Relativistic oscillator model with spin for nucleon resonances
I.V. Uvarov
A.P. Yaroshenko
{\sl 72 Gagarin avenue, Dniepropetrovsk 49010, Ukraine}
The relativistic three-body problem is approached via the extension of the $SL(2,C)$ group to the $Sp(4,C)$ one. In terms of $Sp(4,C)$ spinors, a Dirac-like equation with three-body kinematics is composed. After introducing the linear in coordinates interaction, it describes the spin-1/2 oscillator. For this system, the exact energy spectrum is derived and then applied to fit the Regge trajectories of baryon N-resonances. Having only two parameters, the model is in overall agreement with the experiment.
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\large
\title{\textbf{Relativistic oscillator model with spin for nucleon resonances}}

% revised 22.09.2011

\author{ D.A. Kulikov\thanks{kulikov\_d\_a@yahoo.com, kulikov@dsu.dp.ua},
I.V. Uvarov, A.P. Yaroshenko
\\
\\
{\sl Theoretical Physics Department, FFEKS, Dniepropetrovsk
National
University }\\
{\sl 72 Gagarin avenue, Dniepropetrovsk 49010, Ukraine} }

\date{}
\maketitle



\begin{abstract}
The relativistic three-body problem is approached via the
extension of the $SL(2,C)$ group to the $Sp(4,C)$ one. In terms of
$Sp(4,C)$ spinors, a Dirac-like equation with three-body
kinematics is composed. After introducing the linear in
coordinates interaction, it describes the spin-1/2 oscillator. For
this system, the exact energy spectrum is derived and then applied
to fit the Regge trajectories of baryon N-resonances. Having only
two parameters, the model is in overall agreement with the
experiment.
\end{abstract}

%\newpage

% main text
\section{Introduction}
\label{part1}

%The harmonic oscillator occupies an important place in nuclear and
%particle physics. In hadron physics

A relativistic equation for the symmetric quark model with
harmonic interaction was proposed by Feynman, Kislinger and
Ravndal (FKR) \cite{feynman} as far back as 1971. Since their
work, the concept of relativistic oscillator has been used for
describing the spectra of both the  ordinary hadrons
\cite{mitra,toki,ishida,bohm} and glueballs \cite{semay}. Of
course, these models are purely phenomenological ones. But, due to
its exact solvability and simplicity of the spectrum with levels
grouped into shells, the relativistic oscillator provides a
convenient first approximation in hadron systematics. It is
especially important for light-quark baryons where one should cope
with a large amount of experimental data on the excited states.

%the energy levels develop a shell structure which manifests itself
%experimentally in a series of resonance regions.

The original FKR model for baryons is based on the mass squared
operator  \cite{feynman}
\begin{equation}\label{eq0}
\mathcal{K}_\mathrm{FKR}=3(p_1^2+p_2^2+p_3^2)+\frac{1}{36}\Omega^2[(x_1-x_2)^2+
(x_2-x_3)^2+(x_3-x_1)^2]+C
\end{equation}
constructed from the four-momenta $p_i$ ($i=1,2,3$) of three
quarks and the conjugate positions $x_i$. Since the corresponding
eigenvalues are a succession of integers times $\Omega$, the mass
squared grows linearly with the angular momentum in general
agreement with experiment, but the price paid is the high
degeneracy of the spectrum. This degeneracy has been removed in
further algebraic approaches such as the interacting boson model
\cite{toki} and the stringlike collective model \cite{bijker},
which distinguish between excitations of different kinds.


% Kirchbach \textit{et al.}

It should be stressed that all the above models are formulated
assuming spinless quarks and thus they classify baryons according
to the orbital angular momentum $L$. However, in relativity, only
the total angular momentum $J$ and not its parts is defined. A
relativistic description of the light-baryon excitation spectrum
in terms of $J$ and parity was obtained in Ref.\cite{o4} by
employing the Lorentz group representations of the
Rarita-Schwinger type. Instead of the mass operator with an
explicit interaction like Eq.~(\ref{eq0}), there the Hamiltonian
as a function of the Casimir operators of the symmetry group is
postulated. From the representations involved, the authors of
Ref.\cite{o4} infer that their model has the spin content given by
$J=L\pm 1/2$.


% and place their model in the quark-diquark picture \cite{diquark}.


The question arises as to whether it is possible to incorporate
spin into the baryon model in an alternative and straightforward
way, by taking the "square root" from a second-order oscillator
operator. It is known that this is true in the case of the
one-body Klein-Gordon equation where one ends up with the Dirac
equation linear in both the momentum and the position, the
so-called Dirac oscillator model \cite{moshinsky}.


The purpose of this work is to construct a relativistic three-body
oscillator model based on an appropriate Dirac-like equation with
interaction. Our consideration makes use of the connection between
the Lorentz symmetry and the $SL(2,C)$ symmetry in the spinor
space. We apply the approach for composing relativistic wave
equations that starts with the extension of the $SL(2,C)$ group to
the symplectic $Sp(4,C)$ one \cite{kulikov}. In Ref.\cite{kulikov}
the fermion-boson problem was studied and an exactly solvable
model for the two-body oscillator with spin-1/2 was offered. In
the present work we generalize this model to the three-body case
and show that the exact solvability survives, but now the spectrum
has the $(J+N)$-degeneracy, which resembles the one seen in the
nucleon resonance spectrum.


The plan of the work is as follows. In Section 2 we perform the
symplectic space-time extension to obtain the relativistic
three-body kinematics. In Section 3 the three-body system with the
oscillator interaction involved through generalized momenta linear
in coordinates is studied. Section 4 is devoted to incorporating
the spin in the preceding results. Here we consider the Dirac-like
equation with the interaction and derive the corresponding energy
spectrum. This spectrum is applied in Section 5 to the description
of the nucleon excitations. Our conclusion is given in Section 6.




\section{Three-body kinematics based on the extension of the $SL(2,C)$
group} \label{part2}

In this Section we sketch out the procedure of the symplectic
space-time extension and apply it to construct a relativistic
operator with the three-body kinematics in spirit of the FKR
operator (\ref{eq0}).

Let us recall that the construction of relativistic wave equations
in the Minkowski space relies on the symmetry with the
$Sp(2,C)\equiv SL(2,C)$ group, which governs the transformations
of two-component Weyl spinors.  It is a universal covering group
for the homogeneous Lorentz group $SO(1,3)$. As a consequence,
there exists a one-to-one correspondence between $Sp(2,C)$
Hermitian spin-tensors of second rank and Minkowski four-vectors.
It allows one to parametrize the four-momentum of a relativistic
particle by the $Sp(2,C)$ Hermitian spin-tensor and to write down
the Dirac equation in terms of the Weyl spinors \cite{landau}.

In order to describe few-particle systems, we extend the
symplectic $Sp(2,C)$ group to the $Sp(4,C)$ one. This is the
minimal extension that preserves a non-degenerate antisymmetric
bilinear form $\eta_{\alpha\beta}=-\eta_{\beta\alpha}$
($\alpha,\beta=1,2,3,4$) in the spinor space.

Consider the $Sp(4,C)$ Hermitian spin-tensor
$\mathcal{P}_{\alpha\bar{\alpha}}$ (hereafter bared indices refer
to complex conjugate spinors). According to our previous analysis
\cite{kulikov}, it can be decomposed into four Minkowski
four-momenta as
\begin{equation}\label{eq8}
\mathcal{P}=I\otimes\sigma^m w_m+\tau^1\otimes\sigma^m p_m+
\tau^2\otimes\sigma^m u_m+\tau^3\otimes\sigma^m q_m
\end{equation}
where $w_m$, $p_m$, $u_m$, $q_m$ $(m= 0,1,2,3)$ are the Minkowski
four-momenta, and the following representation with $2\times 2$
unit matrix $I$ and the Pauli matrices $\tau^i$ is used
\begin{eqnarray}\label{eq4}
\sigma^0=I, \quad \sigma^1=\tau^1, \quad \sigma^2=\tau^2, \quad
\sigma^3=\tau^3.
\end{eqnarray}
Note that the second factor in the direct matrix products in
Eq.~(\ref{eq8}) is the $Sp(2,C)$ momentum spin-tensor, while the
first one is due to the group extension.

It should be stressed that the description of a three-body system
requires one time-like and nine space-like variables, whereas the
$Sp(4,C)$ momentum spin-tensor has sixteen components. However, we
are able to decrease the number of the independent components by
introducing subsiduary conditions in a $Sp(4,C)$-invariant form.

For deriving such conditions, we multiply the $Sp(4,C)$ momentum
spin-tensor $\mathcal{P}_{\alpha\bar{\alpha}}$ by the transposed
one $\tilde{\mathcal{P}}^{\bar{\alpha}\beta}$, to obtain the
Klein-Gordon-like operator
\begin{equation}\label{eq8a}
\mathcal{K}\equiv\mathcal{P}\tilde{\mathcal{P}}=w^2+p^2-u^2+q^2+\sum^{5}_{A=1}\gamma_A
K^A
\end{equation}
where $w^2=(w^0)^2-\mathbf{w}^2$, $p^2=(p^0)^2-\mathbf{p}^2$ etc,
$\gamma_A$ are direct products of the Pauli matrices, and $K^A$
are quadratic forms with respect to the four-momenta.

Five quantities $K^A$ are components of a complex vector that
transforms according to the representation $SO(5,C)\subset
Sp(4,C)$. To restore the diagonal form of the Klein-Gordon
operator, we put $K^A=0$ on wave functions. Such a condition is
invariant under the $Sp(4,C)$ group transformations because if a
vector equals to zero in one frame, then it equals to zero in all
frames.

Being written in terms of the four-momenta, the imposed equality
$K^A=0$ reads
\begin{eqnarray}\label{eq8b}
&&wp+pw=0,\qquad wq+qw=0,\qquad up+pu=0,\qquad uq+qu=0, \nonumber \\
&&u^m w^n + w^n u^m - u^n w^m - w^m u^n-\epsilon^{mnkl}(p_k q_l +
q_l p_k)=0
\end{eqnarray}
where $\epsilon^{mnkl}$ is the totally antisymmetric tensor
($\epsilon^{0123}=+1$).


These conditions imply that either one or three of the
four-momenta $w_m$, $p_m$, $u_m$ and $q_m$ must transform as axial
vectors. Let $u_m$ be the sole axial vector. Then we may connect
the remaining four-momenta with the four-momenta $p_{1}^m$,
$p_{2}^m$ and $p_{3}^m$ of the constituent particles in the
standard manner \cite{feynman}
\begin{equation}\label{eq9}
w^m=p_{1}^m+p_{2}^m+p_{3}^m ,\quad
p^m=\sqrt{\frac{3}{2}}\;(p_{1}^m-p_{2}^m), \quad
q^m=\frac{1}{\sqrt{2}}\;(p_{1}^m+p_{2}^m-2p_{3}^m).
\end{equation}

Supposing the total four-momentum $w_m$ to be conserved, for an
arbitrary four-vector $a_m$ we can introduce its transverse and
longitudinal, with respect to $w_m$, parts
\begin{equation}\label{eq185}
a_{\bot}^m=(g^{mn}-w^m w^n/w^2)a_n, \qquad a_{\parallel}^m=(w^m
w^n/w^2)a_n
\end{equation}
where $g^{mn}=diag(1,-1,-1,-1)$ is the Minkowski metrics.

With this notation, the subsiduary conditions (\ref{eq8b}) are
reduced to
\begin{equation}\label{eq19}
p^{m}_{\parallel}=0, \qquad q^{m}_{\parallel}=0, \qquad u_{\bot
m}=\frac{1}{w^2}\epsilon_{mnkl}w^n p^k q^l.
\end{equation}
From the first two equalities it becomes evident that the relative
time variables are  removed, as it is necessary for the three-body
problem \cite{droz06}. The last equality shows that the axial
$u_m$ is an auxiliary quantity and only its longitudinal  part
$u^{m}_{\parallel}$ remains independent.

Upon inserting Eqs.~(\ref{eq9}) and (\ref{eq19}), the
Klein-Gordon-like operator (\ref{eq8a}) takes the form
\begin{equation}\label{eq10}
\mathcal{K}=3(p_1^2+p_2^2+p_3^2)-u^{2}_{\parallel}
+\frac{1}{w^2}[p^{2}_{\bot}q^{2}_{\bot}-(p_{\bot}q_{\bot})^2],
\end{equation}
to be compared with the kinetic part of the FKR operator
(\ref{eq0}). We see that the yet undetermined quantity
$u^{2}_{\parallel}$ substitutes the additive constant $C$
introduced in the FKR model to account for such effects as the
difference of quark masses. In our model, intended to describe the
nucleon resonances only, we put $u^{m}_{\parallel}=0$. The last
term in Eq.~(\ref{eq10}) has no analog in the FKR operator. But
this term vanishes in the non-relativistic limit where $w^2$
containing the rest energy dominates over space-like
$p^{m}_{\bot}$ and $q^{m}_{\bot}$.

Thus, within the approach based on the extension of the $SL(2,C)$
group, the kinematics of three non-interacting particles is
described by the Klein-Gordon-like operator (\ref{eq10})
supplemented with the subsiduary conditions (\ref{eq19}). Mention
that the two-body kinematics can be obtained now by imposing the
further restriction $q^{m}=0$ (see Ref.\cite{kulikov} for
details).


\section{Oscillator interaction}
\label{part3}

Now we are going to include the interaction in the description.
This can be made by replacing the four-momenta of particles by the
generalized momenta ($p_i^m\rightarrow\pi_i^m=p_i^m-A_i^m$,
$i=1,2,3$), so that each particle is in an external potential of
the others. We assume that both the Klein-Gordon-like operator
(\ref{eq10}) (properly symmetrized) and the subsiduary conditions
(\ref{eq19}) are subject to these replacements.

%kglike (\ref{eq10})

Because the generalized momenta do not, generally, commute with
each other, the question on the compatibility arises. In the
language of the Dirac's quantum mechanics with constraints,
Eqs.~(\ref{eq19}) and (\ref{eq10}) are the first-class
constraints. Then a sufficient condition of the compatibility
implies that their mutual commutators do vanish without producing
second-class constraints.

We choose the simplest generalized momenta that meet the above
compatibility requirement and satisfy the subsiduary conditions of
the same form (\ref{eq8b}) as in the non-interacting case. Namely,
this is the interaction linear in the coordinates of particles in
spirit of the Dirac oscillator model \cite{moshinsky}
\begin{eqnarray}\label{eq20}
&&\pi^m_1=p^m_1-\frac{\lambda}{3\sqrt{3}}
(x^m_{2\bot}-x^m_{3\bot}), \quad
\pi^m_2=p^m_2-\frac{\lambda}{3\sqrt{3}} (x^m_{3\bot}-x^m_{1\bot}),
\nonumber \\
&&\phantom{\pi^m_1=p^m_1-} \pi^m_3=p^m_3-\frac{\lambda}{3\sqrt{3}}
(x^m_{1\bot}-x^m_{2\bot})
\end{eqnarray}
where $\lambda$ is a coupling constant.

In terms of the relative momenta Eq.~(\ref{eq20})  translates into
\begin{equation}\label{eq20a}
p^m\rightarrow P^m=p^m-\lambda y_{\bot}^m, \qquad q^m\rightarrow
Q^m=q^m+\lambda x_{\bot}^m
\end{equation}
with the relative positions defined as the Jacobi coordinates
\begin{equation}\label{eq21}
x^m=\frac{x_{1}^m-x_{2}^m}{\sqrt{6}}, \qquad
y^m=\frac{x_{1}^m+x_{2}^m-2x_{3}^m}{3\sqrt{2}},
\end{equation}
which obey $[p^m,x^n]=\mathrm{i}g^{mn}$,
$[q^m,y^n]=\mathrm{i}g^{mn}$.

As a consequence, the following commutation relation holds
 \begin{equation}\label{eq21a}
[P^m,Q^n]=2\mathrm{i}\lambda (g^{mn}- w^m w^n/w^2)
\end{equation}
that resembles the commutator of the generalized momenta for a
charged particle in the magnetic field. Associated with the Landau
levels, the latter system is indeed tightly connected with the
harmonic oscillator.

%and results in the energy spectrum similar to the Landau levels
%\cite{landauV3}. Note also that, with the generalized momenta
%(\ref{eq20a}) inserted, the subsiduary conditions (\ref{eq8b})
%remain unaltered in the form and satisfy the compatibility
%requirement as before.

%Now let us show that the included interaction is the oscillator
%one similar to the interaction in the model of the Dirac
%oscillator \cite{moshinsky}. Remind that the Dirac oscillator is
%obtained from the one-particle Dirac Hamiltonian by the
%non-minimal substitution for the momentum, $\mathbf{p}\rightarrow
%\mathbf{p}-\mathrm{i}m \omega\beta \mathbf{x}$, where $\omega$ is
%the oscillator frequency and $\beta$ is the Dirac matrix.
%Inserting Eq.~(\ref{eq20a}) into the $Sp(4,C)$ momentum spin-tensor
%(\ref{eq8}) and rearranging its terms, so that each relative
%momentum should be accompanied with the corresponding coordinate,
%we get
%\begin{eqnarray}\label{eq22}
%P&=&\Sigma^0\otimes\sigma^m w_m+
%\Sigma^1\otimes\sigma^m(p_m-\mathrm{i}\lambda \Sigma^2\otimes I
%x_m) \\ \nonumber &+&\Sigma^2\otimes\sigma^m r_m
%+\Sigma^3\otimes\sigma^m (q_m-\mathrm{i}\lambda \Sigma^2\otimes I
%y_m).
%\end{eqnarray}
%It is seen that the expressions in parentheses in the last
%equation involve the Hermitian matrix times imaginary unit,
%similarly to the non-minimal substitution used in the Dirac
%oscillator model.


To separate oscillator degrees of freedom in our three-body
system, let us consider its three-dimensional reduction. Although
it may be performed in a covariant manner, by using the
decomposition (\ref{eq185}), we prefer a more illustrative
approach and pass to the center-of-mass (CM) frame in which
$\mathbf{w}=0$. Then $E=w^0$ is the total energy and the dynamics
of the relative motion is described by the three-dimensional
coordinates $\mathbf{x}_{\bot}=\mathbf{x}$ and
$\mathbf{y}_{\bot}=\mathbf{y}$.

From Eq.~(\ref{eq19}) it follows that $P^0=Q^0=0$ and the
Klein-Gordon-like operator (\ref{eq10}), rewritten through the
vectors of generalized relative momenta $\mathbf{P}$ and
$\mathbf{Q}$, becomes
\begin{equation}\label{eq210}
-\mathcal{K}=\mathbf{P}^2+\mathbf{Q}^2-E^2+\frac{1}{E^2}(\mathbf{Q}\times\mathbf{P})^2.
\end{equation}
If one now introduces the creation and annihilation operators
\begin{eqnarray}\label{eq230}
\mathbf{c}=\frac{\mathbf{Q}+\mathrm{i}\mathbf{P}}{2\sqrt{\lambda}},
\qquad
\mathbf{c}^{\dag}=\frac{\mathbf{Q}-\mathrm{i}\mathbf{P}}{2\sqrt{\lambda}},
\end{eqnarray}
%which satisfy
%\begin{equation}\label{eq232}
%[c_a , c_b^{\dag}]=\delta_{ab}, \qquad [c_a , c_b]=0, \qquad
%[c_a^{\dag} , c_b^{\dag}]=0 \quad (a,b,c=1,2,3),
%\end{equation}
the first two terms in the right-hand side of Eq.~(\ref{eq210})
can be identified with the operator counting oscillator quanta,
namely,
\begin{equation}\label{eq240}
\mathbf{P}^2+\mathbf{Q}^2=4\lambda\left(\mathbf{c}^{\dag}\cdot\mathbf{c}
+\frac{3}{2}\right).
\end{equation}
As for the last term in Eq.~(\ref{eq210}), it commutes with
$\mathbf{P}^2+\mathbf{Q}^2$ and thus represents a relativistic
correction that does not change the number of quanta.

It should be emphasized that the model under consideration
exploits its degrees of freedom in a different way than the FKR
model. Within the latter, all six degrees of freedom associated
with the relative motion are of the harmonic oscillator nature. In
contrast to this, our model obviously possesses only one
three-dimensional oscillator mode through the operators
$\mathbf{c}$ and $\mathbf{c}^{\dag}$.

To reveal the three other degrees of freedom, we look at the
orbital angular momentum operator for the system,
$\mathbf{x}\times\mathbf{p}+\mathbf{y}\times\mathbf{q}$. In the
presence of the interaction, its elements can be rearranged as
\begin{equation}\label{eq242}
\mathbf{x}\times\mathbf{p}+\mathbf{y}\times\mathbf{q}=\mathbf{M}+\mathbf{N}
\end{equation}
where
\begin{equation}\label{eq244}
\mathbf{M} =
\frac{1}{2\lambda}(\mathbf{q}+\lambda\mathbf{x})\times
(\mathbf{p}-\lambda\mathbf{y})\equiv\frac{1}{2\lambda}\mathbf{Q}\times\mathbf{P},
\qquad \mathbf{N}
=-\frac{1}{2\lambda}(\mathbf{q}-\lambda\mathbf{x})\times
(\mathbf{p}+\lambda\mathbf{y})
\end{equation}
both obey the algebra for angular momenta, $[L_a ,
L_b]=\mathrm{i}\epsilon_{abc}L_c$ $(a,b,c=1,2,3)$, $\mathbf{L}
=\mathbf{M}$ or $\mathbf{N}$. But $\mathbf{M}$ also satisfies the
commutation relations with $\mathbf{P}$ and $\mathbf{Q}$ thought
as a linear momentum and a position respectively
\begin{equation}\label{eq246}
[M_a , P_b]=\mathrm{i}\varepsilon_{abc}P_c, \qquad [M_a ,
Q_b]=\mathrm{i}\varepsilon_{abc}Q_c,
\end{equation}
whereas $\mathbf{N}$ commutes with  $\mathbf{P}$ and $\mathbf{Q}$
and, hence, is conserved. This last property of $\mathbf{N}$
indicates that there must exist rotational degrees of freedom not
coupled to the oscillator mode. Their dynamics can be determined
by picking up an eigenstate of $\mathbf{N}$ squared and of its
projection. We will elaborate more on this point in the next
Section.

Keeping in mind the application to the baryon spectrum, we treat
the above partition of the degrees of freedom as due to the
emergence of a diquark, two-quark cluster, in the interacting
three-quark system. It is known that existence of diquarks is
supported by various models (see Ref.\cite{diquark} for review)
and the quark-diquark picture provides a simpler classification of
the excited nucleon states \cite{anisovichb}. From this point of
view, the three oscillator degrees of freedom of our system in the
CM frame correspond to the interaction between a quark and a
diquark, while the three remaining ones to rotational excitations.




\section{Oscillator with spin and its energy spectrum}
\label{part4}

In order to describe baryons, we shall incorporate spin in the
preceding results. We take the "square root"\; from the
Klein-Gordon-like operator to obtain the Dirac-like equation. Then
the analytical formulae for eigenenergies of the three-body system
with spin are derived.

\subsection{Dirac-like three-body equation}

Consider the system with spin equal to $1/2$. In practice, this
will be a baryon consisting of a spin-1/2 quark and a spinless (or
"good" in the standard terminology) diquark. The wave function of
this system can be represented by a pair of Weyl spinors. Using
the four-component $Sp(4,C)$ Weyl spinors $\varphi_{\alpha}$,
$\bar{\chi}^{\bar{\alpha}}$ and the momentum spin-tensor
$\mathcal{P}_{\alpha\bar{\alpha}}$ given by Eq.~(\ref{eq8}), one
may compose the wave equation
\begin{equation}\label{eq7}
\mathcal{P}_{\alpha\bar{\alpha}}\bar{\chi}^{\bar{\alpha}}=m\varphi_{\alpha},
\qquad
\tilde{\mathcal{P}}^{\bar{\alpha}\alpha}\varphi_{\alpha}=m\bar{\chi}^{\bar{\alpha}},
\end{equation}
with  $m$ being a mass parameter.

Assuming the same oscillator interaction (\ref{eq20a}) as in the
previous Section, in the CM frame Eq.~(\ref{eq7}) reduces to
\begin{eqnarray}\label{eq250}
&&\left(E-\tau^1\otimes\boldsymbol\tau\cdot\mathbf{P}
-\tau^3\otimes\boldsymbol\tau\cdot\mathbf{Q}-\frac{2\lambda}{E}\tau^2\otimes
\boldsymbol\tau\cdot\mathbf{M}\right)\bar{\chi}=m\varphi, \nonumber \\
&&\left(E+\tau^1\otimes\boldsymbol\tau\cdot\mathbf{P}
+\tau^3\otimes\boldsymbol\tau\cdot\mathbf{Q}-\frac{2\lambda}{E}\tau^2\otimes
\boldsymbol\tau\cdot\mathbf{M}\right)\varphi=m\bar{\chi},
\end{eqnarray}
which can be brought conveniently to a Hamiltonian form. By
introducing
\begin{equation}\label{eq252}
\Psi=\left(\begin{array}{c}
       \psi_{+} \\
       \psi_{-} \\
     \end{array}\right)=
     \left(\begin{array}{c}
       (\bar{\chi}+\varphi)/\sqrt{2} \\
       (\bar{\chi}-\varphi)/\sqrt{2} \\
     \end{array}\right),
\end{equation}
we get the equation with the energy-dependent Dirac-like
Hamiltonian
\begin{equation}\label{eq254}
H\Psi=E\Psi, \,\, H=(\tau^1\otimes\boldsymbol\tau\cdot\mathbf{P}
+\tau^3\otimes\boldsymbol\tau\cdot\mathbf{Q})\left(\begin{array}{cc}
       0 & 1\\
       1 & 0\\
     \end{array}\right)+m\left(\begin{array}{cr}
       1 & 0\\
       0 & -1\\
     \end{array}\right)+\frac{2\lambda}{E}\,\tau^2\otimes\boldsymbol\tau\cdot\mathbf{M}.
\end{equation}

Now it is crucial to find a complete set of operators that commute
with the Hamiltonian and among themselves. Their eigenvalues will
supply us with quantum numbers labeling a state of the system.

It is easy to verify that the total angular momentum
\begin{equation}\label{eq256}
\mathbf{J}=\mathbf{x}\times\mathbf{p}+\mathbf{y}\times\mathbf{q}+I\otimes\frac{\boldsymbol{\tau}}{2}\equiv
\mathbf{M}+\mathbf{N}+I\otimes\frac{\boldsymbol{\tau}}{2}
\end{equation}
commutes with $H$. Moreover, the operators
$\mathbf{J}\cdot\mathbf{N}$ and $\mathbf{N}^2$ commute with both
$H$ and $\mathbf{J}^2$. This amounts to say that, apart from the
total spin $J$ defined by $\mathbf{J}^2\Psi=J(J+1)\Psi$, there
must exist other good quantum numbers associated with rotational
degrees of freedom.

Physically, only whole system rotations could be defined in the
presence of interaction and the sole preferred direction in the CM
frame should be specified by $\mathbf{J}$. To account for this
fact, we "froze"\, unphysical rotations, by picking up only those
states for which $\mathbf{N}$ is parallel to $\mathbf{J}$ and
thereby produces no new preferred direction. Namely, among the
solutions to the eigenvalue equations
\begin{equation}\label{eq257}
\mathbf{N}^2\Psi=N(N+1)\Psi,\qquad
\left(\mathbf{M}+I\otimes\boldsymbol{\tau}/2\right)^2\Psi=j(j+1)\Psi,
\end{equation}
we select the states with $j=J+N$. According to (\ref{eq256}),
these states satisfy the condition of parallelism,
$(\mathbf{J}\cdot\mathbf{N})\Psi=-JN\Psi$, where the minus sign is
necessary since $+JN$ does not belong to the set of eigenvalues of
$(\mathbf{J}\cdot\mathbf{N})$. It should be added that $N$ may
attain only integer values $0,1,...$ and not half-integer ones
because $\mathbf{N}$ is the differential operator in the
coordinate space.

%The conservation of $\mathbf{N}$ implies that the energy spectrum
%of the model possesses an "accidental" degeneracy.

The set of mutually commuting operators also includes the total
angular momentum projection $J_3$, the spin-orbit coupling
operator $\kappa$ and the constant matrix $\varsigma$ given by
\begin{equation}\label{eq258}
\kappa=I\otimes(\boldsymbol\tau\cdot\mathbf{M}+I)\left(\begin{array}{cr}
1 & 0\\
      0 & -1\\
     \end{array}\right),\qquad \varsigma=\tau^2\otimes I \left(\begin{array}{cr}
1 & 0\\
      0 & -1\\
     \end{array}\right).
\end{equation}

The matrix $\varsigma$ reflects the doubling of the number of
spinor components versus the ordinary Dirac equation. It can be
checked that, by applying the projectors $(1\pm \varsigma)/2$, the
three-body equation (\ref{eq254}) is split into two separate ones
for two four-component Dirac bispinors. In view of this doubling
it is temptimg to interpret the symmetry with respect to the
unitary transformation generated by $\varsigma$ as a remnant of
the isospin symmetry between the proton and the neutron.
Remarkably, the symmetry is not broken by the oscillator
interaction.

Now let us derive an explicit oscillator-type equation for our
system. This can be made by taking the square of the Hamiltonian
$H$ and then subtracting an appropriate term linear in $H$, to
diagonalize the matrix. Namely, we evaluate $[H^2-(4/E)\lambda
\varsigma \kappa H]\Psi$, to get the second-order equation
\begin{equation}\label{eq259}
(\mathbf{P}^2+\mathbf{Q}^2)\psi_{\pm}=\left[E^2-\left( m-
\frac{2\lambda\varsigma}{E}\right)^2 \pm 2\lambda\varsigma
+\frac{4\lambda^2 \kappa^2}{E^2} \right]\psi_{\pm}
\end{equation}
that describes a harmonic oscillator with the additional
spin-orbit interaction, which enters through the operator
$\kappa$.




\subsection{Energy spectrum}

We are in the position to calculate the energy spectrum of the
system. First, it should be noticed that in the last equation the
operator $\varsigma$, whose eigenvalues are $\pm 1$, is
accompanied with the coupling constant $\lambda$. We therefore
absorb $\varsigma$ into the definition of $\lambda$ and look for
two branches of the spectrum corresponding to $\lambda>0$ and
$\lambda<0$, respectively.

%consider the cases of  for $\lambda>0$ and $\lambda<0$.

The problem merely reduces to expressing the eigenvalues of the
operators involved in Eq.~(\ref{eq259}) in terms of the observed
spin-parity $J^P$. The left-hand side is just the operator
(\ref{eq240}) counting the number of the oscillator quanta. In
view of the algebra (\ref{eq246}), this number can be partitioned
in the standard manner for the isotropic oscillator \cite{elliott}
as $(2n+M)$ where $n=0,1,...$ is the radial quantum number and
$M=0,1,...$ is the orbital quantum number defined by
$\mathbf{M}^2\psi=M(M+1)\psi$, with $\psi$ standing for one of
$\psi_{+}$ and $\psi_{-}$ from the decomposition (\ref{eq252}).

Since $\mathbf{M}^2$ is not conserved, $\psi_{+}$ and $\psi_{-}$
refer to different values of $M$. From Eqs.~(\ref{eq256}) and
(\ref{eq258}) we deduce that these values and also the parity are
unambiguosly determined by the conserved total angular momentum
$J$ and the eigenvalue $\kappa=\pm (J+N+1/2)$ of the spin-orbit
coupling operator as
\begin{equation}\label{eq260}
M=J+N\mp\frac{\kappa}{2|\kappa|}, \qquad
P=(-1)^{J+N-\kappa/2|\kappa|}
\end{equation}
where the upper (lower) sign inside $M$ refers to $\psi_{+}$
($\psi_{-}$).

Collecting (\ref{eq240}), (\ref{eq259}) and (\ref{eq260}) we
arrive at the dispersion relations
\begin{eqnarray}\label{eq270}
\left(E-\frac{2|\lambda|(J+N+1/2)}{E}\right)^2-\left( m-
\frac{2\lambda}{E}\right)^2=4|\lambda|\left(2n+\frac{|\lambda|-\lambda}{2|\lambda|}\right),
\; P=(-1)^{J+N-1/2}, %\;  \kappa=+ (J+N+\frac12)
\nonumber \\
\left(E-\frac{2|\lambda|(J+N+1/2)}{E}\right)^2-\left( m-
\frac{2\lambda}{E}\right)^2=4|\lambda|\left(2n+\frac{|\lambda|+\lambda}{2|\lambda|}\right),
\; P=(-1)^{J+N+1/2}. \nonumber \\ %\;  \kappa=- (J+N+\frac12)
\end{eqnarray}
Here the first and second lines were obtained by using the
equations for $\psi_{+}$ and $\psi_{-}$, respectively.

The ground-state ($n=0$) solutions for which one of the components
vanishes need a special care. As seen from the structure of the
Hamiltonian (\ref{eq254}), such solutions are obtained by setting
$(\tau^1\otimes\boldsymbol\tau\cdot\mathbf{P}
+\tau^3\otimes\boldsymbol\tau\cdot\mathbf{Q})\psi=0$ and, if
present, must have energy in agreement with the general formulae
(\ref{eq270}).

%\begin{equation}\label{eq270}
%(\tau^1\otimes\boldsymbol\tau\mathbf{P}
%+\tau^3\otimes\boldsymbol\tau\mathbf{Q})\psi=0
%\end{equation}

In the case of $\lambda>0$, the last equation admits a
normalizable solution for $\psi_{+}$ and the corresponding ground
state with $\psi_{-}=0$ is characterized by
\begin{equation}\label{eq280}
\left(E-\frac{2\lambda(J+N+1/2)}{E}\right)-\left( m-
\frac{2\lambda}{E}\right)=0, \quad P=(-1)^{J+N-1/2} \quad
(\lambda>0)
\end{equation}
that agrees with the first line of Eqs.~(\ref{eq270}) with $n=0$.

In the case of $\lambda<0$, there exists the ground state with
$\psi_{+}=0$ possessing
\begin{equation}\label{eq290}
\left(E-\frac{2|\lambda|(J+N+1/2)}{E}\right)+\left( m-
\frac{2\lambda}{E}\right)=0, \quad P=(-1)^{J+N+1/2} \quad
(\lambda<0)
\end{equation}
that falls into the second line of Eqs.~(\ref{eq270}).

The explicit solutions to Eqs.~(\ref{eq270}) written in the form
of the Regge trajectories for the bound states of the three-body
system are, for $\lambda>0$,
%\begin{eqnarray}\label{eq300}
%&&J=\frac{E^2}{2\lambda}-N-\frac{1}{2}-\sqrt{\frac{2nE^2}{\lambda}
%+\left(\frac{mE}{2\lambda}-1\right)^2},
%\quad P=(-1)^{J+N-1/2}, \nonumber \\
%&&J=\frac{E^2}{2\lambda}-N-\frac{1}{2}-\sqrt{\frac{(2n+1)E^2}{\lambda}
%+\left(\frac{mE}{2\lambda}-1\right)^2}, \quad P=(-1)^{J+N+1/2},
%\end{eqnarray}
%and, for $\lambda<0$,
%\begin{eqnarray}\label{eq310}
%&&J=\frac{E^2}{2|\lambda|}-N-\frac{1}{2}-\sqrt{\frac{(2n+1)E^2}{|\lambda|}
%+\left(\frac{mE}{2 \lambda}-1\right)^2},
%\quad P=(-1)^{J+N-1/2}, \nonumber \\
%&&J=\frac{E^2}{2|\lambda|}-N-\frac{1}{2}-\sqrt{\frac{2nE^2}{|\lambda|}
%+\left(\frac{mE}{2 \lambda}-1\right)^2}, \quad P=(-1)^{J+N+1/2}.
%\end{eqnarray}
$$
\begin{array}{llr}
{\displaystyle
  J=\frac{E^2}{2\lambda}-N-\frac{1}{2}-\sqrt{\frac{2nE^2}{\lambda}
+\left(\frac{mE}{2\lambda}-1\right)^2}, } & P=(-1)^{J+N-1/2},  & (32a) \\
{\displaystyle
 J=\frac{E^2}{2\lambda}-N-\frac{1}{2}-\sqrt{\frac{(2n+1)E^2}{\lambda}
+\left(\frac{mE}{2\lambda}-1\right)^2},} & P=(-1)^{J+N+1/2},  & (32b) \rule{0pt}{10mm} \\
\end{array}
$$
and, for $\lambda<0$,
$$
\begin{array}{llr}
{\displaystyle
  J=\frac{E^2}{2|\lambda|}-N-\frac{1}{2}-\sqrt{\frac{(2n+1)E^2}{|\lambda|}
+\left(\frac{mE}{2 \lambda}-1\right)^2}, } & P=(-1)^{J+N-1/2},  & (33a) \\
{\displaystyle
 J=\frac{E^2}{2|\lambda|}-N-\frac{1}{2}-\sqrt{\frac{2nE^2}{|\lambda|}
+\left(\frac{mE}{2 \lambda}-1\right)^2}, } & P=(-1)^{J+N+1/2}.  & (33b) \rule{0pt}{10mm} \\
\end{array}
$$

It is evident that these two sets transform one into each other
under the simultaneous change $m\rightarrow -m$,
$\lambda\rightarrow -\lambda$ and $P\rightarrow -P$. The inversion
of $P$ does not matter because only the relative parity can be
defined for fermions. Thus the situation resembles a classical
model of symmetry breakdown: a point-like classical particle
moving on the line, under the sole influence of the w-shaped
potential $V (x) = (x-a)^2 (x+a)^2$. In this model there are two
positions of stable equilibrium, $x=a$ and $x=-a$, which are
transmuted one into each other by the parameter redefinition,
$a\rightarrow -a$. However, only one of them has to be selected to
get a physical picture.

In our three-body model we have a hint how to choose between the
branches with $\lambda>0$ and  $\lambda<0$. The ground-state
solutions stemming from the general formulae (32) and (33) must be
consistent with those given by Eqs.~(\ref{eq280}) and
(\ref{eq290}), respectively. This requires checking the sign when
taking the square root in (32) and (33) with $n=0$. Notice that
for all relevant energies ($E$ is high enough to yield $J>0$) the
sign of $(mE/2\lambda-1)$ coincides with that of $m/\lambda$. As a
consequence, the $\lambda>0$ solutions turn out to be consistent
if $m>0$, whereas the $\lambda<0$ solutions agree if $m<0$. In the
other cases, one should take an opposite sign in front of the
square root, for the states with $n=0$, that is unnatural. We
therefore fix, for definiteness, $m>0$ and treat only the
$\lambda>0$ solutions (32) as the physical ones.


It is worth adding that the obtained spectrum is similar to that
of the two-body oscillator with spin which we considered in
Ref.~\cite{kulikov} (see Eqs.~(20) there, note that $m$ and
$\lambda$ were rescaled). The main difference is the presence of
the new quantum number $N$ that has no analog in the two-body
case. Thus, if the two-body oscillator of Ref.~\cite{kulikov} may
be viewed as a quark-diquark model with a rigid diquark, the
present three-body treatment accounts for some extra excitations
deforming the diquark.

Since the derived Regge trajectories are nearly linear in the
squared energy and have the same slope in asymptotics, we may
apply them to describe the spectrum of nucleon resonances.


%for our three-body system,


\section{Application. Regge trajectories of nucleon resonances}
\label{part5}

We shall describe the $N$-resonance states and omit the
$\Delta$-resonances. The reason is that in the quark-diquark
picture the ground state $\Delta(1232)$ as well as its radial
excitations correspond to the spin $S=3/2$ and thus can hardly be
described within the Dirac-like equation we use.

To start with, consider the nucleon $N(940)$, the lightest state
having $J^P={1/2}^{+}$. From  Eq.~(32a) or, equivalently, from
Eq.~(\ref{eq280}) it follows that the energy of the ground state
with $n=0$, $N=0$ and $J^P={1/2}^{+}$ is $E=m$, {\it i.e.}, the
value of the mass parameter $m=0.940\,$GeV is unambiguously
determined by the nucleon mass. The remaining slope parameter
$\lambda=0.345\,$GeV$^2$ was chosen so as to fit the nucleon Regge
trajectory that also contains the well-established states
$N_{5/2+}(1680)$ and $N_{9/2+}(2220)$. We did not distinguish
between this trajectory and that for the negative-parity states
$N_{3/2-}(1520)$ and $N_{7/2-}(2190)$. Likewise, all other
trajectories were thought to contain the states with $J=1/2$,
$3/2$, $5/2$,... and the alternating parity.

It should be pointed out that within our model the Regge
trajectories can be obtained in three different ways. First, one
can get the opposite-parity states with $J^P=1/2^-$, $3/2^+$,
$5/2^-$,..., by switching from Eq.~(32a) to Eq.~(32b). Next, there
exist radial excitations with $n=1,2,...$ Last, one should
consider the Regge trajectories with $N=1,2,...$ which are
obtained by shifting the $N=0$ trajectories down to 1,2,... units
in $J$. It is the trajectories of this third type, generated from
the nucleon trajectory, that correspond to lighter states and thus
shall comprise most of the experimental points.

The calculated Regge trajectories are presented in Figs.~1 and 2.
In Fig.~1 we plot the trajectories of $1/2^+$, $3/2^-$,... states:
the nucleon Regge trajectory, its successor with $N=2$ and the
first radially excited trajectory with $n=1$, $N=0$. Fig.~2 shows
the opposite-parity states: the nucleon successors with $N=1$ and
$N=3$ along with the ground-state trajectory ($n=0$, $N=0$) that
was calculated using the opposite-parity formula (32b). The
experimental masses are taken from Particle Data Group \cite{pdg}.
We display the experimental uncertainties if they are high enough.
The unclear states for which the approximate masses are only known
are depicted by empty circles.

% $N_{3/2+}(1900)$, $N_{3/2-}(2080)$, $N_{1/2-}(2090)$, $N_{1/2+}(2100)$ and $N_{5/2-}(2200)$

From Figs.~1 and 2 one can see that on the plotted trajectories
all the $J=1/2$ and $J=3/2$ states below $2100$~MeV correspond to
certain experimentally observed resonances. In particular, the
$N=2$ and $N=3$ trajectories seem to contain $N_{3/2-}(2080)$ and
$N_{1/2-}(2090)$ respectively. In its turn, the radially excited
trajectory in Fig.~1 passes through $N_{1/2+}(2100)$.

On the other hand, several states drop out of our systematics and
we view them as including a different diquark configuration or no
diquark at all. These are the Roper resonance $N_{1/2+}(1440)$,
 $N_{5/2-}(1675)$ and $N_{9/2-}(2250)$. Actually, the position of the Roper resonance is
a longstanding problem since both the conventional three-quark
model \cite{Isgur} and quark-diquark scheme \cite{anisovichb}
treat it as the radial excitation which is unexpected to be
lighter than the first negative-parity excited state
$N_{3/2-}(1535)$. As for $N_{5/2-}(1675)$ and $N_{9/2-}(2250)$,
whose experimental masses are close to those of the
positive-parity states with the same spin, their absence indicates
that our model does not imply the parity doubling  -- the occurrence of degenerate opposite-parity
energy levels due to the chiral symmetry restoration
\cite{DoublGlozmanCohen2,jaffe,
DoublAfonin}. Nevertheless, the model accidentally
gives us the approximate parity doublets: $N_{3/2+}(1900)$ and
$N_{3/2-}(2080)$, $N_{1/2+}(2100)$ and $N_{1/2-}(2090)$.

To complete the comparison, we list the low-lying states predicted
by the present model as well as by some other models in Table 1.
When comparing these results, one should keep in mind that the
number of fitting parameters ranges from two in the present,
AdS/QCD \cite{forkel} and Skyrme \cite{MK} models to seven in the
relativized model \cite{CI}.

Inspecting Table 1, one observes degenerate states with increasing
$J$ among the model predictions. The high degeneracies that occur
within the present approach and the AdS/QCD model \cite{forkel}
deserve some explanation. Within the latter model there exists
$(L+n)$-degeneracy where $L$ and $n$ are the orbital and radial
quantum numbers respectively. This implies that intrinsic orbital
and spin angular momenta can be assigned to the observed states --
the assumption that is feasible since the spin-orbital coupling is
small for baryons. In its turn, our oscillator model involves the
conservation of the total angular momentum part $N$, which is
different from the orbital and spin ones (see Eqs.~(\ref{eq244})
and (\ref{eq256})). Let us stress that this conservation emerges
only after the inclusion of the oscillator interaction. The
resulting $(J+N)$-degeneracy is a bit weaker than the
$(L+n)$-degeneracy of the AdS/QCD approach. For example, our model
predicts that $N_{5/2-}(2200)$ should be substantially heavier
than $N_{1/2-}(2090)$, $N_{3/2-}(2080)$ and $N_{7/2-}(2190)$ which
are degenerate in both models.

\begin{table}\label{tab1}
\caption{Comparison between the calculated masses of nucleon
resonances and the experimental masses \cite{pdg}. This work:
Eqs.~(32), FK: AdS/QCD model by Forkel and Klempt \cite{forkel}
(values are cited from \cite{klempt}), CI: Capstick and Isgur
\cite{CI}, BIL: Bijker, Iachello and Leviatan \cite{bijker}, S:
quark-diquark model by Santopinto \cite{santopinto}, BnA and BnB:
Bonn model \cite{Bn}, MK: Skyrme model by Karliner and Mattis
\cite{MK}.}
\begin{tabular}{ c c c c c c c c c c}
\hline
Resonance &  Exp  & This work  & FK & CI & BIL & S & BnA & BnB & MK\\
\hline
$N(940)$ & $940$ & $940$   & $943$ & $960$ & 939 & 940 & $939$ & $939$ & $1190$\\
$N_{1/2-}(1535)$ & $1535\pm10$ & $1424$  & $1516$ & $1460$ & $1563$ & $1538$ & $1435$ & $1470$ & $1478$\\
$N_{1/2-}(1650)$ & $1660\pm18$ & $1705$ & $1628$ & $1535$ & 1683 & 1675 & $1660$ & $1767$ \\
$N_{1/2+}(1710)$ & $1710\pm30$ & $1735$  & $1735$ & $1770$ & $1683$ & $1640$ & $1729$ & $1778$ & $1427$\\
$N_{1/2-}(2090)$ & $\approx 2090$ & $1983$  & $2102$ & $2135$ & & & $2200$ & $2180$\\
$N_{1/2+}(2100)$ & $\approx 2100$  & $2098$ & $2017$ & $1975$ & & & $2127$ & $2177$\\

$N_{3/2-}(1520)$ & $1520\pm5$ & $1424$   & $1516$ & $1495$ & $1563$ & $1538$ & $1476$ & $1485$ & $1715$\\
$N_{3/2+}(1720)$ & $1725\pm25$ & $1735$  & $1735$ & $1795$ & $1737$ & $1675$& $1688$ & $1762$ & $1982$\\
$N_{3/2+}(1900)$ & $\approx 1900$ & $2005$  & $1926$ & $1870$ & & & $1899$ & $1904$\\
$N_{3/2-}(2080)$ & $\approx 2080$ & $1983$   & $2102$ & $2125$ & & & $2079$ & $2095$\\

$N_{5/2+}(1680)$ & $1685\pm5$ & $1735$  & $1735$ & $1770$ & $1737$ & $1675$ & $1723$ & $1718$ & $1823$\\
$N_{5/2-}(2200)$ & $\approx 2200$  & $2252$ & $2102$ & $2234$ & & & $2185$ & $2217$\\
$N_{7/2-}(2190)$ & $2150\pm50$ & $1983$  & $2102$ & $2090$ & $2140$ & & $2093$ & $2100$ & $2075$\\
$N_{9/2+}(2220)$ & $2250\pm50$ & $2196$  & $2265$ & $2327$ & $2271$ & &$2221$ & $2221$ & $2327$\\
\hline
\end{tabular}
%\vskip 10pt
%\begin{flushleft}
%\end{flushleft}





\end{table}


\section{Conclusion}
\label{part6}

In this work the exactly solvable three-body oscillator model with
the spin-1/2 content has been constructed by employing the
extension of the $SL(2,C)$ group to the $Sp(4,C)$ one. The
Dirac-like equation for the $Sp(4,C)$ spinors incorporates the
ordinary relativistic kinematics, but in the presence of
interaction differs significantly from the equations of the other
three-body oscillator models, in particular, of the FKR model
\cite{feynman}. The main feature is that the present model
includes only one three-dimensional oscillator mode, whereas the
remaining degrees of freedom of relative motion are spent to get
the rotational excitations. The corresponding quantum number $N$
goes as the addition to the total spin $J$, so that the energy
spectrum possesses the $(J+N)$-degeneracy. The application to the
nucleon resonance mass spectrum has shown that such a model is in
overall agreement with the experimental data.

%The model is based on the Dirac-like equation for the $Sp(4,C)$
%spinors, which involves the interaction linear in the coordinates.
%Its main feature is that the degrees of freedom of relative motion
%are so distributed that only three of them refer to the oscillator
%mode whereas the three remaining ones are spent to get the
%rotational excitations. Since the corresponding quantum number $N$
%goes as the addition to the total spin $J$, the energy spectrum
%acquires the $(J+N)$-degeneracy, also predicted in the AdS/QCD
%theory and seen in the nucleon resonance spectrum.



\newpage

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\begin{figure}[htbp]
\centerline{\includegraphics[scale=0.5]{fignuclplus1.eps}}
\label{fig1} \small \caption{Nucleon resonance Regge trajectories
that start with $J^P={1/2}^{+}$, calculated using Eq.~(32a) with
$m=0.940\,$GeV, $\lambda=0.345\,$GeV$^2$. The solid lines are
obtained with the quantum number values $n=0$, $N=0$ and $N=2$,
the dashed line corresponds to $n=1$, $N=0$. The experimental
masses and errors are taken from \cite{pdg}. \hfill} \vspace{20pt}
\centerline{\includegraphics[scale=0.5]{fignuclminus1.eps}}
\label{fig2} \small \caption{Nucleon resonance Regge trajectories
that start with $J^P={1/2}^{-}$, calculated using the same
parameter values  as for Fig.~1. The solid lines are obtained from
Eq.~(32a) with $n=0$, $N=1$ and $N=3$, the dashed line results
from Eq.~(32b) with $n=0$, $N=0$. \hfill}
\end{figure}



\end{document}

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