The pentaquark spectrum from Fermi statistics

Abstract

We study compact hidden charm pentaquarks in the Born–Oppenheimer approximation, previously introduced for tetraquarks, assuming the heavy pair to be in a color octet. We show that Fermi statistics applied to the complex of the three light quarks, also in color octet, requires S-wave pentaquark ground states to consist of three octets of flavour-SU(3)$_f$, two with spin 1/2 and one with spin 3/2, in line with the observed, strangeness $S=0,-1$, spectrum. Additional lines corresponding to decays into $J/\psi +\Sigma$ and $J/\psi +\Xi$ are predicted. In the language of non-relativistic SU(6), ground state pentaquarks form either a ${\varvec{56}}$ or a ${\varvec{20}}$ representation, distinguished by presence or absence of pentaquarks decaying in the spin 3/2 decuplet, e.g. in $J/\psi +\Delta ^{++}$. Observation of a strangeness $S=-2$ or isospin $I=3/2$ pentaquarks would be a clear signature of compact, QCD bound pentaquarks.

1 Introduction

After the LHCb observation in 2015 [1], hidden charm pentaquark, compact or molecular, have received considerable attention. Most recently, three lines are observed for strangeness $S=0$ final state $J/\psi +p$, and two, possibly three, lines for $S=-1$, $J/\psi +\Lambda$ [2].

Loosely bound molecular pentaquark interpretations can be found in [3,4,5,6,7,8,9,10,11]. Kinematical effects are discussed in [12,13,14,15]. An early discussion of compact pentaquarks can be found in [16,17,18,19,20,21,22,23,24].

To describe compact, hidden charm or beauty pentaquarks, the Born–Oppenheimer (BO) approximation is particularly appropriate since it improves when the mass of the heavy constituents increases with respect to the mass of the light ones, see [25,26,27]. In this framework, one considers the heavy constituents (c and $\bar{c}$) as color sources at rest, with fixed relative distance R, and computes the lowest energy E(R) of the three light quarks, either analytically, in perturbation theory with techniques borrowed from molecular physics [25], or non-perturbatively, in lattice QCD [28].

Besides distance and spin, one has to specify the color quantum numbers of the sources. For colored quarks, the natural choice is to assume the heavy quark-antiquark pair to be in a color octet, coupled to the light quarks to make an overall color singlet

$$\begin{aligned} \mathcal{P}_{BO}=[{\bar{c}}({\varvec{x}}_{\bar{c}})\lambda ^A c({\varvec{x}}_c)] B^A \end{aligned}$$

(1)

(sum over $A=1,\dots , 8$ understood). $B^A$ describes the complex of the three light quarks, which live in a color octet as well. Specializing to a proton-like pentaquark, we specify $B^A=B^A[u(x_1),u(x_2), d(x)]$ in terms of the light quark coordinates.

In this paper, we address the restrictions posed by Fermi statistics to the complex of the three light quarks in (1), considering exchange of color, coordinates, flavor and spin. We summarize flavor and spin in representations of non-relativistic SU(6) $\supset$ SU(3)$_f\,\otimes \,$SU(2)$_\textrm{spin}$ [30].

We show that the simplest hypothesis of complete symmetry under coordinates, which would imply pentaquarks in the mixed-symmetry $\textbf{70}$ representation of SU(6), is excluded.

The BO approximation allows mixed symmetry in the coordinates also for ground state, S-wave, pentaquarks by distributing quarks in orbitals around the fixed sources. We show that, in this case, one obtains a consistent solution for the SU(6) representations $\mathbf{{56}}$ and $\mathbf{{20}}$ but not for the $\mathbf{{70}}$. Pentaquarks in SU(3) flavour octets have been considered in [31].

The elimination of the $\textbf{70}$ greatly reduces spin and flavor multiplicity of ground state pentaquarks. Including the spin of the $c-\bar{c}$ pair, we obtain, for either $\mathbf{{56}}$ or $\mathbf{{20}}$, three octets of flavour-SU(3)$_f$, two with spin 1/2 and one with spin 3/2. Indeed three pentaquark states have been observed in $S=0$ and perhaps $S=-1$ channels [2]. A pentaquark in the $J/\psi p\bar{p}$ channel has been later reported in [32]. In the spectrum shown there is no sign of the $P_c(4312)$ seen in [2]. This could be an indication that both 56 and 20 are realized in the hadron spectrum.

Additional lines corresponding to pentaquarks decaying into $J/\psi +\Sigma ~(S=-1)$ and $J/\psi +\Xi ~(S=-2)$ are predicted.

The two alternatives, $\textbf{56}$ and $\textbf{20}$, are distinguished by the presence or absence of pentaquarks decaying into spin 3/2 resonances, e.g.: $\mathcal{P}_{(\Delta ,S=0)}^{++} \rightarrow J/\psi + \Delta ^{++}\rightarrow J/\psi + p+\pi ^+$, which applies to ${\varvec{56}}$ only.

The observation of a strangeness $S=-2$ or isospin $I=3/2$ pentaquarks would be clear signatures of compact, QCD bound pentaquarks.

The plan of the paper is as follows. We discuss the properties of three light quarks operators in Sect 2. Section 3 illustrates the way to deal with the representations of the group of three objects, $S_3$, for color, coordinates, flavor and spin. In Sect. 4 we present the essentials of the BO approximation for pentaquarks and derive a consistency condition on QCD couplings of light to heavy quarks, for quarks distributed in different BO orbitals. In Sect. 5 we illustrate the conditions required by Fermi statistics on the (flavour $\otimes$  spin) wave function of light quarks and in Sect. 6 compute the resulting SU(6) wave functions. Finally, in Sect. 7 we presents our results and illustrate the prospects of future calculations of pentaquark mass spectra in the BO approximation. More technical details are contained in four Appendices.

2 Basic light quark operators

We describe pentaquarks with operators of the generic form given in Eq. (1)

$$\begin{aligned} \mathcal{P}_{BO}=[{\bar{c}}({\varvec{x}}_{\bar{c}})\lambda ^A c({\varvec{x}}_c)] B^A \end{aligned}$$

$B^A$ is obtained from two basic octets, constructed in turn from antisymmetric or symmetric diquark operators.

Specialising to proton-like tetraquarks

• 1. the antisymmetric octet is

  $$\begin{aligned}{} & {} P^A(u_1,d | u_2)=\bar{\theta }(u_1 d)\lambda ^A u_2= u_1^a d^b\, (\epsilon _{abc}\, \lambda ^A_{cd})\, u_2^d\nonumber \\{} & {} \bar{\theta }(q q^\prime )_a=\epsilon _{abc }\, q^b q^{\prime c} \end{aligned}$$

  (2)

• 2. the diquark symmetric octet is

  $$\begin{aligned}{} & {} \Phi ^A(u_1,d | u_2)=\phi ^{ad}\, (\epsilon _{abc}\, \lambda ^A_{cd})\, u_2^b\nonumber \\{} & {} \phi ^{ab}(u,d)=(u^a d^b+u^b d^a) \end{aligned}$$

  (3)

We register here a few relevant Fierz transformations

$$\begin{aligned} T= & {} |(\bar{Q} Q)_{\varvec{8}} (\bar{\theta }q)_{\varvec{8}}\rangle _{\varvec{1}}=\sqrt{\frac{2}{3}} |(Qq)_{\bar{\varvec{3}}} (\bar{Q} \bar{\theta })_{\varvec{3}}\rangle _1 -\frac{1}{\sqrt{3}}|(Qq)_{\varvec{6}} (\bar{Q} \bar{\theta })_{{\bar{\varvec{6}}}}\rangle _1 \end{aligned}$$

(4)

$$\begin{aligned}= & {} \sqrt{\frac{8}{9}}|(\bar{Q} q)_{\varvec{1}}(\bar{\theta }Q)_{\varvec{1}}\rangle -\frac{1}{\sqrt{9}}|(\bar{Q} q)_{\varvec{8}}(\bar{\theta }Q)_{\varvec{8}}\rangle _{\varvec{1}}, \end{aligned}$$

(5)

where $\bar{\theta }$ is a generic ${\bar{\varvec{3}}}$. For a generic $\phi \in {{\varvec{6}}}$, the Fierz relations read

$$\begin{aligned} T= & {} |(\bar{Q} Q)_{\varvec{8}} (\phi q)_{\varvec{8}}\rangle _{\varvec{1}}=|(\bar{Q} q)_{\varvec{8}} (\phi Q)_{\varvec{8}}\rangle _{\varvec{1}}\nonumber \\= & {} |(Qq)_{\bar{\varvec{3}}} (\phi \bar{Q})_{\varvec{3}}\rangle _{\varvec{1}} \end{aligned}$$

(6)

which follow from the SU(3)$_c$ composition rules: ${{\varvec{6}}}\otimes {\varvec{3}}={\varvec{8}}\oplus {\varvec{10}}$ and ${{\varvec{6}}}\otimes {\bar{\varvec{3}}}={\varvec{3}}\oplus {\varvec{15}}$.

Given the form of $B^A$ in terms of $P^A$ and $\Phi ^A$, relations (4) to (6) allow us to find the QCD couplings qQ and $q\bar{Q}$. We define

$$\begin{aligned} g^2_{cq}=\alpha _s \lambda _{cq}\qquad \lambda _{cq}=\frac{1}{2}(C_2({\varvec{R}})-8/3) \end{aligned}$$

(7)

$C_2({\varvec{R}})$ is the quadratic Casimir operator¹ of the color representation ${\varvec{R}}$ of the pair cq. If the pair is in a superposition of two or more SU(3)$_c$ representations with amplitudes a, b, $\dots$ we use [25]

$$\begin{aligned}{} & {} T= a\,|(c q)_{{\varvec{R}}_1}\dots \rangle _{\varvec{1}}+b\,|(c q)_{{\varvec{R}}_2}\dots \rangle _{\varvec{1}}+\cdots \nonumber \\{} & {} \lambda _{cq }=a^2\lambda _{cq}({\varvec{R}_1})+b^2\lambda _{cq}({\varvec{R}_2})+\cdots . \end{aligned}$$

(8)

Exchanging quarks. Since $P^A$ and $\Phi ^A$ are the only octet operators with the given three quarks, we must be able to express $P^A(u_1,d | u_2)$ in terms of octets made by $u_1 u_2$ and d, i.e. find a and b such that

$$\begin{aligned} P^A(u_1 d | u_2)= a~P^A(u_1 u_2 | d)+ b~\Phi ^A(u_1 u_2 | d) \end{aligned}$$

(9)

Relation (9) is what we need to move quarks around and bring a given pair together. However, operators $P,~\Phi$ are not equally normalised. Starting from a and b, we compute the normalisation factors to obtain a relation between normalised kets:

$$\begin{aligned} |(u_1 d)_{{\bar{3}}},u_2\rangle _{\varvec{8}}=\alpha |(u_1 u_2)_{{\bar{3}}}, d\rangle _{\varvec{8}} + \beta |(u_1 u_2)_{6}, d \rangle _{\varvec{8}} \end{aligned}$$

(10)

with $\alpha ^2 +\beta ^2 =1$. A simple calculation leads to the transformation table

Table 1 Transformation table for quark rearrangements inside $P^A$ and $\Phi ^A$ color octets

3 The group $S_3$ and its representations

The group $S_3$ of the six permutations of three elements can be seen as the group of symmetries of an equilateral triangle by thinking of these as permuting the three vertices. This group consists of the identiy transformation corresponding to three cycles of length one, two rotations by $120^\circ$ and $240^\circ$, which are permutations corresponding to cycles of length three (e.g. the vertex $a\rightarrow b\rightarrow c\rightarrow a$) and three reflections in the three altitudes of the triangle, each consisting of a cycle of length two combined with a cycle length one leaving one vertex unchanged.

The two dimensional representation  is given by [33]

$$\begin{aligned}{} & {} E=\begin{pmatrix} 1&{}0\\ 0&{}1\\ \end{pmatrix}\qquad D(\sigma _1)=\frac{1}{2}\begin{pmatrix} -1&{}-\sqrt{3}\\ \sqrt{3}&{}-1\\ \end{pmatrix} \nonumber \\{} & {} D(\sigma _2)=\frac{1}{2}\begin{pmatrix} -1&{}\sqrt{3}\\ sqrt{3}&{}-1\\ \end{pmatrix} \end{aligned}$$

(11)

and

$$\begin{aligned}{} & {} D(\tau _1)=\begin{pmatrix} -1&{}0\\ 0&{}1\\ \end{pmatrix}\qquad D(\tau _2)=\frac{1}{2} \begin{pmatrix} 1&{}-\sqrt{3}\\ sqrt{3}&{}-1\\ \end{pmatrix}\nonumber \\{} & {} D(\tau _3)=\frac{1}{2} \begin{pmatrix} 1&{}\sqrt{3}\\ \sqrt{3}&{}-1\\ \end{pmatrix} \end{aligned}$$

(12)

Having in mind the equilateral triangle

we see that $D(\sigma _1)$ is a anti-clockwise rotation by $120^\circ$ of the position vectors of the vertices of the triangle (taken from the center), and $D(\sigma _2)$ is a rotation by $-120^\circ$. $D(\tau _1)$ represents a reflection (through the y axis). A rotation $D(\sigma _1)$ changes $a\mapsto b\mapsto c\mapsto a$. If after this rotation a $D(\tau _1)$ reflection is done, which amounts to $c\rightleftarrows a$, we get $D(\tau _1)D(\sigma _1)=D(\tau _3)$, which corresponds to a $b\rightleftarrows c$ reflection on the original triangle. Similarly $D(\tau _1)D(\sigma _2)=D(\tau _2)$, corresponding to a $a\rightleftarrows b$ reflection.

Let us name the eigenvectors of the reflection $D(\tau _1)$ as

$$\begin{aligned} D(\tau _1)M^\lambda =M^\lambda \qquad D(\tau _1)M^\rho =-M^\rho \end{aligned}$$

(14)

Therefore

$$\begin{aligned}{} & {} D(\tau _2)M^\lambda =-\frac{\sqrt{3}}{2}M^\rho -\frac{1}{2}M^\lambda ; \nonumber \\{} & {} D(\tau _3)M^\lambda =\frac{\sqrt{3}}{2}M^\rho -\frac{1}{2}M^\lambda \end{aligned}$$

(15)

$$\begin{aligned}{} & {} D(\tau _2)M^\rho =\frac{1}{2}M^\rho -\frac{\sqrt{3}}{2}M^\lambda ; \nonumber \\{} & {} D(\tau _3)M^\rho =\frac{1}{2}M^\rho +\frac{\sqrt{3}}{2}M^\lambda \end{aligned}$$

(16)

Two different mixed representations of $S_3$ acting on different variables, $M_1$ and $M_2$, may combine to an A (anti-symmetric), S (symmetric) or M (mixed) representation, according to the scheme [35]

$$\begin{aligned} S= & {} \frac{1}{\sqrt{2}}(M_1^\rho M_2^\rho +M_1^\lambda M_2^\lambda ) A=\frac{1}{\sqrt{2}}(M_1^\rho M_2^\lambda - M_1^\lambda M_2^\rho )\nonumber \\ M^\rho= & {} \frac{1}{\sqrt{2}}(M_1^\rho M_2^\lambda + M_1^\lambda M_2^\rho ) \quad M^\lambda =\frac{1}{\sqrt{2}}(M_1^\rho M_2^\rho - M_1^\lambda M_2^\lambda )\nonumber \\ \end{aligned}$$

(17)

For color we define

$$\begin{aligned}{} & {} |(q_1q_2)_{\bar{\varvec{3}}},q_3\rangle _{\varvec{8}}= M^\rho \end{aligned}$$

(18)

$$\begin{aligned}{} & {} |(q_1q_2)_{{\varvec{6}}},q_3\rangle _{\varvec{8}}= M^\lambda , \end{aligned}$$

(19)

where $q_1,q_2,q_3$ correspond to the vertices in the triangle a, b, c.

Equations (15) and (16) reproduce the results in Table 1. The left side columns of Table 1 correspond to $D(\tau _3)$, or $b\rightleftarrows c$ reflection, whereas the right side columns corresponds to $D(\tau _2)$, or $a\rightleftarrows c$ reflection.

Following [35], we list explicitly in Appendix B the building blocks of mixed and symmetric representations with respect to color, flavour, spin and coordinates, restricting to proton-like states (flavour octet or decuplet).

4 Born–Oppenheimer approximation: a QCD consistency condition

We follow the perturbative scheme illustrated in [27] for hydrogen molecules and ions in QED and in [25] for hidden charm hadrons in QCD. The starting point is the interaction of each light particle with the fixed sources, following the instructions given in Sect. 2.

Fig. 1

Full lines indicate interactions that make the orbitals. For the three light quarks to be in the same orbital, it is necessary that they carry the same color charge with respect to both c and $\bar{c}$

• 1. In the simplest one orbital scheme, one solves, analytically or numerically [27], the Scrödinger equation of q in the presence of the static sources. The orbital is the corresponding ground state with wave function, $\psi _0(x_q)$ and energy $\epsilon _0$. We represent the orbital in Fig. 1 with an ellipse around the heavy sources. Similarly to what done in atomic physics, we put the other light quarks in the same orbital, corresponding to the ground state wave function

  $$\begin{aligned}{} & {} \Psi _0(x_{u_d},x_{u_1},x_{u_2})=\psi _0(x_{d})\psi _0(x_{u_1}) \psi _0(x_{u_2}) \nonumber \\{} & {} E_0=3\epsilon _0 \end{aligned}$$

  (20)

  $\Psi _0$ is symmetric under the exchange of quark coordinates. In the atomic physics language, we attribute occupation number 3 to the orbital. Denoting by $V_\textrm{res}$ the sum of the light-to-light interactions, that did not intervene in the construction of the orbital, the BO potential, to first order in $V_{res}$, is

  $$\begin{aligned} V_\textrm{BO}=E_0+ \langle \Psi |V_\textrm{res}|\Psi \rangle +V_{c\bar{c}}(R) \end{aligned}$$

  (21)

  where $V_{c\bar{c}}(R)$ is the QCD interaction between c and $\bar{c}$. $V_{BO}$ is obviously a function of R and is the potential of the Schrödinger equation of the $c\bar{c}$ system.

• 2. One can also consider a two orbitals scheme with two separate Schrödinger equations: orbital $q-c$ (denoted by $\phi$) and orbital $q-\bar{c}$ (denoted by $\psi$). In Fig. 2a two light quarks sit in $\phi$ and one in $\psi$, the opposite in Fig. 2b. We have two distinct possibilities for the ground state

  $$\begin{aligned}{} & {} \Psi _0^{(a)}=\phi _0(x_d)\phi _0(x_{u_1})\psi _0(x_{u_2}) \nonumber \\{} & {} \Psi _0^{(b)}=\phi _0(x_d)\psi _0(x_{u_1})\psi _0(x_{u_2}) \end{aligned}$$

  (22)

Fig. 2

Two different possibilities for quark occupation distributed in two orbitals. In a two quarks sit in the c orbital (denoted by $\phi$); in b two quarks sit in the $\bar{c}$ orbital (denoted by $\psi$). Full lines indicate interactions that make the orbitals, dotted lines are examples of the additional interactions taken into account to first order perturbation theory. The dot-dashed line represents the Born–Oppenheimer potential. For consistency, as discussed in Sect. 4, light quarks in the same orbital must have the same interaction with c and $\bar{c}$

A consistency condition. QED charges of protons and electrons are fixed constants. This is not the case for two-body QCD charges, which depend on the superpositions of the relative color representations in which the pair occurs, as indicated in Sect. 2. This leads to a consistency condition, namely that: quarks of different flavor in the same orbital (as in Fig. 1) must share the same QCD coupling to the heavy quarks at the center of the orbital

$$\begin{aligned} \lambda _{{cq}}=\lambda _{{cq^\prime }}\qquad \lambda _{\bar{c} q}=\lambda _{\bar{c} q^\prime } \end{aligned}$$

(23)

with possibly $\lambda _{cq}\ne \lambda _{\bar{c} q}$.

We shall see that this condition is not trivially satisfied in the pentaquark.

5 Three light quark operators and Fermi statistics

Color singlet baryons.

S-wave, color singlet baryons are fully antisymmetric under color exchange and fully symmetric under coordinate exchange. Therefore quark (flavour $\otimes$ spin) must be symmetric.

If we summarise spin and flavour quantum numbers with representations of SU(6) $\supset$ SU(3)$_f\, \otimes$  SU(2)$_\textrm{spin}$, this is the ${\varvec{56}}$ representation of non-relativistic SU(6) [29], with content:

$$\begin{aligned} {\varvec{56}:}~({\varvec{8}},1/2) \oplus ({\varvec{10}}, 3/2) \end{aligned}$$

(24)

A different case is that of excited, negative parity baryons: color is fully antisymmetric, but coordinates are in a mixed state (two quarks in S-wave and one quark in P-wave), see Refs. [34, 35]. The three quarks must form a mixed-symmetry representation in flavour and spin, to obtain full symmetry when combined with coordinates. In SU(6) language, negative parity baryons are in the 70 representation, which decomposes as

$$\begin{aligned} {\varvec{70}:}~({\varvec{1}},1/2)\oplus ({\varvec{8}},3/2) \oplus ({\varvec{8}},1/2) \oplus ({\varvec{10}},1/2). \end{aligned}$$

(25)

The third three-quark SU(6) representation, the fully antisymmetric

$$\begin{aligned} \mathbf{{20}}=(\textbf{1},3/2)\oplus (\textbf{8},1/2) \end{aligned}$$

(26)

is forbidden by Fermi statistics for both ground state and P wave baryons.

Pentaquarks. S-wave pentaquarks have the three light quarks in color octet, distributed in one or more orbitals. Total antisymmetry under quark exchange required by Fermi statistics may be reached in different ways, summarized in Table 2.

The simplest possibility is to assume complete symmetry of the coordinates. In this case, full antisymmetry under quark exchange requires the light quark complex to have mixed symmetry under spin and flavour exchange, to be combined with colour to a totally antisymmetric state, first row of Table 2: the three light quarks must form a ${\varvec{70}}$ representation [31], with the flavour-spin content reported in Eq. (25).

However, one needs to take into account the consistency condition stated in Sect. 4.

• 1. In Born–Oppenheimer parlance, symmetry under coordinate exchanges means that light quarks populate a single orbital, Fig. 1. Accordingly, they must share the same QCD coupling to c and to $\bar{c}$

  $$\begin{aligned} \lambda _{cu}=\lambda _{cd}\qquad \lambda _{\bar{c} u}=\lambda _{\bar{c} d} \end{aligned}$$

  (27)

  with possibly $\lambda _{cq}\ne \lambda _{\bar{c} q}$. An explicit calculation for the case of the $({\varvec{8}}, 3/2)\subset {\varvec{70}}$, see Appendix A, shows however that the couplings of d and u quarks are different: light quarks in a color octet cannot populate a single orbital.

• 2. The next simplest possibility, in analogy with $L=1$ baryons, is to arrange light quarks in two orbitals,² one around c and the other around $\bar{c}$, Fig. 2a or b. Combining color mixed-symmetry with coordinate mixed-symmetry, one may obtain color-coordinate symmetry of type A, S, or M. To obey Fermi statistics, these possibilities must be associated to flavor-spin symmetries of type S, A or M, respectively, namely to the SU(6) representations ${\varvec{56}}$, ${\varvec{20}}$ or $\mathbf{{70}}$ (second to fourth rows of Table 2).

We analyse the three cases in Sect. 6, and show that only ${\varvec{56}}$ and ${\varvec{20}}$ obey the consistency condition stated in Sect. 4.

6 Octets and decuplets with light quarks in two orbitals

We give in this Section the explicit form of the light quarks operators corresponding to the SU(6) representations 56, 20 and 70, with quarks distributed in two orbitals denoted by $\phi$ and $\psi$ and mixed-symmetry in the coordinates (see (B7) and rows 2 to 4 in Table 2).

We give a complete discussion of the spin 1/2 octet, $(\textbf{8},1/2)$ of ${\varvec{56}}$ and report the results for the other cases, referring to Appendix C for details.

In the following, we use the following abbreviated notations.

$$\begin{aligned} \begin{aligned}&[q_1,q_2]q_3 \Leftrightarrow ((q_1,q_2)_{\bar{\varvec{3}}_c} q_3)_{\varvec{8}_c}\\&(q_1,q_2)q_3 \Leftrightarrow ((q_1,q_2)_{{\varvec{6}}_c} q_3)_{\varvec{8}_c}\\&[q_1,q_2]_{s}\vee (q_1,q_2)_{s} \Leftrightarrow (q_1,q_2)_{\bar{\varvec{3}}_c,S=s}\vee (q_1,q_2)_{\varvec{6}_c,S=s}\\&\{[q_1,q_2] q_3\}_s \vee \{(q_1,q_2) q_3\}_s \Leftrightarrow ([q_1,q_2] q_3)_{S=s}\\&\qquad \quad \vee ((q_1,q_2) q_3)_{S=s} \end{aligned} \end{aligned}$$

Whenever we write $(c\bar{c})$ we mean color octet $(c\bar{c})_{\varvec{8}}$

6.1 The spin 1/2 octet of 56

Following Eq. (17), the fully antisymmetric wave function in color and coordinate is

$$\begin{aligned} \mathcal{P}= & {} (\bar{c} c)\times \frac{1}{\sqrt{2}} [M^\lambda N^\rho - M^\rho N^\lambda ]\nonumber \\= & {} (c\bar{c})\times \frac{1}{\sqrt{2}} \left\{ (q_1,q_2) q_3 \, N^\rho -[q_1,q_2]q_3 \,N^\lambda \right\} \end{aligned}$$

(28)

where N is the mixed representation for coordinates. For color $\textbf{6}$ or $\bar{\varvec{3}}$, when $q_1$ and $q_3$ are in $\phi$ and $q_2$ in $\psi$ we introduce the abbreviation

$$\begin{aligned}{} & {} (q_1,q_2)q_3\, \phi (1)\psi (2)\phi (3)\equiv (q^\phi _1,q^\psi _2) q_3^\phi \nonumber \\{} & {} [q_1,q_2]q_3\, \phi (1)\psi (2)\phi (3)\equiv [q^\phi _1,q^\psi _2] q_3^\phi \end{aligned}$$

(29)

We obtain

$$\begin{aligned} \mathcal{P}= & {} (\bar{c} c)\times \frac{-n}{\sqrt{2}}\left\{ \frac{1}{\sqrt{2}}\left( (q^\phi _1,q^\psi _2) q_3^\phi -(q^\psi _1,q^\phi _2) q_3^\phi \right) \right. \nonumber \\{} & {} \left. + \frac{1}{\sqrt{6}} \left( [q^\phi _1,q^\psi _2] q^\phi _3+[q^\psi _1,q^\phi _2] q^\phi _3-2[q^\phi _1,q^\phi _2] q^\psi _3\right) \right\} \nonumber \\ \end{aligned}$$

(30)

with some normalization n. Antisymmetry of Eq. (30) under the exchange (1, 2) is explicit. To check the case (1, 3) one has to use Table 1. A tedious but direct calculation shows that indeed:

$$\begin{aligned}{} & {} (1,3)\mathcal{P}=-\mathcal{P} \end{aligned}$$

(31)

Using Table 1, we bring all $q^\phi$ inside the parentheses and obtain the more symmetric form (dropping an irrelevant overall sign):

Table 2 To get a totally anti-symmetric (A) state, we need $M\times M\rightarrow A$ or $M\times M\rightarrow S$ depending on the SU(6) or coordinate entries. In the last row we require that the mixed symmetries for color and coordinate are composed in a mixed symmetry which, combined with spin-flavor SU(6), makes a totally anti-symmetric state

$$\begin{aligned} \mathcal{P}=(\bar{c} c)\times \frac{n}{\sqrt{3}}~\Big \{[q_1^\phi ,q_2^\phi ] q_3^\psi + \text {cyclic permutations in 1,2,3}\Big \}\nonumber \\ \end{aligned}$$

(32)

We have to combine Eq. (32) with the flavour-spin fully symmetric expression obtained from the mixed-symmetry templates, Eqs. (B3) and (B5) in Appendix B to obtain:

$$\begin{aligned} S({\varvec{56}})=\frac{1}{\sqrt{2}}[F^\rho \chi ^\rho +F^\lambda \chi ^\lambda ] \end{aligned}$$

(33)

The $\varvec{d_3=d}$ condition. We may use (32), (33) and Table 1 to bring the $d_i$ in the same position.One obtains the superposition of three replicas which are cyclical permutations of the assignments of u and d flavours to quarks 1,2,3. In practical calculations, we may decide that the d flavour is in quark $q_3$, keep the term containing $d_3=d$ and discard the rest. To obtain a normalised state we have to multiply by $\sqrt{3}$.

In a way, the condition

$$\begin{aligned} d_1=d_2=0\qquad d_3=d \end{aligned}$$

(34)

is analogous to fixing the gauge in a gauge invariant theory and restrict to one of the (infinitely many) orbits generated by gauge transformations. In our case, the group is the group of quark permutations and there are only three orbits.

Applying the condition $d=d_3$ to (33) only the second term in (33) is non vanishing and we obtain³

$$\begin{aligned} S({\varvec{56}})=-\frac{1}{\sqrt{3}}~\{\{u_1, u_2\}_1 d\}_{1/2} \end{aligned}$$

(35)

or more explicitely

$$\begin{aligned} S({\varvec{56}})=\frac{1}{\sqrt{3}}\Big [\frac{u_1^\uparrow u_2^\downarrow +u_1^\downarrow u_2^\uparrow }{\sqrt{2}}~d^\uparrow -\sqrt{2}~u_1^\uparrow u_2^\uparrow d^\downarrow \Big ] \end{aligned}$$

(36)

Combining (35) and (30), the operator that creates $\mathcal{P}_{\varvec{56}}$ octet takes the proton-like form (the extension to the other members of the octet is given in Appendix D)

$$\begin{aligned} \mathcal{P}^{\varvec{ 56}}_\mathbf{8,1/2}= & {} \sqrt{3}~ \mathcal{P}\otimes S({\varvec{56}}) \nonumber \\= & {} (\bar{c} c)\times \frac{n}{\sqrt{2}}\left\{ \frac{1}{\sqrt{2}}\left( (u^\phi _1,u^\psi _2 )_1 d^\phi -(u^\psi _1,u^\phi _2)_1 d^\phi \right) \right. \nonumber \\{} & {} \left. + \frac{1}{\sqrt{6}} \left( [u^\phi _1,u^\psi _2]_1 d^\phi +[u^\psi _1,u^\phi _2]_1d^\phi -2[u^\phi _1,u^\phi _2]_1d^\psi \right) \right\} \nonumber \\ \end{aligned}$$

(37)

It is convenient to have an expression which puts together quarks that belong to the same orbital. This is obtained by using (35) with (32) and further expressing the wave function in terms of the total spin of the pair inside orbital $\phi$

$$\begin{aligned} \mathcal{P}^{\varvec{ 56}}_\mathbf{8,1/2}= & {} \sqrt{3}~ \mathcal{P}\otimes S({\varvec{56}})\nonumber \\= & {} (\bar{c} c)\times \frac{n}{\sqrt{3}}\Big \{[u_1^\phi ,u_2^\phi ]_1 d^\psi -\frac{1}{2}[d^\phi ,u_1^\phi ]_1 u_2^\psi \nonumber \\{} & {} -\frac{1}{2}[ u_2^\phi ,d^\phi ]_1u_1^\psi -\frac{\sqrt{3}}{2}[d^\phi ,u_1^\phi ]_0 u_2^\psi +\frac{\sqrt{3}}{2}[u_2^\phi ,d^\phi ]_0 u_1^\psi ~\Big \}\nonumber \\ \end{aligned}$$

(38)

Color couplings of $d^\phi$ and $d^\psi$ can be easily calculated from the above formulas

$$\begin{aligned}{} & {} \lambda _{d^\phi c}=\lambda _{u_1^\phi c}=-\frac{7}{18}\qquad \lambda _{d^\phi \bar{c}}=\lambda _{u_1^\phi \bar{c}}= -\frac{2}{18}\nonumber \\{} & {} \lambda _{d^\psi c}=\lambda _{u_1^\psi c}= -\frac{2}{18}\qquad \lambda _{d^\psi \bar{c}}=\lambda _{u_1^\psi \bar{c}}=-\frac{7}{18} \end{aligned}$$

(39)

This operator describes a state with the orbitals, $\phi$ and $\psi$, populated according to:

$$\begin{aligned} \phi ( qq), \psi (q) \end{aligned}$$

We have a second possibility where $\psi$ has occupation number two. In total

$$\begin{aligned} \begin{aligned}&{\varvec{A}}({\varvec{56}}){} & {} c-\phi ( qq){} & {} \bar{c}- \psi (q){} & {} \mathrm{Fig.~2}a\\&{\varvec{B}}({\varvec{56}}){} & {} c- \phi (q){} & {} \bar{c}- \psi ( qq){} & {} \mathrm{Fig.~2}b \end{aligned} \end{aligned}$$

The first possibility has lower energy, $-7/18<-2/18$, in the one gluon exchange approximation.

6.2 The spin 3/2 decuplet of $\textbf{56}$.

We register the case with $I_3=+1/2$:

$$\begin{aligned} {\Delta }^{+}_{10}= & {} (\bar{c} c)\times \frac{n}{\sqrt{3}} \Big \{\{[u^\phi _1, u^\phi _2]d^\psi \}_{3/2}\nonumber \\{} & {} +\{[d^\phi , u^\phi _1]u_2^\psi \}_{3/2}+\{[u_2^\phi , d^\phi ]u_1^\psi \}_{3/2} \Big \} \end{aligned}$$

(40)

Color couplings $u_3^{\phi ,\psi }-c/\bar{c}$ are the same as in (39).

6.3 The spin 1/2 octet of 20

We find

$$\begin{aligned} \mathcal{P}^{\varvec{20}}_{\varvec{8,1/2}}= & {} (\bar{c} c)\times \frac{n}{\sqrt{2}} \left\{ \frac{ [ u_1^\phi , u_2^\psi ]_0 d^\phi - [ u_1^\psi , u_2^\phi ]_0 d^\phi }{\sqrt{2}}\right. \nonumber \\{} & {} \left. -\frac{( u_1^\phi ,u^\psi _2)_0 d^\phi +( u_1^\psi ,u^\phi _2)_0 d^\phi -2(u_1^\phi ,u^\phi _2)_0 d^\psi }{\sqrt{6}}\right\} \nonumber \\ \end{aligned}$$

(41)

or, equivalently:

$$\begin{aligned} \mathcal{P}^{\varvec{20}}_{\varvec{8,1/2}}= & {} (\bar{c} c)\times \frac{n}{\sqrt{3}} \Big [ \{( u_1^\phi ,u_2^\phi )_0 d^\psi \}_{1/2} -\frac{1}{2}\{(d^\phi , u_1^\phi )_0 u_2^\psi \}_{1/2}\nonumber \\{} & {} -\frac{1}{2}\{( d^\phi , u_2^\phi )_0 u_1^\psi \}_{1/2}-\frac{\sqrt{3}}{2}\{( d^\phi ,u_1^\phi )_1u_2^\psi \}_{1/2}\nonumber \\{} & {} +\frac{\sqrt{3}}{2}\{ (d^\phi , u_2^\phi )_1u_1^\psi \}_{1/2}\Big ]\nonumber \\ \end{aligned}$$

(42)

Not surprisingly, (42C10) is obtained from (38) with two simultaneous exchanges of the symmetry characters

$$\begin{aligned} \bar{\varvec{3}}_c \rightarrow \varvec{6}_c\qquad \text {diquark~spin:}~1\rightleftarrows 0\qquad \text {isospin~unchanged}\nonumber \\ \end{aligned}$$

(43)

Thus, both octets have the same antisymmetry under total exchange of quark quantum numbers, as required by Fermi statistics.

We find the couplings

$$\begin{aligned}{} & {} \lambda _{ d^\phi c}=\lambda _{ u_1^\phi c}=-\frac{5}{18}\qquad \lambda _{d^\phi {\bar{c}}}= \lambda _{u_1^\phi {\bar{c}}}=-\frac{10}{18}\\{} & {} \lambda _{d^\psi c}=\lambda _{u_1^\psi c}=-\frac{4}{18}\qquad \lambda _{d^\psi {\bar{c}}}=\lambda _{u_1^\psi {\bar{c}}}=+\frac{1}{18} \end{aligned}$$

There is only one possible combination of orbitals, corresponding to Fig. 2b.

$$\begin{aligned} {{\textbf {B}}}({{\textbf {20}}}) \qquad c- \phi (q) \qquad {\bar{c}}- \psi ( qq) \qquad {\text {Fig. 2b}} \end{aligned}$$

(44)

6.4 The spin 3/2 octet of 70

We obtain

(45)

$J_3=3/2$ all spins up. To bring quarks in the same orbital together, we use Table 1 to find:

$$\begin{aligned} \mathcal{P}^{\varvec{70}}_{\varvec{8,3/2}}= & {} n\frac{1}{\sqrt{2}}\times \left\{ \frac{[u_1^\phi ,d^\phi ] u_2^\psi +[d^\phi , u_2^\phi ] u_1^\psi -2[u_1^\phi ,u_2^\phi ] d^\psi }{\sqrt{6}}\right. \nonumber \\{} & {} \left. -\frac{(u_1^\phi ,d^\phi ) u^\psi _2-(d^\phi , u_2^\phi ) u_1^\psi }{\sqrt{2}}\right\} \end{aligned}$$

(46)

d couplings are the same as $(\varvec{8},1/2)$ of $\varvec{56}$

$$\begin{aligned}{} & {} \lambda _{d^\phi c}=-\frac{14}{36}\qquad \lambda _{d^\phi \bar{c}}= -\frac{2}{18}\\{} & {} \lambda _{d^\psi c}= -\frac{4}{36}\qquad \lambda _{d^\psi \bar{c}}=-\frac{7}{18} \end{aligned}$$

but u couplings do not agree: we find

$$\begin{aligned}{} & {} \lambda _{u_1^\phi c} =-\frac{11}{36}\quad \lambda _{u_1^\phi \bar{c}}=-\frac{8}{18}\lambda _{u_1^\psi c}=-\frac{7}{36}\quad \lambda _{u_1^\psi \bar{c}}=-\frac{1}{18} \end{aligned}$$

The conclusion is that the $\mathbf{{70}}$ is incompatible even with 2 orbitals. A similar disagreement is found between the couplings of the spin 1/2 octet of 70.

7 Results and perspectives

We summarize here our results, valid for the spin 1/2 octet of either ${\varvec{56}}$ or ${\varvec{20}}$.

• 1. Combining the spin 1/2 of light quarks with $c\bar{c}$ total spin $S_c=s_c+s_{\bar{c}}=0,1$, we obtain three pentaquark octets with spin compositions: 2$({\varvec{8}},~1/2)+({\varvec{8}},~3/2)$.

• 2. All $S=0$ states can decay into a final state containing a $J/\psi +p$. Conservation of $S_c$, which would forbid the final $J/\psi$ for $S_c=0$, is broken by light-heavy hyperfine interactions in the two orbitals. The same applies to strange pentaquarks.

• 3. We expect therefore three pentaquark lines for both $S=0$ and $S=-1$, with

  $$\begin{aligned} \mathcal{P}_{(S=0)}\rightarrow J/\psi +p \qquad \mathcal{P}_{(\Lambda ,S=-1)} \rightarrow J/\psi +\Lambda \end{aligned}$$

  (47)

• 4. Other predicted pentaquarks are

  $$\begin{aligned}{} & {} \mathcal{P}^+_{(\Sigma ,S=-1)} \rightarrow J/\psi +\Sigma ^+ \rightarrow J/\psi +p +\pi ^0\nonumber \\{} & {} \mathcal{P}_{(\Xi ,S=-2)}^- \rightarrow J/\psi + \Xi ^- \rightarrow J/\psi +\Lambda + \pi ^- \end{aligned}$$

  (48)

• 5. The two alternatives, ${\varvec{56}}$ or ${\varvec{20}}$, are distinguished by presence or absence of pentaquarks decaying into spin 3/2 resonances, e.g.: $\mathcal{P}_{(\Delta ,S=0)}^{++} \rightarrow J/\psi + \Delta ^{++}\rightarrow J/\psi + p+\pi ^+$, which applies to ${\varvec{56}}$ only.

• 6. Taking one-gluon exchange couplings, Sect. 6, one would conclude for a ground state in the 20 representation. However, the one gluon exchange approximation is not that precise and we consider both cases equally plausible for the ground state, to be decided by presence or absence of the SU(3)$_f$ decuplet.

Before closing, we comment on the possibility to compute the mass spectrum of pentaquarks. In our two-orbitals scheme there are three ground state candidates, namely

We start from the first case, $\varvec{56}^{a}$.

A glance at Eq. (5) shows that orbitals at large R are in triality zero, color configurations $\mathbf{{8}}_c-\mathbf{{ 8}}_c$ or $\mathbf{{1}}_c-\mathbf{{1}}_c$. As argued in [36], soft gluons may screen colors in the $\mathbf{{8}}_c-\mathbf{{ 8}}_c$ configuration and the BO potential vanishes at infinity [25]. For tetraquarks this behaviour is supported by the recent lattice QCD calculation of [28]. The pentaquark goes asymptotically into a superposition of charmed baryon-anticharmed meson states. The pentaquark at intermediate distances, is a kind of molecular state bound by QCD interactions. In this case, the mass spectrum would be fully computable, similarly to the doubly charm tetraquark mass [37,38,39].

Orbitals in the second case, $\varvec{56}^{b},~\varvec{20}^b$, are shown in Fig. 2b. The configuration reminds closely the compact pentaquark: $[cu](\bar{c} [ud])$ proposed in [25]. Orbitals are in color confined configurations ${\varvec{\bar{3}}}_c-{\varvec{ 3}}_c$ or ${\varvec{6}}_c-{\varvec{\bar{6}}}_c$, see (4). We must add to the BO potential a string potential rising to infinity for $R\rightarrow \infty$  [25]. As it happens for charmonia, pentaquark masses contain one undetermined constant and one can compute only mass differences with respect to ground state, i.e. those due to the hyperfine interactions.

Appendices

Appendix A: The case of one orbital

For the spin 3/2 octet ($J_3=1/2,~J=3/2$) we factorise the symmetric spin part (i.e. assume all spins up and ignore spin altoghether) and use (B3)

$$\begin{aligned} \mathcal{P}_{\varvec{8,3/2,}} \text {1-orbit}^{\varvec{70}}= & {} \frac{1}{\sqrt{2}}[M^\rho F^\lambda -M^\lambda F^\rho ]\otimes \chi ^S\nonumber \\= & {} -\frac{1}{2}\Big \{\frac{1}{\sqrt{3}}\left( [ u_1,d_2]u_3+[d_1,u_2] u_3\right. \nonumber \\{} & {} \left. -2 [ u_1, u_2] d_3\right) +\left( (u_1,d_2) u_3-(d_1,u_2) u_3\right) \Big \}\nonumber \\ \end{aligned}$$

(A1)

Fixing the gauge $\varvec{d=d_3}.$ As illustrated in Sect. 6, we may use the gauge condition that d flavour is in quark $q_3$, keep the term containing $d_3=d$ and discarding the rest. To obtain a normalised state we multiply by $\sqrt{3}$.

Applied to (A1) the gauge condition $d_3=d$ leads to the simple result (all spins up):

$$\begin{aligned} \mathcal{P}_{\varvec{8,3/2,}}\text {1-orbit}^{\varvec{70}}(d_3=d)=[u_1, u_2] d \end{aligned}$$

(A2)

We find

$$\begin{aligned}{} & {} \lambda _{dc}=\frac{2}{3}\left( - \frac{2}{3}\right) +\frac{1}{3}\left( \frac{1}{3}\right) =-\frac{2}{6}\nonumber \\{} & {} \lambda _{d\bar{c}}=\frac{8}{9}\left( -\frac{4}{3}\right) +\frac{1}{9}\left( +\frac{1}{6}\right) =-\frac{7}{6} \end{aligned}$$

(A3)

Next, using Table 1, we find

$$\begin{aligned}{} & {} [ u_1, u_2] d =+\frac{1}{2}[d, u_2]u_1-\frac{\sqrt{3}}{2} (d, u_2) u_1 \end{aligned}$$

(A4)

and

$$\begin{aligned}{} & {} \lambda _{u_1c}=\frac{1}{4}\left[ \frac{2}{3}\left( -\frac{2}{3}\right) +\frac{1}{3}\left( +\frac{1}{3}\right) \right] +\frac{3}{4}\left( -\frac{2}{3}\right) =-\frac{7}{12}\nonumber \\{} & {} \lambda _{u_1\bar{c}}=\frac{1}{4}\left[ \frac{8}{9}\left( -\frac{4}{3}\right) +\frac{1}{9} \left( +\frac{1}{6}\right) \right] +\frac{3}{4}\left( +\frac{1}{6}\right) =-\frac{2}{12}\nonumber \\ \end{aligned}$$

(A5)

Appendix B: Templates of mixed and symmetric representations of $S_3$

For color we use the definitions

$$\begin{aligned}{} & {} M^\lambda =\left[ (q_1,q_2)_{\varvec{6}} q_3\right] _{\varvec{8}} \end{aligned}$$

(B1)

$$\begin{aligned}{} & {} M^\rho =\left[ (q_1,q_2)_{\bar{\varvec{3}}} q_3\right] _{\varvec{8}} \end{aligned}$$

(B2)

In the case of flavor, octet or decuplet, we have

$$\begin{aligned}{} & {} F^\rho =\frac{1}{\sqrt{2}} (u_1d_2-d_1u_2)u_3\nonumber \\{} & {} F^\lambda =-\frac{1}{\sqrt{6}}\left[ (u_1d_2+d_1u_2)u_3 -2 u_1 u_2 d_3\right] ~\mathrm{(octet)} \end{aligned}$$

(B3)

$$\begin{aligned}{} & {} F^S=\frac{1}{\sqrt{3}}(d_1u_2u_3+u_1d_2u_3+u_1u_2d_3)~\mathrm{(decuplet)} \end{aligned}$$

(B4)

and similarly for spin

$$\begin{aligned} \chi ^\rho= & {} \frac{1}{\sqrt{2}} (\uparrow \downarrow -\downarrow \uparrow )\uparrow \nonumber \\ \chi ^\lambda= & {} -\frac{1}{\sqrt{6}}\left[ (\uparrow \downarrow + \downarrow \uparrow )\uparrow \right. \nonumber \\{} & {} \left. -2(\uparrow \uparrow ) \downarrow \right] ~(\mathrm spin~1/2) \end{aligned}$$

(B5)

$$\begin{aligned} \chi ^S= & {} \frac{1}{\sqrt{3}}(\uparrow \uparrow \downarrow +\uparrow \downarrow \uparrow + \downarrow \uparrow \uparrow )~(\mathrm spin~3/2) \end{aligned}$$

(B6)

For coordinates, $\phi ,\psi$ refer to Born–Oppenheimer orbitals, Fig. 2, and

$$\begin{aligned} N^\rho= & {} n~\frac{1}{\sqrt{2}}~\left[ \phi (1)\psi (2)-\phi (2)\psi (1)\right] \phi (3)\nonumber \\ N^\lambda= & {} -n~\frac{1}{\sqrt{6}}~\Big \{[\phi (1)\psi (2)+\phi (2)\psi (1)]\phi (3)\nonumber \\{} & {} -2\phi (1)\phi (2)\psi (3)\Big \}\nonumber \\ n= & {} \frac{1}{\sqrt{1-S^2}}\qquad \text {and}\nonumber \\ S= & {} \int ~d1~ \phi (1)\psi (1)<1, ~(\phi ,~\psi =~\textrm{real}) \end{aligned}$$

(B7)

Appendix C: Light quarks in two orbitals

1.1 1. The decuplet off 56

The simplest case is to couple the state (30) with the $(\textbf{10},3/2)$ of $\textbf{56}$, with wave function

$$\begin{aligned} B_{10}= u^{\uparrow } u^{\uparrow }u^{\uparrow } \end{aligned}$$

(C1)

We find:

$$\begin{aligned} {\Delta }^{++}_{10}= & {} (\bar{C} C)\times \frac{N}{\sqrt{2}}\left\{ \frac{1}{\sqrt{2}} (U^{\Phi }_1U^{ \Psi }_2-U^{ \Psi }_1 U^{ \Phi }_2) U^{ \Phi }_3\right. \nonumber \\{} & {} \left. + \frac{1}{\sqrt{6}} \left( [U^{ \Phi }_1 U^{ \Psi }_2+U^{ \Psi }_1 U^{ \Phi }_2] U^{ \Phi }_3-2[U^{ \Phi }_1 U^{ \Phi }_2] U^{ \Psi }_3\right) \right\} \nonumber \\= & {} (\bar{C} C)\times \frac{N}{\sqrt{3}} \times \Big \{[U^\Phi _1, U^\Phi _2] U^\Psi _3+[U_3^\Phi , U^\Phi _1] U_2^\Psi \nonumber \\{} & {} +[U_2^\Phi , U^\Phi _3] U_1^\Psi \Big \} \end{aligned}$$

(C2)

Color couplings $u_3^{\phi ,\psi }-c/\bar{c}$ are the same as in (39). We register also the case with $I_3=+1/2$:

$$\begin{aligned} {\Delta }^{+}_{10}= & {} (\bar{c} c)\times \frac{n}{\sqrt{3}} \Big \{\{[u^\phi _1, u^\phi _2]d^\psi \}_{3/2}\nonumber \\{} & {} +\{[d^\phi , u^\phi _1]u_2^\psi \}_{3/2}+\{[u_2^\phi , d^\phi ]u_1^\psi \}_{3/2} \Big \} \end{aligned}$$

(C3)

1.2 2. The octet of 20

Following Eq. (17), the fully symmetric wave function in color and coordinate is

$$\begin{aligned} \mathcal{P}= & {} (\bar{c} c)\times \frac{1}{\sqrt{2}} [M^\lambda N^\lambda + M^\rho N^\rho ] \nonumber \\= & {} (c\bar{c}) \times \frac{1}{\sqrt{2}} \left\{ (q_1,q_2) q_3\, N^\lambda +[q_1,q_2] q_3\, N^\rho \right\} \end{aligned}$$

(C4)

Introducing coordinate orbitals as before, we obtain (subscript S for symmetric)

$$\begin{aligned} \mathcal{P}_{S}= & {} (\bar{c} c)_{\varvec{8}}\times \frac{n}{\sqrt{2}} \times \left\{ -\frac{1}{\sqrt{6}}\left( (q^\phi _1,q^\psi _2) q_3^\phi +( q^\psi _1,q^\phi _2) q_3^\phi \right. \right. \nonumber \\{} & {} \left. \left. -2( q^\phi _1,q^\phi _2) q_3^\psi \right) + \frac{1}{\sqrt{2}} \left( [q^\phi _1,q^\psi _2] q^\phi _3-[ q^\psi _1,q^\phi _2] q^\phi _3\right) \right\} \nonumber \\ \end{aligned}$$

(C5)

Using Table 1 we bring together quarks with the same orbitals and find, not surprisingly

$$\begin{aligned} \mathcal{P}_{S}= & {} (\bar{c} c)\times \frac{n}{\sqrt{3}}\nonumber \\{} & {} \times \Big [(q_1^\phi ,q_2^\phi )q_3^\psi +\text {cyclic permutations in 1,2,3}\Big ] \end{aligned}$$

(C6)

The full proton-like, octet state of $\varvec{20}$. We have to combine (C5) or (C6) with the flavour-spin fully antisymmetric expression obtained by combining the mixed symmetry templates in Sect. 3. Chosing again the gauge $d=d_3$ the antisymmetric combination is

$$\begin{aligned} A(\mathbf{{20}})=\frac{1}{\sqrt{3}}~\{ \{u_1, u_2\}_0 d \}_{1/2} \end{aligned}$$

(C7)

with $\lfloor u_1, u_2\rceil$ meaning spin zero pair. Using (C5), we find

$$\begin{aligned} \mathcal{P}^{\varvec{20}}_{\varvec{8,1/2}}= & {} \sqrt{3} \mathcal{P}_{S} A(\mathbf{{20}}) =(\bar{c} c)\nonumber \\{} & {} \times \frac{n}{\sqrt{2}} \left\{ \frac{ [ u_1^\phi , u_2^\psi ]_0 d^\phi - [ u_1^\psi , u_2^\phi ]_0 d^\phi }{\sqrt{2}}\right. \nonumber \\{} & {} \left. -\frac{( u_1^\phi ,u^\psi _2)_0 d^\phi +( u_1^\psi ,u^\phi _2)_0 d^\phi -2(u_1^\phi ,u^\phi _2)_0 d^\psi }{\sqrt{6}}\right\} \nonumber \\ \end{aligned}$$

(C8)

We find the couplings repoprted before Eq. (44)

$$\begin{aligned}{} & {} \lambda _{ d^\phi c}=-\frac{5}{18}\qquad \lambda _{d^\phi {\bar{c}}}=-\frac{10}{18}\\{} & {} \lambda _{d^\psi c}=-\frac{4}{18}\qquad \lambda _{d^\psi {\bar{c}}}=+\frac{1}{18} \end{aligned}$$

and one checks that the $u_1^{\phi ,\psi }-c/\bar{c}$ couplings coincide with these.

There is only one possible combination of orbitals, corresponding to Fig. 2b.

$$\begin{aligned} {\varvec{B}}({\varvec{20}}) \quad c- \psi (q) \quad {\bar{c}}- \phi ( qq) \qquad {\text {Fig.~2}b} \end{aligned}$$

(C9)

Separating the orbitals. As for the ${\varvec{56}}$, it is convenient to have an expression which puts together quarks that belong to the same orbital. This is obtained by using (C6) with (C7) and rearranging quark spins so as to put into evidence the total spin inside each orbitalo. We obtain:

$$\begin{aligned} \mathcal{P}^{\varvec{ 20}}_\mathbf{8,1/2}= & {} \sqrt{3}~ \mathcal{P}\otimes A({\varvec{20}}) =(\bar{c} c)\times \frac{n}{\sqrt{3}} \Big \{\{( u_1^\phi ,u_2^\phi )_0 d^\psi \}_{1/2}\nonumber \\{} & {} -\frac{1}{2}\{(d^\phi , u_1^\phi ) u_2^\psi \}_{1/2}-\frac{1}{2}\{( d^\phi , u_2^\phi )_0 u_1^\psi \}_{1/2}-\nonumber \\{} & {} -\frac{\sqrt{3}}{2}\{( d^\phi ,u_1^\phi )u_2^\psi \}_{1/2}+\frac{\sqrt{3}}{2}\{ (d^\phi , u_2^\phi )u_1^\psi \}_{1/2}\Big \}\nonumber \\ \end{aligned}$$

(C10)

Not surprisingly, (42C10) is obtained from (38) with two simultaneous exchanges of the symmetry characters:

$$\begin{aligned} \bar{\varvec{3}}_c \rightarrow \varvec{6}_c\qquad \text {diquark~spin:}~1\leftrightarrows 0 \end{aligned}$$

(C11)

Going from (38) to (42C10), the diquark isospin remains unchanged. Thus, both octets have the same symmetry under total exchange of quark quantum numbers, which is the required Fermi antisymmetry.

1.3 3. The spin 3/2 octet of 70

Color and Coordinates mixed (M) wave functions. Starting from Eqs. (19) and (18) together with (B7), we have to construct

$$\begin{aligned}{} & {} CC^\rho =\frac{1}{\sqrt{2}}~[M^\rho N^\lambda +M^\lambda N^\rho ] \end{aligned}$$

(C12)

$$\begin{aligned}{} & {} CC^\lambda =\frac{1}{\sqrt{2}}~[M^\rho N^\rho -M^\lambda N^\lambda ]\nonumber \\ \end{aligned}$$

(C13)

We find:

$$\begin{aligned} CC^\rho= & {} n~\frac{1}{2\sqrt{3}}\Big \{\sqrt{3}\left( (q^\phi _1,q^\psi _2) q^\phi _3-(q^\psi _1,q^\phi _2) q^\phi _3 \right) \nonumber \\{} & {} - [q^\phi _1,q^\psi _2] q^\phi _3-[q^\psi _1,q^\phi _2] q^\phi _3+2[q^\phi _1,q^\phi _2]q^\psi _3 \Big \}\nonumber \\ \end{aligned}$$

(C14)

and

$$\begin{aligned} CC^\lambda= & {} n~\frac{1}{2\sqrt{3}}\Big \{\sqrt{3}\left( [q^\phi _1,q^\psi _2] q^\phi _3-[q^\psi _1,q^\phi _2] q^\phi _3 \right) \nonumber \\{} & {} + (q^\phi _1,q^\psi _2) q^\phi _3+(q^\psi _1,q^\phi _2) q^\phi _3-2(q^\phi _1,q^\phi _2)q^\psi _3 \Big \}\nonumber \\ \end{aligned}$$

(C15)

If we want the spin 3/2 octet we have to combine the above expressions with

$$\begin{aligned} F^\rho= & {} \frac{1}{\sqrt{2}} (u_1d_2-d_1u_2)u_3\nonumber \\ F^\lambda= & {} -\frac{1}{\sqrt{6}}\left[ (u_1d_2+d_1u_2)u_3 -2 u_1 u_2 d_3\right] ~\mathrm{(octet)} \end{aligned}$$

(C16)

to obtain the full antisymmetric combination (all spins are up)

$$\begin{aligned} A=\frac{1}{\sqrt{2}} [CC^\rho F^\lambda -CC^\lambda F^\rho ] \end{aligned}$$

(C17)

The gauge fixing condition $d_3=d$, Eq. (34), gives $F^\rho =0,~F^\lambda \propto u_1u_2d$ and one obtains

$$\begin{aligned} \mathcal{P}^{\varvec{70}}_{\varvec{8,3/2}}= & {} n\frac{1}{\sqrt{2}}\Bigg \{\frac{(u^\phi _1,u^\psi _2) d^\phi -(u^\psi _1,u^\phi _2) d^\phi }{\sqrt{2}}\nonumber \\{} & {} -\frac{[u^\phi _1,u^\psi _2] d^\phi +[u^\psi _1,u^\phi _2] d^\phi -2[u^\phi _1,u^\phi _2] d^\psi }{\sqrt{6}}\Bigg \} \end{aligned}$$

$J_3=3/2$ all spins up. To bring quarks in the same orbital together, we use Table 1 to find:

$$\begin{aligned} \mathcal{P}^{\varvec{70}}_{\varvec{8,3/2}}= & {} n\frac{1}{\sqrt{2}}\times \Big \{ \frac{[u_1^\phi ,d^\phi ] u_2^\psi +[d^\phi , u_2^\phi ] u_1^\psi -2[u_1^\phi ,u_2^\phi ] d^\psi }{\sqrt{6}}\nonumber \\{} & {} -\frac{(u_1^\phi ,d^\phi ) u^\psi _2-(d^\phi , u_2^\phi ) u_1^\psi }{\sqrt{2}}\Big \} \end{aligned}$$

(C18)

d couplings are the same as $\varvec{8,1/2}$ of $\varvec{56}$

$$\begin{aligned}{} & {} \lambda _{d^\phi c}=-\frac{14}{36}\qquad \lambda _{d^\phi \bar{c}}= -\frac{2}{18}\\{} & {} \lambda _{d^\psi c}= -\frac{4}{36}\qquad \lambda _{d^\psi \bar{c}}=-\frac{7}{18} \end{aligned}$$

but u couplings do not agree: we find

$$\begin{aligned}{} & {} \lambda _{u^\phi _1c}=-\frac{11}{36}\qquad \lambda _{u^\phi _1\bar{c}}=-\frac{8}{18}\\{} & {} \lambda _{u^\psi _1c}=-\frac{7}{36}\qquad \lambda _{u^\psi _1\bar{c}}=-\frac{1}{18} \end{aligned}$$

The conclusion is that the $\mathbf{{70}}$ is incompatible even with 2 orbitals. A similar disagreement is found between the couplings of the spin 1/2 octet of 70.

Appendix D: The full spin 1/2 octet of 56

We start from the proton wave function (38) (for the sake of clarity, we indicate explicitly the spin of diquarks in the squared parentheses)

$$\begin{aligned} {P}= & {} \frac{1}{\sqrt{3}} \Big \{[ u_1^\phi ,u_2^\phi ]_1 d^\psi -\frac{1}{2}[d^\phi ,u_1^\phi ]_1 u_2^\psi -\frac{1}{2} [ u_2^\phi ,d^\phi ]_1 u_1^\psi \nonumber \\{} & {} -\frac{\sqrt{3}}{2}[ d^\phi ,u_1^\phi ]_0 u_2^\psi +\frac{\sqrt{3}}{2}[ u_2^\phi ,d^\phi ]_0 u_1^\psi ~\Big \} \end{aligned}$$

(D 1)

• Replacing $d\rightarrow s$ we obtain the $\Sigma ^+$

  $$\begin{aligned} \Sigma ^+= & {} \frac{1}{\sqrt{3}} \Big \{[ u_1^\phi ,u_2^\phi ]_1 s^\psi -\frac{1}{2}[s^\phi ,u_1^\phi ]_1 u_2^\psi -\frac{1}{2} [ u_2^\phi ,s^\phi ]_1 u_1^\psi \nonumber \\{} & {} -\frac{\sqrt{3}}{2}[ s^\phi ,u_1^\phi ]_0 u_2^\psi +\frac{\sqrt{3}}{2}[ u_2^\phi ,s^\phi ]_0 u_1^\psi ~\Big \} \end{aligned}$$

  (D 2)

• The substitution $u\rightarrow d$ leads to $\sqrt{2}\Sigma ^0$ with ($I^-$ is the isospin lowering operator):

  $$\begin{aligned} {\Sigma ^0}= & {} \frac{I^-\Sigma ^+}{\sqrt{2}}=\frac{1}{\sqrt{3}}\times \Big \{\frac{[ d^\phi ,u^\phi ]_1 s^\psi +[ u^\phi ,d^\phi ]_1 s^\psi }{\sqrt{2}}\nonumber \\{} & {} - \frac{1}{2}\Big [\frac{ [s^\phi ,d^\phi ]_1 u^\psi +[s^\phi ,u^\phi ]_1 d^\psi }{\sqrt{2}}\nonumber \\{} & {} +\frac{ [ s^\phi , d^\phi ]_1 u^\psi +[ s^\phi , u^\phi ]_1 d^\psi }{\sqrt{2}}\Big ]+\nonumber \\{} & {} -\frac{\sqrt{3}}{2}\Big [\frac{ [ s^\phi ,d^\phi ]_0 u^\psi -[d^\phi ,s^\phi ]_0 u^\psi }{\sqrt{2}}\nonumber \\{} & {} +\frac{ [ s^\phi ,u^\phi ]_0 d^\psi -[u^\phi ,s^\phi ]_0 d^\psi }{\sqrt{2}}\Big ]~\Big \} \end{aligned}$$

  (D 3)

• The rising operator of Vspin in octet space (corresponding to $u\rightarrow s$) is represented by the matrix

  $$\begin{aligned} V^+=\begin{pmatrix} 0&{}0&{}0\\ 0&{}0&{}0\\ 1&{}0&{}0 \end{pmatrix} \end{aligned}$$

  (D 4)

  and it acts on the octet baryon matrix B according to the commutator: $B\rightarrow \big [V^+,B\big ]$. Specialising to the proton, we find

  $$\begin{aligned} \frac{\big [V^+,P\big ]}{\sqrt{2}}=\Sigma ^0_V=-\frac{\sqrt{3}}{2}\Lambda -\frac{1}{2}\Sigma ^0. \end{aligned}$$

  (D 5)

  Application of $V^+=u\rightarrow s$ brings the state (D 1D 2) into $\sqrt{2}\Sigma ^0_V$:

  $$\begin{aligned} \Sigma _V= & {} \frac{1}{\sqrt{3}}\times \Big \{\frac{[ s^\phi ,u^\phi ]_{1}d^\psi +[u^\phi ,s^\phi ]_1d^\psi }{\sqrt{2}}\nonumber \\{} & {} - \frac{1}{2}\Big [\frac{[d^\phi ,s^\phi ]_1 u^\psi + [s^\phi ,d^\phi ]_1 u^\psi }{\sqrt{2}} \nonumber \\{} & {} + \frac{[d^\phi ,u^\phi ]_1 s^\psi +[u^\phi ,d^\phi ]_1 s^\psi }{\sqrt{2}}\Big ]-\nonumber \\{} & {} -\frac{\sqrt{3}}{2}\Big [\frac{[d^\phi ,s^\phi ]_0 u^\psi -[s^\phi ,d^\phi ]_0 u^\psi }{\sqrt{2}}\nonumber \\{} & {} - \frac{[u^\phi ,d^\phi ]_0 s^\psi -[d^\phi ,u^\phi ]_0 s^\psi }{\sqrt{2}} \Big ]~\Big \} \end{aligned}$$

  (D 6)

  Using (D 5), we find

  $$\begin{aligned} \Lambda= & {} -\frac{1}{\sqrt{3}}\big (2\Sigma ^0_V+\Sigma ^0)= \frac{1}{\sqrt{3}}\nonumber \\{} & {} \times \Big \{\frac{-\sqrt{3}}{2} \Big [\frac{[s^\phi ,u^\phi ]_1 d^\psi +[u^\phi ,s^\phi ]_1d^\psi }{\sqrt{2}}\nonumber \\{} & {} -\frac{[s^\phi ,d^\phi ]_1 u^\psi +[d^\phi ,s^\phi ]_1 u^\psi }{\sqrt{2}}\Big ]\nonumber \\{} & {} -\frac{1}{2}\Big [\frac{[s^\phi ,d^\phi ]_0 u^\psi - [ d^\phi ,s^\phi ]_0 u^\psi }{\sqrt{2}}\nonumber \\{} & {} - \frac{[s^\phi ,u^\phi ]_0 d^\psi -[u^\phi ,s^\phi ]_0 d^\psi }{\sqrt{2}}\Big ] \nonumber \\{} & {} -\frac{[ u^\phi ,d^\phi ]_0 s^\psi -[d^\phi ,u^\phi ]_0 s^\psi }{\sqrt{2}}\Big \} \end{aligned}$$

  (D 7)

  which, correctly, is pure isoscalar.

• With one more step, from (D 6) we find $\Xi ^-$:

  $$\begin{aligned} \Xi ^-= & {} \frac{1}{\sqrt{3}}\times \Big \{[ s_1^\phi ,s_2^\phi ]_1 d^\psi -\frac{1}{2}[d^\phi ,s_1^\phi ]_1 s_2^\psi \nonumber \\{} & {} -\frac{1}{2}[ d^\phi ,s_2^\phi ]_1 s_1^\psi - \frac{\sqrt{3}}{2}[d^\phi ,s_1^\phi ]_0 s_2^\psi \nonumber \\{} & {} +\frac{\sqrt{3}}{2}[s_2^\phi ,d^\phi ]_0 s_1^\psi \Big \} \end{aligned}$$

  (D 8)

  which could also be obtained from (D 1D 2) by the double change of name $u_1\rightarrow s_1,~u_2\rightarrow s_2$.