Papers in Physics, vol. 3, art. 030003 (2011)

Received: 6 June 2011, Accepted: 13 July 2011
Edited by: D. Restrepo
Reviewed by: J. H. Muñoz, Universidad del Tolima, Ibagué, Colombia;

and Centro Brasileiro de Pesquisas Fisica
Licence: Creative Commons Attribution 3.0
DOI: 10.4279/PIP.030003

www.papersinphysics.org

ISSN 1852-4249

Calculation of almost all energy levels of baryons

Mario Everaldo de Souza 1∗

It is considered that the effective interaction between any two quarks of a baryon can be
approximately described by a simple harmonic potential. The problem is firstly solved in
Cartesian coordinates in order to find the energy levels irrespective of their angular mo-
menta. Then, the problem is also solved in polar cylindrical coordinates in order to take
into account the angular momenta of the levels. Comparing the two solutions, a correspon-
dence is made between the angular momenta and parities for almost all experimentally
determined levels. The agreement with the experimental data is quite impressive and, in
general, the discrepancy between calculated and experimental values is below 5%. A cou-
ple of levels of ∆, N, Σ±, and Ω present discrepacies between 6.7% and 12.5% [N(1655),
N(1440), N(1675), N(1685), N(1700), N(1710), N(1720), N(1990), N(2600), ∆(1700),
∆(2000), ∆(2300), Σ±(1189), Λ(1520), Ω(1672) and Ω(2250)].

I. Introduction

There are several important works that deal with
the calculation of the energy levels of baryons. One
of the most important ones is the pioneering work
of Gasiorowicz and Rosner [1] which has calcula-
tion of baryon energy levels and magnetic moments
of baryons using approximate wavefunctions. An-
other important work is that of Isgur and Karl [2]
which strongly suggests that non-relativistic quan-
tum mechanics can be used in the calculation of
baryon spectra. Other very important attempts
towards the understanding of baryon spectra are
the works of Capstick and Isgur [3], Bhaduri et al.
[4] Murthy et al. [5], Murthy et al. [6] and Stas-
sat et al. [7]. Still another important work that
attempts to describe baryon spectra is the recent
work of Hosaka, Toki and Takayama [8] that makes

use of a non-central harmonic potential (called by
the authors the deformed oscillator ) and is able to
describe many levels. This present work describes
many more levels and is more consistent in the char-
acterization of angular momenta and parities of lev-
els. It is an updated version of the pre-print of Ref.
[9].

II. The approximation for the effec-
tive potential

The effective potential between any two quarks of a
baryon is not known and thus a couple of different
potentials can be found in the literature. Of course,
the effective potential is the result of the attractive
and repulsive forces of QCD and is completely jus-
tified because, as it is well known that the strong

∗E-mail: mariodesouza.ufs@gmail.com

1 Universidade Federal de Sergipe, Departamento de F́ısica, Av. Marechal Rondon, s/n, Campus Universitário, Jardim
Rosa Elze 49100-000, São Cristovão, Brazil.

030003-1



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

force becomes repulsive for very short distances,
and thus repulsion and attraction can form a poten-
tial well that can be approximated with a harmonic
potential about the equilibrium point. Taking into
consideration the work of Isgur and Karl [2] about
the use of non-relativistic quantum mechanics, and
considering that the three quarks of a baryon are al-
ways on a plane, we consider that the system can be
approximately described by three non-central and
non-relativistic linear harmonic potentials. This is
a calculation quite different from those found in
the literature and explains almost all energy levels
of baryons.

III. Calculation in Cartesian coordi-
nates and comparison with ex-
perimental data

The initial calculation, in which we have used
Cartesian coordinates, does not, of course, consider
the angular momentum of the system, that is, it
does not take into account the symmetries of the
system. This calculation is important for the iden-

tification of the energy levels given by the experi-
mental data, and for the assignment of the angular
momenta later on. Also, it allows the prediction of
many energy levels. Since each oscillator has two
degrees of freedom, the energy of the system of 3
quarks is given by [10]

En,m,k =hν1(n + 1) + hν2(m + 1)

+ hν3(k + 1) (1)

where n,m,k = 0, 1, 2, 3, 4, . . . Of course, we iden-
tify hν1, hν2, hν3 with the ground states of the
corresponding energy levels of baryons, and thus
hν1, hν2, hν3 are equal to the masses of constituent
quarks. Since we do not take isospin into account,
we cannot distinguish between N and ∆ states, or
between Σ and Λ states. The experimental values
for the baryon levels were taken from Particle Data
Group (Nakamura et al. [11]). The masses of con-
stituent quarks are taken as mu = md = 0.31 GeV,
ms = 0.5 GeV, mc = 1.7 GeV, mb = 5 GeV, and
mt = 174 GeV. We have, thus, the following formu-
las (see Table 1) for the energy levels of all known
baryons up to now:

Baryons Formulas for the energy levels (in GeV)
N, ∆−, ∆++ En,m,k = 0.31(n + m + k + 3)
Λ0, Σ+, Σ0, Σ− En,m,k = 0.31(n + m + 2) + 0.5(k + 1)
Ξ0, Ξ− En,m,k = 0.31(n + 1) + 0.5(m + k + 2)
Ω− En,m,k = 0.5(n + m + k + 3)
Λ+c , Σ

+
c , Σ

++
c , Σ

0
c En,m,k = 0.31(n + m + 2) + 1.7(k + 1)

Ξ0c, Ξ
+
c En,m,k = 0.31(n + 1) + 0.5(m + 1) + 1.7(k + 1)

Ω0c En,m,k = 0.5(n + m + 2) + 1.7(k + 1)
Xcc En,m,k = 0.31(n + 1) + 1.7(m + k + 2)
Λ0b En,m,k = 0.31(n + m + 2) + 5(k + 1)
Ξ0b , Ξ

−
b En,m,k = 0.31(n + 1) + 0.5(m + 1) + 5(k + 1)

Ω−b En,m,k = 0.5(n + m + 2) + 5(k + 1)

Table 1: Formulas for most energy levels of all baryons.

In Tables 1 to 11, EC is the calculated value
by the above formulas, EM is the measured value
and the error is given by Error = 100% ×|EM −
EC|/EC .

Within the scope of our simple calculation, many
levels are degenerate, of course. Further calcula-
tions, taking into account spin-orbit and spin-spin

effects, should lift part of the degeneracy. We no-
tice that these effects are quite complex. States
such as 1.70(N)D13 and 1.70(∆)D33 clearly show
that isospin does not play an important role in the
splitting of the levels. In general, the Error is be-
low 5%.

030003-2



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

State(n,m,k) EC (GeV) EM (GeV) Error(%) L2I,2J Parity
0, 0, 0 0.93 0.938(N) 0.9 P11 +
n + m + k = 1 1.24 1.232(∆) 0.6 P33 +
n + m + k = 2 1.55 1.44(N) 7.1 P11 +
n + m + k = 2 1.55 1.52(N) 1.9 D13 −
n + m + k = 2 1.55 1.535(N) 1.0 S11 −
n + m + k = 2 1.55 1.6(∆) 3.1 P33 +
n + m + k = 2 1.55 1.62(∆) 4.5 S31 −
n + m + k = 2 1.55 1.655(N) 6.7 S11 −
n + m + k = 2 1.55 1.675(N) 8.1 D15 −
n + m + k = 2 1.55 1.685(N) 8.7 F15 +
n + m + k = 2 1.55 1.70(N) 9.7 D13 −
n + m + k = 2 1.55 1.70(∆) 9.7 D33 −
n + m + k = 2 1.55 1.72(N) 11.0 P13 +
n + m + k = 3 1.86 1.71(N) 8.1 P11 +
n + m + k = 3 1.86 1.90(N) 2.2 P13 +
n + m + k = 3 1.86 1.90(∆) 2.2 S31 −
n + m + k = 3 1.86 1.905(∆) 2.4 F35 +
n + m + k = 3 1.86 1.91(∆) 2.7 P31 +
n + m + k = 3 1.86 1.92(∆) 3.2 P33 +
n + m + k = 3 1.86 1.93(∆) 3.8 D35 −
n + m + k = 3 1.86 1.94(∆) 4.3 D33 −
n + m + k = 3 1.86 2.0(N) 7.5 F15 +
n + m + k = 4 2.17 1.95(∆) 10.1 F37 +
n + m + k = 4 2.17 1.99(N) 8.3 F17 +
n + m + k = 4 2.17 2.00(∆) 7.8 F35 +
n + m + k = 4 2.17 2.08(N) 4.1 D13 −
n + m + k = 4 2.17 2.09(N) 3.7 S11 −
n + m + k = 4 2.17 2.10(N) 3.2 P11 +
n + m + k = 4 2.17 2.15(∆) 0.9 S31 −
n + m + k = 4 2.17 2.19(N) 0.9 G17 −
n + m + k = 4 2.17 2.20(N) 1.4 D15 −
n + m + k = 4 2.17 2.20(∆) 1.4 G37 −
n + m + k = 4 2.17 2.22(N) 2.3 H19 +
n + m + k = 4 2.17 2.225(N) 2.5 G19 −
n + m + k = 4 2.17 2.3(∆) 6.0 H39 +
n + m + k = 5 2.48 2.35(∆) 5.2 D35 −
n + m + k = 5 2.48 2.39(∆) 3.6 F37 +
n + m + k = 5 2.48 2.40(∆) 3.2 G39 −
n + m + k = 5 2.48 2.42(∆) 2.4 H3,11 +
n + m + k = 6 2.79 2.60(N) 6.8 I1,11 −
n + m + k = 6 2.79 2.70(N) 3.2 K1,13 +
n + m + k = 6 2.79 2.75(∆) 1.4 I3,13 −
n + m + k = 7 3.10 2.95(∆) 4.8 K3,15 +
n + m + k = 7 3.10 3.10(N) 0 L1,15 −
n + m + k = 8 3.21 ? ? ? ?

Table 2: Energy levels of baryons N and ∆.

030003-3



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

State (n,m,k) EC (GeV) EM (GeV) Error (%) L2I,2J Parity
0, 0, 0 1.12 1.116(Λ) 0.4 P01 +
0, 0, 0 1.12 1.189(Σ±) 6.2 P11 +
0, 0, 0 1.12 1.193(Σ0) 6.5 P11 +
n + m = 1,k = 0 1.43 1.385(Σ) 3.2 P13 +
n + m = 1,k = 0 1.43 1.405(Λ) 1.7 S01 −
n + m = 1,k = 0 1.43 1.48(Σ) 3.5 ? ?
0, 0, 1 1.62 1.52(Λ) 6.2 D03 −
0, 0, 1 1.62 1.56(Σ) 3.7 ? +
0, 0, 1 1.62 1.58(Σ) 2.5 D13 −
0, 0, 1 1.62 1.60(Λ) 1.2 P01 +
0, 0, 1 1.62 1.62(Σ) 0 S11 −
0, 0, 1 1.62 1.66(Σ) 2.5 P11 +
0, 0, 1 1.62 1.67(Λ) 3.1 S01 −
n + m = 2,k = 0 1.74 1.67(Σ) 4.0 D13 −
n + m = 2,k = 0 1.74 1.69(Λ) 2.9 D03 −
n + m = 2,k = 0 1.74 1.69(Σ) 2.9 ? ?
n + m = 2,k = 0 1.74 1.75(Σ) 0.6 S11 −
n + m = 2,k = 0 1.74 1.77(Σ) 1.7 P11 +
n + m = 2,k = 0 1.74 1.775(Σ) 2.0 D15 −
n + m = 2,k = 0 1.74 1.80(Λ) 3.4 S01 −
n + m = 2,k = 0 1.74 1.81(Λ) 4.0 P01 +
n + m = 2,k = 0 1.74 1.82(Λ) 4.6 F05 +
n + m = 2,k = 0 1.74 1.83(Λ) 5.2 D05 −
n + m = 1,k = 1 1.93 1.84(Σ) 4.7 P13 +
n + m = 1,k = 1 1.93 1.88(Σ) 2.6 P11 +
n + m = 1,k = 1 1.93 1.89(Λ) 2.1 P03 +
n + m = 1,k = 1 1.93 1.915(Σ) 0.8 F15 +
n + m = 1,k = 1 1.93 1.94(Σ) 0.5 D13 −
n + m = 3,k = 0 2.05 2.00(Λ) 2.5 ? ?
n + m = 3,k = 0 2.05 2.00(Σ) 2.5 S11 −
n + m = 3,k = 0 2.05 2.02(Λ) 1.5 F07 +
n + m = 3,k = 0 2.05 2.03(Σ) 1.0 F17 +
n + m = 3,k = 0 2.05 2.07(Σ) 1.0 F15 +
n + m = 3,k = 0 2.05 2.08(Σ) 1.5 P13 +
0, 0, 2 2.12 2.10(Σ) 0.9 G17 −
0, 0, 2 2.12 2.10(Λ) 0.9 G07 −
0, 0, 2 2.12 2.11(Λ) 0.5 F05 +
n + m = 2,k = 1 2.24 2.25(Σ) 0.5 ? ?
n + m = 4,k = 0 2.36 2.325(Λ) 1.5 D03 −
n + m = 4,k = 0 2.36 2.35(Λ) 0.4 H09 +
n + m = 1,k = 2 2.43 2.455(Σ) 1.0 ? ?
n + m = 3,k = 1 2.55 2.585(Λ) 1.4 ? ?

Table 3: Energy levels of Σ and Λ.

030003-4



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

State (n,m,k) EC (GeV) EM (GeV) Error (%) L2I,2J Parity

0, 0, 3 2.62 2.62(Σ) 0 ? ?
n + m = 5,k = 0 2.67 ? ? ? ?
n + m = 2,k = 2 2.74 ? ? ? ?
n + m = 4,k = 1 2.86 ? ? ? ?
n + m = 1,k = 3 2.93 ? ? ? ?
n + m = 6,k = 0 2.98 3.00(Σ) 0.7 ? ?
n + m = 3,k = 2 3.05 ? ? ? ?
n + m = 0,k = 4 3.12 ? ? ? ?
n + m = 5,k = 1 3.17 3.17(Σ) 0 ? ?
n + m = 2,k = 3 3.24 ? ? ? ?
n + m = 2,k = 3 3.29 ? ? ? ?

Table 3 (Cont.): Energy levels of Σ and Λ.

State (n,m,k) EC (GeV) EM (GeV) Error (%) L2I,2J Parity
0, 0, 0 1.31 1.315(Ξ0) 0.5 P11 +
0, 0, 0 1.31 1.321(Ξ−) 0.8 P11 +
1, 0, 0 1.62 1.53 5.6 P13 +
1, 0, 0 1.62 1.62 0 ? ?
1, 0, 0 1.62 1.69 4.3 ? ?
n = 0,m + k = 1 1.81 1.82 0.6 D13 −
2, 0, 0 1.93 1.95 1.0 ? ?
n = 1,m + k = 1 2.12 2.03 4.2 ? ?
n = 1,m + k = 1 2.12 2.12 0 ? ?
n = 3,m = k = 0 2.24 2.25 0.5 ? ?
n = 0,m + k = 2 2.31 2.37 2.6 ? ?
n = 2,m + k = 1 2.43 ? ? ? ?
n = 4,m = k = 0 2.55 2.5 2.0 ? ?
n = 1,m + k = 2 2.62 ? ? ? ?

Table 4: Energy levels of Ξ.

State (n,m,k) EC (GeV) EM (GeV) Error (%)
0, 0, 0 1.5 1.672 11.17
n + m + k = 1 2.0 2.25 12.5
n + m + k = 2 2.5 2.38 4.8
n + m + k = 2 2.5 2.47 1.2
n + m + k = 3 3.0 ? ?

Table 5: Energy levels of Ω.

State (n,m,k) EC (GeV) EM (GeV) Error (%)
0, 0, 0 2.32 2.285 1.5
n + m = 1,k = 0 2.63 2.594 0.1
n + m = 1,k = 0 2.63 2.625 0.2
n + m = 2,k = 0 2.94 ? ?

Table 6: Energy levels of Λc.

030003-5



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

State (n,m,k) EC (GeV) EM (GeV) Error (%)
0, 0, 0 2.51 2.46(Ξ+c ) 2.0
0, 0, 0 2.51 2.47(Ξ0c) 1.6
1, 0, 0 2.82 2.79 1.1
1, 0, 0 2.82 2.815 0.2
0, 1, 0 3.01 2.93 2.7
0, 1, 0 3.01 2.98 1.0
0, 1, 0 3.01 3.055 1.5
2, 0, 0 3.13 3.08 1.6
2, 0, 0 3.13 3.123 0.2
1, 1, 0 3.32 ? ?
3, 0, 0 3.44 ? ?

Table 7: Energy levels of Ξc.

State (n,m,k) EC (GeV) EM (GeV) Error (%)
0, 0, 0 2.7 2.704 0.2
n + m = 1,k = 0 3.2 ? ?
n + m = 2,k = 0 3.7 ? ?

Table 8: Energy levels of Ωc.

State (n,m,k) EC (GeV) EM (GeV) Error (%)
0, 0, 0 5.62 5.6202 0.004
n + m = 1,k = 0 5.93 ? ?
n + m = 2,k = 0 6.24 ? ?

Table 9: Energy levels of Λ0b .

State (n,m,k) EC (GeV) EM (GeV) Error (%)
0, 0, 0 5.81 5.79(Ξ0b ) 0.2
0, 0, 0 5.81 5.79(Ξ−b ) 0.2
1, 0, 0 6.12 ? ?
0, 1, 0 6.31 ? ?

Table 10: Energy levels of Ξb.

State (n,m,k) EC (GeV) EM (GeV) Error (%)
0, 0, 0 6.0 6.071 1.2
n + m = 1,k = 0 6.5 ? ?
n + m = 2,k = 0 7.0 ? ?

Table 11: Energy levels of Ωb.

030003-6



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

We can predict the energy levels of many heavy
baryons, probably already found by the LHC or to
be found in the near future. There are, for example,
the baryon levels (in GeV):

• scc, En,m,k = 0.5(n + 1) + 1.7(m + k + 2);

• ccc, En,m,k = 1.7(n + m + k + 3);

• ccb, En,m,k = 1.7(n + m + 2) + 5(k + 1);

• cbb, En,m,k = 1.7(n + 1) + 5(m + k + 2);

• etc.

IV. Calculation in polar cylindri-
cal coordinates and comparison
with experimental data

In order to take into account angular momentum
and parity, we have to use spherical or polar coor-
dinates. Since the 3 quarks of a baryon are always
in a plane, we can use polar coordinates and choose
the Z axis perpendicular to this plane. Now the
eigenfunctions are eigenfunctions of the orbital an-
gular momentum. Thus, we have three oscillators
in a plane and we consider them to be indepen-
dent. Using again the non-relativistic approxima-
tion, the radial Schrödinger equation for the sta-
tionary states of each oscillator is given by [12, 13]

[
−
h̄2

2µ

(
∂2

∂ρ2
+

1

ρ

∂

∂ρ
−
mz
ρ2

)
+

1

2
µω2ρ2

]
REm(ρ) = EREm(ρ) (2)

where mz is the quantum number associated with
Lz, µ is the reduced mass of the oscillator, and
ω is the oscillator frequency. Therefore, we have
three independent oscillators with orbital angular
momenta ~L1, ~L2 and ~L3 whose Z components are
Lz1, Lz2 and Lz3. Of course, the system has total
orbital angular momentum ~L = ~L1 + ~L2 + ~L3 and
each ~Li has a quantum number li associated with
it. The eigenvalues of the energy levels are given
by [12, 13]

E =(2r1 + |mz1| + 1)hν1 + (2r2 + |mz2| + 1)hν2
+ (2r3 + |mz3| + 1)hν3 (3)

in which ri = 0, 1, 2, . . . and it is a radial quantum
number, and |mzi| = 0, 1, . . . , li. Comparing equa-
tion (3) with equation (1), we have n = 2r1 +|mz1|;
m = 2r2 +|mz2|; k = 2r3 +|mz3|. Let us recall that
if we have three angular momenta ~L1, ~L2 and ~L3
associated to the quantum numbers l1, l2 and l3,
the total orbital angular momentum ~L is described
by the quantum number L given by

l1 + l2 + l3 ≥ L ≥ ||l1 − l2|− l3| (4)

where l1 ≥ |mz1|; l2 ≥ |mz2|; l3 ≥ |mz3|.
Because the three quarks are on a plane, only ri

and mzi are good quantum numbers, that is, li are
not good quantum numbers and their possible val-
ues are found indirectly by means of mzi due to the
condition li ≥ mzi. This means that the upper val-
ues of li cannot be found from the model, and as a
consequence, the upper value of L cannot be found
either. We only determine the values of L compar-
ing the experimental results of the energies of the
baryon states with the energy values calculated by
Enmk. This is a limitation of the model. The other
models have many limitations too. For example,
in the Deformed Oscillator Model some quantum
numbers are not good either and are only approxi-
mate and there is not a direct relation between N
and L where N is the total quantum number. In
a certain way, a baryon is a tri-atomic molecule of
three quarks and thus some features of molecules
may show up and that is indeed the case.

Taking into account spin, we form the total angu-
lar momentum ~J = ~L+ ~S whose quantum numbers
are J = L± s where s = 1/2, 3/2. As we will see,
we will be able to describe almost all baryon levels.

As in the case of the rotational spectra of tri-
atomic molecules [14], due to the couplings of the
different angular momenta, it is expected that there
should exist a minimum value of J = K for the to-
tal angular momentum and, thus, J should have
the possible values J = K,K + 1,K + 2,K + 3, . . ..
But in the case of baryons, this feature does not
always appear to happen.

i. Baryons N and Λ

We will classify the levels according to Table 1 and
take J = L± 1/2 or J = L± 3/2(∆).

030003-7



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

a. Level (n = m = k = 0; 0.93 GeV)

The first state of N is the state (n = m = k = 0)
with energy 0.93 GeV. Therefore, in this case l1 =
l2 = l3 = 0 and thus L = 0. This is the positive
parity state P11

L N ∆ Parity
0 0.938P11 +

b. Level (n = m = k = 1; 1.24 GeV)

This is the first state of Λ. As n + m + k = 1, we
have 2r1 +|mz1|+ 2r2 +|mz2|+ 2r3 +|mz3| = 1, and
thus |mz1|+ |mz2|+ |mz3| = 1, and l1 + l2 + l3 ≥ 1,
and we can choose the sets |mz1| = 1, |mz2| =
|mz3| = 0; |mz2| = 1, |mz1| = |mz3| = 0; |mz3| = 1,
|mz1| = |mz2| = 0, and l1 = 1, l2 = l3 = 0 or
l2 = 1, l1 = l3 = 0 or still l3 = 1, l1 = l2 = 0 which
produce L ≥ 0 (ground state) and thus the level

L N ∆ Parity
0 1.232P33 +

c. Level (n = m = k = 2; 1.55 GeV)

In this case n = m = k = 2 = 2r1 + |mz1| + 2r2 +
|mz2|+2r3 +|mz3|. This means that |mz1|+|mz2|+
|mz3| = 2, 0 and we have the sets of possible values
of l1, l2, l3

l1, l2, l3 2, 0, 0 0, 2, 0 0, 0, 2 1, 1, 0
L 2 2 2 0, 1, 2

l1, l2, l3 1, 0, 1 0, 1, 1 0, 0, 0
L 0, 1, 2 0, 1, 2 0

in which the second row presents the values of L
that satisfy the condition l1 + l2 + l3 ≥ 2, 0. As 2 is
a lower bound, we can also have L = 3. There are,
therefore, the following possible states

L N ∆ Parity
0 1.44P11 1.6P33 +
1 1.535S11; 1.655S11 1.62S31 −

1.52D13; 1.7D13 −
2 1.72P13 ? +

1.685F15
3 1.675D15 1.70D33 −

d. Level (n = m = k = 3; 1.86 GeV)

Since n = m = k = 3 = 2r1 + |mz1|+ 2r2 + |mz2|+
2r3 + |mz3|, |mz1| + |mz2| + |mz3| = 3, 1, and thus
l1 + l2 + l3 ≥ 3, 1. We have, therefore, the pos-
sibilities L = 4, 3, 2, 1, 0 because of the condition
L ≥ ||l1 − l2|− l3| and we can arrange the levels in
the form

L N ∆ Parity
0 1.71P11 1.91P31 +
1 1.90S31 −

1.93D35 −
2 1.90P13 ? +

2.0F15 +
3 1.92P33, 1.94D33 −
4 1.905F35 +

e. Level (n = m = k = 4; 2.17 GeV)

This energy level is split in many close levels. Fol-
lowing what we have done above n = m = k =
4 = 2r1 + |mz1| + 2r2 + |mz2| + 2r3 + |mz3|, which
yields |mz1| + |mz2| + |mz3| = 4, 2, 0, and thus
l1 + l2 + l3 ≥ 4, 2, 0. We have therefore for L the
possible values L = 6, 5, 4, 3, 2, 1, 0 and the follow-
ing assignments

L N ∆ Parity
0 2.10P11 ? +
1 2.09S11 2.15S31 −

2.08D13
2 1.95F37 +
3 2.20D15 ? −

2.19G17 −
4 2.22H19 2.00F35 +
5 2.225G19 2.20G37 −
6 2.30H39 +

f. Level (n = m = k = 5; 2.48 GeV)

Doing as above n = m = k = 5 = 2r1+|mz1|+2r2+
|mz2|+2r3 +|mz3|, and thus |mz1|+|mz2|+|mz3| =
5, 3, 1. That is, l1 + l2 + l3 ≥ 5, 3, 1, and so we may
have L = 5, 4, 3, 2, 1, 0 because of the conditions
L ≥ ||l1 − l2|− l3| and li ≥ |mzi|. Experimentally,
though, we note that K = 1, and hence we have
the possible arrangement of levels

030003-8



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

L N ∆ Parity
1 2.35D35 −
2 2.39F37 +
3 2.40G39 −
4 2.42H3,11 +
5 ? −

g. Level (n = m = k = 6; 2.79 GeV)

We have n = m = k = 6 = 2r1 + |mz1| + 2r2 +
|mz2| + 2r3 + |mz3| and so |mz1| + |mz2| + |mz3| =
6, 4, 2, 0 and thus l1 + l2 + l3 ≥ 6, 4, 2, 0, and L can
be L = 6, 5, 4, 3, 2, 1, 0, but, from the experimental
values, we note that K = 5 and so there are the
possible states

L N ∆ Parity
5 2.60I1,11 2.75I3,13 −
6 2.70K1,13 ? +

h. Level (n = m = k = 7; 3.10 GeV)

From n = m = k = 7 = 2r1 + |mz1| + 2r2 +
|mz2|+2r3 +|mz3| we obtain |mz1|+|mz2|+|mz3| =
7, 5, 3, 1 and hence l1 + l2 + l3 ≥ 7, 5, 3, 1, and thus
the possible values for L are L = 7, 6, 5, 4, 3, 2, 1, 0,
but we note that K = 7. Therefore, we have the
list of states

L N ∆ Parity
6 2.95K3,15 +
7 3.10L1,15 ? −

ii. Baryons Σ and Λ

We will classify the levels according to Table 2.
Again J = L± 1/2.

a. Level (n = m = k = 0; 1.12 GeV)

In this state l1 = l2 = l3 = 0 and thus L = 0 and
we have the state

L Σ Λ Parity
0 1.189(Σ±)P11; 1.116P01 +

1.193(Σ0)P11

b. Level (n = m = k = 1; 1.43 GeV)

From n = m = 1,k = 0 we obtain 2r1 + |mz1| +
2r2 + |mz2| = 1 and 2r3 + |mz3| = 0 which make
|mz1| + |mz2| = 1 and |mz3| = 0. That is, we have
the condition l1 + l2 ≥ 1, l3 ≥ 0 which allows us to
have the possibilities L = 0, 1, 2 and the states

L Σ Λ Parity
0 +
1 ? 1.405S01 −
2 1.358P13 ? +

Thus, the most probable values of L for the state
1.48(Σ) are L = 0, 1. Maybe K = 1 in this case
and thus L = 0 may be suppressed.

c. Level (0, 0, 1; 1.62 GeV)

For n = m = 0 and k = 1 we have |mz1| =
|mz2| = 0 and |mz3| = 1. That is, we have the
condition l1 ≥ 0, l2 ≥ 0, l3 ≥ 1 which allows us to
choose l1 = l2 = 0, l3 = 1; l1 = l3 = 1, l2 = 0;
l1 = 0, l2 = l3 = 1, and thus L ≥ 0, 1, 2, and the
states

L Σ Λ Parity
0 1.66P11 1.60P01 +
1 1.62S11 1.67S01 −

1.58D13 1.52D03 −
2 ? ? +

d. Level (n + m = 2,k = 0; 1.74 GeV)

In this case n + m = 2 = 2r1 + |mz1| + 2r2 + |mz2|
and k = 2r3 + |mz3| = 0, and thus we obtain
|mz1| + |mz2| = 2, 0 and |mz3| = 0. Thus, we have
the conditions l1 + l2 ≥ 2, 0 and l3 ≥ 0. We can
then choose l1 = l2 = l3 = 0; l1 = l2 = 1, l3 = 0;
l1 = 2, l2 = l3 = 0; l2 = 2, l1 = l3 = 0; l1 = 3, l2 =
l3 = 0, and we may have thus L = 0, 1, 2, 3 and the
assignments

L Σ Λ Parity
0 1.77P11 1.81P01 +
1 1.75S11 1.80S01 −

1.67D13 1.69D03 −
2 ? 1.82F05 +
3 1.775D15 1.83D05 −

We can then say that the state 1.69(Σ) is prob-
ably a F15 state.

030003-9



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

e. Level (n + m = 1,k = 1; 1.93 GeV)

We have n + m = 1 = 2r1 + |mz1| + 2r2 + |mz2|
and k = 2r3 + |mz3| = 1, from which we ob-
tain |mz1| + |mz2| = 1 and |mz3| = 1. Hence,
we have the condition l1 + l2 ≥ 1 and l3 ≥ 1.
We can then have the sets l1 = 1, l2 = 0, l3 = 1;
l1 = 0, l2 = 1, l3 = 1. Both yield L ≥ 2, 1, 0 and
thus we identify the states

L Σ Λ Parity
0 1.88P11 ? +
1 1.94D13 ? −
2 1.84P13 1.89P03 +

1.915F15 +

f. Level (n + m = 3,k = 0; 2.05 GeV)

With n + m = 3 = 2r1 + |mz1| + 2r2 + |mz2| and
k = 2r3 + |mz3| = 0 we obtain |mz1| + |mz2| = 3, 1
and |mz3| = 0, and thus the conditions l1 +l2 ≥ 3, 1
and l3 ≥ 0 which yield L ≥ 4, 3, 2, 1, 0, and the pos-
sible identification taking into account that maybe
K = 1

L Σ Λ Parity
1 2.00S11 ? −
2 2.08P13 ? +

2.07F15
3 ? ? −
4 2.03F17 2.02F07 +

g. Level (0, 0, 2; 2.12 GeV)

In this case n = 0 = 2r1 + |mz1|, m = 0 =
2r2 + |mz2| and k = 2 = 2r3 + |mz3| and thus
|mz1| = |mz2| = 0 and |mz3| = 2, 0. Hence, we
have the condition l1 ≥ 0, l2 ≥ 0 and l3 ≥ 2, 0.
We can then choose the sets l1 = l2 = l3 = 0;
l1 = 0, l2 = 0, l3 = 2; l1 = 0, l2 = 0, l3 = 3;
l1 = l2 = l3 = 1; l1 = l2 = 1, l3 = 0 which make
L ≥ 3, 2, 1, 0, and probably K = 2. Hence, we have
the possible states

L Σ Λ Parity
2 ? 2.11F05 +
3 2.10G17 2.10G07 −

h. Level (n + m = 4,k = 0; 2.36 GeV)

From n+m = 4 = 2r1 +|mz1|+2r2 +|mz2| and k =
2r3 +|mz3| = 0 we obtain |mz1|+|mz2| = 4, 2, 0 and

|mz3| = 0, and thus the conditions l1 + l2 ≥ 4, 2, 0
and l3 ≥ 0 which produce L ≥ 4, 3, 2, 1, 0, and the
possible identification

L Σ Λ Parity
0
1 2.325D03 −
2 ?
3 ?
4 2.35H09 +

Probably in this case K = 1 and the levels with
L = 2, 3 are missing just because of a lack of ex-
perimental data. It appears that there is no state
of Σ.

V. Discussion and conclusion

One can immediately ask about the spin degrees
of freedom of the three quarks since the spin-spin
interaction makes a contribution to the mass. We
can say that we took care of part of it because the
formulas of the energy levels depend on the three
parameters hν1, hν2 and hν3 which are assigned
according to the masses of the constituent quarks
which have already taken into account the spin-
spin interaction because the masses of constituent
quarks are in perfect agreement with the ground
state levels of baryons. Of course, the spin-spin in-
teraction contribution depends on the energy level
as is well known from the bottomonium spectrum,
for example. But, as it is seen in the spectrum
of bottomonium, the spin-spin contribution dimin-
ishes as the energy of the level increases. In bot-
tomonium, the difference between the energies of
ηb(1S) and Υ(1S) is about 69.4 MeV, while be-
tween ηb(2S) and Υ(2S) it is about 36.3 MeV, and
between ηb(3S) and Υ(3S) it is about 25.2 MeV,
where we have used, for the energies of ηb(2S) and
ηb(3S), the predicted values from reference [15],
9987.0 MeV and 10330 MeV, respectively. In the
case of baryons, the spin-spin interaction varies
from 15 MeV to 30 MeV for levels of N, Σ, Ξ and Λ
[16]. Therefore, we observe that the spin-spin inter-
action is of the order of magnitude of the splitting
beween neighboring levels. For example, the mea-
sured energy of the D13 level of N is 1.52 MeV,
while our calculated value is 1.55 MeV, and thus
the difference is 0.03 GeV= 30 MeV which is of the
order of the spin-spin interaction. And that is why

030003-10



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

there are large discrepancies in the calculation of
the lowest levels of Ω because in this case all quark
spins are parallel and thus, the total spin-spin con-
tribution is larger than in other baryons in which
two spins are up and the other spin is down. For
the lowest state of Ω, the discrepancy is about 1.672
GeV − 1.5 GeV = 0.172 GeV = 172 MeV. This is
actually the worse calculation. But we either con-
sider the mass of constituent quarks or we try to
find tentative values for the masses of quarks like
is done in QCD models which use a quite different
range of arbitrary quark masses. The use of the
constituent quark mass is completely justifiable in
our case because we do not attempt to calculate at
all the splitting between neighboring baryon levels.
Such calculation can be made in the future upon
improving the present model.

We only addressed the angular momenta of
N, ∆, Σ and Λ due to a lack of experimental data
for the other baryons. Of course, the state n = m =
l = 0 is missing for the ∆ particle because this cor-
responds to the ground state of the nucleon.

We notice that the simple model above describes
almost all energy levels of baryons. The splitting
for a certain L is quite complex. Sometimes, there
is almost no dependence on spin, such as, for ex-
ample, the states of Σ with L = 2, 2.08P13 and
2.07F15. On the other hand, the states of Σ with
the same L = 2, 1.84P13 and 1.915F15, present a
strong spin-orbit dependence. It can just be a mat-
ter of obtaining more accurate experimental results.

It is important to observe that part of the split-
ting is primarily caused by the spin-orbit interac-
tion and is very complex because, in some cases,
it appears to be the normal spin-orbit and, in
other cases, it appears to be the inverted (negative)
spin-orbit. In the simple model above, the oscil-
lators were considered approximately independent
but there may exist some coupling among them
and this can contribute to the splitting of levels.
As we have discussed in the first paragraph, part
of the splitting should be attributed to the spin-
spin interaction which was not taken into account
in a detailed way. Of course, part of it was consid-
ered inside the values of the three parameters hν1,
hν2 and hν3 which are taken as the masses of the
three constituent quarks of a given baryon. It is
important to observe that the discrepancy between
calculated and measured values diminishes as the
energy increases. This fact shows that the splitting

is mainly caused by the spin-spin interaction.
Another important conclusion is that with the

simple model above we cannot calculate the values
of K, and from the above results we note that it is
a quite difficult task because there appears to exist
no pattern with respect to this. As in the case of
triatomic molecules, the values of K are found from
the experimental data.

As we notice, in the above tables the increase in
the energy of levels allows the existence of higher
values of L (and J). This is an old fact and is so
because equation (3) has a linear dependence on
|lzi|.

For experimentalists, the classifications above
are very important and can help them in the pre-
diction of energies and angular momenta of levels.
An old version of this work that appeared in Ref.
[9] predicted the energies of all levels which have
lately been reported, and this is a very important
fact. For example, for Ξc it predicted the levels (on
page 8 of [9]) with energies 2.82, 3.01 and 3.13, and
since 2002 the following corresponding levels of Ξc
have been found: 2.815, (2.93; 2.98; 3.055); (3.08;
3.123).

As it is well known, the first order correction
term of anharmonicity in an oscillator for each de-
gree of freedom is of the form

∆E = A

(
p +

1

2

)2
(5)

where A is a constant and p is a non-negative inte-
ger (p = 0, 1, 2, 3, . . .). Therefore, the calculated en-
ergies of levels with high quantum numbers would
be away from the experimental values. This is
not observed above and, thus, the anharmonicity
should be quite low. For example, for n+m+k = 7
of N we obtain that the experimental and calcu-
lated values are the same (3.10 GeV). In the case
of Σ, we have the same kind of behavior because for
(n+m = 5,k = 1) we also have the same calculated
and experimental value for Σ (3.17 GeV).

The assignments of the angular momenta for
some few levels are only reasonable attempts. It
is the case, for example, of the level 2.0F15 of N
which can belong to either the (n + m + k = 3)
or to the (n + m + k = 4) levels. We chose the
former because 2.0 is closer to 1.86 than to 2.17.
For the level 1.99F17 we chose the (n + m + k = 4)
level because it appears that the highest value of J

030003-11



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

for the level (n + m + k = 3) is 5. It is a strange
feature that the level (n + m + k = 5) only con-
tains ∆′s . Having in mind what has been justified
above, we chose the 2.35D35 level of ∆ belonging to
(n+m+k = 5) as 2.35 is closer to 2.48 than to 2.17.
The level 2.60I1,11 of N was assigned as belonging
to (n + m + k = 6) because its energy is between
2.55 GeV and 2.75 GeV. The level 1.74D13 of Σ was
chosen as belonging to (n + m = 2,k = 0) because
(0, 0, 1) already has a D13 level for Σ(1.58). We
made similar considerations in the choice of the lev-
els 1.83(Λ)D05, 1.84(Σ)P13 and 1.94(Σ)D13. These
ambiguities will be settled either with data with
smaller widths or with a more improved model.

Some levels are not described by the simple ap-
proximation above. It is the case, for instance,
of Ξ(1530)P13 which is probably a composite of
Ξ(0, 0, 0) ≡ Ξ(1.31) with a pion excitation (that
is, it is a hadronic molecule). Its decay is actually
Ξ(1.31)π. In the same way, the state Σc(2455) ap-
pears to be a composite state of Λ+c (2285) and a
pion excitation. The same appears to hold for the
other known states of Λc.

As a whole, the model describes quite well the
baryonic spectra but it is far from describing the
detailed splitting which appears to be quite com-
plex and may depend on the spin-spin interaction.
It does not provide a way of calculating the values
of K. With the acquisition of more data from other
baryons, we may be able to find more patterns and
to improve the model. Due to the complexity of the
problem, we will probably have to go back and forth
several times in the improvements of the model as
it has been done in the description of the molecu-
lar spectrum of molecules. But it is still the only
model that describes almost all levels of baryons in
a consistent way and is able to predict the energies
of levels yet to be found experimentally.

Acknowledgements - I thank the comments of
the referee Prof. José Muñoz.

[1] S Gasiorowicz, J L Rosner, Hadron spectra and
quarks, Am. J. Phys. 49, 954 (1981).

[2] N Isgur, G Karl, P-wave baryons in the quark
model, Phys. Rev. D 18, 4187 (1978).

[3] S Capstick, N Isgur, Baryons in a relativized
quark model with chromodynamic, Phys. Rev.
D 34, 2809 (1986).

[4] R K Bhaduri, B K Jennings, J C Wadding-
ton, Rotational bands in the baryon spectrum,
Phys. Rev. D 29, 2051 (1984).

[5] M V N Murthy, M Dey, R K Bhaduri, Rota-
tional bands in the baryon spectrum. II, Phys.
Rev. D 30, 152 (1984).

[6] M V N Murthy, M Brack, R K Bhaduri, B K
Jennings, The spin-orbit puzzle in the spectra
and deformed baryon model, Z. Phys. C 29,
385 (1985).

[7] P Stassart, F Stancu, J-M Richard, L Theuβl,
On the scalar meson exchange in the baryon
spectra, J. Phys. G: Nucl. Partic. 26, 397
(2000).

[8] A Hosaka, H Toki, M Takayama, Baryon spec-
tra in deformed oscillator quark model, Mod.
Phys. Lett. 13, 1699 (1998).

[9] M E de Souza, Calculation of the energy lev-
els and sizes of baryons with a noncentral
harmonic potential, arXiv:hep-ph/0209064v1
(2002).

[10] M E de Souza, The energies of baryons, In:
Proceedings of the XIV Brazilian National
Meeting of the Physics of Particles and Fields,
Eds. A J da Silva, A Suzuki, C Dobrigkeit, C
Z de Vasconcelos, C Wotzacek, R F Ribeiro, S
R de Oliveira, S A Dias, Pag. 331, Sociedade
Brasileira de F́ısica, São Paulo (1993).

[11] K Nakamura et al.(Particle Data Group), Re-
view of particle physics, J. Phys. G: Nucl. Par-
tic. 37, 075021 (2010).

[12] R Shankar, Principles of quantum mechanics,
Plenum Press, New York (1994).

[13] W Greiner, J A Maruhn, Nuclear models,
Springer, Berlin (1996).

[14] G Herzberg, Infrared and Raman spectra,
Van Nostrand Heinhold Company, New York
(1945).

[15] L. Bai-Qing and C. Kuang-Ta, Bottomonium
spectrum with screeened potential, Commun.
Theor. Phys. 52, 653 (2009).

030003-12



Papers in Physics, vol. 3, art. 030003 (2011) / M. E. de Souza

[16] H Hassanabadi, A A Rajabi, Determination
of the potential coefficients of the baryons and
the effect of spin and isospin potential on their
energy, In: 11th Internation Conference on

meson-Nucleon Physics and Structure of the
Nucleon, Eds. H Machner, S Krewald, Pag.
128, IKP, Forschungzentrum Jülich (2007).

030003-13