---
author:
- |
    J. P. B. C. de Melo$^{(a)}$, A. E. A. Amorim$^{(b)}$, Lauro Tomio$^{(c)}$ and T.Frederico$^{(d)}$\
    $^{(a)}$ Division de Physique Théorique, Institut de Physique Nucléaire, 91406 Orsay CEDEX, and Laboratoire de Physique Théorique et Particules Elémentaires,\
    Universté Pierre et Marie Curie, 4 Place Jussieau, 75252 Paris CEDEX 05, France\
    $^{(b)}$ Faculdade de Tecnologia de Jahu, CEETPS, Jahu, Brasil.\
    $^{(c)}$ Instituto de Física Teórica, UNESP\
    01405 São Paulo, SP, Brasil\
    $^{(d)}$ Dept. de Física, Instituto Tecnológico da Aeronáutica,CTA\
    12.228-900 S. José dos Campos, SP, Brasil
title: '**Relativistic Bound States in 2+1 and 1+1 Dimensions in the Null-Plane [^1]**'
---

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\#1[/\#1]{}

The Faddeev-like equation for the component of the three-boson vertex for a relativistic contact interaction is [@fred92] $$\begin{aligned}
v(q^\mu)=2\tau(M_2) \int \frac{d^nk}{(2\pi)^n}
\frac{i}{k^2-M^2+i{\varepsilon}} \frac{i}{(P_3-q-k)^2-M^2+i{\varepsilon}}v(k^\mu),
\label{1}\end{aligned}$$ where $P^\mu_3=(M_{3B},0,0,0)$ is the three-boson four-momentum in the center of mass system, $n$ is the dimension of the space-time, the mass of the two-boson subsystem is given by $M_2^2=(P_3-q)^2$ and the single boson mass is $M$. The factor 2 comes from symmetrization of the total vertex.

The total three-boson vertex is the sum of three Faddeev components in which each boson is spectator once [@fred92]. The two-boson scattering amplitude, $\tau(M_2)$, enters in the kernel of the integral equation for the vertex in Eq.(\[1\]). It is easily obtained as: $$\begin{aligned}
\tau^{(n)}(M_2)= \left\{i\lambda^{-1} - B^{(n)}(M_2)\right\}^{-1}, 
\label{2}\end{aligned}$$ where $\lambda$ is the coupling constant of the zero-range interaction, $M_2$ is the mass of the two boson system and $B(M_2)$ is the kernel of the integral equation for the scattering amplitude $$\begin{aligned}
 B^{(n)}(M_2)=-\int \frac{d^nk}{(2\pi)^n}
\left\{\left(k^2-M^2+i{\varepsilon}\right)\left((P-k)^2-M^2+i{\varepsilon}\right)
\right\}^{-1},
\label{3}\end{aligned}$$ where $M$ is the boson mass and $P^\mu=(M_2,0,0,0)$ .

The value of $\lambda$ is chosen such that the two-boson system has one bound-state. The scattering amplitude, Eq.(\[2\]), has a pole at the bound-state mass $(M_{2B})$, which yields $$\begin{aligned}
i\lambda^{-1}= B^{(n)}(M_{2B}). 
\label{4}\end{aligned}$$

In two dimensions the two-boson scattering amplitude for $M_2 \ < \ 2M$ is $$\begin{aligned}
\tau^{(2)}(M_2) = - 2\pi i \left\{ 
\frac{atan\left(2\beta(M_{2B})\right)^{-1}}
{M^{2}_{2B}\beta(M_{2B})}
 - \frac{atan\left(2\beta(M_{2})\right)^{-1}} 
{M^{2}_{2}\beta(M_{2})}
\right\}^{-1} \ ,
\label{5}\end{aligned}$$ where $ \beta(M_2)=\sqrt{\frac{M^2}{M_{2}^2}-\frac14}$ .

In three dimensions the scattering amplitude is $$\begin{aligned}
\tau^{(3)}(M_2)=- 8\pi i \left\{ M^{-1}_{2B} 
ln\left(\frac{2M+M_{2B}}{2M-M_{2B}} 
\right)
-M^{-1}_{2} ln\left(\frac{2M+M_{2}}{2M-M_{2}} 
\right)
\right\}^{-1} \ .
\label{6} \end{aligned}$$ For our purpose of the bound-state calculation is enough to know $\tau^{(n)}(M_2)$ for $M_2 < 2M$ .

The momentum variables in the integral equation are the momenta in the null-plane for an on-mass-shell particle, $q^+$ and $q_\perp$ [@fred92]. The transversal momentum is needed in three space-time dimensions.

Let us discuss the limits of the variables $y=\frac{q^+}{M_{3B}}$ and $q_\perp$. In 1+1 space-time dimensions only the momentum fraction is enough to describe the spectator boson. The mass of the two-boson subsystem must be real and in 1+1 dimensions it implies $$\begin{aligned}
(M_2)^2\ = \ (M_{3B}-q^+)\left(M_{3B}-\frac{
M^2}{q^+}\right) \ > \ 0 \ . 
\label{7}\end{aligned}$$ From the above inequality follows $ 1\ > \ y \ > \ \frac{M^2}{M^2_{3B}} \ . $

In 2+1 dimensions, we deduce the range of values of the perpendicular momentum allowed by the reality of the mass of the two-boson subsystem. Then $$\begin{aligned}
(M_2)^2\ = \ (M_{3B}-q^+)\left(M_{3B}-\frac{q^2_\perp
+M^2}{q^+}\right)-q^2_\perp \ > \ 0 \ . 
\label{8}\end{aligned}$$ Solving the inequality for $q^2_\perp$ , we obtain $ q^2_\perp\ < \ (1-y)(M_{3B}^2y-M^2) \ . $ The limits for $y$ are $ 1 \ > \  y \ > \ \frac{M^2}{M^2_{3B}} \ , $ and the lower bound comes from $q^2_\perp \ > \ 0$.

The equation for the Faddeev component of the vertex in 1+1 space-time dimensions is obtained as the result of the $k^-$ integration in the momentum loop of Eq.(\[1\]). We also use Eq.(\[5\]) and the limit in the internal momentum fraction $x$ $$\begin{aligned}
v(y)=\frac{i}{2\pi}\tau^{(2)}(M_2) \int^{1-y}_\frac{M^2}{M^2_{3B}}\frac{dx}{x(1-y-x)}
\frac{v(x)}{M^2_{3B}-M^2_{03}},
\label{9}\end{aligned}$$ where $(M_2)^2$ is given by Eq.(\[7\]) and the free mass of the virtual three-boson state in 1+1 dimensions is: $$\begin{aligned}
M^2_{03}=
\frac{M^2}{x}+
\frac{M^2}{y} + \frac{M^2}{1-y-x} \ .
\label{10}\end{aligned}$$

The equation for the Faddeev component of the vertex in 2+1 space-time dimensions is found after the $k^-$ integration of in the momentum loop of Eq.(\[1\]), $$\begin{aligned}
v(y,\vec q_\perp)=\pi^{-2}
\tau^{(3)}(M_2)
\int^{1-y}_\frac{M^2}{M^2_{3B}}\frac{dx}{x(1-y-x)}
\int^{k_\perp^{max}}_{-k_\perp^{max}}d^2k_\perp
\frac{v(x,\vec k_\perp)}{M^2_{3B}-M^2_{03}},
\label{11}\end{aligned}$$ where $M_2$ is given by Eq.(\[8\]), as well as $ k_\perp^{max}=\sqrt{(1-x)(M_{3B}^2x-M^2)}$ . The mass of the virtual three-boson state is: $$\begin{aligned}
 M^2_{03}=
\frac{k^2_\perp+M^2}{x}+
\frac{q^2_\perp+M^2}{y} + \frac{(q+k)^2_\perp +M^2}{1-y-x} \ .
\label{12}\end{aligned}$$

The dependence of $v$ on $q^-$ is not specified because the spectator boson is on mass-shell. $q^+$ and $\vec q_\perp$ describe the spectator boson propagation. The relativistic equations in 1+1 and 2+1 dimensions, have a lower bound for the mass of the three-boson system which comes from the limits on the $x$ integration and the condition $y>\frac{M^2}{M^2_{3B}}$ which implies $ M_{3B}> \sqrt{2}M \ . $ The same limit was obtained in 3+1 dimensions in [@fred92].

The Faddeev component of the ground state vertex, in 2+1 dimensions is rotationally symmetric in the x-y plane in Eq.(\[11\]). The mass of the boson gives the scale of the system. Here, the solution is presented for $M=1$. In Fig.(1), the numerical results for the ground-state binding energies $(E_{3B}=M_{2B}+M-M_{3B})$ of the three-boson system are shown in two and three-dimensions. In the nonrelativistic limit, for $E_{2B}=0$, the results approach the well-known values [@dodd].

In summary, we give an example of how null-plane dynamics can be elaborated. We develop a zero-range model of the three-boson bound state in the null-plane and solve numerically the dynamical equation for the ground-state in 1+1 and 2+1 space-time dimensions.

This work was supported by Conselho Nacional de Desenvolvimento e Pesquisa - CNPq and Fundação de Amparo a pesquisa do Estado de São Paulo - FAPESP. J. P. B. C. de Melo is a FAPESP-Brazil fellow (contract 97/13902-8).

[99]{} T.Frederico, Phys. Lett. [**B282**]{}, (1992) 409; W.R.B. de Araújo, J.P.B.C. de Melo and T.Frederico, Phys.Rev. [**C52**]{}, (1995) 2733. L.R. Dodd, J. Math. [**11**]{}, (1970) 207.

\[fig1\]

[^1]: To appear in “Proceedings VI Hadrons 1998”, Florianópolis, Santa Catarina, Brazil
