# Properties of doubly heavy baryons in QCD

T. M. Aliev<EMAIL_ADDRESS>Department of Physics, Middle East Technical
University, Ankara, 06800, Turkey  S. Bilmis  Department of Physics, Middle
East Technical University, Ankara, 06800, Turkey  TUBITAK ULAKBIM, Ankara,
06510, Turkey

###### Abstract

The study of the properties of doubly heavy baryons represents a promising
area in particle physics. It can provide us with information about
Cabibbo–Kobayashi–Maskawa (CKM) matrix elements and the low energy dynamics of
QCD. They have a very rich phenomenology. The investigation of weak,
electromagnetic, and strong decays has a vital importance for understanding
the dynamics of doubly heavy baryons. The main ingredients of such studies are
the spectroscopic parameters, the strong coupling constants, and the
transition form factors. For calculations of these quantities, non-
perturbative methods are needed. One of these methods is the QCD sum rules. In
the present work, we review our studies on the properties of the doubly heavy
baryons within the sum rules method, focusing mainly on the mass and strong
coupling constants of the doubly heavy baryons. In addition, the radiative
decays of doubly heavy baryons are estimated using the vector dominance model.
We also make few remarks on the semileptonic decays of the doubly heavy
baryons.

##  I Introduction

The constituent quark model has been a powerful tool for studying the hadron
spectroscopy [1]. The model predicts the existence of hadrons containing
single, double, and triple heavy quarks. Even though many baryons with single
heavy quarks (even the excited states) like $\Omega_{b}$, $\Xi_{b}$,
$\Sigma_{b}$, $\Lambda_{b}$, $\Omega_{c}$, $\Xi_{c}$, and $\Lambda_{c}$ have
been discovered by several flavor and accelerator experiments like CDF, CLEO,
BABAR, BELLE and LHCb [2], search for the baryons involving two heavy quarks
ended with no observation [3, 4, 5] for a long time. This has been a long-
standing puzzle for the quark model. In 2002, SELEX collaboration made a
breakthrough with the discovery of $\Xi_{ccd}^{+}(3520)$ in $pD^{+}K^{-}$
decay channel [6] and the confirmation was also done by the same experiment in
2005 [7]. Moreover, in 2017, LHCb collaboration announced the discovery of
$\Xi_{ccu}^{++}(3621)$ via
$\Xi_{c}^{++}\rightarrow\Lambda_{c}^{+}K^{-}\pi^{+}\pi^{+}$ decay [8]. This
state was also confirmed by LHCb collaboration in 2018 in the decay
$\Xi_{cc}^{++}\rightarrow\Xi_{c}^{+}\pi^{+}$ with $5.9\sigma$ statistical
significance [9]. Experimental studies have been conducted to unveil the
characteristic properties of this new baryon [10, 11, 12] as well as searching
new doubly heavy baryons [13].

Many theoretical studies inspired by these discoveries also have been carried
out to determine the characteristic properties like the masses, lifetimes,
strong coupling constants, and decay widths of doubly heavy baryons by
investigating the weak, electromagnetic, and strong decays of these baryons
within various models [14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 26,
27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45,
46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60]. Investigating the
possible decays of these baryons can give crucial information about the
dynamics of these processes. While the transition form factors can be
determined by studying the weak and electromagnetic decays, coupling constants
can be calculated for analyzing the strong decays.

In the quark model, the baryons are described by quark-diquark picture.
Diquarks form a system either with spin-zero or spin-$1$. For this reason, the
total spin of the diquark system may be either spin $1/2$ or $3/2$. The
lowest-lying doubly heavy baryons with $J^{P}=1/2^{+}$ form the SU$(3)$
triplet $\Xi^{++}_{QQ}(QQu)$, $\Xi_{QQ}(QQd)$, and $\Omega_{QQ}(QQs)$. The
bottom-charm baryons form two sets of SU$(3)$ triplets
($\Xi_{bc}$,$\Omega_{bc}$) and ($\Xi^{\prime}_{bc}$,$\Omega^{\prime}_{bc}$).
The difference between these sets is that while the total spin of the $bc$
system is equal to one for the first set, it is equal to zero for the second
case. It is also possible that mixing exists between these two sets. The
doubly heavy baryons $J^{P}=3/2^{+}$ can decay into the lowest-lying ones
either radiatively or with pion emission. In Table 1, the baryons with two
heavy quarks and with corresponding quantum numbers are represented.

The weak decays of doubly heavy baryons to the single heavy baryons can be
described by spin-$1/2$ to spin-$1/2$, spin-$1/2$ to spin-$3/2$, spin-$3/2$ to
spin-$1/2$, and spin-$3/2$ to spin-$3/2$ transitions.

It is a well-known fact that weak transitions take place via charged or
neutral currents.

  * •

The spin-$1/2$ to spin-$1/2$ ($3/2$) transition with charged currents is
described by $c\rightarrow d,s$ or $b\rightarrow u,c$ processes.

  * •

The spin-$1/2\rightarrow$ spin-$1/2$ ($3/2$) transition via flavour changing
neutral current (FCNC) is described by $c\rightarrow u$ or $b\rightarrow d,s$
transitions.

All these processes are described by the transition matrix elements between
initial and final states. The formation of hadrons from quarks takes place at
a large distance which belongs to the nonperturbative domain of QCD. For this
reason, for describing these transitions, we need some nonperturbative
methods. Among all nonperturbative methods, the sum rules method occupies a
special place because it is based on QCD Lagrangian.

Baryon | $J^{P}$ | Baryon | $J^{P}$
---|---|---|---
$\Xi_{QQ}$ | $1/2^{+}$ | $\Omega_{QQ}$ | $1/2^{+}$
$\Xi^{*}_{QQ}$ | $3/2^{+}$ | $\Omega^{*}_{QQ}$ | $3/2^{+}$
$\Xi^{\prime}_{bc}$ | $1/2^{+}$ | $\Omega^{\prime}_{bc}$ | $1/2^{+}$
$\Xi_{bc}$ | $1/2^{+}$ | $\Omega_{bc}$ | $1/2^{+}$
$\Xi^{*}_{bc}$ | $3/2^{+}$ | $\Omega^{*}_{bc}$ | $3/2^{+}$
Table 1: Baryons containing two heavy quarks with corresponding quantum
numbers are represented. Baryons with prime means that this state is
antisymmetric with respect to the flavors, and diquark has spin-zero.

The semileptonic decays of baryons with two heavy quarks take place through
the weak current. The weak decays of a doubly heavy baryon to a single heavy
baryon transition within the QCD sum rules [60], the light cone version of sum
rules [61, 62] as well as in the framework of the light-front quark model [37,
30, 29] are comprehensively studied.

In this work, we review our studies [61, 14, 16, 17, 63, 64, 65] where we
estimate the mass and decay constants as well as strong coupling constants of
the doubly heavy baryons with light mesons. Moreover, radiative widths are
also estimated by using the vector-meson dominance hypothesis. The paper is
organized as follows. In Section II, we briefly discuss the sum rules method.
In Section III, we represent the details of the calculations for the mass and
decay constants within QCD sum rules. In Section IV, we discuss the
determination of the strong coupling constants for the doubly heavy baryons
with light pseudo-scalar and vector mesons. Section V is devoted to the our
estimation of the radiative decays of doubly heavy baryons by using the vector
dominance model and the last section contains our summary and conclusion.

##  II Sum rules method

The QCD sum rules were first proposed in [66] for mesons and later generalized
to baryons in [67]. This method is powerful for studying the properties of
hadrons. Various aspects of this method are discussed in many review papers
(see for example [68] and [69]). The main advantage of the sum rule is that it
is based on the fundamental QCD Lagrangian. Investigating the bound state
problem in QCD, this method starts with short distance relations and then
includes the nonperturbative effects of QCD vacuum to probe the large
distance. The correlation function is the essence of the sum rules. It is
constructed in terms of the interpolating currents, which carry the same
quantum numbers as the hadrons under consideration. The mass and decay
constants of the hadrons are usually studied within the so-called two-point
correlation function. On the other hand, the strong coupling constants and
form factors are investigated within three-point sum rules.

A generic two and three-point correlation function can be written as

$\displaystyle\Pi(p)$ $\displaystyle=i\int
d^{4}xe^{ipx}\bra{0}\mathcal{T}\left\\{j(x)j^{\dagger}(0)\right\\}\ket{0}$ (1)
$\displaystyle\Pi(p,q)$ $\displaystyle=i^{2}\int
d^{4}xd^{4}ye^{ipx+iqy}\bra{0}\mathcal{T}\left\\{j(x)j(0)j^{\dagger}(y)\right\\}\ket{0}~{}.$
(2)

Moreover, the following correlation function is introduced in so-called light-
cone sum rules (LCSR).

$\Pi(p)=i\int
d^{4}xe^{ipx}\bra{0}\mathcal{T}\left\\{j(x)j^{\dagger}(0)\right\\}\ket{M(q)}~{},$
(3)

where $M$ is the corresponding meson, baryon, or photon.

The LCSR was introduced in [70] in order to solve some drawbacks of three-
point sum rules (see [71]). The correlation function is calculated in terms of
hadrons and quark-gluon degrees of freedom. Matching these two
representations, one can obtain the desired sum rules, which allow us to
express the properties of hadrons in terms of the quarks and gluons. From the
QCD side, the correlation function is calculated with the help of operator
product expansion (OPE), which can be represented as a multiplication of the
so-called Wilson coefficients and vacuum expectation values of the local
operators. In two and three-point sum rules, the OPE is carried out with
respect to the dimensions of the local operators.

In light cone sum rules, the OPE is carried out with respect to the twist,
which is equal to the difference of the dimension of operators and their
spins. In this method, instead of local condensates, which is the main
ingredient of the traditional QCD sum rules, they appear as distribution
amplitudes (DAs). These DAs contain all the information about the
nonperturbative effects.

##  III Calculation of the mass and decay constants

The mass and decay constants of the baryons can be studied in the framework of
the two-point sum rules. For this purpose, we introduce the following two-
point sum rules.

$\Pi(p)=i\int
d^{4}xe^{ipx}\bra{0}\mathcal{T}\left\\{j(x)j^{\dagger}(0)\right\\}\ket{0}$ (4)

To calculate this correlation function, we need to know the form of
interpolating current. For the spin-$\nicefrac{{1}}{{2}}$ double heavy baryon,
we now describe the details of the derivation of the form of the interpolating
current. Baryons in quark model is described with the help of diquark-quark
picture. A diquark is represented by two heavy quarks. Hence, the
interpolating field of the diquark should be similar to the interpolating
current for a meson, i.e.

$J_{\text{meson}}=\bar{q}\Gamma q~{},$ (5)

where $\Gamma$ is the arbitrary Dirac matrix. In the quark model, baryons
consist of three quarks. Therefore $\bar{q}$ should be replaced by its charge-
conjugation. The charge-conjugated state is related to the $\bar{q}$ as

$q^{C}=C\bar{q}^{T}~{},$ (6)

where $C$ is the charge conjugation matrix. From the above equation, one gets

$\bar{q}=q^{T}C~{}.$ (7)

As a result, the interpolating field for a diquark can be written as

$J_{diquark}=q^{T}Cq~{}.$ (8)

The interpolating current for the doubly heavy baryons is constructed by
adding a light quark field to the diquark composed of two heavy quarks.
Besides, the interpolating current must also be a color singlet. Hence, the
general form of the interpolating current for doubly heavy baryons can be
written in the following form,

$J=\varepsilon^{abc}\left(Q^{aT}C\Gamma_{1}Q^{b}\right)\Gamma_{2}q^{c}~{},$
(9)

where $a$, $b$, $c$ are color indices and $\Gamma_{1}$, $\Gamma_{2}$ are the
Dirac matrices from the full set
$\Gamma_{i}=\mathbb{1},\gamma_{5},\gamma_{\mu},\gamma_{5}\gamma_{\mu},\sigma_{\mu\nu}$.
Having found the form of the interpolating current, the next problem is the
determination of $\Gamma_{1}$ and $\Gamma_{2}$ matrices. First, let us
consider the diquark part of the interpolating current

$\varepsilon^{abc}\left(Q^{aT}C\Gamma_{1}Q^{b}\right)~{}.$ (10)

After taking the transpose of this equation, we get

$\left(\varepsilon^{abc}Q^{aT}C\Gamma_{1}Q^{b}\right)^{T}=-\varepsilon^{abc}Q^{bT}\Gamma_{1}^{T}C^{T}Q^{a}~{},$
(11)

where the fermion nature of the quark fields are taken into account. Since
$C^{T}=C^{-1}=C^{\dagger}=-C$, we obtain

$\left(\varepsilon^{abc}Q^{aT}C\Gamma_{1}Q^{b}\right)^{T}=\varepsilon^{abc}Q^{bT}C\left(C\Gamma_{1}^{T}C^{-1}\right)Q^{a}~{}.$
(12)

Moreover, using the relation

$C\Gamma_{1}C^{-1}=\begin{cases}\Gamma_{1}&\text{for
}\Gamma_{1}={\mathbb{1},\gamma_{5},\gamma_{5}\gamma_{\mu}}\\\
-\Gamma_{1}&\text{for }\Gamma_{1}={\gamma_{\mu},\sigma_{\mu\nu}}\\\
\end{cases}$ (13)

we get the following form for the diquark part of the interpolating current

$\displaystyle\left(\varepsilon^{abc}Q^{aT}C\Gamma_{1}Q^{b}\right)^{T}=\begin{cases}-\varepsilon^{abc}Q^{aT}C\Gamma_{1}Q^{b}&\text{for
}\mathbb{1},\gamma_{5},\gamma_{5}\gamma_{\mu}\\\
\varepsilon^{abc}Q^{aT}C\Gamma_{1}Q^{b}&\text{for
}\gamma_{\mu},\sigma_{\mu\nu}~{}.\\\ \end{cases}$ (14)

Since left side of this equation is represented by a $1\times 1$ transposed
matrix, it should be equal to itself. Hence,
$\Gamma_{1}=\mathbb{1},\gamma_{5}\text{ and }\gamma_{5}\gamma_{\mu}$
structures are ruled out.

Second, let us consider the spin of the baryons. For the baryon in diquark
quark picture a light quark with spin $S=\nicefrac{{1}}{{2}}$ and
$S_{z}=\pm\nicefrac{{1}}{{2}}$ is attached to the diquark. Since we consider
the doubly heavy baryon with $J=\nicefrac{{1}}{{2}}$, the diquark part should
have zero spin. This implies that either $\Gamma_{1}=\mathbb{1}$ or
$\Gamma_{1}=\gamma_{5}$. This fact allows us to write the interpolating
current in two possible forms

$\displaystyle j_{1}$
$\displaystyle=\varepsilon^{abc}\left(Q^{aT}Cq^{b}\right)\Gamma_{3}Q^{c}$ (15)
$\displaystyle j_{2}$
$\displaystyle=\varepsilon^{abc}\left(Q^{aT}C\gamma_{5}q^{b}\right)\Gamma_{4}Q^{c}~{}.$

The form of the $\Gamma_{3}$ and $\Gamma_{4}$ can be determined by analyzing
the Lorentz structures and parity consideration. The currents $j_{1}$ and
$j_{2}$ must be a Lorentz scalar. This leads that $\Gamma_{3}$ and
$\Gamma_{4}$ must be either $\mathbb{1}\text{ or }\gamma_{5}$.

To determine the final form of $\Gamma_{3}$ and $\Gamma_{4}$, let us consider
the parity transformation. Applying the parity transformation to each quark
field, we get

$\displaystyle j_{1}^{\prime}$
$\displaystyle=\varepsilon^{abc}\left[\left(\gamma_{0}Q^{a}\right)^{T}C\gamma_{0}q^{b}\right]\Gamma_{3}\gamma_{0}Q^{c}=\varepsilon^{abc}\left(Q^{aT}Cq^{b}\right)\Gamma_{3}\gamma_{0}Q^{c}$
(16) $\displaystyle j_{2}^{\prime}$
$\displaystyle=\varepsilon^{abc}\left[\left(\gamma_{0}Q^{a}\right)^{T}C\gamma_{0}q^{b}\right]\Gamma_{3}\gamma_{0}Q^{c}=\varepsilon^{abc}\left(Q^{aT}C\gamma_{5}q^{b}\right)\Gamma_{4}\gamma_{0}Q^{c}~{}.$

On the other side, applying the parity transformation to the interpolating
currents, we obtain

$\displaystyle j_{1}^{\prime}$
$\displaystyle=\varepsilon^{abc}\left(Q^{aT}Cq^{b}\right)\gamma_{0}\Gamma_{3}Q^{c}$
(17) $\displaystyle j_{2}^{\prime}$
$\displaystyle=\varepsilon^{abc}\left(Q^{aT}C\gamma_{5}q^{b}\right)\gamma_{0}\Gamma_{4}Q^{c}~{}.$

Comparing Eqs. (16) and (17), we find

$\displaystyle-\Gamma_{3}\gamma_{0}$ $\displaystyle=\gamma_{0}\Gamma_{3}$ (18)
$\displaystyle\Gamma_{4}\gamma_{0}$ $\displaystyle=\gamma_{0}\Gamma_{4}~{}.$

From these results, we get $\Gamma_{3}=\gamma_{5}$ and
$\Gamma_{4}=\mathbb{1}$. Hence, we obtain the two possible forms of the
interpolating current of double heavy baryon with spin-$\nicefrac{{1}}{{2}}$

$\displaystyle j_{1}$
$\displaystyle=\varepsilon^{abc}\left(Q^{aT}Cq^{b}\right)\gamma_{5}Q^{c}$ (19)
$\displaystyle j_{2}$
$\displaystyle=\varepsilon^{abc}\left(Q^{aT}C\gamma_{5}q^{b}\right)Q^{c}~{}.$

The linear combination also describes the interpolating current of a
spin-$\nicefrac{{1}}{{2}}$ doubly heavy baryon, i.e.

$J=2\varepsilon^{abc}\left\\{\left(Q^{aT}Cq^{b}\right)\gamma_{5}Q^{c}+\beta\left(Q^{aT}C\gamma_{5}q^{b}\right)Q^{c}\right\\},$
(20)

where $\beta$ is an arbitrary parameter.

By using the Fierz expansion, this current can be written as

$\displaystyle
J=\frac{1-\beta}{2}\varepsilon^{abc}\left(Q^{aT}C\gamma_{\mu}Q^{b}\right)\gamma_{5}\gamma^{\mu}q^{c}+\frac{1+\beta}{4}\varepsilon^{abc}\left(Q^{aT}C\sigma_{\mu\nu}Q^{b}\right)\gamma_{5}\sigma^{\mu\nu}q^{c}~{}.$
(21)

The $\beta=-1$ case is called the Ioffe current in the literature. This form
of the current was introduced by Ioffe in [67] to study the mass of ordinary
baryons within QCD sum rules. In order to construct the sum rules, the
correlation function is calculated in two different regions, namely, in terms
of hadrons (so-called hadronic side) and in terms of quark and gluons in deep
Euclidean region (so-called theoretical part).

Having determined the form of the interpolating current for the baryon in
diquark-quark picture, let us calculate the correlation function from the
hadronic side. The correlation function from the hadronic side can be
calculated with the help of the dispersion relation, which connects it by its
imaginary part

$\Pi(p^{2})=-\int ds\frac{\rho(s)}{p^{2}-s+i\varepsilon}+\text{ ...}~{},$ (22)

where dots denote the possible subtraction terms. The term $\rho(s)$ is called
the spectral density and is equal to the imaginary part of $\Pi$, i.e.
$\rho(s)=\imaginary\Pi(s)$. The main reason for the dispersion relation being
used instead of $\Pi$ itself is that calculations are easier.

Now let us turn our attention to the calculation of the correlation function
from QCD side. The correlation function is calculated in deep Euclidean region
with the help of operator product expansion (OPE). To calculate the spectral
density, we follow the procedure below. Let us consider the correlation
function

$\Pi\left(p^{2}\right)=\int\frac{\rho(s)}{s-p^{2}}ds~{}.$ (23)

As a first step, we take the Borel transformation of both sides with respect
to the variable $p^{2}$, and we get

$\Pi^{\mathcal{B}}=\int\rho(s)e^{\frac{-s}{M^{2}}}ds~{}.$ (24)

Denoting $\frac{1}{M^{2}}$ by $\tau$ and performing Borel transformation over
the variable $\tau$ by using

$\lim_{\tau\to
M^{-2}}\mathcal{B}e^{-s\tau}=\delta\left(\frac{1}{M^{2}}-s\right)$ (25)

and integrating over $s$, we get $\rho=\tilde{\tilde{\Pi}}$, in which
$\tilde{\tilde{\Pi}}$ means double Borel transformation. This observation is
essential for the calculation of spectral density, i.e. the spectral density
$\rho$ can be obtained by applying double Borel transformation to the
correlation function. With this information, we present the spectral density
calculation for one term as an example here. Using Eqs. (4) and (21) as well
as Wick’s theorem for the correlation function we get

$\begin{split}\Pi\left(p^{2}\right)=&i\varepsilon^{abc}\varepsilon^{a^{\prime}b^{\prime}c^{\prime}}\displaystyle\int
d^{4}xe^{ipx}\times\\\
\bra{0}\Bigg{\\{}&-\gamma_{5}S^{cb^{\prime}}_{Q}(x)\widetilde{S}^{ba^{\prime}}_{q}(x)S^{ac^{\prime}}_{Q}(x)\gamma_{5}+\gamma_{5}S^{cc^{\prime}}_{Q}(x)\gamma_{5}\Tr[S^{ab^{\prime}}_{Q}(x)\widetilde{S}^{ba^{\prime}}_{q}(x)]\\\
+&\beta\Bigg{[}-\gamma_{5}S^{cb^{\prime}}_{Q}(x)\gamma_{5}\widetilde{S}^{ba}_{q}(x)S^{ac^{\prime}}_{Q}(x)-S^{cb^{\prime}}_{Q}(x)\widetilde{S}^{ba^{\prime}}_{q}(x)\gamma_{5}S^{ac^{\prime}}_{Q}(x)\gamma_{5}\\\
+&\gamma_{5}S^{cc^{\prime}}_{Q}(x)\Tr[S^{ab^{\prime}}_{Q}(x)\gamma_{5}S^{ab^{\prime}}_{Q}(x)\gamma_{5}\widetilde{S}^{ba^{\prime}}_{q}(x)]+S^{cc^{\prime}}_{Q}(x)\gamma_{5}\Tr
S^{ab^{\prime}}_{Q}(x)\widetilde{S}^{ba^{\prime}}_{q}(x)\gamma_{5}]\Bigg{]}\\\
+&\beta^{2}\Bigg{[}-S^{cb^{\prime}}_{Q}(x)\gamma_{5}\widetilde{S}^{ba^{\prime}}_{q}(x)\gamma_{5}S^{ac^{\prime}}_{Q}(x)+S^{cc^{\prime}}_{Q}(x)\Tr[S^{ba^{\prime}}_{q}(x)\gamma_{5}\widetilde{S}^{ab^{\prime}}_{Q}(x)\gamma_{5}]\Bigg{]}\Bigg{\\}}\ket{0}~{},\end{split}$
(26)

where $\widetilde{S}=CS^{T}C$. The expressions of the light ($S_{q}$) and
heavy quark ($S_{Q}$) propagators are needed to calculate the spectral
density. These expressions are

$\displaystyle S_{q}=$
$\displaystyle\frac{i\not{x}}{2\pi^{2}x^{4}}-\frac{m_{q}}{4\pi^{2}x^{2}}-\frac{\expectationvalue{\bar{q}q}}{12}\left(1-\frac{im_{q}}{4}\not{x}\right)-\frac{x_{0}^{2}}{192}m_{0}^{2}\expectationvalue{\bar{q}q}\left(1-\frac{im_{q}}{6}\not{x}\right)$
(27) $\displaystyle S_{Q}=$
$\displaystyle\frac{im_{Q}^{2}}{\left(2\pi\right)^{2}}\left[i\not{x}\frac{K_{2}\left(m_{Q}\sqrt{-x^{2}}\right)}{\left(\sqrt{-x^{2}}\right)^{2}}+\frac{K_{1}\left(m_{Q}\sqrt{-x^{2}}\right)}{\sqrt{-x^{2}}}\right]~{},$

where $K_{i}$ are the modified Bessel functions of the second kind.

Having the expressions for light and heavy quark propagators, we would like to
present the details of the calculations of the spectral density for one of the
term above and consider the following generic term as an example

$I=\displaystyle\int
d^{4}xe^{ipx}\frac{1}{\left(x^{2}\right)^{n_{1}}}\frac{K_{n_{2}}\left(m_{Q}\sqrt{-x^{2}}\right)}{\left(\sqrt{-x^{2}}\right)^{n_{2}}}\frac{K_{n_{3}}\left(m_{Q}\sqrt{-x^{2}}\right)}{\left(\sqrt{-x^{2}}\right)^{n_{3}}}~{}.$
(28)

To perform the calculations, we need to transform from Minkowskian space to
the Euclidean space, i.e. $-x^{2}\rightarrow x_{E}^{2}$,
$d^{4}x\rightarrow-d^{4}x_{E}$, and $ipx\rightarrow-ip_{E}x_{E}$. Then using
the Schwinger representation,

$\frac{1}{\left(x_{E}^{2}\right)^{n}}=\frac{1}{\Gamma(n)}\displaystyle\int_{0}^{\infty}dt\,t^{n-1}\,e^{-t_{1}x_{E}^{2}}$
(29)

and integral representation for the Bessel function

$\frac{K_{n}\left(m_{Q}\sqrt{-x^{2}}\right)}{\left(\sqrt{-x^{2}}\right)^{n}}=\frac{2^{n-1}}{m_{Q}^{n}}\displaystyle\int_{0}^{\infty}dt_{1}\,t_{1}^{n-1}\,e^{-x_{E}^{2}t_{1}-\frac{m_{Q}^{2}}{4t_{1}}}$
(30)

for $I$, we get

$\displaystyle I=$
$\displaystyle\left(-1\right)^{n}\frac{2^{n_{2}-1}}{\left(m_{Q}\right)^{n_{2}}}\frac{2^{n_{3}-1}}{\left(m_{Q}\right)^{n_{3}}}\int
dt_{1}\int dt_{2}\int dt_{3}t_{1}^{n_{1}-1}\,t_{2}^{n_{2}-1}\,t_{3}^{n_{3}-1}$
(31) $\displaystyle\int
d^{4}x_{E}e^{-\left(t_{1}+t_{2}+t_{3}\right)x_{E}^{2}-\frac{m_{Q}^{2}}{4t_{2}}-\frac{m_{Q}}{4t_{3}}-ip_{E}x_{E}}~{}.$

Integrating over $d^{4}x_{E}$, we get

$\displaystyle I=$ $\displaystyle Ai\pi^{2}\int
dt_{1}\,dt_{2}\,dt_{3}t_{1}^{n_{1}-1}\,t_{2}^{n_{2}-1}\,t_{3}^{n_{3}-1}$ (32)
$\displaystyle\frac{1}{\left(t_{1}+t_{2}+t_{3}\right)^{2}}e^{\frac{-p_{E}^{2}}{4\left(t_{1}+t_{2}+t_{3}\right)}-\frac{m_{Q}^{2}}{4t_{2}}-\frac{m_{Q}^{2}}{4t_{3}}}~{},$

where $\displaystyle
A=\left(-1\right)^{n_{1}}\frac{1}{\Gamma\left(n_{1}\right)}\frac{2^{n_{2}+n_{3}-2}}{\left(m_{Q}\right)^{n_{2}+n_{3}}}$
.

Now let us introduce new variables

$\displaystyle t_{1}$ $\displaystyle=t\left(1-x-y\right)$ (33) $\displaystyle
t_{2}$ $\displaystyle=tx$ $\displaystyle t_{3}$ $\displaystyle=ty~{}.$

Then $dt_{1}dt_{2}dt_{3}\rightarrow dxdydtt^{2}$. With this transformation, we
get

$\displaystyle I=Ai\pi^{2}\int$ $\displaystyle
dx\,dy\,dt\,e^{-\frac{p_{E}^{2}}{4t}-\frac{m_{Q}^{2}}{4tx}-\frac{m_{Q}^{2}}{4ty}}t^{n_{1}+n_{2}+n_{3}-3}x^{n_{2}-1}y^{n_{3}-1}\left(1-x-y\right)^{n_{1}-1}~{}.$
(34)

Once we denote
$\displaystyle\frac{1}{4t}\left[p_{E}^{2}+\frac{m_{Q}^{2}}{x}+\frac{m_{Q}^{2}}{y}\right]=\frac{\alpha}{4t}=t^{\prime}$,
with $\displaystyle\alpha=p_{E}^{2}+\frac{m_{Q}^{2}}{x}+\frac{m_{Q}^{2}}{y}$,
then $\displaystyle dt=\frac{dt^{\prime}}{4t^{{}^{\prime}2}}\alpha$ and Eq.
(32) turns out to be

$\int
dt^{\prime}\frac{1}{4t^{{}^{\prime}2}}\alpha\left(\frac{\alpha}{4t^{{}^{\prime}2}}^{\sum
n_{i}-3}\right)e^{-t^{\prime}}=\frac{\alpha^{\sum n_{i}-2}}{4^{\sum
n_{i}-2}}\Gamma\left(2-\sum n_{i}\right)~{}.$ (35)

As a result, we get

$I=Ai\pi^{2}\frac{\Gamma\left(2-\sum n_{i}\right)}{4^{\sum n_{i}-2}}\int
dx\,dyx^{n_{2}-1}y^{n_{3}-1}\left(1-x-y\right)^{n_{1}-1}\frac{1}{\alpha^{2-\sum
n_{i}}}~{}.$ (36)

For $\alpha^{2-\sum n_{i}}$, we use the Schwinger representation once more.

$\frac{1}{\alpha^{2-\sum n_{i}}}=\frac{1}{\Gamma\left(\alpha^{2-\sum
n_{i}}\right)}\int dtt^{1-\sum
n_{i}}e^{-t\left[\frac{m_{Q}^{2}}{x}+\frac{m_{Q}^{2}}{y}+p_{E}^{2}\right]}$
(37)

Now performing Borel transformation with respect to the variable $p_{E}^{2}$,
i.e. $\displaystyle e^{-tp_{E}^{2}}\rightarrow\delta\left(\sigma-t\right)$,
where $\sigma=1/M^{2}$ and after integrating over $t$ we get

$\displaystyle I^{\mathcal{B}}=$ $\displaystyle Ai\pi^{2}\frac{1}{4^{\sum
n_{i}-2}}\int dx\,dyx^{n_{2}-1}y^{n_{3}-1}\left(\sigma\right)^{1-\sum
n_{i}}e^{-\sigma\left(\frac{m_{Q}^{2}}{x}+\frac{m_{Q}^{2}}{y}\right)}~{}.$
(38)

Using the Schwinger representation for $\displaystyle\frac{1}{\sigma^{\sum
n_{i}-1}}$ for $I^{B}$, we get

$\displaystyle I^{\mathcal{B}}=\frac{Ai\pi^{2}}{4^{\sum n_{i}-2}}$
$\displaystyle\frac{1}{\Gamma\left(\sum n_{i}-1\right)}\int dx\int dy\int
d\alpha\alpha^{\sum
n_{i}-2}e^{-\alpha\sigma}e^{-\sigma\left[\frac{m_{Q}^{2}}{x}+\frac{m_{Q}^{2}}{y}\right]}~{}.$
(39)

After performing the second Borel transformation with respect to the variable
$\sigma$, we get

$\displaystyle I^{\mathcal{BB}}=$ $\displaystyle\frac{Ai\pi^{2}}{4^{\sum
n_{i}-2}}\frac{1}{\Gamma\left(\sum n_{i}-1\right)}\int dx\int dy\int
d\alpha\alpha^{\sum
n_{i}-2}\delta\left(s-\alpha-\frac{m_{Q}^{2}}{x}-\frac{m_{Q}^{2}}{y}\right)$
(40) $\displaystyle=$ $\displaystyle\frac{Ai\pi^{2}}{4^{\sum
n_{i}-2}}\frac{1}{\Gamma\left(\sum n_{i}-1\right)}\int dx\int
dy\left(s-\frac{m_{Q}^{2}}{x}-\frac{m_{Q}^{2}}{y}\right)^{\sum n_{i}-2}~{}.$

This expression is the spectral density for $I$. The bounds of $x$ and $y$
variables are

$\displaystyle y$
$\displaystyle\geq\frac{m_{Q}^{2}}{s-\nicefrac{{m_{Q}^{2}}}{{x}}}s-\frac{m_{Q}^{2}}{x}-\frac{m_{Q}^{2}}{y}$
$\displaystyle\geq 0~{}.$ (41)

Since the upper bound of $y$ is the $1-x$, then the bounds of $y$ become

$\frac{m_{Q}^{2}}{s-\nicefrac{{m_{Q}^{2}}}{{x}}}\leq y\leq 1-x~{}.$ (42)

On the other hand, the bounds of $x$ can be obtained from

$\frac{m_{Q}^{2}}{s-\nicefrac{{m_{Q}^{2}}}{{x}}}\leq 1-x$ (43)

which leads to

$x=\frac{s\pm\sqrt{s^{2}-4s\,m_{Q}^{2}}}{2s}=\frac{\left(1\pm\not{v}\right)}{2},$
(44)

where $\displaystyle v=\sqrt{1-4m_{Q}^{2}/s}$ is the velocity of the heavy
quark.

Now we present the expression of the correlation function from the hadronic
side. Inserting the full set of states carrying the same quantum numbers as
the hadron under consideration we get

$\Pi=\sum_{i}\frac{(\not{p}+m_{i})}{m_{i}^{2}-p^{2}},$ (45)

where summation is performed over full set of states. From here, it follows
that the interpolating current used in Eq. (20) interacts not only with the
ground state but also with the excited states carrying the same quantum
numbers. In many cases, while the ground state is narrow, excited states are
broader. Therefore, the spectral density can be parameterized as a single pole
of the ground state plus a smooth continuum corresponding to the contributions
of the higher states, i.e.

$\rho(s)=\lambda^{2}\delta(s-m^{2})+\rho^{\text{cont}}(s)~{}.$ (46)

In the above equation, $\lambda$ is called the decay constant (residue), and
it is defined as

$\lambda u(p)=\bra{0}j\ket{\text{ground state baryon}}~{}.$ (47)

The contribution of the higher states is determined with the help of the
quark-hadron duality ansatz, which means that the $\rho^{\text{cont}}$
coincides with the OPE part of the spectral density starting from some
effective threshold $s_{0}$. In other words, the spectral density for the
continuum can be written as

$\rho^{cont}(s)=\rho^{OPE}\theta\left(s-s_{0}\right)~{}.$ (48)

As we already noted, the mass sum rules are obtained by matching the two
representations of the correlation function.

$\Pi^{OPE}=\Pi^{had}$ (49)

To enhance the contribution of the ground state as well as suppress the
contributions of higher states and higher dimension operators from OPE part,
Borel transformation is applied to both sides of Eq. (49). After the Borel
transformation, all the polynomial terms disappear. The Borel transformation
for the expression $\frac{1}{\left(m^{2}-p^{2}\right)^{n}}$ is given by

$\frac{1}{\left(m^{2}-p^{2}\right)^{n}}\xrightarrow{\mathcal{B}_{M^{2}}}\frac{1}{\left(n-1\right)!\left(M^{2}\right)^{n-1}}e^{\frac{-m^{2}}{M^{2}}}\,\,(\text{
n{\hbox{>}}0 })~{}.$ (50)

After performing the Borel transformation in both representations of the
correlation function, we get

$\lambda^{2}e^{\displaystyle\nicefrac{{-m^{2}}}{{M^{2}}}}=\displaystyle\int_{s_{min}}^{s_{0}}dse^{\displaystyle\nicefrac{{-s}}{{M^{2}}}}\rho(s)~{}.$
(51)

To derive this expression, the quark-hadron duality ansatz is employed.

The mass of low lying state hadron can be obtained by taking derivatives with
respect to $\frac{-1}{M^{2}}$ from Eq. (51) and dividing the result to Eq.
(51). After this operation, we get the following expression for the ground
state mass

$m^{2}=\frac{\displaystyle\int_{s_{min}}^{s_{0}}\rho(s)se^{\frac{-s}{M^{2}}}ds}{\displaystyle\int_{s_{min}}^{s_{0}}\rho(s)e^{\frac{-s}{M^{2}}}ds}~{}.$
(52)

Performing the similar calculations for all the remaining terms, one can
obtain the expressions of the spectral density (for the expressions of the
spectral density, see [65].) Using the explicit expression of the spectral
density from Eq. (26), one can determine the mass of the ground state baryon.

In addition to the input parameters such as heavy quark mass and various
condensates, the mass sum rules contain three auxiliary parameters: Borel mass
parameter $M^{2}$, continuum threshold $s_{0}$, and auxiliary parameter
$\beta$. Any physically measurable quantity should be independent of these
parameters. Therefore it is necessary to find so-called “working regions” of
these parameters, where the physical quantity (in our case mass) exhibits very
weak dependency to them.

The working region of Borel mass is determined in the following way. The upper
bound of $M^{2}$ is determined by the fact that the pole contribution should
be larger than the continuum and the higher states contributions. Therefore,
we demand that the ratio

$\text{P.C.}=\frac{\displaystyle\int_{s_{min}}^{s_{0}}\rho(s)\,e^{-s/M^{2}}ds}{\displaystyle\int_{s_{min}}^{\infty}\rho(s)\,e^{-s/M^{2}}ds}$
(53)

should be larger than $1/2$.

The lower bound of $M^{2}$ is obtained from the condition that the OPE series
should be convergent, i.e. the perturbative contribution should be larger than
the nonperturbative one. Besides, the continuum threshold is determined from
the condition that the differentiated sum rules reproduce the measured mass of
the lowest state hadron with, say, $10\%$ accuracy. Having the working regions
of $M^{2}$ and $s_{0}$ with the conditions mentioned, we are left with to
determine the parameter $\beta$. Analyzing the dependency of the mass on
$\beta$, the working region where the mass is almost independent of the
variation is obtained.

A similar analysis was conducted for spin-$3/2$ doubly heavy baryons in [63].
The main difference, in this case, is the form of the interpolating current.
In the quark-diquark picture, the diquark field has spin $1$. After performing
the numerical analysis, the obtained values for the mass and residues of
doubly heavy baryons with spin-$1/2$ and spin-$3/2$ are collected in Tables 2
and 3, respectively. For completeness, in these tables, we present the
predictions for the mass and residues obtained from the other approaches.

Baryon | [65] | [23] | [29] | [30] | [60] | [61] | [62] |  $\lambda$ [65]
---|---|---|---|---|---|---|---|---
$\Xi_{bb}$ | 9.96 | 9.78 | 10.17 | 9.94 | 10.202 | - | - | 0.44
$\Omega_{bb}$ | 9.97 | 9.85 | 10.32 | 9.99 | 10.359 | - | - | 0.45
$\Xi_{bc}$ | 6.72 | 6.75 | - | 6.80 | 6.933 | 7.053 | - | 0.28
$\Omega_{bc}$ | 6.75 | 7.02 | - | 6.864 | 7.088 | 7.148 | - | 0.29
$\Xi_{cc}$ | 3.72 | 4.26 | 3.57 | 3.52 | 3.620 | 3.676 | 3.5189 | 0.16
$\Omega_{cc}$ | 3.73 | 4.25 | 3.71 | 3.53 | 3.778 | 3.787 | - | 0.18
$\Xi^{{}^{\prime}}_{bc}$ | 6.79 | 6.95 | - | - | 6.963 | 7.062 | - | 0.30
$\Omega^{{}^{\prime}}_{bc}$ | 6.80 | 7.02 | - | - | 7.116 | 7.151 | - | 0.31
Table 2: The mass and residue of doubly heavy baryons with spin $\nicefrac{{1}}{{2}}$ in units GeV. Baryon |  $\not{q}g_{\mu\nu}$ [63] |  $g_{\mu\nu}$ [63] | [72] | [73] | [74] | [75, 76]
---|---|---|---|---|---|---
$\Xi^{*}_{cc}$ | $3.69\pm 0.16$ | $3.72\pm 0.18$ | $3.61\pm 0.18$ | $3.727$ | $3.90\pm 0.10$ | $3.58\pm 0.05$
$\Omega^{*}_{cc}$ | $3.78\pm 0.16$ | $3.78\pm 0.16$ | $3.76\pm 0.17$ | $3.872$ | $3.81\pm 0.06$ | $3.67\pm 0.05$
$\Xi^{*}_{bb}$ | $10.4\pm 1.0$ | $10.3\pm 0.2$ | $10.22\pm 0.15$ | $10.237$ | $10.35\pm 0.08$ | $10.33\pm 1.09$
$\Omega^{*}_{bb}$ | $10.5\pm 0.2$ | $10.4\pm 0.2$ | — | $6.98$ | $8.00\pm 0.26$ | —
$\Xi^{*}_{bc}$ | $7.25\pm 0.20$ | $7.2\pm 0.2$ | — | $7.13$ | $7.54\pm 0.08$ | —
$\Omega^{*}_{bc}$ | $7.3\pm 0.2$ | $7.35\pm 0.25$ | $10.38\pm 0.14$ | $10.389$ | $10.28\pm 0.05$ | $10.38\pm 1.10$
Table 3: The mass spectra of the spin $3/2$ doubly heavy baryons in units of $\rm{GeV.}$ [63]. Baryon |  $\not{q}g_{\mu\nu}$ [63] |  $g_{\mu\nu}$ [63] | [72] | [75]
---|---|---|---|---
$\Xi^{*}_{cc}$ | $0.12\pm 0.01$ | $0.12\pm 0.01$ | $0.070\pm 0.017$ | $0.071\pm 0.017$
$\Omega^{*}_{cc}$ | $0.14\pm 0.02$ | $0.13\pm 0.01$ | $0.085\pm 0.019$ | —
$\Xi^{*}_{bb}$ | $0.22\pm 0.03$ | $0.21\pm 0.01$ | $0.161\pm 0.041$ | $0.111\pm 0.040$
$\Omega^{*}_{bb}$ | $0.25\pm 0.03$ | $0.25\pm 0.02$ | — | —
$\Xi^{*}_{bc}$ | $0.15\pm 0.01$ | $0.15\pm 0.01$ | — | —
$\Omega^{*}_{bc}$ | $0.18\pm 0.02$ | $0.17\pm 0.01$ | $0.199\pm 0.048$ | —
Table 4: The residues of the spin $3/2$ doubly heavy baryons in units of
$\rm{GeV}^{3}$ [63].

The sum rules method allows not only calculating the properties of the ground
state hadron but also the properties of radial or orbital excitations of these
baryons. The modification of the sum rules is required only in the
phenomenological part. Namely, Eq. (46) is modified as

$\rho(s)=\lambda_{0}^{2}\delta\left(s-m_{0}^{2}\right)+\lambda_{1}^{2}\delta\left(s-m_{1}^{2}\right),$
(54)

where $m_{0}\left(m_{1}\right)$ and $\lambda_{0}\left(\lambda_{1}\right)$ are
the mass and residues of the ground (excited) state, respectively.

The numerical calculations are performed likewise for the ground state
baryons, which are explained in detail above and the results for $2S$ and $1P$
excited state baryons are summarized in Table 5. In these tables, we also
presented the results of other approaches on the mass of doubly heavy baryons.
In Table 5, we also present the values of residues of $2S$ and $1P$ states.
The obtained results on the mass and residues can be useful in looking for
these states in future experiments.

Baryon | $\sqrt{s_{0}}(GeV)$ | $M^{2}(GeV^{2})$ | $2S(GeV)$ | $1P(GeV)$
---|---|---|---|---
$\Xi_{cc}$ | $4.3\pm 0.1$ | $\left[4-7\right]$ | $4.03\pm 0.20$ | $4.03\pm 0.20$
$\Omega_{cc}$ | $4.5\pm 0.1$ | $\left[4-7\right]$ | $4.16\pm 0.14$ | $4.16\pm 0.14$
$\Xi_{bb}$ | $11.1\pm 0.1$ | $\left[10-16\right]$ | $10.32\pm 0.10$ | $10.32\pm 0.10$
$\Omega_{bb}$ | $11.2\pm 0.1$ | $\left[10-16\right]$ | $10.37\pm 0.10$ | $10.32\pm 0.10$
$\Xi_{bc}$ | $7.7\pm 0.1$ | $\left[7-11\right]$ | $7.14\pm 0.11$ | $7.14\pm 0.11$
$\Omega_{bc}$ | $7.9\pm 0.1$ | $\left[7-11\right]$ | $7.20\pm 0.11$ | $7.20\pm 0.11$
$\Xi^{{}^{\prime}}_{bc}$ | $7.9\pm 0.1$ | $\left[7-11\right]$ | $7.02\pm 0.07$ | $7.02\pm 0.07$
$\Omega^{{}^{\prime}}_{bc}$ | $7.9\pm 0.1$ | $\left[7-11\right]$ | $7.09\pm 0.07$ | $7.09\pm 0.07$
Table 5: The working domains of continuum threshold and Borel parameters
$M^{2}$ along with the respective mass of $2S$ and $1P$ excitations of the
doubly heavy baryons [64].

##  IV Calculation of the strong coupling constants of doubly heavy baryons
with light mesons

As we already noted, for a deeper understanding of the dynamics of the double
heavy baryons, the investigation of the weak, strong, and electromagnetic
decays of the baryons should be conducted. Up to now, the studies of the
strong decays have been very limited. Hence, detailed studies of strong decays
can receive great interest for the study of the properties of these baryons.

In this part of the mini-review, we study the strong coupling constants of
doubly heavy baryons with light pseudo-scalar and vector mesons within the
light-cone version of sum rules. Moreover, by using these coupling constants
and applying the vector meson dominance model, we estimate the radiative
decays of double heavy baryons.

To determine the strong coupling constants of the doubly heavy baryons with
light mesons within light-cone sum rules, we start with considering the
following correlation function.

$\Pi=i\int d^{4}xe^{ipx}\bra{M(q)}\eta(x)\,\bar{\eta}(0)\ket{0},$ (55)

where $M(q)$ is a meson with momentum $q$, $\eta$ is the interpolating current
of the corresponding doubly heavy baryon. According to the $SU(3)$
classification, there are two types of currents, symmetric and antisymmetric
currents, with respect to the heavy quarks. The interpolating currents
(symmetric and antisymmetric) for doubly heavy baryons with
$J=\nicefrac{{1}}{{2}}$ are

$\displaystyle\eta^{(S)}=$
$\displaystyle\frac{1}{\sqrt{2}}\varepsilon^{abc}\sum_{i=1}^{2}\left\\{\left(Q^{aT}A^{i}_{1}q^{b}\right)A^{i}_{2}Q^{\prime
c}+\left(Q\xleftrightarrow{}Q^{\prime}\right)\right\\}$ (56)
$\displaystyle\eta^{(A)}=$
$\displaystyle\frac{1}{\sqrt{6}}\varepsilon^{abc}\sum_{i=1}^{2}\Big{\\{}2\left(Q^{aT}A^{i}_{1}Q^{\prime
b}\right)A^{i}_{2}q^{c}+\left(Q^{aT}A^{i}_{1}q^{b}\right)A^{i}_{2}Q^{\prime
c}-\left(Q^{\prime aT}A^{i}_{1}q^{b}\right)A^{i}_{2}Q^{c}\Big{\\}},$

where $A_{1}^{1}=C$, $A_{1}^{2}=C\gamma_{5}$ , $A_{2}^{1}=\gamma_{5}$ and
$A_{2}^{2}=\beta\mathbb{1}$ and $S$ and ${A}$ corresponds to symmetric and
antisymmetric currents.

Similar to the mass sum rules calculation, the correlation function is
calculated in two different regions, namely, in hadrons and the deep Euclidean
domain $p^{2}\ll 0$ and $\left(p+q\right)^{2}\ll 0$ by using the OPE which is
carried out over the twist rather than dimensions of operators. Then,
performing Borel transformation in order to suppress the continuum
contributions and enhance the contribution of the ground state and matching
these two results, we get the desired sum rules.

The representation of the correlation function in terms of the hadron is
obtained by inserting a complete set of baryon states having the same quantum
numbers as the interpolating current. Then, isolating the contribution of the
ground state, we get

$\Pi=\frac{\bra{0}\eta\ket{B(p^{\prime})}\bra{M(q)B(p^{\prime})}\ket{B(p)}\bra{B(p)}\bar{\eta}\ket{0}}{\left(p^{\prime
2}-m_{2}^{2}\right)\left(p^{2}-m_{1}^{2}\right)}\hfill+\text{higher
states}~{}.$ (57)

The matrix elements entering to the Eq. (57) for the pseudoscalar meson case
are determined as

$\bra{0}\eta\ket{B(p^{\prime})}=\lambda_{2}u\left(p^{\prime}\right)$ (58)
$\bra{M(q)B(p^{\prime})}\ket{B(p)}=g\bar{u}\left(p^{\prime}\right)i\gamma_{5}u\big{(}p\big{)}~{}.$
(59)

For the vector meson case, the matrix element
$\bra{M(q)B(p^{\prime})}\ket{B(p)}$ is defined as

$\bra{M(q)B(p^{\prime})}\ket{B(p)}=\bar{u}(p^{\prime})\left[f_{1}\gamma_{\mu}+\frac{i\sigma_{\mu\nu}q^{\nu}}{m_{1}+m_{2}}f_{2}\right]u(p)\varepsilon^{\mu}~{},$
(60)

where $f_{1}$ and $f_{2}$ are the relevant form factors of the doubly heavy
baryons with corresponding light vector mesons and $\varepsilon^{\mu}$ is the
polarization vector of meson.

Using Eqs. (57) and (60) and after performing double Borel transformations
over $-p^{2}$ and $-p^{\prime 2}$ for the correlation functions containing
pseudo-scalar and vector mesons,

$\Pi^{(P)}=\lambda_{1}\lambda_{2}ge^{\frac{-(m_{B_{1}}^{2}+m_{B_{2}}^{2})}{2M^{2}}}\hfill(\text{for
}\not{p}\not{q}\gamma_{5}\text{ structure + ...})$ (61)
$\Pi^{(V)}=\lambda_{1}\lambda_{2}\left[\not{p}\not{\varepsilon}\not{q}(f_{1}+f_{2})+2(\varepsilon
p)\not{p}f_{1}+\text{
...}\right]e^{\frac{-(m_{B_{1}}^{2}+m_{B_{2}}^{2})}{2M^{2}}}~{},$ (62)

where dots denote other structures.

Calculating the correlation functions from the QCD side and separating the
structures $\not{p}\not{q}\gamma_{5}$, $\not{p}\not{\varepsilon}\not{q}$ and
$2(\varepsilon p)\not{p}$ and equating them to the corresponding coefficients
of the same structures, one can obtain the sum rules for the strong coupling
constants for the pseudo-scalar and vector mesons. The new and essential
element for the calculation of the theoretical part of the correlation
function is the so-called distribution amplitudes (DA) of light mesons.

The DAs contain all information about the nonperturbative sector of QCD and
are obtained in many works [77, 78, 79, 80, 81]. The main problem is the
calculation of the spectral density to determine the theoretical part of the
correlation function. We here present the details of the calculation for the
generic term.

$I=\int_{0}^{1}du\frac{u^{k}}{\left[p_{1}^{2}\bar{u}+p_{2}^{2}u+m_{M}^{2}\bar{u}u+m^{2}\right]^{n}}~{},$
(63)

where $m_{M}$ is the mass of the hadron or photon (which is zero), and all
momenta are represented in Euclidean space. Using the Schwinger
representation, we get

$I=\int_{0}^{1}du\int_{0}^{\infty}d\alpha\alpha^{n-1}u^{k}\
e^{-\alpha\left[p_{1}^{2}\bar{u}+p_{2}^{2}u+m_{M}^{2}\bar{u}u+m^{2}\right]}~{}.$
(64)

Performing double Borel transformation over
$p_{1}\rightarrow\frac{1}{\sigma_{1}}$ and
$p_{2}\rightarrow\frac{1}{\sigma_{2}}$ we have

$\displaystyle I^{\mathcal{B}}=$
$\displaystyle\int_{0}^{1}\int_{0}^{\infty}\alpha^{n-1}u^{k}e^{-\alpha(m_{M}^{2}\bar{u}u+m^{2})}\delta(\sigma_{1}-\alpha\bar{u})\delta(\sigma_{2}-\alpha
u)$ (65) $\displaystyle=$
$\displaystyle(\sigma_{1}+\sigma_{2})^{n-k-2}\sigma^{k}e^{(-\sigma_{1}+\sigma_{2})}\left(m^{2}+\frac{m_{M}^{2}}{2}\right)+m_{M}^{2}\frac{\sigma_{1}^{2}+\sigma_{2}^{2}}{2(\sigma_{1}+\sigma_{2})}~{}.$

Once we use the integral representation

$\int_{-\infty}^{\infty}e^{-\rho x^{2}-\beta
x}=\displaystyle\sqrt{\frac{\pi}{\alpha}}e^{\frac{-\beta^{2}}{4\rho}}~{}.$
(66)

Eq. (65) can be written as

$\displaystyle
I^{\mathcal{B}}=\int_{-\infty}^{\infty}\frac{d\sigma_{1}d\sigma_{2}}{2\pi}(\sigma_{1}+\sigma_{2})^{n-k-1}\sigma_{2}^{k}e^{-\sigma_{1}\left(m^{2}+\frac{(m_{M}+x_{1})^{2}+x_{2}^{2}}{2}\right)-\sigma_{2}\left(m^{2}+\frac{(m_{M}+x_{2})^{2}+x_{1}^{2}}{2}\right)}~{}.$
(67)

Again using the Schwinger representation for
$\frac{1}{\left(\sigma_{1}+\sigma_{2}\right)^{n-k-1}}$, we get the following
expression for the spectral density

$\rho(s_{1},s_{2})=\left(\frac{d}{ds_{2}}\right)^{k}\frac{1}{2m_{M}\pi}\frac{\Gamma(k)}{\Gamma(2k)}\left(\frac{\sqrt{\Delta}}{m_{M}}\right)^{2k-1}\theta(\Delta)~{},$
(68)

where

$\displaystyle\Delta$
$\displaystyle=\left[-m_{M}^{4}-(s_{1}-s_{2})^{2}+2m_{M}^{2}(-2m^{2}+s_{1}+s_{2})\right]$
(69) $\displaystyle=(t_{2}-s_{2})(s_{1}-t_{1})~{}.$

Omitting the details of the calculations, we present the numerical results for
the strong coupling constants of the light mesons with heavy baryons in Table
6. The interested reader can find the details of the calculations for sum
rules in [16, 14].

| Channel | Strong coupling constant
---|---|---
|  $\Xi_{cc}$$\Xi_{cc}$$\pi$ | $10.3\pm 0.52$
|  $\Xi_{bb}$$\Xi_{bb}$$\pi$ | $12.73\pm 1.29$
$SS$ |  $\Omega_{bb}$$\Xi_{bb}$$K$ | $17.40\pm 1.89$
|  $\Omega_{cc}$$\Xi_{cc}$$K$ | $12.50\pm 0.75$
|  $\Omega_{bc}$$\Xi_{bc}$$K$ | $5.08\pm 0.43$
$AA$ |  $\Xi^{{}^{\prime}}_{bc}$$\Xi^{{}^{\prime}}_{bc}$$\pi$ | $6.85\pm 0.06$
$\Omega^{{}^{\prime}}_{bc}$$\Xi^{{}^{\prime}}_{bc}$$K$ | $7.90\pm 0.16$
$SA$ |  $\Xi^{{}^{\prime}}_{bc}$$\Xi_{bc}$$\pi$ | $1.49\pm 0.10$
$\Omega^{{}^{\prime}}_{bc}$$\Xi_{bc}$$K$ | $2.03\pm 0.16$
Table 6: The numerical values for the strong coupling constants [14]. | Channel | $f_{1}+f_{2}$ | $f_{1}$
---|---|---|---
|  $\Xi_{cc}$$\rightarrow$$\Xi_{cc}$$\rho$ | $32.53\pm 2.30$ | $-25.32\pm 8.49$
|  $\Xi_{bb}$$\rightarrow$$\Xi_{bb}$$\rho$ | $23.66\pm 2.89$ | $-7.69\pm 2.31$
$SS$ |  $\Omega_{bb}$$\rightarrow$$\Xi_{bb}$$K^{*}$ | $22.55\pm 2.63$ | $8.30\pm 2.74$
|  $\Omega_{cc}$$\rightarrow$$\Xi_{cc}$$K^{*}$ | $28.36\pm 1.83$ | $-23.64\pm 9.85$
|  $\Omega_{bc}$$\rightarrow$$\Xi_{bc}$$K^{*}$ | $8.23\pm 0.77$ | $-3.74\pm 1.37$
$AA$ |  $\Xi^{{}^{\prime}}_{bc}$$\rightarrow$$\Xi^{{}^{\prime}}_{bc}$$\rho$ | $-37.62\pm 0.48$ | $-0.40\pm 0.12$
$\Omega^{{}^{\prime}}_{bc}$$\rightarrow$$\Xi^{{}^{\prime}}_{bc}$$K^{*}$ | $-37.98\pm 6.75$ | $-0.40\pm 0.14$
$SA$ |  $\Xi^{{}^{\prime}}_{bc}$$\rightarrow$$\Xi_{bc}$$\rho$ | $1.50\pm 0.31$ | $-0.97\pm 0.30$
$\Omega^{{}^{\prime}}_{bc}$$\rightarrow$$\Xi_{bc}$$K^{*}$ | $2.20\pm 0.40$ | $-1.00\pm 0.36$
Table 7: The numerical values for the strong coupling constants [16].

Now let us calculate the strong coupling constants of the
spin-$\nicefrac{{3}}{{2}}$-spin$\nicefrac{{1}}{{2}}$ doubly heavy baryons with
light vector mesons. The calculation scheme is similar to the calculation of
spin$\nicefrac{{1}}{{2}}$-spin$\nicefrac{{1}}{{2}}$ vector mesons. However, in
this case, some difficulties appear for such types of transitions.

In spin-$\nicefrac{{3}}{{2}}-$spin-$\nicefrac{{1}}{{2}}$ vector strong
couplings, we need the interpolating currents of spin$\nicefrac{{3}}{{2}}$
particles first. Following the same procedure for constructing interpolating
current for spin$\nicefrac{{1}}{{2}}$ doubly heavy baryon states, one can
easily obtain the following expressions for the interpolating currents for
spin$\nicefrac{{3}}{{2}}$ baryons

$\eta_{\mu}=N\,\varepsilon^{abc}\left\\{\left(q^{aT}C\gamma_{\mu}Q^{b}\right)Q^{{}^{\prime}C}+\left(q^{aT}C\gamma_{\mu}Q^{{}^{\prime}b}\right)Q^{c}+\left(Q^{aT}C\gamma_{\mu}Q^{{}^{\prime}b}\right)q^{c}\right\\}~{},$
(70)

where $\displaystyle N=\sqrt{1/3}\left(\sqrt{2/3}\right)$ for identical
(different) heavy quarks. The following matrix elements appear for the
calculation of the phenomenological part,

$\displaystyle\bra{0}\eta_{\mu}\ket{B^{*}\left(p\right)}=$
$\displaystyle\lambda\,u_{\mu}\left(p\right)$ (71)
$\displaystyle\bra{V\left(q\right)B^{*}\left(p_{2}\right)}\ket{B\left(p_{1}\right)}=$
$\displaystyle\bar{u}_{\alpha}\left(p_{2}\right)\bigg{[}g_{1}\left(\varepsilon_{\alpha}\not{q}-q_{\alpha}\not{\varepsilon}\right)\gamma_{5}+g_{2}\left(Pq\varepsilon_{\alpha}-P\varepsilon
q_{\alpha}\right)\gamma_{5}+g_{3}\left(q\varepsilon
q_{\alpha}-q^{2}\varepsilon_{\alpha}\right)\gamma_{5}\bigg{]}u\left(p_{1}\right),$

where $u_{\alpha}\left(u\right)$ is the Rarita–Schwinger (Dirac) spinor,
$\varepsilon_{\mu}$ is the polarization vector of the light vector meson,
$P=(p_{1}+p_{2})$, and $q=p_{1}-p_{2}$. Using the completeness relations for
the Dirac and Rarita–Schwinger (Dirac) spinors with the help of the expression

$\displaystyle\sum_{s}u\left(p\right)\bar{u}\left(p\right)=$
$\displaystyle\not{p}+m$ (72)
$\displaystyle\sum_{s}u_{\alpha}\left(p\right)\bar{u}_{\beta}\left(p\right)=$
$\displaystyle-\left(\not{p}+m\right)\left(g_{\alpha\beta}-\frac{1}{3}\gamma_{\alpha}\gamma_{\beta}+\frac{2}{3}\frac{p_{\alpha}p_{\beta}}{m^{2}}+\frac{p_{\alpha}\gamma_{\beta}-\gamma_{\alpha}p_{\beta}}{3m}\right)$

one can, in principle, calculate the correlation functions from the hadronic
part. However, at this point problems appear, which is generic for spin
$3/2$-spin $1/2$ transitions. The problem is related to the fact that the
interpolating current for spin $3/2$ baryons interacting not only with spin
$3/2$ but also with spin $1/2$ negative parity baryons too. Obviously, these
pollutions should be eliminated. There are two different ways to overcome this
problem. In the first method, one can introduce the projection operator
$P_{3/2}$ which separates the contribution of spin $3/2$ states. The main idea
of the second way for discarding the contributions of $1/2$ states is as
follows. The matrix element for the interpolating current between vacuum and
the particle with negative parity spin $1/2$ states can be written in general
form as

$\bra{0}\eta_{\mu}\ket{B\left(p\right)}=\left(A\gamma_{\mu}+Bp_{\mu}\right)u\left(p\right)~{}.$
(73)

Multiplying both sides with $\gamma^{\mu}$ and using
$\gamma_{\mu}\eta^{\mu}=0$, we get $4A+Bm=0$. From this expression, we obtain
$\displaystyle A=-\frac{m}{4}B$; hence,

$\bra{0}\eta_{\mu}\ket{B\left(p\right)}=B\left(\gamma_{\mu}-\frac{m}{4}p_{\mu}\right)u\left(p\right)~{}.$
(74)

As a result, we observe that the structures proportional to $\gamma_{\mu}$ or
$p_{\mu}$ contain the contributions of spin $1/2$ states only, and therefore
these structures should be discarded. From Eq. (72), we infer that
$g_{\alpha\beta}$ structure contains the contribution of the spin $3/2$ states
solely.

The second problem is that not all Lorentz structures are independent. This
problem can be solved by using the specific order of the Dirac matrices. We
choose the structure $\gamma_{\mu}\not{\varepsilon}\not{q}\not{p}\gamma_{5}$.
Taking into account these remarks and using Eqs. (71) and (72), for the
hadronic part of the correlation function, we obtain

$\begin{split}\Pi_{\mu}^{V}&=\frac{\lambda_{1}\lambda_{2}}{[m_{1}^{2}-\left(p+q\right)^{2}][m_{2}^{2}-p^{2}]}\bigg{[}-g_{1}\left(m_{1}+m_{2}\right)\not{\varepsilon}\not{p}\gamma_{5}q_{\mu}+g_{2}\,\not{q}\not{p}\gamma_{5}\left(p\varepsilon\right)q_{\mu}+g_{3}\,m_{V}^{2}\not{q}\not{p}\gamma_{5}\varepsilon_{\mu}+\parbox{45.00006pt}{other\\\
structures}\bigg{]}.\end{split}$ (75)

The QCD side of the correlation function is calculated in a standard manner by
taking into account the vector meson DAs. The expressions of the DAs for the
vector meson can be found in [80, 77, 79]. The values of the parameters in the
expressions of the DAs for the light vector mesons at renormalization point
$\mu=1~{}\rm{GeV}$ are presented in Table 8. Our results for the the strong
coupling constants of vector mesons with doubly heavy baryons after analyzing
of the corresponding sum rules are presented in Table 9. The obtained results
are crucial in analysis of the nonleptonic decays and can be used in the
further studies of strong decays in the experiments carried out at LHC.

Parameter | $\rho$ | $K^{*}$ | Parameter | $\rho$ | $K^{*}$ | Parameter | $\rho$ | $K^{*}$ | Parameter | $\rho$ | $K^{*}$
---|---|---|---|---|---|---|---|---|---|---|---
$f_{V}[GeV]$ | $0.216$ | $0.220$ | $a_{1}^{\parallel}$ | $0$ | $0.03$ | $\zeta_{3}^{\parallel}$ | $0.030$ | $0.023$ | $\zeta_{4}^{\parallel}$ | $0.07$ | $0.02$
$f_{V}^{T}[GeV]$ | $0.165$ | $0.185$ | $a_{1}^{\perp}$ | $0$ | $0.04$ | $\tilde{\omega}_{3}^{\parallel}$ | $-0.09$ | $-0.07$ | $\zeta_{4}^{\perp}$ | $-0.08$ | $-0.05$
$m_{V}[GeV]$ | $0.770$ | $0.892$ | $a_{2}^{\parallel}$ | $0.15$ | $0.11$ | $\omega_{3}^{\parallel}$ | $0.15$ | $0.10$ | $\tilde{\zeta}_{4}^{\perp}$ | $-0.08$ | $-0.05$
|  |  | $a_{2}^{\perp}$ | $0.14$ | $0.10$ | $\omega_{3}^{\perp}$ | $0.55$ | $0.30$ | $\tilde{\omega}_{4}^{\perp}$ | $-0.03$ | $-0.02$
|  |  |  |  |  | $\kappa_{3}^{\parallel}$ | $0$ | $0$ | $\kappa_{4}^{\parallel}$ | $0$ | $-0.025$
|  |  |  |  |  | $\kappa_{3}^{\perp}$ | $0$ | $0.003$ | $\kappa_{4}^{\perp}$ | $0$ | $0.013$
|  |  |  |  |  | $\lambda_{3}^{\parallel}$ | $0$ | $-0.008$ |  |  |
|  |  |  |  |  | $\lambda_{3}^{\perp}$ | $0$ | $0.035$ |  |  |
|  |  |  |  |  | $\tilde{\lambda}_{3}^{\parallel}$ | $0$ | $-0.025$ |  |  |
Table 8: The numerical values for the parameters in the DAs for the vector mesons $\rho$ and $K^{*}$. The renormalization scale is $\mu=1~{}GeV$ [16]. Transition | General current | Ioffe current
---|---|---
$\Xi^{*}_{cc}$$\rightarrow$$\Xi_{cc}$$\rho^{0}$ | $1.13\pm 0.25$ | $0.11\pm 0.03$ | $7.81\pm 1.83$ | $0.99\pm 0.22$ | $0.1\pm 0.02$ | $6.92\pm 1.62$
$\Xi^{*}_{bb}$$\rightarrow$$\Xi_{bb}$$\rho^{0}$ | $0.76\pm 0.23$ | $0.03\pm 0$ | $15.19\pm 4.64$ | $0.67\pm 0.2$ | $0.02\pm 0$ | $13.45\pm 4.11$
$\Xi^{*}_{bc}$$\rightarrow$$\Xi_{bc}$$\rho^{0}$ | $1.06\pm 0.2$ | $0.05\pm 0.01$ | $14.44\pm 2.84$ | $0.94\pm 0.18$ | $0.05\pm 0.01$ | $12.79\pm 2.51$
$\Xi^{*}_{cc}$$\rightarrow$$\Xi_{cc}$$\omega$ | $1.02\pm 0.23$ | $0.1\pm 0.02$ | $7.1\pm 1.68$ | $0.9\pm 0.2$ | $0.09\pm 0.02$ | $6.29\pm 1.49$
$\Xi^{*}_{bb}$$\rightarrow$$\Xi_{bb}$$\omega$ | $0.69\pm 0.21$ | $0.03\pm 0$ | $13.82\pm 4.25$ | $0.61\pm 0.19$ | $0.02\pm 0$ | $12.24\pm 3.77$
$\Xi^{*}_{bc}$$\rightarrow$$\Xi_{bc}$$\omega$ | $0.97\pm 0.19$ | $0.05\pm 0.01$ | $13.14\pm 2.6$ | $0.86\pm 0.17$ | $0.05\pm 0.01$ | $11.64\pm 2.31$
$\Omega^{*}_{cc}$$\rightarrow$$\Omega_{cc}$$\phi$ | $1.5\pm 0.32$ | $0.5\pm 0.14$ | $9.79\pm 2.5$ | $1.32\pm 0.28$ | $0.45\pm 0.13$ | $8.64\pm 2.22$
$\Omega^{*}_{bb}$$\rightarrow$$\Omega_{bb}$$\phi$ | $1.22\pm 0.35$ | $0.15\pm 0.03$ | $23.9\pm 7.19$ | $1.08\pm 0.31$ | $0.14\pm 0.03$ | $21.15\pm 6.38$
$\Omega^{*}_{bc}$$\rightarrow$$\Omega_{bc}$$\phi$ | $1.47\pm 0.29$ | $0.25\pm 0.04$ | $19.26\pm 4.13$ | $1.3\pm 0.26$ | $0.22\pm 0.03$ | $17.03\pm 3.66$
Table 9: The obtained values of the moduli of the coupling constants $g_{1}$,
$g_{2}$, and $g_{3}$ for the given transitions accompanied by a light vector
meson [17].

##  V The vector dominance model and radiative decays of the doubly heavy
baryons

In the previous section, we have determined the strong coupling constants of
the light vector mesons with spin$\nicefrac{{3}}{{2}}$ and
spin$\nicefrac{{1}}{{2}}$ doubly heavy baryons. Now using the values of the
obtained coupling constants and the vector dominance model (VDM), we estimate
the radiative decay widths of the $\Xi^{*}_{QQ}\rightarrow\Xi_{QQ}\gamma$ and
$\Omega^{*}_{QQ}\rightarrow\Omega_{QQ}\gamma$.

The vector dominance model (VDM) implies that the form factors for the
$B^{*}_{QQ}\rightarrow B_{QQ}\gamma$ transition can be obtained from
$B^{*}_{QQ}\rightarrow B_{QQ}V$ transition form factors by converting the
corresponding vector meson to photon. From the gauge invariance, the matrix
element
${(\varepsilon^{\gamma*})}^{\nu}\bra{B^{*}_{QQ}(p)}j_{\nu}\ket{B_{QQ}(p+q)}$
is parameterized similar to
${(\varepsilon^{V*})}^{\nu}\bra{B^{*}_{QQ}(p)}j_{\nu}\ket{B_{QQ}(p+q)}$ in
terms of three form factors

$\displaystyle(\varepsilon^{\gamma*})^{\nu}\bra{B*_{QQ}(p)}j_{\nu}\ket{B_{QQ}(p+q)}=e\bar{u}^{\alpha}(p,s)\Big{[}$
$\displaystyle
g_{1}^{\gamma}\left(\varepsilon^{\gamma*}_{\alpha}\not{q}-q_{\alpha}\not{\varepsilon}^{\gamma*}\right)+g_{2}^{\gamma}\left(Pq\varepsilon^{\gamma*}_{\alpha}-P\varepsilon^{\gamma*}q_{\alpha}\right)$
(76)
$\displaystyle+g_{3}^{\gamma}\left(q\varepsilon^{\gamma*}q_{\alpha}-q^{2}\varepsilon^{\gamma*}_{\alpha}\right)\Big{]}\gamma_{5}u(p+q)~{}.$

Here $\varepsilon^{\gamma}$ is the polarization vector of the photon field.
For the real photon case, obviously, the last term is equal to zero. To relate
the form factors $g_{i}^{\gamma}$ with $g_{i}$ it is necessary to make the
following replacement

$\varepsilon_{\mu}\rightarrow\sum_{V=\rho,\omega,\phi}e\frac{f_{V}}{m_{V}}C_{V}\varepsilon_{\mu}^{\gamma}$
(77)

and transfer from $q^{2}=m_{V}$ to $q^{2}=0$. Now let us check this ansatz. To
calculate the matrix element
$(\varepsilon^{V*})^{\nu}\bra{B^{*}_{QQ}(p)}j_{\nu}\ket{B_{QQ}(p+q)}$ we use
the VDM ansatz and consider the contributions of intermediate vector mesons.
In this case, we get

$\displaystyle\varepsilon^{\gamma*\nu}\bra{B^{*}_{QQ}(p)}j_{\nu}\ket{B_{QQ}(p+q)}=\sum_{V=\rho,\omega,\phi}\sum_{\lambda}e\frac{i}{q^{2}-m_{V}^{2}}(-i)\varepsilon^{V}_{\lambda}\varepsilon^{\gamma*}f_{V}m_{V}C_{V}\varepsilon^{V*}_{\lambda}\bra{B^{*}_{QQ}(p)}j_{\nu}\ket{B_{QQ}(p+q)}~{},$
(78)

where $q^{2}=0$. In this derivation, the form of the interaction Lagrangian of
a vector meson with the photon is used as
$\mathcal{L}=-ef_{V}m_{V}C_{V}V^{\mu}A_{\mu}$ [29, 30], where

$C_{V}=\begin{cases}\frac{1}{\sqrt{2}}\text{ for }\rho\text{ meson}\\\
\frac{1}{3\sqrt{2}}\text{ for }\omega\text{ meson}\\\ -\frac{1}{3}\text{ for
}\phi\text{ meson}~{}.\end{cases}$ (79)

Once we take into account the quark contents of $\rho$, $\omega$ and $\phi$
mesons, the factor $C_{V}$ can be written in terms of the electric charges of
corresponding quarks

$C_{V}=\begin{cases}\frac{1}{\sqrt{2}}\left(e_{u}-e_{d}\right)\text{ for
}\rho\text{ meson}\\\ \frac{1}{\sqrt{2}}\left(e_{u}+e_{d}\right)\text{ for
}\omega\text{ meson}\\\ e_{s}\text{ for }\phi\text{ meson}~{}.\end{cases}$
(80)

Then we use the definition of the matrix element
$\varepsilon^{\gamma*\nu}_{\lambda}\bra{B^{*}_{QQ}(p)}j_{\nu}\ket{B_{QQ}(p+q)}$
and perform summation over the polarization of the massive vector mesons via

$\sum\varepsilon_{\lambda}^{\mu}\varepsilon_{\lambda}^{*\nu}=-g^{\mu\nu}+\frac{q^{\mu}q^{\nu}}{m_{V}^{2}}~{}.$
(81)

Relying on the fact that in VDM, the vector meson’s 4-momentum is equal to the
photon’s one, and using $\varepsilon^{\gamma}\cdot q=0$, the last term in Eq.
(81) can be neglected. As a result, we get

$\varepsilon^{\gamma*\nu}\bra{B^{*}_{QQ}(p)}j_{\nu}\ket{B_{QQ}(p+q)}=\sum_{V=\rho,\omega,\phi}e\frac{f_{V}}{m_{V}}C_{V}\varepsilon^{\gamma*}\bra{B^{*}_{QQ}(p)}j_{\nu}\ket{B_{QQ}(p+q)}~{}.$
(82)

Comparing the coefficients of $\varepsilon^{\gamma}$ in both sides of Eqs.(71)
and (82), we get the following relation among the two sets of form factors

$g_{i}^{\gamma}=\sum_{i}e\frac{f_{V}}{m_{V}}g_{i}~{}.$ (83)

Explicitly, these relations are written as

$g_{i}^{\gamma}=\begin{cases}e_{s}\frac{f_{\phi}}{m_{\phi}}g_{i}^{\phi}\text{
for }\Omega^{*}_{QQ}\rightarrow\Omega_{QQ}\gamma\\\
\frac{1}{\sqrt{2}}e_{u}\left(\frac{f_{\rho}}{m_{\rho}}g_{i}^{\rho}\frac{f_{\omega}}{m_{\omega}}g_{i}^{\omega}\right)\text{
for }\Xi^{*}_{QQu}\rightarrow\Xi_{QQu}\gamma\\\
\frac{1}{\sqrt{2}}e_{d}\left(-\frac{f_{\rho}}{m_{\rho}}g_{i}^{\rho}\frac{f_{\omega}}{m_{\omega}}g_{i}^{\omega}\right)\text{
for }\Xi^{*}_{QQd}\rightarrow\Xi_{QQd}\gamma~{}.\end{cases}$ (84)

Two remarks are in order. First, we assume that couplings do not change
considerably when going from $q^{2}=m_{V}^{2}$ to $q^{2}=0$. The second remark
is related to the fact that the possibility of the heavy vector mesons
contributing also. However, these contributions are neglected since in the
heavy quark limit, they are proportional to
$m_{J/\psi,\Upsilon}^{\nicefrac{{-3}}{{2}}}$.

From an experimental point of view, instead of the form factors
$g_{i}^{\gamma}$ , the multipole form factors, such as magnetic dipole
($G_{M}$), electric quadrupole ($G_{E}$) and coulomb quadrupole ($G_{C}$) form
factors are more convenient. The relations among $g_{1}^{\gamma}$,
$g_{2}^{\gamma}$, $G_{M}$, $G_{E}$ and $G_{C}$ are given at Eq. (71) and at
$q^{2}=0$, these relations are

$\displaystyle G_{M}=$
$\displaystyle\left(3m^{*}+m\right)\frac{m}{3m^{*}}g_{1}^{\gamma}+\left(m^{*}-m\right)\frac{m}{3}g_{2}^{\gamma}$
(85) $\displaystyle G_{E}=$
$\displaystyle\left(m^{*}-m\right)\frac{m}{3m^{*}}\left(g_{1}^{\gamma}+m^{*}g_{2}^{\gamma}\right)~{}.$

The decay width for $B_{QQ}^{*}\rightarrow B_{QQ}\gamma$ is given as

$\Gamma=\frac{3\alpha}{4}\frac{k_{\gamma}^{3}}{m^{2}}\left(3G_{E}^{2}+G_{M}^{2}\right)~{},$
(86)

where $k_{\gamma}$ is the energy of the photon,
$k_{\gamma}=\frac{m^{*2}-m^{2}}{2m^{*}}$. The results for $G_{E}$ and $G_{M}$
are presented in Table 10.

Transition | $\left|G_{E}\right|$ | $\left|G_{M}\right|$
---|---|---
$\Xi^{*++}_{cc}$$\rightarrow$$\Xi^{++}_{cc}$$\gamma$ | $0.00\pm 0.00$ | $1.78\pm 0.4$
$\Xi^{*+}_{cc}$$\rightarrow$$\Xi^{+}_{cc}$$\gamma$ | $0.00\pm 0.00$ | $0.11\pm 0.02$
$\Xi^{*0}_{bb}$$\rightarrow$$\Xi^{0}_{bb}$$\gamma$ | $0.00\pm 0.00$ | $3.41\pm 1.03$
$\Xi^{*-}_{bb}$$\rightarrow$$\Xi^{-}_{bb}$$\gamma$ | $0.00\pm 0.00$ | $0.22\pm 0.06$
$\Omega^{*+}_{cc}$$\rightarrow$$\Omega^{+}_{cc}$$\gamma$ | $0.00\pm 0.00$ | $0.52\pm 0.11$
$\Xi^{*-}_{bb}$$\rightarrow$$\Omega^{-}_{bb}$$\gamma$ | $0.00\pm 0.00$ | $1.18\pm 0.34$
Table 10: The electric quadrupole and magnetic dipole form factors for the
shown transitions [17].

From the decay width expression, it follows that it is very sensitive to the
mass difference $\Delta m=m^{*}-m$. A tiny change in mass difference leads to
a significant variation for the decay width. For example, consider the decay
width for $\Omega_{cc}^{*}\rightarrow\Omega_{cc}\gamma$ in our case (when
$\Delta m=0.86MeV$) $\Gamma\sim 0.23keV$. Our result is in good agreement with
lattice QCD calculations [82] but considerably different from the other
approaches [11]. One possible reason for this discrepancy is that VDM was not
working quite well for doubly heavy baryon systems. In order to check how VDM
works for doubly heavy baryon systems, it is necessary to calculate the
radiative decays of them directly and compare the results with VDM
predictions.

At the end of this section, we would like to add following remarks. As we
already emphasized that the weak decays of double heavy baryons play a crucial
role in deeply understanding the dynamics of doubly heavy baryons. First of
all, the study of weak decay allows us to determine the CKM matrix elements
more precisely and second, provides us with more useful information about QCD
dynamics at a large distance. The semileptonic decays of doubly heavy baryons
have very rich phenomenology and can be divided into two classes;

  * •

Decays which take place at tree level of the charged current

  * •

Decays which take place at loop level with flavor changing neutral current
(FCNC).

The weak decays of doubly heavy baryons within the light-front quark model are
intensively studied [23, 29, 30]. Meanwhile, the weak decays via charged
currents are studied within QCD sum rules [60] and light-cone sum rules [61,
62], respectively. The weak decays via flavor-changing neutral current (FCNC)
of doubly heavy baryons in the framework of light-cone sum rules have not been
studied yet. These studies are critical for establishing the ”right” picture
of doubly heavy baryons. Because many nonleptonic decays are defined by the
same transition form factors, these results are also quite timely for
nonleptonic decays too.

##  VI Summary and conclusion

The constituent quark model predicts the existence of baryons containing three
quarks. However, although most of the baryons containing one heavy quark have
already been observed, baryons with two or three heavy quarks not being
discovered were a long-standing puzzle for the theory. The exciting news had
been given by SELEX collaboration with the discovery of $\Xi_{cc}$ first, and
confirmation was achieved by the LHCb collaboration after. While the
experimental hunt has been going for discovering other baryons with doubly or
triply heavy quarks, theoretical studies can also be enlightening for
understanding the internal nature of these baryons and the determination of
the fundamental properties as well as QCD parameters.

In this study, the properties of the doubly heavy baryons are investigated
within the QCD sum rules, considering that these baryons are formed by a light
quark attached to a doubly heavy diquark. We summarized the analyses conducted
to determine the mass and strong coupling constants of the doubly heavy
baryons. Moreover, our results obtained within QCD sum rules are compared with
the other approaches.

Since most of the doubly heavy baryons predicted by the quark model have not
been discovered yet, theoretical analysis can play a significant role in
guiding the experiments. The experimental studies of the spectroscopic
parameters and the weak decays of doubly heavy baryons constitute one of the
essential directions in physical program of the colliders. Measurements of
their decay characteristics can significantly extend our knowledge of the
dynamics of these systems. From the theoretical point of view, their studies
help develop new form of the heavy quark effective theory for doubly heavy
baryons. Hopefully, the presented results on mass, strong coupling constants,
and the electromagnetic decays of doubly heavy baryons will be used in further
theoretical studies and will be tested at the LHCb experiment in the near
future.

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