Teruo Kurai
-19-18 Unoki, Ota-ku, Tokyo, Japan.
DOI: 10.4236/jmp.2025.166046   PDF    HTML   XML   87 Downloads   481 Views   Citations

Abstract

The usual pion distribution amplitude is typically constructed using functions based on one of the trial functions from the ‘t Hooft singular integral equation, specially of the form $x\left(1-x\right)$ type. However, these of function are not exact solutions of the ‘t Hooft equation, which means a pion distribution amplitude constructed in this manner may not accurately represent a real pion as described by Light Front QCD. In contrast, we have developed a pion distribution amplitude using solutions that correspond to zero-mass state wave functions of a bound system of 1 + 1 dimensional QCD derived from the ‘t Hooft model. Our distribution amplitude shows $\frac{1-x}{x}$ behavior as x approaches 1. Notably, asymptotic form of our distribution amplitude reveals an intriguing property: peak of our distribution amplitude shifts towards $x=1$ as the coupling constant $g^2$ increases.

Keywords

Distribution Amplitude, ‘t Hooft Model

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1. Introduction

Recently, several surprising experiment results have emerged in hadron physics.

Among these, Dlamini et al. found that the neutral pion’s electromagnetic transition form factor is Gaussian-like rather than Regge-like [1], whereas Huber and Horn reported that the charged pion’s electromagnetic form factor remains Regge-like, despite employing the same error analysis [2] [3]. In the case of proton, Experiments by H1 and Zeus [4] and Atlas [5] have shown that the fundamental proton structure resembles see-quarks in both electron-proton collision and in proton-proton collision. Given these findings, Arrington et al. highlighted that the lightest pseudo-scalar mesons, particularly the pion and kaon, are crucial for deepening our understanding of mass emergence and structural mechanisms in strongly interacting matter. Consequently, unraveling the structure of pions and kaons has become increasingly important. Proper structure functions are essential for interpreting comprehend experimental findings. For instance, Raya et al. demonstrated the valence u-quark of the pion distribution function, which directly related to pion distribution amplitude [6] [7]. This underscores the significance of obtaining an accurate pion distribution amplitude. Several notable efforts to determine the pion distribution amplitude include work by Zhang et al., who proposed a function of the form $x^{{\alpha }_1}{\left(1-x\right)}^{{\alpha }_2}$ [8], Chang et al., who predicted it using by Dyson-Schwinger Equation [9] and Raya et al., who accounted for pion radius effects [7]. The latter two employed the functions of the form ${\left[x\left(1-x\right)\right]}^{\alpha }P\left(x\right)$ type functions. These efforts utilize characteristic functions akin to trial functions used in solutions of ‘t Hooft singular integral equation [10]. To construct distribution amplitude, Light Front QCD (4 dimension) is commonly applied, which makes it, at least analogically, reasonable to use a type of solutions of ‘t Hooft singular integral equation that derived in 2 or 1 + 1 dimensions, given that his equation uses light-corn gauge. However, if we extend this analogy further, aforementioned function types may not suffice. Litvinov et al. [11] presented a method to construct exact analytical solutions to the ‘t Hooft equation, which are not of the form $x^{{\beta }_1}{\left(1-x\right)}^{{\beta }_2}$ type but rather are sum of

functions of the form ${\left(\frac{x}{1-x}\right)}^{{\gamma }_i}$ and combinations of sine and cosine functions of the form $\left(\left({\lambda }_i+\frac{N}{2}\right)\ln \left(\frac{x}{1-x}\right)\right)$ . These wave functions, corresponding to eigenvalues, are identified in the space $v\in \left[-\infty ,\infty \right]$ and fundamentally represented by the residue of simple poles.

In this paper, we assume that the aforementioned analogy applied to the pion distribution amplitude and explore it within the framework of hadronic operator formalism proposed by Suura [12], rather than using solutions from the ‘t Hooft singular integral equation. There are two main reasons for choosing this framework. First, we successfully derived both neutral and charged pion wave functions and their electromagnetic (transition) form factors within Suura’s hadronic operator framework in 3 + 1 dimensions [13]-[15]. Second, we applied Suura’s hadronic operator to the ‘t Hooft model in 1 + 1 dimensions, obtaining mass spectra similar to those derived from ‘t Hooft singular integral equation [16]. As demonstrated in ref. [16], our set of equations is not exactly equivalent to the ‘t Hooft singular integral equation. Nevertheless, we managed to obtain a zero-mass solution, while the ‘t Hooft singular integral equation does not yield zero-mass solution in massless quark case (chiral limit) as shown in ref. [11]. Since Peccei et al. argued that zero-mass solution exists in the large N limit of two dimensional QCD massless quark case [17], we believe that our approach is significant and meaningful.

2. Formulation

As mentioned in Section 1, to construct a pion distribution amplitude, Light Front QCD (LFQCD) is commonly applied. Therefore, we may consider momentum space wave functions of the large N limit two dimensional QCD massless quarks case as a base argument to construct a pion distribution amplitude. We derived configuration space wave functions of this case in the framework of hadronic operator proposed by Suura as shown in ref. [16]. Here we briefly show the derivation of configuration space wave functions in ref. [16] with small revision.

For 1 + 1 dimensional bound system, the Bethe-Salpeter-like amplitude defined by Suura becomes as

$X\left(1,2\right)=\langle 0|q\left(1,2\right)|P\rangle$ (1)

($|0\rangle$ and $|phy\rangle$ are vacuum and physical states, respectively)

where the gauge-invariant operator $q\left(1,2\right)$ is defined as

$q_{\zeta \eta }\left(1,2\right)=T_r^c{q}_{\eta }^{†}\left(x\right)P\exp \left(ig{\displaystyle {\int }_1^2\text{d}x}{A}_1^a\left(x\right)\frac{{\lambda }_a}{2}\right)q_{\zeta }\left(1\right)$ (2)

Here $\zeta$ and $\eta$ denote Dirac indices. $P$ denotes the path ordering and $\frac{{\lambda }_a}{2}$

components are adjoint representation of the SU(N) color gauge group. $T_r^c$ indicates that the trace is taken for color spin a. Above definition is formulated in the time axial gauge ($A_0=0$ gauge). Path is an exactly straight line because we are interested in 1 + 1 dimension case. The equation of motion for $q\left(1,2\right)$ becomes

$i{\partial }_{t}q\left(1,2\right)=-i\alpha \partial \left(2\right)q\left(1,2\right)-q\left(1,2\right)i\alpha \partial \left(1\right)+g{\displaystyle {\int }_1^{2}dx}{q}_E\left(1,2:x\right)$ (3)

where

$q_E\left(1,2:x\right)=T_r^c{q}^{†}\left(2\right)U\left(2,x\right)E^a\left(x\right)U\left(x,1\right)q\left(1\right)$

$U\left(2,1\right)=P\exp \left(ig{\displaystyle {\int }_1^2\text{d}x}{A}_1^a\left(x\right)\frac{{\lambda }_a}{2}\right)$

Note that $E^a\left(x\right)$ and $A^a\left(x\right)$ indicate $E_1^a\left(x\right)$ and $A_1^a$ because we are working in 1 + 1 dimensions.

We employ the metric system and $\gamma$ -matrices in 2 dimensions as follows, which are also employed by Casher et al. [18]

$g^{00}=1,g^{11}=-1$

${\gamma }^0=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\gamma }^1=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right),\alpha ={\gamma }^0{\gamma }^1=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)={\sigma }_3$

After evaluating $q_E\left(1,2:x\right)$ and regrouping the charge vertex term, we obtained the final form of the equation of motion for $q\left(1,2:x\right)$ as

$\begin{array}{c}i{\partial }_{t}q\left(1,2\right)=-i\partial \left(2\right)q\left(1,2\right)-q\left(1,2\right)i\partial \left(1\right)\\ -\frac{g^2}{4}{\displaystyle {\int }_1^2\text{d}z}{\displaystyle {\int }_{-\infty }^{\infty }\text{d}x}\epsilon \left(x-z\right)q\left(1,x\right)q\left(x,2\right)+O\left(\frac{1}{N}\right)\end{array}$ (4)

where $\epsilon \left(x\right)$ is step function defined as $\epsilon \left(x\right)=+1\left(x\ge 0\right)$ and $-1\left(x<0\right)$ .

Recalling $X\left(1,2\right)$ as $X\left(1,2\right)=\langle 0|q\left(1,2\right)|phy\rangle$ ($|0\rangle$ and $|phy\rangle$ are vacuum and physical states, respectively) and taking the new variables as

$G=\frac{1}{2}\left(x\left(1\right)+x\left(2\right)\right)$ and $r=x\left(2\right)-x\left(1\right)$ ,

we consider the following form of $X\left(1,2\right)$ as

$X\left(1,2\right)={\text{e}}^{-i{P}_{0}t}{\text{e}}^{i{P}_{1}G}X\left(r\right)$

After factoring out the phase part, Equation (4) becomes

$\begin{array}{c}{P}_{0}X\left(r\right)=\left\{\frac{1}{2}\alpha P_1,X\left(r\right)\right\}-\left[i\alpha \frac{\partial }{\partial r},X\left(r\right)\right]\\ +\frac{g^2}{8}{\displaystyle {\int }_{-\infty }^{\infty }\text{d}x}{\text{e}}^{i{P}_1\frac{x-x\left(1\right)}{2}}\left(\left|x\left(2\right)-x\right|-\left|x-x\left(1\right)\right|\right)\left[S\left(1,x\right),X\left(x\left(2\right)-x\right)\right]\\ +O\left(\frac{1}{N}\right)\end{array}$ (5)

where $S\left(1,x\right)=\langle 0|q\left(1,x\right)|0\rangle$ .

Recalling the fact that $\alpha =i{\sigma }_3$ and then taking the decomposition

$X\left(1,2\right)=1X_0\left(1,2\right)+\left(i{\sigma }_3\right)X_1\left(1,2\right)+\left({\sigma }_2\right)X_2\left(1,2\right)+\left({\sigma }_1\right)X_3\left(1,2\right)$ (6)

where 1 denotes unit matrix, we obtain the set of equations:

$P_0{X}_0\left(r\right)=i{P}_1{X}_1\left(r\right)$ (7)

$P_0{X}_1\left(r\right)=-i{P}_1{X}_0\left(r\right)+O\left(\frac{1}{N}\right)$ (8)

$W_0{X}_2\left(r\right)=2\frac{\partial X_3}{\partial r}-\frac{g^2}{8\pi }{\displaystyle {\int }_{-\infty }^{\infty }\text{d}{x}^{\prime }}\frac{\left|r-x^{\prime }\right|-\left|x^{\prime }\right|}{x^{\prime }}{X}_3\left(r-x^{\prime }\right)$ (9)

$W_0{X}_3\left(r\right)=-2\frac{\partial X_2}{\partial r}+\frac{g^2}{8\pi }{\displaystyle {\int }_{-\infty }^{\infty }\text{d}{x}^{\prime }}\frac{\left|r-x^{\prime }\right|-\left|x^{\prime }\right|}{x^{\prime }}{X}_2\left(r-x^{\prime }\right)$ (10)

Here we employ the form $S\left(r\right)=\left(i{\sigma }_3\right)P_r\frac{1}{2\pi }\frac{1}{r}$ ($P_r$ denotes Principal Value)

because Schwinger suggested that the propagator function behaves like a free propagator with small values of r (Schwinger parametrization) [19]. We justify this form on Appendix A. Important point is that Equation (9) and Equation (10) are only necessary to consider the mass spectra of this system under the condition of $P_1=0$ (in rest frame) because equations are closed for $X_2$ and $X_3$ under this condition.

Because we are interested in the ‘t Hooft model, we investigate solutions of Equation (9) and Equation (10).

Changing the variables in the integral as $r-x^{\prime }=s$ and using the notation $r=x$ , and taking $X_{\pm }=X_3\pm i{X}_2$ , Equation (9) and Equation (10) change to following two closed forms.

$W_0{X}_{+}\left(x\right)=2i{\partial }_x{X}_{+}\left(x\right)-i\frac{g^2}{8\pi }{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\frac{\left|s\right|-\left|x-s\right|}{x-s}{X}_{+}\left(s\right)$ (11)

$W_0{X}_{-}\left(x\right)=-2i{\partial }_x{X}_{-}\left(x\right)+i\frac{g^2}{8\pi }{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\frac{\left|s\right|-\left|x-s\right|}{x-s}{X}_{-}\left(s\right)$ (12)

First, we seek a solution of $X_{-}\left(x\right)$ (this corresponds to $X_{+}\left(x\right)$ in ref. [16]).

Note that we could obtain the same form as Equation (11) if we take $-W_0$ for $W_0$ of Equation (12).

Thus, once we found a solution of $X_{-}\left(x\right)$ , we could obtain a solution of $X_{+}\left(x\right)$ by replacing $W_0$ to $-W_0$ in a solution of $X_{-}\left(x\right)$ .

Considering absolute value and using step function $\epsilon \left(x\right)$ defined above, integral term becomes as

$\begin{array}{l}{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\frac{\left|s\right|-\left|x-s\right|}{x-s}{X}_{-}\left(s\right)\\ =-{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\epsilon \left(s\right)X_{-}\left(s\right)-x{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\frac{\epsilon \left(s\right)X_{-}\left(s\right)}{s-x}+{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\epsilon \left(x-s\right)X_{-}\left(s\right)\end{array}$

Multiplying $\epsilon \left(x\right)$ on both sides of Equation (12) and considering the new function $F_{-}\left(x\right)$ defined as $F_{-}\left(x\right)=\epsilon \left(x\right)X_{-}\left(x\right)$ and notifying the fact that $X_{-}\left(x\right)=\epsilon \left(x\right)F_{-}\left(x\right)$ (${\left(\epsilon \left(x\right)\right)}^2=1$ ), Equation (12) becomes

$\begin{array}{c}{W}_0{F}_{-}\left(x\right)=-2i{\partial }_x{F}_{-}\left(x\right)+4i\delta \left(x\right)F_{-}\left(x\right)\epsilon \left(x\right)-i\frac{g^2}{8\pi }\epsilon \left(x\right)x{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\frac{F_{-}\left(s\right)}{s-x}\\ +i\frac{g^2}{8\pi }\epsilon \left(x\right){\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}{F}_{-}\left(s\right)\epsilon \left(s\right)\epsilon \left(x-s\right)-i\frac{g^2}{8\pi }\epsilon \left(x\right){\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}{F}_{-}\left(s\right)\end{array}$ (13)

The $\delta$ -function appears because of the fact ${\partial }_x\epsilon \left(x\right)=2\delta \left(x\right)$ .

Here we consider the positive region ($x>0$ region for which $\epsilon \left(x\right)=+1$ ). Taking the derivative with respect to x on both side of Equation (13), Equation (13) becomes

$\begin{array}{c}{W}_0{\partial }_x{F}_{-}\left(x\right)=-2i{\partial }_x^2{F}_{-}\left(x\right)+4i{\partial }_x\left(\delta \left(x\right)F_{-}\left(x\right)\right)-i\frac{g^2}{8\pi }{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\frac{F_{-}\left(s\right)}{s-x}\\ -i\frac{g^2}{8\pi }x{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\frac{\frac{\text{d}{F}_{+}}{\text{d}s}}{s-x}+i\frac{g^2}{4\pi }{F}_{-}\left(x\right)\end{array}$ (14)

To obtain the last term, we use the fact that ${\partial }_x\epsilon \left(x-s\right)=2\delta \left(x-s\right)$ and that $X_{-}\left(x\right)=\epsilon \left(x\right)F_{-}\left(x\right)=F_{-}\left(x\right)$ (for $x>0$ ).

For the singular integral term, we apply the slightly modified Sokhotsky formula [20]:

$\frac{1}{i\pi }{\displaystyle {\int }_{-\infty }^{\infty }\text{d}s}\frac{F_{-}\left(s\right)}{s-x}=a_1{\varphi }_{-}^{(+)}\left(x\right)+a_2{\varphi }_{-}^{(-)}\left(x\right)$ (15)

$F_{-}\left(x\right)=a_1{\varphi }_{-}^{(+)}\left(x\right)-a_2{\varphi }_{-}^{(-)}\left(x\right)$ (16)

where ${\varphi }_{-}^{(+)}\left(x\right)$ is the value result when x asymptotically approaches real axis in the upper-half hemisphere, while ${\varphi }_{-}^{(-)}\left(x\right)$ is the value result when x asymptotically approaches real axis in the lower-half hemisphere, and $a_1$ and $a_2$ are non-zero real constants.

Here, we have modified the original form of the Sokhotsky formula by changing ${\varphi }_{-}^{(+)}$ to $a_1{\varphi }_{-}^{(+)}$ and ${\varphi }_{-}^{(-)}$ to $a_2{\varphi }_{-}^{(-)}$ , which will be used in a later argument.

Note that s and x are real values.

Subsequently, Equation (14) is written as

$\begin{array}{l}{a}_1\left[\frac{{\partial }^2{\varphi }_{-}^{(+)}}{\partial x^2}+\left[\left(\frac{W_0}{2i}\right)-\frac{g^2}{8}\frac{1}{2i}x\right]\frac{\partial {\varphi }_{-}^{(+)}}{\partial x}+\left[-\left(\frac{g^2}{8\pi }\right)-\left(\frac{g^2}{8}\right)\frac{1}{2i}\right]{\varphi }_{-}^{(+)}-2{\partial }_x\left(\delta \left(x\right){\varphi }_{-}^{(+)}\right)\right]\\ =a_2\left[\frac{{\partial }^2{\varphi }_{-}^{(-)}}{\partial x^2}+\left[\left(\frac{W_0}{2i}\right)+\frac{g^2}{8}\frac{1}{2i}x\right]\frac{\partial {\varphi }_{-}^{(-)}}{\partial x}+\left[-\left(\frac{g^2}{8\pi }\right)+\left(\frac{g^2}{8}\right)\frac{1}{2i}\right]{\varphi }_{-}^{(-)}-2{\partial }_x\left(\delta \left(x\right){\varphi }_{-}^{(-)}\right)\right]\end{array}$ (17)

In order to find the solutions for both ${\varphi }_{-}^{(+)}$ and ${\varphi }_{-}^{(-)}$ , we set both sides of Equation (17) to be equal to zero. Because $a_1$ and $a_2$ are non-zero constants, we can independently obtain the differential equation for ${\varphi }_{-}^{(+)}$ and ${\varphi }_{-}^{(-)}$ .

To address the derivative of the $\text{δ}$ -function term, we integrate successively twice from $\infty$ to x. Note that here we consider the range $x>0$ while ${\varphi }_{-}^{(\pm )}\left(\infty \right)=0$ and ${\partial {\varphi }_{-}^{(\pm )}/\partial x|}_{x=\infty }=0$ ; specifically, we are seeking solutions that satisfy the previous conditions. Subsequently, we can see that

${\displaystyle {\int }_{\infty }^x\text{d}{x}^{\prime }}\delta \left(x^{\prime }\right){\varphi }_{-}^{(\pm )}\left(x^{\prime }\right)=0.$

(see Appendix B)

Taking the derivative successively twice, we obtain the following differential equations.

$\frac{{\partial }^2{\varphi }_{-}^{(+)}}{\partial x^2}+\left[\left(\frac{W_0}{2i}\right)-\frac{g^2}{8}\frac{1}{2i}x\right]\frac{\partial {\varphi }_{-}^{(+)}}{\partial x}+\left[-\left(\frac{g^2}{8\pi }\right)-\left(\frac{g^2}{8}\right)\frac{1}{2i}\right]{\varphi }_{-}^{(+)}=0$ (18)

$\frac{{\partial }^2{\varphi }_{-}^{(-)}}{\partial x^2}+\left[\left(\frac{W_0}{2i}\right)+\frac{g^2}{8}\frac{1}{2i}x\right]\frac{\partial {\varphi }_{-}^{(-)}}{\partial x}+\left[-\left(\frac{g^2}{8\pi }\right)+\left(\frac{g^2}{8}\right)\frac{1}{2i}\right]{\varphi }_{-}^{(-)}=0$ (19)

The solutions satisfying the condition that both function and its first derivative at infinity are zero are as follows (the derivation is given in Appendix C).

${\varphi }_{-}^{(+)}\left(x\right)={\text{e}}^{\frac{1}{4}\alpha x^2}{\text{e}}^{-\frac{1}{2}\beta x}{2}^{-\frac{1}{4}-\frac{i}{\pi }}{\left(\sqrt{\alpha }x-\frac{\beta }{\sqrt{\alpha }}\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{\left(\sqrt{\alpha }x-\frac{\beta }{\sqrt{\alpha }}\right)}^2\right)$ (20)

${\varphi }_{-}^{(-)}\left(x\right)={\text{e}}^{-\frac{1}{4}\alpha x^2}{\text{e}}^{-\frac{1}{2}\beta x}{2}^{-\frac{1}{4}+\frac{i}{\pi }}{\left(i\left(\sqrt{\alpha }x+\frac{\beta }{\sqrt{\alpha }}\right)\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(-\frac{1}{2}{\left(\sqrt{\alpha }x+\frac{\beta }{\sqrt{\alpha }}\right)}^2\right)$ (21)

where

$\alpha =\frac{g^2}{8}\frac{1}{2i},\beta =\frac{W_0}{2i}$

$W_{\kappa ,\mu }\left(x\right)$ is the Whittaker function.

As mentioned before, a solution of $X_{+}\left(x\right)$ can obtain by replacing $W_0$ to $-W_o$ in Equation (20) and Equation (21) as

$X_{+}\left(x\right)=a_1{\varphi }_{+}^{(+)}\left(x\right)-a_2{\varphi }_{+}^{(-)}\left(x\right)$

${\varphi }_{+}^{(+)}\left(x\right)={\text{e}}^{\frac{1}{4}\alpha x^2}{\text{e}}^{\frac{1}{2}\beta x}{2}^{-\frac{1}{4}-\frac{i}{\pi }}{\left(\sqrt{\alpha }x+\frac{\beta }{\sqrt{\alpha }}\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{\left(\sqrt{\alpha }x+\frac{\beta }{\sqrt{\alpha }}\right)}^2\right)$ (22)

${\varphi }_{+}^{(-)}\left(x\right)={\text{e}}^{-\frac{1}{4}\alpha x^2}{\text{e}}^{\frac{1}{2}\beta x}{2}^{-\frac{1}{4}+\frac{i}{\pi }}{\left(i\left(\sqrt{\alpha }x-\frac{\beta }{\sqrt{\alpha }}\right)\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(-\frac{1}{2}{\left(\sqrt{\alpha }x-\frac{\beta }{\sqrt{\alpha }}\right)}^2\right)$ (23)

Boundary condition for our case is that $X_{-}=0$ at $x=0$ ($X_{+}=0$ at $x=0$ ) and $X_{-}=0$ at $x=\infty$ ($X_{+}=0$ at $x=\infty$ ). Because both $X_{-}\left(x\right)$ and $X_{+}\left(x\right)$ are represented by linear combination of Whittaker function $W_{\kappa ,\mu }\left(x\right)$ , the conditions at $x=\infty$ are automatically satisfied.

The condition at $x=0$ gives mass spectra other than zero-mass of our description of the ‘t Hooft model as shown in ref. [16]. The condition at $x=0$ gives the following equation.

${\left(-i{\left(\frac{2W_0}{g}\right)}^2\right)}^{-\frac{1}{4}}\left[i^{-\frac{1}{4}}{a}_1{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(-i\frac{1}{2}{\left(\frac{2W_0}{g}\right)}^2\right)-i^{\frac{1}{4}}{a}_2{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(i\frac{1}{2}{\left(\frac{2W_0}{g}\right)}^2\right)\right]=0$ (24)

Because Equation (18) and Equation (19) include zero-mass case, we can obtain mass spectra including zero-mass by Equation (24).

To determine coefficients $a_1$ and $a_2$ , we only need to consider zero-mass case that is $W_0=0$ ($\beta =0$ ).

Setting a value of $W_0=0$ ($\beta =0$ ), Equation (20) and Equation (21) become as

${\varphi }_{-}^{(+)}\left(x\right)=2^{-\frac{1}{4}-\frac{i}{\pi }}{\left({\text{e}}^{-i\frac{\pi }{4}}\sqrt{\left|\alpha \right|}x\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{\text{e}}^{-i\frac{\pi }{2}}\left|\alpha \right|x^2\right)$ (25)

${\varphi }_{-}^{(-)}\left(x\right)=2^{-\frac{1}{4}+\frac{i}{\pi }}{\left({\text{e}}^{i\frac{\pi }{4}}\sqrt{\left|\alpha \right|}x\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{\text{e}}^{i\frac{\pi }{2}}\left|\alpha \right|x^2\right)$ (26)

Recalling that Whittaker functions $W_{\kappa ,\mu }$ of both equations have $\mu$ value of $-\frac{1}{4}$ , and that $W_{\kappa ,\mu }$ is represented by linear combination of $M_{\kappa ,\mu }$ and $M_{\kappa ,-\mu }$ as [21]

$W_{\kappa ,\mu }\left(z\right)=\frac{\text{Γ}\left(-2\mu \right)}{\text{Γ}\left(\frac{1}{2}-\mu -\kappa \right)}{M}_{\kappa ,\mu }\left(z\right)+\frac{\text{Γ}\left(2\mu \right)}{\text{Γ}\left(\frac{1}{2}+\mu -\kappa \right)}{M}_{\kappa ,-\mu }\left(z\right)$ (27)

where $M_{\kappa ,\mu }$ is defined as

$M_{k,\mu }\left(z\right)=z^{\mu +\frac{1}{2}}{\text{e}}^{-\frac{z}{2}}{}_1{}^{}F{}_1\left(\mu -\kappa +1,2\mu +1;z\right)$ (28)

Each $\kappa$ and $\mu$ values of ${\varphi }_{-}^{(+)}$ and ${\varphi }_{-}^{(-)}$ are $\kappa =-\frac{1}{4}-\frac{i}{\pi }$ , $\mu =-\frac{1}{4}$ and $\kappa =-\frac{1}{4}+\frac{i}{\pi }$ , $\mu =-\frac{1}{4}$ , respectively. Because only the first term of $M_{\kappa ,\mu }$ dominates as x asymptotically approaches 0, behavior of ${\varphi }_{-}^{(\pm )}\left(x\right)$ as $x\to 0$ are following.

${\varphi }_{-}^{(+)}\left(x\right)\to 2^{-\frac{1}{4}-\frac{i}{\pi }}\frac{\text{Γ}\left(\frac{1}{2}\right)}{\text{Γ}\left(1+\frac{i}{\pi }\right)}{\left(\frac{1}{2}\right)}^{\frac{1}{4}}$

${\varphi }_{-}^{(-)}\left(x\right)\to 2^{-\frac{1}{4}+\frac{i}{\pi }}\frac{\text{Γ}\left(\frac{1}{2}\right)}{\text{Γ}\left(1-\frac{i}{\pi }\right)}{\left(\frac{1}{2}\right)}^{\frac{1}{4}}$

Therefore, in order to satisfy the condition that $X_{-}=0$ at $x=0$ , it is sufficient for coefficients to set that $a_1=2^{\frac{i}{\pi }}\text{Γ}\left(1+\frac{i}{\pi }\right)$ and $a_2=2^{-\frac{i}{\pi }}\text{Γ}\left(1-\frac{i}{\pi }\right)$ .

These coefficient values are also true for $X_{+}$ case because ${\varphi }_{+}^{(+)}\left(x\right)={\varphi }_{-}^{(+)}\left(x\right)$ and ${\varphi }_{+}^{(-)}\left(x\right)={\varphi }_{-}^{(-)}\left(x\right)$ in the case of zero-mass ($\beta =0$ ).

Therefore, the wave functions for zero-mass case ($\beta =0$ ) become as

$\begin{array}{c}{X}_{\mp }\left(x\right)=2^{-\frac{1}{4}}[\Gamma \left(1+\frac{i}{\pi }\right){\text{e}}^{\frac{1}{4}\alpha x^2}{\left(\sqrt{\alpha }x\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}\alpha x^2\right)\\ -\Gamma \left(1-\frac{i}{\pi }\right){\text{e}}^{-\frac{1}{4}\alpha x^2}{\left(i\sqrt{\alpha }x\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(-\frac{1}{2}\alpha x^2\right)]\end{array}$ (29)

Thus, we obtain the wave function of $X_3\left(x\right)$ as

$\begin{array}{c}{X}_3\left(x\right)=\frac{X_{-}\left(x\right)+X_{+}\left(x\right)}{2}\\ =2^{-1}[\Gamma \left(1+\frac{i}{\pi }\right){\text{e}}^{\frac{1}{4}\alpha x^2}{\left(\frac{1}{2}\sqrt{\alpha }x\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}\alpha x^2\right)\\ -\Gamma \left(1-\frac{i}{\pi }\right){\text{e}}^{-\frac{1}{4}\alpha x^2}{\left(i\sqrt{\alpha }x\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(-\frac{1}{2}\alpha x^2\right)]\end{array}$ (30)

$X_2\left(x\right)=\frac{X_{+}\left(x\right)-X_{-}\left(x\right)}{2i}=0$ (31)

where $\alpha =\frac{g^2}{8}\frac{1}{2i}$ .

We derived a charged pion wave function and an electromagnetic form factor in 3 + 1 dimensions in ref. [13] and showed that a charged pion wave function is represented by ${\chi }_3$ ($\beta \left(i\overrightarrow{\alpha }\cdot \widehat{r}\right)$ component) only because ${\chi }_2$ ($\beta$ component) is zero. Equation (30) and Equation (31) correspond to these results.

3. Distribution Amplitude

In this section, we construct a charged pion distribution amplitude. To achieve this aim, we apply one dimensional Fourier transform to $X_3\left(x\right)$ .

In the case of our $\kappa$ and $\mu$ values, ${\varphi }_{-}^{(+)}\left(x\right)$ and ${\varphi }_{-}^{(-)}\left(x\right)$ can be described as

${\varphi }_{-}^{(+)}\left(x\right)=c_1\frac{{\left(\frac{1}{2}{\text{e}}^{-i\frac{\pi }{2}}\left|\alpha \right|x^2\right)}^{\frac{1}{4}}}{{\left({\text{e}}^{-i\frac{\pi }{4}}\sqrt{\left|\alpha \right|}x\right)}^{\frac{1}{2}}}{}_1{}^{}F{}_1\left(1+\frac{i}{\pi },\frac{1}{2};\frac{1}{2}{\text{e}}^{-i\frac{\pi }{2}}\left|\alpha \right|x^2\right)$

$\begin{array}{c}+c_2\frac{{\left(\frac{1}{2}{\text{e}}^{-i\frac{\pi }{2}}\left|\alpha \right|x^2\right)}^{\frac{3}{4}}}{{\left({\text{e}}^{-i\frac{\pi }{4}}\sqrt{\left|\alpha \right|}x\right)}^{\frac{1}{2}}}{}_1{}^{}F{}_1\left(\frac{3}{2}+\frac{i}{\pi },\frac{3}{2};\frac{1}{2}{\text{e}}^{-i\frac{\pi }{2}}\left|\alpha \right|x^2\right)\\ ={c^{\prime }}_1{}_1{}^{}F{}_1\left(1+\frac{i}{\pi },\frac{1}{2};\frac{1}{2}{\text{e}}^{-i\frac{\pi }{2}}\left|\alpha \right|x^2\right)+{c^{\prime }}_2\sqrt{\left|\alpha \right|}\sqrt{x^2}{}_1{}^{}F{}_1\left(\frac{3}{2}+\frac{i}{\pi },\frac{3}{2};\frac{1}{2}{\text{e}}^{-i\frac{\pi }{2}}\left|\alpha \right|x^2\right)\end{array}$ (32)

${\varphi }_{-}^{(-)}\left(x\right)={d^{\prime }}_1{}_1{}^{}F{}_1\left(1-\frac{i}{\pi },\frac{1}{2};\frac{1}{2}{\text{e}}^{i\frac{\pi }{2}}\left|\alpha \right|x^2\right)+{d^{\prime }}_2\sqrt{\left|\alpha \right|}\sqrt{x^2}{}_1{}^{}F{}_1\left(\frac{3}{2}-\frac{i}{\pi },\frac{3}{2};\frac{1}{2}{\text{e}}^{i\frac{\pi }{2}}\left|\alpha \right|x^2\right)$ (33)

Equation (32) and Equation (33) show ${\varphi }_{-}^{(+)}\left(x\right)$ and ${\varphi }_{-}^{(-)}\left(x\right)$ are clearly even functions. Thus, $X_3\left(x\right)$ is an even function.

$F_3\left(\left|q\right|\right)={\displaystyle {\int }_{-\infty }^{\infty }\text{d}x}{\text{e}}^{-iqx}{X}_3\left(x\right)={\displaystyle {\int }_{-\infty }^0\text{d}x}{\text{e}}^{-iqx}{X}_3\left(x\right)+{\displaystyle {\int }_0^{\infty }\text{d}x}{\text{e}}^{-iqx}{X}_3\left(x\right)$ (34)

For the first term of the second line of Equation (34), by changing the variable $x$ to $-x^{\prime }$ , the first term becomes

$\text{first}\text{term}={\displaystyle {\int }_{\infty }^0-\text{d}{x}^{\prime }}{\text{e}}^{-iq{x}^{\prime }}{X}_3\left(-x^{\prime }\right)={\displaystyle {\int }_0^{\infty }\text{d}{x}^{\prime }}{\text{e}}^{iq{x}^{\prime }}{X}_3\left(x^{\prime }\right)={\displaystyle {\int }_0^{\infty }\text{d}{x}^{\prime }}{\text{e}}^{-i\left|q\right|x^{\prime }}{X}_3\left(x^{\prime }\right)$

For the second line, we recall the fact that $X_3$ is an even function. Recalling the fact that the measured momentum is always $Q^2$ in experiment and $\sqrt{Q^2}=\pm \left|Q\right|$ , we take $q=-\left|q\right|$ in the last line. By taking $q=+\left|q\right|$ in the second term of the second line of Equation (34), Equation (34) becomes

$F_3\left(\left|q\right|\right)=2{\displaystyle {\int }_0^{\infty }\text{d}x}{\text{e}}^{-i\left|q\right|x}{X}_3\left(x\right)$ (35)

Recalling that the distribution amplitude is considered in 4 (3 + 1) dimensions and that using solutions of the ‘t Hooft model obtained in 2 (1 + 1) dimensions is only analogy, we may consider that $qx$ is actually $\left|q\right|\left|x\right|$ in this case. Then we obtain the same result for $F_3\left(\left|q\right|\right)$ without using above argument.

To proceed further, we evaluate the integral as follows.

$F_3^{\left(1\right)}\left(\left|q\right|\right)=\text{Γ}\left(1+\frac{i}{\pi }\right){\displaystyle {\int }_0^{\infty }\text{d}x}{\text{e}}^{-i\left|q\right|x}{\text{e}}^{\frac{1}{4}\alpha x^2}{\left(\frac{1}{2}\sqrt{\alpha }\sqrt{x^2}\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}\alpha x^2\right)$ (36)

$F_3^{\left(2\right)}\left(\left|q\right|\right)=\Gamma \left(1-\frac{i}{\pi }\right){\displaystyle {\int }_0^{\infty }\text{d}x}{\text{e}}^{-i\left|q\right|x}{\text{e}}^{-\frac{1}{4}\alpha x^2}{\left(i\frac{1}{2}\sqrt{\alpha }\sqrt{x^2}\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(-\frac{1}{2}\alpha x^2\right)$ (37)

Recalling that $\alpha =\frac{g^2}{8}\frac{1}{2i}=\left|\alpha \right|{\text{e}}^{-i\frac{\pi }{2}}$ , we change the variables as $x={\text{e}}^{-i\frac{3\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}z$ for Equation (36) and $x={\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}\overline{z}$ for Equation (37), respectively. Then Equation (36) and Equation (37) become

$F_3^{\left(1\right)}\left(\left|q\right|\right)=\Gamma \left(1+\frac{i}{\pi }\right){\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}{\displaystyle {\int }_0^{{\text{e}}^{i\frac{3\pi }{4}\left|\alpha \right|\infty }}\text{d}z}\exp \left(-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}z\left(1-i\right)\right){\text{e}}^{\frac{1}{4}{z}^2}{\left(\frac{1}{2}z\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{z}^2\right)$ (38)

$F_3^{\left(2\right)}\left(\left|q\right|\right)=\Gamma \left(1-\frac{i}{\pi }\right){\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}{\displaystyle {\int }_0^{{\text{e}}^{i\frac{\pi }{4}}\left|\alpha \right|\infty }\text{d}\overline{z}}\exp \left(-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}\overline{z}\left(1+i\right)\right){\text{e}}^{\frac{1}{4}{\overline{z}}^2}{\left(\frac{1}{2}\overline{z}\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{\overline{z}}^2\right)$ (39)

Note that, for Equation (36), ${\text{e}}^{-i\frac{3\pi }{4}}{\left(\frac{1}{2}\sqrt{\left|\alpha \right|}{\text{e}}^{-i\frac{\pi }{4}}{\text{e}}^{-i\frac{3\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}z\right)}^{-\frac{1}{2}}={\text{e}}^{-i\frac{\pi }{4}}{\left(\frac{1}{2}z\right)}^{-\frac{1}{2}}$ .

We consider the integral contour of Figure 1 for Equation (38) and that of Figure 2 for Equation (39), respectively.

Figure 1. Integral contour for Eq. (38).

Figure 2. Integral contour for Eq. (39).

Because there are no poles inside of contour area for both Equation (38) and Equation (39) cases, we obtain $I_1+I_2+I_3=0$ . Thus $I_1=-I_3$ because of $I_2=0$ (see Appendix D). This means that we can change the integral range of both cases from 0 to $\infty$ .

Then, we obtain the Fourier transform of $X_3\left(x\right)$ , namely $F_3\left(\left|q\right|\right)$ , except constants as follows:

$\begin{array}{c}{F}_3\left(\left|q\right|\right)=\Gamma \left(1+\frac{i}{\pi }\right){\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}{\displaystyle {\int }_0^{\infty }\text{d}z}\exp \left(-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}z\left(1-i\right)\right){\text{e}}^{\frac{1}{4}{z}^2}{\left(\frac{1}{2}z\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{z}^2\right)\\ -\Gamma \left(1-\frac{i}{\pi }\right){\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}{\displaystyle {\int }_0^{\infty }\text{d}z}\exp \left(-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}z\left(1+i\right)\right){\text{e}}^{\frac{1}{4}{z}^2}{\left(\frac{1}{2}z\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{z}^2\right)\end{array}$ (40)

The absolute value of $F_3\left(\left|q\right|\right)$ , namely $\left|F_3\left(\left|q\right|\right)\right|$ , is our charged pion distribution amplitude in the range of $\left[0,\infty \right]$ . Note that over all coefficient is unimportant because actual distribution amplitude should be normalized.

In order to construct a charged pion distribution amplitude in commonly used

range, we have to replace $\left|q\right|$ to $\frac{q_1}{1-q_1}$ . Recalling the fact that $q_1$ is

$\text{Momentum}/\text{Maximal Momentum}$ and that $\text{Momentum}=\overline{x}\cdot \text{Maximal}\text{Momentum}$ $\left(\overline{x}\in \left[0,1\right]\right)$ , we obtain our charged pion distribution amplitude as

$\begin{array}{c}{F}_3\left(\overline{x}\right)=\Gamma \left(1+\frac{i}{\pi }\right){\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}{\displaystyle {\int }_0^{\infty }\text{d}z}\exp \left(-\frac{1-i}{\sqrt{2\left|\alpha \right|}}\frac{\overline{x}}{1-\overline{x}}z\right){\text{e}}^{\frac{1}{4}{z}^2}{\left(\frac{1}{2}z\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{z}^2\right)\\ -\Gamma \left(1-\frac{i}{\pi }\right){\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}{\displaystyle {\int }_0^{\infty }\text{d}z}\exp \left(-\frac{1+i}{\sqrt{2\left|\alpha \right|}}\frac{\overline{x}}{1-\overline{x}}z\right){\text{e}}^{\frac{1}{4}{z}^2}{\left(\frac{1}{2}z\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{z}^2\right)\end{array}$(41)

Again we have to take absolute value of $F_3\left(\overline{x}\right)$ and normalizaon is necessary.

In order to observe the behavior of our distribution amplitude, we construct an approximate form of $F_3\left(\left|q\right|\right)$ .

In order to evaluate integral, we divided integral region two parts, namely range 1 is $\left[0,u\right]$ and range 2 is $\left[u,\infty \right]$ .

When $\left|q\right|$ is very large, only the first term of Whittaker function is sufficient to evaluate integral in range 1 and an asymptotic form of $W_{\kappa ,\mu }$ is sufficient to range 2 integral. Because Whittaker function $W_{\kappa ,\mu }$ is defined as Equation (27), integral becomes

$\begin{array}{c}I={\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}\left({\displaystyle {\int }_0^u\text{d}z}\exp \left(-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}z\left(1-i\right)\right)+{\displaystyle {\int }_u^{\infty }\text{d}z}\frac{1}{z^{1+\frac{i}{\pi }}}\exp \left(-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}z\left(1-i\right)\right)\right)\\ -{\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}\left({\displaystyle {\int }_0^u\text{d}z}\exp \left(-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}z\left(1+i\right)\right)+{\displaystyle {\int }_u^{\infty }\text{d}z}\frac{1}{z^{1-\frac{i}{\pi }}}\exp \left(-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}z\left(1+i\right)\right)\right)\end{array}$ (42)

Here we use an asymptotic form of $W_{\kappa ,\mu }\left(z\right)$ described in Equation (43) [21] for the second and the fourth terms.

$W_{\kappa ,\mu }\left(z\right)~{\text{e}}^{-\frac{z}{2}}{z}^{\kappa }\left[1+{\displaystyle \sum }_{n=1}^{\infty }\frac{\left[{\mu }^2-{\left(\kappa -\frac{1}{2}\right)}^2\right]\cdots \left[{\mu }^2-{\left(\kappa -n+\frac{1}{2}\right)}^2\right]}{n!z^n}\right]$ (43)

Note that Γ-function are cancelled out by coefficient of $M_{\kappa ,\mu }$ of Whittaker function $W_{\kappa ,\mu }$ (refer Equation (27).

Changing variable $\frac{z}{\sqrt{2\left|\alpha \right|}}$ to $z^{\prime }$ , integral becomes

$I={\text{e}}^{-i\frac{\pi }{4}}[\left(\sqrt{2}{\displaystyle {\int }_0^{\frac{u}{\sqrt{2\left|\alpha \right|}}}\text{d}{z}^{\prime }}\exp \left(-\left|q\right|z^{\prime }\left(1-i\right)\right)+{\left(\frac{1}{2\left|\alpha \right|}\right)}^{\frac{i}{2\pi }}{\displaystyle {\int }_{\frac{u}{\sqrt{2\left|\alpha \right|}}}^{\infty }\text{d}{z}^{\prime }}\frac{1}{{z^{\prime }}^{1+\frac{i}{\pi }}}\exp \left(-\left|q\right|z^{\prime }\left(1-i\right)\right)\right)$

$-\left(\sqrt{2}{\displaystyle {\int }_0^{\frac{u}{\sqrt{2\left|\alpha \right|}}}\text{d}{z}^{\prime }}\exp \left(-\left|q\right|z^{\prime }\left(1+i\right)\right)+{\left(\frac{1}{2\left|\alpha \right|}\right)}^{-\frac{i}{2\pi }}{\displaystyle {\int }_{\frac{u}{\sqrt{2\left|\alpha \right|}}}^{\infty }\text{d}{z}^{\prime }}\frac{1}{{z^{\prime }}^{1-\frac{i}{\pi }}}\exp \left(-\left|q\right|z^{\prime }\left(1+i\right)\right)\right)]$ (44)

We can obtain the results of integral in Equation (44) by employing the integral formula [22]

${\displaystyle {\int }_0^{\frac{u}{\sqrt{2\left|\alpha \right|}}}\text{d}z}\exp \left(-\left|q\right|z\left(1\mp i\right)\right)=\frac{1}{\left|q\right|\left(1\mp i\right)}\gamma \left(1,\frac{\left|q\right|\left(1\mp i\right)u}{\sqrt{2\left|\alpha \right|}}\right)$ (45)

where $\gamma \left({\lambda }^{\prime },p\right)$ is incomplete gamma function defined as

$\gamma \left({\lambda }^{\prime },p\right)={\displaystyle {\int }_0^p\text{d}t}{\text{e}}^{-t}{t}^{{\lambda }^{\prime }-1}$

and its series expansion is shown as [21]

$\gamma \left({\lambda }^{\prime },p\right)={\text{e}}^{-p}{\displaystyle \sum }_{n=0}^{\infty }\frac{p^{{\lambda }^{\prime }+n}}{{\lambda }^{\prime }\left({\lambda }^{\prime }+1\right)\cdots \left({\lambda }^{\prime }+n\right)}$

For ${\lambda }^{\prime }=1$ case, we can evaluate summation as follows:

$\gamma \left(1,p\right)={\text{e}}^{-p}{\displaystyle \sum }_{n=0}^{\infty }\frac{p^{n+1}}{12\cdots n\left(n+1\right)}$

Taking $n+1=\overline{n}$ , summation begins $\overline{n}=1$ , then summation becomes as

$\text{sum}={\displaystyle \sum }_{\overline{n}=1}^{\infty }\frac{p^{\overline{n}}}{\overline{n}!}={\displaystyle \sum }_{\overline{n}=0}^{\infty }\frac{p^{\overline{n}}}{\overline{n}!}-1={\text{e}}^p-1$

Then, $\gamma \left(1,p\right)=1-{\text{e}}^{-p}$ .

Thus, we obtain integral results as

$I^{\left(1\right)}=\frac{1}{\left|q\right|\left(1\mp i\right)}\left(1-\exp \left(-\frac{\left|q\right|\left(1\mp i\right)u}{\sqrt{2\left|\alpha \right|}}\right)\right)$

$\begin{array}{c}{I}^{\gamma }=\frac{1}{\left|q\right|}\left[\left(\frac{1}{1-i}-\frac{1}{1+i}\right)-\exp \left(-\frac{\left|q\right|u}{\sqrt{2\left|\alpha \right|}}\right)\left(\frac{1}{1-i}\exp \left(\frac{iqu}{\sqrt{2\left|\alpha \right|}}\right)-\frac{1}{1+i}\exp \left(-\frac{i\left|q\right|u}{\sqrt{2\left|\alpha \right|}}\right)\right)\right]\\ =\frac{i}{\left|q\right|}\left[1-\exp \left(-\frac{\left|q\right|u}{\sqrt{2\left|\alpha \right|}}\right)\left(\sin \left(\frac{\left|q\right|u}{\sqrt{2\left|\alpha \right|}}\right)+\cos \left(\frac{\left|q\right|u}{\sqrt{2\left|\alpha \right|}}\right)\right)\right]\end{array}$ (46)

$\begin{array}{c}{I}^{\Gamma }={\displaystyle {\int }_{\frac{u}{\sqrt{2\left|\alpha \right|}}}^{\infty }\text{d}z}\frac{1}{z^{1\pm \frac{i}{\pi }}}\exp \left(-\left|q\right|z\left(1\mp i\right)\right)\\ ={\displaystyle {\int }_{\frac{u}{\sqrt{2\left|\alpha \right|}}}^{\infty }\text{d}z}\frac{1}{z}\exp \left(-\left|q\right|z\left(1\mp i\right)\right)\left(\cos \left(\frac{1}{\pi }\ln \left(z\right)\right)\mp i\sin \left(\frac{1}{\pi }\ln \left(z\right)\right)\right)\end{array}$ (47)

Because sin-function and cosine-function exist inside the integral and integration is taken up to infinity. We can conclude that $I^{\text{Γ}}$ is much smaller than $I^{\gamma }$ . Thus, we neglect contribution of $I^{\text{Γ}}$ to pion distribution amplitude.

For small $\left|q\right|$ case, we consider Equation (40) that is exact form of $F_3\left(\left|q\right|\right)$ .

Equation (40) is rewritten as

$\begin{array}{c}{F}_3\left(\left|q\right|\right)=\Gamma \left(1+\frac{i}{\pi }\right){\text{e}}^{-i\frac{\pi }{4}}\frac{1}{\sqrt{\left|\alpha \right|}}[{\displaystyle {\int }_0^{\infty }\text{d}z}\exp \left(-\frac{\left|q\right|z}{\sqrt{2\left|\alpha \right|}}\left(1-i\right)\right){\text{e}}^{\frac{1}{4}{z}^2}{\left(\frac{1}{2}z\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{z}^2\right)\\ -\Gamma \left(1-\frac{i}{\pi }\right){\text{e}}^{-i\frac{\pi }{4}}{\displaystyle {\int }_0^{\infty }\text{d}z}\exp \left(-\frac{\left|q\right|z}{\sqrt{2\left|\alpha \right|}}\left(1+i\right)\right){\text{e}}^{\frac{1}{4}{z}^2}{\left(\frac{1}{2}z\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{z}^2\right)]\end{array}$ (48)

For integral, we recall an integral representation of Whittaker function $W_{\kappa ,\mu }\left(z\right)$ .

In our case, because the condition $\text{Re}\left(\mu -\kappa +\frac{1}{2}\right)>0$ is satisfied, we can employ the following integral representation for $W_{\kappa ,\mu }\left(z\right)$ [21]:

$W_{\kappa ,\mu }\left(z\right)=\frac{{\text{e}}^{-\frac{z}{2}}{z}^{\kappa }}{\text{Γ}\left(\mu -\kappa +\frac{1}{2}\right)}{\displaystyle {\int }_0^{\infty }\text{d}t}{\text{e}}^{-t}{t}^{\mu -\kappa -\frac{1}{2}}{\left(1+\frac{t}{z}\right)}^{\mu +\kappa -\frac{1}{2}}$ (49)

Recalling that $\mu =-\frac{1}{4},\kappa =-\frac{1}{4}\mp \frac{i}{\pi }$

${\text{e}}^{\frac{1}{4}{z}^2}{\left(\frac{1}{2}z\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}\mp \frac{i}{\pi },-\frac{1}{4}}\left(\frac{1}{2}{z}^2\right)=\frac{1}{\Gamma \left(\frac{1}{2}\pm \frac{i}{\pi }\right)}{2}^{\frac{3}{4}\pm \frac{i}{\pi }}z{\displaystyle {\int }_0^{\infty }\text{d}t}{\text{e}}^{-t}{t}^{-\frac{1}{2}\pm \frac{i}{\pi }}{\left(\frac{1}{z^2+2t}\right)}^{1\pm \frac{i}{\pi }}$ (50)

Integral part of $F_3\left(\left|q\right|\right)$ apart from constant becomes

$\begin{array}{c}{I}_{F3}={\displaystyle {\int }_0^{\infty }\text{d}z}\exp \left(-\frac{\left|q\right|z}{\sqrt{2\left|\alpha \right|}}\left(1\mp i\right)\right)z{\displaystyle {\int }_0^{\infty }\text{d}t}{\text{e}}^{-t}{t}^{-\frac{1}{2}\pm \frac{i}{\pi }}{\left(\frac{1}{z^2+2t}\right)}^{1\pm \frac{i}{\pi }}\\ ={\displaystyle {\int }_0^{\infty }\text{d}t}{\text{e}}^{-t}{t}^{-\frac{1}{2}\pm \frac{i}{\pi }}{\displaystyle {\int }_0^{\infty }\text{d}z}\exp \left(-\lambda z\right)z{\left(z^2+2t\right)}^{-1\mp \frac{i}{\pi }}\end{array}$ (51)

We can refer the following formula for z integral [22].

${\displaystyle {\int }_0^{\infty }\text{d}x}{x}^{2\nu -1}{\left(w^2+x^2\right)}^{\xi -1}{\text{e}}^{-\lambda x}=\frac{w^{2\nu +2\xi -2}}{2\sqrt{\pi }\text{Γ}\left(1-\xi \right)}{G}_{13}^{31}\left(\frac{{\lambda }^2{w}^2}{4}|\begin{array}{ccc}1-\nu & & \\ 1-\xi -\nu & 0& \frac{1}{2}\end{array}\right)$ (52)

In our case, $\nu =1$ , $\xi =\mp \frac{i}{\pi }$ , $\lambda =\frac{\left|q\right|\left(1\mp i\right)}{\sqrt{2\left|\alpha \right|}}$ , $w=\sqrt{2t}$ , z integral becomes

$I_z=\frac{{\left(2t\right)}^{\mp 2\frac{i}{\pi }}}{2\sqrt{\pi }\text{Γ}\left(1\pm \frac{i}{\pi }\right)}{G}_{13}^{31}\left(\frac{{\lambda }^{2}2t}{4}|\begin{array}{ccc}0& & \\ \pm \frac{i}{\pi }& 0& \frac{1}{2}\end{array}\right)$ (53)

Then, $I_{F3}$ becomes apart from constant

$I_3=\frac{2^{-1\mp 2\frac{i}{\pi }}}{\sqrt{\pi }\text{Γ}\left(1\pm \frac{i}{\pi }\right)}{\displaystyle {\int }_0^{\infty }\text{d}t}{\text{e}}^{-t}{t}^{-\frac{1}{2}\mp \frac{i}{\pi }}{G}_{13}^{31}\left(\frac{{\lambda }^{2}2t}{4}|\begin{array}{ccc}0& & \\ \pm \frac{i}{\pi }& 0& \frac{1}{2}\end{array}\right)$ (54)

where $G_{pq}^{mn}$ is Meijer’s function defined as [22]

$G_{pq}^{mn}\left({x|}_{br}^{ar}\right)={\displaystyle \sum }_{h=1}^m\frac{{{\displaystyle \prod }}_{j=1}^m\text{Γ}\left(b_j-b_h\right){{\displaystyle \prod }}_{j=1}^n\text{Γ}\left(1+b_h-a_j\right)}{{{\displaystyle \prod }}_{j=m+1}^q\text{Γ}\left(1+b_h-b_j\right){{\displaystyle \prod }}_{j=n+1}^p\text{Γ}\left(a_j-b_h\right)}{x}^{b_h}$

$\times {}_p{}^{}F{}_{q-1}\left(1+b_h-a_1,\cdots ,1+b_h-a_p;1+b_h-b_1,\cdots ,\ast ,\cdots ,1+b_h-b_q:{\left(-1\right)}^{p-m-n}x\right)$

This definition is given under the condition such that either $p<q$ or $p=q$ and $\left|x\right|<1$ .

In our case, recalling Equation (48), Equation (49) and Equation (54), the first terms of $\left|F_3\left(\left|q\right|\right)\right|$ become apart from real value constants as

$\begin{array}{c}\text{First}\text{term}\text{of}\left|F_3\left(\left|q\right|\right)\right|=\frac{\left|\text{Γ}\left(\frac{i}{\pi }\right)\right|\text{Γ}\left(\frac{1}{2}\right)}{\left|\text{Γ}\left(\frac{1}{2}+\frac{i}{\pi }\right)\right|}{\left(\frac{{\lambda }^{2}2t}{4}\right)}^0-\frac{\left|\text{Γ}\left(-\frac{i}{\pi }\right)\right|\text{Γ}\left(\frac{1}{2}\right)}{\left|\text{Γ}\left(\frac{1}{2}-\frac{i}{\pi }\right)\right|}{\left(\frac{{\lambda }^{2}2t}{4}\right)}^0\\ +\frac{\left|\text{Γ}\left(-\frac{i}{\pi }\right)\right|\left|\text{Γ}\left(\frac{1}{2}-\frac{i}{\pi }\right)\right|\left|\text{Γ}\left(1+\frac{i}{\pi }\right)\right|}{\left|\text{Γ}\left(\frac{1}{2}+\frac{i}{\pi }\right)\right|}\left|{\left(\frac{{\lambda }^{2}2t}{4}\right)}^{\frac{i}{\pi }}\right|\\ -\frac{\left|\text{Γ}\left(\frac{i}{\pi }\right)\right|\left|\text{Γ}\left(\frac{1}{2}+\frac{i}{\pi }\right)\right|\left|\text{Γ}\left(1-\frac{i}{\pi }\right)\right|}{\left|\text{Γ}\left(\frac{1}{2}-\frac{i}{\pi }\right)\right|}\left|{\left(\frac{{\lambda }^{2}2t}{4}\right)}^{-\frac{i}{\pi }}\right|\\ +\frac{\text{Γ}\left(-\frac{1}{2}\right)\left|\text{Γ}\left(-\frac{1}{2}+\frac{i}{\pi }\right)\right|\text{Γ}\left(\frac{3}{2}\right)}{\left|\text{Γ}\left(\frac{1}{2}+\frac{i}{\pi }\right)\right|}{\left(\frac{{\left(\frac{\left|q\right|\left(1-i\right)}{\sqrt{2\left|\alpha \right|}}\right)}^{2}2t}{4}\right)}^{\frac{1}{2}}\\ -\frac{\text{Γ}\left(-\frac{1}{2}\right)\left|\text{Γ}\left(-\frac{1}{2}-\frac{i}{\pi }\right)\right|\text{Γ}\left(\frac{3}{2}\right)}{\left|\text{Γ}\left(\frac{1}{2}-\frac{i}{\pi }\right)\right|}{\left(\frac{{\left(\frac{\left|q\right|\left(1+i\right)}{\sqrt{2\left|\alpha \right|}}\right)}^{2}2t}{4}\right)}^{\frac{1}{2}}\\ =\text{absolute}\text{value}\text{of}\left(-2\pi i\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}\sqrt{2t}\sqrt{\frac{1}{4}+{\left(\frac{1}{\pi }\right)}^2}\right)\end{array}$ (55)

The reason why we take absolute value of ${\left(\frac{{\lambda }^{2}2t}{4}\right)}^{\pm \frac{i}{\pi }}$ is following. This can be rewritten as

${\left(\frac{{\lambda }^{2}2t}{4}\right)}^{\pm \frac{i}{\pi }}=t^{\pm \frac{i}{\pi }}\exp \left(\pm \frac{i}{\pi }\ln \left|\frac{{\lambda }^2}{2}\right|\right)=t^{\pm \frac{i}{\pi }}\left(\cos \left(\frac{1}{\pi }\ln \left|\frac{{\lambda }^2}{2}\right|\right)\pm i\sin \left(\frac{1}{\pi }\ln \left|\frac{{\lambda }^2}{2}\right|\right)\right)$ . Note that $t^{\pm \frac{i}{\pi }}$ term cancels integral part of $t^{\mp \frac{i}{\pi }}$ term when considering only the first

term of Meijer’s G-function. Recalling the fact that we must include $\left|q\right|=0$ case $\left(\lambda =0\right)$ , $\left|F_3\left(\left|q\right|\right)\right|$ become meaningless because $\sin \left(\infty \right),\cos \left(\infty \right)$ do not exist. In addition, the other reason is that, recalling Equation (54), this integral becomes meaningful when t can be taken from 0 to quite large value for a certain fixed ${\lambda }^2$ value ($\left|q\right|$ value). That means ${\lambda }^2$ should be quite small ($\left|q\right|$ quite small) because

of the condition $\left|\frac{{\lambda }^{2}2t}{4}\right|<1$ . Then, integral becomes ${\displaystyle {\int }_0^{\overline{\infty }}\text{d}t}{\text{e}}^{-t}{t}^{-\frac{1}{2}}\sin \left(\frac{1}{\pi }\ln \left|\frac{{\lambda }^2}{2}\right|\right)$ or $\cos \left(\frac{1}{\pi }\ln \left|\frac{{\lambda }^1}{2}\right|\right)\approx 0$ because sin or cos-function rapidly oscillates. Thus, we cannot evaluate this contribution to $\left|F_3\left(\left|q\right|\right)\right|$ properly when we take sin, cos-functional representation. Note that $\overline{\infty }$ denotes vary large value except $\infty$ for any finite ${\lambda }^2$ (quite small), and that t-integral becomes ${\displaystyle {\int }_0^{\overline{\infty }}\text{d}t}{\text{e}}^{-t}{t}^{n+\frac{1}{2}\mp \frac{i}{\pi }-1}\approx \Gamma \left(n+\frac{1}{2}\mp \frac{i}{\pi }\right)$ because n-th term of ${}_1{}^{}F{}_2$ is ${\left(\frac{{\lambda }^2}{2}\right)}^n{t}^n$ . Recalling the fact that we have to take absolute value of Γ-function when considering $\left|F_3\left(\left|q\right|\right)\right|$ as shown later, we notice that this series become well defined when we take the absolute value of ${\left(\frac{{\lambda }^{2}2t}{4}\right)}^{\pm \frac{i}{\pi }}$ because this absolute value is the first term and also it multiplies to all other terms.

Here we employ formula as

$\begin{array}{l}\Gamma \left(iy\right)={\left[\pi /y\sinh \left(\pi y\right)\right]}^{\frac{1}{2}},\Gamma \left(\frac{1}{2}+iy\right)={\left[\pi /\cosh \left(\pi y\right)\right]}^{\frac{1}{2}},\\ \Gamma \left(\frac{1}{2}\pm n+iy\right)={\left(\frac{\pi }{\cosh \left(\pi y\right)}{\displaystyle \prod }_{k=1}^n\left({\left(k-\frac{1}{2}\right)}^2+y^2\right)\right)}^{\frac{1}{2}},\\ \Gamma \left(1+iy\right)={\left[\pi y/\sinh \left(\pi y\right)\right]}^{\frac{1}{2}}\end{array}$

These formula can be obtained by employing the following relations [21].

$\Gamma \left(z\right)\Gamma \left(1-z\right)=\pi /\sin \left(\pi z\right)$ and $\text{Γ}\left(z+\frac{1}{2}\right)\text{Γ}\left(\frac{1}{2}-z\right)=\pi /\cos \left(\pi z\right)$

$\text{Γ}\left(z+1\right)=z\text{Γ}\left(z\right)$ and $\overline{\text{Γ}\left(z\right)}=\text{Γ}\left(\overline{z}\right)$

The reason why taking the absolute value of Γ-functions in Equation (55) is as follows.

Γ-function is defined by Euler as

$\text{Γ}\left(z\right)=\frac{1}{z}{\displaystyle \prod }_{n=1}^{\infty }\left[{\left(1+\frac{1}{n}\right)}^z/\left(1+\frac{z}{n}\right)\right]$

When z is complex, we cannot specify $\text{Γ}\left(z\right)$ value. Thus we have to take the absolute value to specify $\text{Γ}\left(z\right)$ value in the case of complex z. Because distribution amplitude is absolute value of $F_3\left(\left|q\right|\right)$ , we have to take the absolute value of $\text{Γ}\left(z\right)$ to evaluate it.

Although Equation (46) is obtained for large $\left|q\right|$ case, Expansion of Equation (46) becomes

$\begin{array}{c}{I}^{\gamma }=\frac{i}{\left|q\right|}\left[1-\left(1-\left|q\right|\rho +\frac{1}{2}{\left(\left|q\right|\rho \right)}^2\right)\left(1+\left|q\right|\rho -\frac{1}{2}{\left(\left|q\right|\rho \right)}^2\right)\right]\\ =\frac{i}{\left|q\right|}\left[{\left(\left|q\right|\rho \right)}^2+O\left({\left(\left|q\right|\rho \right)}^3\right)\right]=i{\rho }^2\left|q\right|\end{array}$ (56)

where $\rho =\frac{u}{\sqrt{2\left|\alpha \right|}}$ .

Thus, the first term of $I^{\gamma }$ is the same order of $\left|q\right|$ as that of Equation (55) so that we consider Equation (46) as an approximate form of distribution function.

Then, we obtain an approximate form of charged pion distribution amplitude in the range of $\left[0,\infty \right]$ as

$F_3\left(\left|q\right|\right)=\frac{1}{\left|q\right|}\left[1-\exp \left(-\left|q\right|\rho \right)\left(\sin \left(\left|q\right|\rho \right)+\cos \left(\left|q\right|\rho \right)\right)\right]$ (57)

In order to construct distribution amplitude in the range of $\left[0,1\right]$ , we again set $\left|q\right|$ as $\left|q\right|=\frac{q_1}{1-q_1}$ . The range of $q_1$ is $\left[0,1\right]$ and $q_1$ is considered as $\frac{\text{Momentum}}{\text{Maximum Momentum}}$ .

Momentum can be written as Momentum = $\overline{x}$ ·(Max Momentum) where $\overline{x}\in \left[0,1\right]$ .

Thus, Equation (57) is rewritten as

$\left|F_3\left(\overline{x}\right)\right|=F_3\left(\overline{x}\right)=\frac{1-\overline{x}}{\overline{x}}\left[1-{\text{e}}^{-\frac{\overline{x}}{1-\overline{x}}\rho }\left(\sin \left(\frac{\overline{x}}{1-\overline{x}}\rho \right)+\cos \left(\frac{\overline{x}}{1-\overline{x}}\rho \right)\right)\right]$ (58)

The form of Equation (58) is very simple, however, when we compare this with other form of pion distribution amplitude such that ${\varphi }_{\pi }=\frac{1}{B\left(m,n\right)}{x}^m{\left(1-x\right)}^n$ by

Zhang [8] where $B\left(m,n\right)$ is Beta function, and ${\varphi }_{\pi }=1.81{\left[x\left(1-x\right)\right]}^{0.31}\left[1-0.12C_2^{0.81}\left(2x-1\right)\right]$ by Chang [9], and the form of the distribution amplitude without considering pion radius effect by Raya as [7]

$u^P=n_P{x}^2{\left(1-x\right)}^2{\left[1+{\rho }_P{x}^{\frac{{\alpha }_P}{2}}{\left(1-x\right)}^{\frac{{\beta }_P}{2}}+{\gamma }_P{x}^{{\alpha }_P}{\left(1-x\right)}^{{\beta }_P}\right]}^2$

we can consider that the form of Equation (58) is sufficient to observe the basic behavior of our distribution amplitude.

We show behavior of normalized $F_3\left(x\right)$ based on Equation (58) for several $\rho$ values case in Figure 3.

Note that we change the notation $\overline{x}$ to $x$ from now on so that the range of this $x$ is $\left[0,1\right]$ .

We can compare results of Figure 3 to those of Raya et al. [7]. Especially, we can compare our results of $\rho =1.0$ and $\rho =0.5$ cases to Raya’s results of ${\text{Δ}}^2{r}^2=2$ and ${\text{Δ}}^2{r}^2=5$ case, respectively. According to Raya, $\text{Δ}$ is defined as $\Delta =p^{\prime }-p$ where p’ and p are the final, initial meson momenta in the defining scattering process and r is pion radius. For both cases, our peaks are almost same as theirs and maximum values are close but smaller than theirs. Most intriguing point is that, for less than $\rho =1.0$ case, our peaks shift towards $x=1$ . This shift is also shown in Raya’s results as ${\text{Δ}}^2{r}^2$ increases. In our case, the cause of peak shift is basically magnitude of a coupling constant $g^2$ because of the setting that

$\rho =\frac{u}{\sqrt{2\left|\alpha \right|}}=\frac{2\sqrt{2}u}{\sqrt{g^2}}$ . The behavior of $\left|F_3\left(x\right)\right|$ near $x=1$ is that $\left|F_3\left(x\right)\right|\to \frac{1-x}{x}$

at $x\to 1$ . This behavior is also true for the exact form of $\left|F_3\left(x\right)\right|$ case because the evaluation for very large $\left|q\right|$ case is sufficiently accurate.

Figure 3. Distribution amplitude with various $\rho$ values.

Important point is that Raya et al. consider that pion distribution amplitude is directly related to a valence u-quark distribution function of pion [7] and they employ one of their results of distribution amplitude to represent a valence u-quark distribution function of pion [6].

4. Conclusion and Discussion

Applying our solutions of the ‘t Hooft model to construct a pion distribution amplitude, we obtain a pion distribution amplitude $\left|F_3\left(x\right)\right|$ as the absolute value of Equation (41). By exploring its asymptotic form described as Equation (58), we find that our results show each peak shifts towards $x=1$ caused by magnitude

of a coupling constant $g^2$ and that $\left|F_3\left(x\right)\right|\to \frac{1-x}{x}$ as $x\to 1$ . According to the correct Drell-Yan-West relation, for hadron H defined by $n+1$ valence spin $J=\frac{1}{2}$ partons, its leading elastic form factor scales as ${\left(1/Q^2\right)}^n$ [23]:

$q_H\left(x\right)~{\left(1-x\right)}^{\beta }\text{when}x\cong 1,$

$\beta =2n-1+2\Delta S_z\text{where}\Delta S_z=\left|S_z^q-S_z^H\right|$

Because $S_z^H=0$ for pseudo-scalar meson case, $\Delta S_z=\frac{1}{2}$ so that $n=\frac{1}{2}$ for $\beta =1$ case. Thus, its leading elastic form factor scales as ${\left(1/Q^2\right)}^{\frac{1}{2}}$ .

This contradicts the fact that $n=1$ for pseudo-scalar meson, however, we are interested in a leading elastic form factor when we know distribution function so that we used above argument. According to Raya et al. [7], distribution amplitude is directly related to distribution function so that our case indicates that a leading

charged pion form factor scales as ${\left(1/Q^2\right)}^{\frac{1}{2}}$ . This is slightly larger than experimental

results as $1/Q^2$ ; however, Arriora et al. [24] demonstrate slightly larger space-like form factor using the argument of extrapolating from time-like region. Their

upper bound result shows ${\left(1/Q^2\right)}^{\frac{1}{2}}$ behavior. Pasquini et al. [25] and Xie et al.

[26] also show the distribution functions that have ${\left(1-x\right)}^1$ as $x\cong 1$ behavior (actuary exponent is close to 1 because both results fit E615 Drel Yan experiment [27]). Especially, Pasquini et al. show that their pion electromagnetic form factor explains well for experimental data up to $Q^2=10\left({\text{GeV}}^2\right)$ although their distribution function behaves as ${\left(1-x\right)}^1$ as $x\cong 1$ because their form factor is evaluated by the following form:

$F_{\pi ,q\overline{q}}\left(Q^2\right)=\frac{1}{2}{\displaystyle {\int }_0^1\text{d}x}{\varphi }_{\pi }\left(x\right)\exp \left(-a_{q\overline{q}}{Q}^2\frac{1-x}{2x}\right)$ where $a_{q\overline{q}}$ is a parameter.

According to Xie et al., Leading Neutron Deep Inelastic Scattering (LN-DIS) will reveal a structure of one dimensional mesons (1 + 1 dimensional meson) precisely in the range of $x\in \left[0.1,0.9\right]$ . Presently their simulation results are reliable only in the range of $x\ge 0.75$ ; however, these are good to see the behavior of distribution function near $x=1$ . This means that their results show the behavior of distribution function near $x=1$ of one dimensional pion that corresponds to a meson of ‘t Hooft model. Thus, we can consider that the behavior of our result near $x=1$ is consistent to that of their results.

Although we need to work out more detailed analysis of the absolute value of Equation (41) to confirm shifting property, we can claim that it is worthwhile to investigate pion distribution amplitude by using the exact solution of ‘t Hooft model. Here we would like to insist that the exact solution of ‘t Hooft model means not only the solution of ‘t Hooft singular integral equation but any solutions of large N limit two dimensional QCD massless quark case.

Appendix

Appendix A. $S\left(r\right)=\frac{1}{2\pi }Pr\frac{1}{r}$

The Green’s function is defined as

$G\left(x,x^{\prime }\right)=\langle 0|\Psi \left(x\right){\Psi }^{†}\left(x\right)|0\rangle$ (A1)

where Ψ is a fermion field (quark field), and x and x' have finite separation.

This Green’s function satisfies the following equation:

$\left({\alpha }^{\mu }{\partial }_{\mu }-i{\alpha }^{\mu }g{A}_{\mu }\right)G\left(x,x^{\prime }\right)=1\delta \left(x-x^{\prime }\right)$ (A2)

where 1 denotes unit matrix.

Note that the form of Equation (A2) is taken by ${\gamma }^0$ multiplying Dirac equation in the case of massless field. Because Dirac equation with massive case, in our choice of $\gamma$ -matrices, should be

$\left({\alpha }^{\mu }{D}_{\mu }+{\gamma }^{0}m\right)\varphi \left(x\right)=0$ (A3)

Dirac equation with massless quarks case becomes

$i{\alpha }^{\mu }{D}_{\mu }q\left(x\right)=0$ (A4)

where $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }^a\frac{{\lambda }_a}{2}$ .

The solution can be written in the form

$G\left(x,x^{\prime }\right)=G_0\left(x,x^{\prime }\right)\exp \left(\phi \left(x\right)-\phi \left(x^{\prime }\right)\right)$ (A5)

where

${\alpha }^{\mu }{\partial }_{\mu }{G}_0\left(x,x^{\prime }\right)=1\delta \left(x-x^{\prime }\right)$ (A6)

${\alpha }^{\mu }{\partial }_{\mu }\phi \left(x\right)={\alpha }^{\mu }ig{A}_{\mu }^a\left(x\right)\frac{{\lambda }_a}{2}$ (A7)

Equation (A7) means as

${\partial }_{\mu }\phi \left(x\right)=ig{A}_{\mu }^a\left(x\right)\frac{{\lambda }_a}{2}$ (A8)

Now we will find the solution of $G_0\left(x,x^{\prime }\right)$ that is time-independent. Starting equation is following:

${\partial }_x{\overline{G}}_0\left(x,x^{\prime }\right)=\delta \left(x-x^{\prime }\right)$ (A9)

where $G_0\left(x,x^{\prime }\right)={\alpha }^1{\overline{G}}_0\left(x,x^{\prime }\right)$ .

Then ${\overline{G}}_0\left(x,x^{\prime }\right)$ can be obtained as follows:

$\begin{array}{c}{\overline{G}}_0\left(x,x^{\prime }\right)={\displaystyle {\int }_{-\infty }^x\text{d}\overline{x}}\delta \left(\overline{x}-x^{\prime }\right)={\displaystyle {\int }_{-\infty }^x\text{d}\overline{x}}\frac{1}{2\pi }{\displaystyle {\int }_{-\infty }^{\infty }\text{d}p}{\text{e}}^{ip\left(\overline{x}-x^{\prime }\right)}\\ ={\displaystyle {\int }_{-\infty }^x\text{d}\overline{x}}\frac{1}{2\pi }\left[{\displaystyle {\int }_{-\infty }^0\text{d}p}{\text{e}}^{ip\left(\overline{x}-x^{\prime }\right)}+{\displaystyle {\int }_0^{\infty }\text{d}p}{\text{e}}^{ip\left(\overline{x}-x^{\prime }\right)}\right]\end{array}$ (A10)

In the second line, we use the description of $\delta$ -function as $\delta \left(x-x^{\prime }\right)=\frac{1}{2\pi }{\displaystyle {\int }_{-\infty }^{\infty }\text{d}p}\exp \left(ip\left(x-x^{\prime }\right)\right)$ .

For the first integral term in parenthesis, changing variable $p$ to $-P^{\prime }$ , then

${\displaystyle {\int }_{-\infty }^0\text{d}p}{\text{e}}^{ip\left(\overline{x}-x^{\prime }\right)}={\displaystyle {\int }_{\infty }^0-\text{d}{p}^{\prime }}{\text{e}}^{-i{p}^{\prime }\left(\overline{x}-x^{\prime }\right)}={\displaystyle {\int }_0^{\infty }\text{d}{p}^{\prime }}{\text{e}}^{-i{p}^{\prime }\left(\overline{x}-x^{\prime }\right)}$ (A11)

Combining the last line to the second integral in parenthesis, parenthesis terms become

$\begin{array}{c}\text{Parenthesis}=2{\displaystyle {\int }_0^{\infty }\text{d}p}\cos \left(p\left(\overline{x}-x^{\prime }\right)\right)=2\underset{ϵ\to 0}{\lim }{\displaystyle {\int }_{ϵ}^{\infty }\text{d}p}\cos \left(p\left(\overline{x}-x^{\prime }\right)\right)\\ =2\underset{ϵ\to 0}{\lim }\left[-\frac{\sin \left(ϵ\left(\overline{x}-x^{\prime }\right)\right)}{\overline{x}-x^{\prime }}\right]=2\underset{ϵ\to 0}{\lim }\frac{-1}{2i}\frac{{\text{e}}^{iϵ\left(\overline{x}-x^{\prime }\right)}-{\text{e}}^{-iϵ\left(\overline{x}-x^{\prime }\right)}}{\overline{x}-x^{\prime }}\end{array}$ (A12)

Then ${\overline{G}}_0$ becomes

${\overline{G}}_0\left(x,x^{\prime }\right)={\displaystyle {\int }_{-\infty }^x\text{d}\overline{x}}\frac{1}{2\pi }2\left(\frac{-1}{2i}\right)\underset{ϵ\to 0}{\lim }\frac{{\text{e}}^{iϵ\left(\overline{x}-x^{\prime }\right)}-{\text{e}}^{-iϵ\left(\overline{x}-x^{\prime }\right)}}{\overline{x}-x}$ (A13)

Changing variable as $ϵ\left(\overline{x}-x^{\prime }\right)=\overline{\overline{x}}$ , integral becomes

${\overline{G}}_0\left(x,x^{\prime }\right)=\underset{ϵ\to 0}{\lim }{\displaystyle {\int }_{-\infty }^{ϵ\left(x-x^{\prime }\right)}\text{d}\overline{\overline{x}}}\frac{1}{2\pi }2\left(\frac{-1}{2i}\right)\frac{{\text{e}}^{i\overline{\overline{x}}}-{\text{e}}^{-i\overline{\overline{x}}}}{\overline{\overline{x}}}=\left(\frac{-1}{2\pi i}\right){\displaystyle {\int }_{-\infty }^0\text{d}\overline{\overline{x}}}\frac{{\text{e}}^{i\overline{\overline{x}}}-{\text{e}}^{-i\overline{\overline{x}}}}{\overline{\overline{x}}}$ (A14)

Note that the lower limit of integral is taken as $-\infty$ because we can always take $-\infty$ as $-\frac{1}{ϵ}\infty$ .

For the second term of the last line, changing variable as $-\overline{\overline{x}}={\overline{\overline{x}}}^{\prime }$ , second integral term becomes

$-{\displaystyle {\int }_{-\infty }^0-\text{d}{\overline{\overline{x}}}^{\prime }}\frac{{\text{e}}^{i{\overline{\overline{x}}}^{\prime }}}{-{\overline{\overline{x}}}^{\prime }}={\displaystyle {\int }_0^{\infty }\text{d}\overline{\overline{x}}}\frac{{\text{e}}^{i\overline{\overline{x}}}}{\overline{\overline{x}}}$

Combining this to the first term of the last line, we obtain the form of $G_0\left(x,x^{\prime }\right)$ as

${\overline{G}}_0\left(x,x^{\prime }\right)=\left(\frac{i}{2\pi }\right){\displaystyle {\int }_{-\infty }^{\infty }\text{d}\overline{\overline{x}}}\frac{{\text{e}}^{i\overline{\overline{x}}}}{\overline{\overline{x}}}=\left(\frac{i}{2\pi }\right){\displaystyle {\int }_{-\infty }^{\infty }\text{d}x}\frac{{\text{e}}^{i\left(x-x^{\prime }\right)}}{x-x^{\prime }}=\left(\frac{i}{2\pi }\right)Pr\left(\frac{1}{x-x^{\prime }}\right)$ (A15)

For the second line, we change the variable $\overline{\overline{x}}$ to $x-x^{\prime }$ .

Therefore $G_0\left(x,x^{\prime }\right)$ is obtained as

$G_0\left(x,x^{\prime }\right)=\left(i{\alpha }^1\right)\frac{1}{2\pi }Pr\left(\frac{1}{x-x^{\prime }}\right)=\left(i{\sigma }_3\right)\frac{1}{2\pi }Pr\left(\frac{1}{x-x^{\prime }}\right)$ (A16)

Now we will find a solution of Equation (A8). Because $G\left(x,x^{\prime }\right)$ is time independent, Equation (A8) becomes

${\partial }_x\phi \left(x\right)=ig{A}^a\left(x\right)\frac{{\lambda }_a}{2}$ (A17)

where $A^a\left(x\right)\frac{{\lambda }_a}{2}=A_1^a\left(x\right)\frac{{\lambda }_a}{2}$ .

$\phi \left(x\right)=ig{\displaystyle {\int }_{-\infty }^x\text{d}z}{A}^a\left(z\right)\frac{{\lambda }_a}{2}$ (A18)

Then we can describe the Green’s function $G\left(x,x^{\prime }\right)$ as follows:

$G\left(x,x^{\prime }\right)=\left(i{\sigma }_3\right)\frac{1}{2\pi }Pr\left(\frac{1}{x-x^{\prime }}\right)\exp \left(ig{\displaystyle {\int }_{x^{\prime }}^x\text{d}z}{A}^a\left(z\right)\frac{{\lambda }_a}{2}\right)$ (A19)

The vacuum expectation value of our amplitude is given as follows.

$S\left(x,x^{\prime }\right)=\langle 0|q\left(x\right)q^{†}\left(x^{\prime }\right)\exp \left({\displaystyle {\int }_x^{x^{\prime }}\text{d}z}ig{A}^a\left(z\right)\frac{{\lambda }_a}{2}\right)|0\rangle$ (A20)

Inserting $|0\rangle \langle 0|$ between $q^{†}$ and exponent in Equation (A20), we obtain

$S\left(x,x^{\prime }\right)=G\left(x,x^{\prime }\right)\langle 0|\exp {\displaystyle {\int }_x^{x^{\prime }}\text{d}z}ig{A}^a\left(z\right)\frac{{\lambda }_a}{2}|0\rangle$ (A21)

$G\left(x,x^{\prime }\right)$ is given in Equation (A19) and then we notice that the phase factor is cancelled out.

Thus we obtain

$S\left(x,x^{\prime }\right)=\left(i{\sigma }_3\right)\frac{1}{2\pi }Pr\left(\frac{1}{x-x^{\prime }}\right)$

Therefore using the following expression is justified:

$S\left(r\right)=\left(i{\sigma }_3\right)\frac{1}{2\pi }Pr\left(\frac{1}{r}\right)$ (A22)

Appendix B. ${\displaystyle {\int }_{\infty }^{x}d{x}^{\prime }}\delta \left(x^{\prime }\right)f\left(x^{\prime }\right)=0$

We use the definition of the $\delta$ -function as

$\delta \left(z\right)=\{\begin{array}{ll}\frac{1}{2\epsilon }\hfill & -\epsilon \le z\le \epsilon \hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$ when $\epsilon$ approaches 0.

We may conclude that

${\displaystyle {\int }_x^{\infty }\text{d}z}\delta \left(z\right)f\left(z\right)=0$

under setting the condition that x is always larger than $\epsilon$ .

We can always set this condition except exact $x=0$ point.

At $x=0$ , we go back to original equation such as ${\partial }_x\left(\delta \left(x\right)f\left(x\right)\right)=0$ .

At $x=0$ , $f\left(0\right)=\text{const}$ , and in our definition of $\delta \left(x\right)=\underset{\epsilon \to 0}{\lim }\frac{1}{2\epsilon }$ , $\delta$ -function

is quite large, but just number. Thus, $\delta \left(x\right)f\left(x\right)$ is just constant. Thus, derivative of constant with respect to x is 0. Thus at $x=0$ , ${\partial }_x\left(\delta \left(x\right)f\left(x\right)\right)=0$ is satisfied with.

Because integral of 0 is always 0, we can consider that ${\displaystyle {\int }_x^{\infty }\text{d}z}\delta \left(z\right)f\left(z\right)=0$ for $x\ge 0$ .

Appendix C. Derivation of ${\varphi }_{-}^{(\mp )}$

Equation (18) is expressed as follows;

$\frac{{\text{d}}^2{\varphi }_{-}^{(+)}}{\text{d}{x}^2}+\left(\beta -\alpha x\right)\frac{\text{d}{\varphi }_{-}^{(+)}}{\text{d}x}+\left(-\left(\frac{g^2}{8\pi }\right)-\alpha \right){\varphi }_{-}^{(+)}=0$ (A23)

where $\alpha =\frac{g^2}{8}\frac{1}{2i},\beta =\frac{W_0}{2i}$ .

After first taking

${\varphi }_{-}^{(+)}\left(x\right)={\text{e}}^{\frac{1}{4}\alpha x^2}u\left(x\right)$

Equation (A23) becomes

$\frac{{\text{d}}^{2}u}{\text{d}{x}^2}+\beta \frac{\text{d}u}{\text{d}x}+\left[-\frac{1}{2}\alpha -\left(\frac{g^2}{8\pi }\right)-\frac{1}{4}{\left(\alpha x-\beta \right)}^2+\frac{1}{4}{\beta }^2\right]$ (A24)

Next taking $u\left(x\right)={\text{e}}^{-\frac{1}{2}\beta x}v\left(x\right)$ .

Equation (A24) becomes

$\frac{{\text{d}}^{2}v}{\text{d}{x}^2}+\left[-\frac{1}{2}\alpha -\left(\frac{g^2}{8\pi }\right)-\frac{1}{4}{\left(\alpha x-\beta \right)}^2\right]v\left(x\right)=0$ (A25)

Changing a variable as $\frac{1}{2}\left(\alpha x-\beta \right)=z$ , Equation (A25) becomes

$\frac{{\text{d}}^{2}v}{\text{d}{z}^2}+\left[\frac{-\frac{1}{2}\alpha -\left(\frac{g^2}{8\pi }\right)}{\frac{1}{4}{\alpha }^2}-\frac{z^2}{\frac{1}{4}{\alpha }^2}\right]v\left(z\right)=0$ (A26)

Taking $\gamma =\frac{-\frac{1}{2}\alpha -\left(\frac{g^2}{8\pi }\right)}{\frac{1}{4}{\alpha }^2}$ and changing a variable as $t=\frac{2}{\sqrt{\alpha }}z$ ,

Equation (A26) becomes

$\frac{{\text{d}}^{2}v}{\text{d}{t}^2}+\left[\frac{\alpha \gamma }{4}-\frac{t^2}{4}\right]v\left(t\right)=0$ (A27)

By comparing Equation (A27) with the standard form of the parabolic cylinder function (Weber function) [19], we find $\frac{\alpha \gamma }{4}=\overline{\lambda }+\frac{1}{2}$ .

Recalling the definition of $\alpha$ , we obtain $\overline{\lambda }=-1-\frac{2i}{\pi }$ .

The solution of Equation (A27) is then

$v\left(t\right)=D_{\overline{\lambda }}\left(t\right)=2^{\frac{\overline{\lambda }}{2}+\frac{1}{4}}{t}^{-\frac{1}{2}}{W}_{\frac{\overline{\lambda }}{2}+\frac{1}{4},-\frac{1}{4}}\left(\frac{t^2}{2}\right)$ (A28)

Substituting the value obtained for $\overline{\lambda }$ into Equation (A28), we find

$v\left(t\right)=2^{-\frac{1}{4}-\frac{i}{\pi }}{t}^{-\frac{1}{2}}{W}_{-\frac{1}{4}-\frac{i}{\pi },-\frac{1}{4}}\left(\frac{t^2}{2}\right)$ (A29)

For ${\varphi }_{-}^{(-)}$ , Equation (19) is represented by $\alpha$ and $\beta$ such that

$\frac{{\text{d}}^2{\varphi }_{-}^{(-)}}{\text{d}{x}^2}+\left(\beta +\alpha x\right)\frac{\text{d}{\varphi }_{-}^{(-)}}{\text{d}x}+\left(-\frac{g^2}{8\pi }+\alpha \right){\varphi }_{-}^{(-)}=0$ (A30)

By using the same argument applied previously, we obtain

$\frac{{\text{d}}^{2}v}{\text{d}{x}^2}+\left[\frac{\alpha {\gamma }^{\prime }}{4}-\frac{t^2}{4}\right]v\left(t\right)=0$ (A31)

where

${\gamma }^{\prime }=\frac{\frac{1}{2}\alpha -\left(\frac{g^2}{8\pi }\right)}{\frac{1}{4}{\alpha }^2}$

In this case, $\overline{\lambda }=-\frac{2i}{\pi }$ .

Because the solution of the parabolic cylinder function equation are $D_{\overline{\lambda }}\left(t\right)$ and $D_{-\overline{\lambda }-1}\left(it\right)$ [19], we select $D_{-\overline{\lambda }-1}\left(it\right)$ in this case. The solution is then

$v\left(t\right)=D_{-\overline{\lambda }-1}\left(it\right)=2^{-\frac{1}{4}+\frac{i}{\pi }}{\left(it\right)}^{-\frac{1}{2}}{W}_{-\frac{1}{4}+\frac{i}{\pi },-\frac{1}{4}}\left(-\frac{t^2}{2}\right)$ (A32)

Appendix D

The reason why $I_2$ integral of both Equation (38) and Equation (39) vanish is following.

On the contour of $I_2$ , we can set $z=R{\text{e}}^{i\theta }$ . Then Real part of exponents becomes

$\text{Real}\left(\text{exp}\right)=-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}\left(R\cos \theta +R\sin \theta \right)\le 0\left(0\le \theta \le \frac{3\pi }{4}\right)$ for Equation (38)

$\text{Real}\left(\text{exp}\right)=-\frac{\left|q\right|}{\sqrt{2\left|\alpha \right|}}\left(R\cos \theta -R\sin \theta \right)\le 0\left(0\le \theta \le \frac{\pi }{4}\right)$ for Equation (39)

Thus, exponent term is less than 1 on this contour.

Recalling asymptotic behavior of Whittaker function $W_{\kappa ,\mu }\left(z\right)$ in Equation (43), asymptotic behavior of ${\text{e}}^{\frac{1}{4}{z}^2}\frac{1}{\sqrt{z}}{W}_{\kappa ,\mu }\left(\frac{1}{2}{z}^2\right)=T$ ($\left|z\right|$ is large) becomes

$T~\frac{1}{R}\left[1+{\displaystyle \sum }_{n=1}^{\infty }\frac{\left(\frac{1}{16}-{\left(\frac{3}{4}\pm \frac{i}{\pi }\right)}^2\right)\cdots \left(\frac{1}{16}-{\left(\frac{1}{4}\mp \frac{i}{\pi }-n\right)}^2\right)}{n!R^{2n}}\right]\to 0\left(\text{as}R\to \infty \right)$

Therefore, $I_2$ integral vanish for both cases.