Abundant strange sea quarks in the nucleon and flavor asymmetry of the light sea quarks in the octet baryon at small x

Available online at www.sciencedirect.com

Nuclear Physics A 868–869 (2011) 25–52

www.elsevier.com/locate/nuclphysa

Abundant strange sea quarks in the nucleon and flavorasymmetry of the light sea quarks in the octet baryon

at small x

Susumu Koretune

Department of Physics, Shimane University, Matsue, Shimane 690-8504, Japan

Received 20 May 2011; received in revised form 18 August 2011; accepted 19 August 2011

Available online 1 September 2011

Abstract

By regularizing the sum rules from the current anti-commutation relation on the null plane, we firstexplain the relation between the high-energy behavior of the total cross section of pseudo-scalar-nucleonscattering and the small-x limit of the structure function F2 in lepton–nucleon scattering. Then, by assumingthat the ratio of the total cross section of pion–nucleon scattering to that of kaon–nucleon scattering in thehigh-energy region is kept within the high-energy limit, we show that, in the very-small-x region in thenucleon, the strange sea quark becomes more abundant than the up and the down sea quarks. Further, wefind that the regularization-independent finite sum rules that originate from the spontaneous chiral symmetrybreaking of the vacuum take similar values as those where the sea quarks take symmetric values in the very-small-x region. We apply the method to the sea quarks of the octet baryon and give many similar sum rulesas in the nucleon case; we also give the asymmetry of the sea quarks in the very-small-x region.© 2011 Elsevier B.V. All rights reserved.

Keywords: Sum rule; Current algebra; Sea quark; Strange; Octet baryon; Small x

1. Introduction

Sea quarks play a major role in high-energy reactions; hence, the flavor separation of thesea-quark distribution functions is an important subject to be studied. However, the light seaquarks u, d , and s in the small-x region are poorly separated experimentally. The strange seaquark is usually assumed to have a similar x dependence as the up and the down sea quarks but

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26 S. Koretune / Nuclear Physics A 868–869 (2011) 25–52

that is greatly suppressed in comparison, based on the CCFR/NuTeV dimuon experiment abovex ∼ 0.01 [1]. A recent HERMES result shows that the x dependence differs greatly betweenthe strange sea and the iso-scalar non-strange sea in the charged-lepton DIS experiment abovex ∼ 0.01 [2]. The experimental constraint on the strange sea-quark distribution below x ∼ 0.01is very weak.

Now, because the distribution function is in itself a non-perturbative quantity, theoretical dis-cussions from a general point of view are helpful. The Adler sum rule, the Adler–Weisbergersum rule, and similar ones from the equal-time current algebra [3] are old but suitable in thisrespect. These sum rules are essential ingredients in QCD.

In the late 60s, motivated by the experimental discovery of partons at SLAC, the light-conecurrent algebra, which was abstracted from the leading light-cone singularity of the current com-mutation relation in the free quark model, was developed [4]. In the same period, a current algebrabased on canonical quantization on the null plane, which had some technical superiority over theequal-time one was devised [5]. The light-cone current algebra and the null-plane current algebrahave similar quantities, called the bilocal currents. The bilocal currents in each current algebrawere often identified on the null plane as a heuristic method to obtain physical insights; however,they were essentially different quantities. The bilocal current in the light-cone current algebra isdefined as the regular operator where all the light-cone singularities are removed, but no suchmanipulation is applied to the bilocal current in the null-plane current algebra. This distinction isimportant because the scaling violation that led to QCD contradicts this manipulation. In QCD,by considering the causality of the current commutation relation and dispersion relations, theshort-distance expansion was analytically continued to the light-cone expansion, and throughthis, the moment sum rules were obtained [6]. Because of the anomalous dimension, we cannotremove light-cone singularities uniformly from each moment corresponding to a matrix elementof the local operator obtained by the expansion of the bilocal operator. Thus the expansion bythe singular coefficient function multiplied by the regular bilocal current in the light-cone currentalgebra breaks down. These classical moment sum rules were derived only at alternating inte-gers; however, they were soon generalized to include the quantities corresponding to the missingintegers [7], which were related to the non-local quantities. Now, the moments at the lower inte-gers are recognized to be less well predicted by the perturbative approach. In general, the inversetransform of the moment M(n,Q2), which is nothing but the O(4) partial waves apart from akinematical factor [8], gives us the structure function. Then we find that the very-small-x limit iscontrolled by the rightmost singularity in the complex n plane. Away from the limit, the behaviorat small x is determined by this singularity and the contribution from the background integral atRe(n) = L, where L is the smallest number for which the moment converges. It is very difficultto know the precise kinematical regions where this singularity in the complex n plane domi-nates over the contribution from the background integral. Because lower moments depend moreon smaller-x behavior, we see that the lowest moment is closely related to the physics in thenon-perturbative region. The Adler sum rule derived from the current algebra based on canonicalquantization at equal-time or on the null plane corresponds to the sum rule at this lowest integer,i.e., the moment at n = 1. The importance to investigate the sum rules at the wrong signaturedpoint in the lowest moment was suggested in the light-cone current algebra [4], and it was alsostudied in Ref. [9]1 in the context of the hybrid of the light-cone current algebra and the null-plane current algebra. Though these analyses were based on the operator relation, their validity

1 I am grateful to Prof. S.B. Gerasimov for informing me of these papers.

S. Koretune / Nuclear Physics A 868–869 (2011) 25–52 27

was restricted because of the use of the light-cone current algebra. It was at this point that theDeser–Gilbert–Sudarshan representation [10] played an important role. By using this represen-tation, the null-plane current algebra was generalized to the current anti-commutation relation,not as the operator relation, but as the connected matrix element between the stable hadrons [11].In spite of this limitation, we can obtain information about the sea quarks in the hadrons undersuitable assumptions.

The experimental verification of the violation of the Gottfried sum rule [12] gave a clear hinttoward the separation of the sea quarks from the sum rule approach. In the current algebra basedon canonical quantization on the null plane extended to the current anti-commutation relation,the fact that the integral

1∫0

dx

x

{F

ep

2

(x,Q2) − Fen

2

(x,Q2)}, (1)

where q2 = −Q2, x = Q2/2p · q did not take the value 1/3 expected from the Gottfried sumrule [13] was known [14]. The perturbative approach showed that Eq. (1) had a very small Q2

dependence in the next-to-leading order analysis, and hence it deviated from 1/3 [15]. However,this deviation was far too small to explain the experimental value. On the other hand, the anal-ysis in Ref. [14] showed that the regularized sum rule for the structure functions F

ep

2 and Fen2

corresponding to the moment at n = 1 became Q2-dependent non-perturbatively through the reg-ularization. Further, it was shown that if we took the difference as in Eq. (1), the Q2 dependencecanceled out under the plausible assumption that the pomeron coupling is an iso-spin singlet.The magnitude of the remaining finite quantity was left undetermined. Stimulated by the exper-imental determination of the sum in Eq. (1), this magnitude was explicitly given almost modelindependently and shown to agree roughly with the experimental value [16]. This sum rule givenas

1∫0

dx

x

{F

ep

2

(x,Q2) − Fen

2

(x,Q2)}

= 1

3

(1 − 4f 2

K

π

∞∫mKmN

dν

ν2

√ν2 − (mKmN)2

{σK+n(ν) − σK+p(ν)

}), (2)

where σK+N(ν) with N = p or n is the total cross section of the K+N scattering and fK isthe kaon decay constant, was renamed the modified Gottfried sum rule. It clearly shows that thedeviation from 1/3 originates from the spontaneous chiral symmetry breaking of the vacuum.A re-evaluation of the Gottfried sum using neural networks showed that the result in Ref. [12]was underestimated due to the uncertainty in the small-x region [17]. The explicit evaluation inRef. [16] is closer to this re-evaluated value than to the one in Ref. [12]. Because, in the par-ton model the value 1/3 corresponds to a symmetric sea-quark distribution, the deviation from1/3 implies a flavor asymmetry of the sea quarks. Thus the study of the flavor asymmetry ofthe sea quarks is nothing but the study of the hadronic vacuum originating from spontaneouschiral symmetry breaking. The fact that non-perturbative QCD plays a major role is a commonaspect of the various models intended to explain the inaccuracy of the Gottfried sum [18,19]. Inour approach, we propose that the perturbatively predicted Q2 dependence is shielded by largenon-perturbative effects, or more practically, we propose that we can consider it to be negligible


as far as the moment at n = 1 is concerned [14,20]. It is interesting that, in studying higher-orderradiative correction up to O(α2

s ) in Ref. [21], the correction to the moment at n = 1 is found toreceive a suppression factor of O((1/Nc)

2) compared with the moments above n = 2. This typeof correction has been shown to hold in a general MS-like scheme, but it has not been confirmedin other schemes [21]. However, this particular suppression factor may indicate a deep theoreticalreason why the perturbative corrections remain very small and why the non-perturbative contri-bution is important in the n = 1 moment, and it may explain the puzzle discussed in Ref. [22].In the fixed-mass sum rule approach in this paper, the difference between the Adler sum ruleand the modified Gottfried sum rule lies in the x− integration. The former is obtained by the∫ ∞−∞ dx− δ(x−) · · · integration and the latter by the P

∫ ∞−∞

dx−x− · · · integration. Because of this,

compared with the Adler sum rule, the contribution from the antiquark distribution function getsan extra minus sign in the modified Gottfried sum rule [20]. However, we often encounter a di-vergence in the latter type of integral in the process of reaching the fixed-mass sum rule. Hence,the regularization of the sum rule is essential in our approach.

Now, the modified Gottfried sum rule has a physical meaning of the mean I3 of the {[quark]–[antiquark]} in the proton [20]. Though the high-energy behavior of the pion–nucleon total crosssection and that of the kaon–nucleon total cross section are slightly different and breaks thesymmetry relation which is required to regularize the sum rule, the modified Gottfried sum rulewas shown to be unaffected by this breaking of the symmetry relation [23]. This insensitivityto the high-energy behavior may explain the usual belief that the flavor asymmetry of the upand the down sea quarks is a phenomenon restricted to the low and intermediate energy regions.However, other sum rules concerning the mean hypercharge and the mean charge of the light seaquarks are affected. The purpose of this paper is to determine how these sum rules are changed ifthe ratio of the leading high-energy behavior of the pion–nucleon total cross section to that of thekaon–nucleon total cross section persists up to the high-energy limit. We show that the light seaquarks become asymmetric in the very-small-x region and that the strange sea quark becomesmost abundant. Furthermore we extend the analysis to the octet baryon.

In Section 2, we give a brief summary of the regularization of the sum rules and explain howfar they depend on the model. In Section 3, we show that the sea quarks become asymmetric inthe small-x limit. Furthermore, we derive a refined mean hypercharge sum rule, which can beextensively used in the application to the octet baryon. In Section 4, we extend the analysis tothe octet baryon. In Section 5, we show how the charm degree of freedom affects the discussion.The conclusion is given in Section 6. In Appendix A, we give the analytical results of the sumrules for the sea quarks in the octet baryon. In Appendix B, we give the analytical results of thesum rule of the charm sea quark.

2. A brief review of the regularization of the sum rules

The current anti-commutation relation on the null plane under the chiral SU(n) ⊗ SU(n) isgiven as

〈p|{J+a (x), J+

b (0)}|p〉c

∣∣x+=0

= 〈p|{J 5+a (x), J 5+

b (0)}|p〉c

∣∣x+=0

= 1P

(1−

)δ2(�x⊥)[

dabcAc

(p · x, x2 = 0

) + fabcSc

(p · x, x2 = 0

)]p+, (3)

π x


where c means to take the connected matrix element and the state |p〉 is the stable one-particlehadron state. Here we take J

μa (x) = :q(x)γ μ(λa/2)q(x): and

Sμa (x|0) = 1

2

[:q(x)γ μ λa

2q(0): + :q(0)γ μ λa

2q(x):

],

Aμa (x|0) = 1

2i

[:q(x)γ μ λa

2q(0): − :q(0)γ μ λa

2q(x):

],

〈p|Sμa (x|0)|p〉c = pμSa

(p · x, x2) + xμSa

(p · x, x2),

〈p|Aμa (x|0)|p〉c = pμAa

(p · x, x2) + xμAa

(p · x, x2). (4)

These relations were derived by using the Deser–Gilbert–Sudarshan representation [10] and thecurrent algebra on the null plane [11,14]. Then, by using the standard method to derive thefixed-mass sum rule in the null plane formalism [5], we can derive many sum rules. These sumrules correspond to the ones at the wrong-signature point. Typical sum rules are summarizedin Ref. [23] for the cases of n = 3,4,5,6. Many of them need regularization to make themphysically meaningful. This is because we need some superconvergent relations for the validityof relation (3). However, these relations cannot necessarily be satisfied. In such a case, relation(3) only gives us a formal divergent relation. As a remedy against such a divergence, we must, forexample, extend relation (3) to the non-forward direction and ensure that the superconvergencerelation is satisfied. The importance of the regularization of such divergent sum rules in thecurrent algebra was discussed in Ref. [24]. The regularization of the sum rules from the non-forward directions was given in Ref. [25]. Using this idea, the regularization of the divergentsum rules can be performed. We first derive the sum rules in the non-forward direction, and aftertaking out all the singular pieces as we go to the forward direction, we obtain the regularizedsum rules in the forward direction [14]. Then, we find that the same results are obtained byusing the equivalent practical method in the forward direction. In view of the importance of theregularization and for the necessity of the discussion in this paper, we now summarize the sumrules in the n = 3 case. Following the standard method [5], we obtain from Eq. (3) the sum rules[11,14]

g2A(0) + 2f 2

π

π

∞∫νπ

0

dν

ν

(σπ+p(ν) + σπ−p(ν)

)

= 1

2πP

∞∫−∞

dα

α

2√

3

3

(√2A0(α,0) + A8(α,0)

), (5)

[g

pΣ0

A (0)]2 + [

gpΛA (0)

]2 + Up + 2f 2K

π

∞∫νK

0

dν

ν

(σK+p(ν) + σK−p(ν)

)

= 1

2πP

∞∫−∞

dα

α

(2√

6

3A0(α,0) + A3(α,0) −

√3

3A8(α,0)

), (6)

[gnΣ−

A (0)]2 + Un + 2f 2

K

π

∞∫νK

dν

ν

(σK+n(ν) + σK−n(ν)

)
0


= 1

2πP

∞∫−∞

dα

α

(2√

6

3A0(α,0) − A3(α,0) −

√3

3A8(α,0)

), (7)

1∫0

dx

2x

(F

νp

2

(x,Q2) + F

νp

2

(x,Q2))

= 1

2πP

∞∫−∞

dα

α

2√

3

3

(√2A0(α,0) + A8(α,0)

), (8)

1∫0

dx

xF

ep

2

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

√3

9

(2√

2A0(α,0) + √3A3(α,0) + A8(α,0)

), (9)

1∫0

dx

xF en

2

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

√3

9

(2√

2A0(α,0) − √3A3(α,0) + A8(α,0)

), (10)

where σab means the total cross section of the ab scattering at q2 = 0, ν0 means the thresholdin each reaction, Up and Un are the contributions below threshold, and ν = p · q . In the non-forward case, the function Aa(α,0) contains two other components. One is the square of themomentum transfer t , and the other variable is fixed to 0, being the same as in the forward case.Hence the function Aa(α,0) has an extra parameter dependence corresponding to t . Becauseof the assumption that the divergence comes from the flavor singlet, this parameter dependenceeffectively comes from the A0(α,0) terms in Eqs. (5)–(10). The important point of the regular-ization lies in the fact that the rightmost singularity in the complex n plane or J plane is givenby the moving pole or cut such that the sum rules corresponding to sum rules (5)–(10) for somesmall t become convergent. The soft pomeron in Ref. [26] has a clear physical meaning and sat-isfies this criterion; moreover, the model may be realistic in the region below Q2 ∼ 1 GeV2 [27].Therefore, we explain the regularization with this model. However, the same kind of analysis canbe done for more complicated singularities, such as (ln 1/x)nx1−αP (0), where the intercept of thepomeron is αP (0) � 1. We will explain this fact later in this section.

Now, we assume that the leading high-energy behavior of the total cross sections of thepion (q)–nucleon (p) and the kaon (q)–nucleon (p) scattering is given by

{σπ+p(ν) + σπ−p(ν)

} ∼ βπN

(sπ

s0

)αP (0)−1

,

{σK+p(ν) + σK−p(ν)

} ∼ βKN

(sK

)αP (0)−1

, (11)
s0


where s0 = 1 GeV2, αP (0) = 1 + b − ε with b = 0.0808, sπ = m2π + m2

N + 2ν, and sK =m2

K + m2N + 2ν. It should be noted that the following discussion depends only on the fact b > 0.

The parameter ε in the sum rule in the forward direction corresponds to t in the non-forward sumrule. Then we rewrite the integral of the left-hand side of Eq. (5) as

∞∫νπ

0

dν

ν

{(σπ+p(ν) + σπ−p(ν)

) − βπNsb−επ

} +∞∫

νπ0

dν

νβπNsb−ε

π (12)

and the A0(α,0) term on the right-hand side of Eq. (5) as

P

∞∫−∞

dα

α

(A0(α,0) − f (α)

) + P

∞∫−∞

dα

αf (α), (13)

where the functions A0(α,0) and f (α) have a dependence on ε, and the term f (α) is the di-vergent piece when ε = b. From the above, the second term in Eq. (12) becomes (1/(ε − b) +ln(1/2νπ

0 ) + O(ε − b))βπN as ε → b. This pole term must be canceled by the second term inEq. (13). Therefore, we set the finite quantity in the ε → b limit to(

1

2π

2√

6

3P

∞∫−∞

dα

αf (α) − 2f 2

πβπN

π

1

ε − b

)ε→b

= 1

2π

2√

6

3f . (14)

Because the first term in Eq. (13) is finite, we can take the limit of it as ε → 0. Then we define(P

∞∫−∞

dα

α

(A0(α,0) − f (α)

))ε→0

+ f = P

∞∫−∞

dα

αAr

0(α,0), (15)

and obtain the regularized version of sum rule (5):

g2A(0) + 2f 2

π

π

∞∫νπ

0

dν

ν

(σπ+p(ν) + σπ−p(ν) − βπNsb

π

) + ln

(1

2νπ0

)

= 1

2πP

∞∫−∞

dα

α

2√

3

3

(√2Ar

0(α,0) + A8(α,0)). (16)

The important point is that the coefficients of the Ar0(α,0) and the A8(α,0) terms originating

from the symmetry on the right-hand side of sum rule (5) are unchanged. A similar analysis canbe performed for sum rules (6) and (7). We obtain(

1

2π

2√

6

3P

∞∫−∞

dα

αf (α) − 2f 2

KβKN

π

1

ε − b

)ε→b

= 1

2π

2√

6

3f , (17)

corresponding to Eq. (14), and the regularized versions of sum rules (6) and (7). Constraints (14)and (17) lead to the symmetry relation f 2

πβπN = f 2KβKN .

Let us now consider the regularization of sum rules (8)–(10). Because the simple pole structureof the integral f (α) is determined by Eq. (14) or Eq. (17), the left-hand sides of sum rules (8)–(10) must produce the same pole structure. Thus, without losing generality, we can set the small-xbehavior of the structure function to


{F

νp

2

(x,Q2) + F

νp

2

(x,Q2)} ∼ (

Q20/Q

2)αP (0)−1βνN

(Q2,1 − αP (0)

)(2ν)αP (0)−1, (18)

Fep

2

(x,Q2) ∼ (

Q20/Q

2)αP (0)−1βeN

(Q2,1 − αP (0)

)(2ν)αP (0)−1, (19)

with Q20 = 1 GeV2, where all ambiguities are absorbed in the definition of βlN with l = ν or e.

The βlN is expanded near ε ∼ b as

βlN

(Q2,1 − αP (0)

) ∼ β0lN

(Q2) − (ε − b)β1

lN

(Q2) + (ε − b)2β2

lN

(Q2) + O

((ε − b)3).

(20)

We first consider sum rule (9) and rewrite it as

1∫0

dx

x

{F

ep

2

(x,Q2) −

(1

Q2

)b−ε

βeN

(Q2, ε − b

)(2ν)b−ε

}

+1∫

0

dx

xxε−bβeN

(Q2, ε − b

). (21)

From the above, the second term in Eq. (21) is (β0eN (Q2)/(ε − b)−β1

eN (Q2)+O(ε − b)) in thelimit ε → b. Because the first term in Eq. (21) is finite, we can take its limit as ε → 0. Thus, asin Eq. (14), we obtain the constraint(

1

2π

2√

6

9P

∞∫−∞

f (α) − β0eN (Q2)

ε − b

)ε→b

= 1

2π

2√

6

9f , (22)

and the regularized sum rule

Cp =1∫

0

dx

x

{F

ep

2

(x,Q2) − x−bβeN

(Q2,−b

)} − β1eN

(Q2)

= 1

2πP

∞∫−∞

dα

α

√3

9

(2√

2Ar0(α,0) + √

3A3(α,0) + A8(α,0)). (23)

Constraints (14) and (22) lead to the symmetry relation 3πβ0eN (Q2) = 2f 2

πβπN . Thus, the β0eN

becomes Q2 independent. However, it should be noted that the subtracted term on the left-handside of sum rule (23) is x−bβeN(Q2,−b), not x−bβ0

eN .2 The term βeN(Q2,−b) is Q2 dependent.This Q2 dependence is not a phenomenon of the pole behavior at αP (0) > 1 but a reflection ofthe total cross section increase. To make this point clear, let us consider the case

Fep

2

(x,Q2) ∼ (

Q20/Q

2)αP (0)−1βeN

(Q2,1 − αP (0)

)(2ν)αP (0)−1(1 + η

(Q2) ln (1/x)

),

(24)

2 In previous papers, x−bβlN (Q2,−b) is erroneously set to x−bβ0lN

. In particular, in a phenomenological analysis in

Ref. [23], a Q2-dependent piece of the leading behavior in the sea-quark distribution function is missed. Its inclusionyields a more abundant charm sea quark in the small-x region at large Q2, as experiment suggests. However, the analysisof the regularization-independent sum rules is not affected.


with Q20 = 1 GeV2. By taking ε > b, we obtain

1∫0

dx

xβeN

(Q2, ε − b

)xε−b

(1 + η

(Q2) ln (1/x)

)

= η(Q2)β0eN (Q2)

(ε − b)2+ β0

eN (Q2) − η(Q2)β1eN (Q2)

ε − b+ O(1). (25)

Because the high-energy behavior of the πN total cross section, which has the same polestructure, is βπN(s/s0)

b−ε(1 + γ ln (s/s0)) with γ being a constant, we obtain the followingconditions from the sum rules:

η(Q2)β0

eN

(Q2) = A, β0

eN

(Q2) − η

(Q2)β1

eN

(Q2) = B, (26)

where A and B are Q2-independent constants. Thus we find that β0eN (Q2) is Q2 dependent.

After removing the pole contribution, we take the limit as ε → 0 and obtain

1∫0

{F

ep

2

(x,Q2) − βeN

(Q2, ε − b

)xε−b

(1 + η

(Q2) ln (1/x)

)}ε→0

=1∫

0

{F

ep

2

(x,Q2) − βeN

(Q2,−b

)x−b

(1 + η

(Q2) ln (1/x)

)}. (27)

Thus, for b > 0, we find that the leading behavior of the structure function F2(x,Q2) is Q2-dependent, as in the simple pole case. However, even when b = 0, because βeN(Q2,0) =β0

eN (Q2), the term that is constant with respect to s is Q2 dependent even though the coefficientof the (ln (1/x)) term is Q2 independent. The same holds if the compatibility of the (ln (1/x))2

term with unitarity is enforced. The coefficient of the (ln (1/x))2 term is Q2 independent, butthose of the (ln (1/x)) term and the constant term become Q2 dependent. Thus we obtain thatthe most leading term in the pomeron is Q2 independent, but the other terms in the pomeronis Q2 dependent. Therefore, in the small-x region, the structure function becomes large as Q2

increases, but its increase is weakened as we go to the small-x region. Though not exactly thesame, the saturation as we go to large Q2 in the small-x region has been known for a long time[28]. At large Q2, the strong coupling constant αS(Q2) becomes weak, but at the same time, theparton densities of, e.g., the gluons and the sea quarks become dense; therefore we must considerthe recombination of the partons, and the increase of the parton density is moderated. Here, non-perturbative physics, such as the color glass condensate, enters the picture [29]. The sum ruleapproach in general is consistent with such an approach.

Similar regularized sum rules can be derived from sum rule (10), yielding

Cn =1∫

0

dx

x

{Fen

2

(x,Q2) − x−bβeN

(Q2,−b

)} − β1eN

(Q2)

= 1

2πP

∞∫dα

α

√3

9

(2√

2Ar0(α,0) − √

3A3(α,0) + A8(α,0)), (28)

−∞


and from sum rule (8), yielding

1∫0

dx

2x

{F

νp

2

(x,Q2) + F

νp

2

(x,Q2) − x−bβνN

(Q2,−b

)} − 1

2β1

νN

(Q2)

= 1

2πP

∞∫−∞

dα

α

2√

3

3

(√2Ar

0(α,0) + A8(α,0)), (29)

with the symmetry relation πβ0νN(Q2) = 4f 2

πβπN . All the regularized sum rules are regular-ization dependent. These dependent terms are included in the Ar

0(α,0) and subtracted terms.However, they are constrained by the symmetry. This is because we assume that the diver-gent term comes from the pomeron being a flavor singlet, meaning that it comes from theP

∫ ∞−∞

dαα

A0(α,0) term. Therefore, we have the relation βνN(Q2,1 − αP (0)) = 6βeN(Q2,1 −αP (0)). Then if we take a suitable combination such that the Ar

0(α,0) term cancels out, all theregularization-dependent terms cancel. As is stated previously, this result does not depend onthe soft-pomeron assumption. It also holds in the case that αP (0) = 1 with unitarity-satisfyingbehavior. It can be applied as far as there exist some t regions where the singularity changes con-tinuously with respect to t such that the integral converges. Thus, sum rules where regularization-dependent terms cancel out can be considered to hold quite generally. The modified Gottfriedsum rule, given by Eq. (2), is one example of this type of sum rule. Sum rules concerning themean hypercharge and the mean charge of the light sea quarks are other examples of this type.

3. Flavor asymmetry of the light sea quarks in the nucleon

The symmetry relation f 2πβπN = f 2

KβKN is violated phenomenologically. The recent exper-imental value of the ratio of the pion decay constant to the kaon decay constant is f 2

π /f 2K ∼ 0.7

[30]. Hence, using the values βπN = 13.63 and βKN = 11.93 [26], we obtain (f 2π /f 2

K) ·(βπN/βKN) ∼ 0.8. In the old simple pomeron αP (0) = 1 case where the same symmetry relationholds, this value is about 0.85. The recent Particle Date Group review (PDG) fits the total crosssection as Zab + βab(ln (s/s0))

2 by setting βπN = βKN = β , where ab specifies the reactionand Zab is a constant in each reaction [30]. Because βab satisfies the same symmetry relation,we obtain (f 2

π /f 2K) · (βπN/βKN) ∼ 0.7. In all cases, the symmetry relation f 2

πβπN = f 2KβKN is

violated. To improve the situation, we consider the mixing

A0(α,0) = cA0(α,0) + sA8(α,0),

A8(α,0) = −sA0(α,0) + cA8(α,0) (30)

where c = cos θ and s = sin θ , and assume that the contribution to the pomeron comes from theA0(α,0) term. Then the symmetry relation becomes

f 2π

f 2K

βπN

βKN

= 2(√

2c + s)

2√

2c − s,

3β0eN

2√

2c + s= β0

νN

4(√

2c + s). (31)

In the following, we proceed to follow the soft pomeron model and set (f 2π /f 2

K) · (βπN/βKN) ∼0.8, where θ takes a value of θ ∼ −11.4◦. Because the modified Gottfried sum rule is relatedonly to A3(α,0), it is unaffected by this mixing [23]. The same behavior was found in the sumrule for a deuteron target [31]. However, the sum rules related to A8(α,0) changes because it


contains a divergent piece as a result of the mixing (30). Now, as explained in Refs. [23,20], theintegral of the quark distribution function corresponds to a suitable combination of the integralsof the Aa(α,0)s. For example, the right-hand side of sum rule (9) can be decomposed as follows:

1

2πP

∞∫−∞

dα

α

{8

9

(√6

6A0(α,0) + 1

2A3(α,0) +

√3

6A8(α,0)

)

+ 2

9

(√6

6A0(α,0) − 1

2A3(α,0) +

√3

6A8(α,0)

)

+ 2

9

(√6

6A0(α,0) −

√3

3A8(α,0)

)}. (32)

This decomposition corresponds exactly to the parton model. Because diag(1,0,0) =√

66 λ0 +

12λ3 +

√3

6 λ8, diag(0,1,0) =√

66 λ0 − 1

2λ3 +√

36 λ8, and diag(0,0,1) =

√6

6 λ0 −√

33 λ8, where

diag(a, b, c) is a flavor matrix whose diagonal elements are a, b, and c and whose other elementsare zero, and λi for i = 0,1, . . . ,8 is an SU(3) Gell-Mann matrix, we can write the integrals ofthe quark distribution functions as

1∫0

dx

{1

2uv

(x,Q2) + λu

(x,Q2)}

= 1

2πP

∞∫−∞

dα

α

(√6

6A0(α,0) + 1

2A3(α,0) +

√3

6A8(α,0)

),

1∫0

dx

{1

2dv

(x,Q2) + λd

(x,Q2)}

= 1

2πP

∞∫−∞

dα

α

(√6

6A0(α,0) − 1

2A3(α,0) +

√3

6A8(α,0)

),

1∫0

dx{λs

(x,Q2)} = 1

2πP

∞∫−∞

dα

α

(√6

6A0(α,0) −

√3

3A8(α,0)

), (33)

where uv(x,Q2) and dv(x,Q2) are the valence u and d quark distribution functions, respectively,and λi(x,Q2) for i = u,d, s is the i sea-quark distribution function with the anti-sea-quarkdistribution function being defined as λi(x,Q2) = λi(x,Q2). The regularized version of Eq. (33)can be obtained from sum rule (23):

1∫0

dx

{1

2uv

(x,Q2) + λu

(x,Q2) − 3

4x−bβeN

(Q2,−b

)} − 3

4β1

eN

(Q2)

= 1

2πP

∞∫dα

α

(√6

6Ar

0(α,0) + 1

2A3(α,0) +

√3

6A8(α,0)

),

−∞


1∫0

dx

{1

2dv

(x,Q2) + λd

(x,Q2) − 3

4x−bβeN

(Q2,−b

)} − 3

4β1

eN

(Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6Ar

0(α,0) − 1

2A3(α,0) +

√3

6A8(α,0)

),

1∫0

dx

{λs

(x,Q2) − 3

4x−bβeN

(Q2,−b

)} − 3

4β1

eN

(Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6Ar

0(α,0) −√

3

3A8(α,0)

). (34)

We see that the Q2 dependence, which enters through the integral of the regularized sea-quarkdistribution function, is just canceled by the β1

eN (Q2) term. Now, we rewrite the right-hand sidesof Eq. (33) using the mixing relation (30), and by comparing the coefficients of the integrals ofthe A0(α,0) terms, we find that the sea-quark distribution functions should be defined as

λu

(x,Q2) =

(c + s√

2

)λ3

u

(x,Q2),

λd

(x,Q2) =

(c + s√

2

)λ3

d

(x,Q2),

λs

(x,Q2) = (c − √

2s)λ3s

(x,Q2), (35)

where the suffix 3 means SU(3), and λ3u(x,Q2), λ3

d(x,Q2), λ3s (x,Q2) have symmetrical limits:

limx→0

xαP (0)λ3u

(x,Q2) = lim

x→0xαP (0)λ3

d

(x,Q2) = lim

x→0xαP (0)λ3

s

(x,Q2).

The same result can be obtained if we directly rewrite sum rules (5)–(10) using mixing relation(30) and regularize them. However, this discussion is necessary to illustrate convincingly thereason why relation (35) holds. By using the fact that Eq. (8) and the sum of Eqs. (9) and (10)are expressed only in terms of A0(α,0) and A8(α,0), we can rewrite them with mixing rela-tion (30). Then, by taking a suitable combination to cancel out the A0(α,0) contribution, wecan extract A8(α,0). By expressing the structure functions with quark distribution functions andusing Eq. (35), we obtain

1

2πP

∞∫−∞

dα

αA8(α,0)

= 3

2√

6(√

2c − 2s) + 1√3(c − √

2s)

(c + s√

2

)

×1∫dx

{λ3

u

(x,Q2) + λ3

d

(x,Q2) − 2λ3

s

(x,Q2)}. (36)

0


On the other hand, Eq. (5) and the sum of Eqs. (6) and (7) are also expressed only in terms ofA0(α,0) and A8(α,0). Then, by using the same reasoning as we used to reach Eq. (36), we canextract A8(α,0) as

1

2πP

∞∫−∞

dα

αA8(α,0) = 1

2√

6

{(2

√2c − s)Iπ − (

√2c + s)

(I

pK + In

K

)}, (37)

where Iπ is assumed to be smoothly continued from the off-shell quantities at q2 = 0 given inEq. (16) to the on-shell quantities,

Iπ = g2A(0) + 2f 2

π

π

∞∫νπ

0

dν

ν2

{√ν2 − m2

Nm2π

[σπ+p(ν) + σπ−p(ν)

] − νsbβπN

}

+ 2f 2πβπN

πln

(1

2νπ0

)

= 1

2πP

∞∫−∞

dα

α

2√

3

3

(√2Ar

0(α,0) + A8(α,0)), (38)

and IpK and In

K are given by the regularized sum rule similar to Eq. (16) by assuming smoothcontinuation to the on-shell quantities and are rewritten by using the Adler–Weisberger sumrules for the kaon:

IpK = 2 + 2f 2

K

π

∞∫νK

0

dν

ν2

{2√

ν2 − m2Nm2

KσK+p(ν) − νsbβKN

}

+ 2f 2KβKN

πln

(1

2νK0

)

= 1

2πP

∞∫−∞

dα

α

(2√

6

3Ar

0(α,0) + A3(α,0) −√

3

3A8(α,0)

), (39)

InK = 1 + 2f 2

K

π

∞∫νK

0

dν

ν2

{2√

ν2 − m2Nm2

KσK+n(ν) − νsbβKN

}

+ 2f 2KβKN

πln

(1

2νK0

)

= 1

2πP

∞∫−∞

dα

α

(2√

6

3Ar

0(α,0) − A3(α,0) −√

3

3A8(α,0)

). (40)

Here νπ0 = mπmN , νK

0 = mKmN , and σab means the on-shell total cross section of ab scattering.Using the experimental values of the total cross section of the pion–nucleon and that of the kaon–nucleon, Iπ , I

pK , and In

K were determined to be Iπ ∼ 5.17, IpK ∼ 2.39, In

K ∼ 1.61 [23]. Eq. (37)gives us numerical information about A8(α,0) which, combined with that of A3(α,0) through


the modified Gottfried sum rule, makes it possible to study sea quarks in the octet baryon. In thenucleon case, from Eqs. (36) and (37), we obtain

1

3

1∫0

dx{λ3

u

(x,Q2) + λ3

d

(x,Q2) − 2λ3

s

(x,Q2)} ∼ 0.59. (41)

This is a refinement of the mean hypercharge of the light sea quark in the nucleon in the presenceof the flavor asymmetry of the total cross sections of the pseudo-scalar nucleon scattering athigh-energy. It is interesting to note that the value in Eq. (41) and that of the refined mean chargeof the light sea quark in the nucleon, which is given as

1

3

1∫0

dx{2λ3

u

(x,Q2) − λ3

d

(x,Q2) − λ3

s

(x,Q2)} ∼ 0.23, (42)

are almost the same as in the θ = 0◦ case, which has been studied in Ref. [32]. The phe-nomenological sea-quark distributions defined by Eq. (35) show that the strange sea-quarkdistribution becomes larger than the up and down sea-quark distributions in the very-small-xregion. In the region where λ3

u(x,Q2) ∼ λ3d(x,Q2) ∼ λ3

s (x,Q2), this discrepancy results inλs(x,Q2)/λu(x,Q2) ∼ 1.5. To achieve this, because the strange sea quark is suppressed in thelarge- and the medium-x regions above x = 0.01, the behavior of the strange sea-quark distribu-tion at small x is rapid and very different from those of the up and down sea quarks. A numericalstudy in Ref. [32], based on the sum rule at θ = 0◦ corresponding to sum rule (42), suggests that,at Q2 = 1 GeV2, such a phenomenon will be seen near the region from x = 10−4 to x = 10−6.

4. Flavor asymmetry of the light sea quarks in the octet baryon

Now let us consider how the mixing affects the sum rules in the octet baryon [20]. The matrixelements of the octet baryon can be decomposed as(

Ac(p · x,0))αβ

= ifαcβF (p · x,0) + dαcβD(p · x,0), (43)

where α,β are the symmetry indices specifying the octet hadronic state. For the octet baryon,B , we use the notation AB

c (p · x,0); however, for the proton, we continue to use the notationAc(p · x,0). Then, because A3(p · x,0) = F(p · x,0)+D(p · x,0) and A8(p · x,0) = √

3(F (p ·x,0) − D(p · x,0)/3), we obtain

F(p · x,0) = 1

4

(A3(p · x,0) + √

3sA0(p · x,0) + √3cA8(p · x,0)

)(44)

and

D(p · x,0) = 1

4

(3A3(p · x,0) − √

3sA0(p · x,0) − √3cA8(p · x,0)

). (45)

We see that the integrals∫ ∞−∞

dαα

F and∫ ∞−∞

dαα

D diverge because they each include an A0(p ·x,0) term. This means that even the sum rule of the mean I3 might be modified for the octetbaryon, except for the nucleon. For example, let us consider the Σ baryon. In this case, we have

AΣ+3 (p · x,0) = F(p · x,0), AΣ+

8 (p · x,0) = 2√

3D(p · x,0). (46)

3


Because the singlet satisfies AΣ±0 (p · x,0) = A0(p · x,0) the sum rules for the structure function

FeΣ±2 can be transformed as follows:

1∫0

dx

xF eΣ+

2

(x,Q2)

= 1

18πP

∞∫−∞

dα

α

{2√

6AΣ+0 (α,0) + 3AΣ+

3 (α,0) + √3AΣ+

8 (α,0)}

= 1

18πP

∞∫−∞

dα

α

{(2√

6c +√

3

4s

)A0(α,0) + 9

4A3(α,0)

−(

2√

6s −√

3

4c

)A8(α,0)

}, (47)

1∫0

dx

xF eΣ−

2

(x,Q2)

= 1

18πP

∞∫−∞

dα

α

{2√

6AΣ+0 (α,0) − 3AΣ+

3 (α,0) + √3AΣ+

8 (α,0)}

= 1

18πP

∞∫−∞

dα

α

{(2√

6c − 5√

3

4s

)A0(α,0) + 3

4A3(α,0)

−(

2√

6s + 5√

3

4c

)A8(α,0)

}. (48)

Thus we obtain

1∫0

dx

x

{4FeΣ+

2 (x,Q2)

8√

6c + √3s

− 4FeΣ−2 (x,Q2)

8√

6c − 5√

3s

}

= 8

3π

1

(8√

6c + √3s)(8

√6c − 5

√3s)

× P

∞∫−∞

dα

α

{√3(

√2c − s)A3(α,0) + 3

√2A8(α,0)

}, (49)

with the condition

β0eΣ−

β0eΣ+

= 8√

6c − 5√

3s

8√

6c + √3s

, (50)

where β0eΣ± is defined analogously to Eq. (20). The modified Gottfried sum rule is equivalent to

the relation


Table 1The symmetry-breaking factors of the light sea quarks.

B λBu /λ3B

u λBd

/λ3Bd

λBs /λ3B

s

p c + 1√2s c + 1√

2s c − √

2s

n c + 1√2s c + 1√

2s c − √

2s

Σ+ c +√

28 s c − 5

√2

8 s c +√

22 s

Σ0 c −√

24 s c −

√2

4 s c +√

22 s

Σ− c − 5√

28 s c +

√2

8 s c +√

22 s

Ξ− c − √2s c +

√2

2 s c +√

22 s

Ξ0 c +√

22 s c − √

2s c +√

22 s

Λ0 c +√

24 s c +

√2

4 s c −√

22 s

1

2πP

∞∫−∞

dα

αA3(α,0) = 3

2(Cp − Cn) = 1

2

(I

pK − In

K

). (51)

Then, from Eq. (37) and θ ∼ −11.4◦, we find that the right-hand side of Eq. (49) is about 0.13and that the left-hand side is the integral of the difference {FeΣ+

2 /4.7 − FeΣ−2 /5.2}/x . We also

find that the right-hand side of Eq. (50) is about 1.1. By the same kind of argument used to reachEq. (35), we find that the sea-quark distributions should be defined as

λΣ+u

(x,Q2) =

(c +

√2s

8

)λ3Σ+

u

(x,Q2),

λΣ+d

(x,Q2) =

(c − 5

√2s

8

)λ3Σ+

d

(x,Q2),

λΣ+s

(x,Q2) =

(c +

√2s

2

)λ3Σ+

s

(x,Q2), (52)

where λ3Σ+i (x,Q2) for i = u,d, s is defined similarly as in the proton case. By using sum rule

(49) and the fact that F + D = AΣ+3 (α,0) + √

3AΣ+8 (α,0)/2 = A3(α,0), we find

1

3

1∫0

dx{λ3Σ+

u

(x,Q2) + λ3Σ+

d

(x,Q2) − 2λ3Σ+

s

(x,Q2)} ∼ −0.42, (53)

1

3

1∫0

dx{2λ3Σ+

u

(x,Q2) − λ3Σ+

d

(x,Q2) − λ3Σ+

s

(x,Q2)} ∼ −0.25. (54)

These are very similar relations to the ones in the θ = 0◦ case [20]. Similar statements hold for theother octet baryon. We give symmetry-breaking factors of the light sea quarks in the octet baryonin Table 1. Furthermore, in Table 2, we give numerical values of the mean quantum numbers ofthe symmetric sea-quark distribution λ3B

i for the mixing angle θ = −11.4◦; we also provide thevalue for the θ = 0◦ case in parentheses. The analytical results are summarized in Appendix A.


Table 2The mean quantum number of the light sea quarks. The main reported value is for θ = −11.4◦ , while the value inparentheses corresponds to the θ = 0◦ case.

B 〈I3〉〈Y 〉〈Q〉12

∫ 10 dx {λ3B

u − λ3Bd

} 13

∫ 10 dx {λ3B

u + λ3Bd

− 2λ3Bs } 1

3

∫ 10 dx {2λ3B

u − λ3Bd

− λ3Bs }

p −0.065 (−0.055) 0.59 (0.56) 0.23 (0.23)

n 0.065 (0.055) 0.59 (0.56) 0.36 (0.34)

Σ+ −0.043 (−0.054) −0.42 (−0.34) −0.25 (−0.22)

Σ0 0 (0) −0.41 (−0.34) −0.21 (−0.17)

Σ− 0.043 (0.054) −0.42 (−0.34) −0.16 (−0.11)

Ξ− −0.47 (−0.45) −0.23 (−0.23) −0.59 (−0.56)

Ξ0 0.47 (0.45) −0.23 (−0.23) 0.36 (0.34)

Λ0 0 (0) 0.36 (0.34) 0.18 (0.17)

5. Effect of the charm sea quark

Now let us take into account the charm degree of freedom. In this case, the flavor group is theSU(4) ⊗ SU(4), and we take the mixing matrix to be(

A0(α,0)

A8(α,0)

A15(α,0)

)=

(c2 0 −s20 1 0s2 0 c2

)(c1 −s1 0s1 c1 00 0 1

)(A0(α,0)

A8(α,0)

A15(α,0)

), (55)

where c1 = cos θ1, s1 = sin θ1, c2 = cos θ2, and s2 = sin θ2. Then, because Eq. (43) can be usedfor the octet baryon, we obtain A15(α,0) = √

6D/3. By assuming that the contribution to thepomeron comes from the A0(α,0) term in the sum rules for the SU(4) case given in Ref. [23],we obtain the symmetry relations

f 2π

f 2K

βπN

βKN

=√

2c2c1 + 2√

33 s1 +

√6

3 s2c1√2c2c1 −

√3

3 s1 +√

63 s2c1

,

βeN

5√

2c1c2 + √3s1 − √

6c1s2= βνN

36√

2c1c2, (56)

and

πβ0νN = 8f 2

π

√2c1c2√

2c1c2 + 2√

33 s1 +

√6

3 c1s2

· βπN, (57)

with the condition s1 = −2√

2c1s2. Hence, the angles θ1 and θ2 are not independent. By usingthe same kind of discussion used to reach Eq. (35), we consider the coefficient of A0(α,0) forthe flavor matrices

diag(1,0,0,0), diag(0,1,0,0), diag(0,0,0,1), and diag(0,0,0,1),

and we find that we should define the sea quark distributions as

λu

(x,Q2) =

(c1c2 +

√6s1

)λ4

u

(x,Q2),

4


λd

(x,Q2) =

(c1c2 +

√6

4s1

)λ4

d

(x,Q2),

λs

(x,Q2) =

(c1c2 − 3

√6

4s1

)λ4

s

(x,Q2),

λc

(x,Q2) =

(c1c2 +

√6

4s1

)λ4

c

(x,Q2), (58)

where λ4u(x,Q2), λ4

d(x,Q2), λ4s (x,Q2), and λ4

c(x,Q2) have symmetrical limits:

limx→0

xαP (0)λ4u

(x,Q2)

= limx→0

xαP (0)λ4d

(x,Q2) = lim

x→0xαP (0)λ4

s

(x,Q2) = lim

x→0xαP (0)λ4

c

(x,Q2).

Using the same value of (f 2π /f 2

K) · (βπN/βKN) ∼ 0.8 as in the SU(3) case, we obtain θ1 ∼−10.3◦ and θ2 ∼ 3.67◦. In the region where λ4

u(x,Q2) ∼ λ4d(x,Q2) ∼ λ4

s (x,Q2) ∼ λ4c(x,Q2),

we find λs(x,Q2)/λu(x,Q2) ∼ 1.5. The fact that θ2 is small means the contamination from theA15(α,0) to the pomeron is very small. Then, from the sum rules in the SU(4) case, we obtain

1∫0

dx{λ4

u

(x,Q2) − λ4

d

(x,Q2)} = −0.126. (59)

Compared with the value in the SU(3) case, given as

1∫0

dx{λ3

u

(x,Q2) − λ3

d

(x,Q2)} = −0.131, (60)

we find that the symmetric sea-quark distribution in the SU(4) case is scaled by about0.131/0.126 ∼ 1.04 with respect to that in the SU(3) case. In fact, from Eqs. (35) and (58),we find

c1c2 +√

64 s1

c + s√2

∼ c1c2 − 3√

64 s1

c − √2s

∼ 1.04. (61)

Thus, from the sum rule in the SU(3) case, which is given as

1∫0

dx{λ3

u

(x,Q2) − λ3

s

(x,Q2)} = 0.81, (62)

we obtain

1∫0

dx{λ4

u

(x,Q2) − λ4

s

(x,Q2)} = 0.78, (63)

by multiplying by the scale factor. Using this expression, we obtain the following description ofthe charm sea quark distribution from the sum rule in the SU(4) case:


1∫0

dx{λ4

c

(x,Q2) − λ4

s

(x,Q2)} = 1.48. (64)

This value is again similar to the θ = 0◦ case.3 The charm sea quark is suppressed in the regionabove x = 0.01 due to the mass effect; the strange sea quark is also suppressed in about the sameregion. Both quark distribution functions become large as we go to the small-x region. However,because the strange sea quark is larger than the charm sea quark in the very-small-x region,compared with the strange sea quark, the charm sea quark must be enhanced in the small-xregion to satisfy Eq. (64).

6. Conclusions

The symmetry relation necessary to derive the modified Gottfried sum rule is violated phe-nomenologically. Thus we introduce the mixing for the connected matrix element of the anti-symmetric bilocal current. With the mixing between the singlet and the eighth member of theoctet, we can modify the symmetry relation such that it is well-satisfied phenomenologically.Because the modified Gottfried sum rule is given by the third member of the octet, its form isunchanged under this mixing. However, the sum rules related to the eighth member of the octetare changed. In this paper, we first derive these modified sum rules related to the eighth memberof the octet. Then, we find that the phenomenological sea-quark distribution functions should bedefined such that the symmetric ones in the very-small-x region are multiplied by the symmetry-breaking factors, which depend on the mixing angle. In this way, we find, by using a typicalmixing angle determined phenomenologically, that in the very-small-x region, the ratio of thestrange sea-quark distribution function to the one for the up sea-quark, λs(x,Q2)/λu(x,Q2), isabout 1.5, while the ratio λd(x,Q2)/λu(x,Q2) is 1. This fact is unchanged if we take the charmsea quark into account. Thus, the strange sea quark is most abundant in the very-small-x region.We also find that the ratio λs(x,Q2)/λc(x,Q2) is 1.5 in the very-small-x region. Thus, becausethe charm sea quark is suppressed above x = 0.01 due to the mass effect, the abundance of thecharm sea quark must be enhanced in the small-x region to satisfy the sum rule expressed byEq. (64). Finally, by using the sum rule for the modified eighth member of the octet togetherwith the modified Gottfried sum rule, we extend the analysis to the octet baryon, and derivethe sum rules for the sea quarks in the octet baryon together with their symmetry-breaking fac-tors. We find that the regularization-independent sum rules, which have physical meanings of themean I3, Y , and Q of the symmetric light sea quarks, take similar values as in the case where themixing angle is zero.

Appendix A. Light sea quarks of the octet baryon

The sum rules for the sea quarks of each baryon can be obtained directly in the following way.By the same reasoning used to reach Eq. (33), we obtain the relations

3 The values 0.78 and 1.48 obtained from a simple scale-factor argument are roughly the same as the ones obtaineddirectly by the sum rules in SU(4), i.e., Eqs. (B.8) and (B.9), which yield 0.74 and 1.44 respectively for the explicitvalues of θ1 and θ2 in the text.


1∫0

dx

{1

2uB

v

(x,Q2) + λB

u

(x,Q2)}

= 1

2πP

∞∫−∞

dα

α

(√6

6AB

0 (α,0) + 1

2AB

3 (α,0) +√

3

6AB

8 (α,0)

),

1∫0

dx

{1

2dBv

(x,Q2) + λB

d

(x,Q2)}

= 1

2πP

∞∫−∞

dα

α

(√6

6AB

0 (α,0) − 1

2AB

3 (α,0) +√

3

6AB

8 (α,0)

),

1∫0

dx

{1

2sBv

(x,Q2) + λB

s

(x,Q2)}

= 1

2πP

∞∫−∞

dα

α

(√6

6AB

0 (α,0) −√

3

3AB

8 (α,0)

), (A.1)

where the valence-quark distribution is non-zero only when it reproduces the quantum numberof the baryon. For example, for the Σ+ baryon, we set

1∫0

dx uΣ+v

(x,Q2) = 2,

1∫0

dx dΣ+v

(x,Q2) = 0,

1∫0

dx sΣ+v

(x,Q2) = 1. (A.2)

Further, the singlet satisfies AB0 (α,0) = A0(α,0). Here we give the analytical results for

Σ±,Σ0,Λ0,Ξ−,Ξ0, because the ones for the nucleon are already given in Eqs. (36), (37),and (51).

A.1. Σ+ baryon

By using relation (46), we rewrite the right-hand side of Eq. (A.1) in terms of A0(α,0),F(α,0), and D(α,0). Then, with use of relations (44) and (45), we express it in terms of A0(α,0),A8(α,0), and A3(α,0) and obtain the relations

1 +1∫

0

dx λΣ+u

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6

(c +

√2s

8

)A0(α,0) + 3

8A3(α,0)

+√

3(c − 4

√2s)A8(α,0)

),

24


1∫0

dx λΣ+d

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6

(c − 5

√2s

8

)A0(α,0) + 1

8A3(α,0)

− 5√

3

24

(c + 4

√2s

5

)A8(α,0)

),

1

2+

1∫0

dx λΣ+s

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6

(c + s√

2

)A0(α,0) − 1

2A3(α,0) +

√3

6(c − √

2s)A8(α,0)

).

(A.3)

The regularization-independent sum rules can be obtained by taking suitable combinations suchthat the A0(α,0) term cancels out. Thus we take the combinations

(c − 5

√2s

8

)(1 +

1∫0

dx λΣ+u

(x,Q2)) −

(c +

√2s

8

) 1∫0

dx λΣ+d

(x,Q2) (A.4)

and (c + s√

2

) 1∫0

dx λΣ+d

(x,Q2) −

(c − 5

√2s

8

)(1

2+

1∫0

dx λΣ+s

(x,Q2)). (A.5)

We define a3 and a8 as

a3 = 1

2πP

∞∫−∞

dα

αA3(α,0), a8 = 1

2πP

∞∫−∞

dα

αA8(α,0), (A.6)

where their values are determined by Eqs. (37) and (51) through the estimates of Iπ , IpK , and In

K

from Eqs. (38)–(40). In this way, we find the sum rules for the sea quarks to be

(c − 5

√2s

8

)(c +

√2s

8

) 1∫0

dx{λ3Σ+

u

(x,Q2) − λ3Σ+

d

(x,Q2)}

= −(

c − 5√

2s

8

)+ 1

4

((c − √

2s)a3 + √3a8

),

(c − 5

√2s

8

)(c + s√

2

) 1∫0

dx{λ3Σ+

d

(x,Q2) − λ3Σ+

s

(x,Q2)}

= 1(

c − 5√

2s)

+ 1((5c − 2

√2s)a3 − 3

√3a8

). (A.7)

2 8 8


A.2. Σ− baryon

The sum rules of the sea quarks for the Σ− baryon can be obtained by setting λΣ−u = λΣ+

d ,

λΣ−d = λΣ+

u , and λΣ−s = λΣ+

s .

A.3. Σ0 baryon

In this case, AΣ0

8 (α,0) = 2√

33 D(α,0) and AΣ0

3 (α,0) = 0. Thus we obtain

1

2+

1∫0

dx λΣ0

u

(x,Q2)

= 1

2+

1∫0

dx λΣ0

d

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6

(c −

√2s

4

)A0(α,0) + 1

4A3(α,0) −

√3

12(c + 2

√2s)A8(α,0)

),

1

2+

1∫0

dx λΣ0

s

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6

(c + s√

2

)A0(α,0) − 1

2A3(α,0) +

√3

6(c − √

2s)A8(α,0)

).

(A.8)

Then, by taking the combination

(c + s√

2

)(1

2+

1∫0

dx λΣ0

d

(x,Q2)) −

(c −

√2s

4

)(1

2+

1∫0

dx λΣ0

s

(x,Q2)), (A.9)

we obtain

(c + s√

2

)(c −

√2s

4

) 1∫0

dx{λ3Σ0

d

(x,Q2) − λ3Σ0

s

(x,Q2)}

= −3√

2s

8+ 1

4(3ca3 − √

3a8), (A.10)

together with

1∫0

dx{λ3Σ0

u

(x,Q2) − λ3Σ0

d

(x,Q2)} = 0. (A.11)


A.4. Λ0 baryon

In this case, AΛ0

8 (α,0) = − 2√

33 D(α,0), and AΛ0

3 (α,0) = 0. Thus we obtain

1

2+

1∫0

dx λΛ0

u

(x,Q2)

= 1

2+

1∫0

dx λΛ0

d

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6

(c +

√2s

4

)A0(α,0) − 1

4A3(α,0) +

√3

12(c − 2

√2s)A8(α,0)

),

1

2+

1∫0

dx λΛ0

s

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6

(c − s√

2

)A0(α,0) + 1

2A3(α,0) −

√3

6(c + √

2s)A8(α,0)

).

(A.12)

Then, by taking the combination

(c − s√

2

)(1

2+

1∫0

dx λΛ0

d

(x,Q2)) −

(c +

√2s

4

)(1

2+

1∫0

dx λΛ0

s

(x,Q2)), (A.13)

we obtain

(c − s√

2

)(c +

√2s

4

) 1∫0

dx{λ3Λ0

d

(x,Q2) − λ3Λ0

s

(x,Q2)}

= 3√

2s

8− 1

4(3ca3 − √

3a8), (A.14)

together with

1∫0

dx{λ3Λ0

u

(x,Q2) − λ3Λ0

d

(x,Q2)} = 0. (A.15)

A.5. Ξ− baryon

In this case, AΞ−3 (α,0) = −F(α,0) + D(α,0) and AΞ−

8 (α,0) = −√3F(α,0) −

√3

3 D(α,0).Thus we obtain


1∫0

dx λΞ−u

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6(c − √

2s)A0(α,0) −√

3

3

(c +

√2s

2

)A8(α,0)

),

1

2+

1∫0

dx λΞ−d

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6

(c +

√2s

2

)A0(α,0) − 1

2A3(α,0) +

√3

6(c − √

2s)A8(α,0)

),

1 +1∫

0

dx λΞ−s

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√6

6

(c + s√

2

)A0(α,0) + 1

2A3(α,0) +

√3

6(c − √

2s)A8(α,0)

).

(A.16)

Then, by taking the combinations

1∫0

dx λΞ−d

(x,Q2) −

1∫0

dx λΞ−s

(x,Q2) (A.17)

and (c + s√

2

) 1∫0

dx λΞ−u

(x,Q2) − (c − √

2s)

(1

2+

1∫0

dx λΞ−d

(x,Q2)), (A.18)

we obtain(c + s√

2

) 1∫0

dx{λΞ−

d

(x,Q2) − λΞ−

s

(x,Q2)} = 1

2− a3, (A.19)

(c + s√

2

)(c − √

2s)

1∫0

dx{λΞ−

u

(x,Q2) − λΞ−

d

(x,Q2)}

= 1

2

((c − √

2s) + (c − √2s)a3 − √

3a8). (A.20)

A.6. Ξ0 baryon

The sum rules of the sea quarks for the Ξ0 baryon can be obtained by setting λΞ0

u = λΞ−d ,

λΞ0 = λΞ−u , and λΞ0

s = λΞ−s .
d


Appendix B. Charm sea quark

Let us first consider the condition A15(α,0) = √6D/3 obtained from Eq. (43). Using the

relations A3(α,0) = F + D and A8(α,0) = √3(F − D/3), we obtain the relation

A3(α,0) −√

3

3A8(α,0) = 2

√6

3A15(α,0). (B.1)

Using mixing relation (55), we rewrite Eq. (B.1) under the condition s1 = −2√

2c1s2. Then, wefind that the A0(α,0) term cancels out from both sides of Eq. (B.1) and obtain

A3(α,0) +√

3

3(2

√2s1s2 − c1)A8(α,0) = 2

√6c2

3A15(α,0). (B.2)

By defining

a15 = 1

2πP

∞∫−∞

dα

αA15(α,0), (B.3)

we obtain

a15 =√

2

4c2

(√3a3 + (2

√2s1s2 − c1)a8

). (B.4)

Now, by the same reasoning used to reach Eq. (33), we obtain the sea quarks in the nucleon inthe SU(4) case:

1∫0

dx

{1

2uv

(x,Q2) + λu

(x,Q2)}

= 1

2πP

∞∫−∞

dα

α

(√2

4A0(α,0) + 1

2A3(α,0) +

√3

6A8(α,0) +

√6

12A15(α,0)

),

1∫0

dx

{1

2dv

(x,Q2) + λd

(x,Q2)}

= 1

2πP

∞∫−∞

dα

α

(√2

4A0(α,0) − 1

2A3(α,0) +

√3

6A8(α,0) +

√6

12A15(α,0)

),

1∫0

dx λs

(x,Q2) = 1

2πP

∞∫−∞

dα

α

(√2

4A0(α,0) −

√3

3A8(α,0) +

√6

12A15(α,0)

),

1∫0

dx λc

(x,Q2) = 1

2πP

∞∫−∞

dα

α

(√2

4A0(α,0) −

√6

4A15(α,0)

). (B.5)


By using the mixing relation, we obtain

1 +1∫

0

dx λu

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√2

4

(c1c2 +

√6s1

4

)A0(α,0) + 1

2A3(α,0)

−√

2

4

(s1c2 −

√6

3c1 +

√3

3s1s2

)A8(α,0) +

√2

4

(−s2 +

√3

3c2

)A15(α,0)

),

1

2+

1∫0

dx λd

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√2

4

(c1c2 +

√6s1

4

)A0(α,0) − 1

2A3(α,0)

−√

2

4

(s1c2 −

√6

3c1 +

√3

3s1s2

)A8(α,0) +

√2

4

(−s2 +

√3

3c2

)A15(α,0)),

1∫0

dx λs

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√2

4

(c1c2 − 3

√6s1

4

)A0(α,0)

−√

2

4

(s1c2 + 2

√6

3c1 +

√3

3s1s2

)A8(α,0) +

√2

4

(−s2 +

√3

3c2

)A15(α,0)

),

1∫0

dx λc

(x,Q2)

= 1

2πP

∞∫−∞

dα

α

(√2

4

(c1c2 +

√6s1

4

)A0(α,0)

−√

2

4(s1c2 − √

3s1s2)A8(α,0) −√

2

4(s2 + √

3c2)A15(α,0)

). (B.6)

Then, by taking suitable combinations such that the A0(α,0) term cancels out, we obtain theregularization-independent sum rules as follows:

(c1c2 +

√6

4s1

) 1∫dx

{λ4

u

(x,Q2) − λ4

d

(x,Q2)} = −1

2+ a3, (B.7)

0


(c1c2 − 3

√6

4s1

)(c1c2 +

√6

4s1

) 1∫0

dx{λ4

u

(x,Q2) − λ4

s

(x,Q2)}

= −(

c1c2 − 3√

6

4s1

)(1 − a3

2

)+ 1

2(√

3c2 + s2)a8

−√

3

2s1

(−s2 +

√3

3c2

)a15, (B.8)

(c1c2 +

√6

4s1

) 1∫0

dx{λ4

u

(x,Q2) − λ4

c

(x,Q2)}

= −1 + 1

2a3 +

√3

6(c1 − 2

√2s1s2)a8 +

√6

3c2a15. (B.9)

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Documents

Abundant strange sea quarks in the nucleon and flavor asymmetry of the light sea quarks in the octet baryon at small x