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

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 ${\bar{p}}d \rightarrow e^+e^-n$) 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.

Appendices

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

$$\begin{aligned} V(r)=-V_0\frac{\exp (-\frac{r}{r_0})}{1-\exp (-\frac{r}{r_0})}=-\alpha \mu \frac{\exp (-\mu r)}{1-\exp (-\mu r)}, \end{aligned}$$

(39)

($V_0=\alpha \mu$, $r_0 =1/\mu \leftrightarrow \alpha =V_0r_0$, $\mu =1/r_0$), 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

$$\begin{aligned} \psi _k(r)=\frac{1}{\sqrt{4\pi }}\frac{\chi _k(r)}{r}. \end{aligned}$$

with

$$\begin{aligned} \chi (r)=N e^{-\kappa r}(1-e^{-\mu r}), \end{aligned}$$

(40)

where $\kappa \!=\!\frac{1}{2}(\alpha m\!-\!\mu )\!=\!\sqrt{B_1m}$ 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. [16], Problem 68):

$$\begin{aligned} B_1=\vert E_1\vert =\frac{1}{4m}(\alpha m-\mu )^2. \end{aligned}$$

(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 [17]:

$$\begin{aligned} f_{Jost}(k)= & {} \frac{\Gamma \left( 1 + 2iA\right) }{\Gamma \left( 1+iA - \sqrt{B^2-A^2}\right) } \nonumber \\{} & {} \times \frac{1}{\Gamma \left( 1+iA + \sqrt{B^2-A^2}\right) }. \end{aligned}$$

(42)

where we denoted:

$$\begin{aligned} A=\sqrt{mE} r_0=k r_0=\frac{k}{\mu },\, B^2=mV_0 r^2_0=\alpha \frac{m}{\mu }. \end{aligned}$$

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

As well known:

$$\begin{aligned} \psi _{k}(r=0)=\frac{1}{f_{Jost}(-k)}. \end{aligned}$$

(43)

1.1 Appendix 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

$$\begin{aligned} V(r)=-V_0\exp \left( -\frac{r}{r_0}\right) \end{aligned}$$

(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 [17], Ch. 14.

The bound state wave function reads:

$$\begin{aligned} \chi (r)=AJ_{{\tilde{\nu }}}\left[ \alpha \exp \left( -\frac{r}{2r_0}\right) \right] , \end{aligned}$$

(45)

where ${\tilde{\nu }}=2r_0\sqrt{m\vert E\vert },\,\alpha =2r_0\sqrt{mV_0}$ and $J_{{\tilde{\nu }}}$ is the Bessel function of the first kind.

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

$$\begin{aligned} J_{2r_0\sqrt{m\vert E\vert }}\left( 2r_0\sqrt{mV_0}\right) =0 \end{aligned}$$

(46)

The Jost function f(k) reads¹ [18]:

$$\begin{aligned} f_{Jost}(k)= & {} \left( \frac{\alpha }{2}\right) ^{-\nu }J_{\nu }(\alpha )\Gamma (\nu +1) \nonumber \\= & {} \exp [-ir_0k\log (r_0^2mV_0)] \nonumber \\{} & {} \times J_{2ir_0k}\left( 2r_0\sqrt{mV_0}\right) \Gamma (2ir_0k+1). \end{aligned}$$

(47)

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

$$\begin{aligned} \psi _{k}(r=0)=\frac{\exp \left[ -ir_0k\log \left( r_0^2mV_0\right) \right] }{J_{-2ir_0k}\left( 2r_0\sqrt{mV_0}\right) \Gamma (-2ir_0k+1)}. \end{aligned}$$

(48)

1.1 Appendix 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 $fm^{-1/2}$):

$$\begin{aligned} \chi _d(r)=0.7023J_{\nu }(y), \end{aligned}$$

(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 $fm^{-1/2}$):

$$\begin{aligned} \chi _b(r)=1.175J_{\nu }(y), \end{aligned}$$

(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:

$$\begin{aligned} \vec {x}_3= & {} \vec {r}_1-\vec {r}_2,\quad \vec {y}_3=\frac{2}{\sqrt{3}}\left( \frac{\vec {r}_1+\vec {r}_2}{2}-\vec {r}_3\right) , \nonumber \\ \vec {R}= & {} \frac{\vec {r}_1+\vec {r}_2+\vec {r}_3}{3}. \end{aligned}$$

(51)

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

$$\begin{aligned} \vec {x}_1= & {} \vec {r}_2-\vec {r}_3,\quad \vec {y}_1=\frac{2}{\sqrt{3}}\left( \frac{\vec {r}_2+\vec {r}_3}{2}-\vec {r}_1\right) , \nonumber \\ \vec {R}= & {} \frac{\vec {r}_1+\vec {r}_2+\vec {r}_3}{3}. \end{aligned}$$

(52)

The set $\vec {x}_3,\vec {y}_3$ is expressed via $\vec {x}_1,\vec {y}_1$:

$$\begin{aligned} \vec {x}_3=-\frac{1}{2}(\vec {x}_1+\sqrt{3}\vec {y}_1), \quad \vec {y}_3=\frac{1}{2}(\sqrt{3}\vec {x}_1-\vec {y}_1). \end{aligned}$$

(53)

Inverse relations:

$$\begin{aligned} \vec {r}_1= & {} \vec {R}+\frac{1}{2}\vec {x}_3+\frac{1}{2\sqrt{3}}\vec {y}_3, \nonumber \\ \vec {r}_2= & {} \vec {R}-\frac{1}{2}\vec {x}_3+\frac{1}{2\sqrt{3}}\vec {y}_3,\quad \vec {r}_3=\vec {R}-\frac{1}{\sqrt{3}}\vec {y}_3. \end{aligned}$$

(54)

Substituting Eqs. (54) in the scalar product, we find:

$$\begin{aligned} \vec {k}_1\!\cdot \!\vec {r}_1+\vec {k}_2\!\cdot \!\vec {r}_2+\vec {k}_3\!\cdot \!\vec {r}_3= \vec {q}_3\!\cdot \!\vec {x}_3+\vec {p}_3\!\cdot \!\vec {y}_3+\vec {P}\!\cdot \!\vec {R}, \end{aligned}$$

where

$$\begin{aligned} \vec {q}_3= & {} \frac{1}{2}(\vec {k}_1-\vec {k}_2),\quad \vec {p}_3=\frac{1}{\sqrt{3}}\left( \frac{\vec {k}_1+\vec {k}_2}{2}-\vec {k}_3\right) , \nonumber \\ \vec {P}= & {} \vec {k}_1+\vec {k}_2+\vec {k}_3. \end{aligned}$$

(55)

and similarly for another set:

$$\begin{aligned} \vec {q}_1= & {} \frac{1}{2}(\vec {k}_2-\vec {k}_3),\quad \vec {p}_1=\frac{1}{\sqrt{3}}\left( \frac{\vec {k}_2+\vec {k}_3}{2}-\vec {k}_1\right) , \nonumber \\ \vec {P}= & {} \vec {k}_1+\vec {k}_2+\vec {k}_3. \end{aligned}$$

(56)

which provides:

$$\begin{aligned} \vec {k}_1\!\cdot \!\vec {r}_1+\vec {k}_2\!\cdot \!\vec {r}_2+\vec {k}_3\!\cdot \!\vec {r}_3= \vec {q}_1\!\cdot \!\vec {x}_1+\vec {p}_1\!\cdot \!\vec {y}_1+\vec {P}\!\cdot \!\vec {R}, \end{aligned}$$

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:

$$\begin{aligned} \vec {p}_1= & {} \frac{1}{\sqrt{3}}\left( \frac{\vec {p}_p+\vec {p}_{{\bar{p}}}}{2}-\vec {p}_n\right) , \nonumber \\ \vec {p}_3= & {} \frac{1}{\sqrt{3}}\left( \frac{\vec {p}_n+\vec {p}_p}{2}-\vec {p}_{{\bar{p}}}\right) . \end{aligned}$$

(57)

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

$$\begin{aligned} \vec {p}_1=-\frac{\sqrt{3}}{2}\vec {p}_n,\quad \vec {p}_3=-\frac{\sqrt{3}}{2}\vec {p}_{{\bar{p}}}. \end{aligned}$$

(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 $\vert \exp (2i\delta )\vert =1$, that is

$$\begin{aligned} \vert 1+2if_3(p_3)\vert =1. \end{aligned}$$

(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

The last line of this table contains the values $\vert \exp (2i\delta )\vert =\vert 1+2if_3(p_3)\vert$. We see that $\vert \exp (2i\delta )\vert \approx 1$ 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 $\vert \exp (2i\delta )\vert$ 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:

$$\begin{aligned} \vec {j}=\frac{i\hbar }{2m}(\Psi \vec {\nabla }_{x_i}\Psi ^*-\Psi ^*\vec {\nabla }_{x_i}\Psi +\Psi \vec {\nabla }_{y_i}\Psi ^*-\Psi ^*\vec {\nabla }_{y_i}\Psi ),\nonumber \\ \end{aligned}$$

(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

$$\begin{aligned} \oint \vec {j} d\vec {S}=0, \end{aligned}$$

(61)

where $d\vec {S}$ is element of surface in the 6D space. We integrate over the sphere of large radius $\rho =\sqrt{\vec {x}^2+\vec {y}^2}\rightarrow \infty$, 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:

$$\begin{aligned} \vert 1+2if_3(p_3)\vert ^2=1-\frac{4p_1}{p_3}\vert f_1(p_1)\vert ^2. \end{aligned}$$

(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 $\vert \epsilon _b\vert =55$ MeV, $V_{0{\bar{p}}p}=270.6$ MeV and $r_0=0.6061$ fm

In the last two lines of the Table 2 we see that the values $\vert 1+2if_3(p_3)\vert ^2$ and $1-\frac{4p_1}{p_3}\vert f_1(p_1)\vert ^2$ coincide with each other within the precision better than 0.1%.