# Axial meson contribution to proton form factors

N.Korchagin<EMAIL_ADDRESS>Institute of Modern Physics, Chinese
Academy of Sciences, Lanzhou 730000, People Republic of China  A.Radzhabov
<EMAIL_ADDRESS>Matrosov Institute for System Dynamics and Control Theory SB
RAS, 664033, Irkutsk, Russia  Irkutsk State University, 664003, Irkutsk,
Russia

###### Abstract

The effect of axial meson exchange on extraction of proton electromagnetic
form factors in $ep$ scattering is calculated. Taking as an example
$f_{1}(1285)$ meson exchange, it is quantitatively shown how it affects
Rosenbluth and polarization transfer methods. The contribution of an axial
meson is negligible in comparison with the traditional box-type two photon
exchange and could not be responsible for difference of form factors extracted
by different techniques. However, results support the idea that meson exchange
could explain deviation of individual polarizations.

###### pacs:

##  I Introduction

Understanding of the internal structure of proton is a cornerstone of the
theory of strong interaction. Effectively the proton composition is encoded in
form factors. In electron-proton scattering one can measure electric $G_{E}$
and magnetic $G_{M}$ form factors (also called as Sachs form factors). The
first is related with electric charge distribution and the second is with
magnetization.

There are two techniques to extract $G_{E,M}$. Both utilize the one photon
exchange(OPE) approximation. First method uses unpolarized $ep$ cross section
measurements. It exploits the linearity of the cross-section in OPE
approximation with respect to some kinematic variable $\varepsilon$(meaning
will be discussed later) to extract $G_{E,M}$. This method is known as
Rosenbluth technique(proposed by Rosenbluth [1]).

Second method is used in experiments with polarized beams. It was proposed by
Akhiezer[2] and later developed in works [3, 4]. Longitudinally polarized
electron is scattered off the unpolarized nucleon and the recoil nucleon gets
a polarization $\vec{e}+N\to e+\vec{N}$. The ratio of transverse to
longitudinal polarizations of the recoil nucleon is proportional to
$G_{E}/G_{M}$. This technique is also called polarization transfer method.

Surprisingly behavior of form factors obtained by different methods disagrees.
In Rosenbluth technique $G_{E}$ and $G_{M}$ fall with the same rate with
$Q^{2}$. The ratio $\mu_{p}G_{E}/G_{M}$ is close to 1, while from polarization
transfer experiments this ratio decreases with $Q^{2}$. For more details we
refer the reader to reviews [6, 5, 7].

It is common belief that discrepancy originates from higher $\alpha$ order
corrections. They can be divided on real(bremsstrahlung) and virtual: vacuum
polarization, self-energy corrections, vertex corrections and two photon
exchange(TPE). For polarization transfer bremsstrahlung is small in comparison
with TPE effects [8]. Due to the nature of measuring the ratio of cross
sections, corrections are almost canceled. For the available $Q^{2}$ range
bremsstrahlung gives less than $1\%$ correction to $G_{E}/G_{M}$. Meanwhile
TPE contribution grows significantly at high $Q^{2}$, especially at high beam
energy. On the other hand, for Rosenbluth method, bremsstrahlung is
significant and it is considered during data analysis. However there are
concerns about its completeness: additional higher order corrections and large
logarithms which are not included to analysis could be important [9, 10].

Commonly TPE is represented as box and crossed-box diagrams, Fig.1(a,b).
Contributions with elastic and inelastic intermediate states could resolve
inconsistency between form factors extracted by Rosenbluth and polarization
transfer methods [11]. However, other polarization observables tend to
disagree with most theoretical models. For example measurement of the
longitudinal recoil polarization $P_{L}$ has noticeable deviation at forward
scattering angles from OPE and TPE corrections worsen consistency[12, 13].
Moreover, only part of such calculation for TPE can be done in a model
independent way. A structure of intermediate hadron states influence
results(for more details see reviews [14, 15]).

Figure 1: Possible TPE diagrams. Box, cross-box and the meson exchange
diagrams. Blobs represent non-point-like vertices.

There is an additional TPE contribution from exchange of a single $C$-even
meson, which is connected to the lepton by two photons, Fig.1(c). Such
diagrams were examined in [16, 17]. Pseudoscalar meson contribution is exactly
zero for the cross section of unpolarized particles. Authors of [17] are
focused at MUSE experiment kinematics, i.e. small beam energies and $Q^{2}$.
In such case the primary candidate for the exchanged particle is
$\sigma$-meson. Here we concentrate on the high $Q^{2}$ kinematics region
where the deviation appears between cross section and polarized data.

The contribution of axial-vector mesons was proposed long time ago by Drell
and Sullivan [18] as a possible explanation for the deviation of theory from
experiment for the hyperfine splitting. In the recent paper [19] contribution
of axial mesons in muonic hydrogen spectroscopy was calculated. The effect is
significant for the Zemach radii although small for the charge radius. Most
important is that the meson correction is the same order as TPE for $1S$
level. This fact inspired us to calculate the effect of axial mesons on
extraction of form factors from elastic $ep$ scattering. We repeat analytical
results of [20] and also estimate quantitatively axial-vector contribution.

This work is organized in the following way. In the next section the formalism
of OPE approximation is briefly discussed. In the section III the contribution
of axial vector exchange to polarization transfer and cross section is
calculated. Then, the section IV discusses the calculation of effective
electron-meson and proton meson couplings. The last section V presents the
numerical results for the new contribution.

##  II Elastic lepton-proton scattering formalism

We consider elastic lepton-proton scattering $l(k)+p(p)\to
l(k^{\prime})+p(p^{\prime})$. The exchanged photon or meson has momenta

$q=p^{\prime}-p=k-k^{\prime},\qquad q^{2}=-Q^{2}.$ (1)

The full amplitude $\mathcal{M}$ is the sum of the photon exchange
$\mathcal{M}_{\gamma}$ and axial meson exchange $\mathcal{M}_{A}$ amplitudes

$\mathcal{M}^{2}=|\mathcal{M}_{\gamma}+\mathcal{M}_{A}|^{2}\approx|\mathcal{M}_{\gamma}|^{2}+2\text{Re}\mathcal{M}_{\gamma}\mathcal{M}_{A}^{*}.$
(2)

$|\mathcal{M}_{A}|^{2}$ is suppressed by a higher orders of $\alpha$.
$\mathcal{M}_{\gamma}$ and $\mathcal{M}_{A}$ are represented by the left and
right Feynman diagrams in Fig. 2 respectively. The meson-lepton coupling is
treated as an effective axial vertex with the coupling $g_{eA}$. This coupling
is calculated in the section IV.

Figure 2: One photon and meson exchange contributions. Blobs represent non
point-like vertices.

Amplitudes are expressed in terms of lepton and hadron currents. Vector
currents are

$\displaystyle j^{(\gamma)}_{\mu}$
$\displaystyle=Q_{l}e\bar{u}(k^{\prime})\gamma_{\mu}u(k),$ (3) $\displaystyle
J^{(\gamma)}_{\mu}$
$\displaystyle=Q_{h}e\bar{U}(p^{\prime})\Big{[}\gamma_{\mu}F_{1}(Q^{2})-\frac{\sigma_{\mu\nu}q_{\nu}}{2M_{p}}F_{2}(Q^{2})\Big{]}U(p),$
(4)

where $Q_{l},Q_{h}$ are lepton and nucleon charges in units of electron
charge(i.e. $-1$ for electron, $+1$ for proton), $e$ is the electron
charge($e^{2}/4\pi=\alpha_{QED}$),
$\sigma_{\mu\nu}=\frac{1}{2}[\gamma_{\mu},\gamma_{\nu}]$, $M_{p}$ is the
proton mass. $F_{1}$ and $F_{2}$ are Dirac and Pauli form factors.

For the axial meson exchange the lepton and hadron axial currents are

$\displaystyle j^{(A)}_{\mu}$
$\displaystyle=g_{eA}(Q^{2})\bar{u}(k^{\prime})\gamma_{\mu}\gamma_{5}u(k),$
(5) $\displaystyle J^{(A)}_{\mu}$
$\displaystyle=g_{pA}(Q^{2})\bar{U}(p^{\prime})\gamma_{\mu}\gamma_{5}U(p),$
(6)

where $g_{eA}(Q^{2})$ is the effective coupling of a lepton with an axial
meson and $g_{pA}(Q^{2})$ is the proton-meson coupling. Further the $Q^{2}$
dependency of couplings is omitted for shortness.

Figure 3: Breit frame kinematics.

It is convenient to do calculation in Breit frame (reader can find pedagogical
introduction in this technique in lecture notes [21]). This frame is chosen in
such a way that the photon energy is $0$. The photon 3-momentum is chosen
along $+z$ direction $\vec{q}=(0,0,q_{B})$, scattering is in the $xz$-plane
and the electron mass is neglected. Momenta in this frame are

$\displaystyle k$
$\displaystyle=(E_{e},\frac{q_{B}}{2}\cot\frac{\theta_{B}}{2},0,\frac{q_{B}}{2}),$
$\displaystyle k^{\prime}$
$\displaystyle=(E_{e},\frac{q_{B}}{2}\cot\frac{\theta_{B}}{2},0,-\frac{q_{B}}{2}),$
$\displaystyle E_{e}^{2}=\frac{Q^{2}}{4\sin^{2}\theta_{B}/2},$ (7)
$\displaystyle p$ $\displaystyle=(E_{p},0,0,-q_{B}/2),$ $\displaystyle
p^{\prime}$ $\displaystyle=(E_{p},0,0,q_{B}/2),$ $\displaystyle
E_{p}^{2}=M_{p}^{2}(1+\tau),$ (8)

where $\tau=Q^{2}/4M_{p}^{2}$ and $\theta_{B}$ is the electron scattering
angle in Breit frame. The $\theta_{B}$ and the Lab frame scattering angle
$\theta_{e}$ are related as

$\cot^{2}(\theta_{B}/2)=\frac{\cot^{2}(\theta_{e}/2)}{1+\tau}.$ (9)

It is convenient to introduce a following dimensionless variable $\nu$:

$\nu=\frac{p\cdot
k}{M_{p}^{2}}-\tau=\frac{s-u}{4M_{p}^{2}}=\frac{E_{e}E_{p}}{M_{p}^{2}}=\frac{E_{1}+E_{2}}{2M_{p}}=\sqrt{\tau(1+\tau)}\sqrt{\frac{1+\varepsilon}{1-\varepsilon}},$
(10)

where $E_{1,2}$ are initial and final electron energies in Lab frame

$E_{1}=M_{p}(\nu+\tau),\qquad E_{2}=E_{1}-2M_{p}\tau=M_{p}(\nu-\tau),$ (11)

and $\varepsilon$ is

$\varepsilon=\frac{1}{1+2(1+\tau)\tan^{2}(\theta_{e}/2)}.$ (12)

The $\varepsilon$ is often erroneously called in literature as a degree of
longitudinal polarization. In fact, it is a degree of linear polarization [4].

The contribution of meson exchange in unpolarized and polarized scattering is
calculated in the next section in the described framework.

##  III Axial vector contribution

The OPE amplitude is well known and it has a simple form in Breit frame [21]:

$\displaystyle\overline{|\mathcal{M}_{\gamma}|}^{2}$
$\displaystyle=\frac{1}{4}\sum_{\text{spin}}j^{(\gamma)}_{\mu}(j^{(\gamma)}_{\mu^{\prime}})^{*}J^{(\gamma)}_{\nu}(J^{(\gamma)}_{\nu^{\prime}})^{*}\frac{g^{\mu\nu}g^{\mu^{\prime}\nu^{\prime}}}{Q^{4}}$
(13)
$\displaystyle=(Q_{l}^{2}Q_{h}^{2})e^{4}\frac{4M_{p}^{2}}{Q^{2}}\Big{(}(G_{E}^{2}+\tau
G_{M}^{2})\cot^{2}(\theta_{B}/2)+2\tau G_{M}^{2}\Big{)}$ (14)
$\displaystyle=(Q_{l}^{2}Q_{h}^{2})e^{4}\frac{1}{\tau(1+\tau)}\Big{(}G_{E}^{2}+\frac{\tau}{\varepsilon}G_{M}^{2}\Big{)}\cot^{2}(\theta_{e}/2).$
(15)

The last line corresponds to transition to the Lab frame. Electric and
magnetic form factors $G_{E,M}$ are related with Dirac and Pauli form factors
as

$G_{E}=F_{1}-\tau F_{2},\qquad G_{M}=F_{1}+F_{2}.$ (16)

Calculation for interference between OPE and meson exchange amplitudes is
straightforward and gives

$\displaystyle 2\text{Re}\mathcal{M}_{\gamma}\mathcal{M}_{A}^{\dagger}$
$\displaystyle=2\frac{1}{4}\sum_{\textrm{spin}}j_{\mu}^{(A)}(j_{\mu^{\prime}}^{(\gamma)})^{*}J_{\nu}^{(A)}(J_{\nu^{\prime}}^{(\gamma)})^{*}\frac{g_{\mu\nu}}{q^{2}}\frac{g_{\mu^{\prime}\nu^{\prime}}-q_{\mu^{\prime}}q_{\nu^{\prime}}/q^{2}}{q^{2}-m_{A}^{2}}$
(17)
$\displaystyle=2(Q_{l}Q_{h}e^{2})\frac{g_{eA}g_{pA}}{(Q^{2}+m_{A}^{2})Q^{2}}L^{(A\gamma)}_{\mu\nu}W^{(A\gamma)}_{\mu\nu}$
(18)
$\displaystyle=(Q_{l}Q_{h}e^{2})\frac{g_{eA}g_{pA}}{(Q^{2}+m_{A}^{2})Q^{2}}16E_{e}E_{p}Q^{2}G_{M}$
(19)
$\displaystyle=(Q_{l}Q_{h}e^{2})\frac{g_{eA}g_{pA}}{(Q^{2}+m_{A}^{2})}\Big{(}\frac{8M_{p}Q}{\sin(\theta_{B}/2)}\sqrt{(1+\tau)}G_{M}\Big{)},$
(20)

where $m_{A}$ is the mass of an axial meson. Notice that
$L^{(A\gamma)}_{\mu\nu}W^{(A\gamma)}_{\mu\nu}$ is spin averaged and does not
contain couplings.

The cross section is related with the amplitude in Lab frame as

$\frac{d\sigma}{d\Omega}=\frac{|\mathcal{M}|^{2}}{64\pi^{2}}\left(\frac{E_{2}}{E_{1}}\right)^{2}\frac{1}{M_{p}^{2}}.$
(21)

Historically the $ep$ cross section is represented as a correction to the Mott
cross section, which is a cross section for scattering of spin $1/2$ point-
like particles

$\frac{d\sigma_{M}}{d\Omega}=\frac{\alpha^{2}}{Q^{2}}\left(\frac{E_{2}}{E_{1}}\right)^{2}\cot^{2}\left(\frac{\theta_{e}}{2}\right).$
(22)

The final formula for cross-section is

$\frac{d\sigma}{d\Omega}=\frac{d\sigma_{M}/d\Omega}{\varepsilon(1+\tau)}\Bigg{(}\varepsilon
G_{E}^{2}+\tau
G_{M}^{2}+\frac{(Q_{l}Q_{h})g_{eA}g_{pA}}{2\alpha\pi}\frac{\tau\nu(1-\varepsilon)}{\tau+m_{A}^{2}/4M_{p}^{2}}G_{M}\Bigg{)}.$
(23)

The part in parenthesis is so-called reduced cross section. The last term
corresponds to the meson exchange contribution. It behaves as
$\sqrt{1-\varepsilon^{2}}$ in comparison with the OPE part $(\varepsilon
G_{E}^{2}+\tau G_{M}^{2})$. At $\varepsilon\to 1$, which corresponds to
forward scattering, the interference correction goes to $0$. At
$\varepsilon\to 0$ correction is maximum. Our result is similar to the old
calculation of Drell and Sullivan [18] but differs by factor 2 from Eq. (8) in
[20].

As the next step the meson exchange contribution in polarized scattering is
calculated. The setup is following: there is scattering of a massless,
longitudinally polarized electron $k$ with helicity $\lambda$ on the
unpolarized proton $p$. The recoil proton $p^{\prime}$ receives polarization,
described by the vector $a^{\mu}$. In the OPE approximation, the product of
lepton and hadron tensors in Breit frame is

$\displaystyle L^{(\gamma\gamma)}_{\mu\nu}W^{(\gamma\gamma)}_{\mu\nu}(a)$
$\displaystyle=\frac{1}{4}\text{Tr}[\cancel{k}^{\prime}\gamma_{\mu}\cancel{k}(\lambda\gamma_{5})\gamma_{\nu}]$
$\displaystyle\times\text{Tr}[(\cancel{p}^{\prime}+M_{p})(1-\gamma_{5}\cancel{a})\Big{(}\gamma_{\mu}F_{1}-\frac{\sigma_{\mu\rho}q_{\rho}}{2M_{p}}F_{2}\Big{)}(\cancel{p}+M_{p})\Big{(}\gamma_{\nu}F_{1}-\frac{\sigma_{\rho\nu}q_{\rho}}{2M_{p}}F_{2}\Big{)}]$
$\displaystyle=\frac{16\lambda M_{p}\tau}{1+\tau}G_{M}\Big{[}E_{e}E_{p}(a\cdot
q)\big{(}G_{E}-G_{M}\big{)}+M_{p}^{2}(1+\tau)G_{E}(a\cdot(k+k^{\prime}))\Big{]},$
(24)

where summation over the polarization of the final lepton and average over the
polarization of the initial proton is preformed. The 4-vector $a^{\mu}$ is
related with the rest frame polarization vector $\xi^{\mu}=(0,\vec{\xi})$ by
boost:

$a_{\mu}=\Big{(}\frac{\vec{\xi}\cdot\vec{p}}{M_{p}},\vec{\xi}+\frac{(\vec{\xi}\cdot\vec{p})\vec{p}}{M_{p}(E_{p}+M_{p})}\Big{)}.$
(25)

The products $(a\cdot q)$ and $(a\cdot(k+k^{\prime}))$ for different
polarizations in Breit frame are

$\displaystyle P_{x}:$ $\displaystyle(a_{x}\cdot q)=0,$
$\displaystyle(a_{x}\cdot(k+k^{\prime}))$
$\displaystyle=-q_{B}\cot(\theta_{B}/2),$ $\displaystyle P_{z}:$
$\displaystyle(a_{z}\cdot q)=-\frac{q_{B}}{M_{p}}E_{p},$
$\displaystyle(a_{z}\cdot(k+k^{\prime}))$
$\displaystyle=\frac{q_{B}}{M_{p}}E_{e},$ (26) $\displaystyle P_{y}:$
$\displaystyle(a_{y}\cdot q)=0,$ $\displaystyle(a_{y}\cdot(k+k^{\prime}))$
$\displaystyle=0.$

For $P_{x}$ and $P_{z}$ polarizations the well known formulas [3, 4] are
reproduced

$\displaystyle L^{(\gamma\gamma)}_{\mu\nu}W^{(\gamma\gamma)}_{\mu\nu}(P_{x})$
$\displaystyle=-4\lambda M_{p}Q^{3}G_{M}G_{E}\cot(\theta_{B}/2)=-32\lambda
M_{p}^{4}\tau\sqrt{\frac{2\varepsilon\tau}{1-\varepsilon}}G_{E}G_{M},$ (27)
$\displaystyle L^{(\gamma\gamma)}_{\mu\nu}W^{(\gamma\gamma)}_{\mu\nu}(P_{z})$
$\displaystyle=2\lambda\frac{Q^{4}G_{M}^{2}}{\sin(\theta_{B}/2)}=32\lambda
M_{p}^{4}\tau^{2}\sqrt{\frac{1+\varepsilon}{1-\varepsilon}}G_{M}^{2}.$ (28)

For $P_{y}$ polarization the amplitude is zero in OPE approximation.

The interference between photon and meson exchange amplitudes is

$L^{(A\gamma)}_{\mu\nu}W^{(A\gamma)}_{\mu\nu}(a)=16\lambda M_{p}\Big{[}(a\cdot
q)\big{(}(E_{e}^{2}+M_{p}^{2}\tau)G_{M}-E_{e}^{2}G_{E}\big{)}-E_{e}E_{p}G_{E}(a\cdot(k+k^{\prime}))\Big{]}.$
(29)

It is worthwhile to note the absence of $g_{pA}$ because it is the common
factor “outside” $L^{(A\gamma)}_{\mu\nu}W^{(A\gamma)}_{\mu\nu}$ in Eq. (17).
Using Eq. (III) for longitudinal and transverse polarization one can get

$\displaystyle L^{(A\gamma)}_{\mu\nu}W^{(A\gamma)}_{\mu\nu}(P_{x})$
$\displaystyle=-8\lambda
Q^{2}M_{p}^{2}\frac{\sqrt{1+\tau}}{\sin(\theta_{B}/2)}\cot(\theta_{B}/2)G_{E}$
(30) $\displaystyle=-8\lambda
Q^{2}M_{p}^{2}\frac{\sqrt{2\varepsilon(1+\varepsilon)(1+\tau)}}{1-\varepsilon}G_{E},$
(31) $\displaystyle L^{(A\gamma)}_{\mu\nu}W^{(A\gamma)}_{\mu\nu}(P_{z})$
$\displaystyle=4\lambda
M_{p}Q^{3}\sqrt{1+\tau}(1+\cot^{2}(\theta_{B}/2))G_{M}$ (32)
$\displaystyle=8\lambda M_{p}Q^{3}\frac{\sqrt{1+\tau}}{1-\varepsilon}G_{M}.$
(33)

The interference does not give contribution to $P_{y}$ polarization because
$L^{(A\gamma)}_{\mu\nu}W^{(A\gamma)}_{\mu\nu}(P)$ depends on the same scalar
products as in OPE. Also only imaginary part of TPE gives rise of $P_{y}$.

The final result for polarization transfer is

$\displaystyle\frac{d\sigma}{d\Omega}(P_{x})$
$\displaystyle=-\lambda\frac{\sigma_{M}/d\Omega}{\varepsilon(1+\tau)}\sqrt{2\varepsilon(1-\varepsilon)\tau}\Big{[}G_{E}G_{M}+(Q_{l}Q_{h})\frac{g_{eA}g_{pA}G_{E}}{4\alpha\pi}\frac{\nu}{\tau+m_{A}^{2}/4M_{p}^{2}}\Big{]},$
(34) $\displaystyle\frac{d\sigma}{d\Omega}(P_{z})$
$\displaystyle=\lambda\frac{\sigma_{M}/d\Omega}{\varepsilon(1+\tau)}\tau\sqrt{1-\varepsilon^{2}}\Big{[}G_{M}^{2}+(Q_{l}Q_{h})\frac{g_{eA}g_{pA}G_{M}}{2\alpha\pi(1+\varepsilon)}\frac{\nu}{(\tau+m_{A}^{2}/4M_{p}^{2})}\Big{]}.$
(35)

One can rewrite it in the form as in [20] and result for $P_{x}$ polarization
has an additional factor $\sqrt{\frac{1+\tau}{\tau}}$.

The all contributions to the cross section and polarization transfer is
obtained. To get numerical result and compare with experiments it is necessary
to know the effective couplings $g_{eA}$ and $g_{pA}$. In the next section
that couplings are estimated.

##  IV Effective couplings

This section is dedicated to calculation of the effective couplings of an
axial meson to the electron $g_{eA}$ and to the proton $g_{pA}$. $f_{1}(1285)$
is taken as an example of axial meson.

Figure 4:  The effective lepton-meson coupling is given by the triangle loop.

The meson-photons vertex can be parameterized in the form[22]:

$\displaystyle\mathcal{T}^{\mu\nu}_{\alpha}=$
$\displaystyle\,ie^{2}\varepsilon_{\rho\sigma\tau\alpha}\,\Bigg{\\{}R^{\mu\rho}(l_{1},l_{2})R^{\nu\sigma}(l_{1},l_{2})\,(l_{1}-l_{2})^{\tau}\,\nu\,F^{(0)}(l_{1}^{2},l_{2}^{2})$
$\displaystyle+\,R^{\nu\rho}(l_{1},l_{2})\left(l_{1}^{\mu}-\frac{l_{1}^{2}}{\nu}l_{2}^{\mu}\right)l_{1}^{\sigma}\,l_{2}^{\tau}\,F^{(1)}(l_{1}^{2},l_{2}^{2})$
$\displaystyle+\,R^{\mu\rho}(l_{1},l_{2})\left(l_{2}^{\nu}-\frac{l_{2}^{2}}{\nu}l_{1}^{\nu}\right)l_{2}^{\sigma}\,l_{1}^{\tau}\,{F^{(1)}}(l_{2}^{2},l_{1}^{2})\Bigg{\\}},$
(36)

where superscripts in form factors $F^{(0)}$ and $F^{(1)}$ indicate a helicity
state of an axial meson. $F^{(0)}$ is symmetric under $l_{1}\leftrightarrow
l_{2}$. Normalization factor $1/m_{A}^{2}$ which appears in [22] is included
to form factors. Note that this equation is for the case when a meson is on
the mass-shell. The off-shellness is considered as additional $q^{2}$
dependency of $F^{(i)}$ form factors and discussed further. Other ingredients
of the Eq. (IV) are

$\displaystyle\nu$ $\displaystyle=(l_{1}\cdot l_{2}),$ (37) $\displaystyle X$
$\displaystyle=(l_{1}\cdot l_{2})^{2}-l_{1}^{2}l_{2}^{2},$ (38) $\displaystyle
R^{\mu\nu}(l_{1},l_{2})$
$\displaystyle=-g^{\mu\nu}+\frac{1}{X}\,\bigl{\\{}\nu\left(l_{1}^{\mu}\,l_{2}^{\nu}+l_{2}^{\mu}\,l_{1}^{\nu}\right)-l_{1}^{2}\,l_{2}^{\mu}\,l_{2}^{\nu}-l_{2}^{2}\,l_{1}^{\mu}\,l_{1}^{\nu}\bigr{\\}},$
(39) $\displaystyle l_{1\mu}R^{\mu\nu}(l_{1},l_{2})$
$\displaystyle=l_{2\mu}R^{\mu\nu}(l_{1},l_{2})=l_{1\nu}R^{\mu\nu}(l_{1},l_{2})=l_{2\nu}R^{\mu\nu}(l_{1},l_{2})=0.$
(40)

The effective electron-meson coupling is determined by the loop integral, Fig.
4

$\displaystyle\bar{u}(k_{2})\Big{[}e^{2}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\mathcal{T}_{a}^{\mu\nu}\left(g_{ab}-\frac{q_{a}q_{b}}{q^{2}}\right)\gamma_{\mu}(\cancel{l}+m)\gamma_{\nu}}{l_{1}^{2}l_{2}^{2}(l^{2}-m^{2})}\Big{]}u(k_{1}).$
(41)

Any arbitrary expression can be decomposed onto a Dirac matrix basis,

$\displaystyle
A=a_{0}I+a_{5}\gamma_{5}+v_{\mu}\gamma^{\mu}+a_{\mu}\gamma_{5}\gamma^{\mu}+T_{\mu\nu}\sigma^{\mu\nu},$
(42)

where coefficients determined by

$\begin{split}a_{0}=\frac{1}{4}\mathrm{Tr}_{D}A,\quad
a_{5}=\frac{1}{4}\mathrm{Tr}_{D}(\gamma^{5}A),\quad
v^{\mu}=\frac{1}{4}\mathrm{Tr}_{D}(\gamma^{\mu}A),\\\
a_{\mu}=-\frac{1}{4}\mathrm{Tr}_{D}(\gamma^{5}\gamma^{\mu}A),\quad
T^{\mu\nu}=-\frac{1}{8}\mathrm{Tr}_{D}(\sigma^{\mu\nu}A).\end{split}$ (43)

The only nonzero projection for axial-vector vertex is $a_{\mu}$. As result
one could write the meson-lepton effective interaction as
$g_{eA}(Q^{2})\bar{u}(k^{\prime})\gamma^{5}\gamma_{\mu}u(k)$, where

$\displaystyle g_{eA}(Q^{2})$
$\displaystyle=ie^{4}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{N}{l_{1}^{2}l_{2}^{2}(l^{2}-m^{2})},$
(44) $\displaystyle N$ $\displaystyle=2\Big{(}\frac{(l\cdot
q)}{q^{2}}\nu(l_{1}^{2}-l_{2}^{2})F^{(0)}(Q^{2},l_{1}^{2},l_{2}^{2})+(l\cdot
l_{1})\frac{X}{\nu}{F^{(1)}}(Q^{2},l_{2}^{2},l_{1}^{2})+(l\cdot
l_{2})\frac{X}{\nu}F^{(1)}(Q^{2},l_{1}^{2},l_{2}^{2})\Big{)}.$ (45)

To proceed one needs to know form factors $F^{(i)}$. At present we have only
few experimental data from L3 Collaboration on the transition form factor [23,
24] for $f_{1}(1285)$. In this work the same parameterization as in [19] is
used. Namely, with the L3 data one can parameterize $F^{(0)}$ as

$\displaystyle F^{(0)}(Q^{2},l_{1}^{2},l_{2}^{2})=$ $\displaystyle
F^{(0)}(Q^{2},0,0)F(l_{1}^{2})F(l_{2}^{2}),$ (46) $\displaystyle
F(l_{i}^{2})=$
$\displaystyle\left(\frac{\Lambda^{2}}{\Lambda^{2}-l_{i}^{2}}\right)^{2}.$
(47)

For the numerical estimate we fix the slope of form factors according to the
L3 data to $\Lambda_{f_{1}(1285)}=1.04$ GeV. Effect of off-shellness is
considered by

$F^{(0)}(Q^{2},0,0)=F^{(0)}(M_{A},0,0)e^{-(Q^{2}+M_{A}^{2})/M_{A}^{2}}.$ (48)

The values of the form factors can be fixed from the L3 data using the
relations given by the nonrelativistic quark model calculating the triangle
anomaly diagram:

$F^{(0)}(M_{A}^{2},0,0)=-F^{(1)}(M_{A}^{2},0,0),\qquad\tilde{\Gamma}_{\gamma^{*}\gamma^{*}}=\frac{\pi\alpha^{2}M_{A}^{5}}{12}[F^{(1)}(M_{A}^{2},0,0)]^{2},$
(49)

where $\tilde{\Gamma}_{\gamma^{*}\gamma^{*}}$ is a meson decay width.
Moreover, the nonrelativistic quark model requires the sign of $F^{(0)}$ is
positive and

$F^{(0)}_{f_{1}(1285)}(M_{f_{1}}^{2},0,0)=(0.266\pm 0.043)~{}\text{GeV}^{-2}$
(50)

If we use the result of nonrelativistic quark model that $F^{(0)}=-F^{(1)}$,
then second and third parts of Eq. (45) cancel out. We proceed assuming that
contribution of $F^{(1)}$ part is small.

Figure 5: The effective coupling of axial $f_{1}(1285)$ meson to electron
$g_{ef_{1}}$ and muon $g_{\mu f_{1}}$ as function of $Q^{2}$. Coupling of
$\sigma$ meson from [17] is shown for comparison.

Now the integration of Eq. (A) could be performed. Details are given in
Appendix A. Final result for effective coupling of $f_{1}(1285)$ meson is
shown in Fig. 5 together with the result of $\sigma$ meson coupling from [17]
for comparison. We also calculated the coupling to muon, which turns out to be
smaller.

The coupling to electron at very low $Q^{2}$ is same order of magnitude as
$\sigma$ contribution in [17]. Therefore it is important for precise low
momenta scattering experiments for extraction of proton radius. Contribution
of others axial mesons as $f_{1}(1420)$ and $a_{1}(1260)$ summed up and total
effect will be more significant. Notice that in our calculation we neglect
masses of external leptons, and our results for low $Q^{2}$ region should be
treated as an estimation.

The last required unknown number is coupling to proton $g_{pA}$. It was
estimated in [25] as $g_{f_{1}N}=-(14.6\div 17.2)$. Substituting all numbers
into Eq. (23) and (34) one gets the final full result. In the next section it
will be compared with OPE approximation.

##  V Results and discussion

Rosenbluth method is based on the linearity of the reduced cross section in
OPE with $\epsilon$. At $\epsilon=0$, $\sigma_{R}=\tau G_{M}^{2}$, and at
$\epsilon=1$, $\sigma_{R}=\tau G_{M}^{2}+G_{E}^{2}$. It is necessary to check
how meson exchange contribution violates this linear dependency. From Eq. (23)
one can see that at $\epsilon=0$ contribution is biggest and disappears at
$\epsilon=1$.

TPE correction $\delta_{\gamma\gamma}$ to OPE cross section commonly presented
as

$\frac{d\sigma}{d\Omega}=\frac{d\sigma_{OPE}}{d\Omega}(1+\delta_{\gamma\gamma}).$
(51)

Figure 6: Left: TPE correction to OPE cross section
$(1+\delta_{\gamma\gamma})$ from Eq. (51) as function of $\varepsilon$ at
different $Q^{2}$. Right: same as function of $Q^{2}$ at different
$\varepsilon$.

Fig. 6 demonstrates the correction produced by $f_{1}(1285)$ exchange. One can
see that it has maximum at $Q^{2}=3~{}\text{GeV}^{2}$. Correction to OPE is of
order of $0.1\%$ Box-like diagrams typically give correction on the level of
few percents. Our result is significantly smaller and can not affect
extraction of form factors by Rosenbluth method.

In polarization transfer experiments ratio of form factors in OPE
approximation is proportional to ratio of transverse to longitudinal
polarization:

$R_{OPE}=\mu_{p}\frac{G_{E}}{G_{M}}=-\mu_{p}\sqrt{\frac{\tau(1+\varepsilon)}{2\varepsilon}}\frac{P_{T}}{P_{L}}.$
(52)

Figure 7: TPE correction to the ratio of transverse and longitudinal
polarizations $R$. Right: as function of $Q^{2}$ at different $\varepsilon$,
left: as function of $\varepsilon$ at different $Q^{2}$.

Correction to this ratio can be represented as $R/R_{OPE}$, where $R$ is a
ratio of polarizations calculated considering meson exchange. This correction
is shown in Fig. 7. It is of order of $0.05\%$, which is not enough to
influence form factors obtained from experimental data. Finally, one can
conclude that axial meson exchange could not affect form factors extracted by
Rosenbluth and polarization transfer techniques.

Figure 8: TPE correction to OPE polarization transfer. Left: longitudinal,
right: transverse. Data for longitudinal polarization are from [12],[13](only
statistical error).

Deviation of individual polarizations is also subject of interest since
longitudinal polarization was measured experimentally in [12]. Corrections are
depicted in Fig. 8. It also shows comparison with experimental points for
longitudinal polarization at $Q^{2}=2.49~{}\mathrm{GeV}^{2}$. One can see that
result is close to data points. Contributions of other mesons will add up
rising the theoretical line and could explain data. However it requires more
detailed analysis.

One can notice from Eq. (23) and Eq. (34) that corrections to cross section
depend on form factor itself. The sensitivity of results is checked by using
different form factors. The comparison with dipole form factor and Kelly
fit[26] is performed and difference is negligible.

Figure 9: Possible mechanism for real photon production from meson exchange
which can affect bremsstrahlung.

The axial meson exchange could not be responsible for difference of form
factors extracted by Rosenbluth and polarization transfer techniques. However
it also could affect extraction indirectly by contribution to final state
radiation. It is interesting to consider a diagram shown in Fig. 9. It is not
suppressed by additional QED constant in comparison with diagrams considered
in this work. However it supressed by aditional $\alpha_{QED}$ in respect to
usual radiative corrections.

##  VI Conclusion

Contribution of axial-vector meson exchange to unpolarized $ep$ cross section
and polarization transfer is calculated. For this purpose, effective lepton-
meson coupling was estimated.

The coupling of electron to $f_{1}(1285)$ is same order of magnitude as to
scalar $\sigma$ meson: $g_{ef_{1}}\approx 10^{-6}$. It changes sign at very
low $Q^{2}\approx 0.2~{}\mathrm{GeV}^{2}$ and has maximum at
$1~{}\mathrm{GeV}^{2}$. Meanwhile, coupling $f_{1}(1285)$-muon is much
smaller.

Correction to OPE for unpolarized cross section is of order $0.1\%$, which is
significantly smaller than the current experimental precision. Correction has
the maximum at $\varepsilon=0$ and $Q^{2}\approx 3~{}\mathrm{GeV}$. For
polarization ratio correction has the similar scale: $0.05\%$. More
interesting result is for individual polarizations. It has correct qualitative
behavior and decrease difference between theory and experiment for
longitudinal polarization. Consideration of other mesons could explain data.

Finally, one can conclude that contributions of axial mesons exchanges could
not be responsible for difference of form factors extracted by Rosenbluth and
polarization transfer methods. However results support the idea that meson
exchange could be behind deviation from OPE of individual polarizations.

###### Acknowledgements.

Authors thank [N.I. Kochelev], a great mentor and scientist, who proposed the
idea of this work. Authors also thank A.A. Osipov for useful discussions. The
study was supported by the National Natural Science Foundation of China,
No.11875296 and the Chinese Academy of Sciences President’s International
Fellowship Initiative via Grants No. 2020PM0073.

##  Appendix A Calculation of $g_{eA}$

This appendix is dedicated to details of lepton-meson coupling calculation.
$g_{eA}$ is the product of the following integral

$\displaystyle g_{eA}=$ $\displaystyle
ie^{4}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{N}{l_{1}^{2}l_{2}^{2}(l^{2}-m^{2})},$
(53) $\displaystyle N=$ $\displaystyle 2\Big{(}\frac{((k+l_{1})\cdot
q)}{q^{2}}\nu(l_{1}^{2}-l_{2}^{2})F^{(0)}(l_{1}^{2},l_{2}^{2})$
$\displaystyle+((k+l_{1})\cdot
l_{1})\frac{X}{\nu}{F^{(1)}}(l_{2}^{2},l_{1}^{2})+((k+l_{1})\cdot
l_{2})\frac{X}{\nu}F^{(1)}(l_{1}^{2},l_{2}^{2})\Big{)}.$ (54)

In the NRQM $F^{(0,1)}$ form factors are related with Adler’s form factors as

$\displaystyle F^{(0)}(l_{1}^{2},l_{2}^{2})=-A_{4},\quad
F^{(1)}(l_{1}^{2},l_{2}^{2})=\frac{\nu}{X}(l_{2}^{2}-\nu)A_{4},\quad{F^{(1)}}(l_{2}^{2},l_{1}^{2})=\frac{\nu}{X}(l_{1}^{2}-\nu)A_{4}.$
(55)

In such case terms with $F^{(1)}$ cancels out. Therefore one can assume that
the main contribution comes from $F^{(0)}$ term. Thus considering it and
rewriting Eq. (54) in terms of momenta shown in Fig.4, the numerator $N$
becomes

$\displaystyle N=$ $\displaystyle\frac{2}{q^{2}}(l\cdot p-l^{2}-k_{1}\cdot
k_{2})(l\cdot q)^{2}F^{(0)},$ (56)

where $p=k_{1}+k_{2}$ and $p\cdot q=0$. Form factor is parameterized as

$\displaystyle
F^{(0)}(l_{1}^{2},l_{2}^{2})=\left(\frac{\Lambda^{2}}{\Lambda^{2}-l_{1}^{2}}\right)^{2}\left(\frac{\Lambda^{2}}{\Lambda^{2}-l_{2}^{2}}\right)^{2}.$
(57)

The full expression for lepton-meson coupling is

$\displaystyle g_{eA}(Q^{2})=$ $\displaystyle
ie^{4}F^{(0)}(M_{A}^{2},0,0)e^{-(Q^{2}+M_{A}^{2})/M_{A}^{2}}\times$
$\displaystyle\int\frac{d^{4}l}{(2\pi)^{4}}\frac{2(l\cdot p-l^{2}-k_{1}\cdot
k_{2})(l\cdot
q)^{2}}{q^{2}(l^{2}-m^{2})(l-k_{1})^{2}(l-k_{2})^{2}}\left(\frac{\Lambda^{2}}{\Lambda^{2}-l_{1}^{2}}\right)^{2}\left(\frac{\Lambda^{2}}{\Lambda^{2}-l_{2}^{2}}\right)^{2}.$
(58)

Note that Eq. (54) is for a meson on the mass-shell. The effect of off-
shellness is taken into account by the factor
$\exp(-(Q^{2}+M_{A}^{2})/M_{A}^{2})$. $F^{(0)}(M_{A}^{2},0,0)$ is form factor
normalization.

The integral could be calculated using following general scheme: use Laplace
transformation, do shift in integration momenta $l$ to make integral Gaussian
and take it. Then, integral over $\alpha$-s is transformed to relative
coordinates $L$ and $x_{i}$. $L$ integration is also Gaussian.
Multidimensional integral over $x_{i}$ is done numerically.

First step is to apply following integral representation to propagators

$\frac{1}{(p_{i}^{2}-m^{2}+i\varepsilon)^{k}}=\frac{(-i)^{k}}{\Gamma(k)}\int_{0}^{\infty}d\alpha_{i}e^{i\alpha_{i}(p_{i}^{2}-m^{2}+i\varepsilon)}\alpha_{i}^{k-1}.$
(59)

As result one gets the exponent in power of

$\displaystyle\alpha_{1}l_{1}^{2}+\alpha_{2}l_{2}^{2}+\alpha_{3}l^{2}+\alpha_{4}l_{1}^{2}+\alpha_{5}l_{2}^{2}+B$
$\displaystyle=\bar{a}_{1}l_{1}^{2}+\bar{a}_{2}l_{2}^{2}+\alpha_{3}l^{2}+B$
(60) $\displaystyle=\Delta l^{2}+l\cdot(\bar{a}_{1}k_{1}+\bar{a}_{2}k_{2})+B,$
(61)
$\displaystyle\bar{a}_{1}=\alpha_{1}+\alpha_{4};\qquad\bar{a}_{2}=\alpha_{2}+\alpha_{5},\qquad
B=-\alpha_{3}m^{2}-(\alpha_{4}+\alpha_{5})\Lambda^{2},\qquad\Delta=\sum_{i}\alpha_{i},$
(62)

where $B$ denotes the part independent of $l$. Transition to integral
representation adds $(-i)^{7}=i$ as a common factor.

Further, one needs to do momentum shift and take Gaussian integral over $l$.
Shifting $l\to l-V$ by vector
$V=\frac{\bar{a}_{1}k_{1}+\bar{a}_{2}k_{2}}{\Delta}$, keeping even power of
$l$ and applying $l^{\mu}l^{\nu}=g^{\mu\nu}l^{2}/4$, numerator becomes

$\displaystyle\left(\frac{l^{2}q^{2}}{4}+(V\cdot q)^{2}\right)\left(V\cdot
p-l^{2}-V^{2}+\frac{q^{2}}{2}\right)-l^{2}(V\cdot q)^{2}.$ (63)

After shifting the expression in exponent becomes

$\displaystyle\Delta
l^{2}+\frac{1}{\Delta}(\bar{a}_{1}\bar{a}_{2}q^{2}+\bar{a}_{1}\alpha_{3}k_{1}^{2}+\bar{a}_{2}\alpha_{3}k_{2}^{2})+B.$
(64)

In massless limit of initial and final leptons it simplified to

$\displaystyle\Delta l^{2}+\frac{1}{\Delta}(\bar{a}_{1}\bar{a}_{2}q^{2})+B.$
(65)

Full result is proportional to

$\displaystyle\propto\int\\!\\!\\!d^{4}l\,\frac{2}{q^{2}}\Bigg{[}\left(\frac{l^{2}q^{2}}{4}+(V\cdot
q)^{2}\right)\left(V\cdot p-l^{2}-V^{2}+\frac{q^{2}}{2}\right)-l^{2}(V\cdot
q)^{2}\Bigg{]}e^{i\Delta
l^{2}}\exp\Big{[}\frac{i}{\Delta}(\bar{a}_{1}\bar{a}_{2}q^{2}+B)\Big{]}.$ (66)

After factoring out same powers of $l$ one gets three terms:

$\displaystyle\propto l^{0}:$ $\displaystyle\quad I_{1}=\frac{2}{q^{2}}(V\cdot
q)^{2}(V\cdot p-V^{2}+\frac{q^{2}}{2}),$ (67) $\displaystyle\propto l^{2}:$
$\displaystyle\quad I_{2}=\frac{1}{2}(V\cdot
p-V^{2}+\frac{q^{2}}{2})-\frac{4}{q^{2}}(V\cdot q)^{2},$ (68)
$\displaystyle\propto l^{4}:$ $\displaystyle\quad I_{3}=-\frac{1}{2}.$ (69)

Integration over $l$ can be performed after Wick rotation $l^{0}=il^{0}_{E}$
and for every term gives

$\displaystyle I_{1}\int d^{4}le^{i\Delta
l^{2}}=I_{1}\frac{-i\pi^{2}}{\Delta^{2}},$ (70) $\displaystyle I_{2}\int
d^{4}l\,l^{2}e^{i\Delta l^{2}}=I_{2}\frac{2\pi^{2}}{\Delta^{3}},$ (71)
$\displaystyle I_{3}\int d^{4}l\,l^{4}e^{i\Delta
l^{2}}=I_{3}\frac{i6\pi^{2}}{\Delta^{4}},$ (72)

Next step is to transform integrals over $\alpha$ using standard technique

$\displaystyle\int_{0}^{\infty}\prod_{i}d\alpha_{i}$ $\displaystyle=\int
dLL^{4}\prod_{i}dx_{i}\delta(1-\sum_{i}x_{i})$ (73)
$\displaystyle=\int\displaylimits_{0}^{\infty}L^{4}dL\int\displaylimits_{0}^{1}\\!dx_{1}\int\displaylimits_{0}^{1-x_{1}}\\!dx_{2}\int\displaylimits_{0}^{1-x_{1}-x_{2}}dx_{3}\int\displaylimits_{0}^{1-x_{1}-x_{2}-x_{3}}dx_{4},$
(74) $\displaystyle\Delta=L,\qquad\alpha_{i}=x_{i}L.$ (75)

In such case expressions modify as following

$\displaystyle V=\bar{x}_{1}k_{1}+\bar{x}_{2}k_{2},$
$\displaystyle\bar{x}_{1}=x_{1}+x_{4},$
$\displaystyle\bar{x}_{2}=x_{2}+x_{5},\qquad$ (76) $\displaystyle
V^{2}=-q^{2}\bar{x}_{1}\bar{x}_{2},$ $\displaystyle(V\cdot
q)^{2}=\frac{q^{4}}{4}(\bar{x}_{2}-\bar{x}_{1})^{2},$ $\displaystyle(V\cdot
p)=-\frac{q^{2}}{2}(\bar{x}_{1}+\bar{x}_{2}),$ (77) $\displaystyle[V\cdot
p-V^{2}+\frac{q^{2}}{2}]$
$\displaystyle=\frac{q^{2}}{2}[1-\bar{x}_{1}-\bar{x}_{2}+2\bar{x}_{1}\bar{x}_{2}],$
(78) $\displaystyle I_{1}$
$\displaystyle=\frac{q^{4}}{4}(\bar{x}_{2}-\bar{x}_{1})^{2}[1-\bar{x}_{1}-\bar{x}_{2}+2\bar{x}_{1}\bar{x}_{2}],$
(79) $\displaystyle I_{2}$
$\displaystyle=\frac{q^{2}}{4}[1-\bar{x}_{1}-\bar{x}_{2}+2\bar{x}_{1}\bar{x}_{2}]-q^{2}(\bar{x}_{2}-\bar{x}_{1})^{2}.$
(80)

Now one has to calculate the sum of three convergent integrals over $L$:

$\displaystyle-\frac{e^{4}\Lambda^{8}}{(2\pi)^{4}}x_{4}x_{5}\left(I_{1}(-i)\pi^{2}\int
dLe^{iLA}L^{4}+I_{2}2\pi^{2}\int dLe^{iLA}L^{3}+I_{3}i6\pi^{2}\int
dLe^{iLA}L^{2}\right),$ (81) $\displaystyle
A=\bar{x}_{1}\bar{x}_{2}q^{2}-x_{3}m^{2}-(x_{4}+x_{5})\Lambda^{2}.$ (82)

This integrals is easy to calculate using

$\displaystyle\int_{0}^{\infty}\\!\\!dL\,e^{-iAL}L^{n}=(-i)^{n+1}\frac{n!}{A^{n+1}}.$
(83)

The final result is three integrals: the first one is

$\displaystyle\frac{e^{4}\Lambda^{8}\pi^{2}}{(2\pi)^{4}}\int\\{dx_{i}\\}\frac{4!I_{1}x_{4}x_{5}}{A^{5}}=6\alpha_{\mathrm{QED}}^{2}\Lambda^{8}Q^{4}\int\\{dx_{i}\\}\frac{x_{4}x_{5}(\bar{x}_{2}-\bar{x}_{1})^{2}[1-\bar{x}_{1}-\bar{x}_{2}+2\bar{x}_{1}\bar{x}_{2}]}{(x_{3}m^{2}+(x_{4}+x_{5})\Lambda^{2}+Q^{2}\bar{x}_{1}\bar{x}_{2})^{5}},$
(84)

the second integral

$\displaystyle\frac{e^{4}\Lambda^{8}\pi^{2}}{(2\pi)^{4}}\int\\{dx_{i}\\}\frac{-2\cdot
3!I_{2}x_{4}x_{5}}{A^{4}}=12\alpha_{\mathrm{QED}}^{2}\Lambda^{8}Q^{2}\int\\{dx_{i}\\}x_{4}x_{5}\frac{1/4[1-\bar{x}_{1}-\bar{x}_{2}+2\bar{x}_{1}\bar{x}_{2}]-(\bar{x}_{2}-\bar{x}_{1})^{2}}{(x_{3}m^{2}+(x_{4}+x_{5})\Lambda^{2}+Q^{2}\bar{x}_{1}\bar{x}_{2})^{4}},$
(85)

and the third integral

$\displaystyle\frac{e^{4}\Lambda^{8}\pi^{2}}{(2\pi)^{4}}\int\\{dx_{i}\\}\frac{12I_{3}x_{4}x_{5}}{A^{3}}=\alpha_{\mathrm{QED}}^{2}\Lambda^{8}\int\\{dx_{i}\\}\frac{-6x_{4}x_{5}}{(x_{3}m^{2}+(x_{4}+x_{5})\Lambda^{2}+Q^{2}\bar{x}_{1}\bar{x}_{2})^{3}},$
(86)

where $q^{2}=-Q^{2}$ was used. Coupling $g_{eA}$ proportionals to the sum of
that three integrals. Integration over $x$ was done numerically. Result is
shown in Fig.5.

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