higher, the decays of this pentaquark to these final states are also forbidden
by phase space.

In the same approximation, the diquark-diquark-antiquark operators of the
other states from table 3, containing the heavy-diquark with $S_{hd}=1$, after
color reconnection (the spin structure is determined by eq. (123)) and making
the color indices explicit, are as follows:

$\displaystyle\boldsymbol{\Psi}_{1}^{H(2)}(x,y,z)=\frac{1}{3}\left(\tilde{c}^{i}(x)\,\sigma_{2}\right)\left(c_{i}(y)\,\sigma_{2}\,\boldsymbol{\sigma}\,q_{k}(y)\right)d_{0}^{k}(z),$
(139)
$\displaystyle\boldsymbol{\Psi}_{2}^{H(2)}(x,y,z)=\frac{1}{3}\left(\tilde{c}^{i}(x)\,\sigma_{2}\right)\left(c_{k}(y)\,\sigma_{2}\,\boldsymbol{\sigma}\,q_{i}(y)\right)d_{0}^{k}(z).$
(140)

These operators, being the direct product of a spinor and a vector, are
required to be decomposed into two irreducible representations, corresponding
to the spin $J=1/2$ and $J=3/2$ pentaquarks. The projection onto these states
is done after Fierz transform is performed, for which one uses the relation
Ali:2019roi :

$\left(\sigma_{2}\right)^{\alpha}_{\beta}\left(\sigma_{2}\,\sigma_{a}\right)^{\gamma}_{\delta}=\frac{1}{2}\left[\left(\sigma_{2}\right)^{\alpha}_{\delta}\left(\sigma_{2}\,\sigma_{a}\right)^{\gamma}_{\beta}+\left(\sigma_{2}\,\sigma_{a}\right)^{\alpha}_{\delta}\left(\sigma_{2}\right)^{\gamma}_{\beta}+i\,\varepsilon_{abc}\left(\sigma_{2}\sigma_{b}\right)^{\alpha}_{\delta}\left(\sigma_{2}\sigma_{c}\right)^{\gamma}_{\beta}\right].$
(141)

With this, the states (139) and (140) are reduced to:

$\displaystyle\boldsymbol{\Psi}_{1}^{H(2)}(x,y,z)=\frac{1}{2}\left\\{\left(\tilde{c}(x)\,\sigma_{2}\,\boldsymbol{\sigma}\,c(y)\right)\left(q(y)\,\sigma_{2}\,d_{0}(z)\right)+\left(\tilde{c}(x)\,\sigma_{2}\,c(y)\right)\left(q(y)\,\sigma_{2}\,\boldsymbol{\sigma}\,d_{0}(z)\right)\right.$
$\displaystyle\hskip
71.13188pt+i\left.\left[\left(\tilde{c}(x)\,\sigma_{2}\,\boldsymbol{\sigma}\,c(y)\right)\times\left(q(y)\,\sigma_{2}\,\boldsymbol{\sigma}\,d_{0}(z)\right)\right]\right\\},$
(142)
$\displaystyle\boldsymbol{\Psi}_{2}^{H(2)}(x,y,z)=-\frac{1}{2}\left\\{\left(\tilde{c}(x)\,\sigma_{2}\,\boldsymbol{\sigma}\,q(y)\right)\left(c(y)\,\sigma_{2}\,d_{0}(z)\right)+\left(\tilde{c}(x)\,\sigma_{2}\,q(y)\right)\left(c(y)\,\sigma_{2}\,\boldsymbol{\sigma}\,d_{0}(z)\right)\right.$
$\displaystyle\hskip
88.20354pt+i\left.\left[\left(\tilde{c}(x)\,\sigma_{2}\,\boldsymbol{\sigma}\,q(y)\right)\times\left(c(y)\,\sigma_{2}\,\boldsymbol{\sigma}\,d_{0}(z)\right)\right]\right\\},$
(143)

where the quark-antiquark and quark-diquark combinations are properly
normalized in the color space.

Neglecting as above the position dependence, the quark operators (142) and
(143) of the pentaquarks with the light quark $q=u$ and light diquark
$d_{0}=\left[u\,C\,\gamma_{5}\,d\right]$ can be rewritten in terms of the
hadronic states:

$\displaystyle\boldsymbol{\Psi}_{1}^{H(2)}=\frac{i}{\sqrt{2}}\,p\left\\{a^{\prime}\,\eta_{c}\,\boldsymbol{\sigma}+b^{\prime}\,\mathbf{J}\boldsymbol{/\psi}-ic^{\prime}\left[\boldsymbol{\sigma}\times\mathbf{J}\boldsymbol{/\psi}\right]\right\\},$
(144)
$\displaystyle\boldsymbol{\Psi}_{2}^{H(2)}=-\frac{i}{\sqrt{2}}\,\Lambda_{c}^{+}\left\\{A^{\prime}\,\bar{D}^{0}\,\boldsymbol{\sigma}+B^{\prime}\,\mathbf{\bar{D}^{*0}}-iC^{\prime}\left[\boldsymbol{\sigma}\times\mathbf{\bar{D}^{*0}}\right]\right\\},$
(145)

where $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$ ($a^{\prime}$, $b^{\prime}$
and $c^{\prime}$) are non-perturbative coefficients associated with different
barrier penetration amplitudes. They are equal in the limit of naive Fierz
coupling. These operators are the linear combination of two operators
corresponding to baryons with spins $J=1/2$ and $J=3/2$. They can be separated
by using the equation:

$\boldsymbol{\Psi}_{1,2}^{H(2)}=\left[\boldsymbol{\Psi}_{1,2}^{H(2)}-\frac{1}{3}\left(\boldsymbol{\Psi}_{1,2}^{H(2)}\,\boldsymbol{\sigma}\right)\boldsymbol{\sigma}\right]+\frac{1}{3}\left(\boldsymbol{\Psi}_{1,2}^{H(2)}\,\boldsymbol{\sigma}\right)\boldsymbol{\sigma}=\boldsymbol{\Psi}_{1,2}^{H(3/2)}+\Psi_{1,2}^{H(1/2)}\,\boldsymbol{\sigma},$
(146)

with the following result:

$\displaystyle\Psi_{1}^{H(1/2)}=\frac{i}{\sqrt{2}}\,p\left\\{a^{\prime}\,\eta_{c}+\frac{b^{\prime}}{3}\left(\boldsymbol{\sigma}\,\mathbf{J}\boldsymbol{/\psi}\right)\right\\},$
(147)
$\displaystyle\boldsymbol{\Psi}_{1}^{H(3/2)}=\frac{i\sqrt{2}}{3}\,p\left\\{b^{\prime}\,\mathbf{J}\boldsymbol{/\psi}-2ic^{\prime}\left[\boldsymbol{\sigma}\times\mathbf{J}\boldsymbol{/\psi}\right]\right\\},$
(148)
$\displaystyle\Psi_{2}^{H(1/2)}=-\frac{i}{\sqrt{2}}\,\Lambda_{c}^{+}\left\\{A^{\prime}\,\bar{D}^{0}+\frac{B^{\prime}}{3}\,\left(\boldsymbol{\sigma}\,\mathbf{\bar{D}^{*0}}\right)\right\\},$
(149)
$\displaystyle\boldsymbol{\Psi}_{2}^{H(3/2)}=-\frac{i\sqrt{2}}{3}\,\Lambda_{c}^{+}\left\\{B^{\prime}\,\mathbf{\bar{D}^{*0}}-2iC^{\prime}\left[\boldsymbol{\sigma}\times\mathbf{\bar{D}^{*0}}\right]\right\\}.$
(150)

This analysis shows that the unflavored pentaquark with the spin-parity
$J^{P}=3/2^{-}$ is mainly decaying either to $J/\psi\,p$ final state, in which
it was observed due to our assignment, or to $\Lambda_{c}^{+}\,\bar{D}^{*0}$
state.

If instead we assign $q=s$, but keeping the light diquark $d_{0}$ as above, we
get two states with spins $S_{hd}=1$ from table 11 and the corresponding
operators are as follows:

$\displaystyle\Psi_{1}^{H(1/2)}=\frac{i}{\sqrt{2}}\,\Lambda\left\\{a^{\prime}_{s}\,\eta_{c}+\frac{b^{\prime}_{s}}{3}\left(\boldsymbol{\sigma}\,\mathbf{J}\boldsymbol{/\psi}\right)\right\\},$
(151)
$\displaystyle\boldsymbol{\Psi}_{1}^{H(3/2)}=\frac{i\sqrt{2}}{3}\,\Lambda\left\\{b^{\prime}_{s}\,\mathbf{J}\boldsymbol{/\psi}-2ic^{\prime}_{s}\left[\boldsymbol{\sigma}\times\mathbf{J}\boldsymbol{/\psi}\right]\right\\},$
(152)
$\displaystyle\Psi_{2}^{H(1/2)}=-\frac{i}{\sqrt{2}}\,\Lambda_{c}^{+}\left\\{A^{\prime}_{s}\,\bar{D}_{s}^{0}+\frac{B^{\prime}_{s}}{3}\,\left(\boldsymbol{\sigma}\,\mathbf{\bar{D}_{s}^{*0}}\right)\right\\},$
(153)
$\displaystyle\boldsymbol{\Psi}_{2}^{H(3/2)}=-\frac{i\sqrt{2}}{3}\,\Lambda_{c}^{+}\left\\{B^{\prime}_{s}\,\mathbf{\bar{D}_{s}^{*0}}-2iC^{\prime}_{s}\left[\boldsymbol{\sigma}\times\mathbf{\bar{D}_{s}^{*0}}\right]\right\\}.$
(154)

It is easy to recognize that the main decays in which a singly-strange
$J^{P}=3/2^{-}$ pentaquark should be searched, are the $J/\psi\,\Lambda$ and
$\Lambda_{c}^{+}\,\bar{D}_{s}^{*0}$ final states. Analysis of the other
pentaquark states can be done along similar lines.

From the foregoings in this section, it is evident that the color
representations and color reconnections, on one hand, and localization of the
diquarks and quarks, on the other, are essential to disentangle the dynamics
(production and decays) of multiquark hadrons. We have worked out some
illustrative cases here, in the color-antitriplet diquark representation,
imposing heavy-quark symmetry constraints in the decays of $b$-baryons.

We end this section with a remark about the composition, localisation, and
bindings of the various building blocks of a pentaquark in our model. In the
compact diquark picture which we follow, the quarks in a diquark are bound and
not free, and they have a hadronic size, which could be as large as several
tenths of a Fermi. This applies to both the light
$[q^{\prime}q^{\prime\prime}]$-diquark as well as the heavy-light
$[qc]$-diquark, In this respect, the compact pentaquark model differs from the
hadrocharmonium model Eides:2019tgv , in which the light degrees of freedom
are considered as a cloud. We implicitly assume that the diquark sizes are
smaller than that of the triquark, which in turn is smaller than the
pentaquark size. So, the compact pentaquark model is essentially a three-scale
problem. The light diquark is localized at some distance from the doubly-heavy
triquark and an overlap of their wave functions can be described within the
one-gluon-exchange approach. The pentaquark stabilization is essentially
determined by the triquark stabilization, which is accomplished by assuming
that there is a barrier/bound-state-effect which reduces the probability of
the $\bar{c}$ and the charmed quark in the heavy $[qc]$-diquark to annihilate
each other. This barrier is essential, as otherwise the $c$\- and
$\bar{c}$-quarks will immediately annihilate. The triquark topology is not too
different than that of a tetraquark consisting of $[qc]$\- and
$[\bar{q}^{\prime}\bar{c}]$-diquarks, where the tetraquark is stabilized due
to a barrier between the diquark and the anti-diquark shells Maiani:2019cwl ;
Maiani:2019lpu .

As a final remark, we note that in the diquark models light diquarks
$[qq^{\prime}]$ are ubiquitous. They are present in baryons, such as
$\Lambda_{b}$, and the hidden-charm pentaquarks, discussed in detail in our
paper, but also in the excited baryons, such as $\Omega_{c}^{*}=c\\{ss\\}$
Karliner:2017kfm , and in doubly-heavy tetraquarks, such as
${\bar{b}\bar{b}}[qq^{\prime}]$, discussed in a number of papers on stable
tetraquarks Karliner:2017qjm ; Eichten:2017ffp ; Francis:2016hui ;
Bicudo:2017szl ; Junnarkar:2017sey . How their hadronic sizes impact on
properties of a given multiquark state depends on an underlying dynamical
model. In the constituent diquark model, which we use, the size of the light
diquark influences various parameters of the effective Hamiltonian. We have
ignored the differences which emerge from intrinsically different embeddings
of the diquarks in the baryons, tetraquarks and pentaquarks. This is an
approximation, as the hadrons in question are not expected to have a universal
hadronic size, and it limits the precision of the pentaquark masses presented
here. A detailed dynamical scheme for pentaquarks remains to be developed — an
issue which we hope to address in the future.

##  7 Conclusions

We have presented the mass spectrum of the hidden-charm pentaquark states
$(c\bar{c}qq^{\prime}q^{\prime\prime})$, where $q$, $q^{\prime}$, and
$q^{\prime\prime}$ are light $u$-, $d$,- and $s$-quarks, using isospin
symmetry for the $u$\- and $d$-quarks. In doing this, we have used an
effective Hamiltonian, based on a doubly-heavy triquark — light diquark
picture, shown in fig. 1. Apart from the constituent diquark and quark masses,
the Hamiltonian incorporates dominant spin-spin, spin-orbit, orbital and
tensor interactions. As the first step, we present analytical expressions for
the masses of the states having well-defined quantum numbers, making explicit
the transformations needed in the relations among various angular momentum
basis vectors. This formalism is the main theoretical contribution of our
paper.

For the numerical estimates of the pentaquark masses, we use Model II by
Maiani et al. Maiani:2014aja , developed to calculate the $X$, $Y$, and $Z$
tetraquark mass spectrum in the compact diquark model, in which only the
dominant spin-spin couplings are taken into account. Since not all parameters
in the effective Hamiltonian can be uniquely determined, as there are
currently only three pentaquark states observed so far, and also their spin-
parity assignments are not known, we use the results of the known heavy
tetraquarks and heavy baryons to fix them. Due to the paucity of data on
pentaquarks, and the inference on the model parameters from other systems, it
is difficult to be quantitatively precise. A remark about the accuracy on our
mass estimates due to the neglect of the diquark size effects is in order
here. While we have neglected this, it has been estimated recently for the
double-bottom hadrons (baryons and tetraquarks) by taking into account gluon
exchanges between the spectator quarks and the two heavy quarks in the
diquark. Using the non-relativistic potential model to estimate the matrix
element of the resulting dominant operator leads to a contribution $\Delta
M_{\Xi_{bbq}}\simeq 30$ MeV, with a similar contribution to the mass shift of
the corresponding tetraquark An:2018cln . The corresponding effect for the
hidden-charm tetraquarks and pentaquarks is not yet at hand, but is presumably
of a similar order of magnitude. Thus, there is an intrinsic diquark-size
dependent uncertainty in our mass estimates, which we hope can be estimated in
an improved dynamical diquark framework. In view of this, the masses of the
pentaquarks in the compact diquark approach presented here, while being in the
right ball-park of the current experimental measurements, require an improved
theoretical framework and more data to be precise.

The resulting pentaquark spectrum is very rich, as the states have both spin-0
and spin-1 light diquarks. Since the current experimental effort is dominated
by searches in $\Lambda_{b}$-decays, we impose heavy quark symmetry, in which
the light-diquark spin in $\Lambda_{b}$-baryon is a good quantum number. This,
in turn, implies that only pentaquarks with the spin-0 light diquarks are
anticipated in $\Lambda_{b}$-decays. This reduces the spectrum of the
observable states greatly, yet there are more states present in the spectrum
than have been observed, and hence we are not able to assign the $J^{P}$
quantum numbers to the observed states uniquely. We have argued, based on
theoretical consistency, why some spin-parity assignments for the new
pentaquarks can be ruled out in the compact diquark model. Yet, by virtue of
the fact that both $S$\- and $P$-state pentaquarks are anticipated in such
models, predicting both positive- and negative-parity states, the spectrum
provides a generic discrimination between the compact pentaquarks and the ones
foreseen in hadron molecules, in which the observed pentaquarks have all the
negative parity. Thus, it is crucial to determine the $J^{P}$ quantum numbers
of the observed pentaquarks. Likewise, the remaining states in the pentaquark
spectrum, whose masses we have presented, should be searched for. We find that
some of them may lie in mass below their strong thresholds, defined by
$J/\psi\,p$ (or $\eta_{c}\,p$). Hence, they will decay via the $c\bar{c}$-pair
annihilation into light hadrons and into a lepton pair. Thus, the mode
$p\,\mu^{+}\mu^{-}$ may reveal the presence of these additional pentaquark
states, and we urge the LHCb collaboration to search for them.

The hidden-charm pentaquarks having a strange quark in their valence
composition are at present an experimentally virgin territory. While they can
be produced also promptly, like their non-strange counterparts, only search
strategies based on weak decays of the $b$-baryons with a strange quark are
expected to bear fruits. The strange $b$-baryons which decay weakly are
$\Xi_{b}^{0}\,(bus)$, $\Xi_{b}^{-}\,(bds)$, and $\Omega_{b}^{-}\,(bss)$. Of
these, the decays of $\Omega_{b}^{-}\,(bss)$ will produce pentaquarks with a
spin-1 light diquark, in accordance with the heavy quark symmetry, thus they
allow to investigate the $SU(3)_{F}$-decuplet pentaquark sector. However, the
production of such $b$-baryons is suppressed due to the excitation of the
$s\bar{s}$-pair from the vacuum (for producing $\Xi_{b}^{0}$ and $\Xi_{b}^{-}$
in a $b$-quark initiated jet) and twice suppressed (for the
$\Omega_{b}^{-}$-baryon). As an example, the product branching ratio ${\cal
B}(\Omega_{b}^{-}\to J/\psi\,\Omega^{-})\,{\cal
B}(b\to\Omega_{b}^{-})=\left(2.9^{+1.1}_{-0.8}\right)\times 10^{-6}$
Tanabashi:2018oca . This, compared with the product branching ratio ${\cal
B}(\Lambda_{b}^{0}\to J/\psi\,\Lambda^{0})\,{\cal
B}(b\to\Lambda_{b}^{0})=(5.8\pm 0.8)\times 10^{-5}$, is about 20 times less
probable. Thus, the discovery of hidden-charm strange pentaquarks requires a
lot more data, but they remain a firm prediction, as in the compact diquark
model complete $SU(3)_{F}$-multiplets are anticipated. Hence, their non-
observation in experiments would seriously undermine such models. In the
foreseeable future, experiments at the LHC, in particular LHCb, remain our
best hope for their discovery. In the long-term, the entire hidden-charm
pentaquark sector can be studied at a Tera-$Z$ factory.

###### Acknowledgements.

We would like to thank Vladimir Galkin, Marek Karliner, Luciano Maiani,
Antonello Polosa, Tomasz Skwarnicki, and Sheldon Stone for helpful
discussions. A. A. and A. P. acknowledge support and warm hospitality of the
Theoretical Physics Division at IHEP (Beijing, China) where an initial part of
this research was performed, and Wei Wang for the hospitality at the
Laboratory of Particle Physics of the Shanghai Jiao-Tong University, where it
was completed. A. P. thanks the Theory Group at DESY for their kind
hospitality, where the major part of this work was done. A. P. acknowledges
financial support by the German Academic Exchange Service (DAAD), by the
Russian Foundation for Basic Research and National Natural Science Foundation
of China for the research project No. 19-52-53041, and by the “YSU Initiative
Scientific Research Activity” (Project No. AAAA-A16-116070610023-3).

##  Appendix A Corrections due to spin-spin interactions between the
constituents of doubly-heavy triquark and light diquark

The matrix elements of the operators in eq. (5) have in general six different
coupling strengths. For the pentaquarks, which contain the light $u$\- and
$d$-quarks only, isospin symmetry among the light quarks can be used to reduce
the number of couplings:
$(\tilde{\mathcal{K}}_{cq^{\prime}})_{\bar{3}}=(\tilde{\mathcal{K}}_{cq^{\prime\prime}})_{\bar{3}}=(\tilde{\mathcal{K}}_{cq})_{\bar{3}}$,
$(\tilde{\mathcal{K}}_{qq^{\prime}})_{\bar{3}}=(\tilde{\mathcal{K}}_{qq^{\prime\prime}})_{\bar{3}}=(\tilde{\mathcal{K}}_{qq})_{\bar{3}}$,
and
$\tilde{\mathcal{K}}_{\bar{c}q^{\prime}}=\tilde{\mathcal{K}}_{\bar{c}q^{\prime\prime}}=\tilde{\mathcal{K}}_{\bar{c}q}$.
So, the expression for $H_{SS}^{t-ld}$ in the effective Hamiltonian
responsible for the spin-spin interactions between the (anti)quarks in the
diquark and triquark, given in eq. (5), is greatly simplified:

$\displaystyle
H_{SS}^{t-ld}=2\,\tilde{\mathcal{K}}_{\bar{c}q}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{ld})+2\,(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{ld})+2\,(\tilde{\mathcal{K}}_{qq})_{\bar{3}}\,(\mathbf{S}_{q}\cdot\mathbf{S}_{ld}),$
(155)

where the definition of the light-diquark spin operator,
$\mathbf{S}_{ld}=\mathbf{S}_{q^{\prime}}+\mathbf{S}_{q^{\prime\prime}}$, is
used. We start by calculating the matrix elements of the spin-spin operators
from the first term,
$2\,\tilde{\mathcal{K}}_{\bar{c}q}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{ld})$,
which contains the spins of the charm antiquark and light diquark. Again, it
is convenient to transform the vector basis. To do so, we denote the
pentaquark state $|S_{hd},S_{t},L_{t};S_{ld},L_{ld};S,L\rangle_{J}$ as
$|S_{ld},\,(S_{\bar{c}},\,S_{hd})_{S_{t}};\,S\rangle$ and apply the
transformation (27):

$\displaystyle|S_{ld},\,(S_{\bar{c}},\,S_{hd})_{S_{t}};\,S\rangle=\sum_{\tilde{S}}(-1)^{S_{ld}+S_{\bar{c}}+S_{hd}+S}\sqrt{(2\tilde{S}+1)\,(2S_{t}+1)}$
$\displaystyle\hskip
93.89409pt\times\left\\{\begin{array}[]{ccc}S_{ld}&S_{\bar{c}}&\tilde{S}\\\
S_{hd}&S&S_{t}\end{array}\right\\}|(S_{ld},\,S_{\bar{c}})_{\tilde{S}},\,S_{hd};\,S\rangle.$
(158)

The required matrix elements can be written in the form:

$\displaystyle{}_{J}\langle
S_{hd},S_{t}^{\prime},L_{t};S_{ld},L_{ld};S,L|\,2\,\tilde{\mathcal{K}}_{\bar{c}q}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{ld})\,|S_{hd},S_{t},L_{t};S_{ld},L_{ld};S,L\rangle_{J}$
(159) $\displaystyle\hskip
19.91692pt=\tilde{\mathcal{K}}_{\bar{c}q}\,(-1)^{2S+1}\sqrt{(2S_{t}+1)\,(2S_{t}^{\prime}+1)}\,\sum_{\tilde{S}}\left(2\tilde{S}+1\right)$
$\displaystyle\hskip
19.91692pt\times\left[\tilde{S}\left(\tilde{S}+1\right)-S_{ld}\left(S_{ld}+1\right)-\frac{3}{4}\right]\left\\{\begin{array}[]{ccc}S_{ld}&1/2&\tilde{S}\\\
S_{hd}&S&S_{t}\end{array}\right\\}\left\\{\begin{array}[]{ccc}S_{ld}&1/2&\tilde{S}\\\
S_{hd}&S&S_{t}^{\prime}\end{array}\right\\},$ (164)

where the charm-antiquark spin, $S_{\bar{c}}=1/2$, is substituted.

We calculate now the matrix elements of the last two operators in eq. (155).
Let us start from the operator
$2\,(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{ld})$
which works on the wave functions of the charm quark in the heavy diquark and
light diquark. To find it, we need to couple the heavy and light diquark first
and then couple the light diquark with the charm quark — the constituent of
heavy-light diquark. This requires a double transformation of the original
pentaquark vector state $|S_{hd},S_{t},L_{t};S_{ld},L_{ld};S,L\rangle_{J}$,
which we denote as $|S_{ld},\,(S_{hd},\,S_{\bar{c}})_{S_{t}};\,S\rangle$, to
the vector state $|(S_{ld},\,S_{hd})_{S_{1}},\,S_{\bar{c}};\,S\rangle$; in the
second transformation these basis states, for simplicity called as
$|S_{ld},\,(S_{c},\,S_{q})_{S_{hd}};\,S_{1}\rangle$, are changed to the
required ones, $|(S_{ld},\,S_{c})_{S_{2}},\,S_{q};\,S_{1}\rangle$.
Specifically:

$\displaystyle|S_{ld},\,(S_{hd},\,S_{\bar{c}})_{S_{t}};\,S\rangle=\sum_{S_{1}}(-1)^{S_{ld}+S_{hd}+S_{\bar{c}}+S}\sqrt{(2S_{1}+1)\,(2S_{t}+1)}$
$\displaystyle\hskip
105.27519pt\times\left\\{\begin{array}[]{ccc}S_{ld}&S_{hd}&S_{1}\\\
S_{\bar{c}}&S&S_{t}\end{array}\right\\}|(S_{ld},\,S_{hd})_{S_{1}},\,S_{\bar{c}};\,S\rangle$
(167) $\displaystyle\hskip
19.91692pt=\sum_{S_{1}}(-1)^{S_{ld}+S_{hd}+S_{\bar{c}}+S}\sqrt{(2S_{1}+1)\,(2S_{t}+1)}\left\\{\begin{array}[]{ccc}S_{ld}&S_{hd}&S_{1}\\\
S_{\bar{c}}&S&S_{t}\end{array}\right\\}|S_{ld},\,(S_{c},\,S_{q})_{S_{hd}};\,S_{1}\rangle$
(170) $\displaystyle\hskip
19.91692pt=\sum_{S_{1},S_{2}}(-1)^{2S_{ld}+S_{hd}+S_{\bar{c}}+S+S_{c}+S_{q}+S_{1}}\sqrt{(2S_{1}+1)\,(2S_{t}+1)\,(2S_{hd}+1)\,(2S_{2}+1)}$
$\displaystyle\hskip
48.36967pt\times\left\\{\begin{array}[]{ccc}S_{ld}&S_{hd}&S_{1}\\\
S_{\bar{c}}&S&S_{t}\end{array}\right\\}\left\\{\begin{array}[]{ccc}S_{ld}&S_{c}&S_{2}\\\
S_{q}&S_{1}&S_{hd}\end{array}\right\\}|(S_{ld},\,S_{c})_{S_{2}},\,S_{q};\,S_{1}\rangle.$
(175)

The matrix elements of interest are as follows:

$\displaystyle{}_{J}\langle
S_{hd}^{\prime},S_{t}^{\prime},L_{t};S_{ld},L_{ld};S,L|\,2\,(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{ld})\,|S_{hd},S_{t},L_{t};S_{ld},L_{ld};S,L\rangle_{J}$
$\displaystyle\hskip
19.91692pt=(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\,(-1)^{S_{hd}+S_{hd}^{\prime}+2S+1}\sqrt{(2S_{hd}+1)\,(2S_{hd}^{\prime}+1)\,(2S_{t}+1)\,(2S_{t}^{\prime}+1)}\,$
$\displaystyle\hskip
19.91692pt\times\sum_{S_{1},S_{2}}\left(2S_{1}+1\right)\left(2S_{2}+1\right)\left[S_{2}\left(S_{2}+1\right)-S_{ld}\left(S_{ld}+1\right)-\frac{3}{4}\right]$
$\displaystyle\hskip
19.91692pt\times\left\\{\begin{array}[]{ccc}S_{ld}&S_{hd}&S_{1}\\\
1/2&S&S_{t}\end{array}\right\\}\left\\{\begin{array}[]{ccc}S_{ld}&S_{hd}^{\prime}&S_{1}\\\
1/2&S&S_{t}^{\prime}\end{array}\right\\}\left\\{\begin{array}[]{ccc}S_{ld}&1/2&S_{2}\\\
1/2&S_{1}&S_{hd}\end{array}\right\\}\left\\{\begin{array}[]{ccc}S_{ld}&1/2&S_{2}\\\
1/2&S_{1}&S_{hd}^{\prime}\end{array}\right\\},$ (184)

where the quark and antiquark spins, $S_{c}=S_{q}=S_{\bar{c}}=1/2$, are
substituted. Note that the matrix elements of the last operator in (155)
$2\,(\tilde{\mathcal{K}}_{qq})_{\bar{3}}\,(\mathbf{S}_{q}\cdot\mathbf{S}_{ld})$
differ from the above matrix elements by the replacement
$(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\to(\tilde{\mathcal{K}}_{qq})_{\bar{3}}$
and the interchange $S_{c}=1/2\leftrightarrow S_{q}=1/2$. It means that, up to
the coupling strengths, they coincide. So, the contribution of these two
operators enters the matrix elements with the factor
$(\tilde{\mathcal{K}}_{cq})_{\bar{3}}+(\tilde{\mathcal{K}}_{qq})_{\bar{3}}$.

We continue by considering the spin-spin interaction between the strange
hidden-charm triquark and unflavored (i. e., having $u$\- and $d$-quarks)
light diquark. In this case, the expression given in eq. (5) of the effective
Hamiltonian is again simplified in the isospin-symmetry limit:

$\displaystyle
H_{SS}^{t-ld}=2\,\tilde{\mathcal{K}}_{\bar{c}q}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{ld})+2\,(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{ld})+2\,(\tilde{\mathcal{K}}_{sq})_{\bar{3}}\,(\mathbf{S}_{s}\cdot\mathbf{S}_{ld}),$
(185)

where $\mathbf{S}_{s}$ and
$\mathbf{S}_{ld}=\mathbf{S}_{q^{\prime}}+\mathbf{S}_{q^{\prime\prime}}$ are
the spin operators of the $s$-quark in the triquark and the light-diquark,
respectively. Comparing (185) with (155), it is easy to recognize that they
differ by the factor entering the last terms,
$(\tilde{\mathcal{K}}_{sq})_{\bar{3}}$ and
$(\tilde{\mathcal{K}}_{qq})_{\bar{3}}$. So, the matrix elements (159) and
(184) remain unchanged, but to get the contribution of the last operator in
(185) one should simply change the coupling:
$(\tilde{\mathcal{K}}_{qq})_{\bar{3}}\to(\tilde{\mathcal{K}}_{sq})_{\bar{3}}$.
Taking into account the statement in the paragraph above, the second and third
terms in (185) are equal (modulo couplings) and contribute with the
coefficient
$(\tilde{\mathcal{K}}_{cq})_{\bar{3}}+(\tilde{\mathcal{K}}_{sq})_{\bar{3}}$ in
the mass formulae.

The same simple form of the spin-spin interaction is also hold for the hidden-
charm pentaquark with the strangeness $S=-2$ which consists of the unflavored
doubly-heavy triquark and light diquark containing two $s$-quarks:

$\displaystyle
H_{SS}^{t-ld}=2\,\tilde{\mathcal{K}}_{\bar{c}s}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{ld})+2\,(\tilde{\mathcal{K}}_{cs})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{ld})+2\,(\tilde{\mathcal{K}}_{sq})_{\bar{3}}\,(\mathbf{S}_{q}\cdot\mathbf{S}_{ld}).$
(186)

So, the matrix elements (159) and (184), up to spin-spin couplings, remain
unchanged and their contributions to the mass formulae are accompanied by the
couplings $\tilde{\mathcal{K}}_{\bar{c}s}$ and
$(\tilde{\mathcal{K}}_{cs})_{\bar{3}}+(\tilde{\mathcal{K}}_{sq})_{\bar{3}}$,
respectively.

Similar simple effective Hamiltonian for the spin-spin interactions is present
in the hidden-charm pentaquarks with the strangeness $S=-3$, which contains
three $s$-quarks:

$\displaystyle
H_{SS}^{t-ld}=2\,\tilde{\mathcal{K}}_{\bar{c}s}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{ld})+2\,(\tilde{\mathcal{K}}_{cs})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{ld})+2\,(\tilde{\mathcal{K}}_{ss})_{\bar{3}}\,(\mathbf{S}_{s}\cdot\mathbf{S}_{ld}).$
(187)

The matrix elements (159) and (184) can be safely used in this case also and
their contributions are entering the mass formulae with the couplings
$\tilde{\mathcal{K}}_{\bar{c}s}$ and
$(\tilde{\mathcal{K}}_{cs})_{\bar{3}}+(\tilde{\mathcal{K}}_{ss})_{\bar{3}}$,
respectively.

The next flavor state which corresponds to the hidden-charm strange pentaquark
consists of the the unflavored doubly-heavy triquark and strange light
diquark. The corresponding spin-spin effective Hamiltonian (5) does not
simplified and all the six terms, each of which is accompanied by its own
coupling, should be considered separately:

$\displaystyle H_{SS}^{t-ld}$ $\displaystyle=$ $\displaystyle
2\,\tilde{\mathcal{K}}_{\bar{c}q}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{q^{\prime}})+2\,(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{q^{\prime}})+2\,(\tilde{\mathcal{K}}_{qq})_{\bar{3}}\,(\mathbf{S}_{q}\cdot\mathbf{S}_{q^{\prime}})$
$\displaystyle+$ $\displaystyle
2\,\tilde{\mathcal{K}}_{\bar{c}s}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{s})+2\,(\tilde{\mathcal{K}}_{cs})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{s})+2\,(\tilde{\mathcal{K}}_{sq})_{\bar{3}}\,(\mathbf{S}_{q}\cdot\mathbf{S}_{s}),$

where we identify $q^{\prime\prime}=s$ and apply the isospin symmetry to the
spin-spin couplings.

A calculation of these operators requires a recoupling of four angular
momenta. In general, this transformation scheme is expressed through the
$9j$-symbol which is defined as follows Edmonds:1957 :

$\displaystyle\langle(j_{1},j_{2})_{j_{12}},(j_{3},j_{4})_{j_{34}},j|(j_{1},j_{3})_{j_{13}},(j_{2},j_{4})_{j_{24}},j\rangle$
$\displaystyle\hskip
48.36967pt=\sqrt{(2j_{12}+1)\,(2j_{34}+1)\,(2j_{13}+1)\,(2j_{24}+1)}\left\\{\begin{array}[]{ccc}j_{1}&j_{2}&j_{12}\\\
j_{3}&j_{4}&j_{34}\\\ j_{13}&j_{24}&j\end{array}\right\\}.$ (192)

To demonstrate its necessity, let us consider the first term in (A). In our
basis of states the charm antiquark belongs to the doubly-heavy triquark while
the light $q^{\prime}$-quark is inside the light diquark. The original
pentaquark vector state $|S_{hd},S_{t},L_{t};S_{ld},L_{ld};S,L\rangle_{J}$,
which we denote for simplicity as
$|(S_{\bar{c}},\,S_{hd})_{S_{t}},(S_{q^{\prime}},\,S_{s})_{S_{ld}};\,S\rangle$,
should be transformed to the new state with the recoupling momenta
$|(S_{\bar{c}},\,S_{q^{\prime}})_{S_{\bar{c}q^{\prime}}},\,(S_{hd},\,S_{s})_{S_{hd,s}};\,S\rangle$,
and the $9j$-symbol is responsible for this transform:

$\displaystyle|(S_{\bar{c}},\,S_{hd})_{S_{t}},(S_{q^{\prime}},\,S_{s})_{S_{ld}};\,S\rangle=\sum_{S_{\bar{c}q^{\prime}},S_{hd,s}}\sqrt{(2S_{t}+1)\,(2S_{ld}+1)\,(2S_{\bar{c}q^{\prime}}+1)\,(2S_{hd,s}+1)}$
$\displaystyle\hskip
105.27519pt\times\left\\{\begin{array}[]{ccc}S_{\bar{c}}&S_{hd}&S_{t}\\\
S_{q^{\prime}}&S_{s}&S_{ld}\\\
S_{\bar{c}q^{\prime}}&S_{hd,s}&S\end{array}\right\\}|(S_{\bar{c}},\,S_{q^{\prime}})_{S_{\bar{c}q^{\prime}}},\,(S_{hd},\,S_{s})_{S_{hd,s}};\,S\rangle.$
(196)

So, the matrix element of the operator
$2\,\tilde{\mathcal{K}}_{\bar{c}q}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{q^{\prime}})$
sandwiched between the states with different triquark $S_{t}^{(\prime)}$ and
light-diquark $S_{ld}^{(\prime)}$ vector states can be written as a double sum
(here, we come back to the original notation for the vector state):

$\displaystyle{}_{J}\langle
S_{hd},S_{t}^{\prime},L_{t};S_{ld}^{\prime},L_{ld};S,L|2\,\tilde{\mathcal{K}}_{\bar{c}q}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{q^{\prime}})|S_{hd},S_{t},L_{t};S_{ld},L_{ld};S,L\rangle_{J}$
$\displaystyle\hskip
19.91692pt=\tilde{\mathcal{K}}_{\bar{c}q}\sqrt{(2S_{t}+1)\,(2S_{t}^{\prime}+1)\,(2S_{hd}+1)\,(2S_{hd}^{\prime}+1)}\sum_{S_{\bar{c}q^{\prime}},S_{hd,s}}(2S_{\bar{c}q^{\prime}}+1)\,(2S_{hd,s}+1)\,$
$\displaystyle\hskip
19.91692pt\times\left[S_{\bar{c}q^{\prime}}(S_{\bar{c}q^{\prime}}+1)-\frac{3}{2}\right]\left\\{\begin{array}[]{ccc}1/2&S_{hd}&S_{t}^{\prime}\\\
1/2&1/2&S_{ld}^{\prime}\\\
S_{\bar{c}q^{\prime}}&S_{hd,s}&S\end{array}\right\\}\left\\{\begin{array}[]{ccc}1/2&S_{hd}&S_{t}\\\
1/2&1/2&S_{ld}\\\ S_{\bar{c}q^{\prime}}&S_{hd,s}&S\end{array}\right\\},$ (203)

where the spins $S_{\bar{c}}=S_{q^{\prime}}=S_{s}=1/2$ of quarks were
subsituted. In a similar manner one can get the contribution of the operator
$2\,\tilde{\mathcal{K}}_{\bar{c}s}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{s})$
with the result:

$\displaystyle{}_{J}\langle
S_{hd},S_{t}^{\prime},L_{t};S_{ld}^{\prime},L_{ld};S,L|2\,\tilde{\mathcal{K}}_{\bar{c}s}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{s})|S_{hd},S_{t},L_{t};S_{ld},L_{ld};S,L\rangle_{J}$
$\displaystyle\hskip
19.91692pt=\tilde{\mathcal{K}}_{\bar{c}s}\sqrt{(2S_{t}+1)\,(2S_{t}^{\prime}+1)\,(2S_{hd}+1)\,(2S_{hd}^{\prime}+1)}\sum_{S_{\bar{c}s},S_{hd,q^{\prime}}}(2S_{\bar{c}s}+1)\,(2S_{hd,q^{\prime}}+1)\,$
$\displaystyle\hskip
19.91692pt\times\left[S_{\bar{c}s}(S_{\bar{c}s}+1)-\frac{3}{2}\right]\left\\{\begin{array}[]{ccc}1/2&S_{hd}&S_{t}^{\prime}\\\
1/2&1/2&S_{ld}^{\prime}\\\
S_{\bar{c}s}&S_{hd,q^{\prime}}&S\end{array}\right\\}\left\\{\begin{array}[]{ccc}1/2&S_{hd}&S_{t}\\\
1/2&1/2&S_{ld}\\\ S_{\bar{c}s}&S_{hd,q^{\prime}}&S\end{array}\right\\}.$ (210)

Up to the coupling, $\tilde{\mathcal{K}}_{\bar{c}s}$, this equation coincides
with (203), so their common contribution contains the factor
$\tilde{\mathcal{K}}_{\bar{c}q}+\tilde{\mathcal{K}}_{\bar{c}s}$.

With the other four operators the situation is a little bit more complicated
as spin-spin interactions relate quarks from different diquarks, moreover, the
spin of the heavy diquark is coupled with the spin of the charm antiquark
inside the triquark system. So, at the first step one needs to decouple the
heavy diquark from the charm antiquark and couple it with the light diquark
which generates the tetraquark system. If we denote the vector state as
$|S_{ld},\,(S_{hd},\,S_{\bar{c}})_{S_{t}};\,S\rangle$, we need to recouple in
a way to get the new basis
$|(S_{ld},\,S_{hd},)_{S_{T}}\,S_{\bar{c}};\,S\rangle$ which is possible to do
with the help of the transformation (27):

$\displaystyle|S_{ld},\,(S_{hd},\,S_{\bar{c}})_{S_{t}};\,S\rangle=\sum_{S_{T}}(-1)^{S_{ld}+S_{hd}+S_{\bar{c}}+S}\sqrt{(2S_{T}+1)\,(2S_{t}+1)}$
$\displaystyle\hskip
93.89409pt\times\left\\{\begin{array}[]{ccc}S_{ld}&S_{hd}&S_{T}\\\
S_{\bar{c}}&S&S_{t}\end{array}\right\\}|(S_{ld},\,S_{hd})_{S_{T}},\,S_{\bar{c}};\,S\rangle.$
(213)

At the second step, it is necessary to recouple quark spins inside the
tetraquark. So, for this transformation the $9j$-symbol (192) is required. To
do this, the introduced pentaquark vector state
$|(S_{ld},\,S_{hd})_{S_{T}},\,S_{\bar{c}};\,S\rangle$ is more convenient to
rewrite in the form
$|(S_{c},\,S_{q})_{S_{hd}},\,(S_{q^{\prime}},\,S_{s})_{S_{ld}};\,S_{T}\rangle$
and then perform the transformations:

$\displaystyle|(S_{c},\,S_{q})_{S_{hd}},\,(S_{q^{\prime}},\,S_{s})_{S_{ld}};\,S_{T}\rangle=\sum_{S_{cq^{\prime}},S_{sq}}\sqrt{(2S_{hd}+1)\,(2S_{ld}+1)\,(2S_{cq^{\prime}}+1)\,(2S_{sq}+1)}$
$\displaystyle\hskip
105.27519pt\times\left\\{\begin{array}[]{ccc}S_{c}&S_{q}&S_{hd}\\\
S_{q^{\prime}}&S_{s}&S_{ld}\\\
S_{cq^{\prime}}&S_{sq}&S_{T}\end{array}\right\\}|(S_{c},\,S_{q^{\prime}})_{S_{cq^{\prime}}},\,(S_{q},\,S_{s})_{S_{sq}};\,S_{T}\rangle.$
(217)

Now, we can sandwich the operators
$2\,(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{q^{\prime}})$
and
$2\,(\tilde{\mathcal{K}}_{sq})_{\bar{3}}\,(\mathbf{S}_{s}\cdot\mathbf{S}_{q})$
by the vector states with varying spins of the heavy, $S_{hd}$, and light,
$S_{ld}$, diquarks as well as of the doubly-heavy triquark, $S_{t}$. The
resulting matrix element of these two operators is the triple sum:

$\displaystyle{}_{J}\langle
S_{hd}^{\prime},S_{t}^{\prime},L_{t};S_{ld}^{\prime},L_{ld};S,L|2\,(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{q^{\prime}})+2\,(\tilde{\mathcal{K}}_{sq})_{\bar{3}}\,(\mathbf{S}_{s}\cdot\mathbf{S}_{q})|S_{hd},S_{t},L_{t};S_{ld},L_{ld};S,L\rangle_{J}$
$\displaystyle\hskip
14.22636pt=\left[(\tilde{\mathcal{K}}_{cq})_{\bar{3}}+(\tilde{\mathcal{K}}_{sq})_{\bar{3}}\right]\sqrt{(2S_{t}+1)\,(2S_{t}^{\prime}+1)\,(2S_{hd}+1)\,(2S_{hd}^{\prime}+1)\,(2S_{ld}+1)\,(2S_{ld}^{\prime}+1)}$
$\displaystyle\hskip
14.22636pt\times(-1)^{S_{hd}+S_{hd}^{\prime}+S_{ld}+S_{ld}^{\prime}+2S+1}\\!\\!\\!\\!\sum_{S_{T},S_{cq^{\prime}},S_{sq}}\\!\\!\\!\\!(2S_{T}+1)\,(2S_{cq^{\prime}}+1)\,(2S_{sq}+1)\,\left[S_{cq^{\prime}}(S_{cq^{\prime}}+1)-\frac{3}{2}\right]$
$\displaystyle\hskip
14.22636pt\times\left\\{\begin{array}[]{ccc}S_{ld}^{\prime}&S_{hd}^{\prime}&S_{T}\\\
1/2&S&S_{t}^{\prime}\end{array}\right\\}\left\\{\begin{array}[]{ccc}S_{ld}&S_{hd}&S_{T}\\\
1/2&S&S_{t}\end{array}\right\\}\left\\{\begin{array}[]{ccc}1/2&1/2&S_{hd}^{\prime}\\\
1/2&1/2&S_{ld}^{\prime}\\\
S_{cq^{\prime}}&S_{sq}&S_{T}\end{array}\right\\}\left\\{\begin{array}[]{ccc}1/2&1/2&S_{hd}\\\
1/2&1/2&S_{ld}\\\ S_{cq^{\prime}}&S_{sq}&S_{T}\end{array}\right\\}.$ (228)

The matrix elements of the rest two operators
$2\,(\tilde{\mathcal{K}}_{cs})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{s})$
and
$2\,(\tilde{\mathcal{K}}_{qq})_{\bar{3}}\,(\mathbf{S}_{q}\cdot\mathbf{S}_{q^{\prime}})$
are calculated similarly with the following result:

$\displaystyle{}_{J}\langle
S_{hd}^{\prime},S_{t}^{\prime},L_{t};S_{ld}^{\prime},L_{ld};S,L|2\,(\tilde{\mathcal{K}}_{cs})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{s})+2\,(\tilde{\mathcal{K}}_{qq})_{\bar{3}}\,(\mathbf{S}_{q}\cdot\mathbf{S}_{q^{\prime}})|S_{hd},S_{t},L_{t};S_{ld},L_{ld};S,L\rangle_{J}$
$\displaystyle\hskip
14.22636pt=\left[(\tilde{\mathcal{K}}_{cs})_{\bar{3}}+(\tilde{\mathcal{K}}_{qq})_{\bar{3}}\right]\sqrt{(2S_{t}+1)\,(2S_{t}^{\prime}+1)\,(2S_{hd}+1)\,(2S_{hd}^{\prime}+1)\,(2S_{ld}+1)\,(2S_{ld}^{\prime}+1)}$
$\displaystyle\hskip
14.22636pt\times(-1)^{S_{hd}+S_{hd}^{\prime}+S_{ld}+S_{ld}^{\prime}+2S+1}\\!\\!\\!\\!\sum_{S_{T},S_{cs},S_{qq^{\prime}}}\\!\\!\\!\\!(2S_{T}+1)\,(2S_{cs}+1)\,(2S_{qq^{\prime}}+1)\,\left[S_{cs}(S_{cs}+1)-\frac{3}{2}\right]$
$\displaystyle\hskip
14.22636pt\times\left\\{\begin{array}[]{ccc}S_{ld}^{\prime}&S_{hd}^{\prime}&S_{T}\\\
1/2&S&S_{t}^{\prime}\end{array}\right\\}\left\\{\begin{array}[]{ccc}S_{ld}&S_{hd}&S_{T}\\\
1/2&S&S_{t}\end{array}\right\\}\left\\{\begin{array}[]{ccc}1/2&1/2&S_{hd}^{\prime}\\\
1/2&1/2&S_{ld}^{\prime}\\\
S_{cs}&S_{qq^{\prime}}&S_{T}\end{array}\right\\}\left\\{\begin{array}[]{ccc}1/2&1/2&S_{hd}\\\
1/2&1/2&S_{ld}\\\ S_{cs}&S_{qq^{\prime}}&S_{T}\end{array}\right\\}.$ (239)

In general, eqs. (228) and (239) are the matrices which mix all the states
having the same total spin, $S$, and the same parity.

The last flavor state which corresponds to the hidden-charm strange pentaquark
with the strangeness $S=-2$ consists of the the strange doubly-heavy triquark
and strange light diquark. The corresponding spin-spin effective Hamiltonian
(5) does not get simplified in this case also, and all the six terms, each of
which is accompanied by its own coupling, should be considered separately:

$\displaystyle H_{SS}^{t-ld}$ $\displaystyle=$ $\displaystyle
2\,\tilde{\mathcal{K}}_{\bar{c}q}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{q})+2\,(\tilde{\mathcal{K}}_{cq})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{q})+2\,(\tilde{\mathcal{K}}_{sq})_{\bar{3}}\,(\mathbf{S}_{s}\cdot\mathbf{S}_{q})$
$\displaystyle+$ $\displaystyle
2\,\tilde{\mathcal{K}}_{\bar{c}s}\,(\mathbf{S}_{\bar{c}}\cdot\mathbf{S}_{s^{\prime}})+2\,(\tilde{\mathcal{K}}_{cs})_{\bar{3}}\,(\mathbf{S}_{c}\cdot\mathbf{S}_{s^{\prime}})+2\,(\tilde{\mathcal{K}}_{ss})_{\bar{3}}\,(\mathbf{S}_{s}\cdot\mathbf{S}_{s^{\prime}}),$

where we make the substitution $q=s$, $q^{\prime}=q$, and
$q^{\prime\prime}=s^{\prime}$. The above consideration holds also here, and
one can use the expressions (210), (228), and (239) with the obvious
replacements
$(\tilde{\mathcal{K}}_{sq})_{\bar{3}}\to(\tilde{\mathcal{K}}_{ss})_{\bar{3}}$
in (228) and
$(\tilde{\mathcal{K}}_{qq})_{\bar{3}}\to(\tilde{\mathcal{K}}_{sq})_{\bar{3}}$
in (239).

The contributions discussed in this appendix are relevant for the Model I by
Maiani et al. Maiani:2004vq , where such (subdominant) spin-spin interactions
are included in the effective Hamiltonian. We presented the formalism here.
They are neglected in the numerical analysis given in this paper, as we use
the Model II Maiani:2014aja  for our mass estimates.

##  Appendix B Mass derivations for the $P$-wave pentaquark states

###  B.1 The states with $J^{P}=3/2^{+}$ and triquark spin $S_{t}=3/2$

The mass matrix of the states with the spin-parity $J^{P}=3/2^{+}$ and
triquark spin $S_{t}=1/2$ is as follows:

$\displaystyle\tilde{M}^{P1}_{J=3/2}=M_{0}+\frac{1}{2}\,(\mathcal{K}_{cq})_{\bar{3}}+\frac{1}{2}\,(\mathcal{K}_{q^{\prime}q^{\prime\prime}})_{\bar{3}}+\frac{1}{2}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)+B$
$\displaystyle\hskip
48.36967pt+\frac{1}{15}\,A_{t}\left(\begin{array}[]{ccc}25&25\sqrt{2}&0\\\
25\sqrt{2}&-22&9\sqrt{6}\\\
0&9\sqrt{6}&-63\end{array}\right)+\frac{1}{15}\,A_{ld}\left(\begin{array}[]{ccc}-10&25\sqrt{2}&0\\\
25\sqrt{2}&-8&9\sqrt{6}\\\ 0&9\sqrt{6}&-42\end{array}\right)$ (247)
$\displaystyle\hskip
48.36967pt+b\left(\begin{array}[]{ccc}-10/3&\sqrt{2}/4&-3\sqrt{3}/5\\\
\sqrt{2}/4&-1256/375&81\sqrt{6}/100\\\
-3\sqrt{3}/5&81\sqrt{6}/100&-372/125\end{array}\right).$ (251)

Before calculating the mass spectrum, we make the three matrices above to be
traceless:

$\displaystyle\tilde{M}^{P1}_{J=3/2}=M_{0}+\frac{1}{2}\,(\mathcal{K}_{cq})_{\bar{3}}+\frac{1}{2}\,(\mathcal{K}_{q^{\prime}q^{\prime\prime}})_{\bar{3}}+\frac{1}{2}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)+B-\frac{4}{3}\left(A_{t}+A_{ld}\right)-\frac{3622}{1125}\,b$
$\displaystyle\hskip
48.36967pt+\frac{1}{15}\,A_{t}\left(\begin{array}[]{ccc}45&25\sqrt{2}&0\\\
25\sqrt{2}&-2&9\sqrt{6}\\\
0&9\sqrt{6}&-43\end{array}\right)+\frac{1}{15}\,A_{ld}\left(\begin{array}[]{ccc}10&25\sqrt{2}&0\\\
25\sqrt{2}&12&9\sqrt{6}\\\ 0&9\sqrt{6}&-22\end{array}\right)$ (258)
$\displaystyle\hskip
48.36967pt+b\left(\begin{array}[]{ccc}-128/1125&\sqrt{2}/4&-3\sqrt{3}/5\\\
\sqrt{2}/4&-146/1125&81\sqrt{6}/100\\\
-3\sqrt{3}/5&81\sqrt{6}/100&274/1125\end{array}\right).$ (262)

There is a standard procedure to find the eigenvalues, $\lambda_{i}$,
$i=1,\,2,\,3$, of the traceless symmetric $(3\times 3)$ matrix. Denoting such
a matrix as

$A=\left(\begin{array}[]{ccc}A_{11}&A_{12}&A_{13}\\\ A_{12}&A_{22}&A_{23}\\\
A_{13}&A_{23}&A_{33}\\\ \end{array}\right),$ (263)

its characteristic equation

$\left(\lambda-\lambda_{1}\right)\left(\lambda-\lambda_{2}\right)\left(\lambda-\lambda_{3}\right)=\lambda^{3}-I_{1}\lambda^{2}-3I_{2}\lambda-2I_{3}=0,$
(264)

is expressed in terms of three rotational invariants:

$I_{1}=\operatorname{Tr}(A)=A_{11}+A_{22}+A_{33}=\lambda_{1}+\lambda_{2}+\lambda_{3}=0,$
(265)
$I_{2}=\frac{1}{3}\left[A_{12}^{2}+A_{13}^{2}+A_{23}^{2}-A_{11}A_{22}-A_{11}A_{33}-A_{22}A_{33}\right]=-\frac{1}{3}\left(\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3}\right),$
(266)
$I_{3}=\frac{1}{2}\det(A)=\frac{1}{2}\left(A_{11}A_{22}A_{33}+2A_{12}A_{13}A_{23}-A_{11}A_{23}^{2}-A_{22}A_{13}^{2}-A_{33}A_{12}^{2}\right)=\frac{1}{2}\,\lambda_{1}\lambda_{2}\lambda_{3}.$
(267)

Since real eigenvalues are of our interest, two non-trivial invariants,
$I_{2}$ and $I_{3}$, must satisfy the following conditions:

$I_{2}>0,\qquad I_{2}^{3}>I_{3}^{2}.$ (268)

If these conditions are fulfilled, we can define the angle:

$\phi=\frac{1}{3}\,\arccos\sqrt{\frac{I_{3}^{2}}{I_{2}^{3}}},$ (269)

through which the sorted eigenvalues ($\lambda_{1}>\lambda_{2}>\lambda_{3}$)
can be written as follows:

$\lambda_{1}=2\sqrt{I_{2}}\cos(\phi),\qquad\lambda_{2}=-2\sqrt{I_{2}}\cos(\pi/3+\phi),\qquad\lambda_{3}=-2\sqrt{I_{2}}\cos(\pi/3-\phi).$
(270)

After these eigenvalues are obtained, the masses are calculated easily:

$m^{P1}_{11,12,13}=M_{0}+\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}+r_{ld}\right)+B-\frac{4}{3}\left(A_{t}+A_{ld}\right)-\frac{3622}{1125}\,b+\lambda_{3,2,1}.$
(271)

###  B.2 The states with $J^{P}=3/2^{+}$ and triquark spin $S_{t}=1/2$

To derive the expressions for the masses of the four states with the spin-
parity $J^{P}=3/2^{+}$ and triquark spin $S_{t}=1/2$ from table 5, one needs
to find the eigenvalues of the following non-diagonal symmetric $(4\times 4)$
matrix:

$\displaystyle
M^{P1}_{J=3/2}=M_{0}+\frac{1}{2}\,(\mathcal{K}_{cq})_{\bar{3}}\left(\begin{array}[]{cccc}-3&0&0&0\\\
0&-3&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\
\end{array}\right)+\frac{1}{2}\,(\mathcal{K}_{q^{\prime}q^{\prime\prime}})_{\bar{3}}$
(276) $\displaystyle\hskip
31.29802pt+\frac{1}{2}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(\begin{array}[]{cccc}0&0&\sqrt{3}&0\\\
0&0&0&\sqrt{3}\\\ \sqrt{3}&0&-2&0\\\ 0&\sqrt{3}&0&-2\\\
\end{array}\right)+B-\frac{1}{3}\,A_{t}\left(\begin{array}[]{cccc}1&-2\sqrt{5}&0&0\\\
-2\sqrt{5}&2&0&0\\\ 0&0&1&-2\sqrt{5}\\\ 0&0&-2\sqrt{5}&2\end{array}\right)$
(285) $\displaystyle\hskip
31.29802pt+\frac{2}{3}\,A_{ld}\left(\begin{array}[]{cccc}2&\sqrt{5}&0&0\\\
\sqrt{5}&-2&0&0\\\ 0&0&2&\sqrt{5}\\\
0&0&\sqrt{5}&-2\end{array}\right)+\frac{1}{30}\,b\left(\begin{array}[]{cccc}-40&3\sqrt{5}&0&0\\\
3\sqrt{5}&-16&0&0\\\ 0&0&-40&3\sqrt{5}\\\
0&0&3\sqrt{5}&-16\end{array}\right).$ (294)

Noting that the last three matrices are block-diagonal and their non-trivial
$(2\times 2)$ blocks are identical. A diagonalization of these matrices is
reduced to the diagonalization of the blocks:

$\displaystyle-\frac{A_{t}}{3}\left(\begin{array}[]{cc}1&-2\sqrt{5}\\\
-2\sqrt{5}&2\end{array}\right)+\frac{2}{3}\,A_{ld}\left(\begin{array}[]{cc}2&\sqrt{5}\\\
\sqrt{5}&-2\end{array}\right)+\frac{b}{30}\left(\begin{array}[]{cc}-40&3\sqrt{5}\\\
3\sqrt{5}&-16\end{array}\right)$ (301) $\displaystyle\hskip
93.89409pt\Longrightarrow-\frac{A_{t}}{2}-\frac{14}{15}\,b+\left(\begin{array}[]{cc}\mu^{(3/2)}&0\\\
0&-\mu^{(3/2)}\end{array}\right),$ (304)

where the following parameter is introduced:

$\mu^{(3/2)}=\frac{1}{30}\,\sqrt{\left(5A_{t}+40A_{ld}-12b\right)^{2}+5\left(20A_{t}+20A_{ld}+3b\right)^{2}}.$
(305)

Coming back to the $(4\times 4)$ mass matrix (294), it is necessary to point
out that the transformation to the diagonal form of the sum of the last three
matrices in (294) can be done by using the orthogonal matrix:

$a(\alpha)=\left(\begin{array}[]{cccc}\cos\alpha&\sin\alpha&0&0\\\
-\sin\alpha&\cos\alpha&0&0\\\ 0&0&\cos\alpha&\sin\alpha\\\
0&0&-\sin\alpha&\cos\alpha\end{array}\right),$ (306)

with some fixed value of the angle $\alpha$. Under such a transformation, the
two matrices in the first row in (294) remain unchanged. Hence, the $(4\times
4)$ mass matrix (294) can be presented in the form:

$\displaystyle
M^{P1}_{J=3/2}=M_{0}-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}\right)+B-\frac{A_{t}}{2}-\frac{14}{15}\,b+\frac{1}{2}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)$
$\displaystyle\hskip
19.91692pt\times\left(\begin{array}[]{cccc}1-r_{hd}+\tilde{\mu}^{(3/2)}&0&\sqrt{3}&0\\\
0&1-r_{hd}-\tilde{\mu}^{(3/2)}&0&\sqrt{3}\\\
\sqrt{3}&0&r_{hd}-1+\tilde{\mu}^{(3/2)}&0\\\
0&\sqrt{3}&0&r_{hd}-1-\tilde{\mu}^{(3/2)}\end{array}\right),\qquad$ (311)

where
$\tilde{\mu}^{(3/2)}=2\mu^{(3/2)}/(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c})$.
Applying again the orthogonal transformations, the last matrix can be
diagonalized, and we get the following set of masses:

$m^{P1}_{7}=M_{0}+B-\frac{A_{t}}{2}-\frac{14}{15}\,b-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}+2\sqrt{3+(1-r_{hd})^{2}}\right)-\mu^{(3/2)},$
(312)
$m^{P1}_{8}=M_{0}+B-\frac{A_{t}}{2}-\frac{14}{15}\,b-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}+2\sqrt{3+(1-r_{hd})^{2}}\right)+\mu^{(3/2)},$
(313)
$m^{P1}_{9}=M_{0}+B-\frac{A_{t}}{2}-\frac{14}{15}\,b-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}-2\sqrt{3+(1-r_{hd})^{2}}\right)-\mu^{(3/2)},$
(314)
$m^{P1}_{10}=M_{0}+B-\frac{A_{t}}{2}-\frac{14}{15}\,b-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}-2\sqrt{3+(1-r_{hd})^{2}}\right)+\mu^{(3/2)}.$
(315)

###  B.3 The states with $J^{P}=1/2^{+}$ and triquark spin $S_{t}=1/2$

For the first four states with the spin-parity $J^{P}=1/2^{+}$ and triquark
spin $S_{t}=1/2$ from table 5, the mass matrix is again the non-diagonal
symmetric $(4\times 4)$ matrix:

$\displaystyle
M^{P1}_{J=1/2}=M_{0}+\frac{1}{2}\,(\mathcal{K}_{cq})_{\bar{3}}\left(\begin{array}[]{cccc}-3&0&0&0\\\
0&-3&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\
\end{array}\right)+\frac{1}{2}\,(\mathcal{K}_{q^{\prime}q^{\prime\prime}})_{\bar{3}}$
(320) $\displaystyle\hskip
31.29802pt+\frac{1}{2}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(\begin{array}[]{cccc}0&0&\sqrt{3}&0\\\
0&0&0&\sqrt{3}\\\ \sqrt{3}&0&-2&0\\\ 0&\sqrt{3}&0&-2\\\
\end{array}\right)+B+\frac{1}{3}\,A_{t}\left(\begin{array}[]{cccc}2&2\sqrt{2}&0&0\\\
2\sqrt{2}&-5&0&0\\\ 0&0&2&2\sqrt{2}\\\ 0&0&2\sqrt{2}&-5\end{array}\right)$
(329) $\displaystyle\hskip
31.29802pt+\frac{2}{3}\,A_{ld}\left(\begin{array}[]{cccc}-4&\sqrt{2}&0&0\\\
\sqrt{2}&-5&0&0\\\ 0&0&-4&\sqrt{2}\\\
0&0&\sqrt{2}&-5\end{array}\right)+\frac{1}{30}\,b\left(\begin{array}[]{cccc}-16&21\sqrt{2}&0&0\\\
21\sqrt{2}&-46&0&0\\\ 0&0&-16&21\sqrt{2}\\\
0&0&21\sqrt{2}&-46\end{array}\right).$ (338)

The rotational matrix (306) with some fixed value of the angle $\alpha$
transforms the sum of three matrices in the second line to the diagonal form.
As in the case of $J^{P}=3/2^{+}$ pentaquarks, the diagonalization procedure
can be applied to the $(2\times 2)$ blocks:

$\displaystyle\frac{A_{t}}{3}\left(\begin{array}[]{cc}2&2\sqrt{2}\\\
2\sqrt{2}&-5\end{array}\right)+\frac{2}{3}\,A_{ld}\left(\begin{array}[]{cc}-4&\sqrt{2}\\\
\sqrt{2}&-5\end{array}\right)+\frac{b}{30}\left(\begin{array}[]{cc}-16&21\sqrt{2}\\\
21\sqrt{2}&-46\end{array}\right)$ (345) $\displaystyle\hskip
93.89409pt\Longrightarrow-\frac{A_{t}}{2}-3A_{ld}-\frac{31}{30}\,b+\left(\begin{array}[]{cc}\mu^{(1/2)}&0\\\
0&-\mu^{(1/2)}\end{array}\right),$ (348)

where the following parameter is introduced:

$\mu^{(1/2)}=\frac{1}{30}\,\sqrt{25\left(7A_{t}+2A_{ld}+3b\right)^{2}+2\left(20A_{t}+20A_{ld}+21b\right)^{2}}.$
(349)

After this transformation is done, the $(4\times 4)$ mass matrix (338) takes
the form:

$\displaystyle
M^{P1}_{J=1/2}=M_{0}-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}\right)+B-\frac{A_{t}}{2}-3A_{ld}-\frac{31}{30}\,b+\frac{1}{2}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)$
$\displaystyle\hskip
31.29802pt\times\left(\begin{array}[]{cccc}1-r_{hd}+\tilde{\mu}^{(1/2)}&0&\sqrt{3}&0\\\
0&1-r_{hd}-\tilde{\mu}^{(1/2)}&0&\sqrt{3}\\\
\sqrt{3}&0&r_{hd}-1+\tilde{\mu}^{(1/2)}&0\\\
0&\sqrt{3}&0&r_{hd}-1-\tilde{\mu}^{(1/2)}\end{array}\right),$ (354)

where
$\tilde{\mu}^{(1/2)}=2\mu^{(1/2)}/(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c})$.
The last matrix can be diagonalized by using the orthogonal transformations
and we get the following set of masses:

$\displaystyle m^{P1}_{1}=M_{0}+B-\frac{A_{t}}{2}-3A_{ld}-\frac{31}{30}\,b$
$\displaystyle\hskip
31.29802pt-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}+2\sqrt{3+(1-r_{hd})^{2}}\right)-\mu^{(1/2)},$
(355) $\displaystyle
m^{P1}_{2}=M_{0}+B-\frac{A_{t}}{2}-3A_{ld}-\frac{31}{30}\,b$
$\displaystyle\hskip
31.29802pt-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}+2\sqrt{3+(1-r_{hd})^{2}}\right)+\mu^{(1/2)},$
(356) $\displaystyle
m^{P1}_{3}=M_{0}+B-\frac{A_{t}}{2}-3A_{ld}-\frac{31}{30}\,b$
$\displaystyle\hskip
31.29802pt-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}-2\sqrt{3+(1-r_{hd})^{2}}\right)-\mu^{(1/2)},$
(357) $\displaystyle
m^{P1}_{4}=M_{0}+B-\frac{A_{t}}{2}-3A_{ld}-\frac{31}{30}\,b$
$\displaystyle\hskip
31.29802pt-\frac{1}{4}\left(\mathcal{K}_{\bar{c}q}+\mathcal{K}_{\bar{c}c}\right)\left(2+r_{hd}-r_{ld}-2\sqrt{3+(1-r_{hd})^{2}}\right)+\mu^{(1/2)}.$
(358)

##  Appendix C $\chi^{2}$-analysis of orbitally-excited
$\Omega_{c}^{*}$-baryons

Existing experimental data on the masses of orbitally-excited
$\Omega_{c}^{*}$-baryons were obtained by the LHCb Aaij:2017nav  and Belle
Yelton:2017qxg  collaborations and collected in table 15.

Baryon |  $J^{P}$ Karliner:2017kfm  | LHCb | Belle
---|---|---|---
$\Omega_{c}(3000)$ | $1/2^{-}$ | $3000.4\pm 0.2\pm 0.1$ | $3000.7\pm 1.0\pm 0.2$
$\Omega_{c}(3050)$ | $1/2^{-}$ | $3050.2\pm 0.1\pm 0.1$ | $3050.2\pm 0.4\pm 0.2$
$\Omega_{c}(3066)$ | $3/2^{-}$ | $3065.5\pm 0.1\pm 0.3$ | $3064.9\pm 0.6\pm 0.2$
$\Omega_{c}(3090)$ | $3/2^{-}$ | $3090.2\pm 0.3\pm 0.5$ | $3089.3\pm 1.2\pm 0.2$
$\Omega_{c}(3119)$ | $5/2^{-}$ | $3119.0\pm 0.3\pm 0.9$ | $\cdots$
Table 15:  Experimental data from the LHCb Aaij:2017nav  and Belle
Yelton:2017qxg  collaborations. (Masses are given in MeV.) All the
measurements have the uncertainty ${}^{+0.3}_{-0.5}$ MeV coming from the mass
of the ground-state $\Xi_{c}^{+}$-baryon, which is common in both analysis.
The spin-parity, $J^{P}$, assignment assumes that all the states are
orbitally-excited baryons as suggested in Karliner:2017kfm .

The theoretical expressions for the orbitally-exciting $\Omega_{c}^{*}$-baryon
masses follow from the effective Hamiltonian Karliner:2017kfm 888 The
coefficients of the spin-orbit, $a_{1}$ and $a_{2}$, and spin-spin, $c$,
interactions in the spin-dependent part of the Hamiltonian, $V_{\rm SD}$,
differ by a factor 2 from the ones defined in Karliner:2017kfm ; Ali:2017wsf
.:

$\displaystyle H_{\rm eff}$ $\displaystyle=$ $\displaystyle
m_{c}+m_{\\{ss\\}}+2\kappa_{ss}\left({\bf S}_{s_{1}}\cdot{\bf
S}_{s_{2}}\right)+\frac{B_{\mathcal{Q}}}{2}\,{\bf L}^{2}+V_{\rm SD},$ (359)
$\displaystyle V_{\rm SD}$ $\displaystyle=$ $\displaystyle 2a_{1}\left({\bf
L}\cdot{\bf S}_{\\{ss\\}}\right)+2a_{2}\left({\bf L}\cdot{\bf
S}_{c}\right)+b\,\frac{S_{12}}{4}+2c\left({\bf S}_{\\{ss\\}}\cdot{\bf
S}_{c}\right).$

In eq. (359), $m_{c}$ and $m_{\\{ss\\}}$ are the masses of the $c$-quark and
the spin-1 $\\{ss\\}$-diquark, respectively, $\kappa_{ss}$ is the spin-spin
coupling of the quarks in the diquark, ${\bf L}$ is the orbital angular
momentum of the $\Omega_{c}^{*}$-baryon, and $B_{\mathcal{Q}}$ is the orbital
coupling. The coefficients $a_{1}$ and $a_{2}$ are the strengths of the spin-
orbit terms involving the spin of the diquark ${\bf S}_{\\{ss\\}}$ and the
charm-quark spin ${\bf S}_{c}$, respectively, $c$ is the strength of the spin-
spin interaction between the diquark and the charm quark, and $S_{12}/4$
represents the tensor interaction, defined by

$\frac{S_{12}}{4}=Q({\bf S}_{\\{ss\\}},{\bf S}_{c})=3\left({\bf
S}_{\\{ss\\}}\cdot{\bf n}\right)\left({\bf S}_{c}\cdot{\bf
n}\right)-\left({\bf S}_{\\{ss\\}}\cdot{\bf S}_{c}\right),$ (360)

where ${\bf n}={\bf r}/r$ is the unit vector in the direction from charged
quark to the doubly-strange diquark.

One should remember that several required internal parameters are fixed:
charm-quark spin $S_{c}=1/2$, diquark spin $S_{\\{ss\\}}=1$, and
$\Omega_{c}^{*}$-baryon orbital angular momentum $L=1$. So, in the $L-S$
coupling scheme, two possible values of the total spin $S=1/2$ and $S=3/2$,
after their coupling to $L$, allow to get five $P$-wave states: two with the
total angular momentum $J=1/2$, two with $J=3/2$ and the last one with $J=5/2$
(see table 15). In this scheme the spin-dependent part, $V_{\rm SD}$, is
represented by the block-diagonal matrix, each block of which corresponds to
the states with a fixed value of $J$ Karliner:2017kfm :

$\Delta\mathcal{M}_{1/2}=\frac{1}{3}\left(\begin{array}[]{cc}2\left(a_{2}-4a_{1}\right)&2\sqrt{2}\left(a_{2}-a_{1}\right)\\\
2\sqrt{2}\left(a_{2}-a_{1}\right)&-5\left(2a_{1}+a_{2}\right)\end{array}\right)+\frac{b}{\sqrt{2}}\left(\begin{array}[]{rr}0&1\\\
1&-\sqrt{2}\end{array}\right)+c\left(\begin{array}[]{rr}-2&0\\\
0&1\end{array}\right),$ (361)
$\Delta\mathcal{M}_{3/2}=\frac{1}{3}\left(\begin{array}[]{cc}4a_{1}-a_{2}&2\sqrt{5}\left(a_{2}-a_{1}\right)\\\
2\sqrt{5}\left(a_{2}-a_{1}\right)&-2\left(2a_{1}+a_{2}\right)\end{array}\right)+\frac{b}{10}\left(\begin{array}[]{rr}0&-\sqrt{5}\\\
-\sqrt{5}&8\end{array}\right)+c\left(\begin{array}[]{rr}-2&0\\\
0&1\end{array}\right),$ (362)
$\Delta\mathcal{M}_{5/2}=2a_{1}+a_{2}-\frac{b}{5}+c.$ (363)

After diagonalizing the matrices and adding the common mass term

$M_{0}^{(\Omega_{c})}=m_{c}+m_{\\{ss\\}}+\frac{\kappa_{ss}}{2}+B_{\mathcal{Q}},$
(364)

we get the set of five mass formulae:

$\displaystyle
m^{(1/2)}_{1}=M_{0}^{(\Omega_{c})}-\frac{1}{2}\left(6a_{1}+a_{2}+b+c\right)$
$\displaystyle\hskip
31.29802pt-\frac{1}{6}\sqrt{\left(2a_{1}+7a_{2}+3b-9c\right)^{2}+2\left(4a_{1}-4a_{2}-3b\right)^{2}},$
(365) $\displaystyle
m^{(1/2)}_{2}=M_{0}^{(\Omega_{c})}-\frac{1}{2}\left(6a_{1}+a_{2}+b+c\right)$
$\displaystyle\hskip
31.29802pt+\frac{1}{6}\sqrt{\left(2a_{1}+7a_{2}+3b-9c\right)^{2}+2\left(4a_{1}-4a_{2}-3b\right)^{2}},$
(366) $\displaystyle
m^{(3/2)}_{1}=M_{0}^{(\Omega_{c})}-\frac{1}{10}\left(5a_{2}-4b+5c\right)$
$\displaystyle\hskip
31.29802pt-\frac{1}{30}\sqrt{\left(40a_{1}+5a_{2}-12b-45c\right)^{2}+5\left(20a_{1}-20a_{2}+3b\right)^{2}},$
(367) $\displaystyle
m^{(3/2)}_{2}=M_{0}^{(\Omega_{c})}-\frac{1}{10}\left(5a_{2}-4b+5c\right)$
$\displaystyle\hskip
31.29802pt+\frac{1}{30}\sqrt{\left(40a_{1}+5a_{2}-12b-45c\right)^{2}+5\left(20a_{1}-20a_{2}+3b\right)^{2}},$
(368) $\displaystyle
m^{(5/2)}=M_{0}^{(\Omega_{c})}+2a_{1}+a_{2}-\frac{b}{5}+c.$ (369)

There are five unknown variables
$\\{M_{0}^{(\Omega_{c})},\,a_{1},\,a_{2},\,b,\,c\\}$ in these five equations.
If there are five experimentally measured states which can be assigned with
the $P$-wave $\Omega_{c}$-baryons, all five variables, in general, can be
determined but their values do not always satisfy physical requirements.

After the assignment is done as specified in table 15, it is possible to
perform a $\chi^{2}$-analysis of the data. We start from the LHCb data alone
as all five states are present there and then add the Belle measurements which
allow to find $\chi^{2}_{\rm min}$ for four degrees of freedom. The best-fit
values and $1\sigma$ uncertainties of free parameters are presented in table
16. There are no free degrees of freedom for the fit based on the LHCb data
alone (the first row in table 16) but combining them with the Belle data
results $\chi^{2}_{\rm min}/{\rm ndf}=0.87/4$.

| $M_{0}^{(\Omega_{c})}$ | $a_{1}$ | $a_{2}$ | $b$ | $c$
---|---|---|---|---|---
LHCb | $3079.89\pm 0.40$ | $13.47\pm 0.14$ | $12.86\pm 0.38$ | $13.48\pm 0.54$ | $2.00\pm 0.22$
Combined | $3079.80\pm 0.39$ | $13.45\pm 0.13$ | $12.94\pm 0.36$ | $13.30\pm 0.48$ | $2.01\pm 0.20$
Table 16:  The $\chi^{2}$-analysis of the orbitally-excited
$\Omega_{c}^{*}$-baryon masses based on the measurements by the LHCb and Belle
collaborations. This analysis is performed based on the LHCb data alone and on
the combined data set from both the LHCb and Belle collaborations presented in
table 15. All values of parameters are given in MeV. The combine fit results
$\chi^{2}_{\rm min}/{\rm ndf}=0.87/4$.

The correlation matrices in both cases are as follows:

$R_{\rm LHCb}=\left(\begin{array}[]{rrrrr}1&0.71&0.43&-0.32&0.37\\\
&1&0.21&-0.31&0.08\\\ &&1&-0.73&0.53\\\ &&&1&-0.18\\\
&&&&1\end{array}\right),\quad R_{\rm
comb}=\left(\begin{array}[]{rrrrr}1&0.77&0.53&-0.44&0.45\\\
&1&0.33&-0.38&0.18\\\ &&1&-0.75&0.62\\\ &&&1&-0.32\\\
&&&&1\end{array}\right).$ (370)

The common mass $M_{0}^{(\Omega_{c})}$ is determined quite precisely. With the
values of the input parameters from subsec. 5.1: $m_{c}=(1710\pm 10)$ MeV,
$m_{\\{ss\\}}=(1099\pm 15)$ MeV, and $\mathcal{K}_{ss}=(23\pm 2)$ MeV, the
strength of the orbital interaction can be estimated as:

$B_{\mathcal{Q}}=(259\pm 18)~{}\mathrm{MeV}.$ (371)

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