x)γ CiET (x)Cγ iSbfγ γ 0
h | p n | iE abc defh | 5 µ d ν ad µ u ν 5 | iE
(25)
with
iSab(x) = 0 T(qa(x)q¯b(0)) 0
q h | | i
ixˆ ix mδab
= δab + α (te)abG˜αργ γ + χa(x)χ¯b(0) , (26)
2π2x4 8π2x2 e ρ 5 − 4π2x2 h q q i0
iEab(x) = 0 T(ua(x)d¯b(0) 0
E
h | | i
ig x
= δab λ F˜ γ + χa(x)χ¯b(0) (27)
2√2 8π2x2 λl l h u d iE
1 i
(m m ) (log( x2Λ2/4)+2γ )γ σ Fρν + F γ γ xˆx ,
− u − d 32π2 − E 5 ρν 16π6x2 ρµ 5 µ ρ
(cid:26) (cid:27)
where iEab(x) is calculated by eq.(12) with the first order perturbation, F˜ = 1ǫ Fρω,
µν 2 µνρω
and Λ2 is an infrared cut off parameter.
χa(x)χ¯b(0) in eq.(27) expresses non-perturbative condensate under the external
h u d iE
field, while other terms are coupled directly to the quark propagator. To calculate the
nonperturbative terms, we shall compare it with χa(x)χ¯ (0) which is expanded as
h q q i0
χa(x)χ¯b(0) = 0 qa(x)q¯b(0) 0 (28)
0
h i h | | i
δab δabx2
= 0 q¯q 0 0 g q¯σ Gq 0 + (29)
s
−12h | | i− 192 h | · | i ···
In the case above, we have retained only the Lorentz scalar operators. In contrast,
χa(x)χb(0) is expanded only by the Lorenz tensor terms corresponding to the induced
h u d iE
condensates eq.(22) eq.(24). Hence
−
1 x x
χa(x)χ¯b(0) = δ γ σ u¯γ σ d ρ ωγ σ u¯γ σ D D d +
h u d iE −24 ab 5 µνh 5 µν iE − 48 5 µνh 5 µν ρ ω iE ···
1 g
= δabγ σ Fµν(m m )χ(0)
5 µν u d
−242√2 −
g 1
¯
γ σ dd u¯u
−2√23225 5 µνh − i0
(κ(0) ξ(0))x2F (2κ(0)+ξ(0))x x F + (30)
µν µ ω νω
× − − ···
n o
7
We turn to carry out OPE for eq.(25) with diagrams Fig. 2(a)-(j). For the chiral odd
structures, we impose m = m . This induces P = P = g /2 in eq.(20) due to g = 0.
u d 1 2 A T
−
The chiral odd structure can be applied to estimate the axial-charge (g ) with m = m .
A u d
On the other hand, the chiral even structures in eq.(25) are evaluated up to linear in
(m m ), which leads to g sum rule.
u d T
−
Let us examine each contribution in Fig.2 more closely
The coefficient of F with the chiral odd structure given in Fig.2(a) reads
µν
g 6
Fig.2(a) = x γ F˜ . (31)
−2√2π6x8 λ l λl
The coefficient of F αsG2 with the chiral odd structure given in Fig.12(b) reads
µνh π i
g 1 α
Fig.2(b) = x F˜ γ sG2 , (32)
−2√232π4x4 λ λl l π
(cid:28) (cid:29)0
where αsG2 = 0.012(GeV2) is called gluon condensate, whose value is determined from
π 0
the anDalysis Eof heavy quark system based on QSR. The coefficient of F q¯q 2 with the
µν
h i
chiral odd structure given in Fig.12(c) reads
g q¯q 2
Fig.2(c) = h i0 x F˜ γ . (33)
−2√212π2x2 λ λl l
The coefficient of (m m )F coupled directly to the propagator given in Fig.2(d) reads
u d µν
−
g 3
Fig.2(d) = i x F˜ γ xˆ(m m ). (34)
− 2√22π6x8 λ λl l u − d
Fig.2(f)alsogivesacoefficient of(m m )F . However, itcauses theinfrareddivergence
u d µν
−
which must be absorbed into the same dimensional operator shown in Fig.2(e).
¯
The coefficient of dγ σ u , which is dimension 3 and is given in Fig.2(e), reads
5 µν E
h i
1
¯
Fig.2(e) = i γ xˆσ xˆ dγ σ u . (35)
π4x8 5 ρω h 5 ρω iE
¯
The coefficient of dd u¯u F given in Fig.2(g) reads
0 µν
h − i
g 1
Fig.2(g) = i x F˜ xˆγ d¯d u¯u . (36)
− 2√22π4x6 λ λl lh − i0
The coefficient of g d¯γ G u and g ǫµνρω d¯G u given in Fig.12(h) and (i) read
s 5 µν E s ρω E
h i h i
γ σ
Fig.2(h) and (i) = i 5 ρω [g d¯γ Gρωu +ig ǫρωµν d¯G u ] (37)
12π4x4 sh 5 iE s h µν iE
γ (γ x γ x )xˆ 1
+ 5 ρ ω − ω ρ [g d¯γ Gρωu + g iǫρωµν d¯G u ], (38)
4π4x6 sh 5 iE 2 s h µν iE
¯
where the contribution of Fig.2(h) is zero bacause 0 du 0 = 0 and in Fig.2(i) the gluon
h | | i
field emitted by a soft quark interacts W+ field via quark condensate.
The coefficient of g d¯γ G u and g ǫµνρω d¯G u given in Fig.2(j) read
s 5 µν E s ρω E
h i h i
8
γ (γ x γ x )xˆ
Fig.2(j) = i 5 ρ ω − ω ρ xˆ[g ǫρωµν d¯G u 2ig d¯γ Gρωu ] (39)
− 8π4x4 s h µν iE − sh 5 iE
γ σ
i 5 ρωg d¯γ Gρωu , (40)
− 4π4x4 sh 5 iE
where the gluon field emitted by a hard quark interacts W+ field via quark condensate.
In summary, we obtain the following formula
g
Π (p) = F [Q γ (pˆσ +σ pˆ)+ Q γ σ +Q iγ (γ p γ p )pˆ (m m )],
E 2√2 µν 1 5 µν µν { 2 5 µν 3 5 µ ν − ν ν } u − d
(41)
1 1 αsG2 q¯q 2
Q = − p2log( p2) h π i h i0, (42)
1 32π4 − − 64π2 p2 − 6p4
1 χ(0) 1 κ(0) ξ(0)
Q = p2log( p2) C log( p2), (43)
2 − − 16π4 − 24π2!− m 8π2 − 6π2 − 12π2! −
1 χ(0) C
Q = log( p2) + m , (44)
3 − 32π4 − 12π2! 8π2p2
¯
where we have used a relation dd u¯u = C (m m ) which will be discussed below.
0 m u d
h − i −
V. THE ESTIMATE OF THE QUARK AND INDUCED CONDENSATES
Before setting up the sum rules, we need to estimate the magnitude of the quark
and induced condensates. For the quark condensate, we utilize Finite Energy Sum Rules
(FESR) [27,28] for nucleon mass, where we look for the optimal quark condensate which
reproduce the nuclon mass within the standard values of the condensate q¯q (1GeV2) =
0
h i
(225 25MeV)3.
− ±
From ref. [28], we get the FESR for nucleon as follows:
S3 α 128
64π4λ2 = N +2π2 sG2 S + π4 q¯q 2, (45)
N 3 h π i0 N 3 h i0
32 α
64π4λ2 M = 8π2 q¯q S2 + π4 q¯q sG2 , (46)
N N − h i π 9 h i0h π 0i0
S4 α 128 α
64π4λ2 M2 = N +π2 sG2 S2 π4 q¯q 2 sS , (47)
N N 4 h π i0 N − 9 h i0 π N
where λ is defined above, and S is the continuum threshold of nucleon sum rules.
N N
Solving eq.(45)-(47) numerically, we get results in Table 1. Hence we will utilize the
following numbers in the anaylses of g sum rule later;
T
q¯q (GeV3) = ( 0.2185)3, S (GeV2) = 1.6, λ (GeV3) = 0.0188. (48)
N N
h i −
q¯q ( 0.250GeV)3 ( 0.230GeV)3 ( 0.2185GeV)3 ( 0.210GeV)3
0
h i − − − −
S (GeV2) 2.27 1.84 1.60 1.42
N
λ (GeV3) 0.0296 0.0224 0.0188 0.0162
N
M (GeV) 1.15 1.024 0.940 0.872
N
9
Table 1: S ,λ ,M obtained from Eq.(45) (47) with four different values of q¯q .
N N N 0
∼ h i
Now we turn to the calculation of induced condensates with the help of QSR.
We first expand eq.(22) in terms of W+ up to first order.
g
d¯γ σ u = i d4x 0 T(d¯γ σ u(0)u¯γ γ d(x)) 0 W+(x) (49)
h 5 µν iE − 2√2 h | 5 µν ρ 5 | i
Z
g
= i d4x Π (x)W+(x),
− 2√2 µν,ρ ρ
Z
¯
where Π (x) = 0 T(dγ σ u(0)u¯γ γ d(x)) 0 . (50)
µν,ρ 5 µν ρ 5
h | | i
To estimate Π (x), we expand eq. (50) in terms of the local operators up to dimension
µν,ρ
5, whose diagrams are Fig.3(a)-(c), and retain the terms proportional to quark mass.
Then we get the following equation:
Π (q) = ( q g +q g )(m m )χ(q2) (51)
µν,ρ µ ρν ν ρµ u d
− −
with
3 1 m2
χ(q2) = log( q2)+ + 0 C , (52)
8π2 − q2 3q4! m
where the first, second and third term on the right hand side correspond to Fig.3(a), (b)
and (c), respectively. m2 = 0.8(GeV2) is defined by 0 g q¯σ Gq 0 = m2 0 q¯q 0 [30].
0 h | s · | i 0h | | i
Using eq.(50) and eq.(51), we reach the result defined in eq.(22):
g
¯
dγ σ u = F (0)(m m )χ(0), (53)
5 µν E µν u d
h i 2√2 −
where we replace F (x) by F (0) bacause the field strength is assumed to be constant.
µν µν
For the phenomenological side in eq.(52), we assume that χ(q2) is saturated by a me-
1
son with mass 1260(MeV), which is the lowest state coupled to both pseudovector and
pseudotensor states, and the continuum starting at S :
χ
1 3
Im χ(s) = f δ(s m2 ) θ(s S ). (54)
π χ − a1 − 8π2 − χ
Then, we get
f 3 Λ2 θ(s S )
χ χ
χ(0) = − ds, (55)
m2 − 8π2 s m2
a1 Z0 − a1
where Λ2 = 1GeV2 is taken as a characteristic scale of seperating the perturbative and
¯
the non-perturbative part in dγ σ u . Matching eq.(52) and eq.(54) using FESR, we
5 µν E
h i
get two sum rules,
3
n = 0, S +C = f , (56)
8π2 χ m χ
3 m2
n = 1, S2 + 0 = f m2 . (57)
16π2 χ 3 χ a1
f is rewritten as
χ
10