# Chiral Properties of Baryon Fields with Flavor $SU(3)$ Symmetry

Hua-Xing Chen1,2<EMAIL_ADDRESS>V. Dmitrašinović3
<EMAIL_ADDRESS>Atsushi Hosaka1<EMAIL_ADDRESS>Keitaro Nagata4
<EMAIL_ADDRESS>Shi-Lin Zhu2<EMAIL_ADDRESS>1Research Center for
Nuclear Physics, Osaka University, Ibaraki 567–0047, Japan
2Department of Physics, Peking University, Beijing 100871, China
3 Vinča Institute of Nuclear Sciences, lab 010, P.O.Box 522, 11001 Beograd,
Serbia
4 Department of Physics, Chung-Yuan Christian University, Chung-Li 320, Taiwan

###### Abstract

We investigate chiral properties of local (non-derivative) fields of baryons
consisting of three quarks with flavor $SU(3)$ symmetry. We construct
explicitly independent local three-quark fields belonging to definite Lorentz
and flavor representations. Chiral symmetry is spontaneously broken and
therefore the baryon fields can have different chiral representations. It
turns out that the allowed chiral representations are strongly correlated with
the Lorentz group representations due to the color and spatial structure of
the local three-quark fields. We discuss some implications of the allowed
chiral symmetry representations on phenomenological Lagrangians with chiral
$U(3)_{\rm L}\otimes U(3)_{\rm R}$ symmetry.

baryon, chiral symmetry, interpolating field

###### pacs:

14.20.-c, 11.30.Rd, 11.40.Dw

##  I Introduction

As the chiral symmetry of QCD is spontaneously broken, $SU(N_{f})_{L}\otimes
SU(N_{f})_{R}\rightarrow SU(N_{f})_{V}$ ($N_{f}$ being the number of flavors),
the observed hadrons are classified by the residual symmetry group
representations of $SU(N_{f})_{V}$. The full chiral symmetry may then
conveniently be represented by its non-linear realization and this broken
symmetry plays a dynamical role in the presence of the Nambu-Goldstone bosons
and their interactions.

Yet, as pointed out by Weinberg Weinberg:1969hw , there are situations when it
makes sense to consider algebraic aspects of chiral symmetry, i.e. the chiral
multiplets of hadrons. Such hadrons may be classified in linear
representations of the chiral symmetry group with some representations mixing.
One such situation becomes realistic in the symmetry restored phase which is
expected at high temperatures and/or densities Hatsuda:1994pi . If hadrons
belong to certain representations of the chiral symmetry group, certain
physical properties such as the axial coupling constants are determined by
this symmetry. Therefore, the question as to what chiral representations,
possibly with mixing, the hadrons belong to is of fundamental interest lee72 ;
lee81 ; Jido:2001nt .

Another point of relevance is that the chiral representation can be used as a
measure of the internal structure of hadrons. For instance, for a $\bar{q}q$
spin-one mesons, the possible chiral representations are
$(\mathbf{8},\mathbf{1})$ and $(\mathbf{3},\mathbf{\bar{3}})$ and their left-
right conjugates for flavor octet mesons. As a matter of fact, for the
multiquark hadrons, the allowed chiral representations can be more
complicated/higher dimensional with increasing number of quarks and
antiquarks. Hence the study of chiral representations may provide some hints
to the structure of hadrons, extending possibly beyond the minimal constituent
picture Jido:1997yk ; Jido:1999hd ; Beane:2002td ; Benmerrouche:1989uc ;
Haberzettl:1998rw ; DeTar:1988kn .

Motivated by this argument, we have recently performed a complete
classification of baryon fields constructed from three quarks in the local
form with two light flavors (the so-called $SU(2)$ sector) Nagata:2007di .
Such baryon fields are used as interpolators for the study of two-point
correlation functions in the QCD sum rule approach and in the lattice QCD
Ioffe:1981kw ; Chung:1981cc ; Espriu:1983hu ; Lee:2002jb ; Leinweber:2004it ;
Zanotti:2003fx . Strictly speaking, however, the chiral structure of an
interpolator does not directly reflect that of the physical state when chiral
symmetry is spontaneously broken. But the minimal configuration of three
quarks provides at least a guide to the simplest expectations for baryons. Any
deviation from such a simple structure may be an indication of higher Fock-
space components, such as the multi-quark ones Cohen:1996sb .

Another reason for such a study of chiral classifications is related to the
number of independent fields. In principle, the correlation functions should
contain all information about physical states when computed exactly.
Practically, however, one must rely on some approximation, and it has been
observed in previous studies, that the results may depend significantly on the
choice of the interpolators, which are generally taken as linear combinations
of the independent ones Espriu:1983hu ; Jido:2001nt ; Chen:2006hy .

In this paper, we perform a complete classification of baryon fields written
as local products (without derivatives) of three quarks according to chiral
symmetry group $SU(3)_{L}\otimes SU(3)_{R}$. This is an extension of our
previous work for the case of flavor $SU(2)$ Nagata:2007di . Technically, the
$SU(3)$ algebra introduces more complications, which makes insight less at
work. Hence, here we attempt to explore a rather technical aspect which
enables one to perform systematic classification. We derive general
transformation rules for baryon fields for the classification, while maximally
utilizing the Fierz transformations in order to implement the Pauli principle
among the quarks. The objective of $SU(3)$ baryon fields provides a simple but
suitable exercise how the method works. It can be extended to systems with
more complex hadron fields containing more quarks Chen:2006hy .

As in the previous paper Nagata:2007di , we first establish the classification
under the flavor $SU(3)$ symmetry, and then investigate the properties under
chiral symmetry group. The method is based essentially on the tensor method
for $SU(3)$, while the Fierz method for the Pauli principle associated with
the structure in the color, flavor, Lorentz (spin) and orbital spaces is
utilized when establishing the independent fields. It turns out that for local
three-quark fields, the Pauli principle puts a constraint on the structure of
the Lorentz and chiral representations. This leads essentially to the same
permutation symmetry structure as in the case of flavor $SU(2)$.

This paper is organized as follows. In Section II, we establish the
independent local baryon interpolating fields, and investigate their flavor
$SU(3)$ symmetry properties. In Section III, we investigate the properties of
the baryon fields under chiral symmetry transformations $SU(3)_{L}\otimes
SU(3)_{R}$. We find that both flavor and chiral symmetry properties are
related to the structure of the Lorentz group. Eventually, in Section IV, we
find that this can be explained by the Pauli principle for the left and right
handed quarks, which puts a constraint on permutation symmetry properties of
three-quarks. Some complicated formulae are shown in appendices.

##  II Flavor Symmetries of Three-Quark Baryon Fields

Local fields for baryons consisting of three quarks can be generally written
as

$\displaystyle
B(x)\sim\epsilon_{abc}\left(q^{aT}_{A}(x)C\Gamma_{1}q^{b}_{B}(x)\right)\Gamma_{2}q^{c}_{C}(x)\,,$
(1)

where $a,b,c$ denote the colour and $A,B,C$ the flavour indices,
$C=i\gamma_{2}\gamma_{0}$ is the charge-conjugation operator,
$q_{A}(x)=(u(x),\;d(x),\;s(x))$ is the flavour triplet quark field at location
$x$, and the superscript $T$ represents the transpose of the Dirac indices
only (the flavour and colour $SU(3)$ indices are not transposed). The
antisymmetric tensor in color space $\epsilon_{abc}$, ensures the baryons’
being color singlets. The space-time coordinate $x$ does nothing with our
studies, and we shall omit it. The matrices $\Gamma_{1,2}$ are Dirac matrices
which describe the Lorentz structure. With a suitable choice of $\Gamma_{1,2}$
and taking a combination of indices of $A,B$ and $C$, the baryon operators are
defined so that they form an irreducible representation of the Lorentz and
flavour groups, as we shall show in this section.

We employ the tensor formalism for flavour $SU(3)$ a la Okubo Okubo:1961jc ;
OSN83 ; MMOkubo65 ; Hara65 ; SchU65  for the quark field $q$, although the
explicit expressions in terms of $up$, $down$ and $strange$ quarks are usually
employed in lattice QCD and QCD sum rule studies. We shall see that the tensor
formulation simplifies the classification of baryons into flavour multiplets
and leads to a straightforward (but generally complicated) derivation of Fierz
identities and chiral transformations of baryon operators. This is in contrast
with the $N_{f}=2$ case where we explicitly included isospin into the
$\Gamma_{1,2}$ matrices and thus produced isospin invariant/covariant objects
Nagata:2007di . The reason for this switch is that in the $N_{f}=3$ case the
baryons form octets and decuplets, rather than doublets and quartets, but the
octet and decuplet projection operators cannot be easily introduced into this
formalism.

###  II.1 Flavour $SU(3)_{f}$ decomposition for baryons

For the sake of notational completeness, we start with some definitions. The
quarks of flavor $SU(3)$ form either the contra-variant ($\mathbf{3}$) or the
covariant ($\mathbf{\bar{3}}$) fundamental representations. They are
distinguished by either upper or lower index as

$\displaystyle q^{A}\in q$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}u\\\ d\\\ s\end{array}\right)\,,$ (5)
$\displaystyle q_{A}\in q^{\dagger}$ $\displaystyle=$
$\displaystyle(u^{*},\;d^{*},\;s^{*})\,.$

The two conjugate fundamental representations transform under flavor $SU(3)$
transformations as

$\displaystyle q$ $\displaystyle\to$ $\displaystyle\exp(i{\vec{\lambda}\over
2}\vec{a}){q}\,,$ (6) $\displaystyle q^{\dagger}$ $\displaystyle\to$
$\displaystyle{q^{\dagger}}\exp(-i{\vec{\lambda}\over 2}\vec{a})\,,$

where $a_{N}$ ($N=1,\cdots,8$) are the octet of $SU(3)_{F}$ group parameters
and $\lambda^{N}$ are the eight Gell-Mann matrices. Since the latter are
Hermitian, we may replace the transposed matrices with the complex conjugate
ones. The set of eight
${\bar{\lambda}}^{N}=-(\lambda^{N})^{T}=-(\lambda^{N})^{*}$ matrices form the
generators of the irreducible $\mathbf{\bar{3}}$ representation.

Now for three quarks, we show flavor $SU(3)$ irreducible decomposition
$\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}=\mathbf{1}\oplus\mathbf{8}\oplus\mathbf{8}\oplus\mathbf{10}$
explicitly in terms of three quarks. It can be done by making suitable
permutation symmetry representations of three-quark products
$q_{A}q_{B}q_{C}$.

  1. 1.

The totally antisymmetric combination which forms the singlet,

$\displaystyle\Psi_{[ABC]}={\cal
N}\left(q_{A}q_{B}q_{C}+q_{B}q_{C}q_{A}+q_{C}q_{A}q_{B}-q_{B}q_{A}q_{C}-q_{A}q_{C}q_{B}-q_{C}q_{B}q_{A}\right)\,.$
(7)

The normalization constant here is ${\cal N}=1/\sqrt{6}$. In the quark model
this corresponds to $\Lambda(1405)$. In order to represent this totally
antisymmetric combination, we can use the totally antisymmetric tensor
$\epsilon^{ABC}$. Then the flavor singlet baryon field $\Lambda$ can be
written as:

$\Lambda\equiv\epsilon^{ABC}\epsilon_{abc}\left(q^{aT}_{A}C\Gamma_{1}q^{b}_{B}\right)\Gamma_{2}q^{c}_{C}\,.$
(8)

  2. 2.

The totally symmetric combination which forms the decuplet,

$\displaystyle\Psi_{\\{ABC\\}}={\cal
N}\left(q_{A}q_{B}q_{C}+q_{B}q_{C}q_{A}+q_{C}q_{A}q_{B}+q_{B}q_{A}q_{C}+q_{A}q_{C}q_{B}+q_{C}q_{B}q_{A}\right)\,.$
(9)

The normalization constant depends on the set of quarks for baryons: for
$q_{A},q_{B},q_{C}=u,d,s,{\cal N}=1/\sqrt{6}$, while it is 1/6 for
$q_{A},q_{B},q_{C}=uuu$. In order to represent this totally symmetric flavor
structure, we introduce the totally symmetric tensor $S_{P}^{ABC}$
($P=1,\cdots,10$). Then the flavor decuplet baryon field $\Delta$ can be
written as:

$\Delta^{P}\equiv
S_{P}^{ABC}\epsilon_{abc}\left(q^{aT}_{A}C\Gamma_{1}q^{b}_{B}\right)\Gamma_{2}q^{c}_{C}\,.$
(10)

The non-zero components of $S_{P}^{ABC}$ ($=1$) are summarized in Table 1. The
rest of components are just zero, for instance, $S_{1}^{112}=0$.

Table 1: Non-zero components of $S_{P}^{ABC}($=1$)$ $P$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
---|---|---|---|---|---|---|---|---|---|---
$ABC$ | 111 | 112 | 113 | 122 | 123 | 133 | 222 | 223 | 233 | 333
Baryons | $\Delta^{++}$ | $\Delta^{+}$ | $\Sigma^{*+}$ | $\Delta^{0}$ | $\Sigma^{*0}$ | $\Xi^{*0}$ | $\Delta^{-}$ | $\Sigma^{*-}$ | $\Xi^{*-}$ | $\Omega^{-}$
  3. 3.

The two mixed symmetry tensors of the $\rho$ and $\lambda$ types are defined
by

$\displaystyle\Psi_{[A\\{B]C\\}}^{\rho}$ $\displaystyle=$ $\displaystyle{\cal
N}\left(2q_{A}q_{B}q_{C}-q_{B}q_{C}q_{A}-q_{C}q_{A}q_{B}-2q_{B}q_{A}q_{C}+q_{A}q_{C}q_{B}+q_{C}q_{B}q_{A}\right)\,,$
(11) $\displaystyle\Psi_{\\{A[B\\}C]}^{\lambda}$ $\displaystyle=$
$\displaystyle{\cal
N}\left(2q_{A}q_{B}q_{C}-q_{B}q_{C}q_{A}-q_{C}q_{A}q_{B}+2q_{B}q_{A}q_{C}-q_{A}q_{C}q_{B}-q_{C}q_{B}q_{A}\right)\,.$

Here the two symbols in $\\{~{}\\}$ are first symmetrized and then the symbols
in $[~{}]$ are anti-symmetrized. The normalization constant depends again on
the number of different kinds of terms. The correspondence of the octet fields
of (11) and the physical ones can be made first by taking the following
combinations

$\displaystyle N_{8\rho}^{N}$ $\displaystyle=$
$\displaystyle\epsilon^{ABD}(\bm{\lambda}^{N})_{DC}\Psi^{\rho}_{[A\\{B]C\\}}\,,$
(12) $\displaystyle N_{8\lambda}^{N}$ $\displaystyle=$
$\displaystyle\epsilon^{BCD}(\bm{\lambda}^{N})_{DA}\Psi^{\lambda}_{\\{A[B\\}C]}\,.$

This kind of “double index” ($DC$ for $N^{N}_{8\rho}$ and $DA$ for
$N^{N}_{8\lambda}$) notation for the baryon flavour has been used by Christos
Christos:1986us . In our discussions, we shall use the following form for the
flavor octet baryon field

$\displaystyle N^{N}$ $\displaystyle\equiv$
$\displaystyle\epsilon^{ABD}(\bm{\lambda}^{N})_{DC}\epsilon_{abc}\left(q^{aT}_{A}C\Gamma_{1}q^{b}_{B}\right)\Gamma_{2}q^{c}_{C}\,.$
(13)

It is of the $\rho$ type. But after using Fierz transformations to interchange
the second and the third quarks, the transformed one contains $\lambda$ type
also, as we shall show in the following. The octet of physical baryon fields
are then determined by

$\displaystyle N^{1}\pm
iN^{2}\sim\Sigma^{\mp}\,,\;\;\;N^{3}\sim\Sigma^{0}\,,\;\;\;N^{8}\sim\Lambda\,,$
(14) $\displaystyle N^{4}\pm iN^{5}\sim\Xi^{-},p\,,\;\;\;N^{6}\pm
iN^{7}\sim\Xi^{0},n\,,$

or put into the $3\times 3$ baryon matrix

$\displaystyle\mathfrak{B}=\left(\begin{array}[]{c c
c}{\Sigma^{0}\over\sqrt{2}}+{\Lambda^{8}\over\sqrt{6}}&\Sigma^{+}&p\\\
\Sigma^{-}&-{\Sigma^{0}\over\sqrt{2}}+{\Lambda^{8}\over\sqrt{6}}&n\\\
\Xi^{-}&\Xi^{0}&-{2\over\sqrt{6}}\Lambda^{8}\end{array}\right)\,.$ (18)

###  II.2 Counting the (in)dependent fields

In this section we investigate independent baryon fields for each Lorentz
group representation which is formed by three quarks. The irreducible
decomposition of the Lorentz group is done as

$\displaystyle\left((\frac{1}{2},0)\oplus(0,\frac{1}{2})\right)^{3}\sim\left((\frac{1}{2},0)\oplus(0,\frac{1}{2})\right)\oplus\left((1,\frac{1}{2})\oplus(\frac{1}{2},1)\right)\oplus\left((\frac{3}{2},0)\oplus(0,\frac{3}{2})\right)\,,$
(19)

where we have ignored the multiplicity on the right hand side. The three
representations are described by the Dirac spinor field, the Rarita-
Schwinger’s vector spinor field and the antisymmetric tensor spinor field,
respectively. In order to establish independent fields when combined with
color, flavor, and Lorentz (spin) degrees of freedom, we employ the method of
Fierz transformations which are essentially equivalent to the use of the Pauli
principle for three quarks. Here we demonstrate the essential idea for the
simplest case for the Dirac spinor, $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$.
Other cases are briefly explained in Appendices A and B.

####  II.2.1 Flavor singlet baryon

Let us start with writing down five baryon fields which contain a diquark
formed by five sets of Dirac matrices,
$1,\gamma_{5},\gamma_{\mu},\gamma_{\mu}\gamma_{5}$ and $\sigma_{\mu\nu}$,

$\displaystyle\begin{array}[]{l}\Lambda_{1}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}Cq_{B}^{b})\gamma_{5}q_{C}^{c}\,,\\\
\Lambda_{2}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}C\gamma_{5}q_{B}^{b})q_{C}^{c}\,,\\\
\Lambda_{3}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}C\gamma_{\mu}\gamma_{5}q_{B}^{b})\gamma^{\mu}q_{C}^{c}\,,\\\
\Lambda_{4}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}C\gamma_{\mu}q_{B}^{b})\gamma^{\mu}\gamma_{5}q_{C}^{c}\,,\\\
\Lambda_{5}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}C\sigma_{\mu\nu}q_{B}^{b})\sigma_{\mu\nu}\gamma_{5}q_{C}^{c}\,.\end{array}$
(25)

Among these five fields, we can show that the fourth and fifth ones vanish,
$\Lambda_{4,5}=0$. This is due to the Pauli principle between the first two
quarks, and can be verified, for instance, by taking the transpose of the
diquark component and compare the resulting three-quark field with the
original expressions Christos:1986us . The Pauli principle can also be used
between the first and the third quarks, so we construct the primed fields
where the second and the third quarks are interchanged, for instance,

$\displaystyle\Lambda^{\prime}_{1}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}Cq_{C}^{c})\gamma_{5}q_{B}^{b}\,.$

Now expressing $\Lambda_{i}$ in terms of the Fierz transformed fields
$\Lambda^{\prime}_{i}$, we find the following relations (see Appendix C),

$\displaystyle\begin{array}[]{l}\Lambda_{1}=-{1\over
4}\Lambda^{\prime}_{1}-{1\over 4}\Lambda^{\prime}_{2}-{1\over
4}\Lambda^{\prime}_{3}\,,\\\ \Lambda_{2}=-{1\over
4}\Lambda^{\prime}_{1}-{1\over 4}\Lambda^{\prime}_{2}+{1\over
4}\Lambda^{\prime}_{3}\,,\\\
\Lambda_{3}=-\Lambda^{\prime}_{1}+\Lambda^{\prime}_{2}+{1\over
2}\Lambda_{3}^{\prime}\,.\end{array}$ (29)

On the other hand, by changing the indices $B,C$ and $b,c$, for instance,

$\displaystyle\Lambda^{\prime}_{1}$ $\displaystyle=$
$\displaystyle\epsilon_{acb}\epsilon^{ACB}(q_{A}^{aT}Cq_{B}^{b})\gamma_{5}q_{C}^{c}$
$\displaystyle=$
$\displaystyle\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}Cq_{B}^{b})\gamma_{5}q_{C}^{c}\,,$

we see that the primed fields are just the corresponding unprimed ones,
$\Lambda^{\prime}_{i}=\Lambda_{i}$. Consequently, we obtain three homogeneous
linear equations whose rank is just one, and we find the following solution

$\Lambda_{3}=4\Lambda_{2}=-4\Lambda_{1}\,,\Lambda_{4}=\Lambda_{5}=0\,.$

We see that there is only one non-vanishing independent field, which in the
quark model corresponds to the odd-parity $\Lambda(1405)$.

####  II.2.2 The flavour decuplet baryons

Among the five decuplet baryon fields formed by the five different
$\gamma$-matrices, only two are non-zero:

$\displaystyle\begin{array}[]{l}\Delta^{P}_{4}=\epsilon_{abc}S_{P}^{ABC}(q_{A}^{aT}C\gamma_{\mu}q_{B}^{b})\gamma^{\mu}\gamma_{5}q_{C}^{c}\,,\\\
\Delta^{P}_{5}=\epsilon_{abc}S_{P}^{ABC}(q_{A}^{aT}C\sigma_{\mu\nu}q_{B}^{b})\sigma_{\mu\nu}\gamma_{5}q_{C}^{c}\,.\end{array}$
(32)

Performing the Fierz transformation and with the relation
$\Delta^{P\prime}_{i}=-\Delta^{P}_{i}$
($\epsilon_{acb}S_{P}^{ACB}=-\epsilon_{abc}S_{P}^{ABC}$), we find that there
is only a trivial (null) solution to the homogeneous linear equations.
Therefore, the Dirac baryon fields (fundamental representation of the Lorentz
group) formed by three quarks can not survive the flavor decuplet.

####  II.2.3 The flavor octet baryon fields

Let us start once again with five fields, which have three potentially non-
zero ones

$\displaystyle\begin{array}[]{l}N_{1}^{N}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}Cq_{B}^{b})\gamma_{5}q_{C}^{c}\,,\\\
N_{2}^{N}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{5}q_{B}^{b})q_{C}^{c}\,,\\\
N_{3}^{N}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{\mu}\gamma_{5}q_{B}^{b})\gamma^{\mu}q_{C}^{c}\,,\\\
N_{4}^{N}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{\mu}q_{B}^{b})\gamma^{\mu}\gamma_{5}q_{C}^{c}=0\,,\\\
N_{5}^{N}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\sigma_{\mu\nu}q_{B}^{b})\sigma_{\mu\nu}\gamma_{5}q_{C}^{c}=0\,.\\\
\end{array}$ (38)

These octet baryon fields have been studied in Refs Ioffe:1981kw ;
Espriu:1983hu ; Chung:1981cc , where the independent ones are clarified. As
before, we perform the Fierz rearrangement to obtain five equations with the
primed fields, while $N_{4}^{N\prime}$ and $N_{5}^{N\prime}$ are not zero. For
the first three equations, $N_{1,2,3}$ on the left hand side should be
expressed by the primed fields. To this end, we can use the Jacobi identity

$\displaystyle\epsilon^{ABD}\lambda_{DC}^{N}+\epsilon^{BCD}\lambda_{DA}^{N}+\epsilon^{CAD}\lambda_{DB}^{N}=0\,,$
(39)

which can be used to relate the original fields $N^{N}_{i}$ and primed ones
$N^{N\prime}_{i}$, for instance,

$\displaystyle\left(\epsilon^{ABD}\lambda_{DC}^{N}+\epsilon^{BCD}\lambda_{DA}^{N}+\epsilon^{CAD}\lambda_{DB}^{N}\right)(q_{A}^{aT}Cq_{B}^{b})\gamma_{5}q_{C}^{c}=0\,,$

from which we find

$\displaystyle N_{1}^{N\prime}=-{1\over 2}N_{1}^{N}\,,$

and the same relations for $N_{2,3}^{N}$. There are no relations between
$N_{4,5}^{N}$ and $N_{4,5}^{N\prime}$. Altogether, we have five equations. The
equations related to $N_{4}^{N}$ and $N_{5}^{N}$ are also necessary because
the corresponding primed ones are not zero. They can be solved to obtain the
following solutions:

${2\over
3}N_{4}^{N\prime}=N^{N}_{3}=N^{N}_{1}-N^{N}_{2}\,,N_{5}^{N\prime}=-3(N^{N}_{1}+N^{N}_{2})\,,$

which indicates that there are two independent octet fields, for instance,
$N^{N}_{1}$ and $N^{N}_{2}$. Thus we have shown the same result just as in the
two-flavour case Nagata:2007di . In the following sections we shall show that
the difference between the two fields $N_{1}$ and $N_{2}$ lies in their chiral
properties: $N^{N}_{1}-N^{N}_{2}$ together with $\Lambda$ belong to
$(\mathbf{\bar{3}},~{}\mathbf{3})\oplus(\mathbf{3},~{}\mathbf{\bar{3}})$, and
the other $N^{N}_{1}+N^{N}_{2}$ belongs to
$(\mathbf{8},~{}\mathbf{1})\oplus(\mathbf{1},~{}\mathbf{8})$.

There are two ways to construct the octet baryon fields. One is done already
as shown in Eqs. (38), whose flavor structure is the same as the $\rho$ type
baryon field $N^{N}_{8\rho}$ in Eqs. (12):

$\displaystyle\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}\longrightarrow(\mathbf{3}\otimes\mathbf{3})\otimes\mathbf{3}\longrightarrow\mathbf{\bar{3}}\otimes\mathbf{3}\longrightarrow\mathbf{8}_{\rho}\,.$
(40)

The other $\lambda$ type baryon field $N^{N}_{8\lambda}$ is complicated when
used straightforwardly:

$\displaystyle\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}\longrightarrow(\mathbf{3}\otimes\mathbf{3})\otimes\mathbf{3}\longrightarrow\mathbf{6}\otimes\mathbf{3}\longrightarrow\mathbf{8}_{\lambda}\,.$
(41)

Therefore, we use another way based on

$\displaystyle\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}\longrightarrow\mathbf{3}\otimes(\mathbf{3}\otimes\mathbf{3})\longrightarrow\mathbf{3}\otimes\mathbf{\bar{3}}\longrightarrow\mathbf{8}^{\prime}_{\rho}\,.$
(42)

This contains partly $\mathbf{8}_{\lambda}$, and it is easily to verify that
(40) and (42) compose a full description of octet baryon which is also fully
described by using (40) and (41). The way $\mathbf{8}_{\rho}$ leads to octet
fields $N^{N}_{i}$, and the other way $\mathbf{8}^{\prime}_{\rho}$ leads to
other five ones

$\displaystyle\begin{array}[]{l}\widetilde{N}_{1}^{N}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}(q_{A}^{aT}Cq_{B}^{b})\gamma_{5}q_{C}^{c}\,,\\\
\widetilde{N}_{2}^{N}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}(q_{A}^{aT}C\gamma_{5}q_{B}^{b})q_{C}^{c}\,,\\\
\widetilde{N}_{3}^{N}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}(q_{A}^{aT}C\gamma_{\mu}\gamma_{5}q_{B}^{b})\gamma^{\mu}q_{C}^{c}\,,\\\
\widetilde{N}_{4}^{N}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}(q_{A}^{aT}C\gamma_{\mu}q_{B}^{b})\gamma^{\mu}\gamma_{5}q_{C}^{c}\,,\\\
\widetilde{N}_{5}^{N}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}(q_{A}^{aT}C\sigma_{\mu\nu}q_{B}^{b})\sigma_{\mu\nu}\gamma_{5}q_{C}^{c}\,.\end{array}$
(48)

However, these fields can be related to the previous ones by changing the
flavor and color indices $B,C$ and $b,c$:

$\widetilde{N}_{i}^{N}=-N_{i}^{N\prime}\,.$

In nearly all the cases, the octet baryon fields from the second way can be
related to the ones from the first way. Therefore, we shall omit the
discussion of the second octet. One exception which concerns the chiral
representation
$(\mathbf{\bar{3}},~{}\mathbf{3})\otimes(\mathbf{6},~{}\mathbf{3})$ is
discussed in Appendix D.

###  II.3 A short summary of independent baryon fields

Properties of baryons fields expressed by the Rarita-Schwinger fields with one
Lorentz index and those of the antisymmetric tensor-spinor fields with two
Lorentz indices are discussed in Appendices A and B, respectively. Here let us
make a short summary for independent baryon fields for all cases constructed
by three quarks. For simplicity, here we suppress the antisymmetric tensor in
color space $\epsilon_{abc}$, since it appears in all baryon fields in the
same manner. Furthermore, it is convenient to introduce a “tilde-transposed”
quark field $\widetilde{q}$ as follows

$\displaystyle\widetilde{q}=q^{T}C\gamma_{5}\,.$ (49)

As we have shown already, for Dirac fields without Lorentz index, there are
one singlet field $\Lambda$ and two octet fields $N^{N}_{1}$ and $N^{N}_{2}$:

$\displaystyle\Lambda_{1}$ $\displaystyle=$
$\displaystyle\epsilon^{ABC}(\widetilde{q}_{A}\gamma_{5}q_{B})\gamma_{5}q_{C}\,,$
$\displaystyle N^{N}_{1}$ $\displaystyle=$
$\displaystyle\epsilon^{ABD}\lambda_{DC}^{N}(\widetilde{q}_{A}\gamma_{5}q_{B})\gamma_{5}q_{C}\,,$
$\displaystyle N^{N}_{2}$ $\displaystyle=$
$\displaystyle\epsilon^{ABD}\lambda_{DC}^{N}(\widetilde{q}_{A}q_{B})q_{C}\,.$

For the Rarita-Schwinger fields with one Lorentz index, we would consider one
singlet, three octet and one decuplet fields:

$\displaystyle\Lambda_{1\mu}$ $\displaystyle=$
$\displaystyle\epsilon^{ABC}(\widetilde{q}_{A}\gamma_{5}q_{B})\gamma_{\mu}q_{C}\,,$
$\displaystyle N^{N}_{1\mu}$ $\displaystyle=$
$\displaystyle\epsilon^{ABD}\lambda_{DC}^{N}(\widetilde{q}_{A}\gamma_{5}q_{B})\gamma_{\mu}q_{C}\,,$
$\displaystyle N^{N}_{2\mu}$ $\displaystyle=$
$\displaystyle\epsilon^{ABD}\lambda_{DC}^{N}(\widetilde{q}_{A}q_{B})\gamma_{\mu}\gamma_{5}q_{C}\,,$
$\displaystyle N^{N}_{3\mu}$ $\displaystyle=$
$\displaystyle-\epsilon^{ABD}\lambda_{DC}^{N}(\widetilde{q}_{A}\gamma_{\mu}q_{B})\gamma_{5}q_{C}\,,$
$\displaystyle\Delta^{P}_{5\mu}$ $\displaystyle=$ $\displaystyle-
S_{P}^{ABC}(\widetilde{q}_{A}\gamma_{\mu}\gamma_{5}q_{B})q_{C}\,.$

However, we find that $\Lambda_{1\mu}=\gamma_{\mu}\gamma_{5}\Lambda$,
$N^{N}_{1\mu}=\gamma_{\mu}\gamma_{5}N^{N}_{1}$ and
$N^{N}_{2\mu}=\gamma_{\mu}\gamma_{5}N^{N}_{2}$. So, there are two non-
vanishing independent fields: one octet field $N^{N}_{\mu}$ and one decuplet
field $\Delta_{\mu}$. By using the projection operator:

$\displaystyle P^{3/2}_{\mu\nu}=(g_{\mu\nu}-{1\over
4}\gamma_{\mu}\gamma_{\nu})\,,$ (50)

they can be written as

$\displaystyle N^{N}_{\mu}=P^{3/2}_{\mu\nu}N^{N}_{3\nu}$ $\displaystyle=$
$\displaystyle-(g_{\mu\nu}-{1\over
4}\gamma_{\mu}\gamma_{\nu})\epsilon^{ABD}\lambda_{DC}^{N}(\widetilde{q}_{A}\gamma_{\mu}q_{B})\gamma_{5}q_{C}$
$\displaystyle=$ $\displaystyle N^{N}_{3\mu}+{1\over
4}\gamma_{\mu}\gamma_{5}(N^{N}_{1}-N^{N}_{2})\,,$
$\displaystyle\Delta^{P}_{\mu}=P^{3/2}_{\mu\nu}\Delta^{P}_{5\nu}$
$\displaystyle=$ $\displaystyle-(g_{\mu\nu}-{1\over
4}\gamma_{\mu}\gamma_{\nu})S_{P}^{ABC}(\widetilde{q}_{A}\gamma_{\mu}\gamma_{5}q_{B})q_{C}$
$\displaystyle=$ $\displaystyle\Delta^{P}_{5\mu}\,.$

For tensor fields with two antisymmetric Lorentz indices, we would have one
singlet, three octet and two decuplet fields:

$\displaystyle\Lambda_{1\mu}$ $\displaystyle=$
$\displaystyle\epsilon^{ABC}(\widetilde{q}_{A}\gamma_{5}q_{B})\sigma_{\mu\nu}\gamma_{5}q_{C}\,,$
$\displaystyle N^{N}_{3\mu\nu}$ $\displaystyle=$
$\displaystyle-\epsilon^{ABD}\lambda_{DC}^{N}(\widetilde{q}_{A}\gamma_{\mu}q_{B})\gamma_{\nu}q_{C}+(\mu\leftrightarrow\nu)\,,$
$\displaystyle N^{N}_{10\mu\nu}$ $\displaystyle=$
$\displaystyle\epsilon^{ABD}\lambda_{DC}^{N}(\widetilde{q}_{A}\gamma_{5}q_{B})\sigma_{\mu\nu}\gamma_{5}q_{C}\,,$
$\displaystyle N^{N}_{11\mu\nu}$ $\displaystyle=$
$\displaystyle\epsilon^{ABD}\lambda_{DC}^{N}(\widetilde{q}_{A}q_{B})\sigma_{\mu\nu}q_{C}\,,$
$\displaystyle\Delta^{P}_{2\mu\nu}$ $\displaystyle=$ $\displaystyle-
S_{P}^{ABC}(\widetilde{q}_{A}\gamma_{\mu}\gamma_{5}q_{B})\gamma_{\nu}\gamma_{5}q_{C}+(\mu\leftrightarrow\nu)\,,$
$\displaystyle\Delta^{P}_{7\mu\nu}$ $\displaystyle=$ $\displaystyle
S_{P}^{ABC}(\widetilde{q}_{A}\sigma_{\mu\nu}\gamma_{5}q_{B})\gamma_{5}q_{C}\,.$

But in this case, we can show that there is only one non-vanishing field
$\Delta_{\mu\nu}$:

$\displaystyle\Delta^{P}_{\mu\nu}=\Gamma^{\mu\nu\alpha\beta}\Delta^{P}_{7\mu\nu}$
$\displaystyle=$
$\displaystyle\Gamma^{\mu\nu\alpha\beta}S_{P}^{ABC}(\widetilde{q}_{A}\sigma_{\mu\nu}\gamma_{5}q_{B})\gamma_{5}q_{C}$
$\displaystyle=$ $\displaystyle\Delta^{P}_{7\mu\nu}-{i\over
2}\gamma_{\mu}\gamma_{5}\Delta^{P}_{5\nu}+{i\over
2}\gamma_{\nu}\gamma_{5}\Delta^{P}_{5\mu}\,,$

where

$\displaystyle\Gamma^{\mu\nu\alpha\beta}=(g^{\mu\alpha}g^{\nu\beta}-{1\over
2}g^{\nu\beta}\gamma^{\mu}\gamma^{\alpha}+{1\over
2}g^{\mu\beta}\gamma^{\nu}\gamma^{\alpha}+{1\over
6}\sigma^{\mu\nu}\sigma^{\alpha\beta})\,.$ (51)

##  III Chiral Transformations

In this section, we establish the chiral transformation properties of the
baryon fields which we have obtained in the previous section. Technically,
this requires somewhat complicated algebra. However, the final result will be
understood by making the left and right handed decomposition, as we will
perform in the next section.

Let us start with the chiral transformation properties of quarks which are
given by the following equations:

$\displaystyle\bf{U(1)_{V}}$ $\displaystyle:$ $\displaystyle
q\to\exp(i{\lambda^{0}\over 2}a_{0})q=q+\delta q\,,$
$\displaystyle\bf{SU(3)_{V}}$ $\displaystyle:$
$\displaystyle{q}\to\exp(i{\vec{\lambda}\over
2}\vec{a}){q}=q+\delta^{\vec{a}}q\,,$ (52) $\displaystyle\bf{U(1)_{A}}$
$\displaystyle:$ $\displaystyle q\to\exp(i\gamma_{5}{\lambda^{0}\over
2}b_{0})q=q+\delta_{5}q\,,$ $\displaystyle\bf{SU(3)_{A}}$ $\displaystyle:$
$\displaystyle{q}\to\exp(i\gamma_{5}{\vec{\lambda}\over
2}\vec{b}){q}=q+\delta_{5}^{\vec{b}}q\,,$

where $\lambda^{0}=\sqrt{2/3}\mathbf{1}$, $a^{0}$ is an infinitesimal
parameter for the $U(1)_{V}$ transformation, $\vec{a}$ the octet of
$SU(3)_{V}$ group parameters, $b^{0}$ an infinitesimal parameter for the
$U(1)_{A}$ transformation, and $\vec{b}$ the octet of the chiral
transformations.

The $U(1)_{V}$ chiral transformation is trivial which picks up a phase factor
proportional to the baryon number. The $U(1)_{A}$ chiral transformation is
slightly less trivial, and the baryon fields are transformed as

$\displaystyle\delta_{5}\Lambda$ $\displaystyle=$
$\displaystyle-i\gamma_{5}\sqrt{1\over 6}b^{0}\Lambda\,,$
$\displaystyle\delta_{5}(N^{N}_{1}-N^{N}_{2})$ $\displaystyle=$
$\displaystyle-i\gamma_{5}\sqrt{1\over 6}b^{0}(N^{N}_{1}-N^{N}_{2})\,,$
$\displaystyle\delta_{5}(N^{N}_{1}+N^{N}_{2})$ $\displaystyle=$ $\displaystyle
i\gamma_{5}\sqrt{3\over 2}b^{0}(N^{N}_{1}+N^{N}_{2})\,,$
$\displaystyle\delta_{5}N^{N}_{\mu}$ $\displaystyle=$ $\displaystyle
i\gamma_{5}\sqrt{1\over 6}b^{0}N^{N}_{\mu}\,,$ (53)
$\displaystyle\delta_{5}\Delta^{P}_{\mu}$ $\displaystyle=$ $\displaystyle
i\gamma_{5}\sqrt{1\over 6}b^{0}\Delta^{P}_{\mu}\,,$
$\displaystyle\delta_{5}\Delta^{P}_{\mu\nu}$ $\displaystyle=$ $\displaystyle
i\gamma_{5}\sqrt{3\over 2}b^{0}\Delta^{P}_{\mu\nu}\,.$

We note that the combinations of $N_{1}^{N}\pm N_{2}^{N}$ form different
representations.

Under the vector chiral transformation, the baryon fields are transformed as

$\displaystyle\delta^{\vec{a}}\Lambda$ $\displaystyle=$ $\displaystyle 0\,,$
$\displaystyle\delta^{\vec{a}}N^{N}_{1}$ $\displaystyle=$
$\displaystyle-a^{M}f^{NMO}N^{O}_{1}\,,$
$\displaystyle\delta^{\vec{a}}N^{N}_{2}$ $\displaystyle=$
$\displaystyle-a^{M}f^{NMO}N^{O}_{2}\,,$
$\displaystyle\delta^{\vec{a}}N^{N}_{\mu}$ $\displaystyle=$
$\displaystyle-a^{M}f^{NMO}N^{N}_{\mu}\,,$ (54)
$\displaystyle\delta^{\vec{a}}\Delta^{P}_{\mu}$ $\displaystyle=$
$\displaystyle{3i\over 2}a^{M}g_{7}^{PMQ}\Delta^{Q}_{\mu}\,,$
$\displaystyle\delta^{\vec{a}}\Delta^{P}_{\mu\nu}$ $\displaystyle=$
$\displaystyle{3i\over 2}a^{M}g_{7}^{PMQ}\Delta^{Q}_{\mu\nu}\,,$

where $f^{ABC}$ is the standard antisymmetric structure constant of $SU(3)$,
and $g^{ABC}_{7}$ is defined in Table 2. Eqs. (III) show nothing but the
flavor charge of the baryons. For example, we can show explicitly:

$\displaystyle\delta^{a3}p=+{i\over 2}a_{3}p\,,\delta^{a3}n=-{i\over
2}a_{3}n\,,\delta^{a3}\Delta^{++}={3i\over 2}a_{3}\Delta^{++}\,\cdots$

The transformation rule under the axial-vector chiral transformations are
rather complicated as they are no longer conserved and reflect the internal
structure of baryons. To start with, we have the axial transformation of the
three-quark baryon fields such as

$\displaystyle\delta_{5}^{\vec{b}}\Lambda$ $\displaystyle=$
$\displaystyle\epsilon_{abc}\epsilon^{ABC}\Big{(}(q_{A}^{aT}Cq_{B}^{b})\gamma_{5}(\delta_{5}^{\vec{b}}q_{C}^{c})+(q_{A}^{aT}C(\delta_{5}^{\vec{b}}q_{B}^{b}))\gamma_{5}q_{C}^{c}+((\delta_{5}^{\vec{b}}q_{A}^{aT})Cq_{B}^{b})\gamma_{5}q_{C}^{c}\Big{)}\,.$

The calculation is complicated, but rather straightforward. Here, we show
therefore the final result of the axial transformation:

$\displaystyle\delta_{5}^{\vec{b}}\Lambda$ $\displaystyle=$
$\displaystyle{i\over 2}\gamma_{5}b^{N}(N^{N}_{1}-N^{N}_{2})\,,$
$\displaystyle\delta_{5}^{\vec{b}}(N^{N}_{1}-N^{N}_{2})$ $\displaystyle=$
$\displaystyle{4i\over
3}\gamma_{5}b^{N}\Lambda+i\gamma_{5}b^{M}d^{NMO}(N^{O}_{1}-N^{O}_{2})\,,$
$\displaystyle\delta_{5}^{\vec{b}}(N^{N}_{1}+N^{N}_{2})$ $\displaystyle=$
$\displaystyle-\gamma_{5}b^{M}f^{NMO}(N^{O}_{1}+N^{O}_{2})\,,$
$\displaystyle\delta_{5}^{\vec{b}}N^{N}_{\mu}$ $\displaystyle=$ $\displaystyle
i\gamma_{5}b^{M}(d^{MNO}-{2i\over
3}f^{MNO})N^{O}_{\mu}+i\gamma_{5}b^{M}g_{3}^{MNP}\Delta^{P}_{\mu}\,,$ (55)
$\displaystyle\delta_{5}^{\vec{b}}\Delta^{P}_{\mu}$ $\displaystyle=$
$\displaystyle-2i\gamma_{5}b^{M}g_{5}^{PMO}N^{O}_{\mu}+{i\over
2}\gamma_{5}b^{M}g_{7}^{PMQ}\Delta^{Q}_{\mu}\,,$
$\displaystyle\delta_{5}^{\vec{b}}\Delta^{P}_{\mu\nu}$ $\displaystyle=$
$\displaystyle{3i\over 2}\gamma_{5}b^{M}g_{7}^{PMQ}\Delta^{Q}_{\mu\nu}\,.$

The coefficients $d^{ABC}$ are the standard symmetric structure constants of
$SU(3)$. For completeness, we show the following equation which define the $d$
and $f$ coefficients

$\displaystyle\lambda^{N}_{AB}\lambda^{M}_{BC}$ $\displaystyle=$
$\displaystyle(\lambda^{N}\lambda^{M})_{AC}={1\over
2}\\{\lambda^{N},\lambda^{M}\\}_{AC}+{1\over 2}[\lambda^{N},\lambda^{M}]_{AC}$
$\displaystyle=$ $\displaystyle{2\over
3}\delta^{NM}\delta_{AC}+(d^{NMO}+if^{NMO})\lambda^{O}_{AC}\,.$

Furthermore, the following formulae define the coefficients $g_{3}$, $g_{5}$
and $g_{7}$, which are proved by using $Mathematica$, a software good at
matrix calculation:

$\displaystyle\epsilon^{ADE}\lambda^{N}_{DB}\lambda^{M}_{EC}=g_{1}^{NMO}\epsilon^{ABD}\lambda^{O}_{DC}+g_{2}^{NMO}\epsilon^{ACD}\lambda^{O}_{DB}+g_{3}^{NMP}S_{P}^{ABC}+g_{4}^{NM}\epsilon^{ABC}\,,$
(57) $\displaystyle
S_{Q}^{ABD}\lambda^{M}_{DC}=g_{5}^{QMO}\epsilon^{ABD}\lambda^{O}_{DC}+g_{6}^{QMO}\epsilon^{ACD}\lambda^{O}_{DB}+g_{7}^{QMP}S_{P}^{ABC}+g_{8}^{QM}\epsilon^{ABC}\,,$

where indices $A\sim E$ take values 1, 2 and 3, $N$, $M$ and $O$ $1,\cdots,8$,
and $P$ and $Q$ $1,\cdots,10$. The coefficients $g_{3}$, $g_{5}$ and $g_{7}$
are listed in Table 2, where we use “0” instead of “10”. Other coefficients
can be related to $d$, $f$, $g_{3}$, $g_{5}$ and $g_{7}$:

$\displaystyle g_{1}^{MNO}$ $\displaystyle=$ $\displaystyle-d^{MNO}-{i\over
3}f^{MNO}\,,$ $\displaystyle g_{2}^{MNO}$ $\displaystyle=$ $\displaystyle
d^{MNO}-{i\over 3}f^{MNO}\,,$ $\displaystyle g_{4}^{MN}$ $\displaystyle=$
$\displaystyle-{1\over 3}\delta^{MN}\,,$ (58) $\displaystyle g_{6}^{QMO}$
$\displaystyle=$ $\displaystyle-2g_{5}^{QMO}\,,$ $\displaystyle g_{8}^{MN}$
$\displaystyle=$ $\displaystyle 0\,.$

Let us explain Eqs. (57) a bit more. The quantities on the left hand side have
three indices $A$, $B$ and $C$, and therefore, they are regarded as direct
products of three fundamental representations of $SU(3)$:
$\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}$. They can be decomposed into
irreducible components by applying the four kinds of operators:
$\epsilon_{ABC}$, $\epsilon^{ABD}\lambda^{O}_{DC}$,
$\epsilon^{ACD}\lambda^{O}_{DB}$ and $S_{P}^{ABC}$, which correspond to
$\mathbf{1}$, $\mathbf{8}$, $\mathbf{8}$ and $\mathbf{10}$ of $SU(3)$,
respectively.

Eqs. (III) and (III) imply that $\Lambda$ and $N^{N}_{1}-N^{N}_{2}$ are
together combined into one chiral multiplet, and $N^{N}_{\mu}$ and
$\Delta^{P}_{\mu}$ are together combined into another chiral multiplet. While
$N^{N}_{1}+N^{N}_{2}$ and $\Delta^{P}_{\mu\nu}$ are transformed into
themselves under chiral transformation. In our following discussion, we will
find that $\Lambda$ and $N^{N}_{1}-N^{N}_{2}$ belong to the chiral
representation
$(\mathbf{\bar{3}},\mathbf{3})\oplus(\mathbf{3},\mathbf{\bar{3}})$,
$N^{N}_{1}+N^{N}_{2}$ belongs to the chiral representation
$(\mathbf{8},\mathbf{1})\oplus(\mathbf{1},\mathbf{8})$, $N^{N}_{\mu}$ and
$\Delta^{P}_{\mu}$ belong to the chiral representation
$(\mathbf{6},\mathbf{3})\oplus(\mathbf{3},\mathbf{6})$, and
$\Delta^{P}_{\mu\nu}$ belongs to the chiral representation
$(\mathbf{10},\mathbf{1})\oplus(\mathbf{1},\mathbf{10})$. We show several
examples of the axial-vector chiral transformation:

$\displaystyle\delta_{5}^{b3}p_{-}={i\over
2}\gamma_{5}b_{3}p_{-}\,,\delta_{5}^{b3}p_{+}={i\over
2}\gamma_{5}b_{3}p_{+}\,,\delta_{5}^{b3}p_{\mu}={5i\over
6}\gamma_{5}b_{3}p_{\mu}-{4i\over 3}\gamma_{5}b_{3}\Delta^{+}_{\mu}\,,$

where $p_{-}$ belongs to the octet baryon fields $N_{1}^{N}-N_{2}^{N}$,
$p_{+}$ belongs to $N^{N}_{1}+N^{N}_{2}$, and $p_{\mu}$ belongs to
$N^{N}_{\mu}$ (see Eqs. (14)).

Table 2: $g$-coefficients defined by Eqs. (57) $g_{3}$ | 133, 138, 144, 146, 254, 256, 272, 279, 439, 463, 468, 573, 578, 612, 619, 636 | $-1/3$
---|---|---
| 162, 169, 313, 318, 349, 366, 414, 416, 524, 526, 643, 648, 722, 729, 753, 758 | $1/3$
| 154, 179, 215, 233, 246, 269, 328, 359, 376, 424, 455, 478, 516, 563, 622, 658, 712, 743, 765 | $-i/3$
| 125, 156, 172, 238, 244, 262, 323, 426, 473, 514, 539, 545, 568, 629, 653, 675, 719, 736, 748 | $i/3$
| 183, 686, 818, 835, 849 | $-1/\sqrt{3}$ | 167, 251, 277, 411, 570, 640 | $-1$ | 342, 364 | $-2/3$
| 188, 385, 489, 813, 866 | $1/\sqrt{3}$ | 141, 460, 521, 617, 727, 750 | $1$ | 432, 634 | $2/3$
| 283, 288, 589, 876 | $-i/\sqrt{3}$ | 177, 421, 470, 511, 560, 627 | $-i$ | 352, 374 | $-2i/3$
| 786, 823, 828, 859 | $i/\sqrt{3}$ | 151, 241, 267, 650, 717, 740 | $i$ | 532, 734 | $2i/3$
$g_{5}$ | 125, 141, 227, 261, 313, 346, 357, 414, 425, 614, 625 | $1/6$ | 318, 668, 881, 984 | $1/2\sqrt{3}$
| 663, 716, 727, 813, 846, 857, 927, 943, 961, 057, 064 |  | 381, 686, 818, 948 | $-1/2\sqrt{3}$
| 114, 152, 216, 272, 331, 364, 375, 441, 452, 636, 641 | $-1/6$ | 382, 678, 882, 985 | $i/2\sqrt{3}$
| 652, 761, 772, 831, 864, 875, 916, 934, 972, 046, 075 |  | 328, 687, 828, 958 | $-i/2\sqrt{3}$
| 115, 124, 217, 226, 332, 347, 365, 424, 451, 615, 642 | $i/6$ | 234, 436 | $1/3$
| 673, 726, 771, 823, 856, 874, 953, 962, 971, 065, 074 |  | 243, 463 | $-1/3$
| 142, 151, 262, 271, 323, 356, 374, 415, 442, 624, 637 | $-i/6$ | 253, 473, 512, 554, 567 | $i/3$
| 651, 717, 762, 832, 847, 865, 917, 926, 935, 047, 056 |  | 235, 437, 521, 545, 576 | $-i/3$
| 583 | $1/\sqrt{3}$ | 538 | $-1/\sqrt{3}$
$g_{7}$ | 112, 143, 232, 245, 263, 315, 362, 448, 465, 619 | $1/3$ | 214, 333, 346, 412, 513, 518 | $2/3$
| 636, 665, 714, 768, 815, 844, 916, 945, 046, 069 |  | 542, 549, 564, 566, 643, 869, 968 |
| 434, 939 | $-1/3$ | 838 | $-2/3$
| 372, 675, 724, 825, 854, 926, 955, 056, 079 | $i/3$ | 422, 523, 552, 574, 653, 978 | $2i/3$
| 122, 153, 255, 273, 325, 458, 475, 629, 778 | $-i/3$ | 224, 356, 528, 559, 576, 879 | $-2i/3$
| 131, 211, 341, 417, 640, 867, 960 | $1$ | 181, 282, 484, 787 | $1/\sqrt{3}$
| 737 | $-1$ | 686, 989 | $-1/\sqrt{3}$
| 221, 351, 877 | $i$ | 080 | $-2/\sqrt{3}$
| 427, 650, 970 | $-i$ |  |

##  IV Chiral multiplets/representations

So far, we have performed classifications without explicitly taking into
account the left- and right-handed components of the quark fields. However, it
does not require great imagination to see that the chiral properties are also
conveniently studied in that language, since chiral symmetry is defined as the
symmetries upon each chiral field. Hence, we define the left- and right-handed
(chiral or Weyl representation) quark fields as

$\displaystyle L\equiv q_{L}={1-\gamma_{5}\over
2}q\mbox{,~{}~{}~{}~{}~{}and}~{}~{}~{}~{}~{}R\equiv q_{R}={1+\gamma_{5}\over
2}q\,.$ (59)

They form the fundamental representations of both the Lorentz group and the
chiral group,

$\displaystyle L:$ $\displaystyle{\rm
Lorentz:}~{}~{}(\frac{1}{2},0)\,,~{}~{}~{}~{}{\rm Chiral:}~{}~{}(3,1)\,,$
$\displaystyle R:$ $\displaystyle{\rm
Lorentz:}~{}~{}(0,\frac{1}{2})\,,~{}~{}~{}~{}{\rm Chiral:}~{}~{}(1,3)\,.$

It is convenient first to note that $\gamma$-matrices are classified into two
categories; chiral-even and chiral-odd classes. The chiral-even
$\gamma$-matrices survive forming diquarks with identical chiralities, while
the chiral-odd ones form diquarks from quarks with opposite chiralities. The
chiral-even and -odd $\gamma$-matrices are

chiral-even: $\displaystyle 1,\gamma_{5},\sigma_{\mu\nu}\,,$ chiral-odd:
$\displaystyle\gamma_{\mu},\gamma_{\mu}\gamma_{5}\,.$

Therefore, we have six non-vanishing diquarks in the chiral representations,

$\displaystyle\left.\begin{array}[]{l}L^{T}CL=-L^{T}C\gamma_{5}L\\\
R^{T}CR=+R^{T}C\gamma_{5}R\end{array}\right\\}$
$\displaystyle(0,0)\oplus(0,0)\,,~{}~{}~{}~{}~{}(\mathbf{\bar{3}},~{}\mathbf{1})\oplus(\mathbf{1},~{}\mathbf{\bar{3}})\,,$
(62)
$\displaystyle\left.\begin{array}[]{l}L^{T}C\gamma_{\mu}\gamma_{5}R=+L^{T}C\gamma_{\mu}R\\\
R^{T}C\gamma_{\mu}\gamma_{5}L=-R^{T}C\gamma_{\mu}L\end{array}\right\\}$
$\displaystyle(\frac{1}{2},\frac{1}{2})\oplus(\frac{1}{2},\frac{1}{2})\,,~{}~{}~{}~{}(\mathbf{3},~{}\mathbf{3})\oplus(\mathbf{3},~{}\mathbf{3})\,,$
(65) $\displaystyle\left.\begin{array}[]{l}L^{T}C\sigma_{\mu\nu}L\\\
R^{T}C\sigma_{\mu\nu}R\end{array}\right\\}$
$\displaystyle(1,0)\oplus(0,1)\,,~{}~{}~{}~{}~{}~{}~{}(\mathbf{6},~{}\mathbf{1})\oplus(\mathbf{1},~{}\mathbf{6})\,,$
(68)

where we have indicated the Lorentz and chiral representations of the
diquarks.

For three quarks, we have

$\displaystyle(L+R)^{3}\to\left\\{\begin{array}[]{l}LLL~{}~{}~{}~{}(\frac{1}{2},0)\oplus(\frac{3}{2},0)\,,~{}~{}~{}~{}(\mathbf{1},\mathbf{1})\oplus(\mathbf{8},\mathbf{1})\oplus(\mathbf{8},\mathbf{1})\oplus(\mathbf{10},\mathbf{1})\\\
LLR~{}~{}~{}~{}(0,\frac{1}{2})\oplus(1,\frac{1}{2})\,,~{}~{}~{}~{}(\mathbf{\bar{3}},\mathbf{3})\oplus(\mathbf{6},\mathbf{3})\end{array}\right.$
(71)

and together with the terms where $L$ and $R$ are exchanged. Now we discuss
the independent fields in terms of the chiral representations. Once again, for
illustration we will discuss here the case of the simplest Lorentz
representation $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ for the Dirac fields.

###  IV.1 Independent $(LL)L$ fields

The $(LL)L$ must belong to one of the following chiral representations:
$(\mathbf{1},\mathbf{1})\oplus(\mathbf{8},\mathbf{1})\oplus(\mathbf{8},\mathbf{1})\oplus(\mathbf{10},\mathbf{1})$.
For each chiral representation, there is one flavor representation available.

For $(\mathbf{1},~{}\mathbf{1})\to\mathbf{1_{f}}$, there are apparently two
non-zero fields

$\displaystyle\begin{array}[]{l}\Lambda_{L1}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}CL_{B}^{b})\gamma_{5}L_{C}^{c}\,,\\\
\Lambda_{L2}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}C\gamma_{5}L_{B}^{b})L_{C}^{c}\,,\\\
\Lambda_{L3}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}C\gamma_{\mu}\gamma_{5}L_{B}^{b})\gamma^{\mu}L_{C}^{c}=0\,,\end{array}$
(75)

where $\Lambda^{L}_{3}$ vanishes because $\gamma_{\mu}\gamma_{5}$ is chiral-
odd

$L^{T}C\gamma_{\mu}\gamma_{5}L=0\,.$

After performing the Fierz transformation to relate $\Lambda_{Li}$ and
$\Lambda^{\prime}_{Li}$ as we have done before, and solving the coupled
equations, we find the solution that all such fields vanish.

For $(\mathbf{10},~{}\mathbf{1})\to\mathbf{10_{f}}$, we would have again two
non-zero components:

$\displaystyle\begin{array}[]{l}\Delta^{P}_{L4}=\epsilon_{abc}S_{P}^{ABC}(L_{A}^{aT}C\gamma_{\mu}L_{B}^{b})\gamma^{\mu}\gamma_{5}L_{C}^{c}\,,\\\
\Delta^{P}_{L5}=\epsilon_{abc}S_{P}^{ABC}(L_{A}^{aT}C\sigma_{\mu\nu}L_{B}^{b})\sigma^{\mu\nu}\gamma_{5}L_{C}^{c}\,.\end{array}$
(78)

Performing the Fierz transformation to relate $\Delta^{P}_{Li}$ and
$\Delta^{P\prime}_{Li}$, we obtain the solution that all such $(LL)L$ fields
vanish.

Finally for $(\mathbf{8},~{}\mathbf{1})\to\mathbf{8_{f}}$, we may consider
once again two non-zero fields to start with

$\displaystyle\begin{array}[]{l}N^{N}_{L1}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}CL_{B}^{b})\gamma_{5}L_{C}^{c}\,,\\\
N^{N}_{L2}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}C\gamma_{5}L_{B}^{b})L_{C}^{c}\,.\end{array}$
(81)

Applying the Fierz transformation to relate $N^{N}_{Li}$ and
$N^{N\prime}_{Li}$, we obtain the solution

$N^{N}_{L2}=N^{N}_{L1}\,.$

Therefore, there is only one independent $(LL)L$ $\mathbf{8_{f}}$ field.

###  IV.2 Independent $(LL)R$ fields

The chiral representations of $(LL)R$ are
$(\mathbf{\bar{3}},~{}\mathbf{3})\oplus(\mathbf{6},~{}\mathbf{3})$. We will
study them separately in the following.

For $(\mathbf{\bar{3}},~{}\mathbf{3})\to\mathbf{1_{f}}$, there appears to
exist two non-zero components among the five fields,

$\displaystyle\begin{array}[]{l}\Lambda_{M1}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}CL_{B}^{b})\gamma_{5}R_{C}^{c}\,,\\\
\Lambda_{M2}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}C\gamma_{5}L_{B}^{b})R_{C}^{c}\,,\\\
\Lambda_{M3}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}C\gamma_{\mu}\gamma_{5}L_{B}^{b})\gamma^{\mu}R_{C}^{c}=0\,,\\\
\Lambda_{M4}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}C\gamma_{\mu}L_{B}^{b})\gamma^{\mu}\gamma_{5}R_{C}^{c}=0\,,\\\
\Lambda_{M5}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}C\sigma_{\mu\nu}L_{B}^{b})\sigma^{\mu\nu}\gamma_{5}R_{C}^{c}=0\,,\end{array}$
(87)

where $M$ (mixed) indicates that the fields contain both left and right handed
quarks. Performing the Fierz transformation to relate $\Lambda_{Mi}$ and
$\Lambda^{\prime}_{Mi}$, we obtain the following relations

$\Lambda^{\prime}_{M4}=-\Lambda^{\prime}_{M3}=-2\Lambda_{M2}=2\Lambda_{M1}\,.$

We may consider other ten combinations formed by $(LR)$ and $(RL)$ diquarks,
$(LR)L$ and $(RL)L$. However, they can be related to the above ones of $(LL)R$
by a rearrangement of indices as well as the Fierz transformation, for
instance,

$\displaystyle\Lambda_{M6}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}CR_{B}^{b})\gamma_{5}L_{C}^{c}=\Lambda_{M1}^{\prime}\,.$
(88)

Therefore, we have only one independent field.

For the chiral representation
$(\mathbf{6},~{}\mathbf{3})\rightarrow\mathbf{10_{f}}$, we can write five
fields containing diquarks formed by five Dirac matrices. However, we can show
that after performing the Fierz transformation all fields vanish. Therefore,
this representation can not support three-quark fields.

The baryon fields of chiral representations
$(\mathbf{\bar{3}},~{}\mathbf{3})\rightarrow\mathbf{8_{f}}$ can be formed

$\displaystyle\begin{array}[]{l}N^{N}_{M1}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}CL_{B}^{b})\gamma_{5}R_{C}^{c}\,,\\\
N^{N}_{M2}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}C\gamma_{5}L_{B}^{b})R_{C}^{c}\,,\\\
N^{N}_{M3}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}C\gamma_{\mu}\gamma_{5}L_{B}^{b})\gamma^{\mu}R_{C}^{c}=0\,,\\\
N^{N}_{M4}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}C\gamma_{\mu}L_{B}^{b})\gamma^{\mu}\gamma_{5}R_{C}^{c}=0\,,\\\
N^{N}_{M5}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}C\sigma_{\mu\nu}L_{B}^{b})\sigma^{\mu\nu}\gamma_{5}R_{C}^{c}=0\,,\end{array}$
(94)

where we see that there are two non-zero fields. Applying the Fierz
transformation, we can verify that there is only one independent field with
the following relations

$N^{N\prime}_{M4}=-N^{N\prime}_{M3}=-2N^{N}_{M2}=2N^{N}_{M1}\,.$

Another chiral representation
$(\mathbf{6},~{}\mathbf{3})\rightarrow\mathbf{8_{f}}$ can be constructed by
the combinations similar to (94), for instance,

$\displaystyle
N^{N}_{(6,3)1}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}\\{(L_{A}^{aT}CL_{B}^{b})\gamma_{5}R_{C}^{c}+(L_{B}^{aT}CL_{A}^{b})\gamma_{5}R_{C}^{c}\\}\,.$
(95)

After similar algebra we can verify that all these fields vanish.

###  IV.3 A short summary of chiral representations

To summarize this section, we find that possible chiral representations for
Dirac spinor baryon fields without Lorentz index are:

$\displaystyle\Lambda$ $\displaystyle=$
$\displaystyle\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}CL_{B}^{b})\gamma_{5}R_{C}^{c}+\epsilon_{abc}\epsilon^{ABC}(R_{A}^{aT}CR_{B}^{b})\gamma_{5}L_{C}^{c}$
$\displaystyle=$ $\displaystyle\Lambda_{M1}+(L\leftrightarrow R)\,,$
$\displaystyle N^{N}_{1}-N^{N}_{2}$ $\displaystyle=$ $\displaystyle
2\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}CL_{B}^{b})\gamma_{5}R_{C}^{c}+2\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(R_{A}^{aT}CR_{B}^{b})\gamma_{5}L_{C}^{c}$
$\displaystyle=$ $\displaystyle 2N^{N}_{M1}+(L\leftrightarrow R)\,,$
$\displaystyle N^{N}_{1}+N^{N}_{2}$ $\displaystyle=$ $\displaystyle
2\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}CL_{B}^{b})\gamma_{5}L_{C}^{c}+2\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(R_{A}^{aT}CR_{B}^{b})\gamma_{5}R_{C}^{c}$
$\displaystyle=$ $\displaystyle 2N^{N}_{L1}+(L\leftrightarrow R)\,.$

So we can see that the fields $\Lambda$ and $N^{N}_{1}-N^{N}_{2}$ has a type
of $LLR\oplus RRL$, and belong to the chiral representation
$(\mathbf{\bar{3}},\mathbf{3})\oplus(\mathbf{3},\mathbf{\bar{3}})$; while the
field $N^{N}_{1}+N^{N}_{2}$ has a type of $LLL\oplus RRR$, and belongs to the
chiral representation $(\mathbf{8},\mathbf{1})\oplus(\mathbf{1},\mathbf{8})$.

The chiral properties of Rarita-Schwinger fields $\Big{(}{\rm
Lorentz~{}rep.}(\mathbf{1},\mathbf{1\over 2})\oplus(\mathbf{1\over
2},\mathbf{1})\Big{)}$ are listed in Appendix D. We summarize the results
here:

$\displaystyle N^{N}_{\mu}$ $\displaystyle=$ $\displaystyle
2\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}C\gamma_{\mu}\gamma_{5}R_{B}^{b})\gamma_{5}L_{C}^{c}+2\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(R_{A}^{aT}C\gamma_{\mu}\gamma_{5}L_{B}^{b})\gamma_{5}R_{C}^{c}$
$\displaystyle+$ $\displaystyle{1\over
2}\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}CL_{B}^{b})\gamma_{\mu}R_{C}^{c}+{1\over
2}\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(R_{A}^{aT}CR_{B}^{b})\gamma_{\mu}L_{C}^{c}\,,$
$\displaystyle\Delta^{P}_{\mu}$ $\displaystyle=$ $\displaystyle
2\epsilon_{abc}S_{P}^{ABC}(L_{A}^{aT}C\gamma_{\mu}R_{B}^{b})L_{C}^{c}+2\epsilon_{abc}S_{P}^{ABC}(R_{A}^{aT}C\gamma_{\mu}L_{B}^{b})R_{C}^{c}\,.$
(100)

So we see that $N^{N}_{\mu}$ and $\Delta^{P}_{\mu}$ are of the type $LLR\oplus
RRL$, and belong to the chiral representation
$(\mathbf{6},\mathbf{3})\oplus(\mathbf{3},\mathbf{6})$. The (similar) results
for $\Delta^{P}_{\mu\nu}$, which is of the type $LLL\oplus RRR$, and belongs
to the chiral representation
$(\mathbf{10},\mathbf{1})\oplus(\mathbf{1},\mathbf{10})$, are omitted here.

##  V Axial coupling constants

As a simple application of the present mathematical formalism, we can extract
the (diagonal) axial coupling constants $g_{A}$ of baryons. All information is
contained in Eqs. (III) and (III), from which we can calculate the Abelian
$U(1)_{A}$ axial coupling constant $g^{0}_{A}$ and the non-Abelian
$SU(3)_{V}\times SU(3)_{A}$ diagonal axial coupling constants, $g^{3}_{A}$ and
$g^{8}_{A}$, which can be extracted from the chiral transformations
$\delta_{5}$, $\delta_{5}^{b3}$ and $\delta_{5}^{b8}$, respectively. The
Abelian $g^{0}_{A}$ basically counts the difference between the numbers of
left- and right- handed quarks in a baryon. In general, diagonal elements of
the $SU(3)$ $g_{A}$’s can be decomposed into $F$ and $D$ components, which are
defined by the axial vector current $A_{\mu}^{a}$ ($a=0,1,...8$)

$\displaystyle A^{a}_{\mu}=g_{A}^{F}{\rm
tr}\bar{\mathfrak{B}}\gamma_{\mu}\gamma_{5}[\frac{\lambda_{a}}{2},{\mathfrak{B}}]+g_{A}^{D}{\rm
tr}\bar{\mathfrak{B}}\gamma_{\mu}\gamma_{5}\\{\frac{\lambda_{a}}{2},{\mathfrak{B}}\\}\,,$
(101)

where $\mathfrak{B}$ is the $3\times 3$ baryon matrix, Eq. (18). Therefore, we
have

$\displaystyle A^{3}_{\mu}$ $\displaystyle=$
$\displaystyle(g_{A}^{F}+g_{A}^{D})\Big{(}p^{+}p-n^{+}n\Big{)}$
$\displaystyle+$ $\displaystyle
2g_{A}^{F}\Big{(}(\Sigma^{+})^{+}\Sigma^{+}-(\Sigma^{-})^{+}\Sigma^{-}\Big{)}$
$\displaystyle+$
$\displaystyle(g_{A}^{F}-g_{A}^{D})\Big{(}(\Xi^{0})^{+}\Xi^{0}-(\Xi^{-})^{+}\Xi^{-}\Big{)}\,,$
$\displaystyle A^{8}_{\mu}$ $\displaystyle=$
$\displaystyle(\sqrt{3}g_{A}^{F}-{g_{A}^{D}\over\sqrt{3}})\Big{(}p^{+}p+n^{+}n\Big{)}$
$\displaystyle+$
$\displaystyle{2g_{A}^{D}\over\sqrt{3}}\Big{(}(\Sigma^{+})^{+}\Sigma^{+}+(\Sigma^{-})^{+}\Sigma^{-}\Big{)}$
$\displaystyle+$
$\displaystyle(-\sqrt{3}g_{A}^{F}-{g_{A}^{D}\over\sqrt{3}})\Big{(}(\Xi^{0})^{+}\Xi^{0}+(\Xi^{-})^{+}\Xi^{-}\Big{)}-{2g_{A}^{D}\over\sqrt{3}}(\Lambda^{8})^{+}\Lambda^{8}\,,$

where we omit the Lorentz part. In other words,

$\displaystyle g_{A}^{3}(N)$ $\displaystyle\sim$
$\displaystyle(g_{A}^{F}+g_{A}^{D}){\bf
I_{z}}\,,\;\;\;g_{A}^{3}(\Sigma)\sim{2g_{A}^{D}}{\bf
I_{z}}\,,\;\;\;g_{A}^{3}(\Xi)\sim(g_{A}^{F}-g_{A}^{D}){\bf I_{z}}\,,$ (104)
$\displaystyle g_{A}^{8}(N)$ $\displaystyle\sim$
$\displaystyle\sqrt{3}g_{A}^{F}-{g_{A}^{D}\over\sqrt{3}}\,,\;\;\;g_{A}^{8}(\Sigma)\sim{2g_{A}^{D}\over\sqrt{3}}\,,\;\;\;g_{A}^{8}(\Xi)\sim-\sqrt{3}g_{A}^{F}-{g_{A}^{D}\over\sqrt{3}}\,,\;\;\;g_{A}^{8}(\Lambda)\sim-{2g_{A}^{D}\over\sqrt{3}}\,,$

for the octet parts. The operator $\bf I_{z}$ is the third component of
isospin. While the singlet part $g_{A}^{0}$ contains only the $D$ term and is
trivial.

For the decuplet baryons, the $SU(3)$ coupling constants contain only one
$SU(3)$ irreducible term because the $SU(3)$ Clebsch-Gordan series for
$\mathbf{\bar{10}}\otimes\mathbf{10}\otimes\mathbf{8}$ contains only one
singlet. In order to extract the coupling constants, we first rewrite Eqs.
(III) and (III) in the following form, for all the singlet, octet and decuplet
baryon fields:

  1. 1.

Because $\lambda^{0}_{11}=\lambda^{0}_{22}=\lambda^{0}_{33}$ for $g^{0}_{A}$,
the chiral transformations $\delta_{5}$ are identical for all baryon fields
within the same chiral representation, so we may define $g_{A}^{0}$ by

$\displaystyle\delta_{5}B=i\gamma_{5}{{\lambda^{0}}_{11}b_{0}\over
2}g_{A}^{0}B={i\gamma_{5}b_{0}\over\sqrt{6}}g_{A}^{0}B\,,$ (105)

where $B$ represents the baryon field, such as $\Lambda$ and
$N_{1}^{N}-N^{N}_{2}$ etc.

  2. 2.

For $g^{3}_{A}$, because $\lambda^{3}_{11}=-\lambda^{3}_{22}$, the chiral
transformation $\delta_{5}^{b3}$ is proportional to the isospin value of $\bf
I_{z}$, which is related to $\lambda^{3}/2$. We factor it out from the
definition of $g_{A}^{3}$:

$\displaystyle\delta^{b3}_{5}B=i\gamma_{5}b_{3}g_{A}^{3}{\bf
I_{z}}B+\cdots\,,$ (106)

where dots $\cdots$ on the right hand side contain off-diagonal terms.

  3. 3.

For $g^{8}_{A}$, because $\lambda^{8}_{11}=\lambda^{8}_{22}$, the chiral
transformations $\delta_{5}^{b8}$ is the same for the baryon fields belonging
to one isospin multiplet. We define it to be

$\displaystyle\delta^{b8}_{5}B=i\gamma_{5}{\bm{\lambda^{8}}_{11}b_{8}\over
2}g_{A}^{8}B+\cdots={i\gamma_{5}b_{8}\over 2\sqrt{3}}g_{A}^{8}B+\cdots\,.$
(107)

Table 3: Axial coupling constants $g^{0}_{A}$, $g^{3}_{A}$ and $g^{8}_{A}$. In the last column $\alpha={g_{A}^{D}/(g_{A}^{F}+g_{A}^{D})}$.  $SU(3)_{L}\otimes SU(3)_{R}$ | $SU(3)_{F}$ | $\begin{array}[]{c}{}\hfil\\\ {}\hfil\end{array}$ | $g_{A}^{0}$ | $g_{A}^{3}$ | $g_{A}^{8}$ |  $\alpha$
---|---|---|---|---|---|---
| $\mathbf{1}$ | $\Lambda$ | -1 | – | 0 | –
|  | $N_{-}$ | -1 | 1 | -1 |
$(\mathbf{\bar{3}},~{}\mathbf{3})\oplus(\mathbf{3},~{}\mathbf{\bar{3}})$ |  | $\Sigma_{-}$ | -1 | 0 | 2 |
| $\mathbf{8}$ | $\Xi_{-}$ | -1 | -1 | -1 | 1
|  | $\Lambda_{-}$ | -1 | – | -2 |
|  | $N_{+}$ | 3 | 1 | 3 |
|  | $\Sigma_{+}$ | 3 | 1 | 0 |
$(\mathbf{8},~{}\mathbf{1})\oplus(\mathbf{1},~{}\mathbf{8})$ | $\mathbf{8}$ | $\Xi_{+}$ | 3 | 1 | -3 | 0
|  | $\Lambda_{+}$ | 3 | – | 0 |
|  | $N_{\mu}$ | 1 | 5/3 | 1 |
|  | $\Sigma_{\mu}$ | 1 | 2/3 | 2 |
| $\mathbf{8}$ | $\Xi_{\mu}$ | 1 | -1/3 | -3 | 3/5
|  | $\Lambda_{\mu}$ | 1 | – | -2 |
$(\mathbf{3},~{}\mathbf{6})\oplus(\mathbf{6},~{}\mathbf{3})$ |  | $\Delta_{\mu}$ | 1 | 1/3 | 1 |
|  | $\Sigma^{*}_{\mu}$ | 1 | 1/3 | 0 |
| $\mathbf{10}$ | $\Xi^{*}_{\mu}$ | 1 | 1/3 | -1 | –
|  | $\Omega_{\mu}$ | 1 | – | -2 |
|  | $\Delta_{\mu\nu}$ | 3 | 1 | 3 |
|  | $\Sigma^{*}_{\mu\nu}$ | 3 | 1 | 0 |
$(\mathbf{10},~{}\mathbf{1})\oplus(\mathbf{1},~{}\mathbf{10})$ | $\mathbf{10}$ | $\Xi^{*}_{\mu\nu}$ | 3 | 1 | -3 | –
|  | $\Omega_{\mu\nu}$ | 3 | – | -6 |

The resulting axial coupling constants $g^{0}_{A}$, $g^{3}_{A}$ and
$g^{8}_{A}$ are shown in Table 3, where $\Lambda$ is the (only) singlet field
$\Lambda$; then $N_{-}$, $\Sigma_{-}$, $\Xi_{-}$ and $\Lambda_{-}$ are the
octet fields of the type $N^{N}_{1}-N^{N}_{2}$; the $N_{+}$, $\Sigma_{+}$,
$\Xi_{+}$ and $\Lambda_{+}$ are the octet fields of the type
$N^{N}_{1}+N^{N}_{2}$; the $N_{\mu}$, $\Sigma_{\mu}$, $\Xi_{\mu}$ and
$\Lambda_{\mu}$ are the octet fields $N^{N}_{\mu}$; the $\Delta_{\mu}$,
$\Sigma^{*}_{\mu}$, $\Xi^{*}_{\mu}$ and $\Omega_{\mu}$ are the decuplet fields
$\Delta^{P}_{\mu}$; $\Delta_{\mu\nu}$, $\Sigma^{*}_{\mu\nu}$,
$\Xi^{*}_{\mu\nu}$ and $\Omega_{\mu\nu}$ are the decuplet fields
$\Delta^{P}_{\mu\nu}$.

From the values in Table 3, one can compute the $F$ and $D$ couplings easily.
The resulting $F/D$ ratio,

$\displaystyle\alpha={g_{A}^{D}\over g_{A}^{F}+g_{A}^{D}}\,,$ (108)

is also tabulated in the last column of Table 3. Empirically, $\alpha\sim
0.6$, which is fairly close to the $SU(6)$ quark model value. In the present
formalism we see that only the
$(\mathbf{3},~{}\mathbf{6})\oplus(\mathbf{6},~{}\mathbf{3})$ chiral
multiplet/representation reproduces this value. Previous works have shown that
this value is physically related to the coupling of the nucleon to the
$\Delta(1232)$, as demonstrated in the Adler-Weisberger sum rule adler ;
weissberger . This was also shown algebraically by Weinberg Weinberg:1969hw .
In both cases, saturation of the pion (axial-vector) induced transition from
the nucleon to the $\Delta(1232)$ is essential Donoghue:1977yh . In the
present study, this is realized by the chiral representation which includes
both the nucleon (isospin 1/2) and delta (isospin 3/2) states.

It is also interesting that Table 3 shows that
$g_{A}^{3}(N)=5/3,g_{A}^{0}(N)=1$ for
$(\mathbf{3},~{}\mathbf{6})\oplus(\mathbf{6},~{}\mathbf{3})$, while
$g_{A}^{3}(N)=1,g_{A}^{0}(N)=-1$ for
$(\mathbf{\bar{3}},~{}\mathbf{3})\oplus(\mathbf{3},~{}\mathbf{\bar{3}})$.
$g_{A}^{0}$ corresponds to the so-called nucleon spin value, as measured in
polarized deep inelastic scattering. A suitable superposition of the two
chiral representations may improve the nucleon axial coupling in either the
isovector and/or isosinglet sectors. The importance of such mixing for the
isovector axial coupling constant has been emphasized by Weinberg since the
late 1960-s, Ref. Weinberg:1969hw . Here we have found the same result for the
isovector, as well as extended it to the isosinglet part in a purely algebraic
manner.

##  VI Summary

In this paper we have performed a complete classification of flavor vector and
chiral symmetries, and established several types of independent relativistic
$SU(3)$ baryon interpolating fields. The three-quark fields may belong to one
of several different Lorentz group representations which fact imposes certain
constraints on possible chiral symmetry representations. This is due to the
Pauli principle and has been explicitly verified by the method of Fierz
transformations. As the present results reflect essentially the Pauli
principle, they can be conveniently summarized as shown in Table 4 by using
the permutation symmetry group properties/representations. This table
“explains” also the previous results for the case of isospin $SU(2)_{L}\times
SU(2)_{R}$ Nagata:2007di . From this table we have explicated the effective
role of the Pauli principle in separate sectors of the left- and right-handed
fermions.

Table 4: Structure of allowed three-quark baryon fields. Lorentz | J=Spin | $\begin{array}[]{c}\mbox{Young table}\\\ \mbox{for Chiral rep.}\end{array}$ | Chiral $SU(2)$ | Chiral $SU(3)$ | Flavor $SU(3)$
---|---|---|---|---|---
$(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ | $1/2$ | $\begin{array}[]{c}([21],-)\oplus(-,[21])\\\ ([1],[11])\oplus([11],[1])\end{array}$ | $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ | $\begin{array}[]{c}(8,1)\oplus(1,8)\\\ (3,\bar{3})\oplus(\bar{3},3)\end{array}$ | $\begin{array}[]{c}8\\\ 1,8\end{array}$
$(1,\frac{1}{2})\oplus(\frac{1}{2},1)$ | $3/2$ | $([2],[1])\oplus([1],[2])$ | $(1,\frac{1}{2})\oplus(\frac{1}{2},1)$ | $(6,3)\oplus(3,6)$ |  $8$, $10$
$(\frac{3}{2},0)\oplus(0,\frac{3}{2})$ | $3/2$ | $([3],-)\oplus(-,[3])$ | $(\frac{3}{2},0)\oplus(0,\frac{3}{2})$ | $(10,1)\oplus(1,10)$ | $10$

In the real world, with spontaneous breaking of chiral symmetry, physical
states of pure chiral (axial) symmetry representation do not occur, but in
general they can mix in a state having a definite flavor symmetry. The present
results show that the three-quark structures accommodate only a few (sometimes
just one) chiral representations, for instance, for the total spin 1/2 field
of Dirac spinor, the allowed chiral representations are two having the
structure of Young tableaux $([21],-)$ and ([1],[11]), where $-$ indicates
singlet. The $([21],-)$ representation corresponds respectively to
$(\frac{1}{2},0)$ and $(8,1)$ of $SU(2)$ and $SU(3)$, whereas the $([1],[11])$
corresponds to $(\frac{1}{2},0)$ and $(3,\bar{3})$ of $SU(2)$ and $SU(3)$,
respectively. Note that the chiral representations have the same structure as
those of the Lorentz group. In this way, the Lorentz (spin) and flavor
structures are combined into a structure of total permutation symmetry. As
shown in the computation of $g_{A}$, in general, various couplings depend on
the chiral representations with possible mixing. Such comparison may be useful
for further understanding of the internal structure of hadrons in relation to
chiral symmetry.

## Acknowledgments

H.X.C. is grateful to Monkasho for their support of his stay at the Research
Center for Nuclear Physics where this work was done. V.D and K.N thank Prof H.
Toki for his hospitality during their stay at RCNP. A.H. is supported in part
by the Grant for Scientific Research ((C) No.19540297) from the Ministry of
Education, Culture, Science and Technology, Japan. K.N is supported by the
National Science Council (NSC) of Republic of China under grant No.
NSC96-2119-M-002-001. S.L.Z. was supported by the National Natural Science
Foundation of China under Grants 10625521 and 10721063 and Ministry of
Education of China.

##  Appendix A Rarita-Schwinger fields

In this appendix, we study the properties of Rarita-Schwinger fields, in the
form of

$B_{\mu}(x)\sim\epsilon_{abc}(q^{aT}_{A}(x)C\Gamma_{1}q^{b}_{B}(x))\Gamma_{2}q^{c}_{C}(x)\,,$
(109)

where there are eight possible pairs of $\Gamma_{1}$ and $\Gamma_{2}$,

$\displaystyle(\Gamma_{1},~{}\Gamma_{2})$ $\displaystyle=$
$\displaystyle(\mathbf{1},~{}\gamma_{\mu}),~{}(\gamma_{5},~{}\gamma_{\mu}\gamma_{5}),~{}(\gamma_{\mu}\gamma_{5},~{}\gamma_{5}),~{}(\gamma^{\nu}\gamma_{5},~{}\sigma_{\mu\nu}\gamma_{5}),~{}$
$\displaystyle(\gamma_{\mu},~{}\mathbf{1}),~{}(\gamma^{\nu},~{}\sigma_{\mu\nu}),~{}(\sigma_{\mu\nu},~{}\gamma^{\nu}),~{}(\sigma_{\mu\nu}\gamma_{5},~{}\gamma^{\nu}\gamma_{5}).$

The fields formed by these $(\Gamma_{1},\Gamma_{2})$ pairs are labeled by the
subscript $i=(1,\cdots,8)$ with the ordering of Eq. (A). The discussion is
separated into singlet, decuplet and octet cases.

####  A.0.1 Flavor singlet baryon

For flavor singlet fields, there are four apparently non-zero fields

$\displaystyle\begin{array}[]{l}\Lambda_{1\mu}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}Cq_{B}^{b})\gamma_{\mu}q_{C}^{c}\,,\\\
\Lambda_{2\mu}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}C\gamma_{5}q_{B}^{b})\gamma_{\mu}\gamma_{5}q_{C}^{c}\,,\\\
\Lambda_{3\mu}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}C\gamma_{\mu}\gamma_{5}q_{B}^{b})\gamma_{5}q_{C}^{c}\,,\\\
\Lambda_{4\mu}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}C\gamma^{\nu}\gamma_{5}q_{B}^{b})\sigma_{\mu\nu}\gamma_{5}q_{C}^{c}\,.\end{array}$
(115)

As before, the Fierz transformed fields (primed fields) are just the
corresponding unprimed ones, $\Lambda^{\prime}_{i\mu}=\Lambda_{i\mu}$. On the
other hand, by applying the Fierz rearrangement (see Appendix. C), we obtain
four equations

$\displaystyle\begin{array}[]{l}\Lambda_{1\mu}=-{1\over
4}\Lambda_{1\mu}^{\prime}-{1\over 4}\Lambda_{2\mu}^{\prime}+{1\over
4}\Lambda_{3\mu}^{\prime}-{i\over 4}\Lambda_{4\mu}^{\prime}\,,\\\
\Lambda_{2\mu}=-{1\over 4}\Lambda_{1\mu}^{\prime}-{1\over
4}\Lambda_{2\mu}^{\prime}-{1\over 4}\Lambda_{3\mu}^{\prime}+{i\over
4}\Lambda_{4\mu}^{\prime}\,,\\\ \Lambda_{3\mu}={1\over
4}\Lambda_{1\mu}^{\prime}-{1\over 4}\Lambda_{2\mu}^{\prime}-{1\over
4}\Lambda_{3\mu}^{\prime}-{i\over 4}\Lambda_{4\mu}^{\prime}\,,\\\
\Lambda_{4\mu}={3i\over 4}\Lambda_{1\mu}^{\prime}-{3i\over
4}\Lambda_{2\mu}^{\prime}+{3i\over 4}\Lambda_{3\mu}^{\prime}+{1\over
4}\Lambda_{4\mu}^{\prime}\,.\end{array}$ (120)

Thus we find the following solution

$\Lambda_{1\mu}=-\Lambda_{2\mu}=\Lambda_{3\mu}=-{i\over
3}\Lambda_{4\mu}=\gamma_{\mu}\gamma_{5}\Lambda_{1}\,,~{}~{}~{}~{}~{}\Lambda_{6\mu}=\Lambda_{7\mu}=\Lambda_{8\mu}=0.$

We see that there is only one non-vanishing independent field. However, it has
a structure of $\gamma_{\mu}\Lambda_{i}$ . Therefore, they are all Dirac
fields, and there is no flavor singlet fields of the Rarita-Schwinger type.

####  A.0.2 Flavor decuplet baryon

For flavour decuplet fields, we have four potentially non-zero interpolators

$\displaystyle\begin{array}[]{l}\Delta^{P}_{5\mu}=\epsilon_{abc}S_{P}^{ABC}(q_{A}^{aT}C\gamma_{\mu}q_{B}^{b})q_{C}^{c}\,,\\\
\Delta^{P}_{6\mu}=\epsilon_{abc}S_{P}^{ABC}(q_{A}^{aT}C\gamma^{\nu}q_{B}^{b})\sigma_{\mu\nu}q_{C}^{c}\,,\\\
\Delta^{P}_{7\mu}=\epsilon_{abc}S_{P}^{ABC}(q_{A}^{aT}C\sigma_{\mu\nu}q_{B}^{b})\gamma^{\nu}q_{C}^{c}\,,\\\
\Delta^{P}_{8\mu}=\epsilon_{abc}S_{P}^{ABC}(q_{A}^{aT}C\sigma_{\mu\nu}\gamma_{5}q_{B}^{b})\gamma^{\nu}\gamma_{5}q_{C}^{c}\,.\end{array}$
(125)

As before, the Fierz transformed fields can be related to the corresponding
unprimed ones, $\Delta^{P\prime}_{i\mu}=-\Delta^{P}_{i\mu}$. On the other
hand, by applying the Fierz rearrangement to relate $\Delta_{i\mu}^{N}$ and
$\Delta_{i\mu}^{N\prime}$, we obtain the solution

$\Delta^{P}_{5\mu}=i\Delta^{P}_{6\mu}=-i\Delta^{P}_{7\mu}=i\Delta^{P}_{8\mu}\,.$

There are no Dirac decuplet fields. Therefore, we obtain one extra non-
vanishing field.

####  A.0.3 Flavor octet baryon

To study the octet baryon fields, we start with eight baryon fields:

$\displaystyle\begin{array}[]{l}N^{N}_{1\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}Cq_{B}^{b})\gamma_{\mu}q_{C}^{c}\,,\\\
N^{N}_{2\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{5}q_{B}^{b})\gamma_{\mu}\gamma_{5}q_{C}^{c}\,,\\\
N^{N}_{3\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{\mu}\gamma_{5}q_{B}^{b})\gamma_{5}q_{C}^{c}\,,\\\
N^{N}_{4\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma^{\nu}\gamma_{5}q_{B}^{b})\sigma_{\mu\nu}\gamma_{5}q_{C}^{c}\,,\\\
N^{N}_{5\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{\mu}q_{B}^{b})q_{C}^{c}=0\,,\\\
N^{N}_{6\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma^{\nu}q_{B}^{b})\sigma_{\mu\nu}q_{C}^{c}=0\,,\\\
N^{N}_{7\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\sigma_{\mu\nu}q_{B}^{b})\gamma^{\nu}q_{C}^{c}=0\,,\\\
N^{N}_{8\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\sigma_{\mu\nu}\gamma_{5}q_{B}^{b})\gamma^{\nu}\gamma_{5}q_{C}^{c}=0\,.\end{array}$
(134)

There are four zero fields, but the corresponding Fierz transformed ones are
non-zero. By using the Jacobi identity in Eq. (39), we obtain

$\displaystyle N_{1\mu}^{N\prime}=-{1\over
2}N_{1\mu}^{N}\,,N_{2\mu}^{N\prime}=-{1\over
2}N_{2\mu}^{N}\,,N_{3\mu}^{N\prime}=-{1\over
2}N_{3\mu}^{N}\,,N_{4\mu}^{N\prime}=-{1\over 2}N_{4\mu}^{N}\,.$

Similarly, performing the Fierz transformation to relate $N_{i\mu}^{N}$ and
$N_{i\mu}^{N\prime}$, we obtain the solution

$\displaystyle\begin{array}[]{l}N_{4\mu}^{N}=-iN_{1\mu}^{N}+iN_{2\mu}^{N}-iN_{3\mu}^{N}\,,\\\
N_{5\mu}^{N\prime}=-{1\over 2}N_{1\mu}^{N}+{1\over 2}N_{2\mu}^{N}-{1\over
2}N_{3\mu}^{N}\,,\\\ N_{6\mu}^{N\prime}=-iN_{1\mu}^{N}+iN_{2\mu}^{N}+{i\over
2}N_{3\mu}^{N}\,,\\\ N_{7\mu}^{N\prime}=iN_{1\mu}^{N}+{i\over
2}N_{2\mu}^{N}+iN_{3\mu}^{N}\,,\\\ N_{8\mu}^{N\prime}={i\over
2}N_{1\mu}^{N}+iN_{2\mu}^{N}-iN_{3\mu}^{N}\,.\end{array}$ (140)

Thus we have shown that there are three different kinds of octets. However,
$N_{1\mu}^{N}$ and $N_{2\mu}^{N}$ are nothing but
$\gamma_{\mu}\gamma_{5}N_{1}^{N}$ and $\gamma_{\mu}\gamma_{5}N_{2}^{N}$ (see
Eqs. (38)). Therefore, we only obtain one extra octet baryon field. It is
formed by using the projection operator:

$\displaystyle P^{3/2}_{\mu\nu}=(g_{\mu\nu}-{1\over
4}\gamma_{\mu}\gamma_{\nu})\,,$

as

$\displaystyle N^{N}_{\mu}=P^{3/2}_{\mu\nu}N^{N}_{3\nu}$ $\displaystyle=$
$\displaystyle(g_{\mu\nu}-{1\over
4}\gamma_{\mu}\gamma_{\nu})\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{\mu}\gamma_{5}q_{B}^{b})\gamma_{5}q_{C}^{c}$
$\displaystyle=$ $\displaystyle N^{N}_{3\mu}+{1\over
4}\gamma_{\mu}\gamma_{5}(N^{N}_{1}-N^{N}_{2})\,.$

##  Appendix B Tensor Fields

In this appendix, we study the antisymmetric tensor baryons fields
$J_{\mu\nu}$ with $J_{\mu\nu}=-J_{\nu\mu}$. For the tensor fields, we can form
nine three-quark fields where the possible pairs of $\Gamma_{1}$ and
$\Gamma_{2}$ are

$\displaystyle(\Gamma_{1},~{}\Gamma_{2})$ $\displaystyle=$
$\displaystyle(\gamma_{\mu},~{}\gamma_{\nu}\gamma_{5})-(\mu\leftrightarrow\nu),~{}(\gamma_{\mu}\gamma_{5},~{}\gamma_{\nu})-(\mu\leftrightarrow\nu),~{}$
$\displaystyle\epsilon_{\mu\nu\rho\sigma}(\gamma^{\rho},~{}\gamma^{\sigma}),~{}\epsilon_{\mu\nu\rho\sigma}(\gamma^{\rho}\gamma_{5},~{}\gamma^{\sigma}\gamma_{5}),~{}(\mathbf{1},~{}\sigma_{\mu\nu}\gamma_{5}),~{}(\gamma_{5},~{}\sigma_{\mu\nu}),~{}$
$\displaystyle(\sigma_{\mu\nu},~{}\gamma_{5}),~{}(\sigma_{\mu\nu}\gamma_{5},~{}\mathbf{1}),~{}\epsilon_{\mu\nu\rho\sigma}(\sigma_{\rho
l},\sigma_{\sigma l})\,.$

The fields formed by these $(\Gamma_{1},\Gamma_{2})$ pairs are labeled by the
subscript $i=(1,\cdots,9)$ with the ordering of Eq. (B). The discussion is
separated into singlet, decuplet and octet cases.

####  B.0.1 Flavor singlet baryon

The flavour singlet baryon fields have four potentially non-zero interpolators
among nine fields:

$\displaystyle\begin{array}[]{l}\Lambda_{2\mu\nu}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}C\gamma_{\mu}\gamma_{5}q_{B}^{b})\gamma_{\nu}q_{C}^{c}-(\mu\leftrightarrow\nu)\,,\\\
\Lambda_{4\mu\nu}=\epsilon_{abc}\epsilon^{ABC}\epsilon_{\mu\nu\rho\sigma}(q_{A}^{aT}C\gamma_{\rho}\gamma_{5}q_{B}^{b})\gamma_{\sigma}\gamma_{5}q_{C}^{c}\,,\\\
\Lambda_{5\mu\nu}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}Cq_{B}^{b})\sigma_{\mu\nu}\gamma_{5}q_{C}^{c}\,,\\\
\Lambda_{6\mu\nu}=\epsilon_{abc}\epsilon^{ABC}(q_{A}^{aT}C\gamma_{5}q_{B}^{b})\sigma_{\mu\nu}q_{C}^{c}\,.\end{array}$
(146)

As before, the Fierz transformed fields are just the corresponding unprimed
ones, $\Lambda^{\prime}_{i\mu\nu}=\Lambda_{i\mu\nu}$. On the other hand, by
applying the Fierz rearrangement to relate $\Lambda_{i\mu\nu}$ and
$\Lambda_{\i\mu\nu}^{\prime}$, we obtain the solution:

$\displaystyle
i\Lambda_{2\mu\nu}=\Lambda_{4\mu\nu}=2\Lambda_{5\mu\nu}=-2\Lambda_{6\mu\nu}\,.$

Therefore, there is only one independent field. However, it has a structure of
$\sigma_{\mu\nu}\Lambda_{i}$ . Therefore, there are no extra fields.

####  B.0.2 Flavor decuplet baryon

The flavour decuplet baryon fields have five potentially non-zero
interpolators:

$\displaystyle\begin{array}[]{l}\Delta^{P}_{1\mu\nu}=\epsilon_{abc}S^{ABC}(q_{A}^{aT}C\gamma_{\mu}q_{B}^{b})\gamma_{\nu}\gamma_{5}q_{C}^{c}-(\mu\leftrightarrow\nu)\,,\\\
\Delta^{P}_{3\mu\nu}=\epsilon_{abc}S^{ABC}\epsilon_{\mu\nu\rho\sigma}(q_{A}^{aT}C\gamma_{\rho}q_{B}^{b})\gamma_{\sigma}q_{C}^{c}\,,\\\
\Delta^{P}_{7\mu\nu}=\epsilon_{abc}S^{ABC}(q_{A}^{aT}C\sigma_{\mu\nu}q_{B}^{b})\gamma_{5}q_{C}^{c}\,,\\\
\Delta^{P}_{8\mu\nu}=\epsilon_{abc}S^{ABC}(q_{A}^{aT}C\sigma_{\mu\nu}\gamma_{5}q_{B}^{b})q_{C}^{c}\,,\\\
\Delta^{P}_{9\mu\nu}=\epsilon_{abc}S^{ABC}\epsilon_{\mu\nu\rho\sigma}(q_{A}^{aT}C\sigma_{\rho
l}q_{B}^{b})\sigma_{\sigma l}q_{C}^{c}\,.\end{array}$ (152)

As before, the Fierz transformed fields can be related to the corresponding
unprimed ones, $\Delta^{P\prime}_{i\mu\mu}=-\Delta^{P}_{i\mu\mu}$. On the
other hand, by applying the Fierz rearrangement to relate
$\Delta^{P}_{i\mu\nu}$ and $\Delta_{i\mu\nu}^{P\prime}$, we obtain two
independent fields: $\Delta^{P}_{1\mu\nu}$ and $\Delta^{P}_{7\mu\nu}$:

$\displaystyle\Delta^{P}_{3\mu\nu}=-i\Delta^{P}_{1\mu\nu}\,,\Delta^{P}_{8\mu\nu}=i\Delta^{P}_{1\mu\nu}+\Delta^{P}_{7\mu\nu}\,,\Delta^{P}_{9\mu\nu}=-i\Delta^{P}_{1\mu\nu}-2\Delta^{P}_{7\mu\nu}\,.$

The first one $\Delta^{P}_{1\mu\nu}$ can be related to the Rarita-Schwinger
baryon fields, but the second one $\Delta^{P}_{7\mu\nu}$ can not. Therefore,
we obtain one extra decuplet fields. It is formed by using the projection
operator:

$\displaystyle\Gamma^{\mu\nu\alpha\beta}=(g^{\mu\alpha}g^{\nu\beta}-{1\over
2}g^{\nu\beta}\gamma^{\mu}\gamma^{\alpha}+{1\over
2}g^{\mu\beta}\gamma^{\nu}\gamma^{\alpha}+{1\over
6}\sigma^{\mu\nu}\sigma^{\alpha\beta})\,,$

as

$\displaystyle\Delta^{P}_{\mu\nu}=\Gamma^{\mu\nu\alpha\beta}\Delta^{P}_{7\mu\nu}$
$\displaystyle=$
$\displaystyle\Gamma^{\mu\nu\alpha\beta}\epsilon_{abc}S^{ABC}(q_{A}^{aT}C\sigma_{\mu\nu}q_{B}^{b})\gamma_{5}q_{C}^{c}$
$\displaystyle=$ $\displaystyle\Delta^{P}_{7\mu\nu}-{i\over
2}\gamma_{\mu}\gamma_{5}\Delta^{P}_{5\nu}+{i\over
2}\gamma_{\nu}\gamma_{5}\Delta^{P}_{5\mu}\,.$

####  B.0.3 Flavor octet baryon

To study the octet baryon fields, we start with nine octet baryon fields

$\displaystyle\begin{array}[]{l}N^{N}_{1\mu\nu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{\mu}q_{B}^{b})\gamma_{\nu}\gamma_{5}q_{C}^{c}-(\mu\leftrightarrow\nu)=0\,,\\\
N^{N}_{2\mu\nu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{\mu}\gamma_{5}q_{B}^{b})\gamma_{\nu}q_{C}^{c}-(\mu\leftrightarrow\nu)\,,\\\
N^{N}_{3\mu\nu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}\epsilon_{\mu\nu\rho\sigma}(q_{A}^{aT}C\gamma_{\rho}q_{B}^{b})\gamma_{\sigma}q_{C}^{c}=0\,,\\\
N^{N}_{4\mu\nu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}\epsilon_{\mu\nu\rho\sigma}(q_{A}^{aT}C\gamma_{\rho}\gamma_{5}q_{B}^{b})\gamma_{\sigma}\gamma_{5}q_{C}^{c}\,,\\\
N^{N}_{5\mu\nu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}Cq_{B}^{b})\sigma_{\mu\nu}\gamma_{5}q_{C}^{c}\,,\\\
N^{N}_{6\mu\nu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\gamma_{5}q_{B}^{b})\sigma_{\mu\nu}q_{C}^{c}\,,\\\
N^{N}_{7\mu\nu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\sigma_{\mu\nu}q_{B}^{b})\gamma_{5}q_{C}^{c}=0\,,\\\
N^{N}_{8\mu\nu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(q_{A}^{aT}C\sigma_{\mu\nu}\gamma_{5}q_{B}^{b})q_{C}^{c}=0\,,\\\
N^{N}_{9\mu\nu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}\epsilon_{\mu\nu\rho\sigma}(q_{A}^{aT}C\sigma_{\rho
l}q_{B}^{b})\sigma_{\sigma l}q_{C}^{c}=0\,.\end{array}$ (162)

There are five zero fields, but the Fierz transformed ones are non-zero. By
using the Jacobi identity in Eq. (39), we obtain

$\displaystyle N^{N\prime}_{2\mu\nu}=-{1\over
2}N^{N}_{2\mu\nu}\,,N^{N\prime}_{4\mu\nu}=-{1\over
2}N^{N}_{4\mu\nu}\,,N^{N\prime}_{5\mu\nu}=-{1\over
2}N^{N}_{5\mu\nu}\,,N^{N\prime}_{6\mu\nu}=-{1\over 2}N^{N}_{6\mu\nu}\,.$

Similarly, performing the Fierz transformation to relate $N^{N}_{i\mu\nu}$ and
$N_{i\mu\nu}^{N\prime}$, we find that there are three independent fields
$N_{2\mu\nu}^{N}$, $N_{5\mu\nu}^{N}$ and $N_{6\mu\nu}^{N}$. Here are the
relations:

$\displaystyle\begin{array}[]{l}N^{N}_{4\mu\nu}=-iN^{N}_{2\mu\nu}-N^{N}_{5\mu\nu}+N^{N}_{6\mu\nu}\,,\\\
N^{N\prime}_{1\mu\nu}=-{1\over
2}N^{N}_{2\mu\nu}+iN^{N}_{5\mu\nu}-iN^{N}_{6\mu\nu}\,,\\\
N^{N\prime}_{3\mu\nu}={i\over 2}N^{N}_{2\mu\nu}-{1\over
2}N^{N}_{5\mu\nu}+{1\over 2}N^{N}_{6\mu\nu}\,,\\\
N^{N\prime}_{7\mu\nu}=-{i\over 2}N^{N}_{2\mu\nu}-{1\over
2}N^{N}_{5\mu\nu}\,,\\\ N^{N\prime}_{8\mu\nu}={i\over
2}N^{N}_{2\mu\nu}-{1\over 2}N^{N}_{6\mu\nu}\,,\\\
N^{N\prime}_{9\mu\nu}=-N^{N}_{5\mu\nu}-N^{N}_{6\mu\nu}\,.\end{array}$ (169)

All these three fields can be related to the Dirac spinor and Rarita-Schwinger
fields. Therefore, there are no extra octet fields.

##  Appendix C Fierz Transformation

In this appendix, we list the Fierz transformations used in our calculation,
which are proved by using $Mathematica$ Maruhn:2000af . Here we would like to
show only the change in the structure of Lorentz indices of direct products of
two Dirac matrices under the Fierz rearrangement. Therefore, in the following
equations, we do not include the minus sign which arises from the exchange of
quark fields. The formulae go for the three cases corresponding to the Dirac,
Rarita-Schwinger and tensor fields when applied to three-quark fields.

  1. 1.

Products of two Dirac matrices without Lorentz indices:

$\displaystyle\left(\begin{array}[]{l}\mathbf{1}\otimes\gamma_{5}\\\
\gamma_{\mu}\otimes\gamma^{\mu}\gamma_{5}\\\
\sigma_{\mu\nu}\otimes\sigma^{\mu\nu}\gamma_{5}\\\
\gamma_{\mu}\gamma_{5}\otimes\gamma^{\mu}\\\
\gamma_{5}\otimes\mathbf{1}\end{array}\right)_{ab,cd}=\left(\begin{array}[]{lllll}{1\over
4}&-{1\over 4}&{1\over 8}&{1\over 4}&{1\over 4}\\\ -1&-{1\over 2}&0&-{1\over
2}&1\\\ 3&0&-{1\over 2}&0&3\\\ 1&-{1\over 2}&0&-{1\over 2}&-1\\\ {1\over
4}&{1\over 4}&{1\over 8}&-{1\over 4}&{1\over
4}\end{array}\right)\left(\begin{array}[]{l}\mathbf{1}\otimes\gamma_{5}\\\
\gamma_{\mu}\otimes\gamma^{\mu}\gamma_{5}\\\
\sigma_{\mu\nu}\otimes\sigma^{\mu\nu}\gamma_{5}\\\
\gamma_{\mu}\gamma_{5}\otimes\gamma^{\mu}\\\
\gamma_{5}\otimes\mathbf{1}\end{array}\right)_{ad,bc}$ (185)

  2. 2.

Products of two Dirac matrices with one Lorentz index:

$\displaystyle\left(\begin{array}[]{l}\mathbf{1}\otimes\gamma^{\mu}\\\
\gamma^{\mu}\otimes\mathbf{1}\\\ \gamma_{5}\otimes\gamma_{\mu}\gamma_{5}\\\
\gamma_{\mu}\gamma_{5}\otimes\gamma_{5}\\\
\gamma^{\nu}\otimes\sigma_{\mu\nu}\\\ \sigma_{\mu\nu}\otimes\gamma^{\nu}\\\
\gamma^{\nu}\gamma_{5}\otimes\sigma_{\mu\nu}\gamma_{5}\\\
\sigma_{\mu\nu}\gamma_{5}\otimes\gamma^{\nu}\gamma_{5}\end{array}\right)_{ab,cd}=\left(\begin{array}[]{llllllll}{1\over
4}&{1\over 4}&{1\over 4}&-{1\over 4}&-{i\over 4}&{i\over 4}&{i\over 4}&{i\over
4}\\\ {1\over 4}&{1\over 4}&-{1\over 4}&{1\over 4}&{i\over 4}&-{i\over
4}&{i\over 4}&{i\over 4}\\\ {1\over 4}&-{1\over 4}&{1\over 4}&{1\over
4}&{i\over 4}&{i\over 4}&-{i\over 4}&{i\over 4}\\\ -{1\over 4}&{1\over
4}&{1\over 4}&{1\over 4}&{i\over 4}&{i\over 4}&{i\over 4}&-{i\over 4}\\\
{3i\over 4}&-{3i\over 4}&-{3i\over 4}&-{3i\over 4}&-{1\over 4}&-{1\over
4}&-{1\over 4}&{1\over 4}\\\ -{3i\over 4}&{3i\over 4}&-{3i\over 4}&-{3i\over
4}&-{1\over 4}&-{1\over 4}&{1\over 4}&-{1\over 4}\\\ -{3i\over 4}&-{3i\over
4}&{3i\over 4}&-{3i\over 4}&-{1\over 4}&{1\over 4}&-{1\over 4}&-{1\over 4}\\\
-{3i\over 4}&-{3i\over 4}&-{3i\over 4}&{3i\over 4}&{1\over 4}&-{1\over
4}&-{1\over 4}&-{1\over
4}\end{array}\right)\left(\begin{array}[]{l}\mathbf{1}\otimes\gamma^{\mu}\\\
\gamma^{\mu}\otimes\mathbf{1}\\\ \gamma_{5}\otimes\gamma_{\mu}\gamma_{5}\\\
\gamma_{\mu}\gamma_{5}\otimes\gamma_{5}\\\
\gamma^{\nu}\otimes\sigma_{\mu\nu}\\\ \sigma_{\mu\nu}\otimes\gamma^{\nu}\\\
\gamma^{\nu}\gamma_{5}\otimes\sigma_{\mu\nu}\gamma_{5}\\\
\sigma_{\mu\nu}\gamma_{5}\otimes\gamma^{\nu}\gamma_{5}\end{array}\right)_{ad,bc}$
(210)

  3. 3.

Products of two Dirac matrices with two anti-symmetric Lorentz indices:

$\displaystyle\left(\begin{array}[]{l}\mathbf{1}\otimes\sigma_{\mu\nu}\gamma_{5}\\\
\gamma_{5}\otimes\sigma_{\mu\nu}\\\ \sigma_{\mu\nu}\otimes\gamma_{5}\\\
\sigma_{\mu\nu}\gamma_{5}\otimes\mathbf{1}\\\
\epsilon_{\mu\nu\rho\sigma}\sigma_{\rho l}\otimes\sigma_{\sigma l}\\\
\gamma_{\mu}\otimes\gamma_{\nu}\gamma_{5}-(\mu\leftrightarrow\nu)\\\
\gamma_{\mu}\gamma_{5}\otimes\gamma_{\nu}-(\mu\leftrightarrow\nu)\\\
\epsilon_{\mu\nu\rho\sigma}\gamma_{\rho}\otimes\gamma_{\sigma}\\\
\epsilon_{\mu\nu\rho\sigma}\gamma_{\rho}\gamma_{5}\otimes\gamma_{\sigma}\gamma_{5}\end{array}\right)_{ab,cd}=\left(\begin{array}[]{lllllllll}{1\over
4}&{1\over 4}&{1\over 4}&{1\over 4}&{1\over 4}&{i\over 4}&-{i\over 4}&{1\over
4}&-{1\over 4}\\\ {1\over 4}&{1\over 4}&{1\over 4}&{1\over 4}&{1\over
4}&-{i\over 4}&{i\over 4}&-{1\over 4}&{1\over 4}\\\ {1\over 4}&{1\over
4}&{1\over 4}&{1\over 4}&-{1\over 4}&-{i\over 4}&{i\over 4}&{1\over
4}&-{1\over 4}\\\ {1\over 4}&{1\over 4}&{1\over 4}&{1\over 4}&-{1\over
4}&{i\over 4}&-{i\over 4}&-{1\over 4}&{1\over 4}\\\ 1&1&-1&-1&0&0&0&0&0\\\
-{i\over 2}&{i\over 2}&{i\over 2}&-{i\over 2}&0&0&0&{i\over 2}&{i\over 2}\\\
{i\over 2}&-{i\over 2}&-{i\over 2}&{i\over 2}&0&0&0&{i\over 2}&{i\over 2}\\\
{1\over 2}&-{1\over 2}&{1\over 2}&-{1\over 2}&0&-{i\over 2}&-{i\over 2}&0&0\\\
-{1\over 2}&{1\over 2}&-{1\over 2}&{1\over 2}&0&-{i\over 2}&-{i\over
2}&0&0\end{array}\right)\left(\begin{array}[]{l}\mathbf{1}\otimes\sigma_{\mu\nu}\gamma_{5}\\\
\gamma_{5}\otimes\sigma_{\mu\nu}\\\ \sigma_{\mu\nu}\otimes\gamma_{5}\\\
\sigma_{\mu\nu}\gamma_{5}\otimes\mathbf{1}\\\
\epsilon_{\mu\nu\rho\sigma}\sigma_{\rho l}\otimes\sigma_{\sigma l}\\\
\gamma_{\mu}\otimes\gamma_{\nu}\gamma_{5}-(\mu\leftrightarrow\nu)\\\
\gamma_{\mu}\gamma_{5}\otimes\gamma_{\nu}-(\mu\leftrightarrow\nu)\\\
\epsilon_{\mu\nu\rho\sigma}\gamma_{\rho}\otimes\gamma_{\sigma}\\\
\epsilon_{\mu\nu\rho\sigma}\gamma_{\rho}\gamma_{5}\otimes\gamma_{\sigma}\gamma_{5}\end{array}\right)_{ad,bc}$
(238)

##  Appendix D Chiral properties of Rarita-Schwinger fields

In this appendix, we study the chiral properties of Rarita-Schwinger fields.
As previously described in Section IV, we only need to study the properties of
$(LL)L$, $(LL)R$, $(LR)L$ and $(RL)L$.

###  D.1 Chiral properties of $(LL)L$

The chiral representations of $(LL)L$ are
$(\mathbf{1},~{}\mathbf{1})\oplus(\mathbf{8},~{}\mathbf{1})\oplus(\mathbf{8},~{}\mathbf{1})\oplus(\mathbf{10},~{}\mathbf{1})$.
We will study them separately in the following.

(1) In principle, there are eight possibilities of making the Rarita-Schwinger
fields as shown in Eq. (A). However, the chiral representation
$(\mathbf{1},~{}\mathbf{1})$ has just two non-zero fields:

$\displaystyle\begin{array}[]{l}\Lambda_{L1\mu}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}CL_{B}^{b})\gamma_{\mu}L_{C}^{c}\,,\\\
\Lambda_{L2\mu}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}C\gamma_{5}L_{B}^{b})\gamma_{\mu}\gamma_{5}L_{C}^{c}\,.\end{array}$
(241)

Similarly performing the Fierz transformation to relate $\Lambda_{Li\mu}$ and
$\Lambda^{\prime}_{Li\mu}$, we obtain the solution that all such kind of
fields vanish.

(2) The chiral representation $(\mathbf{10},~{}\mathbf{1})$ has two non-zero
fields:

$\displaystyle\begin{array}[]{l}\Delta^{P}_{L7\mu}=\epsilon_{abc}S_{P}^{ABC}(L_{A}^{aT}C\sigma_{\mu\nu}L_{B}^{b})\gamma^{\nu}L_{C}^{c}\,,\\\
\Delta^{P}_{L8\mu}=\epsilon_{abc}S_{P}^{ABC}(L_{A}^{aT}C\sigma_{\mu\nu}\gamma_{5}L_{B}^{b})\gamma^{\nu}\gamma_{5}L_{C}^{c}\,.\end{array}$
(244)

Similarly performing the Fierz transformation to relate $\Delta^{P}_{Li\mu}$
and $\Delta^{P\prime}_{Li\mu}$, we obtain the solution that all such kind of
fields vanish.

(3) The chiral representation $(\mathbf{8},~{}\mathbf{1})$ has two non-zero
fields:

$\displaystyle\begin{array}[]{l}N^{N}_{L1\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}CL_{B}^{b})\gamma_{\mu}L_{C}^{c}\,,\\\
N^{N}_{L2\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}C\gamma_{5}L_{B}^{b})\gamma_{\mu}\gamma_{5}L_{C}^{c}\,.\end{array}$
(247)

Similarly performing the Fierz transformation to relate $N^{N}_{Li\mu}$ and
$N^{N\prime}_{Li\mu}$, we obtain the solution

$N^{N}_{L2\mu}=N^{N}_{L1\mu}\,.$

Others are just zero. There is only one non-vanishing octet baryon field.

###  D.2 Chiral properties of $(LL)R$, $(LR)L$ and $(RL)L$

The chiral representations of $(LL)R$, $(LR)L$ and $(RL)L$ are
$(\mathbf{\bar{3}},~{}\mathbf{3})\oplus(\mathbf{6},~{}\mathbf{3})$. We will
study them separately in the following.

(1) The chiral representation
$(\mathbf{\bar{3}},~{}\mathbf{3})\rightarrow\mathbf{1_{f}}$ has two non-zero
components:

$\displaystyle\begin{array}[]{l}\Lambda_{M1\mu}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}CL_{B}^{b})\gamma_{\mu}R_{C}^{c}\,,\\\
\Lambda_{M2\mu}=\epsilon_{abc}\epsilon^{ABC}(L_{A}^{aT}C\gamma_{5}L_{B}^{b})\gamma_{\mu}\gamma_{5}R_{C}^{c}\,.\end{array}$
(250)

Similarly performing the Fierz transformation to relate $\Lambda_{Mi\mu}$ and
$\Lambda^{\prime}_{Mi\mu}$, we obtain the solution

$\Lambda_{M1\mu}=-\Lambda_{M2\mu}\,.$

Others are just zero. There is only one non-vanishing field. Others $(LR)L$
and $(RL)L$ can be related to this one.

(2) The chiral representation
$(\mathbf{6},~{}\mathbf{3})\rightarrow\mathbf{10_{f}}$ has two non-zero
components:

$\displaystyle\begin{array}[]{l}\\\
\Delta^{P}_{M7\mu}=\epsilon_{abc}S^{ABC}(L_{A}^{aT}C\sigma_{\mu\nu}L_{B}^{b})\gamma^{\nu}R_{C}^{c}\,,\\\
\Delta^{P}_{M8\mu}=\epsilon_{abc}S^{ABC}(L_{A}^{aT}C\sigma_{\mu\nu}\gamma_{5}L_{B}^{b})\gamma^{\nu}\gamma_{5}R_{C}^{c}\,.\end{array}$
(254)

Others are just zero. Similarly performing the Fierz transformation to relate
$\Delta^{P}_{Mi\mu}$ and $\Delta^{P\prime}_{Mi\mu}$, we obtain the solution

$\Delta^{P}_{M7\mu}=-\Delta^{P}_{M8\mu}\,.$

There is only one non-vanishing field. Others $(LR)L$ and $(RL)L$ can be
related to this one.

(3) The chiral representations
$(\mathbf{\bar{3}},~{}\mathbf{3})\rightarrow\mathbf{8_{f}}$ has only two non-
zero interpolators:

$\displaystyle\begin{array}[]{l}N^{N}_{M1\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}CL_{B}^{b})\gamma_{\mu}R_{C}^{c}\,,\\\
N^{N}_{M2\mu}=\epsilon_{abc}\epsilon^{ABD}\lambda_{DC}^{N}(L_{A}^{aT}C\gamma_{5}L_{B}^{b})\gamma_{\mu}\gamma_{5}R_{C}^{c}\,.\end{array}$
(257)

Similarly performing the Fierz transformation to relate $N^{N}_{Mi\mu}$ and
$N^{N\prime}_{Mi\mu}$, we obtain the solution

$N^{N}_{M1\mu}=-N^{N}_{M2\mu}\,.$

In order to study the chiral representations
$(\mathbf{6},~{}\mathbf{3})\rightarrow\mathbf{8_{f}}$, we need to consider the
second way (see the discussion in Section II.2.3) which leads to four non-zero
fields:

$\displaystyle\begin{array}[]{l}\widetilde{N}^{N}_{M1\mu}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}(L_{A}^{aT}CL_{B}^{b})\gamma_{\mu}R_{C}^{c}\,,\\\
\widetilde{N}^{N}_{M2\mu}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}(L_{A}^{aT}C\gamma_{5}L_{B}^{b})\gamma_{\mu}\gamma_{5}R_{C}^{c}\,,\\\
\widetilde{N}^{N}_{M7\mu}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}(L_{A}^{aT}C\sigma_{\mu\nu}L_{B}^{b})\gamma^{\nu}R_{C}^{c}\,,\\\
\widetilde{N}^{N}_{M8\mu}=\epsilon_{abc}\epsilon^{ACD}\lambda_{DB}^{N}(L_{A}^{aT}C\sigma_{\mu\nu}\gamma_{5}L_{B}^{b})\gamma^{\nu}\gamma_{5}R_{C}^{c}\,.\end{array}$
(262)

By using the Jacobi identity in Eq. (39), we obtain:

$\widetilde{N}^{N}_{M1\mu}={1\over
2}N^{N}_{M1\mu}\,,~{}\widetilde{N}^{N}_{M2\mu}={1\over 2}N^{N}_{M2\mu}\,.$

Similarly performing the Fierz transformation to relate
$\widetilde{N}^{N}_{Mi\mu}$ and $\widetilde{N}^{N\prime}_{Mi\mu}$, we obtain
the solution

$\displaystyle\begin{array}[]{l}\widetilde{N}^{N}_{M2\mu}=-\widetilde{N}^{N}_{M1\mu}=-{1\over
2}N^{N}_{M1\mu}\,,\\\
\widetilde{N}^{N}_{M8\mu}=-\widetilde{N}^{N}_{M7\mu}\,.\end{array}$ (265)

All together there are two non-vanishing independent fields:
$\widetilde{N}^{N}_{M1\mu}$ and $\widetilde{N}^{N}_{M7\mu}$.
$\widetilde{N}^{N}_{M1\mu}$ is related to $N^{N}_{M1\mu}$, and so belongs to
the chiral representation $((\mathbf{\bar{3}},~{}\mathbf{3}))$. However, the
other $\widetilde{N}^{N}_{M7\mu}$ can not be related to $N^{N}_{Mi\mu}$, and
so contains some $(\mathbf{6},~{}\mathbf{3})$ component. other Others $(LR)L$
and $(RL)L$ can be related to $(LL)R$. Chiral properties of the tensor fields
can be also explored in completely the same procedure explained here.
Therefore, we do not show this case any more.

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