2021

These authors contributed equally to this work.

These authors contributed equally to this work.

[2]E.A. Kupriyanova These authors contributed equally to this work.

1]MEPhI, National Research Nuclear University, Kashirskoe Highway 31, Moscow,
115409, Russia

2]Nuclear Physics and Astrophysics Division, Lebedev Physical Institute,
Leninsky Prospect 53, Moscow, 119991, Russia

# On cancellation of non-adiabatic and off-shell effects in the antiproton
annihilation in deuteron

O.D. Dalkarov<EMAIL_ADDRESS>V.A. Karmanov<EMAIL_ADDRESS><EMAIL_ADDRESS>[  [

###### Abstract

As known, some approximate approaches to the hadron scattering from nuclei
work rather well far beyond the limits of their applicability. This was
explained by cancellation of the contributions (non-adiabatic and off-shell
effects) omitted in these approaches. Moreover, in some cases (in particular,
for the reaction $\bar{p}d\to e^{+}e^{-}n$) this cancellation allowed to
derive rather simple analytical formula for the reaction amplitude. Solving
the Faddeev equations, we confirm numerically this formula and, hence, the
cancellations.

###### keywords:

Antiproton annihilation, Proton form factor, Faddeev equations

##  1 Introduction

A few decades ago, when numerical solving the Faddeev equations was not yet
possible and calculation of the hadron scattering on nuclei based on
approximate analytical approaches, it was noticed that some of these
approximate approaches worked surprisingly well in wider domain than initially
expected. As an example, we mention the Glauber approach Glauber , which works
well, often beyond the formal limits of its applicability. An illustration is
given by good description of scattering of the low-energy antiprptons from
nuclei, down to 50 MeV dk  (see for review Ref. dk87  and references therein).
For applicability of this approach, two principal conditions should be
satisfied: (i) straightforwardness of the trajectory of the projectile
particle in nucleus (eikonal approximation); (ii) possibility to neglect the
motion of the intra-nuclear nucleons (adiabatic approximation). In addition,
though the nucleons in the intermediate states are virtual (off-mass-shell),
in these calculations, (iii) the elementary nucleon-nucleon amplitudes were
taken on-mass shell. In the case of the low-energy antiproton scattering, the
condition (i) can be ensured by the fact that the antiproton-nucleon
scattering is strongly forward directed even for relatively low energies.
However, for the first glance, the conditions (ii) and (iii) can be hardly
satisfied.

At that time, these observations triggered the researches aimed to explain the
reasons of these successes. The detail analytical study of the hadron-nucleus
scattering amplitudes was fulfilled KolKondr ; KolKsen ; DK81 ; DKK83 ; Faldt
; Wallace ; Gurvitz . The cancellation of different effects was discovered.
Thus, it was found that the non-adiabatic effects and off-shell effects in the
elementary scattering amplitudes considerably canceled each other. This
explains, why, for example, the Glauber approach Glauber , which just does not
take into account these effects, works so well. It also explains, why an
approximate approach loses its accuracy after attempt to ”improve” it by
incorporating only a part of omitted effects, say, the non-adiabatic effects
only, omitted in the eikonal propagators.

Currently, the problem of scattering of nucleons by light nuclei is
successfully being solved by the numerical treatment of the Faddeev-Yakubovsky
equations (see, e.g., the review RJ  for the equations and references therein
for their solutions). The input is the nucleon-nucleon potential (or the off-
shell amplitudes), the output is the many-channel scattering amplitude, which
includes both the elastic scattering, rearrangements and breakup, if any. This
great and well-deserved success achieved, partially, due to the ”brute force”
of computers, overshadowed the interest in the important question ”how does
this work?” Though all the effects, discussed above, are automatically taken
into account by the Faddeev-Yakubovsky equations, the question ”how does this
work?” , that is, ”what is the role and contribution of the non-adiabatic and
off-shell effects in forming the final many-channel scattering amplitude?” -
still can be asked if one wants to achieve the deep understanding of mechanism
of a nuclear reaction.

Figure 1: Multiple scattering graphs determining amplitude of the reaction
$\bar{p}d\to e^{+}e^{-}n$.

These researches had not only academic, but also crucial practical interest,
related to possibility to extract from the reaction $\bar{p}d\to e^{+}e^{-}n$
the proton form factor in the time-like domain of the momentum transfer
squared $q^{2}>0$, under the $\bar{p}p$ threshold $q^{2}<4m^{2}$. In the works
DK81 ; DKK83 , the scrutinized study of full set of the diagrams, shown in
Fig. 1, contributing to the amplitude $\bar{p}d\to e^{+}e^{-}n$, was carried
out. The amplitude corresponding to the first term of this set, Fig. 1a, has
the form

$M_{u}=\psi_{d}\Gamma_{u},$ (1)

where $\psi_{d}$ is the deuteron wave function, $\Gamma_{u}$ is the
annihilation amplitude $\bar{p}p\to e^{+}e^{-}$, Figure 2,

$\Gamma_{u}=\psi_{k_{u}}(r=0)\Gamma_{0},$ (2)

$\Gamma_{0}$ is the amplitude of annihilation $\bar{p}p\to e^{+}e^{-}$ not
including the initial state $\bar{p}p$ interaction (Fig. 2a), and
$\psi_{k}(r)$ is the continuous spectrum $\bar{p}p$ coordinate space wave
function, corresponding to the eigenvalue $E=k^{2}/m$ ($k$ is the nucleon
momentum in the $\bar{p}p$ c.m. frame, $m$ is the nucleon mass). It takes into
account the initial state $\bar{p}p$ interaction, Fig. 2b. The equation (2)
corresponds to the sum of two graphs shown in Fig. 2.

The subscript ”$u$” (we follow the notations of Refs. DK81 ; DKK83 ) denotes
the $\bar{p}p$ energy found from the energy-momentum conservation laws applied
to the reaction $\bar{p}d\to e^{+}e^{-}n$. It is calculated in the next
section, Eq. (7). This energy, relative to the $\bar{p}p$ mass $2m$, can be
both positive and negative. That is, if the impulse approximation diagram Fig.
1a dominates, this indeed would allow to access the proton form factor under
threshold.

Figure 2: The annihilation amplitude $\bar{p}p\to e^{+}e^{-}$.

In contrast to Fig. 2b, which, besides the annihilation, includes the initial
state $\bar{p}p$ interaction, the antiproton-nucleon interactions in the
reaction $\bar{p}d\to e^{+}e^{-}n$, before the $\bar{p}p$ annihilation, are
determined by the infinite set of the multiple scattering graphs shown in Fig.
1. These graphs are not reduced to a single $\bar{p}p$ interaction. That’s
why, generally speaking, the impulse approximation formula (1), proportional
to the proton form factor, is distorted. However, as it was proved in DK81 ;
DKK83 , the total effect of all the re-scatterings and cancellations mentioned
above and shown in Fig. 1, restores the amplitude in the form of Eq. (1), but
with the shifted and always positive energy $E$ in $\psi_{k}(0)$, called below
”$v$” (defined by Eq. (9)). Namely,

$M_{v}=\psi_{d}\Gamma_{v}$ (3)

(which will be given in more detail in the next section) with

$\Gamma_{v}=\psi_{k_{v}}(r=0)\Gamma_{0}.$ (4)

This approximate formula is not valid for $\bar{p}p$ in the vicinity of the
$\bar{p}p$ bound states. Therefore, our discussion concerns the situation when
the $e^{+}e^{-}$ effective mass is not close to mass of the $\bar{p}p$ bound
states. The accuracy of Eq. (4) was estimated in DK81 ; DKK83  in 30%.

Like in Eq. (2), $\Gamma_{0}$ in Eq. (4) is still the amplitude of
annihilation $\bar{p}p\to e^{+}e^{-}$ not including the initial state
$\bar{p}p$ interaction. It contains the ”internal” proton form factor which,
like in Eq. (2), depends on the true $q^{2}$, defined in the next section by
Eq. (2). $q^{2}$ can take the under-threshold values less than $4m^{2}$. The
argument $q^{2}$ is not shifted. However, though this formula looks like the
impulse approximation (Figure 1a), the factor $\Gamma_{v}$, responsible for
the initial state $\bar{p}p$ interaction, is quite different from $\Gamma_{u}$
in the impulse approximation, Eq. (1). It takes approximately into account an
infinite set of diagrams corresponding to all the $np$ and $\bar{p}p$ re-
scatterings shown in Figure 1. The energy $E$, denoted now, following Refs.
DK81 ; DKK83 , by $v$, should be found not from the conservation laws,
allowing to take the negative values. It takes the value as if the proton is
not bound in the deuteron, but is free DK81 ; DKK83 . That is, as if its four-
momentum were on the nucleon mass shell. $v$ is always positive and therefore
the $np$ and $\bar{p}p$ re-scatterings, resulting in Eq. (4), prevent from
penetration under the $\bar{p}p$ threshold and don’t allow to access the
proton form factor in the time-like domain, except for the close vicinity of
the $\bar{p}p$ bound states, if any.

This was very important and unexpected observation. For the first glance, the
”common sense” intuition refuses to accept it. Therefore, it needs independent
confirmation. The aim of the present work is to check this result not
analyzing the diagrams (in contrast to the way taken in Refs. DK81 ; DKK83 ),
but by direct numerical solving the Faddeev equations. In this our way, the
results of the papers DK81 ; DKK83  will be confirmed with rather good
precision (better than 30% expected in DK81 ; DKK83 ). Similarly to these
papers, we will consider the relatively small antiproton energies. In the
framework of the impulse approximation, the value of the cross section of the
reaction $\bar{p}d\to e^{+}e^{-}n$ for high-energy antiprotons with $p=1.5$
GeV/c was estimated in Helene .

Plan of the present article is the following. In Sect. 2 we give the
analytical formula for the amplitude $\bar{p}d\to e^{+}e^{-}n$ found in Refs.
DK81 ; DKK83 . In Sect. 3 we present the formalism of the Faddeev equations in
the coordinate space. In Sect. 4 the amplitude $\bar{p}d\to e^{+}e^{-}n$ is
calculated by means of the Faddeev formalism. The comparison of the numerical
results, found via the Faddeev equations, with Eqs. (1) and (3) is presented
in Sect. 5. Sect. 6 contains concluding remarks. Some technical details are
included in four Appendices A,B,C,D.

##  2 Refining the amplitudes $M_{u}$ and $M_{v}$

The set of graphs determining the amplitude of the reaction $\bar{p}d\to
e^{+}e^{-}n$ includes, before annihilation $\bar{p}p\to e^{+}e^{-}$, the re-
scatterings of the incident antiproton $\bar{p}$ on nucleons of deuteron and
the re-scatterings of the deuteron nucleons ($np$) on each other. Examples of
these graphs taking into account the $\bar{p}p$ and $pn$ re-scatterings only
(not the $\bar{p}n$ ones) are shown in Figure 1. The approximation omitting
all these re-scatterings is the impulse approximation. It takes into account
only the graph Fig. 1a. We also omit, for two reasons, the Coulomb interaction
between $\bar{p}$ and $p$. (i) Though we consider the low-energy antiprotons,
their energy is not so low for importance of the Coulomb interaction. (ii)
Testing numerically the results of Refs. DK81 ; DKK83 , we have to assume
exactly the same interaction which was taken in these papers. Otherwise, we
will be unable to obtain clear conclusions.

The impulse approximation amplitude reads:

$M(\bar{p}d\to
e^{+}e^{-}n)=\psi_{d}\left(\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}\right)\psi_{k_{u}}(0)\Gamma_{0},$
(5)

where $k_{u}=\sqrt{m\,u}$ and $u$ is the $\bar{p}p$ energy (over or below
$2m$) in this reaction, found from the conservation laws. Its explicit form is
given below in Eq. (7). We work in the c.m. frame
$\vec{p}_{d}+\vec{p}_{\bar{p}}=0$, where the amplitude depends on the momentum
$\vec{p}_{\bar{p}}$ of the incident antiproton and on the momentum
$\vec{p}_{n}$ of outgoing neutron and we assume that the proton and neutron
masses are equal to each other: $m_{p}=m_{n}\equiv m$. In Eq. (5),
$\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}$ is the relative $pn$ momentum in
c.m. frame. Since the deuteron momentum is $\vec{p}_{d}=-\vec{p}_{\bar{p}}$,
the proton momentum is
$\vec{p}_{p}=\vec{p}_{d}-\vec{p}_{n}=-\vec{p}_{\bar{p}}-\vec{p}_{n}$, this
gives for the relative $np$ momentum
$\frac{1}{2}(\vec{p}_{n}-\vec{p}_{p})=\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}$.

The invariant $e^{+}e^{-}$ energy squared (equal, for the graph Fig. 1a, to
the $\bar{p}p$ energy squared) is

$\displaystyle M^{2}_{e^{+}e^{-}}$ $\displaystyle=$ $\displaystyle
q^{2}=(E_{d}+E_{\bar{p}}-E_{n})^{2}-\vec{p}\,^{2}_{n}$ $\displaystyle=$
$\displaystyle\Bigl{(}\sqrt{(2m-|\epsilon_{d}|)^{2}+\vec{p}\,^{2}_{\bar{p}}}+\sqrt{m^{2}+\vec{p}\,^{2}_{\bar{p}}}$
$\displaystyle-$
$\displaystyle\sqrt{m^{2}+\vec{p}\,^{2}_{n}}\Bigr{)}^{2}-\vec{p}\,^{2}_{n}\approx(2m+u)^{2},$

where $\epsilon_{d}$ is the deuteron binding energy and (in the non-
relativistic limit – large $m$):

$u=\frac{3(\vec{p}\,^{2}_{\bar{p}}-\vec{p}\,^{2}_{n})}{4m}-|\epsilon_{d}|.$
(7)

This variable can take the negative under-threshold values.

As mentioned in the Introduction, according to DK81 ; DKK83 , the amplitude
corresponding to infinite sum of graphs shown in Fig. 1, due to cancellation
in all of them the non-adiabatic effects in the nucleon propagators with the
off-mass-shell effects in the elementary amplitudes, obtains the form of the
impulse approximation Eq. (5), but with the shifted value of $k_{u}\to
k_{v}=\sqrt{m\,v}$, where $v$ is the kinetic energy of the $\bar{p}p$ pair
calculated as if the proton were on mass shell. Its explicit expression is
given below by Eq. (9).

We emphasize again: the proton in all the graphs Fig. 1 is, of course, off-
mass-shell. The shift of the argument $u\to v$, resulting to the prescription
$k_{v}=\sqrt{m\,v}$, is due to incorporating the infinite sum of the graphs
Figure 1 and their partial cancellation, as it was proved in DK81 ; DKK83 .

Let us express now the energy $v$ via the momenta $\vec{p}_{\bar{p}}$ and
$\vec{p}_{n}$. The assumption that the proton is on the mass shell means that
we take for its energy not the difference $E_{p}=E_{d}-E_{n}$, but the value
$E^{on}_{p}=\sqrt{m^{2}+\vec{p}\,^{2}_{p}}$. As stated above, the proton
momentum: $\vec{p}_{p}=-\vec{p}_{\bar{p}}-\vec{p}_{n}$. The on-shell proton
energy
$E^{on}_{p}=\sqrt{m^{2}+\vec{p}\,^{2}_{p}}=\sqrt{m^{2}+(\vec{p}_{\bar{p}}+\vec{p}_{n})^{2}}$.
Then the invariant $\bar{p}p$ energy squared is

$\displaystyle M^{2}_{e^{+}e^{-}}$ $\displaystyle=$
$\displaystyle(E^{on}_{p}+E_{\bar{p}})^{2}-(\vec{p}_{p}+\vec{p}_{\bar{p}})^{2}$
$\displaystyle=$
$\displaystyle\left(\sqrt{m^{2}+(\vec{p}_{\bar{p}}+\vec{p}_{n})^{2}}+\sqrt{m^{2}+\vec{p}\,^{2}_{p}}\right)^{2}$
$\displaystyle-$ $\displaystyle\vec{p}\,^{2}_{n}\approx(2m+v)^{2},$

where (for large $m$):

$v=\frac{(\vec{p}_{n}+2\vec{p}_{\bar{p}})^{2}}{4m}>0.$ (9)

Since the Faddeev equations will be solved for the S-wave, for comparison, we
should also take the S-wave of the amplitude (5) and of the amplitude obtained
from (5) by the replacement $u\to v$. That is, we should calculate the
integral $\frac{1}{2}\int_{-1}^{1}\ldots dz$, where $z$ is the cosine of the
angle between the vectors $\vec{p}_{n}$ and $\vec{p}_{\bar{p}}$. In this way,
depending on approximation, we find the S-wave amplitudes:
$M(\bar{p}d\to e^{+}e^{-}n)=M_{u}\Gamma_{0}\\\ $ or
$M(\bar{p}d\to e^{+}e^{-}n)=M_{v}\Gamma_{0},\\\ $ where

$\displaystyle M_{u}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{-1}^{1}dz\,\psi_{d}\left(|\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}|\right)\psi_{k_{u}}(0),$
(10) $\displaystyle M_{v}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{-1}^{1}dz\,\psi_{d}\left(|\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}|\right)\psi_{k_{v}}(0).$
(11)

$z$ enters both in $|\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}|$ and in
$k_{v}$:

$\displaystyle|\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}|$ $\displaystyle=$
$\displaystyle\sqrt{p_{n}^{2}+p_{n}p_{\bar{p}}z+\frac{1}{4}p^{2}_{\bar{p}}},$
(12) $\displaystyle k_{v}=\sqrt{m\,v}$ $\displaystyle=$
$\displaystyle\sqrt{\frac{1}{4}p_{n}^{2}+p_{n}p_{\bar{p}}z+p^{2}_{\bar{p}}},$
(13)

whereas $k_{u}$ does not depend on $z$:

$k_{u}=\sqrt{m\,u}=\sqrt{\frac{3}{4}(\vec{p}\,^{2}_{\bar{p}}-\vec{p}\,^{2}_{n})-m|\epsilon_{d}|}.$
(14)

We would like to stress: according to DK81 ; DKK83 , the amplitude $M_{u}$
corresponds to the impulse approximation - Fig. 1a, whereas the amplitude
$M_{v}$ incorporates approximately all the graphs shown in Fig. 1.

Note that $\psi_{k}(0)=1/f_{Jost}(-k)$, where $f_{Jost}(k)$ is the Jost
function for the $\bar{p}p$ system. In numerical calculations we will use the
Hulthen and exponential potentials, for which $\psi_{k}(0)$ is known
analytically. It is given in Appendices A and B.

##  3 Using Faddeev equations

For the two-body $p\bar{p}$ collision, the amplitude $\Gamma$ of annihilation
$\bar{p}p\to e^{+}e^{-}$ is given by Eq. (4). It corresponds to the following
transparent physics: the incident antiproton and the target proton,
interacting with each other, are forming the two-body continuous spectrum
state described by the wave function $\psi_{k}(r)$. This state itself does not
include annihilation. Under annihilation interaction (taken in the first
order), the continuous spectrum state $\psi_{k}(r)$ turns into the state
containing free $e^{+}e^{-}$ pair (neglecting their Coulomb interaction).
Since the annihilation occurs at small distances, the transition amplitude is
determined by $\psi_{k}(r=0)$ which appears in Eq. (4).

The process $\bar{p}d\to e^{+}e^{-}n$ is determined by the same physics: the
incident antiproton and the proton + neutron in the target deuteron,
interacting with each other, are forming the three-body continuous spectrum
state described by the three-body wave function $\Psi(\vec{x},\vec{y})$
($\vec{x},\vec{y}$ are the Jacobi coordinates). The annihilation turns the
pair $\bar{p}p$ into $e^{+}e^{-}$, i.e., the continuous spectrum state
$\Psi(\vec{x},\vec{y})$ turns into the state containing three free particles
$e^{+}e^{-}n$ (again neglecting the interaction between them). We will see
that since the $\bar{p}p$ annihilation occurs at small distances, the
transition amplitude is determined by $\Psi(\vec{x}=0,p_{n})$, where $\vec{x}$
is the relative coordinate of the $\bar{p}p$ pair and the neutron momentum
$p_{n}$ appears after Fourier transform in the second variable $\vec{y}$. The
$p_{n}$-dependence in $\Psi(\vec{x}=0,p_{n})$ describes the momentum
distribution of neutrons. This distribution is the only new principal element
which differs the three-body problem from the two-body one. Though
technically, the difference between the two-body and three-body problems is
very considerable.

So, to calculate the amplitude of the reaction $\bar{p}d\to e^{+}e^{-}n$ we
should first find the three-body continuous spectrum wave function
$\Psi(\vec{x},\vec{y})$ of the $\bar{p}pn$ system and then use it for
calculating the amplitude of the transition into the $e^{+}e^{-}n$ state. The
three-body wave function is determined by the Faddeev equations with
appropriate boundary (asymptotic) conditions containing the incident and
outgoing deuteron + antiproton (in the case of elastic scattering only), or,
in addition, outgoing baryonium $B$ ($\bar{p}p$ bound state) + neutron. For
simplicity, we will consider the low incident antiproton momenta which are
insufficient for the breakup – to the $\bar{p}pn$ asymptotic state. That is,
$p_{\bar{p}}<79.2$ MeV/c in the deuteron rest (lab.) frame, or
$p_{\bar{p}}<52.8$ MeV/c in the $\bar{p}d$ c.m. frame. If the baryonium
binding energy is greater than the deuteron one, the channel with the
baryonium creation is open for any small momentum of the incident antiproton.

Let us underline: the annihilation transforms one three-body state $\bar{p}pn$
(formed by the strong interaction) into another three-body one $e^{+}e^{-}n$
(containing free particles). The $\bar{p}p$ annihilation can occur at any
moment, not necessary in the asymptotic $\bar{p}pn$ state. Therefore, to
calculate this transition, for the $\bar{p}pn$ system we will need just the
full three-body continuous spectrum wave function $\Psi(\vec{x},\vec{y})$, not
its asymptotic at $y\to\infty$. The asymptotic conditions will be imposed only
to find this full wave function from the Faddeev equations. The asymptotic
will be also used for tests. (i) We will check that the amplitudes
$\bar{p}d\to\bar{p}d$ and $\bar{p}d\to Bn$ extracted from asymptotic satisfy
the unitarity condition involving both these amplitudes. This will confirm the
full solution $\Psi(\vec{x},\vec{y})$ found numerically. (ii) In some
particular cases it turns out that the asymptotic starts rather early and
reproduces with satisfactory precision the full wave function. With the
analytical asymptotic form of the wave function one can find an analytical
formula for the amplitude $\bar{p}d\to e^{+}e^{-}n$. We will check that in
this case the amplitude found numerically is close to the analytical one.
Besides, excluding from this formula the $\bar{p}p$ interaction (except for
the $\bar{p}p$ annihilation), we reproduce the impulse approximation. This
provides another test.

We will work in the coordinate space. The formulation of the Faddeev equations
in the coordinate space see in Ref. RJ . The coordinate space three-body wave
function is represented as sum of two Faddeev components:

$\displaystyle\Psi(\vec{x}_{1},\vec{y}_{1})$ $\displaystyle=$
$\displaystyle\psi_{1}(x_{1},y_{1})+\psi_{3}(x_{3},y_{3})$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{4\pi}}\left(\frac{\chi_{1}(x_{1},y_{1})}{x_{1}y_{1}}+\frac{\chi_{3}(x_{3},y_{3})}{x_{3}y_{3}}\right).$

Since the $\bar{p}n$ interaction is not taken into account (according to Refs.
DK81 ; DKK83 , it does not change the principal results), the Faddeev
component $\psi_{2}(x_{2},y_{2})$ is absent. In each Faddeev component we keep
the S-wave only, therefore any Faddeev component depends on the modulas of the
corresponding Jacobi coordinates defined in the Appendix C. Whereas the
modulas $x_{3},y_{3}$ in
$\psi_{3}(x_{3},y_{3})=\frac{1}{\sqrt{4\pi}}\frac{\chi_{3}(x_{3},y_{3})}{x_{3}y_{3}}$
in Eq. (3) are expressed via the vectors $\vec{x}_{1},\vec{y}_{1}$ by Eqs.
(53) (see Appendix C). This results in the angle integration in the r.h.-sides
of the Faddeev equations (17) below. We introduced in (3) the factor
$\frac{1}{\sqrt{4\pi}}$ in order to have the conventional normalization (21)
for the radial bound state wave functions.

Since $\Psi$ is finite, the functions $\chi_{i}(x,y)$ must satisfy the
conditions

$\chi_{i}(x=0,y)=\chi_{i}(x,y=0)=0.$ (16)

Then the system of equations for the S-wave obtains the form:

$\displaystyle\left[k^{2}+\frac{\partial^{2}}{\partial
x^{2}}+\frac{\partial^{2}}{\partial
y^{2}}-mV_{\bar{p}p}(x)\right]\chi_{1}(x,y)$ (17) $\displaystyle=$
$\displaystyle\frac{1}{2}mV_{\bar{p}p}(x)\int_{-1}^{1}du\frac{xy}{x^{\prime}y^{\prime}}\chi_{3}(x^{\prime},y^{\prime}),$
$\displaystyle\left[k^{2}+\frac{\partial^{2}}{\partial
x^{2}}+\frac{\partial^{2}}{\partial y^{2}}-mV_{pn}(x)\right]\chi_{3}(x,y)=$
$\displaystyle=$
$\displaystyle\frac{1}{2}mV_{pn}(x)\int_{-1}^{1}du\frac{xy}{x^{\prime}y^{\prime}}\chi_{1}(x^{\prime},y^{\prime}),$

where $V_{\bar{p}p}(x),V_{pn}(x)$ are the $\bar{p}p$ and $pn$ potentials,

$\displaystyle x^{\prime}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(x^{2}+2\sqrt{3}xyu+3y^{2})^{1/2},$ $\displaystyle
y^{\prime}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(3x^{2}-2\sqrt{3}xyu+y^{2})^{1/2},$ (18)

and the Faddeev components $\chi_{1,3}$ have the asymptotic:

$\displaystyle\chi_{1}(x_{1},y_{1}\to\infty)$ $\displaystyle=$
$\displaystyle\chi_{b}(x_{1})\frac{1}{p_{3}}f_{1}(p_{1})\exp(ip_{1}y_{1}),$
$\displaystyle\chi_{3}(x_{3},y_{3}\to\infty)$ $\displaystyle=$
$\displaystyle\chi_{d}(x_{3})\frac{1}{p_{3}}[\sin(p_{3}y_{3})$ (20)
$\displaystyle+$ $\displaystyle f_{3}(p_{3})\exp(ip_{3}y_{3})].$

Here $\chi_{b}(x_{1}),\chi_{d}(x_{3})$ are the wave functions of baryonium
($\bar{p}p$) and deuteron both normalized as

$\int_{0}^{\infty}\chi^{2}(r)dr=1.$ (21)

$f_{1}(p_{1})$ is the dimensionless amplitude of the rearrangement
$\bar{p}d\to Bn$. $f_{3}(p_{3})$ in (20) is the dimensionless elastic
$\bar{p}d$ scattering amplitude, related to the phase shift as
$f_{3}(p_{3})=\frac{1}{2i}[\exp(2i\delta)-1]$. If the channel with
rearrangement is closed, the phase $\delta$ is real. $p_{1}$ is the Jacobi
momentum in the $Bn$ channel, $p_{3}$ is the Jacobi momentum in the $\bar{p}d$
channel. In these channels, in the c.m. frame, the Jacobi momenta are related
to the c.m. ones as $p_{1}=\frac{\sqrt{3}}{2}\tilde{p}_{n}$,
$p_{3}=\frac{\sqrt{3}}{2}p_{\bar{p}}$. In the elastic $\bar{p}d$ channel the
momentum $p_{\bar{p}}$ is the incident (and final) antiproton momentum. In the
rearrangement channel $\bar{p}d\to Bn$ the neutron momentum $\tilde{p}_{n}$ is
found from the energy conservation law. It is determined by the incident
antiproton momentum $p_{\bar{p}}$ and by the particle masses participating in
the reaction. Above we used the notation $\vec{p}_{n}$ (or $p_{n}$ without
tilde) for the neutron momentum in the reaction $\bar{p}d\to e^{+}e^{-}n$. It
depends on the effective mass of the $e^{+}e^{-}$ pair and therefore varies.
Whereas, $\tilde{p}_{n}$ is a particular value of $p_{n}$ which the neutron
obtains in the reaction $\bar{p}d\to Bn$, or kinematically equivalent, for the
$e^{+}e^{-}$ effective mass equal to mass of baryonium. One should not confuse
$\tilde{p}_{n}$ and $p_{n}$.

If the $\bar{p}p$ interaction is not enough to create the $\bar{p}p$ bound
state, then

$\displaystyle\chi_{1}(x_{1},y_{1}\to\infty)$ $\displaystyle=$ $\displaystyle
0,$ (22) $\displaystyle\chi_{3}(x_{3},y_{3}\to\infty)$ $\displaystyle=$
$\displaystyle\chi_{d}(x_{3})\frac{1}{p_{3}}[\sin(p_{3}y_{3})$ (23)
$\displaystyle+$ $\displaystyle f_{3}(p_{3})\exp(ip_{3}y_{3})],$

and, as mentioned above, the phase shift corresponding to $f_{3}(p_{3})$ is
real.

The boundary conditions in the variables $x_{1},x_{3}\to\infty$ correspond to
decrease of both functions $\chi_{1}$ and $\chi_{3}$, due to finite size of
the deuteron and baryonium. That is:

$\chi_{1}(x_{1}\to\infty,y_{1})\to 0,\quad\chi_{3}(x_{3}\to\infty,y_{3})\to
0.$ (24)

Equivalently, instead of Eqs. (24), one can impose at $x_{1},x_{3}\to\infty$
the equations (17) and put in them $V(x)=0$ (that we used in our numerical
calculations). For negative energies the solutions of the free equations are
zeroes, that is, are given by Eqs. (24).

If the $\bar{p}p$ interaction is enough strong and creates the $\bar{p}p$
bound state with rather large binding energy (but no excited states, for
simplicity), the final relative $Bn$ momentum $p_{1}$ is rather large. The
relative $\bar{p}d$ momentum $p_{3}$ can be also large. Since the oscillating
factors $\exp(ip_{1}y_{1})$ and $\exp(ip_{3}y_{3})$, extracted in the
asymptotic, are included in $\chi_{1}(x_{1},y_{1}),\,\chi_{3}(x_{3},y_{3})$,
the functions (LABEL:eq61), (20) oscillate. This complicates their numerical
calculations. A way to overcome that is to extract these exponents from
$\chi_{1}(x_{1},y_{1}),\,\chi_{3}(x_{3},y_{3})$. Besides, we have noticed that
extraction of the deuteron and baryonium wave functions improves the accuracy
and convergence of the numerical calculations. That is, we represent the
Faddeev components as:

$\displaystyle\chi_{1}(x_{1},y_{1})$ $\displaystyle=$
$\displaystyle\exp(ip_{1}y_{1})\chi_{b}(x_{1})\tilde{\chi}_{1}(x_{1},y_{1}),$
(25) $\displaystyle\chi_{3}(x_{3},y_{3})$ $\displaystyle=$
$\displaystyle\exp(ip_{3}y_{3})\chi_{d}(x_{3})\tilde{\chi}_{3}(x_{3},y_{3})$
(26) $\displaystyle+$ $\displaystyle\chi_{d}(x_{3})\sin(p_{3}y_{3}).$

We substitute expressions (25), (26) in the equations (22), (23), derive
equations for the functions $\tilde{\chi}_{1}(x,y)$, $\tilde{\chi}_{3}(x,y)$,
and impose on them the boundary conditions

$\tilde{\chi}_{1,3}(x,y=0)=0,\quad\left.\frac{\partial}{\partial
y}\tilde{\chi}_{1,3}(x,y)\right|_{y=y_{max}}=0.$ (27)

Solving numerically the equations for $\tilde{\chi}_{1,3}(x,y)$, we find the
three-body wave function $\Psi(\vec{x},\vec{y})$, Eq. (3).

##  4 Finding the amplitude $\bar{p}d\to e^{+}e^{-}n$

As explained above, the initial state containing $\bar{p}d$, under strong
interaction, turns into a continuous spectrum three-body state $\bar{p}pn$
described by the wave function found in the previous section. Due to the
$\bar{p}p$ annihilation, this state turns into the state of free particles
$e^{+}e^{-}n$. To find the amplitude of this transition in the first order of
the annihilation interaction, we should calculate the matrix element from the
transition operator between the continuous spectrum state $\bar{p}pn$ and the
final one $e^{+}e^{-}n$. This amplitude differs from the two-body one
$\bar{p}p\to e^{+}e^{-}$ by the initial and final states, whereas the
transition operator, corresponding to the annihilation, is the same. Since the
two-body amplitude $\bar{p}p\to e^{+}e^{-}$ is known (it is given by Eq. (2)),
we can easy determine the transition operator leading to this amplitude and
then use this operator in the three-body problem. Looking at the amplitude
(2), we see that this operator has the form
$\hat{O}=\delta^{(3)}(\vec{r})\Gamma_{0}$, where $\vec{r}$ is the relative
distance between $\bar{p}$ and $p$ and $\Gamma_{0}$ is the $\bar{p}p\to
e^{+}e^{-}$ annihilation amplitude not including the initial interaction via
the $\bar{p}p$ potential. It corresponds to annihilation at $\vec{r}=0$. Since
the pair $\bar{p}p$ after annihilation is absent in the final state and the
$e^{+}e^{-}$ interaction in the final state is neglected, the matrix element
$\Gamma$, in the two-body reaction $\bar{p}p\to e^{+}e^{-}$, contains only the
initial $\bar{p}p$ state, described by the continuous spectrum wave function
$\psi_{k}(r)$:

$\Gamma=\Gamma_{0}\int\delta^{(3)}(\vec{r})\psi_{k}(r)d^{3}r=\psi_{k}(r=0)\Gamma_{0}$
(28)

that indeed reproduces Eq. (2).

Similarly, for the three-body reaction $\bar{p}d\to e^{+}e^{-}n$, the initial
state is given by the wave function
$\exp(i\vec{P}\makebox[2.27626pt]{$\cdot$}\vec{R})\Psi(\vec{x},\vec{y})$ with
$\Psi$ defined in (3). The final state contains the (complex conjugated)
neutron plane wave:
$\exp(-i\vec{P}^{\prime}\makebox[2.27626pt]{$\cdot$}\vec{R})\psi_{n}^{*}$,
where
$\psi^{*}_{n}=\exp(-i\vec{p}_{1}\makebox[2.27626pt]{$\cdot$}\vec{y}_{1})$. The
equations (LABEL:eq61), (20), imply that the particle No. 1 is neutron, No. 2
– proton and No. 3 – antiproton. Therefore $\vec{y}_{1}$ is relative
coordinate of neutron and $\bar{p}p$ system (apart the coefficient
$\sqrt{3}/2$), $\vec{p}_{1}$ is the Jacobi momentum of the final neutron, that
justifies the expression for $\psi_{n}$. The factors
$\exp(i\vec{P}\makebox[2.27626pt]{$\cdot$}\vec{R})$ and
$\exp(-i\vec{P}^{\prime}\makebox[2.27626pt]{$\cdot$}\vec{R})$ take into
account the c.m. motion of the initial and final three-body systems. The
center of mass coordinate $\vec{R}$ and the total momentum $\vec{P}$ are
defined in Appendix C, Eqs. (C) and (C) respectively.

In this way, for the amplitude of the transition $\bar{p}d\to e^{+}e^{-}n$ we
find:

$\displaystyle M(\bar{p}d\to
e^{+}e^{-}n)=\int\exp(-i\vec{P}^{\prime}\makebox[2.27626pt]{$\cdot$}\vec{R})\psi_{n}^{*}(y_{1})$
(29) $\displaystyle\times$
$\displaystyle\hat{O}\exp(i\vec{P}\makebox[2.27626pt]{$\cdot$}\vec{R})\Psi(\vec{x}_{1},\vec{y}_{1})d^{3}r_{1}d^{3}r_{2}d^{3}r_{3}.$

We transform the integration to the Jacobi variables:

$d^{3}r_{1}d^{3}r_{2}d^{3}r_{3}=\left(\frac{\sqrt{3}}{2}\right)^{3}d^{3}x_{1}d^{3}y_{1}d^{3}R.$

After integration over $d^{3}R$ we get

$M(\bar{p}d\to
e^{+}e^{-}n)=(2\pi)^{3}\delta^{(3)}(\vec{P}-\vec{P}^{\prime})M_{3}\Gamma_{0},$
(30)

where

$\displaystyle M_{3}$ $\displaystyle=$
$\displaystyle\left(\frac{\sqrt{3}}{2}\right)^{3}\frac{1}{\sqrt{4\pi}}\int
d^{3}x_{1}d^{3}y_{1}e^{-i\vec{p}_{1}\makebox[2.27626pt]{$\cdot$}\vec{y}_{1}}\delta^{(3)}(\vec{x}_{1})$
(31) $\displaystyle\times$
$\displaystyle\left(\frac{\chi_{1}(x_{1},y_{1})}{x_{1}y_{1}}+\frac{\chi_{3}(x_{3},y_{3})}{x_{3}y_{3}}\right).$

We substitute here $x_{3},y_{3}$ from (53) and integrate over $\vec{x}_{1}$ by
means of the delta-function $\delta^{(3)}(\vec{x}_{1})$ that reduces the
variables $\vec{x}_{3},\vec{y}_{3}$ as follows:

$\displaystyle\vec{x}_{3}=-\frac{1}{2}(\vec{x}_{1}+\sqrt{3}\vec{y}_{1})\to-\frac{\sqrt{3}}{2}\vec{y}_{1},$
$\displaystyle\vec{y}_{3}=\frac{1}{2}(\sqrt{3}\vec{x}_{1}-\vec{y}_{1})\to-\frac{1}{2}\vec{y}_{1}.$

We also integrate over the angles of the vector $\vec{y}_{1}$. In this way we
obtain

$\displaystyle M_{3}$ $\displaystyle=$
$\displaystyle\left(\frac{\sqrt{3}}{2}\right)^{3}\frac{\sqrt{4\pi}}{p_{1}}\int_{0}^{\infty}dy_{1}\left[\lim_{x_{1}\to
0}\frac{1}{x_{1}}\chi_{1}(x_{1},y_{1})\right.$ (32) $\displaystyle+$
$\displaystyle\left.\frac{4}{\sqrt{3}y_{1}}\chi_{3}\left(\frac{\sqrt{3}}{2}y_{1},\frac{1}{2}y_{1}\right)\right]\sin(p_{1}y_{1})$
$\displaystyle=$
$\displaystyle\frac{\sqrt{3\pi}}{p_{n}}\int_{0}^{\infty}dy\left[\lim_{x\to
0}\frac{1}{x}\chi_{1}\left(x,\frac{2}{\sqrt{3}}y\right)\right.$
$\displaystyle+$
$\displaystyle\left.\frac{2}{y}\chi_{3}\left(y,\frac{1}{\sqrt{3}}y\right)\right]\sin(p_{n}y).$

Transforming the middle part of Eq. (32) into the r.h.-side, we made the
replacement of the variable $y_{1}=\frac{2}{\sqrt{3}}y$ and took the values
$p_{1}=\frac{\sqrt{3}}{2}p_{n}$, $p_{3}=\frac{\sqrt{3}}{2}p_{\bar{p}}$ from
Eq. (58) in Appendix C.

According to Eq. (LABEL:eq61), the function $\chi(x_{1},y_{1}\to\infty)$
oscillates that complicates the numerical calculation of the first integral in
the r.h.-side of (32). To overcome this difficulty, we represent
$\chi(x_{1},y_{1})$ in the form:
$\chi(x_{1},y_{1})=[\chi(x_{1},y_{1})-\chi^{as}(x_{1},y_{1})]+\chi^{as}(x_{1},y_{1})$,
where $\chi^{as}(x_{1},y_{1})$ is the asymptotic form of $\chi(x_{1},y_{1})$
given by (LABEL:eq61). Then the integral containing the difference
$[\chi(x_{1},y_{1})-\chi^{as}(x_{1},y_{1})]$ is calculated numerically without
problem, whereas the integral containing $\chi^{as}(x_{1},y_{1})$ can be
calculated analytically. Its analytical calculation is carried out in the next
section.

The amplitude $M_{3}$, Eq. (32), is the final result. It determines the full
amplitude $M(\bar{p}d\to e^{+}e^{-}n)$ by Eq. (30) and differs from it by a
factor. $M_{3}$ corresponds to the case, considered in DK81 ; DKK83 , when
both interactions – $\bar{p}p$ and $pn$ – are present (but the $\bar{p}n$
interaction is absent) and is expressed via the three-body continuous spectrum
wave function $\Psi(\vec{x},\vec{y})$ determined by the Faddeev equations.
Since the Faddeev equations include all the potential interactions, except for
annihilation, expression (32) automatically takes into account, in addition to
other re-scatterings, the initial $\bar{p}p$ interaction, which is described
in (28) by the factor $\psi_{k}(r=0)$.

###  4.1 Amplitude $M_{3}$ for asymptotic wave function

In this subsection we consider the case, when the Faddeev components in the
three-body wave function $\Psi(\vec{x},\vec{y})$ are replaced by their
asymptotic determined, in the S-wave, by the equations (LABEL:eq61), (20).
This allows to find analytically the pole in the amplitude related to creation
of baryonium. In some cases, this can also provide good numerical accuracy.

Substituting $\chi_{1},\chi_{3}$ from (LABEL:eq61), (20) into (32), we find
$M_{3}$ in the following form:

$M_{3}=M_{31}+M_{33},$

where

$\displaystyle M_{31}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{3\pi}}{p_{n}}\int_{0}^{\infty}dy\lim_{x\to
0}\frac{1}{x}\chi_{1}(x,\frac{2}{\sqrt{3}}y)\sin(p_{n}y)$ (33)
$\displaystyle=$ $\displaystyle\frac{\sqrt{4\pi}}{p_{n}p_{\bar{p}}}\lim_{x\to
0}\frac{1}{x}\chi_{b}(x)f_{1}\left(\frac{\sqrt{3}}{2}\tilde{p}_{n}\right)$
$\displaystyle\times$
$\displaystyle\int_{0}^{\infty}dy\exp(i\tilde{p}_{n}y)\sin(p_{n}y)$

To calculate the last integral, we regularize it, introducing in the integrand
the factor $\exp(-\epsilon y)$ ($\epsilon>0$), and after integration take the
limit $\epsilon\to 0$. That is,

$\displaystyle\int_{0}^{\infty}dy\exp(-\epsilon
y)\exp(i\tilde{p}_{n}y)\sin(p_{n}y)$ $\displaystyle=$
$\displaystyle\frac{p_{n}}{{p}^{2}_{n}-(\epsilon-i\tilde{p}_{n})^{2}}|_{\epsilon\to
0}=\frac{p_{n}}{{p}^{2}_{n}-\tilde{p}_{n}^{2}}.$

In this way, we find

$M_{31}=\frac{\sqrt{4\pi}}{p_{n}p_{\bar{p}}}\lim_{x\to
0}\frac{1}{x}\chi_{b}(x)f_{1}\left(\frac{\sqrt{3}}{2}\tilde{p}_{n}\right)\frac{p_{n}}{({p}^{2}_{n}-\tilde{p}^{2}_{n})}.$
(34)

The contribution of the amplitude $M_{13}$ originates from the channel with
the baryonium creation $\bar{p}d\to Bn$. We remind that $\tilde{p}_{n}$ is the
neutron momentum in this channel. For given masses, it is fixed by the
conservation laws. Whereas $p_{n}$ is the neutron momentum after annihilation,
that is, in the reaction $\bar{p}d\to e^{+}e^{-}n$. Variation of the effective
mass of the pair $e^{+}e^{-}$ corresponds to variation of $p_{n}$, whereas
$\tilde{p}_{n}$ is the particular value of $p_{n}$ corresponding to the fixed
effective $e^{+}e^{-}$ mass equal to mass of baryonium. According to Eq. (34),
$M_{31}$ vs. $p_{n}$ has the under-threshold pole corresponding to creation of
baryonium. The baryonium creation amplitude $f_{1}$ is found from the Faddeev
equations, which contain the Jacobi momenta. Therefore
$f_{1}=f_{1}(p_{1})=f_{1}\left(\frac{\sqrt{3}}{2}\tilde{p}_{n}\right)$. We
used the relation (58).

Similarly, we find from (32) the contribution $M_{33}$:

$\displaystyle M_{33}$ $\displaystyle=$
$\displaystyle\frac{2\sqrt{3\pi}}{p_{n}}\int_{0}^{\infty}\frac{dy}{y}\chi_{3}\left(y,\frac{1}{\sqrt{3}}y\right)\sin(p_{n}y)$
$\displaystyle=$
$\displaystyle\frac{4\sqrt{\pi}}{p_{n}p_{\bar{p}}}\int_{0}^{\infty}\frac{dy}{y}\chi_{d}(y)\left[\sin\left(\frac{1}{2}p_{\bar{p}}y\right)\right.$
$\displaystyle+$
$\displaystyle\left.f_{3}\left(\frac{\sqrt{3}}{2}p_{\bar{p}}\right)\exp\left(\frac{i}{2}p_{\bar{p}}y\right)\right]\sin(p_{n}y).$

We again assumed that the amplitude $f_{3}$ is known as function of the Jacobi
momentum, that is,
$f_{3}=f_{3}(p_{3})=f_{3}\left(\frac{\sqrt{3}}{2}p_{\bar{p}}\right)$.

###  4.2 Without the $\bar{p}p$ interaction

In the present section we suppose that both the $\bar{p}p$ and $\bar{p}n$
interactions are absent (though the annihilation $\bar{p}p\to e^{+}e^{-}$ is
present). This exactly corresponds to the impulse approximation, Fig. 1a. The
amplitude of reaction $\bar{p}d\to e^{+}e^{-}n$ in the impulse approximation
is obtained from Eqs. (10) or (11) by excluding from them the factor
$\psi_{k}(0)$ responsible for the initial $\bar{p}p$ interaction. In this way,
after omitting also the factor $\Gamma_{0}$, we find:

$\tilde{M}_{S}=\frac{1}{2}\int_{-1}^{1}dz\psi_{d}\left(|\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}|\right).$
(36)

The amplitude obtained above by the Faddeev formalism, after omitting the
$\bar{p}p$ interaction and $\Gamma_{0}$, should coincide with (36). This will
be a test of our calculation.

In absence of the $\bar{p}p$ interaction we have $f=f_{1}=0$. Therefore, from
(34) we get $M_{31}=0$ and from (4.1) we obtain:

$M_{33}=\frac{2\sqrt{\pi}}{p_{n}p_{\bar{p}}}\int_{0}^{\infty}\frac{dy}{y}\chi_{d}(y)\sin\left(\frac{1}{2}p_{\bar{p}}y\right)\sin\left(p_{n}y\right).$
(37)

To prove the coincidence of the amplitude (36) and (37), at first, we express
$\psi_{d}(p)$ in (36) via the coordinate space wave function

$\displaystyle\psi_{d}(p)$ $\displaystyle=$
$\displaystyle\int\psi_{d}(r)\exp(i\vec{p}\vec{r})d^{3}r$ $\displaystyle=$
$\displaystyle\sqrt{4\pi}\int_{0}^{\infty}dr\chi_{d}(r)\frac{\sin(pr)}{p}.$

Then Eq. (36) obtains the form:

$\tilde{M}_{S}=\sqrt{4\pi}\int_{0}^{\infty}dr\chi_{d}(r)\frac{1}{2}\int_{-1}^{1}dz\frac{\sin(|\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}|r)}{|\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}|}.$
(38)

With expression (12) for $|\vec{p}_{n}+\frac{1}{2}\vec{p}_{\bar{p}}|$, the
integral (38) over $z$ is calculated explicitly. The result exactly coincides
with (37). Since Eq. (37) was obtained as a particular case from the general
expression (32), the latter one is also confirmed in this way.

##  5 Numerical results and discussion

The derivation of the result found in Refs. DK81 ; DKK83  – our equation (11)
– is general, it does not rely on any particular model of $np$ and $\bar{p}p$
interaction. Therefore, it should be valid for any $np$ and $\bar{p}p$
potentials. The numerical calculations, presented in this section, were
carried out with three potentials: the Hulthen potential, the exponential one
and the spherically symmetric potential well. We chose them since they provide
the analytical solutions for the two-body bound and scattering state wave
functions. We can take the unrealistic potentials, since we don’t intend to
compare with experimental data, but we will compare the amplitudes found from
the Faddeev equations with Eqs. (1) and (3). Since we will find the
coincidence with Eq. (3), we check this coincidence with three different
potentials, to show that this coincidence is not by chance. That’s quite
enough for our aim. Therefore, the fact that the potentials are unrealistic
does not cast a shadow on our conclusions. The form of potentials was taken
the same for the $np$ and $\bar{p}p$ systems, but the parameters are
different. Depending on the system, we chose the parameters of the potential
to provide the correct deuteron binding energy ($-2.23$ MeV) and the realistic
$\bar{p}p$ binding energy (two cases: $-15$ and $-55$ MeV). As mentioned, for
all these potentials, there are explicit solutions of the two-body problem -
the bound state wave functions, used for solving the Faddeev equations, and
the scattering state wave functions, determining $\psi_{k}(r=0)$ in Eqs. (10),
(11). The results are similar for all three potentials. Therefore, we will
mainly present the results found with the Hulthen potential, though we will
also give some results found with the exponential potential. The Hulthen
potential and corresponding wave functions are presented in Appendix A. The
exponential potential and the corresponding wave functions are given in
Appendix B.

As explained above, to calculate the amplitude of the reaction $\bar{p}d\to
e^{+}e^{-}n$, solving the Faddeev equations, we need full three-body wave
function $\Psi(\vec{x},\vec{y})$, not only its asymptotic. To check this
solution, we take the asymptotic tails of this wave function, extract from
them the elastic ($\bar{p}d\to\bar{p}d$) and rearrangement ($\bar{p}d\to Bn$)
scattering amplitudes and check, whether they satisfy the unitarity
conditions. The details of this procedure and the numerical results are
presented in Appendix D. We consider two cases: (i) when the $\bar{p}p$
interaction is not enough to create baryonium (the elastic scattering only)
and (ii) when the $\bar{p}p$ interaction creates baryonium (the elastic
scattering and rearrangement).

In the case (i), the test of unitarity is reduced to the test that the phase
shift $\delta$, extracted from the scattering amplitude, is real. The results
for two antiproton incident momenta $p_{\bar{p}}=23.9$ MeV/c and
$p_{\bar{p}}=48.3$ MeV/c are given in the Table 1 in Appendix D. They show
that the phase shift is indeed real with high accuracy.

In the case (ii) (two open channels), the test of unitarity is reduced to the
test of the equality (62). The results for the same two antiproton incident
momenta are given in the Table 2 in Appendix D. They show that in this case,
in our numerical calculation, the unitarity is satisfied with accuracy better
than 0.1%.

These tests confirm that our numerical solutions of the Faddeev equations are
correct.

In Figures 3—8, the real and imaginary parts of the amplitude of reaction
$\bar{p}d\to e^{+}e^{-}n$, however, not including the factor $\Gamma_{0}$, for
the incident antiproton momentum $p_{\bar{p}}=48.3$ MeV/c (Figs. 3-7) and
$p_{\bar{p}}=23.9$ MeV/c (Fig. 8) are shown as function of the outgoing
neutron momentum $p_{n}$. The calculations for Figs. 3–4 and Figs. 6–8 were
fulfilled with the Hulthen potential (Appendix A), whereas the curves shown in
Fig. 5 were calculated with the exponential potential (Appendix B). Everywhere
the solid curve is the Faddeev calculation by Eq. (32), the dashed curve is
calculated by the prescription of Refs. DK81 ; DKK83 , in which we keep the
S-wave only, i.e., by Eq. (11), the dot-dashed curve is the impulse
approximation Eq. (10).

The curves in Fig. 3 were calculated for the potential parameter
$V_{0\bar{p}p}=10$ MeV. This interaction is not enough to create the
$\bar{p}p$ bound state. In the three-body $\bar{p}pn$ problem, the elastic
scattering channel $\bar{p}d\to\bar{p}d$ only is open.

Figure 3: (Color online) Amplitude of the reaction $\bar{p}d\to e^{+}e^{-}n$,
divided by the annihilation amplitude $\Gamma_{0}$ for the incident c.m.
antiproton momentum $p_{\bar{p}}=48.3$ MeV/c vs. outgoing c.m. neutron
momentum $p_{n}$. No $\bar{p}p$ bound state.
Upper panel: Real part of the amplitude. Solid curve – calculated via Faddeev
formalism, Eq. (32). Dashed curve – calculated by Eq. (11) with
$k=\sqrt{m\,v}$ given by Eq. (13). Dot-dashed curve – impulse approximation,
calculated by Eq. (10) with $k=\sqrt{m\,u}$ given by Eq. (14). The curves are
indistinguishable from each other.
Lower panel: the same as the upper one but for imaginary part of amplitude.

We see that in Figure 3 (upper panel) all three curves - Faddeev calculation,
Eq. (32), the one calculated by Eq. (11) (with $k_{v}=\sqrt{m\,v}$, given by
Eq. (13)) and impulse approximation – calculated by Eq. (10) (with
$k_{u}=\sqrt{m\,u}$, given by Eq. (14)), are practically indistinguishable
from each other. This is, apparently, a consequence of weak $\bar{p}p$
interaction allowing to neglect re-scattering of $\bar{p}$ on the deuteron
nucleons which just makes the difference between the three cases. Hence, the
diagram Fig. 1a dominates in this case.

In contrast to the reals parts, the imaginary parts of the amplitudes – lower
panel of Fig. 3, calculated by the Faddeev equations and by Eq. (11), visibly
differ from each other. However, the imaginary parts in Fig. 3 are at least 20
times smaller than the real ones. Whereas, the accuracy of Eq. (5) was
estimated in DK81 ; DKK83  in 30%. In impulse approximations, the imaginary
parts are zero. So, the magnitudes of the imaginary parts are beyond of this
accuracy and, therefore, cannot be reproduced, whereas the reals parts are
described by Eq. (11) even with better accuracy than it was expected in DK81 ;
DKK83 . It can be also noticed that the larger the imaginary parts, the better
they are reproduced through the Faddeev equations.

Note that replacing in the full three-body wave function by its asymptotic,
i.e., calculating the sum of amplitudes by Eqs. (33) and (4.1) (the case
considered in Sec. 4.1), we obtain the results, both for real and imaginary
parts (not shown), which are close to the solid curves in Fig. 3. This means
that for these parameters asymptotic of the three-body wave function starts
rather early.

The main result of the present work is illustrated by Figs. 4-8. In these
figures, the case when the $\bar{p}p$ interaction is sufficient for the
formation of baryonium is considered. Figure 4 corresponds to the case when a
$\bar{p}p$ bound state in the potential — baryonium — with energy
$|\epsilon_{b}|=15$ MeV, provided by the potential parameters given in
Appendix A.1. In Fig. 6 the domain of the neutron momentum $p_{n}$ is shown in
which the $\bar{p}p$ bound state clearly manifests itself. Figures 7 and 8
correspond to the baryonium binding energy $|\epsilon_{b}|=55$ MeV. We remind
that everywhere $r_{0}=0.6061$ fm. The curves in Figures 3-7 are calculated
for the incident antiproton momentum $p_{\bar{p}}=48.3$ MeV/c. Figure 8
corresponds to $p_{\bar{p}}=23.9$ MeV/c.

Figure 4: The same as in Figures 3 for $p_{\bar{p}}=48.3$ MeV/c. The
$\bar{p}p$ pair is bound with $|\epsilon_{b}|=15$ MeV.

Figure 5: The same as in Figures 4 but for the exponential potential given in
Appendix B.

Figure 6: The same as in Figures 4 for $p_{\bar{p}}=48.3$ MeV/c, but for
larger domain of $p_{n}$, covering the response from the $\bar{p}p$ bound
state with $|\epsilon_{b}|=15$ MeV.

Figure 7: The same as in Figures 3 and 4 for $p_{\bar{p}}=48.3$ MeV/c. The
$\bar{p}p$ pair is bound with $|\epsilon_{b}|=55$ MeV.

Figure 8: The same as in Figures 3 \- 4 and 7, but for $p_{\bar{p}}=23.9$
MeV/c. The $\bar{p}p$ pair is bound with $|\epsilon_{b}|=55$ MeV.

For different $\bar{p}p$ binding energies and incident momenta of antiproton,
for the Hulthen potential, the real parts of the amplitudes calculated by the
approximate formula derived in DK81 ; DKK83 , i.e., by Eq. (11), with $k_{v}$
given by Eq. (13), are well reproduced by the ”exact” calculations via Faddeev
equations. There is some deviation for the imaginary parts, which, however, is
within the accuracy of Eq. (5), estimated in DK81 ; DKK83 .

One can wonder if the agreement shown in our work is not accidental in the
sense that it may not append for another parameters of the Hulthen potential
or another functional form. To check that, we carried out the calculations
with another potential (the exponential one) given in Appendix B and providing
quite different wave function. As it is seen in Fig. 5, the values of
amplitudes calculated with the exponential potential differs from ones found
with the Hulthen potential, but the relative positions of the curves is
practically the same. This shows that our conclusions remain unchanged under
considerable variation of the $pn$ and $\bar{p}p$ interaction.

The same situation is reproduced for the spherically symmetric potential well
(not shown). Though in the latter case, the deviation of imaginary parts is a
little bit larger. Note that the imaginary parts of the amplitudes calculated
in the impulse approximation Eq. (10) (the dot-dashed curves at the right
panels in all the figures) is zero.

The coincidence of the curves found, following Refs. DK81 ; DKK83 , by Eq.
(11), with the results obtained by the Faddeev equations, is even better than
one can expect from estimation of the precision made in DK81 ; DKK83 . These
two coinciding results - solid and dashed curves - considerably differ from
the impulse approximation - the dot-dashed curve. This shows that the series
of graphs Fig. 1 is indeed important. Whereas, the delicate cancellations
between the non-adiabatic and off-shell effects, really take place. Due to
these cancellations, the sum of this series is reduced to the impulse
approximation amplitude with the shifted argument $k_{u}\to k_{v}$ except for
vicinity of the $\bar{p}p$ bound state, i.e., in the interval $p_{n}=130\div
140$ MeV/c in Fig. 6. In this vicinity the condition (1) from DK81 :
$\frac{1}{3}|\Gamma_{u}-\Gamma_{v}|\gg|\Gamma_{v}|$ is satisfied that, as
indicated in Refs. DK81 ; DKK83 , violates the applicability of Eq. (11).

##  6 Conclusion

The annihilation amplitude $\bar{p}p\to e^{+}e^{-}$ of free $\bar{p}p$ pair is
graphically shown in Fig. 2. According to Eq. (2), it is given by the product
$\psi_{k}(0)\Gamma_{0}$, where the factor $\psi_{k}(0)$ incorporates the
$\bar{p}p$ interaction in the initial state, whereas $\Gamma_{0}$ is the
”pure” annihilation amplitude which does not take into account the initial
state interaction between $\bar{p}$ and $p$. The factor $\psi_{k}(0)$ provides
important information about the $\bar{p}p$ interaction. It was hope that from
the reaction $\bar{p}d\to e^{+}e^{-}n$, where the effective $e^{+}e^{-}$ mass
can take the under-threshold values (less than $2m$), one can extract the
time-like proton form factor in the under-threshold domain. As it follows from
the results of Refs. DK81 ; DKK83 , confirmed by the present work, this hope
is not realized in full measure. As Figure 6 shows, the existence and position
of the $\bar{p}p$ bound states indeed can be found. However, the replacement
in the initial state of proton by deuteron results in the set of much more
complicated graphs, shown in Fig. 1, which, in general, are not reduced to the
factor $\psi_{k}(0)\Gamma_{0}$. As it was shown in Refs. DK81 ; DKK83
analytically, due to cancellations of the non-adiabatic and off-mass-shell
effects in these graphs, the sum of graphs Fig. 1 is reduced to the product
$\psi_{k_{v}}(0)\Gamma_{0}$, where $\Gamma_{0}$ still depends on the momentum
transfer squared $q^{2}$, which can take the under-threshold values
$q^{2}<4m^{2}$, but everywhere, except for the close vicinity of the bound
state, the factor $\psi_{k_{v}}(0)$ is determined by the energy (9) which is
always positive. That is, the positions of the $\bar{p}p$ bound states can be
found in the reaction $\bar{p}d\to e^{+}e^{-}n$. Whereas, the values of the
proton electromagnetic form factor under threshold remain inaccessible, since
the reaction $\bar{p}d\to e^{+}e^{-}n$ provides the factor $\psi_{k_{v}}(0)$,
determining the proton time-like form factor for the $\bar{p}p$ energy $v$
which is not under but above threshold.

To check this result and the cancellation of the non-adiabatic and off-mass-
shell effects, we have calculated numerically the amplitude of the reaction
$\bar{p}d\to e^{+}e^{-}n$, treating the three-body system $\bar{p}pn$ by means
of the Faddeev equations. Everywhere, except for vicinity of singularities
corresponding to the $\bar{p}p$ bound states, the result of this calculation
turned out to be rather close to the amplitude calculated by Eq. (11), derived
in Refs. DK81 ; DKK83 . In this way, we confirm the conclusion of Refs. DK81 ;
DKK83 . Simultaneously, we confirm the delicate cancellations of the non-
adiabatic and off-mass-shell effects in the graphs shown in Fig. 1.

We emphasize again that in Eqs. (5), (10), the $\bar{p}p$ bound states
manifest themselves in the under-threshold behavior of the wave function
$\psi_{k_{u}}(r=0)=\frac{1}{f_{Jost}(-k_{u})}$, where $f_{Jost}(k)$ is the
Jost function. However, everywhere except for close vicinity of the $\bar{p}p$
bound state the re-scatterings and cancellation of the non-adiabatic and off-
shell effects change the value of $\psi_{k_{u}}(0)$, resulting in the shift
$k_{u}\to k_{v}$, where the energy $v$ is always positive that makes the
proton under-threshold form factor inaccessible in this domain in the reaction
$\bar{p}d\to e^{+}e^{-}n$. Closeness of the solid and dashed curves in the
figures shown in Sec. 5 confirms numerically the cancellation of the non-
adiabatic and off-shell effects proved analytically in DK81 ; DKK83 .

In Refs. DK81 ; DKK83  an experimental manifestation of the expression (3) for
the $\bar{p}d\to e^{+}e^{-}n$ amplitude was discussed (Fig. 3 from DK81  and
Fig. 7 from DKK83 ). It requires the experiments in which the neutron is
emitted in two opposite directions relative to momentum of incident
antiproton. Solving Faddeev equations, we restricted our calculations by the
S-waves only that corresponds to the replacement of Eq. (3) by (11).
Therefore, in our approximation, the amplitude is insensitive to the emitted
neutron momentum direction. However, in our simplified calculations, nothing
particular related to the S-wave was used to test eq. (11). We expect it
should be valid for the higher partial waves too. In our opinion, the
calculations presented in this paper and confirming Eq. (3) for the S-wave
increase the interest to the experimental test proposed in DK81 ; DKK83  which
involve the higher partial waves.

What would happen if we could gradually weaken the $\bar{p}p$ interaction so
that the $\bar{p}p$ re-scatterings become unimportant? For example, if
$V_{0}\to 0$ in the Hulthen potential, Eq. (39). We would come in the
situation considered in Sec. 4.2. In this case, the impulse approximation
dominates without any shift $k_{u}\to k_{v}$. This does not prevent from
possibility to access the under-threshold proton form factor. However, the
weak $\bar{p}p$ interaction does not create any bound states or resonances and
does not change significantly the ”bare” form factor $\Gamma_{0}(q^{2})$. It
would be still possible to go down under threshold, but, in this case, there
is no $\bar{p}p$ interaction there. The Jost function (42) (where $B\to 0$)
and the factor $\psi_{k}(0)$ tend to 1. So, the same reason - enough strong
$\bar{p}p$ interaction, which creates the baryonium, prevents from access to
the proton form factor (except for access to $\Gamma_{0}(q^{2})$) outside the
vicinities of the $\bar{p}p$ bound states in the reaction $\bar{p}d\to
e^{+}e^{-}n$. If this reason disappears, access becomes possible, but it will
be access to the desert from point of view of the $\bar{p}p$ interaction.

Acknowledgments The authors are grateful to J. Carbonell and R. Lazauskas for
useful recommendations on solving the Faddeev equations.

##  Appendix A Hulthen potential

For convenience, we will assume that both the proton and neutron and proton
and antiproton interact by means of the Hulthen potential

$V(r)=-V_{0}\frac{\exp(-\frac{r}{r_{0}})}{1-\exp(-\frac{r}{r_{0}})}=-\alpha\mu\frac{\exp(-\mu
r)}{1-\exp(-\mu r)},$ (39)

($V_{0}=\alpha\mu$, $r_{0}=1/\mu\leftrightarrow\alpha=V_{0}r_{0}$,
$\mu=1/r_{0}$), with the same parameter $r_{0}$, however, with different
parameters $V_{0}$. We assume that all these particles have equal masses $m$.
The wave function for the S-wave in the coordinate space is represented as

$\psi_{k}(r)=\frac{1}{\sqrt{4\pi}}\frac{\chi_{k}(r)}{r}.$

with

$\chi(r)=Ne^{-\kappa r}(1-e^{-\mu r}),$ (40)

where $\kappa=\frac{1}{2}(\alpha m-\mu)=\sqrt{B_{1}m}$ and
$N=\sqrt{2\kappa(\kappa+\mu)(2\kappa+\mu)}/\mu.$ This wave function is
normalized by Eq. (21).

The ground state binding energy for two particles of the equal masses $m$
reads (see e.g. flugge , Problem 68:

$B_{1}=|E_{1}|=\frac{1}{4m}(\alpha m-\mu)^{2}.$ (41)

The bound state exists if $\alpha m>\mu$. Everywhere $m$ is not reduced mass,
but mass of one of the particles.

For the Hulthen potential the Jost function $f_{Jost}(k)$ is known in
analytical form Newton :

$\displaystyle f_{Jost}(k)$ $\displaystyle=$
$\displaystyle\frac{\Gamma\left(1+2iA\right)}{\Gamma\left(1+iA-\sqrt{B^{2}-A^{2}}\right)}$
(42) $\displaystyle\times$
$\displaystyle\frac{1}{\Gamma\left(1+iA+\sqrt{B^{2}-A^{2}}\right)}.$

where we denoted:

$A=\sqrt{mE}r_{0}=kr_{0}=\frac{k}{\mu},\;B^{2}=mV_{0}r^{2}_{0}=\alpha\frac{m}{\mu}.$

$\Gamma(z)$ is the gamma-function.

As well known:

$\psi_{k}(r=0)=\frac{1}{f_{Jost}(-k)}.$ (43)

###  A.1 Deuteron and baryonium parameters

For the parameters $\hbar=197.3$ MeV$\cdot$fm, $m=938.3$ MeV, $r_{0}=0.6061$
fm, $\epsilon_{d}=-2.23$ MeV (deuteron), we find $V_{0}=144.7$ MeV.

For the same parameters except for $\epsilon_{b}=-15$ MeV (baryonium), we find
$V_{0}=195.3$ MeV.

For the same parameters except for $\epsilon_{b}=-55$ MeV (baryonium), we find
$V_{0}=270.6$ MeV.

##  Appendix B The exponential potential

In the exponential potential

$V(r)=-V_{0}\exp\left(-\frac{r}{r_{0}}\right)$ (44)

like in the Hulthen one, there are explicit analytical solutions for the
ground and scattering state wave functions, as well as the Jost function
Newton , Ch. 14.

The bound state wave function reads:

$\chi(r)=AJ_{\tilde{\nu}}\left[\alpha\exp\left(-\frac{r}{2r_{0}}\right)\right],$
(45)

where $\tilde{\nu}=2r_{0}\sqrt{m|E|},\;\alpha=2r_{0}\sqrt{mV_{0}}$ and
$J_{\tilde{\nu}}$ is the Bessel function of the first kind.

The eigenvalue equation follows from $\chi(0)=0$. It obtains the form

$J_{2r_{0}\sqrt{m|E|}}\left(2r_{0}\sqrt{mV_{0}}\right)=0$ (46)

The Jost function $f(k)$ reads111The Jost function (B) differs from Eq. (14.9)
in chapter 14, ref. Newton , by the complex conjugation. GW :

$\displaystyle f_{Jost}(k)$ $\displaystyle=$
$\displaystyle\left(\frac{\alpha}{2}\right)^{-\nu}J_{\nu}(\alpha)\Gamma(\nu+1)$
$\displaystyle=$ $\displaystyle\exp[-ir_{0}k\log(r_{0}^{2}mV_{0})]$
$\displaystyle\times$ $\displaystyle
J_{2ir_{0}k}\left(2r_{0}\sqrt{mV_{0}}\right)\Gamma(2ir_{0}k+1).$

The value $\psi_{k}(r=0)$ is given by Eq. (43). Therefore:

$\psi_{k}(r=0)=\frac{\exp\left[-ir_{0}k\log\left(r_{0}^{2}mV_{0}\right)\right]}{J_{-2ir_{0}k}\left(2r_{0}\sqrt{mV_{0}}\right)\Gamma(-2ir_{0}k+1)}.$
(48)

###  B.1 Deuteron and baryonium wave functions

For the parameters $\hbar=197.3$ MeV$\cdot$fm, $m=938.3$ MeV, $r_{0}=0.6061$
fm, $\epsilon_{d}=-2.23$ MeV (deuteron), we find $V_{0}=225.6$ MeV. The
deuteron wave function obtains the form ($r$ in $fm$, $\chi(r)$ in
$fm^{-1/2}$):

$\chi_{d}(r)=0.7023J_{\nu}(y),$ (49)

where $\nu=0.2810,\;y=2.826\exp(-0.8249r)$.

For the same parameters except for $\epsilon_{b}=-15$ MeV (baryonium), we find
$V_{0}=338.4$ MeV. The baryonium wave function reads ($r$ in $fm$, $\chi(r)$
in $fm^{-1/2}$):

$\chi_{b}(r)=1.175J_{\nu}(y),$ (50)

where $\nu=0.7288,\;y=3.462\exp(-0.8249r)$.

Both wave functions are normalized to 1 by Eq. (21).

##  Appendix C Three-body Jacobi coordinates and momenta

Three-body relative coordinates are defined as:

$\displaystyle\vec{x}_{3}=\vec{r}_{1}-\vec{r}_{2},\quad\vec{y}_{3}=\frac{2}{\sqrt{3}}\left(\frac{\vec{r}_{1}+\vec{r}_{2}}{2}-\vec{r}_{3}\right),$
$\displaystyle\vec{R}=\frac{\vec{r}_{1}+\vec{r}_{2}+\vec{r}_{3}}{3}.$ (51)

Two other sets of relative coordinates are obtained by cyclic permutation. In
particular:

$\displaystyle\vec{x}_{1}=\vec{r}_{2}-\vec{r}_{3},\quad\vec{y}_{1}=\frac{2}{\sqrt{3}}\left(\frac{\vec{r}_{2}+\vec{r}_{3}}{2}-\vec{r}_{1}\right),$
$\displaystyle\vec{R}=\frac{\vec{r}_{1}+\vec{r}_{2}+\vec{r}_{3}}{3}.$ (52)

The set $\vec{x}_{3},\vec{y}_{3}$ is expressed via $\vec{x}_{1},\vec{y}_{1}$:

$\vec{x}_{3}=-\frac{1}{2}(\vec{x}_{1}+\sqrt{3}\vec{y}_{1}),\quad\vec{y}_{3}=\frac{1}{2}(\sqrt{3}\vec{x}_{1}-\vec{y}_{1}).$
(53)

Inverse relations:

$\displaystyle\vec{r}_{1}=\vec{R}+\frac{1}{2}\vec{x}_{3}+\frac{1}{2\sqrt{3}}\vec{y}_{3},$
$\displaystyle\vec{r}_{2}=\vec{R}-\frac{1}{2}\vec{x}_{3}+\frac{1}{2\sqrt{3}}\vec{y}_{3},\quad\vec{r}_{3}=\vec{R}-\frac{1}{\sqrt{3}}\vec{y}_{3}.$

Substituting Eqs. (C) in the scalar product, we find:

$\vec{k}_{1}\makebox[2.27626pt]{$\cdot$}\vec{r}_{1}+\vec{k}_{2}\makebox[2.27626pt]{$\cdot$}\vec{r}_{2}+\vec{k}_{3}\makebox[2.27626pt]{$\cdot$}\vec{r}_{3}=\vec{q}_{3}\makebox[2.27626pt]{$\cdot$}\vec{x}_{3}+\vec{p}_{3}\makebox[2.27626pt]{$\cdot$}\vec{y}_{3}+\vec{P}\makebox[2.27626pt]{$\cdot$}\vec{R},$

where

$\displaystyle\vec{q}_{3}=\frac{1}{2}(\vec{k}_{1}-\vec{k}_{2}),\quad\vec{p}_{3}=\frac{1}{\sqrt{3}}\left(\frac{\vec{k}_{1}+\vec{k}_{2}}{2}-\vec{k}_{3}\right),$
$\displaystyle\vec{P}=\vec{k}_{1}+\vec{k}_{2}+\vec{k}_{3}.$ (55)

and similarly for another set:

$\displaystyle\vec{q}_{1}=\frac{1}{2}(\vec{k}_{2}-\vec{k}_{3}),\quad\vec{p}_{1}=\frac{1}{\sqrt{3}}\left(\frac{\vec{k}_{2}+\vec{k}_{3}}{2}-\vec{k}_{1}\right),$
$\displaystyle\vec{P}=\vec{k}_{1}+\vec{k}_{2}+\vec{k}_{3}.$ (56)

which provides:

$\vec{k}_{1}\makebox[2.27626pt]{$\cdot$}\vec{r}_{1}+\vec{k}_{2}\makebox[2.27626pt]{$\cdot$}\vec{r}_{2}+\vec{k}_{3}\makebox[2.27626pt]{$\cdot$}\vec{r}_{3}=\vec{q}_{1}\makebox[2.27626pt]{$\cdot$}\vec{x}_{1}+\vec{p}_{1}\makebox[2.27626pt]{$\cdot$}\vec{y}_{1}+\vec{P}\makebox[2.27626pt]{$\cdot$}\vec{R},$

We will use only these coordinates and momenta. We don’t use any other
combinations called the ”relative” coordinates and momenta. However, we will
also use the particle momenta themselves defined in the c.m. frame of
reaction.

Let us express the Jacobi momenta $\vec{p}_{1}$, $\vec{p}_{3}$ through the
c.m. ones. Let the particle 1 is the neutron and the particle 2 is the proton
and the particle 3 is the antiproton. That is, in the c.m. frame:
$\vec{k}_{1}=\vec{p}_{n},\;\vec{k}_{2}=\vec{p}_{p}$ and
$\vec{k}_{3}=\vec{p}_{\bar{p}}$. In these notations:

$\displaystyle\vec{p}_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{3}}\left(\frac{\vec{p}_{p}+\vec{p}_{\bar{p}}}{2}-\vec{p}_{n}\right),$
$\displaystyle\vec{p}_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{3}}\left(\frac{\vec{p}_{n}+\vec{p}_{p}}{2}-\vec{p}_{\bar{p}}\right).$
(57)

In the c.m. frame:
$\vec{p}_{n}+\vec{p}_{p}+\vec{p}_{\bar{p}}=0\to\vec{p}_{p}=-\vec{p}_{n}-\vec{p}_{\bar{p}}$.
Therefore:

$\vec{p}_{1}=-\frac{\sqrt{3}}{2}\vec{p}_{n},\quad\vec{p}_{3}=-\frac{\sqrt{3}}{2}\vec{p}_{\bar{p}}.$
(58)

##  Appendix D Test of unitarity

The fulfillment of unitarity provides very strong test of the solution of the
Faddeev equations. We take the incident antiproton momentum which is not
enough for the deuteron breakup. We will consider two cases: (i) the
$\bar{p}p$ system has no bound states; (ii) the $\bar{p}p$ system has bound
state (baryonium).

In the case (i), the $\bar{p}d$ scattering is elastic. Hence, the
corresponding phase shift $\delta$ is real. In this case, the test of
unitarity is reduced to the test that $\delta$ is real. The amplitude
$f_{3}(p_{3})$ in Eq. (23) reads $f_{3}(p_{3})=\frac{1}{2i}(\exp(2i\delta)-1)$
(we use the definition of $f$ without momentum $p_{3}$ in the denominator).
Then $\exp(2i\delta)=1+2if_{3}(p_{3})$. We will check that
$|\exp(2i\delta)|=1$, that is

$|1+2if_{3}(p_{3})|=1.$ (59)

The results for $p_{\bar{p}}=23.9,\;48.3$ MeV/c are given in the Table 1.

Table 1: The amplitude $f_{3}(p_{\bar{p}})$ for elastic $\bar{p}d$ scattering for $V_{0\bar{p}p}=10$ MeV and $r_{0}=0.6061$ fm $p_{\bar{p}}$, MeV/c | 23.9 | 48.3
---|---|---
$f_{3}(p_{\bar{p}})$ | $0.0202+0.0004i$ | $0.0370+0.0014i$
$2if_{3}(p_{\bar{p}})+1$ | $0.9992+0.0404i$ | $0.9972+0.0740i$
$|2if_{3}(p_{\bar{p}})+1|$ | $1.0000$ | $0.9999$
|  |

The last line of this table contains the values
$|\exp(2i\delta)|=|1+2if_{3}(p_{3})|$. We see that $|\exp(2i\delta)|\approx 1$
with high accuracy. This proves that $\delta$ is real, that confirms the
validity of the numerical solution of the Faddeev equation.

In the case of open channel with the baryonium creation the value
$|\exp(2i\delta)|$ is smaller than 1. This indicates that the part of the
final particles goes in this channel. Inclusion of this channel should restore
the unitarity condition. This channel is taken into account below.

We deal now with the 6D space (two the 3D Jacobi coordinates
$\vec{X}=(\vec{x}_{i},\vec{y}_{i})$). For $(\vec{x}_{i},\vec{y}_{i})$ one can
chose any pair of the Jacobi coordinates. Unitarity means the conservation of
flax:

$\vec{j}=\frac{i\hbar}{2m}(\Psi\vec{\nabla}_{x_{i}}\Psi^{*}-\Psi^{*}\vec{\nabla}_{x_{i}}\Psi+\Psi\vec{\nabla}_{y_{i}}\Psi^{*}-\Psi^{*}\vec{\nabla}_{y_{i}}\Psi),$
(60)

where $\vec{\nabla}_{x_{i}}=\frac{\partial}{\partial\vec{x}_{i}}$,
$\vec{\nabla}_{y_{i}}=\frac{\partial}{\partial\vec{y}_{i}}$. The conservation
of flax is expressed by the formula

$\oint\vec{j}d\vec{S}=0,$ (61)

where $d\vec{S}$ is element of surface in the 6D space. We integrate over the
sphere of large radius $\rho=\sqrt{\vec{x}^{2}+\vec{y}^{2}}\to\infty$, where
the three-body wave function is known and is given by its asymptotic form. In
this way, Eq. (61) provides a relation (the unitarity condition) between the
amplitudes $f_{3},f_{1}$. It has the form:

$|1+2if_{3}(p_{3})|^{2}=1-\frac{4p_{1}}{p_{3}}|f_{1}(p_{1})|^{2}.$ (62)

If the rearrangement channel is closed, that is, $f_{1}=0$, the equation (62)
turns into (59).

Table 2: The amplitude $f_{3}(p_{3})$ and $f_{1}(p_{1})$ for $|\epsilon_{b}|=55$ MeV, $V_{0\bar{p}p}=270.6$ MeV and $r_{0}=0.6061$ fm $p_{\bar{p}}$, MeV/c | 23.9 | 48.3
---|---|---
$p_{3}$, MeV/c | $20.7$ | $41.8$
$p_{1}$, MeV/c | $223.5$ | $226.3$
$f_{3}(p_{3})$ | $-0.13+0.02i$ | $-0.25+0.07i$
$f_{1}(p_{1})$ | $-0.00-0.01i$ | $-0.02-0.02i$
$|1+2if_{3}(p_{3})|^{2}$ | $0.9886$ | $0.9763$
$1-\frac{4p_{1}}{p_{3}}|f_{1}(p_{1})|^{2}$ | $0.9879$ | $0.9761$
|  |

In the last two lines of the Table 2 we see that the values
$|1+2if_{3}(p_{3})|^{2}$ and $1-\frac{4p_{1}}{p_{3}}|f_{1}(p_{1})|^{2}$
coincide with each other within the precision better than 0.1%.

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