$P_c(4380)$ in a constituent quark model

TL;DR

This study uses a constituent quark model to analyze the properties of a potential pentaquark state, concluding that the observed LHCb pentaquark cannot be a compact multiquark configuration with the specified quantum numbers.

Contribution

It provides a detailed variational analysis of the pentaquark's wave function within a constituent quark model, challenging the compact multiquark interpretation of observed states.

Findings

Ground state is the isolated p and J/psi

Excited state mass is much higher than observed pentaquark

Observed states are unlikely to be compact multiquark configurations

Abstract

The constituent quark model with color-spin hyperfine potential is used to investigate the property of a compact pentaquark configuration with $J^p$=$3/2^-$ and isospin=1/2, which is the most likely quantum number of one of the recently observed exotic baryon states at LHCb. Starting from the characterization of the isospin, color, and spin states for the pentaquark configuration, we construct the total wave function composed of the spatial wave function, which we take to be symmetric and in S-wave, and the four orthogonal isospin $\otimes$ color $\otimes$ spin states that satisfy the Pauli principle. We then use the variational method to find a compact stable configuration. While there are compact configurations where the hyperfine potential is more attractive than the sum of $p$ and $J/\psi$ hyperfine potentials, we find that the ground state is the isolated $p$ and $J/\psi$ state. Furthermore, the mass of the excited state lies far above the observed pentaquark state leading us to conclude that the observed states can not be a compact multiquark configuration with $J^p$=$3/2^-$.

Tables

Table 1. Table 1: Parameters of the Hamiltonian fitted to the baryon and meson masses occurring in the decay channels of the q 3 ​ c ​ c ¯ superscript 𝑞 3 𝑐 ¯ 𝑐 q^{3}c\bar{c} .

$\gamma$	$\kappa$	$a_0$	D	${\kappa }_0$	$\alpha$	$\beta$	$m_u$	$m_c$
1.667${\left(\mathrm{GeV}\right)}^{-1}$	0.107	1.042${\left(\mathrm{GeV}\right)}^{-2}$	0.955 $\mathrm{GeV}$	0.168 $\mathrm{GeV}$	1.224 $\mathrm{GeV}$	1.467	0.302 $\mathrm{GeV}$	1.889 $\mathrm{GeV}$

Table 2. Table 2: Masses of baryons and mesons obtained from the variational method. The third row shows the variational parameter in fm − 2 superscript fm 2 \rm{fm^{-2}} . The fourth row shows the experimental data in GeV GeV \rm{GeV} .

(I,S)	($\frac{1}{2}$,$\frac{1}{2}$) P	($\frac{3}{2}$,$\frac{3}{2}$) $\Delta$	($0$,$\frac{1}{2}$) ${\Lambda }_c$	($1$,$\frac{1}{2}$) ${\Sigma }_c$	($1$,$\frac{3}{2}$) ${\Sigma }_c^{\ast }$	($0$,$0$) ${\eta }_c$	($0$,$1$) $J/\psi$	($\frac{1}{2}$,$0$) $D$	($\frac{1}{2}$,$1$) $D^{\ast }$
Mass	0.972	1.266	2.286	2.459	2.536	2.984	3.115	1.872	2.012
Variational
parameters	a=3.4, b=1.4	a=2.1, b=1.2	a=2.7, b=3.4	a=1.9, b=3.5	a=1.8, b=3.1	a=15.1	a=11	a=4.4	a=3.4
Exp	0.938	1.232	2.286	2.453	2.518	2.983	3.96	1.869	2.01

Table 3. Table 3: The sum of the expectation value of the hyperfine factor of both a baryon and a meson for the possible decay channel with respect to I = 1 / 2 𝐼 1 2 I=1/2 . The third column shows the value for the fitting mass m u subscript 𝑚 𝑢 m_{u} and m c subscript 𝑚 𝑐 m_{c} . (unit: ( GeV ) − 2 superscript GeV 2 \rm{(GeV)^{-2}} )

Decay channel	-	Value
$pJ/\psi$	-$\frac{8}{m_1^{}{}_{}{}^2}$+$\frac{16}{3m_2^{}{}_{}{}^2}$	-86.2
${\Lambda }_c{D}^{\ast }$	-$\frac{8}{m_1^{}{}_{}{}^2}$+$\frac{16}{3m_1{m}_2}$	-78.3
${\Sigma }_c^{\ast }D$	$\frac{8}{3m_1^{}{}_{}{}^2}$-$\frac{32}{3m_1{m}_2}$	10.5
${\Sigma }_c{D}^{\ast }$	$\frac{8}{3m_1^{}{}_{}{}^2}$-$\frac{16}{3m_1{m}_2}$	19.8
${\Sigma }_c^{\ast }{D}^{\ast }$	$\frac{8}{3m_1^{}{}_{}{}^2}$+$\frac{32}{3m_1{m}_2}$	47.9

Table 4. Table 4: The mass of the excited state of the pentaquark with I = 1 / 2 𝐼 1 2 I=1/2 obtained from the variational method, by diagonalizing the matrix element of the Hamiltonian in terms of | R ⟩ ​ | ψ 2 ⟩ ket 𝑅 ket subscript 𝜓 2 |R\rangle|\psi_{2}\rangle , | R ⟩ ​ | ψ 3 ⟩ ket 𝑅 ket subscript 𝜓 3 |R\rangle|\psi_{3}\rangle , and | R ⟩ ​ | ψ 4 ⟩ ket 𝑅 ket subscript 𝜓 4 |R\rangle|\psi_{4}\rangle . Δ B subscript Δ 𝐵 {\Delta}_{B} indicate the binding energy. The units for the energy and variational parameter are GeV GeV \rm{GeV} and fm -2 , respectively.

I=1/2	$q^{3}c\overline{c}$
Mass	4.626
Variational parameters	$a_1$=2.3, $a_2$=1.4, $a_3$=4, $a_2$=3.4
Decay channel	$pJ/\psi$	${\Lambda }_c{D}^{\ast }$	${\Sigma }_c^{\ast }D$	${\Sigma }_c{D}^{\ast }$	${\Sigma }_c^{\ast }{D}^{\ast }$
Threshold	4.088	4.298	4.408	4.471	4.548
${\Delta }_B$	0.538	0.328	0.218	0.155	0.078

Table 5. Table 5: The values of each energy term of the excited state of the pentaquark and the sum of a baryon and a meson in decay channel. Δ Δ \Delta E is the difference between the pentaquark and its decay channel in each term. (unit: Mev Mev \rm{Mev} )

Pentaquark	Kinetic	Comfinement	Hyperfine	Sum
The excited state	1144.3	1238	-52.1
Decay channel	Kinetic	Comfinement	Hyperfine	Sum
$pJ/\psi$	1190.5	745.8	-145.1
$\Delta$E	-46.2	492.2	93
${\Lambda }_c{D}^{\ast }$	1192.7	982.2	-173.1
$\Delta$E	-48.4	255.8	121
${\Sigma }_c^{\ast }D$	1105.3	1055.1	-48.6
$\Delta$E	39	182.9	-3.5
${\Sigma }_c{D}^{\ast }$	1046.5	1102.9	25.8
$\Delta$E	97.8	135.1	-77.9
${\Sigma }_c^{\ast }{D}^{\ast }$	993.1	1157	101.4
$\Delta$E	151.2	81	-153.5

Equations

$$H=\sum _{i=1}^5(m_i+\frac{{\text{p}}_i^2}{2m_i})-\frac{3}{16}\sum _{i<j}^4{\lambda }_i^c{\lambda }_j^c(V_{ij}^C+V_{ij}^{SS}),$$

$$V_{ij}^C=-\frac{\kappa }{r_{ij}}+\frac{(r_{ij}{)}^{1/2}}{a_0}-D.$$

$$V_{ij}^{SS}=\frac{1}{m_i{m}_j{c}^4}\frac{{\hslash }^2{c}^2{\kappa }^{\prime }}{(r_{0ij})}\frac{e^{-(r_{ij}{)}^2/(r_{0ij}{)}^2}}{r_{ij}}{\sigma }_i\cdot {\sigma }_j.$$

$$r_{0ij}=1/(\alpha +\beta \frac{m_i{m}_j}{m_i+m_j}),$$

$${\kappa }^{\prime }={\kappa }_0(1+\gamma \frac{m_i{m}_j}{m_i+m_j}).$$

$${\mathbf{x}}_1^1=\frac{1}{\sqrt{2}}({\text{r}}_1-{\text{r}}_2),{\mathbf{x}}_2^1=\sqrt{\frac{2}{3}}({\text{r}}_3-\frac{1}{2}{\text{r}}_1-\frac{1}{2}{\text{r}}_2),$$

$${\mathbf{x}}_3^1=\frac{1}{\sqrt{2}}({\text{r}}_4-{\text{r}}_5),$$

$${\mathbf{x}}_4^1=\sqrt{\frac{6}{5}}(\frac{1}{3}({\text{r}}_1+{\text{r}}_2+{\text{r}}_3)-\frac{1}{2}({\text{r}}_4+{\text{r}}_5)),$$

$$R^{s_1}=\exp [-a_1({\mathbf{x}}_1^1{)}^2-a_2({\mathbf{x}}_2^1{)}^2-a_3({\mathbf{x}}_3^1{)}^2-a_4({\mathbf{x}}_4^1{)}^2],$$

$${\mathbf{x}}_1^2=\frac{1}{\sqrt{2}}({\text{r}}_1-{\text{r}}_3),{\mathbf{x}}_2^2=\sqrt{\frac{2}{3}}({\text{r}}_2-\frac{1}{2}{\text{r}}_1-\frac{1}{2}{\text{r}}_3),$$

$${\mathbf{x}}_3^2=\frac{1}{\sqrt{2}}({\text{r}}_4-{\text{r}}_5),$$

$${\mathbf{x}}_4^2=\sqrt{\frac{6}{5}}(\frac{1}{3}({\text{r}}_1+{\text{r}}_2+{\text{r}}_3)-\frac{1}{2}({\text{r}}_4+{\text{r}}_5)),$$

$${\mathbf{x}}_1^3=\frac{1}{\sqrt{2}}({\text{r}}_2-{\text{r}}_3),{\mathbf{x}}_2^3=\sqrt{\frac{2}{3}}({\text{r}}_1-\frac{1}{2}{\text{r}}_2-\frac{1}{2}{\text{r}}_3),$$

$${\mathbf{x}}_3^3=\frac{1}{\sqrt{2}}({\text{r}}_4-{\text{r}}_5),$$

$${\mathbf{x}}_4^3=\sqrt{\frac{6}{5}}(\frac{1}{3}({\text{r}}_1+{\text{r}}_2+{\text{r}}_3)-\frac{1}{2}({\text{r}}_4+{\text{r}}_5)),$$

$$R=$$

$$\exp [-a_1({\mathbf{x}}_1^2{)}^2-a_2({\mathbf{x}}_2^2{)}^2-a_3({\mathbf{x}}_3^2{)}^2-a_4({\mathbf{x}}_4^2{)}^2]+$$

$$\exp [-a_1({\mathbf{x}}_1^3{)}^2-a_2({\mathbf{x}}_2^3{)}^2-a_3({\mathbf{x}}_3^3{)}^2-a_4({\mathbf{x}}_4^3{)}^2].$$

$$K.E.$$

$$f(a_1,a_2)=(a_1+a_2)\times$$

$$\{\frac{1}{(a_1{a}_2{)}^{(3/2)}}+\frac{2048a_1{a}_2}{(3{a_1}^2+10a_1{a}_2+3{a_2}^2{)}^{(3/2)}}\}/$$

$$\{\frac{2}{(a_1{a}_2{)}^{(3/2)}}+\frac{256a_1{a}_2}{(3{a_1}^2+10a_1{a}_2+3{a_2}^2{)}^{(3/2)}}\}.$$

$$\displaystyle\begin{tabular}[]{c}$|I^{1/2}_{1}\rangle$=\end{tabular}\begin{tabular}[]{|c|c|}\hline\cr 1&2\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 3\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}=\frac{1}{\sqrt{6}}(2uud-udu-duu),$$

$$\displaystyle\begin{tabular}[]{c}$|I^{1/2}_{2}\rangle$=\end{tabular}\begin{tabular}[]{|c|c|}\hline\cr 1&3\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 2\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}=\frac{1}{\sqrt{2}}(udu-duu).$$

$$[3{]}_C\otimes [3{]}_C\otimes [3{]}_C\otimes [3{]}_C\otimes {\overline{[3]}}_C.$$

$$\displaystyle\begin{tabular}[]{|c|c|}\hline\cr 1&2\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 3\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\vrule\lx@intercol\hfil 4\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}=\{(12)_{6}(34)_{\bar{3}}\}_{3},\quad\begin{tabular}[]{|c|c|}\hline\cr 1&3\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 2\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\vrule\lx@intercol\hfil 4\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}=\{(12)_{6}34\}_{3},$$

$$\displaystyle\begin{tabular}[]{|c|c|}\hline\cr 1&4\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 2\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\vrule\lx@intercol\hfil 3\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}=\{(123)_{1}4\}_{3}.$$

$$\displaystyle|C_{1}\rangle=\begin{tabular}[]{|c|c|}\hline\cr 1&2\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 3\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\vrule\lx@intercol\hfil 4\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}_{3}\otimes(5)_{\bar{3}},\quad|C_{2}\rangle=\begin{tabular}[]{|c|c|}\hline\cr 1&3\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 2\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\vrule\lx@intercol\hfil 4\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}_{3}\otimes(5)_{\bar{3}},$$

$$\displaystyle|C_{3}\rangle=\begin{tabular}[]{|c|c|}\hline\cr 1&4\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 2\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\vrule\lx@intercol\hfil 3\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}_{3}\otimes(5)_{\bar{3}}.$$

$$\displaystyle[3]_{C}\otimes[3]_{C}\otimes[3]_{C}=\begin{tabular}[]{|c|c|}\hline\cr 1&2\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 3\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}_{8}\oplus\begin{tabular}[]{|c|c|}\hline\cr 1&3\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 2\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}_{8}\oplus\begin{tabular}[]{|c|}\hline\cr 1\\ \hline\cr 2\\ \hline\cr 3\\ \hline\cr\end{tabular}_{1},$$

$$[3{]}_C\otimes {\overline{[3]}}_C=[8{]}_C\oplus [1{]}_C.$$

$$\displaystyle|C_{1}\rangle=\begin{tabular}[]{|c|c|}\hline\cr 1&2\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 3\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}_{8}\otimes(45)_{8},\quad|C_{2}\rangle=\begin{tabular}[]{|c|c|}\hline\cr 1&3\\ \cline{1-2}\cr\vrule\lx@intercol\hfil 2\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}_{8}\otimes(45)_{8},$$

$$\displaystyle|C_{3}\rangle=\begin{tabular}[]{|c|}\hline\cr 1\\ \hline\cr 2\\ \hline\cr 3\\ \hline\cr\end{tabular}_{1}\otimes(45)_{1}.$$

$$|C_1⟩=[\{(12{)}_6(34{)}_{\bar{3}}{\}}_3{5}_{\bar{3}}{]}_1=[\{(12{)}_{6}3{\}}_8(45{)}_8{]}_1,$$

$$|C_2⟩=[\{(12{)}_{\bar{3}}34{\}}_3{5}_{\bar{3}}{]}_1=[\{(12{)}_{\bar{3}}3{\}}_8(45{)}_8{]}_1,$$

$$|C_3⟩=[\{(123{)}_{1}4{\}}_3{5}_{\bar{3}}{]}_1=[\{(123{)}_1(45{)}_1{]}_1.$$

$$\displaystyle|S^{3/2}_{1}\rangle=\begin{tabular}[]{|c|c|c|c|}\hline\cr 1&2&3&4\\ \cline{1-4}\cr\vrule\lx@intercol\hfil 5\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular},|S^{3/2}_{2}\rangle=\begin{tabular}[]{|c|c|c|c|}\hline\cr 1&2&3&5\\ \cline{1-4}\cr\vrule\lx@intercol\hfil 4\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular},|S^{3/2}_{3}\rangle=\begin{tabular}[]{|c|c|c|c|}\hline\cr 1&2&4&5\\ \cline{1-4}\cr\vrule\lx@intercol\hfil 3\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular},$$

$$\displaystyle|S^{3/2}_{4}\rangle=\begin{tabular}[]{|c|c|c|c|}\hline\cr 1&3&4&5\\ \cline{1-4}\cr\vrule\lx@intercol\hfil 2\hfil\lx@intercol\vrule\lx@intercol \\ \cline{1-1}\cr\end{tabular}.$$

$$|{\varphi }_1⟩=[\{(12{)}_1{3}_{1/2}{\}}_{3/2}(45{)}_0{]}_{3/2},$$

$$|{\varphi }_2⟩=[\{(12{)}_1{3}_{1/2}{\}}_{3/2}(45{)}_1{]}_{3/2},$$

Full text

I Introduction

After the introduction of the quark model for the baryon and meson GellMann:1964nj and the color quantum number for quarks Han:1965pf , model calculations for hadrons natural led to the possible existence of mutiquark hadrons beyond the normal hadrons Jaffe:1976ig ; Jaffe:1976ih . Indeed, recent experimental findings point to the possible existence of such configurations; these are the $XYZ$ states with the $X(3872)$ being the first of these states observed by the Belle collaboration Choi:2003ue . The $XYZ$ states could be either compact tetraquark states composed of two quarks and two antiquarks or molecular states with their masses close to the relevant two meson thresholds.

Molecular configurations involving heavy mesons were first discussed in Ref. Tornqvist:1993ng where deuteronlike meson-meson bound states were found to exist when a long range pion exchange potential was included with additional short range attraction depending on the mass of the meson. The possible bound states included a $D{\bar{D}}^{*}$ state in the isopin 0 and $J^{PC}=1^{++}$ channel, which is the quantum numbers of the $X(3872)$. After the experimental observation of $X(3872)$, attempts to explain the state in terms of molecular configuration with important contribution coming from the pion exchange potentials still continues to this date  Thomas:2008ja ; Lee:2009hy ; Wang:2013kva ; Baru:2015nea ; Hosaka:2016pey .

Numerous efforts have been made to explain the mass of the charmonium-like state using various other approaches. In a non-relativistic quark model that includes a confining interaction and a short range spin-dependent interaction through the one gluon exchange as well as an effective pion-induced interaction, it was argued that the $X(3872)$ can be a $D{\bar{D}}^{*}$ hadronic resonance with important admixtures of $\rho J/\psi$ and $\omega J/\psi$ states Swanson:2003tb . In Ref. Wong:2003xk , the $X(3872)$ was considered as a weakly bound molecular state found in the combination of $\{D,D^{*}\}$ with $\{\bar{D},{\bar{D}}^{*}\}$ states based on a quark based non-relativistic four-body Hamiltonian with a pairwise interaction.

There are also models that find $X(3872)$ to be a tetraquark system. These include methods based on a diquark-antidiquark model Maiani:2004vq ; Ebert:2005nc , the QCD sum rule Chen:2010ze , and a simple quark model with chromomagnetic interactions Hogaasen:2005jv ; Cui:2006mp ; Buccella:2006fn . In a lattice QCD calculation Padmanath:2015era , it was shown that a candidate for $X(3872)$ with $I=0$ could only be found if both the $\bar{c}c$ and $\bar{D}{\bar{D}}^{*}$ interpolators are included, while no signal was found if diquark-antidiquark and $\bar{D}{\bar{D}}^{*}$ are used without a $\bar{c}c$ component.

Recently, the observation of hidden-charm pentaquark states by the LHCb collaboration Aaij:2015tga , has triggered another wave of works among many researchers. The $J/\psi p$ invariant mass spectrum of ${\Lambda}_{b}$ $\to$ $J/\psi K^{-}p$ revealed hidden-charm pentaquark states, for which the preferred quantum numbers are $J^{p}$=$3/2^{-}$ for $P_{c}(4380)$ and $J^{p}$=$5/2^{+}$ for $P_{c}(4450)$. In fact, even before the discovery was made, possible hidden-charm molecular baryons composed of anti-charmed meson and charmed baryon, such as the of ${\Sigma}_{c}\bar{D}^{*}$ states with $I(J^{p})$=$\frac{1}{2}({\frac{1}{2}}^{-})$, $\frac{1}{2}({\frac{3}{2}}^{-})$, $\frac{3}{2}({\frac{1}{2}}^{-})$, $\frac{3}{2}({\frac{3}{2}}^{-})$, and ${\Sigma}_{c}\bar{D}$ states with $\frac{3}{2}({\frac{1}{2}}^{-})$, were proposed to exist within the one-boson-exchange model Yang:2011wz . The two hidden-charm pentaquark states were also found to be loosely bound ${\Sigma}_{c}\bar{D}^{*}$ and ${\Sigma}_{c}^{*}\bar{D}^{*}$ molecular states, respectively, within a boson exchange interaction model Chen:2015loa . Furthermore, in a meson exchange model He:2015cea , $P_{c}(4380)$ with $J^{p}$=$3/2^{-}$ was produced from ${\Sigma}_{c}^{*}\bar{D}$, while $P_{c}(4450)$ with $J^{p}$=$5/2^{+}$ was produced from ${\Sigma}_{c}\bar{D}^{*}$. More recently, the pentaquarks were identified with structures around the $\Sigma_{c}^{(*)}\bar{D}^{(*)}$ threshold in a quark cluster model Takeuchi:2016ejt .

While molecular pictures for the two pentaquark states are quite likely, one can not rule out the possibility that these states are compact multiquark configurations based on a strong diquark-antidiquark pair Maiani:2015vwa or quark interactions in general Takeuchi:2016nbj . To distinguish these two configurations, it is important to fully explore these two possible scenarios. In this work, we will explore the possibility that one of the pentaquark is a compact multiquark configuration within a constituent quark model based on the color and spin hyperfine potential Bhaduri:1981pn , which is known to reproduce the masses of the normal meson and baryon states. In particular, in order to asses the possibility that the $P_{c}(4380)$ is a compact multiquark state, we will classify the isospin, color, and spin states for the pentaquark system containing a heavy quark and an antiquark with $J^{p}$=$3/2^{-}$ and isospin=1/2 from the view point of the permutation group which is used in characterizing a certain symmetry so that the isospin, color, and spin states can be represented in terms of the irreducible Young-Yamanouchi bases. We will then systematically construct the isospin $\otimes$ color $\otimes$ spin states satisfying the Pauli principle from the coupling scheme appearing in the combination of any two states. We then use the variational method to calculate the ground state mass of the pentaquark with $J^{p}$=$3/2^{-}$ and isospin=1/2.

This paper is organized as follows. In Sec. II, we first introduce the Hamiltonian describing the constituent quark model, and determine the fitting parameters of the model so as to reproduce the mass of the baryons and mesons associated with the thresholds. Then, by using the variational method, we construct the spatial wave function suitable for a baryon and a meson. In Sec. III, we represent the isospin, color, and spin states and then construct the isospin $\otimes$ color $\otimes$ spin states with respect to $I=3/2$ and $I=1/2$ in two independent basis, which can be transformed into each other through an orthonormal matrix. We analyze the numerical results obtained from the variational method in Sec. IV. We finally give a summary of the paper in Sec. V.

II Hamiltonian

To investigate the stability of the pentaquark in the non-relativistic frame work, the Hamiltonian is chosen to take the confinement and hyperfine potential for the color and spin interaction;

[TABLE]

where $m_{i}$’s are the quark masses, $\lambda^{c}_{i}/2$ the color operator of the $i$’th quark for the color SU(3), and $V^{C}_{ij}$ and $V^{SS}_{ij}$ the confinement and hyperfine potential, respectively. The confinement potential is usually composed of the linearizing term as suggested by the lattice gauge theory, and the Coulomb-type potential as derived from the perturbative QCD;

[TABLE]

The hyperfine potential is given to take the following form, including the spin interaction;

[TABLE]

Here, $r_{ij}$ is the distance between interquarks, $\mid\textbf{r}_{i}-\textbf{r}_{j}\mid$, and both $r_{0ij}$ and ${\kappa}^{\prime}$ are chosen to depend on the masses of interquarks, given by

[TABLE]

The hyperfine potential in Eq. (3), which becomes $1/(m_{i}m_{j})$ $\delta(r)$ in the heavy quark mass limit $m_{i}\to\infty$, is chosen to fit the meson and baryon mass splitting with both light and heavy quarks. The parameters in the Hamiltonian are fitted to the baryons and mesons masses by using the variational method Park:2015nha . The fitting parameters are given in Table 1, and the calculated masses in Table 2.

Since we deal with the pentaquark composed of $q(1)q(2)q(3)c(4)\bar{c}(5)$ with $I=1/2$, where the number indicate the position of the constituent quark, the symmetry of the three light quarks should be taken into account to satisfy the Pauli principle because the total wave function must be antisymmetric among the three light quarks. As we are interested in the ground state, a natural choice would be to take the spatial function to be symmetric, which requires the remaining part of the total wave function to be antisymmetric among the three light quarks. We denote the symmetry (antisymmetry) property by $[123]$ ($\{123\}$). In the center of the mass frame, the pentaquark system is reduced into the four-body problem, represented by the four Jacobian coordinates suitable for describing the decay into a baryon and a meson.

We take the spatial function to be a Gaussian which was extensively used with the variational method to handle calculations in many body problem. The four Jacobian coordinates suitable for describing the decay into a baryon and a meson are given by

[TABLE]

where the first and second terms represent a baryon configuration, the third a meson configuration, and the last the relative position vector between the center of mass of a baryon and a meson. The boldface letters stand for the vectors.

We then construct a spatial wave function given by

[TABLE]

where $a_{1}$, $a_{2}$, $a_{3}$, and $a_{4}$ are variational parameters. Since the spatial wave function in Eq. (6) is symmetric only between the particle 1 and 2, we need two additional spatial wave functions so as to satisfy $[123]$ symmetry; one is symmetric between the particle 1 and 3, and the other is symmetric between the particle 2 and 3. The two sets of four Jacobian coordinates are given by

[TABLE]

[TABLE]

By using the two set of four Jacobian coordinates, we construct the two spatial wave functions with either $[13]$ symmetry or $[23]$ symmetry. Combining these spatial functions with a certain symmetry into a linear form, we obtain the spatial function with four variational parameters $a_{1}$, $a_{2}$, $a_{3}$, and $a_{4}$ which is fully symmetric among the particle 1, 2, and 3 as follows;

[TABLE]

The spatial wave function of the pentaquark in Eq. (9) is in a state with total angular moment $L=0$, where both the baryon and meson configurations as well as their relative motion is in the S-wave state. The kinetic energy part coming from Eq. (9) is given as

[TABLE]

Here $\textbf{p}^{2}_{1}+\textbf{p}^{2}_{2}=3\hbar^{2}f(a_{1},a_{2})$, $\textbf{p}^{2}_{3}=3\hbar^{2}a_{3}$, and $\textbf{p}^{2}_{4}=3\hbar^{2}a_{4}$, where $m_{1},m_{2}$ are the light and heavy quark masses respectively, and $\mu=5m_{1}m_{2}/(3m_{1}+2m_{2})$. We present $f(a_{1},a_{2})$ appearing in the kinetic terms of the baryon;

[TABLE]

Hence, for the compact mutiquark state to be stable compared to the separated baryon and meson state, the extra attraction coming from bringing the baryon and meson should be large enough to overcome the extra kinetic energy given by the last term in Eq. (10).

III Isospin $\otimes$ color $\otimes$ spin state of the pentaquark

In this section, we will construct the isospin $\otimes$ color $\otimes$ spin state appropriate for the $q(1)q(2)q(3)Q(4)\bar{Q}(5)$ system with $I=1/2$ and spin=3/2, where the number in the bracket indicates the position of the constituent quark. The component of three identical light quarks of the pentaquark restricts the total wave function to be antisymmetric with respect to the exchange of any pair among the three light quarks due to Pauli principle. When the spatial function of the pentaquark is chosen to be fully symmetric for the three light quarks, the remaining part of the total wave function should be fully antisymmetric. Therefore, as we are interested in the ground state, the symmetry property of the isospin $\otimes$ color $\otimes$ spin state should be taken to be antisymmetric for the particle 1, 2, and 3. We will use $\{123\}$ notation for the antisymmetry property. Young tableau, which represents the irreducible bases of the permutation group, enable us to easily identify the multi-quark configuration with certain symmetry property. In this paper, we will use the Young tableau and the Young-Yamanouchi basis, which corresponds to the Young tableau in describing the states necessary for the pentaquark. In the following subsections, we first start by separately discussing the isospin, color and spin states consisting of five quarks, and then discuss the total wave function.

III.1 Isospin states

In the SU(2) flavor symmetry, it is easy to find that the possible isospin ($I$) states for the three light quarks are 1/2 and 3/2. The Young-Yamanouchi basis corresponding to the $I=1/2$ state is as follows:

[TABLE]

III.2 Color states

For the possible color states, we only consider the color singlets which are assumed to be observables in hadron state. There are several ways of obtaining the color singlets for the pentaquark, coming from the direct product, given by

[TABLE]

We introduce the two methods which are equivalent to each other, but different in the way of combining the irreducible representation of SU(3). First, since the antiquark corresponds to the antitriplet, we can construct the triplet in the direct product, $[3]_{C}\otimes[3]_{C}\otimes[3]_{C}\otimes[3]_{C}$, which corresponds to Young tableau [2,1,1];

[TABLE]

Here, the subscript indicates the irreducible representation of SU(3). Then, we can obtain the three color singlets, combining the triplet in Eq. (30) with the antitriplet of antiquark. We denote the color singlets by,

[TABLE]

Secondly, we can decompose the direct product, $[3]_{C}\otimes[3]_{C}\otimes[3]_{C}$ and $\otimes[3]_{C}\otimes\bar{[3]}_{C}$ into the direct sum of the irreducible representations, respectively, as follows;

[TABLE]

Then, the coupling of either the octet with the octet or the singlet with the singlet in Eq. (49) and Eq. (50) gives the three color singlets of the pentaquark, denoted by,

[TABLE]

It should be noted that the color singlets represented in terms of different Young tableau in Eq. (41) and Eq. (59) are the same in a tensor form. We define the color singlets derived from the above methods, as follows;

[TABLE]

III.3 Spin states

For the spin=3/2 pentaquark case, the spin states are represented in terms of Young tableau [4,1] with four dimension, as follows:

[TABLE]

When we investigate the stability of the pentaquark against the strong decay into a baryon and a meson, it is very convenient to use the spin states related with the decay mode. We denote the four spin states by,

[TABLE]

where the subscript indicates the spin state. Due to the orthonormality of the two sets of spin sates, Eq. (70) and Eq. (71) are related by the following orthogonal transformation:

[TABLE]

III.4 Isospin $\otimes$ color $\otimes$ spin state for $I=1/2$

Since the isospin, color and spin states represented in terms of the Young tableau have a certain symmetry property, we can construct the isospin $\otimes$ color $\otimes$ spin state of the pentaquark which is fully antisymmetric under the exchange of any pair among the particle 1, 2 and 3. For this purpose, depending on how the coupling scheme is implemented, we consider two methods. In the first method, we start from the notation of the color singlets in Eq. (41), and combine the color singlets with spin states by the out product of the permutation group, $S_{4}$, resulting in the color $\otimes$ spin states for the particle 1, 2, 3, and 4. Then, we can easily obtain the isospin $\otimes$ color $\otimes$ spin state with $\{123\}$ symmetry by coupling of the isospin state with the color $\otimes$ spin states. In the second method, we start the notation of the color singlets in Eq. (59), and use the $S_{3}$ permutation group applied on the coupling scheme.

According to the permutation group theory Chen:2002gd , the irreducible basis of $S_{5}$ becomes the irreducible basis of $S_{4}$ as well, irrespective of the particle 5. When we consider the symmetry property for the particle 1, 2, 3, and 4 in coupling scheme, we can identify the spin states in Eq. (70) with the Young-Yamanouchi bases for Young tableau [4] and Young tableau [3,1] without the particle 5;

[TABLE]

It is necessary to show the outer product between Young tableau [2,1,1] of the color singlets in Eq. (41) and Young tableau [3,1] of the spin states in Eq. (85) so that we obtain the color $\otimes$ spin states;

[TABLE]

In addition to this, we should consider the outer product between Young tableau [2,1,1] of the color singlets in Eq. (41) and Young tableau [4] of the spin states in Eq. (85);

[TABLE]

The coupling scheme designed to construct the isospin $\otimes$ color $\otimes$ spin states with the $\{123\}$ symmetry is completed by using the Clebsch-Gordan (CG) coefficient of the permutation group, $S_{n}$, which is factorized into the Clebsch-Gordan (CG) coefficient of $S_{n-1}$ and K matrix Stancu:1999qr , given by,

[TABLE]

where $S$ in the left-hand (right-hand) side is a CG coefficient of $S_{n}$ ($S_{n-1}$). In this work, we take a similar process which was described in Refs Park:2015nha ; Park:2016cmg .

Below, we show the Young-Yamanouchi bases corresponding to Young tableau [2,1,1] which is obtained from the color $\otimes$ spin coupling in Eq. (103);

[TABLE]

[TABLE]

For the case of Young tableau [2,2], which is obtained from the color $\otimes$ spin coupling in Eq. (103), the Young-Yamanouchi bases are as follows;

[TABLE]

[TABLE]

For the case of Young tableau [3,1], which is obtained from the color $\otimes$ spin coupling in Eq. (103), Young-Yamanouchi bases are as follows;

[TABLE]

[TABLE]

For the case of Young tableau [2,1,1], which is obtained from the color $\otimes$ spin coupling in Eq. (111), the Young-Yamanouchi bases are as follows;

[TABLE]

[TABLE]

To find the isospin $\otimes$ color $\otimes$ spin state with $\{123\}$ symmetry, we finally combine the isospin states in Eq. (19) with color $\otimes$ spin states for Young tableau [2,1,1] in Eq. (111) as well as Young tableau [2,1,1], [2,2], and [3,1] in Eq. (103). Therefore, we have four isospin $\otimes$ color $\otimes$ spin states with $\{123\}$ symmetry for $I=1/2$;

[TABLE]

From the notation of the color singlets in Eq. (59) which represents the symmetry of the permutation group, $S_{3}$, we easily see that the $|C_{3}\rangle$ state has the symmetry property with $\{123\}$. For that reason, the isospin $\otimes$ spin state in combining with the $|C_{3}\rangle$ state should be fully symmetric in the exchange of any pair among the particle 1, 2, and 3, and the coupling of $|C_{3}\rangle$ state with the isospin $\otimes$ spin states gives the isospin $\otimes$ color $\otimes$ spin state with $\{123\}$ symmetry. We denote the isospin $\otimes$ spin states satisfying fully symmetry, by

[TABLE]

On the contrary, since both $|S^{3/2}_{1}\rangle$ and $|S^{3/2}_{2}\rangle$ sates in Eq. (70) are fully symmetric in the exchange of any pair among the particle 1, 2, and 3, the isospin $\otimes$ color state in combining with either $|S^{3/2}_{1}\rangle$ or $|S^{3/2}_{2}\rangle$ state should have the opposite symmetry due to the same reason. We denote the isospin $\otimes$ color state satisfying fully antisymmetry, by

[TABLE]

Lastly, we can consider the color $\otimes$ spin states corresponding to Young tableau which are conjugate to that of the isospin states, for the reason why any fully antisymmetric state can be obtained by the coupling of any Young tableau with the conjugate. We denote the color $\otimes$ spin states corresponding to Young tableau [2,1] for the particle 1, 2, and 3, by

[TABLE]

We denote another set of the isospin $\otimes$ color $\otimes$ spin states satisfying fully symmetry, by

[TABLE]

We note that both the states in Eq. (184) and the states in Eq. (255) are orthonormal to each other in four dimension vector space, respectively.

It is worthwhile to mention that from a hadron state point of view $|\psi_{1}\rangle$ accounts for the $(p)_{1}\otimes(J/\psi)_{1}$ state, where the subscript indicates the color state, in a fact that the color part consists of the color singlet of a baryon multiplied by the color singlet of a meson, and the spin part contains a baryon with spin=1/2 multiplied by a meson with spin=1 in Eq. (76). On the other hand, $|\psi_{2}\rangle$ represents the $(p)_{8}\otimes(J/\psi)_{8}$ state as an unphysical state, since the color singlet represents the hidden color, coming from the color octet of a baryon multiplied by the color octet of a meson. The rest corresponds to a unphysical state, resulting from the property of the pentaquark and the Pauli principle.

In a vector space with four dimension where the isospin $\otimes$ color $\otimes$ spin states have the symmetry property with $\{123\}$, there exists orthogonal matrix which transforms the set of Eq. (255) into the set of Eq. (184), given by,

[TABLE]

IV Numerical Results

In this section, we analyze the numerical results performed using the variational method for the Hamiltonian given in Eq. (1). For that purpose, we adopt the trial wave function which consists of the spatial function in Eq. (9) and the isospin $\otimes$ color $\otimes$ spin states obtained from Sec. III. The trial wave function can thus be expanded as follows:

[TABLE]

Before discussing the numerical analysis, it is useful to examine the expectation value of the color spin part of the hyperfine potential, with the spatial dependence factored out, in the matrix form generated by the four independent isospin $\otimes$ color $\otimes$ spin states. This hyperfine matrix is essential in identifying possible attraction in the four configurations. A stable or resonant pentaquark state can only exist if the hyperfine potential of the pentaquark configuration is sufficiently attractive compared to that from the sum of a baryon and a meson. The 4 by 4 matrix form of the expectation value of the hyperfine factor of the pentaquark configuration generated by the isospin $\otimes$ color $\otimes$ spin states in Eq. (184) is given as follows:

[TABLE]

To compare the expectation values of the hyperfine factor of the pentaquark with the corresponding sum of a baryon and a meson, we need to diagonalize -$\langle{\sum}_{i<j}^{5}\frac{1}{m_{i}m_{j}}\lambda_{i}^{c}\lambda_{j}^{c}{\sigma}_{i}\cdot{\sigma}_{j}\rangle$ in Eq. (266) and compare it to the possible decay channels. The diagonalized form of the matrix -$\langle{\sum}_{i<j}^{5}\frac{1}{m_{i}m_{j}}\lambda_{i}^{c}\lambda_{j}^{c}{\sigma}_{i}\cdot{\sigma}_{j}\rangle$ in Eq. (266) can be represented as combinations of terms proportional to $1/{m_{1}}^{2}$, $1/{m_{2}}^{2}$, and $1/(m_{1}m_{2})$, respectively. When the fitting mass $m_{u}$ and $m_{c}$ in Table 1 are used, the ground state is given as

[TABLE]

As can be seen in Table 3, the ground state of the diagonalized hyperfine factor of the pentaquark in Eq. (267) is slightly more attractive than the most attractive $p+J/\psi$ decay channel. This attraction is coming from the term proprotional to $1/m_{1}m_{2}$, which originates from the additional attraction coming from bringing the color octet component of $p$ and $J/\psi$ together, as noted recently in Ref. Takeuchi:2016ejt . However, as we will show below, the attraction is very small and will not compensate for the additional kinetic energy term that arises from making the pentaquark state compact compared to the isolated meson baryon states.

To investigate the mass and the property of the pentaquark with the variational method, we calculate the Schrödinger equation $H|{\Psi}_{\alpha}\rangle=E_{\alpha}|{\Psi}_{\alpha}\rangle$ and diagonalize the $4\times 4$ matrix. We find the ground state to be 4087.6 $\rm{MeV}$, which is the sum of the mass of the $p$ and $J/\psi$ in our model. The wave function is given as

[TABLE]

where the variational parameters are given as $a_{1}=3.4~{}\rm{fm^{-2}}$, $a_{2}=1.4~{}\rm{fm^{-2}}$, $a_{3}=11~{}\rm{fm^{-2}}$ and $a_{4}\sim 0$. The first two parameters and the third parameter correspond to those of the baryon and meson, respectively, while the last shows that the distance between the center of mass of the baryon and the meson approaches infinity. In fact, as we can see from the transformation matrix in Eq. (260), the ground state, $|{\Psi}_{g}\rangle$, for $I=1/2$ is exactly equal to -$(p)_{1}\otimes(J/\psi)_{1}$ corresponding to $|\psi_{1}\rangle$ in Eq. (255), which means that the ground state corresponds to the isolated $p$ and $J/\psi$ state in the relative S-wave.

It is useful to inspect the expectation value of the Hamiltonian for the state $|\psi_{1}\rangle$ to understand why the separated $p$ and $J/\psi$ configuration becomes the ground state. First, the hyperfine potential -$\langle{\sum}_{i<j}^{5}\frac{1}{m_{i}m_{j}}\lambda_{i}^{c}\lambda_{j}^{c}{\sigma}_{i}\cdot{\sigma}_{j}\rangle$ is -$\frac{8}{{m_{1}}^{2}}$+$\frac{16}{3{m_{2}}^{2}}$, which is exactly equal to the sum of the expectation value of the $p$ and $J/\psi$ with the first term (the second) coming from the $p$ ($J/\psi$). Moreover, as discussed before, the lowest eigenvalue of the hyperfine matrix is not so different from this value, suggesting that the attraction in the color octet $p$ and $J/\psi$ is not so strong attraction. As for the confinement potential, as can be seen from Eq. (274)-(279) in the Appendix, the first diagonal components consist of the terms corresponding to the $p$ and $J/\psi$ only. Therefore, the only mass difference between the pentaquark and the $p$ + $J/\psi$ channel comes from the additional kinetic term, which vanishes for the separated $p+J/\psi$ state. Using the last term in Eq. (10), one can estimate the additional kinetic energy to bring the $p$ and $J/\psi$ together. Taking $a_{4}\sim 2~{}\rm{fm}^{-2}$, which corresponds to a separation of about 0.7 fm, one obtains an extra kinetic energy of 200 MeV, making the energy of the compact pentaquark state to be around 4290 MeV. Even if we allow the other three states to mix, which could bring in small additional hyperfine attraction, the additional confining potential will conspire to keep the $(p)_{1}\otimes(J/\psi)_{1}$ state the dominant compact configuration. Obviously, such a compact state would just fall apart into the $p+J/\psi$ state and thus not be stable unless the spatial wave function has a small overlap with the final state $p+J/\psi$ Melikhov:2006ec .

As any configuration generated with $|\psi_{1}\rangle$ is dominated by the fall apart $p+J/\psi$ state, we need to investigate whether the excited state can be compact and quasi-stable. To accomplish this, we consider the $|\psi_{2}\rangle$, $|\psi_{3}\rangle$, and $|\psi_{4}\rangle$ in Eq. (255) without $|\psi_{1}\rangle$. The detailed property of the excited state of this state is given in Table 4. Due to the quantum numbers, except for the $p+J/\psi$ configuration, the excited states can not be written as a sum of a single baryon and meson state. Hence, we find a compact state. However, it can decay into several baryon and meson decay channels and is not stable. As for the color spin part of the potential -$\langle{\sum}_{i<j}^{5}\frac{1}{m_{i}m_{j}}\lambda_{i}^{c}\lambda_{j}^{c}{\sigma}_{i}\cdot{\sigma}_{j}\rangle$, we find that this state has the following form;

[TABLE]

While the diagonalized hyperfine factor are less attractive than that of the $p+J/\psi$ and ${\Lambda}_{c}+D^{*}$ decay channels, it is still more attractive than other decay channels. Nevertheless, the reason why the excited state has energy larger than any decay channel is due to the large contribution from the confining potential. As discussed in the Appendix, the sum of the color matrix are all equal for the four orthonormal states. However, due to the interplay with the kinetic term, the confining part of the potential is most attractive in the $p+J/\psi$ channel. The contributions from the kinetic, confinement and hyperfine interaction terms for the excited pentaquark state as well as separated baryon meson states are summarized in Table5. The large confinement contribution for the pentaquark state can be seen in the Table5. The obtained mass is too large for it to be the one of the recently observed pentaquark states. Moreover, it will decay to all possible baryon meson state and not be stable.

V Summary

To understand the possible quark configuration of the recently observed hidden charm pentaquark state, we systematically construct the isospin $\otimes$ color $\otimes$ spin pentaquark states containing two heavy quark and antiquark with $I=1/2$ and $S=3/2$ that satisfy the Pauli principle. We systematically derive the isospin $\otimes$ color $\otimes$ spin states from the color and spin coupling scheme, which is based on the permutation group property. We found that there are four orthonormal state, one of which is the color, spin and isospin corresponding to the proton and $J/\psi$. Then, by using a spatial trial wave function that is suitable for describing the decay into a baryon and meson state, we perform the variational method to obtain the lowest mass state of the pentaquark with $I=1/2$ and $S=3/2$. We found that the ground state is the isolated $p+J/\psi$ state and that any compact configuration will also be dominated by the same baryon and meson state, which will thus fall apart decay to the ground state. We further calculate the mass with a excited state, involving the other isospin $\otimes$ color $\otimes$ spin states which are orthonormal to the ground state. The mass of the compact exited state is found to be well above all baryon meson decay channel and not stable. We are therefore led to conclude that the recently observed pentaquark state can not be a compact multiquark state within the conventional constituent quark model with only confining and color spin interaction. There could still be intrinsic three or four body quark interaction that might change the situation. Also, hadronic molecular configurations originating from meson exchange can certainly not be handled in the present picture. All such works are topics for future works.

VI Acknowledgments

This work was supported by the Korea National Research Foundation under the grant number KRF-2011-0020333 and KRF-2011-0030621. The work of W.S. Park was supported in part by the Yonsei University Research Fund (Post Doc. Researcher Supporting Program) of 2014 (project no.: 2014-12-0139). The work of S. Cho and W. S. Park was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP) (No. 2016R1C1B1016270).

Appendix A

In this Appendix, we will present the matrix element of $\lambda_{i}^{c}\lambda_{j}^{c}$ (i$$<$$j=1$\sim$5) of the pentaquark in terms of a four dimensional matrix generated by the states $|\psi_{1}\rangle$, $|\psi_{2}\rangle$, $|\psi_{3}\rangle$, and $|\psi_{4}\rangle$ in Eq. (255).

a) ($i$,$j$)=(1,2), (1,3), or (2,3);

[TABLE]

b) ($i$,$j$)=(1,4), (1,5), (2,4), (2,5), (3,4), or (3,5);

[TABLE]

c) ($i$,$j$)=(4,5);

[TABLE]

It is easily seen that $\langle{\sum}_{i<j}^{5}\lambda_{i}^{c}\lambda_{j}^{c}\rangle$= -40/3$I$, where the $I$ is identity matrix.

In the case of a baryon, $\langle{\sum}_{i<j}^{3}\lambda_{i}^{c}\lambda_{j}^{c}\rangle=-8$ coming from the color singlet state $\frac{1}{\sqrt{6}}\epsilon_{ijk}q^{i}(1)q^{j}(2)q^{k}(3))$. For a meson state, $\langle\lambda_{4}^{c}\lambda_{5}^{c}\rangle=-16/3$ with the color state $\bar{q}_{i}(4)q^{i}(5)$. These values are the first diagonal components in the above matrix elements. Hence, as pointed out before, we find that the first diagonal term of $\langle{\sum}_{i<j}^{5}\lambda_{i}^{c}\lambda_{j}^{c}\rangle$ of the pentaquark is just the sum of those of the baryon and meson. In fact, as far as this color matrix is concerned, all the four sum of diagonal matrix elements have the same value. However, depending on the spatial wave function, the matrices for the confining potential will have different weighting factors coming from spatial wave functions and their sum will no longer be proportional to the identity matrix. If the kinetic terms are considered, it is energetically more favorable to maximize the attraction in the $p$ and $J/\psi$ channel, which makes it the most attractive state even for compact configurations.