Possible Relationship between Sea Quarks Distribution Functions and Valence Distribution Functions in Proton

Introduction

After Gell-Mann succeeded to predict the existence of ${\mathrm{\Omega }}^{-}$ by estimating its mass by his eight-fold way model [1], people had believed that baryons including nucleons are consisted by three quarks. However, New Muon Collaboration (NMC) [2], NA51 Collaboration [3] and E866 Collaboration [4] reported ${\overline{d}}^{-}\overline{u}$ asymmetry in proton. These results has stimulated people’s concern for the structure of proton. As results, H1 and Zeus (DESY) [5] and Atlas (CERN) [6] have reported sea quarks (quark-antiquark system) distribution functions, valence quark distribution functions and gluon distribution functions in proton. These results show that proton has more profound structure than that composed by simple three quarks. H1 and Zeus results are obtained by electron-proton collision, while Atlas results are obtained by proton-proton collision. Especially, Atlas results show the following very interesting properties. Besides they found $s\overline{s}$ sea quarks distribution functions denoted as $x\overline{s}$, $x\overline{d}$ value at $x=1\times {10}^{-3}$ looks like equal to the peak value of $x\overline{d}$ and $x\overline{u}$ value at $x=1\times {10}^{-3}$ looks like one half of the peak value of $xu$. In addition, $x\overline{d}$ value at $x=1\times {10}^{-3}$ looks like equal to $x\overline{u}$ value at $x=1\times {10}^{-3}$. Note that Atlas group denote each sea quarks distribution functions as $xu$, $xd$ and $xs$, respectively, and also denote valence u quark and valence d quark distribution functions as $xu$ and $xd$, respectively [6]. These properties indicate that it is possible or rather probable that there are some relationship between sea quarks distribution functions and valence quark distribution functions. In Atlas experiment, there is no valence s quark distribution functions. We try to explain the possibility of existence of sea quark distribution functions and valence u and d quark distribution functions. Pions are composed of $u\overline{d}$, $u\overline{u}$ and $d\overline{d}$ only. Thus, we use pion distribution functions as basis. In this paper, we attempt to show behavior of $\overline{u}$ and $\overline{d}$ sea quarks distribution functions and that of valence u quark and valence d quark distribution functions starting from our charged pion distribution amplitude and functions shown in ref. [7] and ref. [8], respectively. Even in Lattice QCD calculation, recently, Francis et al. shows asymptotic limit of charged pion distribution function is close to 1, ie, $(1-x{)}^{\beta },\beta \approx 1$ [9]. This supports our asymptotic form of charged pion distribution functions.

Formulation

We proposed for description of baryons as composition of bound and unbound state of sea quarks [10]. By following those descriptions, proton is described as composition of a charged pion ${\pi }^{+}(u\overline{d})$ and a neutral pion ${\pi }^0$ (mix of $u\overline{u}$ and $d\overline{d}$ $((u\overline{u}-d\overline{d})/\sqrt{2}))$) besides s unbound sea quarks. Important point is that ${\pi }^{+}$ and ${\pi }^0$ can take either bound state or unbound state repeatedly. For an example of numerical calculation, we showed proton and neutron electromagnetic form factors using pion pair consideration [11]. Note that we used ${\pi }^{+}$ wave function for ${\pi }^0$ wave function in proton case in that paper because we had not obtained ${\pi }^0$ wave function yet at that paper’s publication date. In addition, we obtained a charged pion distribution amplitude [7] and its distribution function [8]. Using those arguments, we attempt to explain the behavior of Atlas experiment results [6].

Although we proposed that proton is composed as ${\pi }^{+}$ and ${\pi }^0$ besides $\mathsf{s}\overline{\mathit{s}}$ in ref. [10], in reality, the distribution functions of $\mathrm{u}\overline{d}$ , uu and $\mathrm{d}\overline{d}$ should be slightly different from those of ${\pi }^{+}$ and ${\pi }^0$ because there is no evidence that shows to exist exact ${\pi }^{+}$ and ${\pi }^0$ in proton. Thus, we use our distribution amplitude and functions of charged pion as basic forms but we deviate slightly to describe the distribution amplitude and functions of $\mathbf{u}\overline{d}$ , $\mathrm{u}\overline{u}$ and $\mathrm{d}\overline{d}$ in proton. Here, we use our approximate forms for distribution amplitude and functions of charged pion. Note that, as we mentioned in ref. [7], our approximate forms are sufficient to compare to other proposed distribution amplitude, thus, also to other proposed distribution functions from the definition as shown in later. Our approximate form of distribution amplitude of charged pion is described as [7]

where $\rho =\frac{u}{\sqrt{2|\alpha |}},|\alpha |=\frac{g^2}{16},x\in [0,1]$

Note that $F_3(x)$ is actually $|F_3(x)|$ .

From Eq. (1), our approximate form of the charged pion distribution functions is described as [8]

We consider that sea quarks states correspond to unbound state. Here, we define unbound state as follows:

Definition:

Unbound state is a state corresponding to a distribution amplitude, that is, exponential $e^{-\rho \frac{x}{1-x}}$ becomes negligibly small because the coupling constant $g^2$ is sufficiently small.

Thus, Eq. (1) gives our distribution amplitude of sea quarks $\overline{d}$ that come from ${\pi }^{+}$ as

Thus, our distribution functions of $\overline{d}$ become

Notation of $f_{{\pi }^{+}}^{sea\overline{d}}(x)$ means sea quarks $\overline{d}$ distribution functions come from ${\pi }^{+}$ .

The behavior of $f_{{\pi }^{+}}^{sea\overline{d}}(x)$ in Eq. (4) shows a little bit larger than that of sea quarks distribution functions $xd$ shown in Atlas experiment results [6]. Therefore, we deviate distribution amplitude slightly from Eq. (1). For this point, we compare above sea quarks distribution functions to our view point of sea quarks distribution functions due to gluon splitting in Appendix A.

To achieve this purpose, we return to the exact form of our distribution amplitude shown in Eq. (40) in ref. [7] as:

To evaluate the integral in Eq. (5), we again divide the integral as ${\int }_0^{\mathrm{\infty }}dz={\int }_0^{u}dz+{\int }_u^{\mathrm{\infty }}dz$ and use the argument for large |q| case shown in ref. [7]. Then, our distribution amplitude is represented by following equation.

Recalling that we obtained our approximate form of pion distribution amplitude using only the first term of $M_{\kappa ,\mu }(z)$ to evaluate the integral in the case of large $|q|$ and also recalling that $W_{\kappa ,\mu }(z)$ is defined as $W_{\kappa ,\mu }(z)=\frac{\mathrm{\Gamma }(-2\mu )}{\mathrm{\Gamma }(\frac{1}{2}-\mu -\kappa )}{M}_{\kappa ,\mu }(z)+\frac{\mathrm{\Gamma }(2\mu )}{\mathrm{\Gamma }(\frac{1}{2}+\mu -\kappa )}{M}_{\kappa ,-\mu }(z)$ [12], a slight deviation can be considered by adding the first term of $M_{\kappa ,-\mu }(z)$ because actual form of $W_{\kappa ,\mu }(z)$ in Eq. (6) is $W_{\kappa ,\mu }\left(\frac{1}{2}{z}^2\right)$.

The definition of $M_{\kappa ,\mu }(z)$ is described as [12]

where ${}_1{F}_1$ is the confluent hypergeometric series defined as [12]

Thus, the first term of $M_{\kappa ,-\mu }\left(\frac{1}{2}{z}^2\right)$ in the case of $\kappa =-\frac{1}{4}\mp \frac{i}{\pi }$ and $\mu =-\frac{1}{4}$ becomes $1$. Thus, Eq. (6) becomes

In Eq. (7), we use the facts that $\mathrm{\Gamma }(1\pm \frac{i}{\pi })={[\pi \left(\frac{\pm 1}{\pi }\right)/\sinh \left(\pi \left(\frac{\pm 1}{\pi }\right)\right)]}^{\frac{1}{2}}={\left(\frac{1}{\sinh \left(1\right)}\right)}^{\frac{1}{2}}$ and $\mathrm{\Gamma }(\frac{1}{2}\pm \frac{i}{\pi })=\frac{\pi }{\cosh \left(\pi \left(\frac{\pm 1}{\pi }\right)\right)}=\frac{\pi }{\cosh \left(1\right)},\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi },\mathrm{\Gamma }(-\frac{1}{2})=-2\sqrt{\pi }(\mathrm{\Gamma }(-n+\frac{1}{2})=\frac{(-1{)}^n{2}^n}{(2n-1)!!}\sqrt{\pi }[11])$ as shown in ref. [7].

Integral of ${\int }_0^{u}dz\exp (-{\mu }_{\mp }z)z$ becomes

where ${\mu }_{\mp }=\frac{|q|}{\sqrt{2|\alpha |}}(1\mp i)$

By following our definition of unbound state, sea quarks states (unbound state), are defined as follows.

Definition: $\sqrt{|\alpha |}\propto \sqrt{g^2}$ is sufficiently small so that the exponential term becomes negligibly small in entire x domain.

Thus, this part gives $\frac{\sqrt{|\alpha |}}{|q{|}^2}⟶\sqrt{|\alpha |}\frac{(1-x{)}^2}{x^2}$ term.

Important point is that the first term of Eq. (8) is the same order of $|q|$ as the first terms of the first and the second terms of Eq. (7). However, including terms that come from this integral breaks the constraint condition of our distribution amplitude, such as $F_3(|q|)⟶$ $x$ when $x⟶0$ as shown in ref. [7] (We obtain this condition by using the integral representation of $W_{\kappa ,\mu }(z))$ when coupling constant $g^2$ is not sufficiently small $(\sqrt{|\alpha |}$ is not smal) so that the exponential term remains. Breaking the constraint condition means that including only the first term of $M_{\kappa ,-\mu }\left(\frac{1}{2}{z}^2\right)$ does not describe the exact pion state. This is plausible because there is no evidence which shows to exist the exact pion in proton.

Recalling that our pion distribution amplitude and functions correspond to the charged pion ${\pi }^{\pm }$ , we also need those of neutral pion ${\pi }^0$ for further consideration. Looking at Eq. (7) and Eq. (8) in ref. [7] and recalling that we are considering the chiral limit case (massless quark case), our equation for ${\chi }_0$ , which corresponds to ${\pi }^0$ , becomes

In rest frame, that is $P_1=0,P_0$ becomes equal to $W_0$ so that Eq. (8) becomes

Thus, for the pion case, that is $W_0=0$ , ${\chi }_0$ cannot be determined.

However, Dlamini et al. [13] shows, for the neutral pion ${\pi }^0$, that t dependence ($t=(q-q^{\prime }{)}^2$, q is virtual photon momenta and $q^{\prime }$ is ${\pi }^0$ momenta) usually parametrized by Regge-like functions is no longer valid when $(-t)$ is larger than 1 GeV${}^2$ and that cross section is represented as d\sigma = \text{const} Q^{-6} \exp(-Bt') ($t^{\prime }=t_{min}-t$) when $Q^2=5.49$ and $8.31{\text{ GeV}}^2$ shown in Fig. 3 of ref. [13].

The results of Dlamini et al. means that when $Q^2$ is smaller than those values and ${\pi }^0$ momenta is small, ${\pi }^0$ behaves as Regge-like function (cross section is described by omitting form of exponential term). In addition, Horn [14] shows, for the charged pion ${\pi }^{\pm }$, that cross section is represented as $d\sigma =\text{const}{Q}^{-6}$. Considering the results of Dlamini et al. together with the results of Horn leads to the fact that cross section of ${\pi }^0$ becomes same description of that of ${\pi }^{\pm }$ when $Q^2$ or ${\pi }^0$ momenta is small. Recalling that $Q^2$ of Atlas experiment is $Q^2=1.9{\text{ GeV}}^2$ and $Q^2$ of H1 and Zeus experiment is $Q^2=10{\text{ GeV}}^2$ and that main range of the distribution functions of sea quarks is less than $x=0.4$, we can consider that ${\pi }^0$ (actually close state of ${\pi }^0$) behaves as Regge-like function in both experiment cases. Therefore, we assume that the distribution amplitude and functions of ${\pi }^0$ in Atlas experiment case are represented as the similar forms that described by our charged pion distribution amplitude and functions. From Eq. (7) and Eq. (8) and using our definition of sea quarks, distribution amplitude of sea quarks is described as:

Recalling that the distribution functions is proportional to $\mathbf{X}$ times the distribution amplitude. Thus, our sea quarks distribution functions $f^{sea}(x)$ are described as

where $\beta =\frac{2^{\frac{3}{4}}\cosh \left(1\right)}{\pi \sqrt{\sinh \left(1\right)}}$

Recalling that our definition of sea quarks is that the coupling constant $g^2$ is sufficiently small so that the exponential term becomes negligibly small for entire $\mathbf{X}$ domain, $\sqrt{|\alpha |}$ is sufficiently small for entire x domain because $|\alpha |$ is proportional to $g^2$ as shown in Eq. (1). Here, we use the following ansatz to satisfy this condition.

Ansatz:

Coupling constant $g^2$ depends on momentum. The property of the coupling constant $g^2$ is that $g^2$ becomes smaller as momentum is larger and also $g^2$ becomes smaller when momentum is smaller.

The first part of property is not so peculiar but the second part is technical. However, the exponential term is multiplied by $(\sin \rho \frac{x}{1-x}+\cos \rho \frac{x}{1-x})$ when calculating integral in Eq. (7) (refer to ref. [7]). Thus, the second term of Eq. (8) becomes const when x goes to 0 so that, for the distribution functions, this term becomes const when x goes to 0. Therefore, in the case of using only the first part of property in the ansatz, the second term of Eq. (8) goes to 0 when x goes to 0 even though $g^2$ becomes constant when x goes to 0. Actually, we need only this property for the adding term (deviation term). In addition, the first term of Eq. (8) becomes constant for the distribution functions when x goes to 0 because of multiplication of x. Thus, the adding term changes only constant value when x goes to 0 because constant term in parenthesis of the integral in Eq. (7) becomes 0 when x goes to 0. In Sec. 4, we check the magnitude of the exponential in the case of using ansatz. According to the argument in Sec.4, the exponential is sufficiently small in most of all region of x except for the region that $\mathbf{X}$ is very close to 0 even though $\sqrt{|\alpha |}$ is constant. The previous paragraphs indicate that the behavior of sea quarks distribution functions would be different form, that is not uniquely decreasing, in the neighbor of $\mathrm{x}=0$ in spite of starting from constant value at $\cdot$ if we use only the first part of property in the ansatz. Recalling that x domain of Atlas experiment results is $\mathrm{\_}$ , it is possible that x domain of Atlas experiment results is outside of the region in which possible improper behavior occurs. Thus, it is meaningful to remove this possible improper behavior simply, by adding the second part of property in the ansatz. Thus, we use the ansatz here.