Strong Coupling Constants of the Doubly Heavy $\Xi_{QQ}$ Baryons with $\pi$ Meson

In order to calculate the strong coupling constants among doubly heavy baryons and light pseudoscalar $\pi$ meson, we start our discussion by considering the following light-cone correlation function:

$\displaystyle\Pi(p,q)=i\int d^{4}xe^{ipx}\left<{\cal P}(q)|{\cal T}\left\{\eta(x)\bar{\eta}(0)\right\}|0\right>~{},$

(1)

where ${\cal P}(q)$ denotes the pseudoscalar mesons of momentum $q$, ${\cal T}$ is the time ordering operator, and $\eta$ represents the interpolating currents of the doubly heavy baryons. The general expressions of the interpolating currents for the spin–1/2 doubly heavy baryons in their symmetric and antisymmetric forms can be written as

	$\displaystyle\eta^{S}$	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{2}}\epsilon_{abc}\Bigg{\{}(Q^{aT}Cq^{b})\gamma_{5}Q^{\prime c}+(Q^{\prime aT}Cq^{b})\gamma_{5}Q^{c}+t(Q^{aT}C\gamma_{5}q^{b})Q^{\prime c}+t(Q^{\prime aT}C\gamma_{5}q^{b})Q^{c}\Bigg{\}},$
	$\displaystyle\eta^{A}$	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{6}}\epsilon_{abc}\Bigg{\{}2(Q^{aT}CQ^{\prime b})\gamma_{5}q^{c}+(Q^{aT}Cq^{b})\gamma_{5}Q^{\prime c}-(Q^{\prime aT}Cq^{b})\gamma_{5}Q^{c}+2t(Q^{aT}C\gamma_{5}Q^{\prime b})q^{c}$		(2)
		$\displaystyle+$	$\displaystyle t(Q^{aT}C\gamma_{5}q^{b})Q^{\prime c}-t(Q^{\prime aT}C\gamma_{5}q^{b})Q^{c}\Bigg{\}},$

where $t$ is an arbitrary auxiliary parameter and the case, $t=-1$ corresponds to the Ioffe current. Here $Q^{(^{\prime})}$ and $q$ stand for the heavy and light quarks respectively; $a$, $b$, and $c$ are the color indices, $C$ stands for the charge conjugation operator and $T$ denotes the transposition. For the doubly heavy baryons with two identical heavy quarks, the antisymmetric form of the interpolating current is zero and we just need to employ the symmetric form, $\eta^{S}$. In the following, we calculate the correlation function in Eq. 1 in two different windows.

To obtain the physical side of correlation function, we insert complete sets of hadronic states with the same quantum numbers as the interpolating currents and isolate the ground state. After performing the integration over four-$x$, we get

$\displaystyle\Pi^{\text{Phys.}}(p,q)=\frac{\langle 0|\eta|B_{2}(p)\rangle\langle B_{2}(p){\cal P}(q)|B_{1}(p+q)\rangle\langle B_{1}(p+q)|\bar{\eta}|0\rangle}{(p^{2}-m_{1}^{2})[(p+q)^{2}-m_{2}^{2}]}+\cdots~{},$

(3)

where dots in the above equation stand for the contribution of the higher states and continuum. To proceed we introduce the matrix elements for spin-1/2 baryons as

	$\displaystyle\langle 0|\eta|B_{2}(p,r)\rangle$	$\displaystyle=$	$\displaystyle\lambda_{B_{2}}u(p,r)~{},$
	$\displaystyle\langle B_{1}(p+q,s)|\bar{\eta}|0\rangle$	$\displaystyle=$	$\displaystyle\lambda_{B_{1}}\bar{u}(p+q,s)~{},$
	$\displaystyle\langle B_{2}(p,r){\cal P}(q)|B_{1}(p+q,s)\rangle$	$\displaystyle=$	$\displaystyle g_{B_{1}B_{2}{\cal P}}\bar{u}(p,r)\gamma_{5}u(p+q,s)~{},$		(4)

where $g_{B_{1}B_{2}{\cal P}}$, representing the strong decay $B_{1}\rightarrow B_{2}{\cal P}$, is the strong coupling constant among the baryons $B_{1}$ and $B_{2}$ as well as the ${\cal P}$ meson, $\lambda_{B_{1}}$ and $\lambda_{B_{2}}$ are the residues of the corresponding baryons and $u(q,s)$ is Dirac spinor with spin $s$. Putting the above equations all together, and performing summation over spins, we get the following representation of the correlator for the phenomenological side:

$\displaystyle\Pi^{\text{Phys.}}(p,q)=\frac{g_{B_{1}B_{2}{\cal P}}\lambda_{B_{1}}\lambda_{B_{2}}}{(p^{2}-m_{B_{2}}^{2})[(p+q)^{2}-m_{B_{1}}^{2}]}[\hbox to0.0pt{/\hss}q\hbox to0.0pt{/\hss}p\gamma_{5}+\cdots~{}],$

(5)

where dots represent other structures come from spin summation as well as the contributions of higher states and continuum. We will use the explicitly presented structure to extract the value of the strong coupling constant, $g_{B_{1}B_{2}{\cal P}}$. We apply the double Borel transformation with respect to the variables $p^{2}_{1}=(p+q)^{2}$ and $p^{2}_{2}=p^{2}$:

	$\displaystyle{\cal B}_{p_{1}}(M_{1}^{2}){\cal B}_{p_{2}}(M_{2}^{2})\Pi^{\text{Phys.}}(p,q)$	$\displaystyle\equiv$	$\displaystyle\Pi^{\text{Phys.}}(M^{2})$		(6)
		$\displaystyle=$	$\displaystyle g_{B_{1}B_{2}{\cal P}}\lambda_{B_{1}}\lambda_{B_{2}}e^{-m_{B_{1}}^{2}/M_{1}^{2}}e^{-m_{B_{2}}^{2}/M_{2}^{2}}\hbox to0.0pt{/\hss}q\hbox to0.0pt{/\hss}p\gamma_{5}~{}+\cdots~{},$

where $M^{2}=M^{2}_{1}M^{2}_{2}/(M^{2}_{1}+M^{2}_{2})$ and the Borel parameters $M^{2}_{1}$ and $M^{2}_{2}$ for the problem under consideration are chosen to be equal as the masses of the initial and final state baryons are the same. Hence $M^{2}_{1}=M^{2}_{2}=2M^{2}$.

In QCD side, the correlation function is calculated in deep Euclidean region with the help of OPE. To proceed, we need to determine the correlation function using the quark propagators and distribution amplitudes of the $\pi$ meson. The $\Pi^{QCD}(p,q)$ can be written in the following general form:

$$\Pi^{\text{QCD}}(p,q)=\Pi\big{(}(p+q)^{2},p^{2}\big{)}\hbox to0.0pt{/\hss}q\hbox to0.0pt{/\hss}p\gamma_{5}+\cdots~{},$$

(7)

where the $\Pi\big{(}(p+q)^{2},p^{2}\big{)}$ is an invariant function that should be calculated in terms of QCD degrees of freedom as well as the parameters inside the DAs.

Inserting, for instance, $\eta^{S}$ in the CF and using the Wick theorem to contract all the heavy quark fields we get the following expression in terms of the heavy quark propagators and $\pi$ meson matrix elements:

	$\displaystyle\Pi^{\text{QCD}}_{\rho\sigma}(p,q)$	$\displaystyle=$	$\displaystyle\frac{i}{2}\epsilon_{abc}\epsilon_{a^{\prime}b^{\prime}c^{\prime}}\int d^{4}xe^{iq.x}\langle\pi(q)|\bar{q}^{c^{\prime}}_{\alpha}(0)q^{c}_{\beta}(x)|0\rangle\Bigg{\{}\Bigg{[}\Big{(}\tilde{S}^{aa^{\prime}}_{Q}(x)\Big{)}_{\alpha\beta}\Big{(}\gamma_{5}S^{bb^{\prime}}_{Q^{\prime}}(x)\gamma_{5}\Big{)}_{\rho\sigma}$		(8)
		$\displaystyle+$	$\displaystyle\Big{(}\gamma_{5}S_{Q^{\prime}}^{bb^{\prime}}(x)C\Big{)}_{\rho\alpha}\Big{(}CS^{aa^{\prime}}_{Q}(x)\gamma_{5}\Big{)}_{\beta\sigma}+t\Big{\{}\Big{(}\gamma_{5}\tilde{S}_{Q}^{aa^{\prime}}(x)\Big{)}_{\alpha\beta}\Big{(}\gamma_{5}S^{bb^{\prime}}_{Q^{\prime}}(x)\Big{)}_{\rho\sigma}$
		$\displaystyle+$	$\displaystyle\Big{(}\tilde{S}_{Q}^{aa^{\prime}}(x)\gamma_{5}\Big{)}_{\alpha\beta}\Big{(}S^{bb^{\prime}}_{Q^{\prime}}(x)\gamma_{5}\Big{)}_{\rho\sigma}+\Big{(}\gamma_{5}S_{Q^{\prime}}^{bb^{\prime}}(x)C\gamma_{5}\Big{)}_{\rho\alpha}\Big{(}CS^{aa^{\prime}}_{Q}(x)\Big{)}_{\beta\sigma}$
		$\displaystyle-$	$\displaystyle\Big{(}S_{Q^{\prime}}^{bb^{\prime}}(x)C\Big{)}_{\rho\alpha}\Big{(}\gamma_{5}CS^{aa^{\prime}}_{Q}(x)\gamma_{5}\Big{)}_{\beta\sigma}\Big{\}}+t^{2}\Big{\{}\Big{(}\gamma_{5}\tilde{S}_{Q}^{aa^{\prime}}(x)\gamma_{5}\Big{)}_{\alpha\beta}\Big{(}S^{bb^{\prime}}_{Q^{\prime}}(x)\Big{)}_{\rho\sigma}$
		$\displaystyle-$	$\displaystyle\Big{(}S_{Q^{\prime}}^{bb^{\prime}}(x)C\gamma_{5}\Big{)}_{\rho\alpha}\Big{(}\gamma_{5}CS^{aa^{\prime}}_{Q}(x)\Big{)}_{\beta\sigma}\Big{\}}\Bigg{]}+\Bigg{(}Q\longleftrightarrow Q^{\prime}\Bigg{)}\Bigg{\}},$

where $\tilde{S}=CS^{T}C$, and $\langle\pi(q)|\bar{q}^{b^{\prime}}_{\alpha}(0)q^{b}_{\beta}(x)|0\rangle$ are the matrix elements for the light quark contents of the doubly heavy baryons. To proceed, we need to know the explicit expression for the heavy quark propagator that is

	$\displaystyle S_{Q}(x)$	$\displaystyle=$	$\displaystyle{m_{Q}^{2}\over 4\pi^{2}}{K_{1}(m_{Q}\sqrt{-x^{2}})\over\sqrt{-x^{2}}}-i{m_{Q}^{2}\hbox to0.0pt{/\hss}{x}\over 4\pi^{2}x^{2}}K_{2}(m_{Q}\sqrt{-x^{2}})$
		$\displaystyle-$	$\displaystyle ig_{s}\int{d^{4}k\over(2\pi)^{4}}e^{-ikx}\int_{0}^{1}du\Bigg{[}{\hbox to0.0pt{/\hss}k+m_{Q}\over 2(m_{Q}^{2}-k^{2})^{2}}G^{\mu\nu}(ux)\sigma_{\mu\nu}+{u\over m_{Q}^{2}-k^{2}}x_{\mu}G^{\mu\nu}\gamma_{\nu}\Bigg{]}+...~{},$

where $K_{1}$ and $K_{2}$ are the modified Bessel functions of the second kind, and $G_{ab}^{\mu\nu}\equiv G_{A}^{\mu\nu}t_{ab}^{A}$ with $A=1,\,2\,\ldots 8$, and $t^{A}=\lambda^{A}/2$, where $\lambda^{A}$ are the Gell-Mann matrices. The first two terms correspond to perturbative or free part and the rest belong to the interacting parts.

The next step is to use the heavy quark propagator and the matrix elements $\langle\pi(q)|\bar{q}^{b^{\prime}}_{\alpha}(0)q^{b}_{\beta}(x)|0\rangle$ in Eq. 8. This leads to different kinds of contributions to the CF. Figs. 1 and 2 are the Feynman diagrams correspond to the leading and next-to-leading order contributions, respectively which are considered in this work. Only matrix elements corresponding to these diagrams are available. To calculate the leading order contribution, the heavy quark propagators are replaced by just their perturbative parts. This contribution can be computed using $\pi$ meson two-particle DAs of twist two and higher.

The next-to-leading order contributions can also be calculated by choosing the gluonic parts in Eq. 2 for one of the heavy quark propagators and leaving the other with its perturbative term. They can be expressed in terms of pion three particles DAs. The terms involving more than one gluon field that proportional to four-particle DAs or more are neglected as they are not available

Now, we concentrate on the strong decay $\Xi_{cc}^{++}\rightarrow\Xi_{cc}^{++}\pi^{0}$ with the aim of calculating the corresponding strong coupling constant $g_{\Xi_{cc}^{++}\Xi_{cc}^{++}\pi^{0}}$. The other channels have similar procedures. To proceed, we replace the heavy quark propagators in 8 by their explicit expression and perform the summation over the color indices by applying the replacement

$$\overline{u}_{\alpha}^{a}(x)u_{\beta}^{a^{\prime}}(0)\rightarrow\frac{1}{3}\delta_{aa^{\prime}}\overline{u}_{\alpha}(x)u_{\beta}(0).$$

(10)

Now, using the expression

$$\overline{u}_{\alpha}(u)u_{\beta}(0)\equiv\frac{1}{4}\Gamma_{\beta\alpha}^{J}\overline{u}(x)\Gamma^{J}u(0),$$

(11)

one can relate the CF to the DAs of the pion with different twists. Here the summation over $J$ runs as

$$\Gamma^{J}=\mathbf{1,\ }\gamma_{5},\ \gamma_{\mu},\ i\gamma_{5}\gamma_{\mu},\ \sigma_{\mu\nu}/\sqrt{2}.$$

(12)

Following the similar way one can calculate the contributions involving the gluon field.

As a result, the CF is found in terms of the QCD parameters as well as the matrix elements

	$\displaystyle\langle\pi^{0}(q)|\overline{u}(x)\Gamma^{J}u(0)|0\rangle,$
	$\displaystyle\langle\pi^{0}(q)|\overline{u}(x)\Gamma^{J}G_{\mu\nu}(vx)u(0)|0\rangle,$		(13)

whose expressions in terms of the wave functions of the pion with different twists are given in the Appendix.

Inserting the expression of the above-mentioned matrix elements in term of wave functions of different twists we get the CF in $x$ space. This is followed by the Fourier and Borel transformations as well as continuum subtraction. To proceed we need to perform the Fourier transformation of the following kind:

	$\displaystyle T_{[~{}~{},\alpha,\alpha\beta]}(p,q)$	$\displaystyle=$	$\displaystyle i\int d^{4}x\int_{0}^{1}dv\int{\cal D}\alpha e^{ip.x}\big{(}x^{2}\big{)}^{n}[e^{i(\alpha_{\bar{q}}+v\alpha_{g})q.x}\mathcal{G}(\alpha_{i}),e^{iq.x}f(u)]$		(14)
		$\displaystyle\times$	$\displaystyle[1,x_{\alpha},x_{\alpha}x_{\beta}]K_{\mu}(m_{Q}\sqrt{-x^{2}})K_{\nu}(m_{Q}\sqrt{-x^{2}}),$

where the expressions in the brackets denote different possibilities arise in the calculations, the blank subscript in the left hand side indicates no indices regarding no $x_{\alpha}$ in the configuration, $\mathcal{G}(\alpha_{i})$ and $f(u)$ represent wave functions coming from the three and two-particle matrix elements and $n$ is a positive integer. The measure

$$\int\mathcal{D}\alpha=\int_{0}^{1}d\alpha_{q}\int_{0}^{1}d\alpha_{\bar{q}}\int_{0}^{1}d\alpha_{g}\delta(1-\alpha_{q}-\alpha_{\bar{q}}-\alpha_{g}),$$

is used in the calculations. To start the Fourier transformation, we use

$$(x^{2})^{n}=(-1)^{n}\frac{d^{n}}{d\beta^{n}}\big{(}e^{-\beta x^{2}}\big{)}\arrowvert_{\beta=0}.$$

(15)

for positive integer $n$ and

$\displaystyle x_{\alpha}e^{iP.x}=(-i)\frac{d}{dP^{\alpha}}e^{iP.x}.$

(16)

We also use the following representation of the Bessel functions $K_{\nu}$ (see also Ref. [44]):

$$K_{\nu}(m_{Q}\sqrt{-x^{2}})=\frac{\Gamma(\nu+1/2)~{}2^{\nu}}{\sqrt{\pi}m_{Q}^{\nu}}\int_{0}^{\infty}dt~{}\cos(m_{Q}t)\frac{(\sqrt{-x^{2}})^{\nu}}{(t^{2}-x^{2})^{\nu+1/2}}.$$

(17)

As an example let us consider the following generic form:

	$\displaystyle{\cal Z}_{\alpha\beta}(p,q)$	$\displaystyle=$	$\displaystyle i\int d^{4}x\int_{0}^{1}dv\int{\cal D}\alpha e^{i[p+(\alpha_{\bar{q}}+v\alpha_{g})q].x}\mathcal{G}(\alpha_{i})\big{(}x^{2}\big{)}^{n}$		(18)
		$\displaystyle\times$	$\displaystyle x_{\alpha}x_{\beta}K_{\mu}(m_{Q}\sqrt{-x^{2}})K_{\nu}(m_{Q}\sqrt{-x^{2}}).$

We substitute Eqs. 15, 16 and 17 into 18. Then to perform the $x$-integration we go to the Euclidean space by Wick rotation and get

	$\displaystyle{\cal Z}_{\alpha\beta}(p,q)$	$\displaystyle=$	$\displaystyle\frac{i\pi^{2}2^{\mu+\nu-2}}{m_{Q_{1}}^{2\mu}m_{Q_{2}}^{2\nu}}\int\mathcal{D}\alpha\int_{0}^{1}dv\int_{0}^{1}dy_{1}\int_{0}^{1}dy_{2}\frac{\partial}{\partial P_{\alpha}}\frac{\partial}{\partial P_{\beta}}\frac{\partial^{n}}{\partial\beta^{n}}$		(19)
		$\displaystyle\times$	$\displaystyle\dfrac{y_{1}^{\mu-1}y_{2}^{\nu-1}}{(y_{1}+y_{2}+\beta)^{2}}e^{-\frac{1}{4}\big{(}\frac{m_{Q_{1}}^{2}}{y_{1}}+\frac{m_{Q_{2}}^{2}}{y_{2}}-\frac{P^{2}}{y_{1}+y_{2}+\beta}\big{)}},$

where $P=p+q(v\alpha_{g}+\alpha_{q})$ . Changing variables from $y_{1}$ and $y_{2}$ to $\rho$ and $z$ as

$$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\rho=y_{1}+y_{2},~{}~{}~{}~{}~{}~{}~{}~{}z=\frac{y_{1}}{y_{1}+y_{2}},$$

(20)

and taking derivative with respect to $P_{\alpha}$ and $P_{\beta}$, we get

	$\displaystyle{\cal Z}_{\alpha\beta}(p,q)$	$\displaystyle=$	$\displaystyle\frac{i\pi^{2}2^{\mu+\nu-4}}{m_{Q_{1}}^{2\mu}m_{Q_{2}}^{2\nu}}\int\mathcal{D}\alpha\int_{0}^{1}dv\int_{0}^{\infty}d\rho\int_{0}^{1}dz\frac{\partial^{n}}{\partial\beta^{n}}z^{\mu-1}\bar{z}^{\nu-1}\frac{\rho^{\nu+\mu-1}}{(\rho+\beta)^{4}}e^{-\frac{1}{4}\big{(}\frac{m_{Q_{1}}^{2}\bar{z}+m_{Q_{2}}^{2}z}{z\bar{z}\rho}+\frac{P^{2}}{\beta+\rho}\big{)}}$		(21)
		$\displaystyle\times$	$\displaystyle\Big{[}p_{\alpha}p_{\beta}+(v\alpha_{g}+\alpha_{q})(p_{\alpha}q_{\beta}+q_{\alpha}p_{\beta})+(v\alpha_{g}+\alpha_{q})^{2}q_{\alpha}q_{\beta}+2(\rho+\beta)g_{\alpha\beta}\Big{]}.$

Now we perform the double Borel transformation using

$${\cal B}_{p_{1}}(M_{1}^{2}){\cal B}_{p_{2}}(M_{2}^{2})e^{b(p+uq)^{2}}=M^{2}\delta(b+\frac{1}{M^{2}})\delta(u_{0}-u)e^{\frac{-q^{2}}{M_{1}^{2}+M_{2}^{2}}},$$

(22)

where $u_{0}=M_{1}^{2}/(M_{1}^{2}+M_{2}^{2})$. After integrating over $\rho$ we have

	$\displaystyle{\cal Z}_{\alpha\beta}(M^{2})$	$\displaystyle=$	$\displaystyle\frac{i\pi^{2}2^{4-\mu-\nu}e^{\frac{-q^{2}}{M_{1}^{2}+M_{2}^{2}}}}{M^{2}m_{Q_{1}}^{2\mu}m_{Q_{2}}^{2\nu}}\int\mathcal{D}\alpha\int_{0}^{1}dv\int_{0}^{1}dz\frac{\partial^{n}}{\partial\beta^{n}}e^{-\frac{m_{Q_{1}}^{2}\bar{z}+m_{Q_{2}}^{2}z}{z\bar{z}(M^{2}-4\beta)}}z^{\mu-1}\bar{z}^{\nu-1}(M^{2}-4\beta)^{\mu+\nu-1}$		(23)
		$\displaystyle\times$	$\displaystyle\delta[u_{0}-(\alpha_{q}+v\alpha_{g})]\Big{[}p_{\alpha}p_{\beta}+(v\alpha_{g}+\alpha_{q})(p_{\alpha}q_{\beta}+q_{\alpha}p_{\beta})+(v\alpha_{g}+\alpha_{q})^{2}q_{\alpha}q_{\beta}$
			$\displaystyle+\frac{M^{2}}{2}g_{\alpha\beta}\Big{]}.$

The next step is to perform the continuum subtraction in order to more suppress the contribution of the higher states and continuum. The subtraction procedures for different systems are described in Ref. [44] in details. When the masses of the initial and final baryonic states are equal, as we stated previously, we set $M_{1}^{2}=M_{2}^{2}=2M^{2}$. In this case, the double spectral density is concentrated near the diagonal $s_{1}=s_{2}$ and reduces to a single representation $s$ (see also Ref. [45] and references therein) and for the continuum subtraction more simple expressions are derived, which are not sensitive to the shape of the duality region. For the case, $M_{1}^{2}=M_{2}^{2}=2M^{2}$ and $u_{0}=1/2$, the subtraction procedure is explained in Ref. [45] in details, which we use in the calculations.

By calculation of all the Fourier integrals and applying the Borel transformation and continuum subtraction, for QCD side of the calculations in Borel Scheme, we get,

$\displaystyle\Pi^{QCD}(M^{2},s_{0},t)=\big{[}\Pi^{(0)}(M^{2},s_{0},t)+\Pi^{(GG)}(M^{2},s_{0},t)\big{]}\hbox to0.0pt{/\hss}q\hbox to0.0pt{/\hss}p\gamma_{5}+\cdots,$

(24)

where the functions $\Pi^{(0)}(M^{2},s_{0},t)$ and $\Pi^{(GG)}(M^{2},s_{0},t)$ are obtained as

	$\displaystyle\Pi^{(0)}(M^{2},s_{0},t)$	$\displaystyle=$	$\displaystyle\dfrac{e^{-\frac{m_{\pi}^{2}}{4M^{2}}}}{96\pi^{2}\sqrt{2}}\int_{0}^{1}dz\frac{1}{z}e^{\frac{-m_{c}^{2}}{M^{2}z\bar{z}}}\Bigg{\{}3f_{\pi}m_{\pi}^{2}m_{c}^{3}(t^{2}-1){\mathbb{A}}(u_{0})$
		$\displaystyle+$	$\displaystyle 2e^{-\frac{4m_{c}^{2}}{M^{2}}}\int_{4m_{c}^{2}}^{s_{0}}dse^{-\frac{s}{M^{2}}}z\bigg{[}3f_{\pi}m_{\pi}^{2}m_{c}(t^{2}-1)\bar{z}{\mathbb{A}}(u_{0})$
		$\displaystyle-$	$\displaystyle 6f_{\pi}m_{c}(s-4m_{c}^{2})(t^{2}-1)\bar{z}\varphi_{\pi}(u_{0})$
		$\displaystyle-$	$\displaystyle\mu_{\pi}(\tilde{\mu}_{\pi}^{2}-1)\big{[}-4m_{c}^{2}t+3(s-4m_{c}^{2})(t-1)^{2}z\bar{z}\big{]}\varphi_{\sigma}(u_{0})\bigg{]}$
		$\displaystyle+$	$\displaystyle 6e^{-\frac{4m_{c}^{2}}{M^{2}}}(t^{2}-1)\int_{4m_{c}^{2}}^{s_{0}}dse^{-\frac{s}{M^{2}}}\int_{0}^{1}dv\int\mathcal{D}\alpha$
		$\displaystyle\times$	$\displaystyle\Bigg{[}f_{\pi}m_{\pi}^{2}m_{c}\delta[u_{0}-(\alpha_{q}+v\alpha_{g})]\bigg{(}-2z(v-1/2){\cal A}_{\parallel}(\alpha_{i})+2\bar{z}{\cal V}_{\perp}(\alpha_{i})$
		$\displaystyle+$	$\displaystyle(2z-3){\cal V}_{\parallel}(\alpha_{i})\bigg{)}+2\mu_{\pi}\delta^{\prime}[u_{0}-(\alpha_{q}+v\alpha_{g})](s-4m_{c}^{2})z\bar{z}(v-1/2){\cal T}(\alpha_{i})\Bigg{]}\Bigg{\}},$

and

	$\displaystyle\Pi^{(GG)}(M^{2},s_{0},t)$	$\displaystyle=$	$\displaystyle\frac{\langle g_{s}^{2}G^{2}\rangle e^{-\frac{m_{\pi}^{2}}{4M^{2}}}}{6912\sqrt{2}\pi^{2}m_{c}M^{6}}\int_{0}^{1}dz\frac{1}{z^{2}\bar{z}^{4}}e^{-\frac{m_{c}^{2}}{M^{2}z\bar{z}}}$		(26)
		$\displaystyle\times$	$\displaystyle\Bigg{\{}\bar{z}^{2}\Bigg{[}-3f_{\pi}m_{\pi}^{2}(1-t^{2})\Big{(}6M^{6}z^{2}\bar{z}^{4}+6M^{4}m_{c}^{2}z\bar{z}^{3}+3M^{2}m_{c}^{4}\bar{z}^{2}-2m_{c}^{6}\Big{)}{\mathbb{A}}(u_{0})$
		$\displaystyle+$	$\displaystyle 4M^{2}m_{c}\bar{z}\Bigg{(}-3f_{\pi}M^{2}m_{c}(1-t^{2})z\big{[}2m_{c}^{2}+M^{2}\bar{z}(5z-3)\big{]}\varphi_{\pi}(u_{0})$
		$\displaystyle+$	$\displaystyle(\tilde{\mu}_{\pi}^{2}-1)\mu_{\pi}\Big{[}2m_{c}^{4}\left(t^{2}+1\right)+[(1+t^{2})(1-4z)-6t](M^{2}m_{c}^{2}z+M^{4}z^{2}\bar{z})\Big{]}\varphi_{\sigma}(u_{0})\Bigg{)}$
		$\displaystyle+$	$\displaystyle 72f_{\pi}M^{8}(1-t^{2})z^{2}\bar{z}^{4}\Big{(}1-e^{-\frac{(s_{0}-4m_{c}^{2})}{M^{2}}}\Big{)}\varphi_{\pi}(u_{0})\Bigg{]}$
		$\displaystyle+$	$\displaystyle 6M^{2}(1-t)\int_{0}^{1}dv\int{\cal D}\alpha\Bigg{[}f_{\pi}m_{\pi}^{2}(1+t)\bar{z}^{3}\delta[u_{0}-(\alpha_{q}+v\alpha_{g})]\Bigg{(}2m_{c}^{4}{\cal V}_{\perp}(\alpha_{i})$
		$\displaystyle+$	$\displaystyle\Big{[}-2m_{c}^{4}+M^{2}m_{c}^{2}(1+2z)z+M^{4}(1+2z)z^{2}\bar{z}\Big{]}{\cal V}_{\parallel}(\alpha_{i})$
		$\displaystyle-$	$\displaystyle(2v-1)\Big{[}m_{c}^{4}+M^{2}m_{c}^{2}(1+2z)z+M^{4}(1+2z)z^{2}\bar{z}\Big{]}{\cal A}_{\parallel}(\alpha_{i})\Bigg{)}$
		$\displaystyle+$	$\displaystyle\mu_{\pi}(1-t)(1+2v)z\bar{z}^{3}\delta^{\prime}[u_{0}-(\alpha_{q}+v\alpha_{g})](M^{2}m_{c}^{3}+M^{4}m_{c}z\bar{z}){\cal T}(\alpha_{i})\Bigg{]}\Bigg{\}}.$

The sum rule for the coupling constant under study is found by matching the coefficients of the structure $\hbox to0.0pt{/\hss}q\hbox to0.0pt{/\hss}p\gamma_{5}$ from both the physical and QCD sides. As a result, we get:

$\displaystyle g_{B_{1}B_{2}{\cal P}}(M^{2},s_{0},t)=\frac{1}{\lambda_{\Xi_{cc}}\lambda_{\Xi_{cc}}}e^{\frac{m_{B_{1}}^{2}}{2M^{2}}}e^{\frac{m_{B_{2}}^{2}}{2M^{2}}}\big{[}\Pi^{(0)}(M^{2},s_{0},t)+\Pi^{(GG)}(M^{2},s_{0},t)\big{]}.$

(27)

As is seen, the sum rules for coupling constants contain the residues of doubly heavy baryons, which are borrowed from Ref.[10]. Similarly, we obtain the sum rules for other strong coupling constants under consideration.