# Study of structure of deuteron from analysis of bremsstrahlung emission in
proton-deuteron scattering in cluster models

K. A. Shaulskyi1<EMAIL_ADDRESS>S. P. Maydanyuk1,2
<EMAIL_ADDRESS>V. S. Vasilevsky3<EMAIL_ADDRESS>(1)Institute for Nuclear Research, National Academy of Sciences of Ukraine,
Kyiv, 03680, Ukraine  (2)Wigner Research Centre for Physics, Budapest, 1121,
Hungary  (3)Bogolyubov Institute for Theoretical Physics, Metrolohichna str.,
14b, Kyiv, 03143, Ukraine

###### Abstract

Purpose: In this paper we investigated emission of bremsstrahlung photons in
the scattering of protons off deuterons within the microscopic cluster models
in a wide region of the beam energy from low energies up to 1.5 GeV. Methods:
Three-cluster model of bremsstrahlung is constructed for such a reaction.
Formalism of the model includes form factor of deuteron which characterizes
dependence of bremsstrahlung cross sections on structure of deuteron. This
gives possibility to investigate the structure of nuclei from analysis of
bremsstrahlung cross sections. Results: We studied dependence of the
bremsstrahlung cross section on the structure of deuteron. We use three
different shapes of the deuteron wave functions. Besides, we also calculate
the cross section by neglecting internal structure of deuteron. Analysis of
dependence of the cross section on such a parameter shows the following. (1)
At beam energies 145 and 195 MeV used in experiments bremsstrahlung cross
section is not sensitive visibly on variations of the shape of the deuteron
wave functions. (2) Stable difference between cross sections calculated with
and without internal structure of deuteron is observed at higher energy of
beam (larger 500 MeV). (3) The spectrum is increased as we pass from
structureless deuteron (the oscillator length $b$=0) to the deuteron discribed
by the shell-model wave function (the realistic oscillator length) inside the
full energy region of the emitted photons. Conclusion: Our cluster model is a
suitable tool to study the structure of deuteron with high enough precision
from bremsstrahlung analysis. We propose new experiments for such an
investigation.

proton deuteron scattering, bremsstrahlung, cluster model, photon, coherent
emission, form factor of deuteron, oscillator basis, tunneling

##  I Introduction

The bremsstrahlung emission of photons accompanying nuclear reactions is an
important topic of nuclear physics and has been attracted a significant
interest of many researchers for a long time (see reviews Amusia.1988.PhysRep
; Pluiko.1987.PEPAN ; Kamanin.1989.PEPAN ; Bertulani.1988.PhysRep ). This is
explained by that the spectra of bremsstrahlung photons are calculated on the
basis of nuclear models which include mechanisms of reactions, interactions
between nuclei, dynamics, and many other physical issues. A lot of aspects of
nuclear processes, such as dynamics of nucleons in the nuclear scattering,
interactions between nucleons, mechanisms of reactions, quantum effects,
deformations of nuclei, properties of hypernuclei in reactions, etc. can be
included to the model describing the bremsstrahlung emission (for example, see
Refs. Maydanyuk.2003.PTP ; Maydanyuk.2006.EPJA  for general properties of
$\alpha$ decay from bremsstrahlung analysis, Ref. Maydanyuk.2009.NPA  for
extraction of information about deformation of nuclei in the $\alpha$ decay
from experimental bremsstrahlung data, Ref. Maydanyuk.2011.JPG  for
bremsstrahlung in the nuclear radioactivity with emission of protons, Ref.
Maydanyuk.2010.PRC  for bremsstrahlung in the spontaneous fission of [252]Cf,
Ref. Maydanyuk.2011.JPCS  for bremsstrahlung in the ternary fission of
[252]Cf, Ref. Maydanyuk_Zhang_Zou.2018.PRC  for bremsstrahlung in the pion-
nucleus scattering from our research, there are many investigations of other
researchers). Note perspectives on studying electromagnetic observables of
light nuclei based on chiral effective field theory Pastore.2008.PRC . The
measurements of photons with analysis provide important information on these
phenomena.

Analysis of bremsstrahlung photons accompanying nuclear reactions gives
possibility to extract additional information on structure of nuclei. Study of
the structure of nuclei on the basis of bremsstrahlung analysis is one of the
most ambitious aims in nuclear physics. Analyzing formalism of models, option
to investigate structure of nuclei exists, in principle, and understandable.
Study of structure of nuclei is one of the most promising research directions,
taking into account that photons can be measured in experiments. However,
during long period of investigations of bremsstrahlung photons in nuclear
physics, systematic study of structure of nuclei has not been realized yet.
One can explain that by difficulty in development of mathematical formalism of
models, importance to reach stability of numerical calculations that is
possible at high precision. Moreover, it turns out that not all available
experimental bremsstrahlung data are well sensitive to structure of nuclei. In
this regards, one can remind investigations of bremsstrahlung emission in
reactions with light nuclei within microscopic two-cluster models
1985NuPhA.443..302B ; 1990PhRvC..41.1401L ; 1990PhRvC..42.1895L ;
1991NuPhA.529..467B ; 1991PhRvC..44.1695L ; 1992NuPhA.550..250B ; Liu.1992.FBS
; 1993nuco.conf..423K ; Dohet-Eraly.2011.JPCS ; Dohet-Eraly.2011.PRC ; Dohet-
Eraly.2013.PRC ; 2013JPhCS.436a2030D ; Dohet-Eraly.2013.PhD ;
2014PhRvC..89b4617D ; 2014PhRvC..90c4611D .

Summarizing all issues mentioned above, we see perspective problem on
realization of such an idea, that is a main aim of this paper. We would like
to understand, which parameters of nuclear structure are more effective to
realize such an investigation. Of course, the best way is to construct this
model on the fully quantum basis, with inclusion of realistic nuclear
interactions which were well tested experimentally. A promising way is cluster
formalism for description of structure of nuclei and nuclear process. So, as a
basis of this research we will develop the fully cluster model in combination
of bremsstrahlung formalism. The most effective process for such study is
proton-deuteron scattering. We focus on construction of such a unified cluster
formalism, analysis of available experimental information about bremsstrahlung
for proton-deuteron scattering. This paper is continuation of our previous
research Maydanyuk_Vasilevsky.2023.PRC , where we developed cluster model in
the folding approximation in study of bremsstrahlung emission in the
scattering of nuclei with the small number of nucleons and we did not analyze
possibility to extract information about structure of nuclei from
bremsstrahlung cross sections.

The paper is organized in the following way. In Sec. II cluster models of
emission of the bremsstrahlung photons in the proton-deuteron scattering is
formulated. Here, we give an explicit form of the operator of the
bremsstrahlung emission, define wave functions of $p+d$ system, calculate
matrix elements of bremsstrahlung emission, define form factors of deuteron
(characterizing its structure), apply the multiple expansion approach for
calculation of matrix elements. In Sec. IV matrix elements of bremsstrahlung
emission in folding approximation are reviewed (following to formalism in Ref.
Maydanyuk_Vasilevsky.2023.PRC ). In Sec. V cross section of the bremsstrahlung
emission of photons is determined and resulting formulas are summarized. In
Sec. VI emission of the bremsstrahlung photons for the proton-deuteron
scattering is studied on the basis of the model above. We analyze role of the
deuteron wave function and its form factor in calculations of the cross
section at different energies of relative motions between the scattered proton
and deuteron. We also describe the experimental bremsstrahlung data for the
proton-deuteron scattering on the basis of the model. Conclusions and
perspectives are summarized in Sec. VII. Operator of bremsstrahlung emission
in three-cluster model is calculated in App. A. Useful details of calculation
of integrals are presented in App. B. Form factor of deuteron in three-cluster
approach is derived in App. C.

##  II Matrix elements of bremsstrahlung emission in three-cluster model

###  II.1 Operator of bremsstrahlung emission in three-cluster model

Consider the translationally invariant interaction of photon with three-
nucleon system

$\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)=\frac{1}{2}\frac{e\hbar}{m_{N}c}\sum_{i=1}^{A=3}\frac{1}{2}\left(1+\widehat{\tau}_{iz}\right)\left[\widehat{\mathbf{\pi}}_{i}^{\ast}\mathbf{A}^{\ast}\left(i\right)+\mathbf{A}^{\ast}\left(i\right)\widehat{\mathbf{\pi}}_{i}^{\ast}\right]$
(1)

where

$\begin{array}[]{lllllll}\vspace{1.0mm}\mathbf{A}^{\ast}\left(i\right)=\mathbf{\varepsilon}^{(\alpha)}\exp\left\\{-i\left(\mathbf{k}_{\gamma}\mathbf{\rho}_{i}\right)\right\\},&\widehat{\mathbf{\pi}}_{i}^{\ast}=i\nabla_{\mathbf{\rho}_{i}},&\mathbf{\rho}_{i}=\mathbf{r}_{i}-\mathbf{R}_{\rm
cm},\\\
\mathbf{R}_{cm}=\displaystyle\frac{1}{A}\displaystyle\sum_{i=1}^{A}\mathbf{r}_{i},&\widehat{\mathbf{\pi}}_{i}=\widehat{\mathbf{p}}_{i}-\widehat{\mathbf{P}}_{cm},&\widehat{\mathbf{P}}_{cm}=\displaystyle\frac{1}{A}\displaystyle\sum_{i=1}^{A}\widehat{\mathbf{p}}_{i}.\end{array}$
(2)

Here, $\mbox{\boldmath$\varepsilon$}^{(\alpha)}$ are unit vectors of _linear_
polarization of the photon emitted
($\mbox{\boldmath$\varepsilon$}^{(\alpha),*}=\mbox{\boldmath$\varepsilon$}^{(\alpha)}$),
$\mathbf{k}_{\gamma}$ is wave vector of the photon and
$w_{\gamma}=k_{\gamma}c=\bigl{|}\mathbf{k}_{\gamma}\bigr{|}c$. Vectors
$\mbox{\boldmath$\varepsilon$}^{(\alpha)}$ are perpendicular to
$\mathbf{k}_{\gamma}$ in Coulomb gauge. We have two independent polarizations
$\mbox{\boldmath$\varepsilon$}^{(1)}$ and
$\mbox{\boldmath$\varepsilon$}^{(2)}$ for the photon with impulse
$\mathbf{k}_{\gamma}$ ($\alpha=1,2$). Also we have properties:

$\begin{array}[]{lclc}\Bigl{[}\mathbf{k}_{\gamma}\times\mbox{\boldmath$\varepsilon$}^{(1)}\Bigr{]}=k_{\gamma}\,\mbox{\boldmath$\varepsilon$}^{(2)},&\Bigl{[}\mathbf{k}_{\gamma}\times\mbox{\boldmath$\varepsilon$}^{(2)}\Bigr{]}=-\,k_{\gamma}\,\mbox{\boldmath$\varepsilon$}^{(1)},&\Bigl{[}\mathbf{k}_{\gamma}\times\mbox{\boldmath$\varepsilon$}^{(3)}\Bigr{]}=0,&\displaystyle\sum\limits_{\alpha=1,2,3}\Bigl{[}\mathbf{k}_{\gamma}\times\mbox{\boldmath$\varepsilon$}^{(\alpha)}\Bigr{]}=k_{\gamma}\,(\mbox{\boldmath$\varepsilon$}^{(2)}-\mbox{\boldmath$\varepsilon$}^{(1)}).\end{array}$
(3)

Let us introduce new variables, namely Jacobi vectors $\mathbf{r}$ and
$\mathbf{q}$

$\begin{array}[]{lllll}\mathbf{r}=\displaystyle\frac{1}{\sqrt{2}}\left(\mathbf{\rho}_{1}-\mathbf{\rho}_{2}\right)=\displaystyle\frac{1}{\sqrt{2}}\left(\mathbf{r}_{1}-\mathbf{r}_{2}\right),\quad\mathbf{q}=\sqrt{\displaystyle\frac{2}{3}}\left(\mathbf{\rho}_{3}-\frac{\mathbf{\rho}_{1}+\mathbf{\rho}_{2}}{2}\right)=\sqrt{\frac{2}{3}}\left(\mathbf{r}_{3}-\displaystyle\frac{\mathbf{r}_{1}+\mathbf{r}_{2}}{2}\right),\\\
\mathbf{q}_{A}=\sqrt{\displaystyle\frac{1}{3}}\left(\mathbf{r}_{1}+\mathbf{r}_{2}+\mathbf{r}_{3}\right)=\sqrt{3}\mathbf{R}_{cm}.\end{array}$

Inverse relations are

$\begin{array}[]{lllll}\mathbf{r}_{1}=\displaystyle\frac{1}{\sqrt{2}}\mathbf{r}-\displaystyle\frac{1}{\sqrt{6}}\mathbf{q}+\displaystyle\frac{1}{\sqrt{3}}\mathbf{q}_{A},&\mathbf{r}_{2}=-\displaystyle\frac{1}{\sqrt{2}}\mathbf{r}-\displaystyle\frac{1}{\sqrt{6}}\mathbf{q}+\displaystyle\frac{1}{\sqrt{3}}\mathbf{q}_{A},&\mathbf{r}_{3}=\displaystyle\sqrt{\frac{2}{3}}\mathbf{q}+\displaystyle\frac{1}{\sqrt{3}}\mathbf{q}_{A}.\end{array}$
(4)

Similar relations can be written for momenta

$\displaystyle\mathbf{\pi}_{\mathbf{r}}$
$\displaystyle=\frac{1}{\sqrt{2}}\left(\mathbf{\pi}_{1}-\mathbf{\pi}_{2}\right),$
$\displaystyle\mathbf{\pi}_{\mathbf{q}}$
$\displaystyle=\sqrt{\frac{2}{3}}\left(\mathbf{\pi}_{3}-\frac{\mathbf{\pi}_{1}+\mathbf{\pi}_{2}}{2}\right),$
$\displaystyle\mathbf{\pi}_{A}$
$\displaystyle=\sqrt{\frac{1}{3}}\left(\mathbf{\pi}_{1}+\mathbf{\pi}_{2}+\mathbf{\pi}_{3}\right).$

Inverse relations

$\displaystyle\mathbf{\pi}_{1}$
$\displaystyle=\frac{1}{\sqrt{2}}\mathbf{\pi}_{\mathbf{r}}\mathbf{-}\frac{1}{\sqrt{6}}\mathbf{\pi}_{\mathbf{q}}\mathbf{+}\frac{1}{\sqrt{3}}\mathbf{\pi}_{A},$
$\displaystyle\mathbf{\pi}_{2}$
$\displaystyle=-\frac{1}{\sqrt{2}}\mathbf{\pi}_{\mathbf{r}}\mathbf{-}\frac{1}{\sqrt{6}}\mathbf{\pi}_{\mathbf{q}}\mathbf{+}\frac{1}{\sqrt{3}}\mathbf{\pi}_{A},$
$\displaystyle\mathbf{\pi}_{3}$
$\displaystyle=\sqrt{\frac{2}{3}}\mathbf{\pi}_{\mathbf{q}}\mathbf{+}\frac{1}{\sqrt{3}}\mathbf{\pi}_{A}.$

Now we fix position of nucleons. We assume that vector $\mathbf{r}$ measures
the distance between proton and neutron which form a deuteron. We also assume
that $\mathbf{r}_{1}$ is a coordinate of the first proton and $\mathbf{r}_{2}$
is a coordinate of a neutron. Vector $\mathbf{r}_{3}$ determines the location
of the second proton. With such definitions, the operator
$\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)$
is [see App. A for details, also we take into account that
$\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{k}_{\gamma}\right)=0$]

$\begin{array}[]{lllll}\vspace{1.5mm}\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)&=&\displaystyle\frac{1}{2}\displaystyle\frac{e\hbar}{m_{N}c}\biggl{\\{}\displaystyle\frac{2}{\sqrt{2}}\exp\left\\{-i\displaystyle\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{r}}^{\ast}\right)\exp\left\\{i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\,-\\\
&-&\sqrt{\displaystyle\frac{2}{3}}\exp\left\\{-i\displaystyle\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\exp\left\\{i\displaystyle\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)+2\sqrt{\displaystyle\frac{2}{3}}\exp\\{-i\sqrt{\displaystyle\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\biggr{\\}}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\biggr{\\}}.\end{array}$
(5)

This is the universal and model-independent form of the operator of
bremsstrahlung emission for a system comprising from two protons and one
neutron. To calculated cross section of bremsstrahlung emission in the process
of a proton scattering from a deuteron, we need to formulate model which
provides a realistic description of the $p+d$ scattering in economical way,
i.e. with minimum of computations but with a reliable output. As the output,
we need to determine wave functions of the $p+d$ scattering at selected
energies of initial and final states of bremsstrahlung emission. For this aim,
we select the resonating group method (RGM), which is the most rigorous and
self-consistent realization of a cluster model. Actually, we will use three
different variants of the RGM: two- and three-cluster variants and so-called
the folding approximation. These three variants of the RGM are explained in
detail in next Section.

##  III Two and three-cluster models of $p+d$ system

Three-nucleon system 3He and its decay channel $p+d$ will be studied in the
framework of two- and three-cluster models. In a two-cluster model, wave
function of the system is

$\Psi=\widehat{\mathcal{A}}\left\\{\phi^{(S)}\left(d,\mathbf{r}\right)\phi\left(p\right)\psi_{E}\left(\mathbf{q}\right)\right\\},$
(6)

where $\phi^{(S)}\left(d,\mathbf{r}\right)$ is the deuteron wave function from
the oscillator shell-model, $\phi\left(p\right)$ is a wave function of proton
represented by its spin and isospin parts, and
$\psi_{E}\left(\mathbf{q}\right)$ is a wave function of relative motion of
proton and deuteron. The antisymmetrization operator $\widehat{\mathcal{A}}$
makes wave functions of the $p+d$ system fully antisymmetric. Three-cluster
model suggests the following form for three-nucleon system

$\Psi=\widehat{\mathcal{A}}\left\\{\phi\left(n\right)\phi\left(p_{1}\right)\phi\left(p_{2}\right)f\left(\mathbf{r,q}\right)\right\\},$
(7)

Wave function $\phi\left(p_{2}\right)f\left(\mathbf{r,q}\right)$ of relative
motion of nucleons has to be determined by solving the Schrödinger equation or
the Faddeev equations.

By assuming that the shape of a deuteron does not change when proton is
approaching, then three-particle wave function can be represented as

$\Psi=\widehat{\mathcal{A}}\left\\{\phi\left(d,\mathbf{r}\right)\phi\left(p\right)\psi_{E}\left(\mathbf{q}\right)\right\\},$
(8)

where $\phi\left(d,\mathbf{r}\right)$ is a wave function of the bound state of
deuteron. The wave function $\phi\left(d,\mathbf{r}\right)$ is a solution of
two-body Schrödinger equation with selected nucleon-nucleon potential.

Note that in both models, in two-cluster and three-cluster, the wave function
of deuteron is assumed to be antisymmetric, than the antisymmetrization
operator $\widehat{\mathcal{A}}$ in Eqs. (6) and (8) consists of the unit
operator and two permutation operators. As the results, the antisymmetrization
operator $\widehat{\mathcal{A}}$ creates in Eqs. (6) and (8) three terms which
are similar to the terms in curly brackets.

If one ignores the full antisymmetrization in Eqs. (6) and (8) by omitting the
operator $\widehat{\mathcal{A}}$, one obtains a simple version of the two- and
three-cluster models which is called a folding approximation or folding model.
In order to avoid bulky expressions, we will use this approximation to present
matrix elements of the
$\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}_{\mu}\right)$
between the initial and final states of the $p+d$ system.

To construct wave functions of the system $p+d$ in different approximations
(models), we need to solve the appropriate Schrödinger equations. For this aim
we employ the algebraic version of the resonating group method (RGM),
formulated in Refs. kn:Fil_Okhr , kn:Fil81 . This version of the RGM uses the
full basis of oscillator functions to expand wave functions of the relative
motion of clusters. As the results, the Schrödinger equation is reduced to a
system of linear algebraic equation for expansion coefficients. Besides, the
algebraic version implements proper boundary conditions in discrete,
oscillator representation.

To study $p+d$ system in three-cluster approximation, we will employ a three-
cluster model developed in Ref. 2009NuPhA.824…37V .

###  III.1 Wave functions of $p+d$ system in the cluster formalism

To calculate matrix elements of the operator
$\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)$
we need to construct wave functions of the system $p+d$. If we neglect the
Pauli principle and employ an adiabatic approximation, then wave function of
the system can be constructed in a separable form

$\Psi\left(\mathbf{r},\mathbf{q}\right)=\phi\left(\mathbf{r}\right)\psi\left(\mathbf{q}\right),$
(9)

where wave function of deuteron $\phi\left(\mathbf{r}\right)$ is a solution of
the two-body Schrödinger equation

$\left(\widehat{H}_{d}-E_{d}\right)\phi\left(\mathbf{r}\right)=0,$ (10)
$\widehat{H}_{d}=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{d\mathbf{r}^{2}}+\widehat{V}_{NN}\left(\mathbf{r}\right)$
(11)

where $m$ is a mass of nucleon. If the nucleon-nucleon potential
${V}_{NN}\left(\mathbf{r}\right)$ is used in the form

$\widehat{V}_{NN}\left(\mathbf{r}\right)=\sum_{S=0,1}\sum_{T=0,1}V_{2S+1,2T+1}\left(\mathbf{r}\right)\widehat{P}_{S}\widehat{P}_{T},$
(12)

where $\widehat{P}_{S}$ ($\widehat{P}_{T}$) is the projection operator
projecting onto the spin $S$ (the isospin $T$) of two-nucleon system, then in
Eq. (11) $\widehat{V}_{NN}\left(\mathbf{r}\right)$ should be replace with the
even component $V_{31}\left(\mathbf{r}\right)$, as deuteron has the spin $S$=1
and the isospin $T$=1.

The wave function describing interaction of proton with deuteron obeys the
following equations

$\left(\widehat{H}_{p}-E_{p}\right)\psi\left(\mathbf{q}\right)=0,$ (13)

where

$\widehat{H}_{p}=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{d\mathbf{q}^{2}}+\widehat{V}_{pd}\left(\mathbf{q}\right)$

and the potential energy $\widehat{V}_{pd}\left(\mathbf{q}\right)$ equals

$\widehat{V}_{pd}\left(\mathbf{q}\right)=\left\langle\phi\left(\mathbf{r}\right)\left|\sum_{i=1,2}\widehat{V}_{NN}\left(\mathbf{r}_{3}-\mathbf{r}_{i}\right)+\sum_{i=1,2}\widehat{V}_{C}\left(\mathbf{r}_{3}-\mathbf{r}_{i}\right)\right|\phi\left(\mathbf{r}\right)\right\rangle.$

Here, integration is performed over vector $\mathbf{r}$, and nucleon-nucleon
$\widehat{V}_{NN}$ and Coulomb $\widehat{V}_{C}$ potentials are involved in
definition of $\widehat{V}_{pd}\left(\mathbf{q}\right)$. Equation (13)
determines both initial $\psi_{E_{i}}\left(\mathbf{q}\right)$ and final
$\psi_{E_{f}}\left(\mathbf{q}\right)$ wave functions of the $p+d$ system.

###  III.2 Matrix elements of bremsstrahlung emission in the cluster formalism

Based on assumptions made, we have got matrix element of transition from
initial to final states

$\left\langle\phi\left(\mathbf{r}\right)\psi_{E_{f}}\left(\mathbf{q}\right)\left|\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)\right|\phi\left(\mathbf{r}\right)\psi_{E_{i}}\left(\mathbf{q}\right)\right\rangle.$

One suggest to calculate this matrix elements on two steps. On the first step,
we calculate the matrix element

$\widehat{\mathcal{H}}_{e}\left(\mathbf{q}\right)=\left\langle\phi\left(\mathbf{r}\right)\left|\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)\right|\phi\left(\mathbf{r}\right)\right\rangle$

by integrating over vector $\mathbf{r}$. By using Eq. (5), we obtain

$\displaystyle\widehat{\mathcal{H}}_{e}\left(\mathbf{q}\right)=\left\langle\phi\left(\mathbf{r}\right)\left|\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)\right|\phi\left(\mathbf{r}\right)\right\rangle$
(14)
$\displaystyle=\frac{1}{2}\frac{e\hbar}{m_{N}c}\biggl{\\{}\frac{2}{\sqrt{2}}\left\langle\phi\left(\mathbf{r}\right)\left|\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{r}}^{\ast}\right)\right|\phi\left(\mathbf{r}\right)\right\rangle\exp\left\\{i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}$
$\displaystyle\mathbf{-}\sqrt{\frac{2}{3}}\left\langle\phi\left(\mathbf{r}\right)\left|\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\right|\phi\left(\mathbf{r}\right)\right\rangle\exp\left\\{i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)$
$\displaystyle+\left.2\sqrt{\frac{2}{3}}\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right\\}$

and then

$\displaystyle\left\langle\phi\left(\mathbf{r}\right)\psi_{E_{f}}\left(\mathbf{q}\right)\left|\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)\right|\phi\left(\mathbf{r}\right)\psi_{E_{i}}\left(\mathbf{q}\right)\right\rangle=$
(15)
$\displaystyle=\frac{1}{2}\frac{e\hbar}{m_{N}c}\biggl{\\{}\frac{2}{\sqrt{2}}\left\langle\phi\left(\mathbf{r}\right)\left|\exp\left\\{-\frac{i}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{r}}^{\ast}\right)\right|\phi\left(\mathbf{r}\right)\right\rangle\left\langle\psi_{E_{f}}\left(\mathbf{q}\right)\left|\exp\left\\{\frac{i}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\right|\psi_{E_{i}}\left(\mathbf{q}\right)\right\rangle$
$\displaystyle-\sqrt{\frac{2}{3}}\left\langle\phi\left(\mathbf{r}\right)\left|\exp\left\\{-\frac{i}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\right|\phi\left(\mathbf{r}\right)\right\rangle\left\langle\psi_{E_{f}}\left(\mathbf{q}\right)\left|\exp\left\\{\frac{i}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right|\psi_{E_{i}}\left(\mathbf{q}\right)\right\rangle$
$\displaystyle+\left.2\sqrt{\frac{2}{3}}\left\langle\psi_{E_{f}}\left(\mathbf{q}\right)\left|\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right|\psi_{E_{i}}\left(\mathbf{q}\right)\right\rangle\right\\}.$

Thus we need to calculate few basic integrals

$\displaystyle\left\langle\phi\left(\mathbf{r}\right)\left|\exp\left\\{-\frac{i}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\right|\phi\left(\mathbf{r}\right)\right\rangle,$
(16)
$\displaystyle\left\langle\phi\left(\mathbf{r}\right)\left|\exp\left\\{-\frac{i}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{r}}^{\ast}\right)\right|\phi\left(\mathbf{r}\right)\right\rangle,$
$\displaystyle\left\langle\psi_{E_{f}}\left(\mathbf{q}\right)\left|\exp\left\\{\frac{i}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\right|\psi_{E_{i}}\left(\mathbf{q}\right)\right\rangle,$
$\displaystyle\left\langle\psi_{E_{f}}\left(\mathbf{q}\right)\left|\exp\left\\{\frac{i}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right|\psi_{E_{i}}\left(\mathbf{q}\right)\right\rangle.$

Note that with such definition of coordinates (II.1), the wave vectors of
initial and final states are defined as $k_{i}=\sqrt{2mE_{i}\,/\,\hbar^{2}}$,
$k_{f}=\sqrt{2mE_{f}\,/\,\hbar^{2}}$, where $E_{i}$ and $E_{f}$ energies are
in MeV and in the center of mass motion.

###  III.3 Introduction of form factors

Introduce the following definitions of _form factors of deuteron_ (in
formalism of three-cluster model):

$\begin{array}[]{lllll}\vspace{1.0mm}F_{1}(\mathbf{k}_{\gamma})&=&\Bigl{\langle}\phi(\mathbf{r})\Bigl{|}\,\exp\\{-\,\displaystyle\frac{i}{\sqrt{2}}\,(\mathbf{k}_{\gamma}\mathbf{r})\Bigr{\\}}\Bigr{|}\,\phi(\mathbf{r})\Bigr{\rangle},\\\
F_{2,\,\alpha}(\mathbf{k}_{\gamma})&=&\Bigl{\langle}\phi(\mathbf{r})\Bigl{|}\,\exp\\{-\,\displaystyle\frac{i}{\sqrt{2}}\,(\mathbf{k}_{\gamma}\mathbf{r})\Bigr{\\}}\,(\varepsilon^{(\alpha)},\mathbf{\pi}_{\mathbf{r}}^{*})\,\Bigr{|}\,\phi(\mathbf{r})\Bigr{\rangle}.\end{array}$
(17)

Then, the matrix element of emission in Eq. (15) is rewritten as

$\begin{array}[]{lllll}\vspace{1.0mm}&\Bigl{\langle}\Psi_{E_{f}}(\mathbf{r},\mathbf{q})\Bigl{|}\,\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha)})\Bigr{|}\,\Psi_{E_{i}}(\mathbf{r},\mathbf{q})\Bigr{\rangle}\;=\\\
\vspace{1.0mm}=&\displaystyle\frac{1}{2}\,\displaystyle\frac{e\,\hbar}{m_{N}c}\Bigl{\\{}\par\displaystyle\frac{2}{\sqrt{2}}\,\Bigl{\langle}\psi_{E_{f}}(\mathbf{q})\Bigl{|}\,\exp\\{\displaystyle\frac{i}{\sqrt{6}}\,(\mathbf{k}_{\gamma}\mathbf{q})\Bigr{\\}}\Bigr{|}\,\phi_{E_{i}}(\mathbf{q})\Bigr{\rangle}\cdot
F_{2,\,\alpha}\;-\\\
\vspace{1.0mm}-&\sqrt{\displaystyle\frac{2}{3}}\,\Bigl{\langle}\psi_{E_{f}}(\mathbf{q})\Bigl{|}\,\exp\\{\displaystyle\frac{i}{\sqrt{6}}\,(\mathbf{k}_{\gamma}\mathbf{q})\Bigr{\\}}\,(\varepsilon^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{*})\,\Bigr{|}\,\psi_{E_{i}}(\mathbf{q})\Bigr{\rangle}\cdot
F_{1}\;+\\\
+&2\,\sqrt{\displaystyle\frac{2}{3}}\,\Bigl{\langle}\varphi_{E_{f}}(\mathbf{q})\Bigl{|}\,\exp\\{-i\,\sqrt{\displaystyle\frac{2}{3}}\,(\mathbf{k}_{\gamma}\mathbf{q})\Bigr{\\}}(\varepsilon^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{*})\,\Bigr{|}\varphi_{E_{i}}(\mathbf{q})\Bigr{\rangle}\Bigr{\\}},\end{array}$
(18)

where

$\begin{array}[]{lllll}\Psi_{E_{i}}(\mathbf{r},\mathbf{q})=\phi\left(\mathbf{r}\right)\psi_{E_{i}}\left(\mathbf{q}\right),\end{array}$
(19)

$\begin{array}[]{lllll}\Psi_{E_{f}}(\mathbf{r},\mathbf{q})=\phi\left(\mathbf{r}\right)\psi_{E_{f}}\left(\mathbf{q}\right).\end{array}$
(20)

We introduce the following notations for matrix elements as

$\begin{array}[]{ll}\mathbf{I}_{1}(\alpha)=\biggl{<}\varphi_{E_{f}}(\mathbf{q})\biggl{|}\,e^{-i\alpha\mathbf{k_{\gamma}q}}\displaystyle\frac{\partial}{\partial\mathbf{q}}\,\biggr{|}\,\varphi_{E_{i}}(\mathbf{q})\biggr{>}_{\mathbf{q}},&I_{2}(\alpha)=\Bigl{<}\varphi_{E_{f}}(\mathbf{q})\Bigl{|}\,e^{-i\alpha\mathbf{k_{\gamma}q}}\,\Bigr{|}\,\varphi_{E_{i}}(\mathbf{q})\Bigr{>}_{\mathbf{q}}.\end{array}$
(21)

Then, the full matrix element (18) can be rewritten as

$\begin{array}[]{lllll}\vspace{1.0mm}&\Bigl{\langle}\Psi_{E_{f}}(\mathbf{r},\mathbf{q})\Bigl{|}\,\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha)})\Bigr{|}\,\Psi_{E_{i}}(\mathbf{r},\mathbf{q})\Bigr{\rangle}\;=\\\
\vspace{1.0mm}=&\displaystyle\frac{1}{2}\,\displaystyle\frac{e\,\hbar}{m_{N}c}\Bigl{\\{}\par\displaystyle\frac{2}{\sqrt{2}}\cdot
I_{2}\Bigl{(}\displaystyle\frac{-1}{\sqrt{6}}\Bigr{)}\cdot
F_{2,\,\alpha}-\par\sqrt{\displaystyle\frac{2}{3}}\,\Bigl{\langle}\psi_{E_{f}}(\mathbf{q})\Bigl{|}\,\exp\\{\displaystyle\frac{i}{\sqrt{6}}\,(\mathbf{k}_{\gamma}\mathbf{q})\Bigr{\\}}\,(\varepsilon^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{*})\,\Bigr{|}\,\psi_{E_{i}}(\mathbf{q})\Bigr{\rangle}\cdot
F_{1}\;+\\\
+&2\,\sqrt{\displaystyle\frac{2}{3}}\,\Bigl{\langle}\varphi_{E_{f}}(\mathbf{q})\Bigl{|}\,\exp\\{-i\,\sqrt{\displaystyle\frac{2}{3}}\,(\mathbf{k}_{\gamma}\mathbf{q})\Bigr{\\}}(\varepsilon^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{*})\,\Bigr{|}\varphi_{E_{i}}(\mathbf{q})\Bigr{\rangle}\Bigr{\\}}.\end{array}$
(22)

Taking into account

$\begin{array}[]{lllll}\mathbf{\pi}_{\mathbf{q}}=-i\hbar\,\mathbf{\displaystyle\frac{d}{dq}},&\mathbf{\pi}_{\mathbf{q}}^{*}=i\hbar\,\mathbf{\displaystyle\frac{d}{dq}},\end{array}$
(23)

we rewrite

$\begin{array}[]{lllll}\Bigl{\langle}\psi_{E_{f}}(\mathbf{q})\Bigl{|}\,\exp\\{\pm
i\,\alpha^{\prime}\,(\mathbf{k}_{\gamma}\mathbf{q})\Bigr{\\}}\,(\varepsilon^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{*})\,\Bigr{|}\,\psi_{E_{i}}(\mathbf{q})\Bigr{\rangle}=i\hbar\,\varepsilon^{(\alpha)}\,\mathbf{I}_{1}(\mp\alpha^{\prime}).\end{array}$
(24)

So, the full matrix element (22) obtains the following form as

$\begin{array}[]{lllll}\vspace{1.0mm}&\Bigl{\langle}\Psi_{E_{f}}(\mathbf{r},\mathbf{q})\Bigl{|}\,\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha)})\Bigr{|}\,\Psi_{E_{i}}(\mathbf{r},\mathbf{q})\Bigr{\rangle}\;=\\\
=&-\,\displaystyle\frac{1}{2}\,\displaystyle\frac{e\,\hbar}{m_{N}c}\Bigl{\\{}\par\displaystyle\frac{2}{\sqrt{2}}\,F_{2,\,\alpha}\cdot
I_{2}\Bigl{(}\displaystyle\frac{-1}{\sqrt{6}}\Bigr{)}-\par
i\hbar\,\sqrt{\displaystyle\frac{2}{3}}\,F_{1}\;\varepsilon^{(\alpha)}\;\mathbf{I}_{1}\Bigl{(}-\displaystyle\frac{1}{\sqrt{6}}\Bigr{)}+\par
2\,i\hbar\,\sqrt{\displaystyle\frac{2}{3}}\,\varepsilon^{(\alpha)}\;\mathbf{I}_{1}\Bigl{(}\sqrt{\displaystyle\frac{2}{3}}\Bigr{)}\Bigr{\\}}.\end{array}$
(25)

Note that vectors $\mathbf{\varepsilon}^{(\alpha)}$ are perpendicular to
$\mathbf{k}_{\gamma}$ in Coulomb gauge. Taking this property into account, we
obtain

$\begin{array}[]{lllll}(\varepsilon^{(\alpha)},\mathbf{k}_{\gamma})=0.\end{array}$
(26)

In the case of zero form factor $F_{2,\,\alpha}=0$, the matrix element is
simplified as

$\begin{array}[]{lllll}\Bigl{\langle}\Psi_{E_{f}}(\mathbf{r},\mathbf{q})\Bigl{|}\,\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha)})\Bigr{|}\,\Psi_{E_{i}}(\mathbf{r},\mathbf{q})\Bigr{\rangle}\;=\par-\,i\,\displaystyle\frac{1}{2}\,\sqrt{\displaystyle\frac{2}{3}}\,\displaystyle\frac{e\,\hbar^{2}}{m_{N}c}\Bigl{\\{}F_{1}\>\varepsilon_{\mu}\;\mathbf{I}_{1}\Bigl{(}-\displaystyle\frac{1}{\sqrt{6}}\Bigr{)}-2\,\varepsilon^{(\alpha)}\;\mathbf{I}_{1}\Bigl{(}\sqrt{\displaystyle\frac{2}{3}}\Bigr{)}\Bigr{\\}}.\end{array}$
(27)

###  III.4 Multipole expansion

In further calculation of Eq. (27) it needs to find integrals (21). Applying
the multipolar expansion, these integrals obtain form [see App. B, Eqs. (67),
(70)]

$\begin{array}[]{ll}\vspace{1mm}\mathbf{I}_{1}(\alpha)=\biggl{<}\varphi_{E_{f}}(\mathbf{q})\biggl{|}\,e^{-i\alpha\mathbf{k_{\gamma}q}}\displaystyle\frac{\partial}{\partial\mathbf{q}}\,\biggr{|}\,\varphi_{E_{i}}(\mathbf{q})\biggr{>}_{\mathbf{q}}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\mbox{\boldmath$\xi$}_{\mu}\,\mu\,\times\Bigl{[}p_{l_{\gamma}\mu}^{M}(\alpha)-i\mu\>p_{l_{\gamma}\mu}^{E}(\alpha)\Bigr{]},\\\
I_{2}(\alpha)=\Bigl{<}\varphi_{E_{f}}(\mathbf{q})\Bigl{|}\,e^{-i\alpha\mathbf{k_{\gamma}q}}\,\Bigr{|}\,\varphi_{E_{i}}(\mathbf{q})\Bigr{>}_{\mathbf{q}}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\Bigl{[}\mu\,\tilde{p}_{l_{\gamma}\mu}^{M}(\alpha)-i\,\tilde{p}_{l_{\gamma}\mu}^{E}(\alpha)\Bigr{]},\end{array}$
(28)

where [see Eqs. (68), (71)]

$\begin{array}[]{lcl}\vspace{3mm}p_{l_{\gamma}\mu}^{M}(\alpha)&=&-I_{M}(0,l_{f},l_{\gamma},1,\mu)\cdot
J_{1}(\alpha,0,l_{f},l_{\gamma}),\\\
\vspace{1mm}p_{l_{\gamma}\mu}^{E}(\alpha)&=&\sqrt{\displaystyle\frac{l_{C}+1}{2l_{\gamma}+1}}\cdot
I_{E}(0,l_{f},l_{\gamma},1,l_{\gamma}-1,\mu)\cdot
J_{1}(\alpha,0,l_{f},l_{\gamma}-1)\;-\\\
&-&\sqrt{\displaystyle\frac{l_{\gamma}}{2l_{\gamma}+1}}\cdot
I_{E}(0,l_{f},l_{\gamma},1,l_{\gamma}+1,\mu)\cdot
J_{1}(\alpha,0,l_{f},l_{\gamma}+1),\end{array}$ (29)
$\begin{array}[]{llll}\vspace{3mm}\tilde{p}_{l_{\gamma}\mu}^{M}(\alpha)=\tilde{I}\,(0,l_{f},l_{\gamma},l_{\gamma},\mu)\cdot\tilde{J}\,(\alpha,0,l_{f},l_{\gamma}),\\\
\tilde{p}_{l_{\gamma}\mu}^{E}(\alpha)=\sqrt{\displaystyle\frac{l_{\gamma}+1}{2l_{\gamma}+1}}\tilde{I}\,(0,l_{f},l_{\gamma},l_{\gamma}-1,\mu)\cdot\tilde{J}\,(\alpha,0,l_{f},l_{\gamma}-1)\;-\sqrt{\displaystyle\frac{l_{\gamma}}{2l_{\gamma}+1}}\tilde{I}\,(0,l_{f},l_{\gamma},l_{\gamma}+1,\mu)\cdot\tilde{J}\,(\alpha,0,l_{f},l_{\gamma}+1),\end{array}$
(30)

and [see Eqs. (69), (72)]

$\begin{array}[]{lllll}J_{1}(\alpha,l_{i},l_{f},n)&=&\displaystyle\int\limits^{+\infty}_{0}\displaystyle\frac{dR_{i}(r,l_{i})}{dr}\>R^{*}_{f}(l_{f},r)\,j_{n}(\alpha\,kr)\;r^{2}dr,\\\
\tilde{J}\,(\alpha,l_{i},l_{f},n)&=&\displaystyle\int\limits^{+\infty}_{0}R_{i}(r)\,R^{*}_{f}(l,r)\,j_{n}(\alpha
k_{\gamma}r)\;r^{2}dr.\end{array}$ (31)

Here, $\mathbf{\xi}_{\mu}$ are _vectors of circular polarization_ with
opposite directions of rotation (see Ref. Eisenberg.1973 , (2.39), p. 42).
Also we have the following properties [see App. B, Eqs. (84), (85)]

$\begin{array}[]{llll}\vspace{0.5mm}\displaystyle\sum\limits_{\alpha=1,2}\mbox{\boldmath$\varepsilon$}^{(\alpha)}\cdot\mathbf{I}_{1}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\mu\,h_{\mu}\,\bigl{(}p_{l_{\gamma},\mu}^{M}+p_{l_{\gamma},-\mu}^{E}\bigr{)},\\\
(\mbox{\boldmath$\varepsilon$}_{\rm x}+\mbox{\boldmath$\varepsilon$}_{\rm
z})\,\displaystyle\sum\limits_{\alpha=1,2}\Bigl{[}\mathbf{I}_{1}\times\mbox{\boldmath$\varepsilon$}^{(\alpha)}\Bigr{]}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\mu\,h_{\mu}\,\bigl{(}p_{l_{\gamma},\mu}^{M}-p_{l_{\gamma},-\mu}^{E}\bigr{)},\end{array}$
(32)

where [see App. B, Eqs. (74)]

$\begin{array}[]{llllll}h_{\pm}=\mp\displaystyle\frac{1\pm
i}{\sqrt{2}},&h_{-}+h_{+}=-i\,\sqrt{2},&\displaystyle\sum\limits_{\mu=\pm
1}\mu\,h_{\mu}=-h_{-}+h_{+}=-\sqrt{2},&\displaystyle\sum\limits_{\alpha=1,2}\mbox{\boldmath$\varepsilon$}^{(\alpha),*}=h_{-1}\mathbf{\xi}_{-1}^{*}+h_{+1}\mathbf{\xi}_{+1}^{*}.\end{array}$
(33)

###  III.5 Case of $l_{i}=0$, $l_{f}=1$, $l_{\gamma}=1$

In a case of $l_{i}=0$, $l_{f}=1$, $l_{\gamma}=1$ integrals (28) are
simplified to [see App. B, Eqs. (86)]

$\begin{array}[]{llllll}\vspace{1mm}\mathbf{I}_{1}=-i\,\sqrt{\displaystyle\frac{3\pi}{2}}\>\displaystyle\sum\limits_{\mu=\pm
1}\mbox{\boldmath$\xi$}_{\mu}\,\mu\,\times\Bigl{[}p_{l_{\gamma}=1,\,\mu}^{M}-i\mu\>p_{l_{\gamma}=1,\,\mu}^{E}\Bigr{]},\\\
I_{2}=-i\,\sqrt{\displaystyle\frac{3\pi}{2}}\>\displaystyle\sum\limits_{\mu=\pm
1}\Bigl{[}\mu\,\tilde{p}_{l_{\gamma}=1,\,\mu}^{M}-i\,\tilde{p}_{l_{\gamma}=1,\,\mu}^{E}\Bigr{]},\par\par\par\end{array}$
(34)

where matrix elements are simplified to [see details in App. B, Eqs. (89)]

$\begin{array}[]{lllllllll}\vspace{2mm}p_{l_{\gamma}\mu}^{M}=0,&p_{l_{\gamma}\mu}^{E}=\displaystyle\frac{1}{6}\sqrt{\displaystyle\frac{1}{\pi}}\cdot
J_{1}(0,1,0)-\displaystyle\frac{47}{240}\sqrt{\displaystyle\frac{1}{2\pi}}\cdot
J_{1}(0,1,2),\\\
\tilde{p}_{1\mu}^{M}(c)=\displaystyle\frac{\mu}{2\sqrt{2\pi}}\cdot\tilde{J}\,(c,0,1,1),&\tilde{p}_{1\mu}^{E}(c)=0.\par\end{array}$
(35)

We substitute these solutions to Eq. (34) and obtain [see App. B, Eqs. (89),
(90)]:

$\begin{array}[]{llllll}\mathbf{I}_{1}&=&-\displaystyle\frac{1}{6}\cdot\sqrt{\displaystyle\frac{3}{2}}\>\displaystyle\sum\limits_{\mu=\pm
1}\mbox{\boldmath$\xi$}_{\mu}\cdot\Bigl{(}J_{1}(0,1,0)-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}(0,1,2)\Bigr{)}.\end{array}$ (36)

Integrals do not depend on vectors of polarization. So, we simplify further:

$\begin{array}[]{llllll}\mathbf{I}_{1}&=&-\displaystyle\frac{1}{6}\cdot\sqrt{\displaystyle\frac{3}{2}}\>\Bigl{(}J_{1}(0,1,0)-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}(0,1,2)\Bigr{)}\cdot\bigl{(}\mbox{\boldmath$\xi$}_{\mu=+1}+\mbox{\boldmath$\xi$}_{\mu=-1}\bigr{)}.\end{array}$
(37)

Also from Eqs. (34) we find

$\begin{array}[]{llllll}I_{2}(\alpha)&=&-i\,\displaystyle\frac{\sqrt{3}}{2}\>\tilde{J}\,(\alpha,0,1,1).\end{array}$
(38)

###  III.6 Action on vectors of polarization

Now we calculate summation over vectors of polarization. We use definition of
vectors of polarizations as in Eqs. (57)–(58) in Ref.
Maydanyuk_Vasilevsky.2023.PRC  (see App. C in that paper for details):

$\begin{array}[]{cc}\mbox{\boldmath$\varepsilon$}^{(1)}=\displaystyle\frac{1}{\sqrt{2}}\,\bigl{(}\mbox{\boldmath$\xi$}_{-1}-\mbox{\boldmath$\xi$}_{+1}\bigr{)},&\mbox{\boldmath$\varepsilon$}^{(2)}=\displaystyle\frac{i}{\sqrt{2}}\,\bigl{(}\mbox{\boldmath$\xi$}_{-1}+\mbox{\boldmath$\xi$}_{+1}\bigr{)},\end{array}$
(39)

and

$\begin{array}[]{llll}\mbox{\boldmath$\varepsilon$}^{(1)}\cdot\bigl{(}\mbox{\boldmath$\xi$}_{\mu=+1}+\mbox{\boldmath$\xi$}_{\mu=-1}\bigr{)}=0,&\mbox{\boldmath$\varepsilon$}^{(2)}\cdot\bigl{(}\mbox{\boldmath$\xi$}_{\mu=+1}+\mbox{\boldmath$\xi$}_{\mu=-1}\bigr{)}=-\,i\,\sqrt{2}.\end{array}$
(40)

On such a basis, from Eq. (37) we find:

$\begin{array}[]{llllll}\mbox{\boldmath$\varepsilon$}^{(1)}\cdot\mathbf{I}_{1}=0,&\mbox{\boldmath$\varepsilon$}^{(2)}\cdot\mathbf{I}_{1}(\alpha)=\par\par
i\;\displaystyle\frac{\sqrt{3}}{6}\cdot\Bigl{(}J_{1}(\alpha,0,1,0)-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}(\alpha,0,1,2)\Bigr{)}.\end{array}$ (41)

Now we can recalculate the matrix element (27) as (at $\mu=1$ it equals to
zero)

$\begin{array}[]{lllll}\vspace{1.0mm}\Bigl{\langle}\Psi_{E_{f}}(\mathbf{r},\mathbf{q})\Bigl{|}\,\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha=2)})\Bigr{|}\,\Psi_{E_{i}}(\mathbf{r},\mathbf{q})\Bigr{\rangle}&=&\displaystyle\frac{\sqrt{2}\,e\,\hbar^{2}}{12\,m_{N}c}\,\Bigl{\\{}F_{1}\,\Bigl{[}J_{1}\,\Bigl{(}-\displaystyle\frac{1}{\sqrt{6}},0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}\,\Bigl{(}-\displaystyle\frac{1}{\sqrt{6}},\,0,1,2\bigr{)}\Bigr{]}\;-\\\
&-&2\,\Bigl{[}J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{2}{3}},0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{2}{3}},\,0,1,2\bigr{)}\Bigr{]}\Bigr{\\}},\end{array}$
(42)

where integrals are defined in Eqs. (31).

###  III.7 Resonating group method

To study structure of two- and -three-cluster systems, we use the algebraic
version of the resonating group method, which was formulated in Refs.
kn:Fil_Okhr , kn:Fil81 . Two main merits (advantages) of the algebraic version
of the RGM are: (i) it employ a full set of oscillator functions to expand
wave functions of relative motion of clusters and, thus, reduces the many-
particle Schrödinger equation a set of linear algebraic equations, and (ii) it
implements proper boundary conditions for bound and continuous-spectrum states
in discrete, oscillator space.

###  III.8 Wave function of deuteron

Wave function of the bound state of deuteron was obtained with the Minnesota
NN potential kn:Minn_pot1 . This potential creates the bound state at
$E_{d}=-2.202$ MeV, which has to be compared with experimental value
$E_{d}=-2.225$ MeV. Wave function of deuteron is shown in Fig. 1.

Figure 1: (Color online) Wave function of deuteron as a function of coordinate
$r$

It has a long exponential tail

$\phi_{E_{d},L=0}\left(\mathbf{r}\right)\approx\exp\left\\{-\kappa
r\right\\}/r,$

where

$\kappa=\sqrt{\frac{2m\left|E_{d}\right|}{\hbar^{2}}}=\sqrt{\frac{2\times
2.202}{41.47}}=0.325879~{}\text{fm}^{-1}.$

Recall that the vector Jacobi $r$ is measured in fm.

It is interesting to note that the function (16) is an exact solution of two-
body problem with the contact interaction

$V(r)=V_{0}\delta\left(r\right).$

This interaction is also called as the zero-range interaction and is widely
used in atomic and nuclear physics (for more details see Ref. kn:DemkovE ). We
will use the normalized to unity function

$\phi\left(r\right)=\sqrt{2\kappa}\exp\left\\{-\kappa r\right\\}/r,$ (43)

to approximate correct wave function of deuteron.

To solve the Schrödinger equations (10) and (13) for deuteron and $p+d$
system, wave functions $\phi_{E_{d},l}\left(r\right)$ and
$\psi_{E,L}\left(q\right)$ are expanded over basis of oscillator functions

$\displaystyle\phi_{E_{d},l}\left(r\right)$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{N\max}C_{n}^{\left(E_{d},l\right)}\Phi_{nl}\left(r,b\right),$
(44) $\displaystyle\psi_{E,L}\left(q\right)$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{N\max}C_{n}^{\left(E,L\right)}\Phi_{nL}\left(q,b\right),$
(45)

where $\Phi_{n}\left(r,b\right)$ is an oscillator function

$\Phi_{nL}\left(r,b\right)=\left(-1\right)^{n}N_{nL}~{}b^{-3/2}\rho^{L}e^{-\frac{1}{2}\rho^{2}}L_{n}^{L+1/2}\left(\rho^{2}\right),\quad\rho=\frac{r}{b}$
(46)

and $b$ is the oscillator length, and

$N_{nL}=\sqrt{\frac{2\Gamma\left(n+1\right)}{\Gamma\left(n+L+3/2\right)}}.$

A set of expansion coefficients $\left\\{C_{n}\right\\}$ can be considered as
deuteron wave function in oscillator representation. In Fig. 2 we show
deuteron wave function in oscillator representation.

Figure 2: (Color online) Wave function of deuteron in oscillator
representation

This wave function was constructed with 200 oscillator functions
($N\max=199$), however as one can see that only a small number of basis
functions (0$\leq n\leq$25) give a noticeable contribution.

In the shell-model approximation, the wave function of the deuteron bound
state is a Gaussian function

$\phi\left(r\right)=\frac{1}{b^{3/2}}\exp\left\\{-\frac{1}{2}\left(\frac{r}{b}\right)^{2}\right\\}.$
(47)

Form factor of deuteron is then

$\left\langle\phi\left(\mathbf{r}\right)\left|\exp\left\\{-\frac{i}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\right|\phi\left(\mathbf{r}\right)\right\rangle=\exp\left\\{-\frac{1}{8}\left(k_{\gamma}b\right)^{2}\right\\}.$
(48)

If we take the deuteron wave function in the form

$\phi\left(r\right)=\sqrt{2\kappa}\exp\left\\{-\kappa r\right\\}/r,$ (49)

then we obtain deuteron form factor

$\left\langle\phi\left(\mathbf{r}\right)\left|\exp\left\\{-\frac{i}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\right|\phi\left(\mathbf{r}\right)\right\rangle=\frac{2\sqrt{2}\kappa}{k_{\gamma}}\operatorname{arctan}\left(\frac{k_{\gamma}}{2\sqrt{2}\kappa}\right).$
(50)

Form factor from Eq. (50) as function of $k_{\gamma}$ demonstrates slower
decreasing comparing to the form factor of the shell mode (50)

##  IV Matrix elements in the folding approximation

Matrix element of bremsstrahlung emission of photons for two $s$-clusters
(i.e., for clusters with $1\leq A_{\alpha}\leq 4$ or for $n$, $p$, $d$,
${}^{3}{\rm H}$, ${}^{3}{\rm He}$, ${}^{4}{\rm He}$) can be written down as
[see Ref. Maydanyuk_Vasilevsky.2023.PRC , for details]

$\begin{array}[]{lllll}\vspace{1.0mm}&\Bigl{\langle}\Psi_{E_{f}l_{f}}\Bigl{|}\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\mbox{\boldmath$\varepsilon$}^{(\alpha)})\Bigr{|}\Psi_{E_{i}l_{i}}\Bigr{\rangle}_{\rm
fold}=\\\
\vspace{1.0mm}=&\displaystyle\frac{e\hbar}{m_{N}c}\biggl{\\{}\sqrt{\displaystyle\frac{A_{2}}{A_{1}\,A}}\biggl{\langle}R_{E_{f}l_{f}}(r)\,Y_{l_{f}m_{f}}(\widehat{\mathbf{r}}_{i})\biggl{|}\exp{-i\,\sqrt{\displaystyle\frac{A_{2}}{A_{1}\,A}}\,(\mathbf{k}_{\gamma},\mathbf{r})}\,(\mbox{\boldmath$\varepsilon$}^{(\alpha)},\hat{\mbox{\boldmath$\pi$}})\biggr{|}R_{E_{i}l_{i}}(r)\,Y_{l_{i}m_{i}}(\widehat{\mathbf{r}}_{i})\biggr{\rangle}\,F_{1}-\\\
-&\sqrt{\displaystyle\frac{A_{1}}{A_{2}\,A}}\biggl{\langle}R_{E_{f}l_{f}}(r)\,Y_{l_{f}m_{f}}(\widehat{\mathbf{r}}_{i})\biggl{|}\exp{i\,\sqrt{\displaystyle\frac{A_{1}}{A_{2}\,A}}\,(\mathbf{k}_{\gamma},\mathbf{r})}\,(\mbox{\boldmath$\varepsilon$}^{(\alpha)},\hat{\mbox{\boldmath$\pi$}})\biggr{|}R_{E_{i}l_{i}}(r)\,Y_{l_{i}m_{i}}(\widehat{\mathbf{r}}_{i})\biggr{\rangle}\,F_{2}\biggr{\\}}.\end{array}$
(51)

In the standard approximation of resonating group method, form factor $F_{n}$
equals ($n=1,2$)

$\begin{array}[]{lllll}F_{n}=&\Bigl{\langle}\Phi_{n}(A_{n})\Bigl{|}F_{0}^{(n)}\Bigr{|}\Phi_{n}(A_{n})\Bigr{\rangle}=Z_{n}\,\exp{-\displaystyle\frac{1}{4}\,\displaystyle\frac{A_{n}-1}{A_{n}}\,(k,b)^{2}},\end{array}$
(52)

with $b$ is oscillator length. Using property (24):

$\begin{array}[]{lllll}\Bigl{\langle}\psi_{E_{f}}(\mathbf{q})\Bigl{|}\,\exp\\{\pm
i\,\alpha^{\prime}\,(\mathbf{k}_{\gamma}\mathbf{q})\Bigr{\\}}\,(\varepsilon^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{*})\,\Bigr{|}\,\psi_{E_{i}}(\mathbf{q})\Bigr{\rangle}=\par
i\hbar\,\varepsilon^{(\alpha)}\,\mathbf{I}_{1}(\mp\alpha^{\prime}),\end{array}$
(53)

matrix element is rewritten as

$\begin{array}[]{lllll}\vspace{1.0mm}\Bigl{\langle}\Psi_{E_{f}l_{f}}\Bigl{|}\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha)})\Bigr{|}\Psi_{E_{i}l_{i}}\Bigr{\rangle}_{\rm
fold}&=&i\,\displaystyle\frac{e\hbar^{2}}{m_{N}c}\,\varepsilon^{(\alpha)}\,\biggl{\\{}\sqrt{\displaystyle\frac{A_{2}}{A_{1}\,A}}\mathbf{I}_{1}\Bigl{(}\sqrt{\displaystyle\frac{A_{2}}{A_{1}\,A}}\Bigr{)}\,F_{1}-\sqrt{\displaystyle\frac{A_{1}}{A_{2}\,A}}\mathbf{I}_{1}\Bigl{(}-\sqrt{\displaystyle\frac{A_{1}}{A_{2}\,A}}\Bigr{)}\,F_{2}\biggr{\\}}.\end{array}$
(54)

Now we take into account property (41)

$\begin{array}[]{llllll}\mbox{\boldmath$\varepsilon$}^{(1)}\cdot\mathbf{I}_{1}=0,&\mbox{\boldmath$\varepsilon$}^{(2)}\cdot\mathbf{I}_{1}(\alpha)=i\>\displaystyle\frac{\sqrt{3}}{6}\cdot\Bigl{(}J_{1}(\alpha,0,1,0)-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}(\alpha,0,1,2)\Bigr{)},\end{array}$

and obtain

$\begin{array}[]{lllll}\vspace{1.5mm}\Bigl{\langle}\Psi_{E_{f}l_{f}}\Bigl{|}\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha)})\Bigr{|}\Psi_{E_{i}l_{i}}\Bigr{\rangle}_{\rm
fold}&=&\displaystyle\frac{\sqrt{3}\,e\hbar^{2}}{6\,m_{N}c}\,\Bigl{\\{}\sqrt{\displaystyle\frac{A_{2}}{A_{1}\,A}}\Bigl{[}J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{A_{2}}{A_{1}\,A}},\,0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\,J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{A_{2}}{A_{1}\,A}},\,0,1,2\Bigr{)}\Bigr{]}\,F_{1}\;-\\\
&-&\sqrt{\displaystyle\frac{A_{1}}{A_{2}\,A}}\Bigl{[}J_{1}\,\Bigl{(}-\sqrt{\displaystyle\frac{A_{1}}{A_{2}\,A}},\,0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\,J_{1}\,\Bigl{(}-\sqrt{\displaystyle\frac{A_{1}}{A_{2}\,A}},\,0,1,2\Bigr{)}\Bigr{]}\,F_{2}\Bigr{\\}}.\end{array}$
(55)

In particular, for proton-deuteron scattering we have (we choose the first
index — for proton: $A_{1}=1$, $F_{1}=F_{\rm p}$, the second index — for
deuteron: $A_{2}=2$, $F_{2}=F_{\rm D}$):

$\begin{array}[]{lllll}\vspace{1.5mm}\Bigl{\langle}\Psi_{E_{f}l_{f}}\Bigl{|}\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha)})\Bigr{|}\Psi_{E_{i}l_{i}}\Bigr{\rangle}_{\rm
fold}&=&\displaystyle\frac{\sqrt{3}\,e\hbar^{2}}{6\,m_{N}c}\,\Bigl{\\{}\sqrt{\displaystyle\frac{2}{3}}\Bigl{[}J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{2}{3}},\,0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\,J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{2}{3}},\,0,1,2\Bigr{)}\Bigr{]}\,F_{\rm
p}\;-\\\
&-&\sqrt{\displaystyle\frac{1}{6}}\Bigl{[}J_{1}\,\Bigl{(}-\sqrt{\displaystyle\frac{1}{6}},\,0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\,J_{1}\,\Bigl{(}-\sqrt{\displaystyle\frac{1}{6}},\,0,1,2\Bigr{)}\Bigr{]}\,F_{\rm
D}\Bigr{\\}}.\end{array}$ (56)

For further analysis it is more convenient to rewrite this solution as

$\begin{array}[]{lllll}\vspace{1.5mm}\Bigl{\langle}\Psi_{E_{f}l_{f}}\Bigl{|}\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha)})\Bigr{|}\Psi_{E_{i}l_{i}}\Bigr{\rangle}_{\rm
fold}&=&\displaystyle\frac{\sqrt{2}\,e\hbar^{2}}{12\,m_{N}c}\,\Bigl{\\{}\Bigl{[}J_{1}\,\Bigl{(}-\sqrt{\displaystyle\frac{1}{6}},\,0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\,J_{1}\,\Bigl{(}-\sqrt{\displaystyle\frac{1}{6}},\,0,1,2\Bigr{)}\Bigr{]}\,F_{\rm
D}\>-\\\
&-&2\,\Bigl{[}J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{2}{3}},\,0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\,J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{2}{3}},\,0,1,2\Bigr{)}\Bigr{]}\,F_{\rm
p}\Bigr{\\}}.\end{array}$ (57)

##  V Definition of cross section of bremsstrahlung emission of photons and
resulting formulas

Cross-section of bremsstrahlung emission of photons is
Maydanyuk_Vasilevsky.2023.PRC

$\begin{array}[]{lllll}\vspace{1.0mm}\displaystyle\frac{d\,\sigma^{(1)}}{d\Omega_{A_{1}}\,d\Omega_{A_{2}}\,d\Omega_{\gamma}}&=&\displaystyle\frac{E_{\gamma}}{(2\pi\hbar)^{4}}\,\Bigl{(}\displaystyle\frac{p_{1f}}{\hbar
c}\Bigr{)}\,\displaystyle\frac{\sin^{2}\theta_{1}\,\sin^{2}\theta_{2}}{sin^{5}(\theta_{1}+\theta_{2})}\;\times\par\displaystyle\frac{1}{2J+1}\displaystyle\sum\limits_{\mu
M_{i}}\Bigl{|}\Bigl{\langle}\Psi_{\tilde{E}\tilde{L}}\Bigl{|}\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha)})\Bigr{|}\Psi_{EL}\Bigr{\rangle}\Bigr{|}^{2},\end{array}$
(58)

where $p_{1}$ is the momentum of the incident nucleus (cluster) $A_{1}$,
$\theta_{1}$ and $\theta_{2}$ are scattering angels of the first and second
clusters in laboratory frame.

We write down final formulas of matrix elements of bremsstrahlung emission in
the proton-deuteron scattering. It turn out that in first approach [we will
call it as _the three-cluster model_ , see Eq. (27)] and in the second
approach [we will call it as _the folding model_ , see Eq. (57)] matrix
elements are the same:

$\begin{array}[]{lllll}\vspace{1.0mm}\Bigl{\langle}\Psi_{E_{f}}(\mathbf{r},\mathbf{q})\Bigl{|}\,\hat{H}_{\gamma}(\mathbf{k}_{\gamma},\varepsilon^{(\alpha=2)})\Bigr{|}\,\Psi_{E_{i}}(\mathbf{r},\mathbf{q})\Bigr{\rangle}&=&\displaystyle\frac{\sqrt{2}\,e\,\hbar^{2}}{12\,m_{N}c}\,\Bigl{\\{}\Bigl{[}J_{1}\,\Bigl{(}-\displaystyle\frac{1}{\sqrt{6}},0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}\,\Bigl{(}-\displaystyle\frac{1}{\sqrt{6}},\,0,1,2\bigr{)}\Bigr{]}\;F_{1}\,-\\\
&-&2\,\Bigl{[}J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{2}{3}},0,1,0\Bigr{)}-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}\,\Bigl{(}\sqrt{\displaystyle\frac{2}{3}},\,0,1,2\bigr{)}\Bigr{]}\Bigr{\\}}.\end{array}$
(59)

Integrals are [see Eqs. (31)]

$\begin{array}[]{lllll}J_{1}(\alpha,l_{i},l_{f},n)&=&\displaystyle\int\limits^{+\infty}_{0}\displaystyle\frac{dR_{i}(r,l_{i})}{dr}\>R^{*}_{f}(l_{f},r)\,j_{n}(\alpha\,k_{\gamma}r)\;r^{2}dr.\par\end{array}$
(60)

##  VI Analysis, numerical calculations

To understand the role of the deuteron structure, we are going to perform
three (four) types of calculations. They are distinguished by the wave
function of the deuteron and are labeled by indexes $0$, ($C$), $S$, and $R$.
The index $0$ means that the deuteron is considered as a structureless
particle and, thus, its internal structure is ignored. (If the wave function
of deuteron is approximated by the case of the contact interaction (43), we
will use the index C). The index $S$ stands for the shell-model approximation
(47) of the wave function of deuteron, and the last case $R$ means that the
realistic wave function (45) of deuteron is involved in calculations.

###  VI.1 Deuteron wave functions and form factor

The key element of our model is a wave function of the deuteron bound state.
This function will determine the interaction of proton and deuteron. Thus we
start our analysis from deuteron wave functions in different model and
approximations. In Figs. 3 and 4 we display wave functions of deuteron
obtained with the Minnesota potential. The shell model (SM) and cluster model
(CM) visually are very similar. However, displaying wave functions in a
logarithmic scale, we see that they have quite different asymptotic behavior.

Figure 3: Wave functions of deuteron obtained in the shell model (SM) and
cluster model (CM). Figure 4: Asymptotic form of the deuteron wave functions
constructed in the shell model (SM) and cluster model (CM).

Deuteron form factors calculated within the shell model and cluster model are
shown in Figs. 5 and 6.

Figure 5: Form factors of deuteron determined in the shell and cluster models
Figure 6: Form factors of deuteron in logarithmic scale calculated in the
shell and cluster models

If the zero-range interaction is used to determine wave function of deuteron
(43), then the deuteron form factor is (see App. C)

$F_{1}(\boldsymbol{k}_{\gamma})=\dfrac{2\sqrt{2}\kappa}{k_{\gamma}}\arctan\biggl(\dfrac{k_{\gamma}}{2\sqrt{2}\kappa}\biggr{missing}).$
(61)

This form factor is shown in Figs. 7.

Figure 7: Form factor $F_{1}$ (in left panel) and square of this form factor
$F_{1}^{2}$ (in right panel) at $\kappa=0.325879$ fm-1.

###  VI.2 Wave functions of $p+d$ system

In Fig. 8 we display phase shifts of the elastic $p+d$ scattering. One can see
that the strongest interaction is observed in the 1/2+ state, where the
nucleus 3He has a bound state. For energy $E>$100 MeV, all displayed phase
shifts are very close to zero. This is an additional indication that potential
of the $p+d$ interaction is weak and that the Born approximation can be used
for this energy range.

Figure 8: Phase shifts of the elastic $p+d$ scattering, obtained with the MP
for different $J^{\pi}$ states

We constructed wave functions of the continuous spectrum states using
diagonalization procedure of the 100$\times$100 matrix of Hamiltonian. Details
and justification of this procedure can be found, for example, in Refs.
2015NuPhA.941..121L , 2023UkrJPh..68..3K . Fig. 9 shows wave functions of 1/2+
states as a function of distance between proton and deuteron. In Fig. 10 we
display wave functions for the 1/2- state. Note that the states 1/2+ and 1/2-
can be connected by dipole transition operator. General features of the
displayed wave functions are that they have large amplitude at relatively
small distances between clusters ($r<$5 fm) and that they slowly decreasing as
$1/r$.

Figure 9: Wave functions of continuous spectrum states in the 1/2+ state of
$p+d$ system Figure 10: Wave functions of relative motion of proton with
respect to deuteron in the 1/2- state

###  VI.3 Different NN potentials

In this section we consider how the shape of nucleon-nucleon potential affects
phase shifts of the elastic $p+d$ scattering. For this aim we involved in our
calculations two new NN potentials. They are the Volkov potential (VP)
kn:Volk65  and modified Hasegawa-Nagata potential (MHNP) potMHN1 ; potMHN2 .
These potentials alongside with the Minnesota potential are often used in
different cluster models. It is demonstrated in Fig. 11, where the even
components $V_{31}$ and $V_{13}$ of three nucleon-nucleon potentials are
displayed, that the MHNP has the largest repulsive core at small distance
between nucleons, the VP has smallest repulsive core and the MP represents
intermediate case among three selected potentials.

Figure 11: The even components $V_{31}$ and $V_{13}$ of the MHNP, MP and VP
nucleon-nucleon potentials as a function of distance between nucleons

In Fig. 12 we display phase shifts of the elastic $p+d$ scattering in the 1/2+
state. We can see that the phase shifts slightly depend on the shape of the
nucleon-nucleon potentials especially at the energy region $E<$50 MeV. At the
energy range 150$<E<$200 differences of phase shifts for different potentials
are less than 20 degrees.

Figure 12: Phase shifts of the $p+d$ scattering in the 1/2+ state calculated
with three nucleon-nucleon potentials

As a results, wave functions of the elastic $p+d$ scattering obtained with
different NN potentials are very close to one other. In Fig. 13 we display
wave functions of $p+d$ system for the energy 147 MeV. The noticeable
difference of wave functions is observed at small distances $r<$ 1 fm.

Figure 13: Wave functions of the $p+d$ system at energy 147 MeV obtained with
three different nucleon-nucleon potentials

Let us consider the 1/2- states in 3He and in the $p+d$ elastic scattering.
The 1/2- states can be connected to the 1/2+ states by the dipole transition
operator. In Fig. 14 we display phase shifts of $p+d$ elastic scattering in
the 1/2- states, calculated with three nucleon-nucleon potentials. At low
energy range, phase shifts exhibit resonance-like behavior, when phase shifts
rapidly growing with increasing of energy $E$. However, amplitudes of growing
are small, besides we estimate that the widths of such resonance states are
larger than 20 MeV and their energies are less then 8 MeV. Thus, such states
cannot be considered as resonance states.

Figure 14: Phase shifts of the elastic $p+d$ scattering in the 1/2- state
obtained with the MP, MHNP and VP

This conclusion can be partially confirmed by behavior of wave functions in
the 1/2- states. In Fig. 15 we display wave functions obtained with three
nucleon-nucleon potentials at the energy $E$=12.7 MeV and in Fig. 16 are shown
wave functions for larger value of the energy $E$=147 MeV. As we can see, the
wave functions at relatively small and large energies have noticeable maxima
at relatively small distances (0.8$<r<$2.5 fm) between proton and deuteron. We
may conclude that maxima of wave functions in the 1/2- state at small
distances are due to interplay between effects of nucleon-nucleon and Coulomb
interactions from one side and effects of the Pauli principle from another
side.

Figure 15: Wave functions of relative motion of proton and deuteron with
energy $E$=12.7 MeV and the total angular momentum $J^{\pi}=$1/2-

Figure 16: Wave functions of relative motion of proton and deuteron with
energy $E$=147 MeV and the total angular momentum $J^{\pi}=$1/2-

###  VI.4 Dependence of the bremsstrahlung cross section on the structure of
deuteron

We are interesting in question if the bremsstrahlung cross section is
dependent on the structure of deuteron. In the previous section we have
discussed several forms of the deuteron wave functions. Two of them are
presented in analytic form, and one of them is obtained numerically by solving
two-body Schrödinger equation with the Minnesota potential. The deuteron wave
function of the oscillator shell model, displayed in Eq. (47), allows us in a
simple way to study effects of the deuteron structure on the bremsstrahlung
cross section. Indeed, this wave function depends on the oscillator length
$b$. Recall, that the oscillator length is selected to minimize the ground
state energy of deuteron with selected potential. If in Eq. (47) put $b$=0,
then we obtain structureless deuteron or, in other words, we disregard of the
internal structure of deuteron.

If to suppose existence of such a dependence, it can be small or even not
visible for analysis. If this is correct, then it is unclear, if this is not
so at other energies. In particular, we would like to find energies, where the
spectra are already dependent visibly on the internal structure of deuteron.

Some information about the internal structure of deuteron is included in its
form factor, which is presented in the folding and cluster approaches. The
matrix element of bremsstrahlung emission in both approaches is defined by Eq.
(59) as ($F_{\rm p}=1$), where we will consider form factor of deuteron in
folding approach given by Eq. (52) [$Z_{D}=1$, $A_{D}=2$ for deuteron]

$\begin{array}[]{lllll}F_{D}=&Z_{D}\,\exp{-\displaystyle\frac{1}{4}\,\displaystyle\frac{A_{D}-1}{A_{D}}\,(k,b)^{2}}=\exp{-\displaystyle\frac{1}{8}\,(k,b)^{2}}\end{array}$
(62)

with the oscillator length $b$.

Results of such calculations at different energies of beam are presented in
Fig. 17.

Figure 17: (Color online) Cross sections of bremsstrahlung emission for
$p+\isotope[2]{H}$ with included structure of deuteron and without it
calculated at energies of beam of 145 MeV ($E_{\rm kin}=98~{}MeV$), 500 MeV
($E_{\rm kin}=333~{}MeV$) and 1.5 GeV ($E_{\rm kin}=1.0~{}GeV$) [Parameters of
calculations: cross section is defined in Eq. (58) and then averaged over all
angles with exception of photon emission angle $\theta$ used in experiments
($\theta=60^{\circ}$, see also Ref. Nakayama.1992.PRC.v45 , Fig. 5), as non-
zero oscillator length for deuteron we choose $b=1.3$ fm, $R_{\rm
max}=20000\,{\rm fm}$ and 2500000 intervals are used in the numerical
integration; time of computer calculations is 2–4 min for 40 points of each
calculated spectrum, kinetic energy $E_{\rm kin}$ of relative motion of proton
and deuteron is used in calculations of bremsstrahlung matrix elements, which
is $E_{\rm kin}=2/3\cdot E_{\rm beam}$]. Here, experimental data at 145 MeV of
beam energy given by black triangles are extracted from Ref.
Clayton.1992.PRC.p1810 .

From such results we conclude the following.

  1. 1.

Difference between cross sections calculated with included structure of
deuteron and without it at the same beam energy becomes visible and stable at
higher energy of beam (larger 500 MeV).

  2. 2.

Calculation with realistic wave functions (with included structure of
deuteron) gives larger cross section of bremsstrahlung than cross section
without inclusion of structure of deuteron.

  3. 3.

Experiment Clayton.1992.PRC.p1810  at used beam energy of 145 MeV is not
effective for such a study (it is demonstrated by cross sections at 1.5 GeV in
comparison with cross section at 145 MeV in this figure). At the same time,
possible new measurement of bremsstrahlung cross sections but at higher
energies (about 0.5–1.5 GeV of beam energy) will allow to extract information
about structure of deuteron (realistic oscillator length, and wave function).

  4. 4.

More precise information about structure of deuteron can be obtained if to
organize unified experiments in measurement of bremsstrahlung cross section at
two different energies of beam (for example, at 145 MeV and 500 MeV). Then,
our model will estimate ratio between two bremsstrahlung cross sections at
such energies and provide value about realistic oscillator length with high
precision.

###  VI.5 Dependence of bremsstrahlung cross section on new form factor of
deuteron

Now we will analyze how much the spectrum is changed if to use new form factor
of deuteron instead of previous calculations. So, we have the matrix element
in form (59) as with form factor of deuteron given by Eq. (61). Note that this
form factor of deuteron does not include the oscillator length. Results of
such calculations with new form factor are presented in Fig. 18.

Figure 18: (Color online) Cross sections of bremsstrahlung emission for
$p+\isotope[2]{H}$ with included new form factor calculated at energies of
beam of 145 MeV ($E_{\rm kin}=98~{}MeV$), 500 MeV ($E_{\rm kin}=333~{}MeV$)
and 1.5 GeV ($E_{\rm kin}=1.0~{}GeV$) [Parameters of calculations: cross
section is defined in Eq. (58), matrix element is defined in Eq. (59), as non-
zero oscillator length for deuteron we choose $b=1.3$ fm, $R_{\rm
max}=20000\,{\rm fm}$ and 2500000 intervals are used in the numerical
integration; time of computer calculations is 2–4 min for 40 points of each
calculated spectrum, kinetic energy $E_{\rm kin}$ of relative motion of proton
and deuteron is used in calculations of bremsstrahlung matrix elements, which
is $E_{\rm kin}=2/3\cdot E_{\rm beam}$]. Here, experimental data at 145 MeV of
beam energy given by black triangles are extracted from Ref.
Clayton.1992.PRC.p1810 .

From these calculations one can see that inclusion of new form factor reduces
full cross section a little. But, general dependence of the cross section on
this form of form factor is observed at higher energies.

##  VII Conclusions and perspectives

In this paper we investigated emission of bremsstrahlung photons in the
scattering of protons off deuterons on the fully cluster basis in a wide
region of the beam energy from low energies till 1.5 GeV. To realize this
investigations, we developed a new model. On the basis of such a model we
obtain the following results:

  * •

It is demonstrated that the matrix elements of bremsstrahlung emission in the
deuteron-proton scattering in the three-cluster formalism coincides with the
corresponding matrix elements in the folding approximation given in Ref.
Maydanyuk_Vasilevsky.2023.PRC .

  * •

Formalism of the model includes form factor of a deuteron which affects
behavior of bremsstrahlung cross sections and reflects the structure of
deuteron and influence of parameters of nucleon-nucleon interactions. This
gives possibility to investigate structure of nuclei and properties of
interactions from analysis of bremsstrahlung cross sections.

  * •

We studied dependence of the bremsstrahlung cross section on structure of
deuteron. We find that the oscillator length $b$, related to the shell-model
description of the deuteron, is convenient parameter for such a study.
Analysis of dependence of the cross section on such a parameter shows the
following. At beam energies used at experiment Clayton.1992.PRC.p1810  the
cross section is not sensitive visibly on variations of oscillator length,
i.e. on the internal structure of deuteron. However, stable difference between
cross sections calculated at zero and non-zero oscillator lengths at the same
beam energy is observed at higher energy of beam (larger 500 MeV). The
spectrum is increased at increasing of the oscillator length inside the full
energy region of the emitted photons. However, the usage of new the deuteron
form factor in the cluster formalism [see Eq. (61)] reduces the bremsstrahlung
cross section a little.

  * •

More precise information about structure of deuteron can be obtained if to
organize unified new experiments in measurement of bremsstrahlung cross
section at two different energies of beam (for example, at 145 MeV and 500 MeV
or above). Then, our model will estimate ratio between two bremsstrahlung
cross sections at such energies and provide information about realistic value
of the oscillator length with high precision.

As a perspective, we see that formalism of our model provides strict basis for
description of the deuteron-proton scattering and emission of virtual photons
in study of dilepton productions in the deuteron-proton scattering (see, for
example, Refs. Ernst.1998.PRC ; Wilson.1998.PRC ). This can be interesting for
further investigations and applications.

## Acknowledgements

Authors are highly appreciated to Prof. Gyorgy Wolf for fruitful discussions
concerning to different aspects of nuclear collisions at high energies, Dr. S.
A. Omelchenko for fruitful discussions concerning to aspects of different
nuclear models for scattering and decays. This work was supported in part by
the Program of Fundamental Research of the Physics and Astronomy Department of
the National Academy of Sciences of Ukraine (Project No. 0117U000239 and
Project No. 0122U000889). S. P. M. thanks the support of OTKA grant K138277.
V.V. is grateful to the Simons foundation for financial support (Award No.
1030284).

##  Appendix A Calculation of operator of bremsstrahlung emission in three-
cluster model

In this Appendix we will calculate operator of emission of bremsstrahlung
photons in three-cluster formalism. We fix position of nucleons. We assume
that vector $\mathbf{r}$ measures the distance between proton and neutron
which form a deuteron. We also assume that $\mathbf{r}_{1}$ is a coordinate of
the first proton and $\mathbf{r}_{2}$ is a coordinate of a neutron. Vector
$\mathbf{r}_{3}$ determines the location of the second proton. Starting from
Eq. (1), this operator is obtained the following form:

$\displaystyle\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)=\frac{1}{2}\frac{e\hbar}{m_{N}c}\left[\widehat{\mathbf{\pi}}_{1}^{\ast}\mathbf{A}^{\ast}\left(1\right)+\mathbf{A}^{\ast}\left(1\right)\widehat{\mathbf{\pi}}_{1}^{\ast}+\widehat{\mathbf{\pi}}_{3}^{\ast}\mathbf{A}^{\ast}\left(3\right)+\mathbf{A}^{\ast}\left(3\right)\widehat{\mathbf{\pi}}_{3}^{\ast}\right]$
$\displaystyle=\frac{1}{2}\frac{e\hbar}{m_{N}c}\left\\{\widehat{\mathbf{\pi}}_{1}^{\ast}\mathbf{\varepsilon}^{(\alpha)}\exp\left\\{-i\left(\mathbf{k}_{\gamma}\mathbf{\rho}_{1}\right)\right\\}+\mathbf{\varepsilon}^{(\alpha)}\exp\left\\{-i\left(\mathbf{k}_{\gamma}\mathbf{\rho}_{1}\right)\right\\}\widehat{\mathbf{\pi}}_{1}^{\ast}\right.$
$\displaystyle+\left.\widehat{\mathbf{\pi}}_{3}^{\ast}\mathbf{\varepsilon}^{(\alpha)}\exp\left\\{-i\left(\mathbf{k}_{\gamma}\mathbf{\rho}_{3}\right)\right\\}+\mathbf{\varepsilon}^{(\alpha)}\exp\left\\{-i\left(\mathbf{k}_{\gamma}\mathbf{\rho}_{3}\right)\right\\}\widehat{\mathbf{\pi}}_{3}^{\ast}\right\\}$
$\displaystyle=\frac{1}{2}\frac{e\hbar}{m_{N}c}\left\\{\left(\mathbf{\varepsilon}^{(\alpha)}\widehat{\mathbf{\pi}}_{1}^{\ast}\right)\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\right.$
$\displaystyle+\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)}\widehat{\mathbf{\pi}}_{1}^{\ast}\right)$
$\displaystyle+\left.\left(\mathbf{\varepsilon}^{(\alpha)}\widehat{\mathbf{\pi}}_{3}^{\ast}\right)\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}+\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)}\widehat{\mathbf{\pi}}_{3}^{\ast}\right)\right\\}$

and then

$\displaystyle\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)$
$\displaystyle=\frac{1}{2}\frac{e\hbar}{m_{N}c}\left\\{\left(\mathbf{\varepsilon}^{(\alpha)},\frac{1}{\sqrt{2}}\mathbf{\pi}_{\mathbf{r}}^{\ast}\mathbf{-}\frac{1}{\sqrt{6}}\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\right.$
$\displaystyle+\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\frac{1}{\sqrt{2}}\mathbf{\pi}_{\mathbf{r}}^{\ast}\mathbf{-}\frac{1}{\sqrt{6}}\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)$
$\displaystyle+\left.\sqrt{\frac{2}{3}}\left(\mathbf{\varepsilon}^{(\alpha)}\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}+\sqrt{\frac{2}{3}}\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)}\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right\\}$

Collecting similar terms we obtain

$\displaystyle\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)=\frac{1}{2}\frac{e\hbar}{m_{N}c}\left\\{\left(\mathbf{\varepsilon}^{(\alpha)},\frac{1}{2}\mathbf{k}_{\gamma}\mathbf{+}\frac{1}{6}\mathbf{k}_{\gamma}\right)\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)+i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\right.$
$\displaystyle+2\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\frac{1}{\sqrt{2}}\mathbf{\pi}_{\mathbf{r}}^{\ast}\mathbf{-}\frac{1}{\sqrt{6}}\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)$
$\displaystyle+\sqrt{\frac{2}{3}}\left(\mathbf{\varepsilon}^{(\alpha)}i\left[-i\sqrt{\frac{2}{3}}\mathbf{k}_{\gamma}\right]\right)\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}$
$\displaystyle+\left.2\sqrt{\frac{2}{3}}\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)}\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right\\}$

or

$\displaystyle\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)=\frac{1}{2}\frac{e\hbar}{m_{N}c}\left\\{\frac{2}{3}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{k}_{\gamma}\right)\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\right.$
$\displaystyle+2\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\frac{1}{\sqrt{2}}\mathbf{\pi}_{\mathbf{r}}^{\ast}\mathbf{-}\frac{1}{\sqrt{6}}\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)$
$\displaystyle+\frac{2}{3}\left(\mathbf{\varepsilon}^{(\alpha)}\mathbf{k}_{\gamma}\right)\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}$
$\displaystyle+\left.2\sqrt{\frac{2}{3}}\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)}\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right\\}$

or

$\displaystyle\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)=\frac{1}{2}\frac{e\hbar}{m_{N}c}\left\\{\frac{2}{3}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{k}_{\gamma}\right)\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\right.$
$\displaystyle+\frac{2}{3}\left(\mathbf{\varepsilon}^{(\alpha)}\mathbf{k}_{\gamma}\right)\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}$
$\displaystyle+2\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\frac{1}{\sqrt{2}}\mathbf{\pi}_{\mathbf{r}}^{\ast}\mathbf{-}\frac{1}{\sqrt{6}}\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)$
$\displaystyle+\left.2\sqrt{\frac{2}{3}}\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right\\}$

Final form of the operator

$\displaystyle\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)=\frac{1}{2}\frac{e\hbar}{m_{N}c}\left\\{\frac{2}{3}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{k}_{\gamma}\right)\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\right.$
(63)
$\displaystyle+\frac{2}{\sqrt{2}}\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\mathbf{+}\right\\}\exp\left\\{\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{r}}^{\ast}\right)$
$\displaystyle+\frac{2}{3}\left(\mathbf{\varepsilon}^{(\alpha)}\mathbf{k}_{\gamma}\right)\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}$
$\displaystyle\mathbf{-}\sqrt{\frac{2}{3}}\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\exp\left\\{\mathbf{+}i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)$
$\displaystyle+\left.2\sqrt{\frac{2}{3}}\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right\\}$

or by taking into account that
$\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{k}_{\gamma}\right)=0$, we
obtain

$\displaystyle\widehat{H}_{e}\left(\mathbf{k}_{\gamma},\mathbf{\varepsilon}^{(\alpha)}\right)=\frac{1}{2}\frac{e\hbar}{m_{N}c}\left\\{{}\right.$
(64)
$\displaystyle+\frac{2}{\sqrt{2}}\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{r}}^{\ast}\right)\exp\left\\{i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}$
$\displaystyle\mathbf{-}\sqrt{\frac{2}{3}}\exp\left\\{-i\frac{1}{\sqrt{2}}\left(\mathbf{k}_{\gamma}\mathbf{r}\right)\right\\}\exp\left\\{i\frac{1}{\sqrt{6}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)$
$\displaystyle+\left.2\sqrt{\frac{2}{3}}\exp\left\\{-i\sqrt{\frac{2}{3}}\left(\mathbf{k}_{\gamma}\mathbf{q}\right)\right\\}\left(\mathbf{\varepsilon}^{(\alpha)},\mathbf{\pi}_{\mathbf{q}}^{\ast}\right)\right\\}.$

##  Appendix B Calculations of integrals

###  B.1 A general case

In this Appendix we calculate integrals (21):

$\begin{array}[]{lllll}\vspace{0.5mm}\mathbf{I}_{1}=\biggl{\langle}\>\Phi_{f}(\mathbf{r})\;\biggl{|}\,e^{-i\,\mathbf{k}_{\gamma}\mathbf{r}}\;\mathbf{\displaystyle\frac{d}{dr}}\biggr{|}\>\Phi_{i}(\mathbf{r})\>\biggr{\rangle}_{\mathbf{r}},\\\
\vspace{0.5mm}I_{2}=\Bigl{\langle}\Phi_{f}(\mathbf{r})\;\Bigl{|}\,e^{i\,c_{p}\,\mathbf{k_{\gamma}}\mathbf{r}}\,\Bigr{|}\,\Phi_{i}(\mathbf{r})\>\Bigr{\rangle}_{\mathbf{r}},\\\
\vspace{0.5mm}I_{3}=\Bigl{\langle}\Phi_{f}(\mathbf{r})\;\Bigl{|}\,e^{-i\,c_{A}\,\mathbf{k_{\gamma}}\mathbf{r}}\,\Bigr{|}\,\Phi_{i}(\mathbf{r})\>\Bigr{\rangle}_{\mathbf{r}},\\\
I_{4}=\Bigl{\langle}\Phi_{f}(\mathbf{r})\;\Bigl{|}\,e^{-i\,c_{A}\,\mathbf{k_{\gamma}}\mathbf{r}}\,V(\mathbf{r})\,\Bigr{|}\,\Phi_{i}(\mathbf{r})\>\Bigr{\rangle}_{\mathbf{r}}.\end{array}$
(65)

Here, to two integrals in Eqs. (21) we have added two new types of integrals
else, which are used in calculations in other problems of bremsstrahlung
emission. $V(\mathbf{r})$ is arbitrary potential function.

We apply multipole expansion of wave function of photons, following to
formalism in Sect. D in Ref. Maydanyuk.2012.PRC  [see Eqs. (29)–(31) and
(24)–(28) in that paper]. Here, we obtain the following formulas for matrix
elements:

$\begin{array}[]{ll}\vspace{1mm}\Bigl{<}k_{f}\Bigl{|}\,e^{-i\mathbf{k_{\gamma}r}}\,\Bigr{|}\,k_{i}\Bigr{>}_{\mathbf{r}}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\Bigl{[}\mu\,\tilde{p}_{l_{\gamma}\mu}^{M}-i\,\tilde{p}_{l_{\gamma}\mu}^{E}\Bigr{]},\\\
\biggl{<}k_{f}\biggl{|}\,e^{-i\mathbf{k_{\gamma}r}}\displaystyle\frac{\partial}{\partial\mathbf{r}}\,\biggr{|}\,k_{i}\biggr{>}_{\mathbf{r}}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\mbox{\boldmath$\xi$}_{\mu}\,\mu\,\times\Bigl{[}p_{l_{\gamma}\mu}^{M}-i\mu\>p_{l_{\gamma}\mu}^{E}\Bigr{]}.\end{array}$
(66)

On the basis of these formulas, we write solutions for integrals (for
simplicity, we study case of $l_{i}=0$).

According to the second formula in Eqs. (66), the first integral is

$\begin{array}[]{ll}\mathbf{I}_{1}=\biggl{<}\Phi_{f}\biggl{|}\,e^{-i\mathbf{k_{\gamma}r}}\displaystyle\frac{\partial}{\partial\mathbf{r}}\,\biggr{|}\,\Phi_{i}\biggr{>}_{\mathbf{r}}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\mbox{\boldmath$\xi$}_{\mu}\,\mu\,\times\Bigl{[}p_{l_{\gamma}\mu}^{M}-i\mu\>p_{l_{\gamma}\mu}^{E}\Bigr{]},\end{array}$
(67)

where [see Eqs. (38) at $l_{i}=0$ in Ref. Maydanyuk.2012.PRC ]

$\begin{array}[]{lcl}\vspace{3mm}p_{l_{\gamma}\mu}^{M}&=&-I_{M}(0,l_{f},l_{\gamma},1,\mu)\cdot
J_{1}(0,l_{f},l_{\gamma}),\\\
\vspace{1mm}p_{l_{\gamma}\mu}^{E}&=&\sqrt{\displaystyle\frac{l_{\gamma}+1}{2l_{\gamma}+1}}\cdot
I_{E}(0,l_{f},l_{\gamma},1,l_{\gamma}-1,\mu)\cdot
J_{1}(0,l_{f},l_{\gamma}-1)\;-\\\
&-&\sqrt{\displaystyle\frac{l_{\gamma}}{2l_{\gamma}+1}}\cdot
I_{E}(0,l_{f},l_{\gamma},1,l_{\gamma}+1,\mu)\cdot
J_{1}(0,l_{f},l_{\gamma}+1)\end{array}$ (68)

and [see Eqs. (39) in Ref. Maydanyuk.2012.PRC ]

$\begin{array}[]{ccl}J_{1}(l_{i},l_{f},n)&=&\displaystyle\int\limits^{+\infty}_{0}\displaystyle\frac{dR_{i}(r,l_{i})}{dr}\>R^{*}_{f}(l_{f},r)\,j_{n}(k_{\gamma}r)\;r^{2}dr.\end{array}$
(69)

For the other integrals from Eqs. (65) one can get similar solutions [those
are directly derived from the first expansion in Eqs. (66), where the other
corresponding radial integrals should be used]:

$\begin{array}[]{lllll}\vspace{0.5mm}I_{2}=\Bigl{\langle}\Phi_{f}\;\Bigl{|}\,e^{i\,c_{p}\,\mathbf{k_{\gamma}}\mathbf{r}}\,\Bigr{|}\,\Phi_{i}\>\Bigr{\rangle}_{\mathbf{r}}=\par\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\Bigl{[}\mu\,\tilde{p}_{l_{\gamma}\mu}^{M}(-c_{p})-i\,\tilde{p}_{l_{\gamma}\mu}^{E}(-c_{p})\Bigr{]},\\\
\vspace{0.5mm}I_{3}=\Bigl{\langle}\Phi_{f}\;\Bigl{|}\,e^{-i\,c_{A}\,\mathbf{k_{\gamma}}\mathbf{r}}\,\Bigr{|}\,\Phi_{i}\>\Bigr{\rangle}_{\mathbf{r}}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\Bigl{[}\mu\,\tilde{p}_{l_{\gamma}\mu}^{M}(c_{A})-i\,\tilde{p}_{l_{\gamma}\mu}^{E}(c_{A})\Bigr{]},\\\
I_{4}=\Bigl{\langle}\Phi_{f}\;\Bigl{|}\,e^{-i\,c_{A}\,\mathbf{k_{\gamma}}\mathbf{r}}\,V(\mathbf{r})\,\Bigr{|}\,\Phi_{i}\>\Bigr{\rangle}_{\mathbf{r}}=\par\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\Bigl{[}\mu\,\breve{p}_{l_{\gamma}\mu}^{M}(c_{A})-i\,\breve{p}_{l_{\gamma}\mu}^{E}(c_{A})\Bigr{]},\end{array}$
(70)

where [see solutions (40) at $l_{i}=0$ in Ref. Maydanyuk.2012.PRC  for
$\tilde{p}_{l_{\gamma}\mu}^{M}$, $\tilde{p}_{l_{\gamma}\mu}^{E}$, Eqs. (F14)
and (F26) in Ref. Maydanyuk_Zhang_Zou.2019.PRC.microscopy  for all matrix
elements]

$\begin{array}[]{llll}\vspace{3mm}\tilde{p}_{l_{\gamma}\mu}^{M}(c)=\tilde{I}\,(0,l_{f},l_{\gamma},l_{\gamma},\mu)\cdot\tilde{J}\,(c,0,l_{f},l_{\gamma}),\\\
\vspace{3mm}\tilde{p}_{l_{\gamma}\mu}^{E}(c)=\sqrt{\displaystyle\frac{l_{\gamma}+1}{2l_{\gamma}+1}}\tilde{I}\,(0,l_{f},l_{\gamma},l_{\gamma}-1,\mu)\cdot\tilde{J}\,(c,0,l_{f},l_{\gamma}-1)\;-\sqrt{\displaystyle\frac{l_{\gamma}}{2l_{\gamma}+1}}\tilde{I}\,(0,l_{f},l_{\gamma},l_{\gamma}+1,\mu)\cdot\tilde{J}\,(c,0,l_{f},l_{\gamma}+1),\\\
\vspace{3mm}\breve{p}_{l_{\gamma}\mu}^{M}(c_{A})=\tilde{I}\,(0,l_{f},l_{\gamma},l_{\gamma},\mu)\cdot\breve{J}\,(c_{A},0,l_{f},l_{\gamma}),\\\
\breve{p}_{l_{\gamma}\mu}^{E}(c_{A})=\sqrt{\displaystyle\frac{l_{\gamma}+1}{2l_{\gamma}+1}}\tilde{I}\,(0,l_{f},l_{\gamma},l_{\gamma}-1,\mu)\cdot\breve{J}\,(c_{A},0,l_{f},l_{\gamma}-1)\;-\sqrt{\displaystyle\frac{l_{\gamma}}{2l_{\gamma}+1}}\tilde{I}\,(0,l_{f},l_{\gamma},l_{\gamma}+1,\mu)\cdot\breve{J}\,(c_{A},0,l_{f},l_{\gamma}+1)\end{array}$
(71)

and [see solutions (41) in Ref. Maydanyuk.2012.PRC  for $\tilde{J}$ and
corresponding angular integral, Eqs. (F13) and (F21) in Ref.
Maydanyuk_Zhang_Zou.2019.PRC.microscopy  for all matrix elements]

$\begin{array}[]{ccl}\tilde{J}\,(c,l_{i},l_{f},n)&=&\displaystyle\int\limits^{+\infty}_{0}R_{i}(r)\,R^{*}_{f}(l,r)\,j_{n}(c\,k_{\gamma}r)\;r^{2}dr,\\\
\breve{J}\,(c_{A},l_{i},l_{f},n)&=&\displaystyle\int\limits^{+\infty}_{0}R_{i}(r)\,R^{*}_{l,f}(r)\,V(\mathbf{r})\,j_{n}(c_{A}\,kr)\;r^{2}dr.\end{array}$
(72)

###  B.2 Linear and circular polarizations of the photon emitted

Rewrite vectors of _linear_ polarization
$\mbox{\boldmath$\varepsilon$}^{(\alpha)}$ through _vectors of circular
polarization_ $\mathbf{\xi}_{\mu}$ with opposite directions of rotation (see
Ref. Eisenberg.1973 , (2.39), p. 42):

$\begin{array}[]{ccc}\mathbf{\xi}_{-1}=\displaystyle\frac{1}{\sqrt{2}}\,\bigl{(}\mbox{\boldmath$\varepsilon$}^{(1)}-i\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{)},&\mathbf{\xi}_{+1}=-\displaystyle\frac{1}{\sqrt{2}}\,\bigl{(}\mbox{\boldmath$\varepsilon$}^{(1)}+i\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{)},&\mathbf{\xi}_{0}=\mbox{\boldmath$\varepsilon$}^{(3)}.\end{array}$
(73)

where

$\begin{array}[]{ccc}h_{\pm}=\mp\displaystyle\frac{1\pm
i}{\sqrt{2}},&h_{-1}+h_{+1}=-i\sqrt{2},&\sum\limits_{\alpha=1,2}\mbox{\boldmath$\varepsilon$}^{(\alpha),*}=h_{-1}\mathbf{\xi}_{-1}^{*}+h_{+1}\mathbf{\xi}_{+1}^{*}.\end{array}$
(74)

We have (in Coulomb gauge at $\mbox{\boldmath$\varepsilon$}^{(3)}=0$)

$\begin{array}[]{cc}\mbox{\boldmath$\varepsilon$}^{(1)}=\displaystyle\frac{1}{\sqrt{2}}\,\bigl{(}\mbox{\boldmath$\xi$}_{-1}-\mbox{\boldmath$\xi$}_{+1}\bigr{)},&\mbox{\boldmath$\varepsilon$}^{(2)}=\displaystyle\frac{i}{\sqrt{2}}\,\bigl{(}\mbox{\boldmath$\xi$}_{-1}+\mbox{\boldmath$\xi$}_{+1}\bigr{)},\end{array}$
(75) $\begin{array}[]{ccc}\displaystyle\sum\limits_{\mu=\pm
1}\xi_{\mu}^{*}\cdot\xi_{\mu}=\displaystyle\frac{1}{2}\,\bigl{(}\mbox{\boldmath$\varepsilon$}^{(1)}-i\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{)}\,\bigl{(}\mbox{\boldmath$\varepsilon$}^{(1)}-i\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{)}^{*}+\displaystyle\frac{1}{2}\,\bigl{(}\mbox{\boldmath$\varepsilon$}^{(1)}+i\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{)}\,\bigl{(}\mbox{\boldmath$\varepsilon$}^{(1)}+i\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{)}^{*}=2.\end{array}$
(76)

Also we will find vectorial products of vectors $\mbox{\boldmath$\xi$}_{\pm
1}$. From Eqs. (73) we obtain

$\begin{array}[]{cc}\mbox{\boldmath$\xi$}_{-1}^{*}=-\mbox{\boldmath$\xi$}_{+1},&\mbox{\boldmath$\xi$}_{+1}^{*}=-\mbox{\boldmath$\xi$}_{-1}.\end{array}$
(77)

From here we obtain vector multiplications as

$\begin{array}[]{lcl}\Bigl{[}\mbox{\boldmath$\xi$}_{-1}\times\mbox{\boldmath$\xi$}_{+1}\Bigr{]}&=&-i\,\bigl{[}\mbox{\boldmath$\varepsilon$}^{(1)}\times\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{]}=-i\,\mbox{\boldmath$\varepsilon$}_{\rm
z},\end{array}$ (78)

$\begin{array}[]{lcllcl}\vspace{2mm}\Bigl{[}\mbox{\boldmath$\xi$}_{-1}^{*}\times\mbox{\boldmath$\xi$}_{+1}\Bigr{]}&=&-\,\Bigl{[}\mbox{\boldmath$\xi$}_{+1}\times\mbox{\boldmath$\xi$}_{+1}\Bigr{]}=0,&\hskip
28.45274pt\Bigl{[}\mbox{\boldmath$\xi$}_{-1}^{*}\times\mbox{\boldmath$\xi$}_{-1}\Bigr{]}&=&-\,\Bigl{[}\mbox{\boldmath$\xi$}_{+1}\times\mbox{\boldmath$\xi$}_{-1}\Bigr{]}=i\,\mbox{\boldmath$\varepsilon$}_{\rm
z},\\\
\Bigl{[}\mbox{\boldmath$\xi$}_{+1}^{*}\times\mbox{\boldmath$\xi$}_{-1}\Bigr{]}&=&-\,\Bigl{[}\mbox{\boldmath$\xi$}_{-1}\times\mbox{\boldmath$\xi$}_{-1}\Bigr{]}=0,&\hskip
28.45274pt\Bigl{[}\mbox{\boldmath$\xi$}_{+1}^{*}\times\mbox{\boldmath$\xi$}_{+1}\Bigr{]}&=&-\,\Bigl{[}\mbox{\boldmath$\xi$}_{-1}\times\mbox{\boldmath$\xi$}_{+1}\Bigr{]}=-i\,\mbox{\boldmath$\varepsilon$}_{\rm
z}.\end{array}$ (79)

###  B.3 Summation over vectors of polarizations

In this section we will calculate multiplications of integrals on vectors of
polarizations. Let’ consider the first integral $\mathbf{I}_{1}$ which has
form [see Eqs. (67)]

$\begin{array}[]{ll}\mathbf{I}_{1}=\biggl{<}\Phi_{f}\biggl{|}\,e^{-i\mathbf{k_{\gamma}r}}\displaystyle\frac{\partial}{\partial\mathbf{r}}\,\biggr{|}\,\Phi_{i}\biggr{>}_{\mathbf{r}}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\mbox{\boldmath$\xi$}_{\mu}\,\mu\,\times\Bigl{[}p_{l_{\gamma}\mu}^{M}-i\mu\>p_{l_{\gamma}\mu}^{E}\Bigr{]}.\end{array}$
(80)

We calculate

$\begin{array}[]{llll}\vspace{1.5mm}\mbox{\boldmath$\varepsilon$}^{(1)}\cdot\mathbf{I}_{1}&=&-\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\frac{1}{\sqrt{2}}\,\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}p_{l_{\gamma}\mu}^{M}-i\,\mu\,p_{l_{\gamma}\mu}^{E}\Bigr{]},\\\
\mbox{\boldmath$\varepsilon$}^{(2)}\cdot\mathbf{I}_{1}&=&-\,\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\frac{i}{\sqrt{2}}\,\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}\mu\,p_{l_{\gamma}\mu}^{M}-i\,p_{l_{\gamma}\mu}^{E}\Bigr{]},\end{array}$
(81)

and summation over vectors of polarization is

$\begin{array}[]{llllll}\displaystyle\sum\limits_{\alpha=1,2}\mbox{\boldmath$\varepsilon$}^{(\alpha)}\cdot\mathbf{I}_{1}&=&-\,\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{{\gamma}+1}}\;\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}\displaystyle\frac{1+i\,\mu}{\sqrt{2}}\,p_{l_{\gamma}\mu}^{M}+\displaystyle\frac{1-i\,\mu}{\sqrt{2}}\,p_{l_{\gamma}\mu}^{E}\Bigr{]}.\end{array}$
(82)

There is property

$\begin{array}[]{llllll}\vspace{1.0mm}\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}\displaystyle\frac{1-i\,\mu}{\sqrt{2}}\,p_{l_{\gamma}\mu}^{E}\Bigr{]}=\par\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}\displaystyle\frac{1+i\,\mu}{\sqrt{2}}\,p_{l_{\gamma},\,-\mu}^{E}\Bigr{]},\\\
\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}\displaystyle\frac{1+i\,\mu}{\sqrt{2}}\,p_{l_{\gamma}\mu}^{M}+\displaystyle\frac{1-i\,\mu}{\sqrt{2}}\,p_{l_{\gamma}\mu}^{E}\Bigr{]}=\par\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}\displaystyle\frac{1+i\,\mu}{\sqrt{2}}\,p_{l_{\gamma}\mu}^{M}+\displaystyle\frac{1+i\,\mu}{\sqrt{2}}\,p_{l_{\gamma},\,-\mu}^{E}\Bigr{]}=\par\displaystyle\sum\limits_{\mu=\pm
1}\,\displaystyle\frac{1+i\,\mu}{\sqrt{2}}\,\Bigl{[}p_{l_{\gamma}\mu}^{M}+p_{l_{\gamma},\,-\mu}^{E}\Bigr{]}.\end{array}$
(83)

Then one can write Eq. (82) as

$\begin{array}[]{llll}\displaystyle\sum\limits_{\alpha=1,2}\mbox{\boldmath$\varepsilon$}^{(\alpha)}\cdot\mathbf{I}_{1}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\mu\,h_{\mu}\,\bigl{(}p_{l_{\gamma},\mu}^{M}+p_{l_{\gamma},-\mu}^{E}\bigr{)}.\end{array}$
(84)

We calculate properties

$\begin{array}[]{llll}(\mbox{\boldmath$\varepsilon$}_{\rm
x}+\mbox{\boldmath$\varepsilon$}_{\rm
z})\,\displaystyle\sum\limits_{\alpha=1,2}\Bigl{[}\mathbf{I}_{1}\times\mbox{\boldmath$\varepsilon$}^{(\alpha)}\Bigr{]}=\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\sum\limits_{\mu=\pm
1}\mu\,h_{\mu}\,\bigl{(}p_{l_{\gamma},\mu}^{M}-p_{l_{\gamma},-\mu}^{E}\bigr{)}.\end{array}$
(85)

###  B.4 Case of $l_{i}=0$, $l_{f}=1$, $l_{\gamma}=1$

In a case of $l_{i}=0$, $l_{f}=1$, $l_{\gamma}=1$ integrals (68), (71) are
simplified to

$\begin{array}[]{llllll}\vspace{0.5mm}\mathbf{I}_{1}=-i\,\sqrt{\displaystyle\frac{3\pi}{2}}\>\displaystyle\sum\limits_{\mu=\pm
1}\mbox{\boldmath$\xi$}_{\mu}\,\mu\,\times\Bigl{[}p_{l_{\gamma}\mu}^{M}-i\mu\>p_{l_{\gamma}\mu}^{E}\Bigr{]},&I_{3}=-i\,\sqrt{\displaystyle\frac{3\pi}{2}}\>\displaystyle\sum\limits_{\mu=\pm
1}\Bigl{[}\mu\,\tilde{p}_{l_{\gamma}\mu}^{M}(c_{A})-i\,\tilde{p}_{l_{\gamma}\mu}^{E}(c_{A})\Bigr{]},\\\
I_{2}=-i\,\sqrt{\displaystyle\frac{3\pi}{2}}\>\displaystyle\sum\limits_{\mu=\pm
1}\Bigl{[}\mu\,\tilde{p}_{l_{\gamma}\mu}^{M}(-c_{p})-i\,\tilde{p}_{l_{\gamma}\mu}^{E}(-c_{p})\Bigr{]},&I_{4}=-i\,\sqrt{\displaystyle\frac{3\pi}{2}}\>\displaystyle\sum\limits_{\mu=\pm
1}\Bigl{[}\mu\,\breve{p}_{l_{\gamma}\mu}^{M}(c_{A})-i\,\breve{p}_{l_{\gamma}\mu}^{E}(c_{A})\Bigr{]},\end{array}$
(86)

where [see from Eqs. (68), (71)]

$\begin{array}[]{lll}\vspace{3mm}p_{l_{\gamma}\mu}^{M}=-I_{M}(0,1,1,1,\mu)\cdot
J_{1}(0,1,1),\\\
\vspace{3mm}p_{l_{\gamma}\mu}^{E}=\sqrt{\displaystyle\frac{2}{3}}\,I_{E}(0,1,1,1,0,\mu)\cdot
J_{1}(0,1,0)-\sqrt{\displaystyle\frac{1}{3}}\,I_{E}(0,1,1,1,2,\mu)\cdot
J_{1}(0,1,2),\\\
\vspace{3mm}\tilde{p}_{l_{\gamma}\mu}^{M}(c)=\tilde{I}\,(0,1,1,1,\mu)\cdot\tilde{J}\,(c,0,1,1),\\\
\vspace{3mm}\tilde{p}_{l_{\gamma}\mu}^{E}(c)=\sqrt{\displaystyle\frac{2}{3}}\tilde{I}\,(0,1,1,0,\mu)\cdot\tilde{J}\,(c,0,1,0)-\sqrt{\displaystyle\frac{1}{3}}\tilde{I}\,(0,1,1,2,\mu)\cdot\tilde{J}\,(c,0,1,2),\\\
\vspace{3mm}\breve{p}_{l_{\gamma}\mu}^{M}(c_{A})=\tilde{I}\,(0,1,1,1,\mu)\cdot\breve{J}\,(c_{A},0,1,1),\\\
\breve{p}_{l_{\gamma}\mu}^{E}(c_{A})=\sqrt{\displaystyle\frac{2}{3}}\tilde{I}\,(0,1,1,0,\mu)\cdot\breve{J}\,(c_{A},0,1,0)-\sqrt{\displaystyle\frac{1}{3}}\tilde{I}\,(0,1,1,2,\mu)\cdot\breve{J}\,(c_{A},0,1,2).\end{array}$
(87)

The angular integrals are calculated in Appendix B in Ref. Maydanyuk.2012.PRC
[see Eqs. (B1)–(B10) in that paper]. Results of calculation of angular
integrals are

$\begin{array}[]{llllll}\vspace{0.8mm}I_{E}\,(0,1,1,1,0,\mu)=\sqrt{\displaystyle\frac{1}{24\pi}},&I_{M}\,(0,1,1,1,\mu)=0,&I_{E}\,(0,1,1,1,2,\mu)=\displaystyle\frac{47}{240}\sqrt{\displaystyle\frac{3}{2\pi}},\\\
\tilde{I}\,(0,1,1,0,\mu)=0,&\tilde{I}\,(0,1,1,1,\mu)=\displaystyle\frac{\mu}{2\sqrt{2\pi}},&\tilde{I}\,(0,1,1,2,\mu)=0,\end{array}$
(88)

and matrix elements (87) are simplified to

$\begin{array}[]{lllllllll}\vspace{2mm}p_{l_{\gamma}\mu}^{M}=0,&p_{l_{\gamma}\mu}^{E}=\displaystyle\frac{1}{6}\sqrt{\displaystyle\frac{1}{\pi}}\cdot
J_{1}(0,1,0)-\displaystyle\frac{47}{240}\sqrt{\displaystyle\frac{1}{2\pi}}\cdot
J_{1}(0,1,2),\\\
\vspace{2mm}\tilde{p}_{1\mu}^{M}(c)=\displaystyle\frac{\mu}{2\sqrt{2\pi}}\cdot\tilde{J}\,(c,0,1,1),&\tilde{p}_{1\mu}^{E}(c)=0,\\\
\breve{p}_{1\mu}^{M}(c_{A})=\displaystyle\frac{\mu}{2\sqrt{2\pi}}\cdot\breve{J}\,(c_{A},0,1,1),&\breve{p}_{1\mu}^{E}(c_{A})=0.\end{array}$
(89)

Now we calculate integrals in Eqs. (89). For $p_{l_{\gamma}=1,\mu}^{M}$ and
$p_{l_{\gamma}=1,\mu}^{E}$ we obtain:

$\begin{array}[]{llllll}\mathbf{I}_{1}=-i\,\sqrt{\displaystyle\frac{3\pi}{2}}\>\displaystyle\sum\limits_{\mu=\pm
1}\mbox{\boldmath$\xi$}_{\mu}\,\mu\,\times\Bigl{[}-i\mu\>p_{l_{\gamma}=1,\,\mu}^{E}\Bigr{]}=\par\par-\,\sqrt{\displaystyle\frac{3\pi}{2}}\>p_{l_{\gamma}=1,\,\mu}^{E}\,(\mbox{\boldmath$\xi$}_{-1}+\mbox{\boldmath$\xi$}_{+1}).\end{array}$
(90)

Taking into account summation of vectors of polarizations (73)

$\begin{array}[]{ccc}\mathbf{\xi}_{-1}=\displaystyle\frac{1}{\sqrt{2}}\,\bigl{(}\mbox{\boldmath$\varepsilon$}^{(1)}-i\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{)},&\mathbf{\xi}_{+1}=-\displaystyle\frac{1}{\sqrt{2}}\,\bigl{(}\mbox{\boldmath$\varepsilon$}^{(1)}+i\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{)},&\mathbf{\xi}_{0}=\mbox{\boldmath$\varepsilon$}^{(3)},\end{array}$

we simplify Eq. (90) as

$\begin{array}[]{llllll}\mathbf{I}_{1}=\par-\,\sqrt{\displaystyle\frac{3\pi}{2}}\>p_{l_{\gamma}=1,\,\mu}^{E}\cdot\bigl{(}-i\,\sqrt{2}\,\mbox{\boldmath$\varepsilon$}^{(2)}\bigr{)}=\par
i\,\sqrt{3\pi}\>p_{l_{\gamma}=1,\,\mu}^{E}\cdot\mbox{\boldmath$\varepsilon$}^{(2)}.\end{array}$
(91)

From Eqs. (81) we calculate products of integrals on vectors of polarizations:

$\begin{array}[]{llll}\vspace{0.5mm}\mbox{\boldmath$\varepsilon$}^{(1)}\cdot\mathbf{I}_{1}&=&\Bigl{\\{}-\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\frac{1}{\sqrt{2}}\,\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}p_{l_{\gamma}\mu}^{M}-i\,\mu\,p_{l_{\gamma}\mu}^{E}\Bigr{]}\Bigr{\\}}_{l_{\gamma}=1}\;=\\\
\vspace{0.5mm}&=&i\,\sqrt{\displaystyle\frac{\pi}{2}}\>\sqrt{\displaystyle\frac{3}{2}}\,\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}p_{l_{\gamma}=1,\mu}^{M}-i\,\mu\,p_{l_{\gamma}=1,\mu}^{E}\Bigr{]}=\par\par
i\,\displaystyle\frac{\sqrt{3\pi}}{2}\,p_{l_{\gamma}=1,\mu}^{E}\,\displaystyle\sum\limits_{\mu=\pm
1}\,(-i\,\mu\,)=0,\end{array}$ (92)
$\begin{array}[]{llll}\vspace{0.5mm}\mbox{\boldmath$\varepsilon$}^{(2)}\cdot\mathbf{I}_{1}&=&\Bigl{\\{}-\,\sqrt{\displaystyle\frac{\pi}{2}}\>\displaystyle\sum\limits_{l_{\gamma}=1}\,(-i)^{l_{\gamma}}\,\sqrt{2l_{\gamma}+1}\;\displaystyle\frac{i}{\sqrt{2}}\,\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}\mu\,p_{l_{\gamma}\mu}^{M}-i\,p_{l_{\gamma}\mu}^{E}\Bigr{]}\Bigr{\\}}_{l_{\gamma}=1}\;=\\\
\vspace{0.5mm}&=&-\,\sqrt{\displaystyle\frac{\pi}{2}}\,\sqrt{\displaystyle\frac{3}{2}}\,\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}\mu\,p_{l_{\gamma}=1,\mu}^{M}-i\,p_{l_{\gamma}=1,\mu}^{E}\Bigr{]}=\par-\,\displaystyle\frac{\sqrt{3\pi}}{2}\,\displaystyle\sum\limits_{\mu=\pm
1}\,\Bigl{[}-i\,p_{l_{\gamma}=1,\mu}^{E}\Bigr{]}\;=\par\par
i\,\sqrt{3\pi}\,p_{l_{\gamma}=1,\mu}^{E}.\end{array}$ (93)

Therefore, we write down the final solutions:

$\begin{array}[]{llllll}\mbox{\boldmath$\varepsilon$}^{(1)}\cdot\mathbf{I}_{1}=0,&\mbox{\boldmath$\varepsilon$}^{(2)}\cdot\mathbf{I}_{1}=\displaystyle\sum\limits_{\alpha=1,2}\mbox{\boldmath$\varepsilon$}^{(\alpha)}\cdot\mathbf{I}_{1}=i\,\sqrt{3\pi}\,p_{l_{\gamma}=1,\mu}^{E}.\end{array}$
(94)

We obtain property

$\begin{array}[]{llllll}\mathbf{I}_{1}=i\,\sqrt{3\pi}\>p_{l_{\gamma}=1,\,\mu}^{E}\cdot\mbox{\boldmath$\varepsilon$}^{(2)}=\par\Bigl{[}\displaystyle\sum\limits_{\alpha=1,2}\mbox{\boldmath$\varepsilon$}^{(\alpha)}\cdot\mathbf{I}_{1}\Bigr{]}\cdot\mbox{\boldmath$\varepsilon$}^{(2)}.\end{array}$
(95)

Taking solution (89) into account, rewritten via raadial integrals:

$\begin{array}[]{lllll}p_{l_{\gamma}\mu}^{E}=\displaystyle\frac{1}{6}\sqrt{\displaystyle\frac{1}{\pi}}\cdot
J_{1}(0,1,0)-\displaystyle\frac{47}{240}\sqrt{\displaystyle\frac{1}{2\pi}}\cdot
J_{1}(0,1,2),\end{array}$

we obtain

$\begin{array}[]{llllll}\vspace{0.5mm}\mathbf{I}_{1}=\mbox{\boldmath$\varepsilon$}^{(2)}\cdot
i\,\displaystyle\frac{\sqrt{3}}{6}\,\Bigl{\\{}J_{1}(0,1,0)-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}(0,1,2)\Bigr{\\}},\\\
\mbox{\boldmath$\varepsilon$}^{(1)}\cdot\mathbf{I}_{1}=0,\\\
\mbox{\boldmath$\varepsilon$}^{(2)}\cdot\mathbf{I}_{1}=\displaystyle\sum\limits_{\alpha=1,2}\mbox{\boldmath$\varepsilon$}^{(\alpha)}\cdot\mathbf{I}_{1}=i\,\displaystyle\frac{\sqrt{3}}{6}\,\Bigl{\\{}J_{1}(0,1,0)-\displaystyle\frac{47}{40}\sqrt{\displaystyle\frac{1}{2}}\cdot
J_{1}(0,1,2)\Bigr{\\}}.\end{array}$ (96)

##  Appendix C Form factor of deuteron in cluster approach

###  C.1 Normalization of the deuteron wave function

The deuteron wave function

$\vspace{0.5mm}\phi(\boldsymbol{r})=A\dfrac{e^{-\kappa
r}}{r}Y_{lm}(\theta,\varphi).$ (97)

where $\kappa=\sqrt{\dfrac{2m|E_{d}|}{\hbar^{2}}}$ and $\boldsymbol{r}$ \-
Jacobi vector of the relative position of the nucleons inside deuteron.

Normalization condition:

$\begin{array}[]{lllll}\vspace{0.5mm}\displaystyle\int|\phi(\boldsymbol{r})|^{2}dV=1,\;A^{2}\int\dfrac{e^{-2\kappa
r}}{r^{2}}Y_{l^{\prime}m^{\prime}}^{*}(\theta,\varphi)Y_{lm}(\theta,\varphi)dV=1,\end{array}$
(98)

$\begin{array}[]{lllll}\vspace{0.5mm}\displaystyle
A^{2}\int_{0}^{\infty}\dfrac{e^{-2\kappa r}}{r^{2}}r^{2}dr\int
Y_{l^{\prime}m^{\prime}}^{*}(\theta,\varphi)Y_{lm}(\theta,\varphi)d\Omega=1,\;\delta_{l,l^{\prime}}\delta_{m,m^{\prime}}A^{2}\int_{0}^{\infty}e^{-2\kappa
r}dr=1,\end{array}$ (99)

$\begin{array}[]{lllll}\vspace{0.5mm}-\dfrac{A^{2}}{2\kappa}e^{-2\kappa
r}\bigg{|}_{r=0}^{r\rightarrow\infty}=1,\;\frac{A^{2}}{2\kappa}=1.\end{array}$
(100)

Thus we have the normalization multiplier:

$\vspace{5mm}A=\sqrt{2\kappa},$ (101)

and the final form of the deuteron wave function:

$\vspace{0.5mm}\phi(\boldsymbol{r})=\sqrt{2\kappa}\dfrac{e^{-\kappa
r}}{r}Y_{l,m}(\theta,\varphi)=\sqrt{\dfrac{\kappa}{2\pi}}\dfrac{e^{-\kappa
r}}{r},$ (102)

where we have chosen $l=0$.

###  C.2 Calculation of the form factors

We have to calculate the following form factors :

$\begin{array}[]{lll}\vspace{0.5mm}F_{1}(\boldsymbol{k}_{\gamma})&=&\biggl{\langle}\phi(\boldsymbol{r})\biggl{|}\exp\\{-\dfrac{i}{\sqrt{2}}(\boldsymbol{k}_{\gamma}\boldsymbol{r})\biggr{\\}}\biggr{|}\phi(\boldsymbol{r})\biggr{\rangle},\\\
F_{2}(\boldsymbol{k}_{\gamma})&=&\biggl{\langle}\phi(\boldsymbol{r})\biggl{|}\exp\\{-\dfrac{i}{\sqrt{2}}(\boldsymbol{k}_{\gamma}\boldsymbol{r})\biggr{\\}}(\varepsilon^{(\alpha)},\pi_{\boldsymbol{r}}^{*})\biggr{|}\phi(\boldsymbol{r})\biggr{\rangle}.\end{array}$
(103)

We calculate the first form factor as

$\begin{array}[]{lll}\vspace{0.5mm}F_{1}(\boldsymbol{k}_{\gamma})=\displaystyle\int\phi^{*}(\boldsymbol{r})\exp\\{-\dfrac{i}{\sqrt{2}}(\boldsymbol{k}_{\gamma}\boldsymbol{r})\biggr{\\}}\phi(\boldsymbol{r})d\boldsymbol{r}=\dfrac{\kappa}{2\pi}\int\dfrac{e^{-2\kappa
r}}{r^{2}}e^{-i/\sqrt{2}\,\boldsymbol{k}_{\gamma}\boldsymbol{r}}r^{2}drd\Omega.\end{array}$
(104)

Now we use the following expansion

$\begin{array}[]{llll}\vspace{0.5mm}&\displaystyle
e^{i\boldsymbol{k}\boldsymbol{r}}=\sum_{l=0}^{\infty}i^{l}j_{l}(kr)\sum_{m=-l}^{m=l}Y_{lm}^{*}(\theta_{k},\varphi_{k})Y_{lm}(\theta_{r},\varphi_{r}),\\\
&\displaystyle
e^{-i/\sqrt{2}\,\boldsymbol{k}_{\gamma}\boldsymbol{r}}=\sum_{l=0}^{\infty}(i^{l})^{*}j_{l}\biggl{(}\dfrac{k_{\gamma}r}{\sqrt{2}}\biggr{)}\sum_{m=-l}^{m=l}Y_{lm}^{*}(\theta_{k},\varphi_{k})Y_{lm}(\theta_{r},\varphi_{r}).\end{array}$
(105)

We will have for our integral

$\begin{array}[]{ll}\vspace{0.5mm}\displaystyle
F_{1}(\boldsymbol{k}_{\gamma})&=\dfrac{\kappa}{2\pi}\displaystyle\sum_{l^{\prime}=0}^{\infty}\bigl{(}i^{l^{\prime}}\bigr{)}^{*}\displaystyle\int_{0}^{\infty}j_{l^{\prime}}\bigl{(}k_{\gamma}r/\sqrt{2}\bigr{)}\dfrac{e^{-2\kappa
r}}{r^{2}}r^{2}dr\;\int
d\Omega\;\sum_{m^{\prime}=-l^{\prime}}^{m=l^{\prime}}Y_{l^{\prime}m^{\prime}}(\theta,\varphi)Y_{l^{\prime}m^{\prime}}^{*}(\theta,\varphi)=\\\
&=\dfrac{\kappa}{2\pi}\displaystyle\int_{0}^{\infty}j_{0}\bigl{(}k_{\gamma}r/\sqrt{2}\bigr{)}e^{-2\kappa
r}dr.\end{array}$ (106)

and for the Bessel spherical functions we have $j_{0}(x)=\sin(x)/x$:

$\begin{array}[]{ll}\vspace{0.5mm}F_{1}(\boldsymbol{k}_{\gamma})&=\dfrac{\sqrt{2}\kappa}{2\pi
k_{\gamma}}\displaystyle\int_{0}^{\infty}\dfrac{\sin\bigl(k_{\gamma}r/\sqrt{2}\bigr{missing})}{r}e^{-2\kappa
r}dr,\end{array}$ (107)

Let us focus on the integral itself

$\begin{array}[]{ll}\vspace{0.5mm}I&=\displaystyle\int_{0}^{\infty}\dfrac{\sin\bigl(k_{\gamma}r/\sqrt{2}\bigr{missing})}{r}e^{-2\kappa
r}dr=\displaystyle\int_{0}^{\infty}\dfrac{\sin at}{t}e^{-t}dt,\end{array}$
(108)

where $t=2\kappa r$, $a=k_{\gamma}/2\sqrt{2}\kappa$.

$\begin{array}[]{lll}\vspace{0.5mm}I(a)&=\displaystyle\int_{0}^{\infty}\dfrac{\sin
at}{t}e^{-t}dt,\;\dfrac{dI}{da}=\dfrac{d}{da}\biggl{[}\displaystyle\int_{0}^{\infty}\dfrac{\sin
at}{t}e^{-t}dt\biggr{]},\;\dfrac{dI}{da}=\displaystyle\int_{0}^{\infty}\cos
at\;e^{-t}dt=I_{1}.\end{array}$ (109)

For the integral $I_{1}$ we have

$\begin{array}[]{ll}\vspace{0.5mm}I_{1}=\displaystyle\int_{0}^{\infty}\cos
at\;e^{-t}dt=-\displaystyle\int_{0}^{\infty}\cos at\;d(e^{-t}),\\\ I_{1}=-\cos
at\;e^{-t}\bigg{|}_{t=0}^{t\rightarrow\infty}+\displaystyle\int_{0}^{\infty}e^{-t}d(\cos
at)=1-a\displaystyle\int_{0}^{\infty}\sin at\;e^{-t}dt,\\\
I_{1}=1+a\displaystyle\int_{0}^{\infty}\sin
at\;d(e^{-t}),\;I_{1}=1+a\biggl{[}\sin
at\;e^{-t}\bigg{|}_{t=0}^{t\rightarrow\infty}-\displaystyle\int_{0}^{\infty}e^{-t}d(\sin
at)\biggr{]},\\\ I_{1}=1-a^{2}\displaystyle\int_{0}^{\infty}\cos
at\;e^{-t}dt=1-a^{2}I_{1},\\\ I_{1}+a^{2}I_{1}=1.\end{array}$ (110)

Finally, we have the following equation for the integral $I_{1}$

$\vspace{0.5mm}I_{1}=\dfrac{1}{1+a^{2}},$ (111)

so for the $I$ we have the following equation

$\begin{array}[]{l}\vspace{3mm}\dfrac{dI}{da}=\dfrac{1}{1+a^{2}},\displaystyle\int
dI=\displaystyle\int\dfrac{da}{1+a^{2}},\\\ I=\arctan a+C.\end{array}$ (112)

Let us come back to the form factor and write down the following

$\vspace{0.5mm}F_{1}(\boldsymbol{k}_{\gamma})=\dfrac{2\sqrt{2}\kappa}{k_{\gamma}}\arctan\biggl(\dfrac{k_{\gamma}}{2\sqrt{2}\kappa}\biggr{missing})+C_{1},$
(113)

to find out $C$ we will use

$\begin{array}[]{l}\vspace{3mm}F_{1}(0)=1,\;\lim_{k_{\gamma}\rightarrow
0}\dfrac{\sqrt{2}\kappa}{2\pi
k_{\gamma}}\arctan\biggl(\dfrac{k_{\gamma}}{2\sqrt{2}\kappa}\biggr{missing})+C_{1}=1,\\\
\lim_{k_{\gamma}\rightarrow
0}\dfrac{2\sqrt{2}\kappa}{k_{\gamma}}\arctan\biggl(\dfrac{k_{\gamma}}{2\sqrt{2}\kappa}\biggr{missing})+C_{1}=1,\
1+C_{1}=1,\\\ C_{1}=0.\end{array}$ (114)

The final result

$\vspace{0.5mm}F_{1}(\boldsymbol{k}_{\gamma})=\dfrac{2\sqrt{2}\kappa}{k_{\gamma}}\arctan\biggl(\dfrac{k_{\gamma}}{2\sqrt{2}\kappa}\biggr{missing}).$
(115)

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