$\sigma$ exchange in the one-boson exchange model involving the ground state octet baryons

One can write down a dispersive representation of the ${\mathbb{B}}\bar{\mathbb{B}}$ scattering amplitude corresponding to Fig. 1(a) as

$\displaystyle\mathcal{M}^{\rm{DR}}_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\bar{{\mathbb{B}}}{\mathbb{B}},0}(s)=\frac{s-4m_{\mathbb{B}}^{2}}{2\pi i}\int_{4M_{\pi}^{2}}^{+\infty}\frac{{\rm{disc}}\left[\mathcal{M}^{\rm{DR}}_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\bar{{\mathbb{B}}}{\mathbb{B}},0}(z)\right]}{(z-s)(z-4m_{\mathbb{B}}^{2})}{\rm{d}}z\,.$

(4)

Here a once-subtracted dispersive integral is employed to facilitate the convergence of the dispersive integral. The threshold $s=4m_{\mathbb{B}}^{2}$ is chosen as the subtraction point, and we set the subtraction constant $\mathcal{M}^{\rm{DR}}(s=4m_{\mathbb{B}}^{2})=0$ as Eq. (3) since the two amplitudes will be matched later.

In order to capture the $\pi\pi$ rescattering in the $\sigma$ region and avoid the interference from other resonances, e.g., the $f_{0}(980)$, the upper limit of the dispersive integral in Eq. (4) is set to $s_{0}=(0.8\ {\rm{GeV}})^{2}$, as in Ref. [39] (see also Ref. [42]). We will investigate the impact of varying the upper limit of the integration on the final result and regard it as a part of the uncertainty of the coupling constants.

Next, let us discuss the discontinuity. Taking into account the unitary relation that is fulfilled by the partial-wave $T$-matrix elements, we can express the discontinuity of the $S$-wave amplitude (here the partial wave refers to that between the pions) along the cut $s\in[4M_{\pi}^{2},+\infty)$ in terms of $T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}(s)$ and $T_{\pi\pi\to\bar{{\mathbb{B}}}{\mathbb{B}},0}(s)$. However, it is crucial to notice that when dealing with systems that involve spins, particularly those containing fermions, kinematical singularities arise [43]. These singularities stem from the definition of the wave functions for the initial and final states. Following Ref. [43], we introduce the kinematical-singularity-free amplitudes,

	$\displaystyle T^{\rm{new}}_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}(s)=\sqrt{s-4m_{\mathbb{B}}^{2}}\,T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}(s)\,,$		(5)
	$\displaystyle T^{\rm{new}}_{\pi\pi\to\bar{{\mathbb{B}}}{\mathbb{B}},0}(s)=\sqrt{s-4m_{\mathbb{B}}^{2}}\,T_{\pi\pi\to\bar{{\mathbb{B}}}{\mathbb{B}},0}(s)\,.$		(6)

Then the unitary relation for the $S$-wave $T$-matrix elements is given by

$\displaystyle{\rm{disc}}\left[\mathcal{M}_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\bar{{\mathbb{B}}}{\mathbb{B}},0}(s)\right]=2i\rho_{\pi}(s)\frac{T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}^{\rm{new}}(s)T^{{\rm{new}}\ *}_{\pi\pi\to\bar{{\mathbb{B}}}{\mathbb{B}},0}(s)}{s-4m_{\mathbb{B}}^{2}}\theta\left(\sqrt{s}-2M_{\pi}\right),$

(7)

where $\rho_{\pi}(s)=\frac{1}{16\pi}\sqrt{\frac{s-4M_{\pi}^{2}}{s}}$ is the two-body phase space factor. Moreover, as we will discuss in detail in Sec. II.2.4, the treatment of kinematical singularity plays a vital role in guaranteeing the self-consistency of the theory.

Furthermore, with the phase conventions outlined in Appendix A and considering the isospin scalar $\pi\pi$ system, we obtain the following relations

	$$\displaystyle T_{\Sigma\bar{\Sigma}\to\pi\pi,0}(s)=-T_{\pi\pi\to\bar{\Sigma}\Sigma,0}(s)\,,$$
	$$\displaystyle T_{\Xi\bar{\Xi}\to\pi\pi,0}(s)=T_{\pi\pi\to\bar{\Xi}\Xi,0}(s)\,,$$
	$$\displaystyle T_{\Lambda\bar{\Lambda}\to\pi\pi,0}(s)=-T_{\pi\pi\to\bar{\Lambda}\Lambda,0}(s)\,,$$
	$$\displaystyle T_{N\bar{N}\to\pi\pi,0}(s)=T_{\pi\pi\to\bar{N}N,0}(s)\,.$$		(8)

Then, we obtain the following discontinuities,

	$$\displaystyle{\rm{disc}}\left[\mathcal{M}_{\Sigma\bar{\Sigma}\to\bar{\Sigma}\Sigma,0}(s)\right]=2i\rho_{\pi}\frac{-\left|T_{\Sigma\bar{\Sigma}\to\pi\pi,0}^{\rm{new}}(s)\right|^{2}}{s-4m_{\Sigma}^{2}}\theta\left(\sqrt{s}-2M_{\pi}\right),$$		(9)
	$$\displaystyle{\rm{disc}}\left[\mathcal{M}_{\Xi\bar{\Xi}\to\bar{\Xi}\Xi,0}(s)\right]=2i\rho_{\pi}\frac{\left|T_{\Xi\bar{\Xi}\to\pi\pi,0}^{\rm{new}}(s)\right|^{2}}{s-4m_{\Xi}^{2}}\theta\left(\sqrt{s}-2M_{\pi}\right),$$		(10)
	$$\displaystyle{\rm{disc}}\left[\mathcal{M}_{\Lambda\bar{\Lambda}\to\bar{\Lambda}\Lambda,0}(s)\right]=2i\rho_{\pi}\frac{-\left|T_{\Lambda\bar{\Lambda}\to\pi\pi,0}^{\rm{new}}(s)\right|^{2}}{s-4m_{\Lambda}^{2}}\theta\left(\sqrt{s}-2M_{\pi}\right),$$		(11)
	$$\displaystyle{\rm{disc}}\left[\mathcal{M}_{N\bar{N}\to\bar{N}N,0}(s)\right]=2i\rho_{\pi}\frac{\left|T_{N\bar{N}\to\pi\pi,0}^{\rm{new}}(s)\right|^{2}}{s-4m_{N}^{2}}\theta\left(\sqrt{s}-2M_{\pi}\right).$$		(12)

To obtain the low-energy $S$-wave amplitudes $T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}(s)$, we need the corresponding chiral baryon-meson Lagrangian. The leading-order (LO) chiral Lagrangian is given by [44],

$$\displaystyle\mathcal{L}_{\mathbb{M}{\mathbb{B}}}^{(1)}=\left\langle\bar{{\mathbb{B}}}(i\not{\mathcal{D}}-m_{0}){\mathbb{B}}\right\rangle+\frac{D}{2}\left\langle\bar{{\mathbb{B}}}\gamma^{\mu}\gamma_{5}\{u_{\mu},{\mathbb{B}}\}\right\rangle+\frac{F}{2}\left\langle\bar{{\mathbb{B}}}\gamma^{\mu}\gamma_{5}\left[u_{\mu},{\mathbb{B}}\right]\right\rangle,$$

(13)

which contains three low-energy constants (LECs), $m_{0}$, $D$ and $F$. Here, $\langle\cdot\rangle$ means trace in the flavor space, the baryon octet are collected in the matrix,

$\displaystyle{\mathbb{B}}=\left(\begin{array}[]{ccc}\frac{1}{\sqrt{2}}\Sigma^{0}+\frac{1}{\sqrt{6}}\Lambda&\Sigma^{+}&p\\ \Sigma^{-}&-\frac{1}{\sqrt{2}}\Sigma^{0}+\frac{1}{\sqrt{6}}\Lambda&n\\ \Xi^{-}&\Xi^{0}&-\frac{2}{\sqrt{6}}\Lambda\end{array}\right),$

(17)

and the chiral vielbein and covariant derivative are given by

$\displaystyle u_{\mu}=iu^{\dagger}\partial_{\mu}u+iu\partial_{\mu}u^{\dagger},\quad\mathcal{D}_{\mu}{\mathbb{B}}=\partial_{\mu}B+\left[\Gamma_{\mu},B\right],\quad\Gamma_{\mu}=\frac{1}{2}\left(u^{\dagger}\partial_{\mu}u+u\partial_{\mu}u^{\dagger}\right),$

(18)

with $u^{2}=U$, $U=\exp\left(i{\sqrt{2}\Phi}/{F_{\pi}}\right)$, where

$\displaystyle\Phi=\left(\begin{array}[]{ccc}\frac{\pi^{0}}{\sqrt{2}}+\frac{\eta}{\sqrt{6}}&\pi^{+}&K^{+}\\ \pi^{-}&-\frac{\pi^{0}}{\sqrt{2}}+\frac{\eta}{\sqrt{6}}&K^{0}\\ K^{-}&\bar{K}^{0}&-\frac{2}{\sqrt{6}}\eta\end{array}\right).$

(22)

Notice that the chiral connection $\Gamma_{\mu}$ containing two pions is a vector, the two pions from that term cannot be in the $S$-wave. As a result, only the $t$- and $u$-channel exchanges depicted in Fig. 2 (a) and (b), which contribute to the LHC part of $T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}(s)$, will be present in the LO calculation. In addition, the LO Lagrangian contains the $\Sigma\Lambda\pi$ coupling terms of the form $\bar{\Lambda}\gamma^{\mu}\gamma_{5}\partial_{\mu}\pi\Sigma$ and $\bar{\Sigma}\gamma^{\mu}\gamma_{5}\partial_{\mu}\pi\Lambda$. Therefore, it is necessary to consider the exchange of $\Sigma$ in the $\Lambda\bar{\Lambda}\to\pi\pi$ process and the exchange of $\Lambda$ in the $\Sigma\bar{\Sigma}\to\pi\pi$ process.

The $\mathcal{O}(p^{2})$ chiral Lagrangian contains the $\bar{{\mathbb{B}}}{\mathbb{B}}\pi\pi$ contact contribution with the $S$-wave pion pair, as illustrated in Fig. 2 (c). At the NLO, the number of LECs increases, and the Lagrangian reads [45, 46],⁴⁴4Here we use the Lagrangian in [46] while the one in Ref. [45] has redundant terms. Note also the ordering of the operators in Ref. [46] is different from that in Refs. [45, 47].

	$\displaystyle\mathcal{L}_{\mathbb{MB}}^{(2)}=$	$\displaystyle\,b_{D}\langle\bar{{\mathbb{B}}}\{\chi_{+},{\mathbb{B}}\}\rangle+b_{F}\langle\bar{{\mathbb{B}}}\left[\chi_{+},{\mathbb{B}}\right]\rangle+b_{0}\langle\bar{{\mathbb{B}}}{\mathbb{B}}\rangle\langle\chi_{+}\rangle$
		$\displaystyle+b_{1}\langle\bar{{\mathbb{B}}}\left[u^{\mu},\left[u_{\mu},{\mathbb{B}}\right]\right]\rangle+b_{2}\langle\bar{{\mathbb{B}}}\{u^{\mu},\{u_{\mu},{\mathbb{B}}\}\}\rangle$
		$\displaystyle+b_{3}\langle\bar{{\mathbb{B}}}\{u^{\mu},\left[u_{\mu},{\mathbb{B}}\right]\}\rangle+b_{4}\langle\bar{{\mathbb{B}}}{\mathbb{B}}\rangle\langle u^{\mu}u_{\mu}\rangle$
		$\displaystyle+ib_{5}\left(\langle\bar{{\mathbb{B}}}\left[u^{\mu},\left[u^{\nu},\gamma_{\mu}\mathcal{D}_{\nu}{\mathbb{B}}\right]\right]\rangle-\langle\bar{{\mathbb{B}}}\overleftarrow{\mathcal{D}}_{\nu}\left[u^{\nu},\left[u^{\mu},\gamma_{\mu}{\mathbb{B}}\right]\right]\rangle\right)$		(23)
		$\displaystyle+ib_{6}\left(\langle\bar{{\mathbb{B}}}\left[u^{\mu},\{u^{\nu},\gamma_{\mu}\mathcal{D}_{\nu}{\mathbb{B}}\}\right]\rangle-\langle\bar{{\mathbb{B}}}\overleftarrow{\mathcal{D}}_{\nu}\{u^{\nu},\left[u^{\mu},\gamma_{\mu}{\mathbb{B}}\right]\}\rangle\right)$
		$\displaystyle+ib_{7}\left(\langle\bar{{\mathbb{B}}}\{u^{\mu},\{u^{\nu},\gamma_{\mu}\mathcal{D}_{\nu}{\mathbb{B}}\}\}\rangle-\langle\bar{{\mathbb{B}}}\overleftarrow{\mathcal{D}}_{\nu}\{u^{\nu},\{u^{\mu},\gamma_{\mu}{\mathbb{B}}\}\}\rangle\right)$
		$\displaystyle+ib_{8}\left(\langle\bar{{\mathbb{B}}}\gamma_{\mu}\mathcal{D}_{\nu}{\mathbb{B}}\rangle-\langle\bar{{\mathbb{B}}}\overleftarrow{\mathcal{D}}_{\nu}\gamma_{\mu}{\mathbb{B}}\rangle\right)\langle u^{\mu}u^{\nu}\rangle+\raisebox{2.15277pt}{\ldots},$

where $\chi_{\pm}=u^{\dagger}\chi u^{\dagger}\pm u\chi^{\dagger}u,\chi=2B_{0}\mathcal{M}$ with $B_{0}$ a constant related to the quark condensate in the chiral limit and $\mathcal{M}$ the light-quark mass matrix. We will use the values of involved LECs from Fit II in Ref. [48], which are $D=0.8$, $F=0.46$, $b_{D}=0.222(20)$, $b_{F}=-0.428(12)$, $b_{0}=-0.714(21)$, $b_{1}=0.515(132)$, $b_{2}=0.148(48)$, $b_{3}=-0.663(155)$, $b_{4}=-0.868(105)$, $b_{5}=-0.643(246)$, $b_{6}=-0.268(334)$, $b_{7}=0.176(72)$, $b_{8}=-0.0694(1638)$.

Using the LO and NLO Lagrangians given in Eqs. (13, 23), we can calculate the tree-level amplitude for the process of ${\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi$ as depicted in Fig. 2. However, in order to determine the final state $\pi\pi$ with $IJ=00$, we need to perform a partial-wave (PW) expansion. The generalized PW expansion of the helicity amplitude for arbitrary spin can be found in Ref. [49]. The final PW amplitude for ${\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi$ reads

$\displaystyle T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,L}(s)=\frac{1}{4\pi}\int{\rm{d}}\Omega\sqrt{\frac{4\pi}{2L+1}}Y^{*}_{L,\lambda_{1}-\lambda_{2}}(\theta,\phi)e^{i(\lambda_{1}-\lambda_{2})\phi}\langle\theta,0;0,0|\hat{T}|0,0;\lambda_{1},\lambda_{2}\rangle\,,$

(24)

where $L$ is the relative orbital angular momentum of the pions, $\lambda_{1}$ and $\lambda_{2}$ are the third components of the helicities of ${\mathbb{B}}$ and $\bar{{\mathbb{B}}}$. The basis is such that ${\mathbb{B}}\bar{\mathbb{B}}$ is expanded in terms of $|\theta_{0},\phi_{0};\lambda_{1},\lambda_{2}\rangle$, and the ${\mathbb{B}}\bar{\mathbb{B}}$ relative momentum is chosen to be along the $z$ axis so that $\theta_{0}=\phi_{0}=0$; $(\theta,\phi)$ are the polar and azimuthal angles of the $\pi\pi$ relative momentum.

For the tree-level $S$-wave amplitude for ${\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi$, the LHC part from the $t$- and $u$-channel baryon exchange is⁵⁵5Here we consider only the baryon exchanges such that the two mesons emitted are two pions since we focus on the correlated $S$-wave two-pion exchange. That is, although we use an SU(3) chiral Lagrangian, the exchanged baryon has the same strangeness as the external ones. The framework may be understood as an SU(2) one for each of the baryons, but with the LECs matched to those in the SU(3) Lagrangian.

	$\displaystyle\hat{A}_{0}^{N}(s)$	$\displaystyle=-\frac{\sqrt{3}(D+F)^{2}m_{N}}{F_{\pi}^{2}}\frac{L(s,m_{N},m_{N},M_{\pi})}{\sqrt{s-4m_{N}^{2}}}\,,$		(25)
	$\displaystyle\hat{A}_{0}^{\Sigma}(s)$	$\displaystyle=-\frac{4\sqrt{2}F^{2}m_{\Sigma}}{F_{\pi}^{2}}\frac{L(s,m_{\Sigma},m_{\Sigma},M_{\pi})}{\sqrt{s-4m_{\Sigma}^{2}}}-\frac{\sqrt{2}D^{2}(m_{\Sigma}+m_{\Lambda})}{3F_{\pi}^{2}}\frac{L(s,m_{\Sigma},m_{\Lambda},M_{\pi})}{\sqrt{s-4m_{\Sigma}^{2}}}\,,$		(26)
	$\displaystyle\hat{A}_{0}^{\Lambda}(s)$	$\displaystyle=\sqrt{\frac{2}{3}}\frac{D^{2}\left(m_{\Lambda}+m_{\Sigma}\right)}{F_{\pi}^{2}}\frac{L(s,m_{\Lambda},m_{\Sigma},M_{\pi})}{\sqrt{s-4m_{\Lambda}^{2}}}\,,$		(27)
	$\displaystyle\hat{A}_{0}^{\Xi}(s)$	$\displaystyle=\frac{\sqrt{3}(D-F)^{2}m_{\Xi}}{F_{\pi}^{2}}\frac{L(s,m_{\Xi},m_{\Xi},M_{\pi})}{\sqrt{s-4m_{\Xi}^{2}}}\,,$		(28)

where

	$\displaystyle L(s,m_{1},m_{2},m)$	$\displaystyle=s-2m_{1}(m_{1}-m_{2})+H_{0}(s,m_{1},m_{2},m)H_{1}(s,m_{1},m_{2},m)\,,$
	$\displaystyle H_{0}(s,m_{1},m_{2},m)$	$\displaystyle=2(m_{1}+m_{2})\left[-2m^{2}m_{1}+2m_{1}(m_{1}-m_{2})^{2}+m_{2}s\right],$
	$\displaystyle H_{1}(s,m_{1},m_{2},m)$	$\displaystyle=\frac{H_{2}^{+}(s,m_{1},m_{2},m)-H_{2}^{-}(s,m_{1},m_{2},m)}{2\sqrt{(s-4m^{2})(s-4m_{1}^{2})}}\,,$
	$\displaystyle H_{2}^{\pm}(s,m_{1},m_{2},m)$	$\displaystyle={\rm{ln}}\left[s-2(m^{2}+m_{1}^{2}-m_{2}^{2})\mp\sqrt{(s-4m^{2})(s-4m_{1}^{2})}\right].$

The contact terms, which are from the NLO Lagrangian and contribute to the RHC part of $T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}(s)$ after taking into account the $\pi\pi$ rescattering, read

	$\displaystyle A_{0}^{N}(s)=\frac{\sqrt{s-4m_{N}^{2}}}{4\sqrt{3}F_{\pi}^{2}}\Bigl{(}$	$\displaystyle 8M_{\pi}^{2}(6b_{0}-3(b_{1}+b_{2}+b_{3}+2b_{4}-b_{D}-b_{F})-2(b_{5}+b_{6}+b_{7}+2b_{8})m_{N})$
		$\displaystyle+4(3(b_{1}+b_{2}+b_{3}+2b_{4})+(b_{5}+b_{6}+b_{7}+2b_{8})m_{N})s\Bigr{)},$		(29)
	$\displaystyle A_{0}^{\Sigma}(s)=\frac{\sqrt{s-4m_{\Sigma}^{2}}}{6\sqrt{2}F_{\pi}^{2}}\Bigl{(}$	$\displaystyle 8M_{\pi}^{2}(9b_{0}-12b_{1}-6b_{2}-9b_{4}+9b_{D}-2(4b_{5}+2b_{7}+3b_{8})m_{\Sigma})$
		$\displaystyle+4(12b_{1}+6b_{2}+9b_{4}+4b_{5}m_{\Sigma}+2b_{7}m_{\Sigma}+3b_{8}m_{\Sigma})s\Bigr{)},$		(30)
	$\displaystyle A_{0}^{\Lambda}(s)=-\frac{\sqrt{s-4m_{\Lambda}^{2}}}{6\sqrt{6}F_{\pi}^{2}}\Bigl{(}$	$\displaystyle 8M_{\pi}^{2}(9b_{0}-6b_{2}-9b_{4}+3b_{D}-4b_{7}m_{\Lambda}-6b_{8}m_{\Lambda})$
		$\displaystyle+4(6b_{2}+9b_{4}+2b_{7}m_{\Lambda}+3b_{8}m_{\Lambda})s\Bigr{)},$		(31)
	$\displaystyle A_{0}^{\Xi}(s)=-\frac{\sqrt{s-4m_{\Xi}^{2}}}{4\sqrt{3}F_{\pi}^{2}}\Bigl{(}$	$\displaystyle 8M_{\pi}^{2}(6b_{0}-3(b_{1}+b_{2}-b_{3}+2b_{4}-b_{D}+b_{F})-2(b_{5}-b_{6}+b_{7}+2b_{8})m_{\Xi})$
		$\displaystyle+4(3(b_{1}+b_{2}-b_{3}+2b_{4})+(b_{5}-b_{6}+b_{7}+2b_{8})m_{\Xi})s\Bigr{)},$		(32)

where the parameter $F_{\pi}$ is the decay constant of the $\pi$ in the chiral limit. Since we use the LECs determined in Ref. [48], we adopt the same value $F_{\pi}=87.1$ MeV [50] for consistency.

Moreover, employing Eq. (5), the tree-level $S$-wave amplitudes for ${\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi$ after eliminating the kinematical singularities read, for the LHC part,

	$\displaystyle\hat{A}_{0}^{N\ {\rm{new}}}(s)$	$\displaystyle=-\frac{\sqrt{3}(D+F)^{2}m_{N}}{F_{\pi}^{2}}L(s,m_{N},m_{N},M_{\pi}),$		(33)
	$\displaystyle\hat{A}_{0}^{\Sigma\ {\rm{new}}}(s)$	$\displaystyle=-\frac{4\sqrt{2}F^{2}m_{\Sigma}}{F_{\pi}^{2}}L(s,m_{\Sigma},m_{\Sigma},M_{\pi})-\frac{\sqrt{2}D^{2}(m_{\Sigma}+m_{\Lambda})}{3F_{\pi}^{2}}L(s,m_{\Sigma},m_{\Lambda},M_{\pi}),$		(34)
	$\displaystyle\hat{A}_{0}^{\Lambda\ {\rm{new}}}(s)$	$\displaystyle=\sqrt{\frac{2}{3}}\frac{D^{2}\left(m_{\Lambda}+m_{\Sigma}\right)}{F_{\pi}^{2}}L(s,m_{\Lambda},m_{\Sigma},M_{\pi}),$		(35)
	$\displaystyle\hat{A}_{0}^{\Xi\ {\rm{new}}}(s)$	$\displaystyle=\frac{\sqrt{3}(D-F)^{2}m_{\Xi}}{F_{\pi}^{2}}L(s,m_{\Xi},m_{\Xi},M_{\pi}),$		(36)

and for the contact term part,

	$\displaystyle A_{0}^{N\ {\rm{new}}}(s)=\frac{s-4m_{N}^{2}}{4\sqrt{3}F_{\pi}^{2}}\Bigl{(}$	$\displaystyle 8M_{\pi}^{2}[6b_{0}-3(b_{1}+b_{2}+b_{3}+2b_{4}-b_{D}-b_{F})-2(b_{5}+b_{6}+b_{7}+2b_{8})m_{N}]$
		$\displaystyle+4[3(b_{1}+b_{2}+b_{3}+2b_{4})+(b_{5}+b_{6}+b_{7}+2b_{8})m_{N}]s\Bigr{)},$		(37)
	$\displaystyle A_{0}^{\Sigma\ {\rm{new}}}(s)=\frac{s-4m_{\Sigma}^{2}}{6\sqrt{2}F_{\pi}^{2}}\Bigl{(}$	$\displaystyle 8M_{\pi}^{2}[9b_{0}-12b_{1}-6b_{2}-9b_{4}+9b_{D}-2(4b_{5}+2b_{7}+3b_{8})m_{\Sigma}]$
		$\displaystyle+4(12b_{1}+6b_{2}+9b_{4}+4b_{5}m_{\Sigma}+2b_{7}m_{\Sigma}+3b_{8}m_{\Sigma})s\Bigr{)},$		(38)
	$\displaystyle A_{0}^{\Lambda\ {\rm{new}}}(s)=-\frac{s-4m_{\Lambda}^{2}}{6\sqrt{6}F_{\pi}^{2}}\Bigl{(}$	$\displaystyle 8M_{\pi}^{2}(9b_{0}-6b_{2}-9b_{4}+3b_{D}-4b_{7}m_{\Lambda}-6b_{8}m_{\Lambda})$
		$\displaystyle+4(6b_{2}+9b_{4}+2b_{7}m_{\Lambda}+3b_{8}m_{\Lambda})s\Bigr{)},$		(39)
	$\displaystyle A_{0}^{\Xi\ {\rm{new}}}(s)=-\frac{s-4m_{\Xi}^{2}}{4\sqrt{3}F_{\pi}^{2}}\Bigl{(}$	$\displaystyle 8M_{\pi}^{2}[6b_{0}-3(b_{1}+b_{2}-b_{3}+2b_{4}-b_{D}+b_{F})-2(b_{5}-b_{6}+b_{7}+2b_{8})m_{\Xi}]$
		$\displaystyle+4[3(b_{1}+b_{2}-b_{3}+2b_{4})+(b_{5}-b_{6}+b_{7}+2b_{8})m_{\Xi}]s\Bigr{)}.$		(40)

We now incorporate the $\pi\pi$ rescattering based on the tree-level amplitude, into the Muskhelishvili-Omnès representation. For the ${\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi$ process, we partition the total $S$-wave kinematical-singularity-free amplitude into the LHC and the RHC parts,

$\displaystyle T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}^{\rm{new}}(s)=R_{{\mathbb{B}},0}^{\rm{new}}(s)+L_{{\mathbb{B}},0}^{\rm{new}}(s).$

(41)

Utilizing the $\pi\pi$ amplitude in the scalar-isoscalar channel $T_{\pi\pi\to\pi\pi,0}(s)=e^{i\delta_{0}(s)}\sin\delta_{0}(s)/\rho_{\pi}(s)$, where $\delta_{0}(s)$ is the $S$-wave isoscalar phase shift, and since there is no overlap between the LHC and RHC for kinematic-singularity-free amplitudes,⁶⁶6The RHC is chosen to be along the positive $s$ axis in the interval $s\geq 4M_{\pi}^{2}$. The LHC is in the interval $\left(-\infty,\left(4m_{\mathbb{B}}^{2}M_{\pi}^{2}-(m_{0}^{2}-m_{\mathbb{B}}^{2}-M_{\pi}^{2})^{2}\right)/{m_{0}^{2}}\right]$ for the $t$- or $u$-channel process of ${\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi$, where $m_{0}$ represents the mass of the exchanged particle. It can be easily proven that $\left(4m_{\mathbb{B}}^{2}M_{\pi}^{2}-(m_{0}^{2}-m_{\mathbb{B}}^{2}-M_{\pi}^{2})^{2}\right)/{m_{0}^{2}}\leq 4M_{\pi}^{2}$. the unitary relation implies,

$\displaystyle{\rm{disc}}\left[R_{{\mathbb{B}},0}^{\rm{new}}(s)\right]=2i\left(R_{{\mathbb{B}},0}^{\rm{new}}(s)+L_{{\mathbb{B}},0}^{\rm{new}}(s)\right)e^{-i\delta_{0}(s)}\sin\delta_{0}(s)\theta(\sqrt{s}-2M_{\pi}).$

(42)

To solve this equation, we first define the Omnès function [51],

$\displaystyle\Omega_{0}(s)\equiv\exp{\left[\frac{s}{\pi}\int_{4M_{\pi}^{2}}^{+\infty}\frac{\delta_{0}(z)}{z(z-s)}{\rm{d}}z\right]}.$

(43)

By using $\Omega_{0}(s\pm i\epsilon)=|\Omega_{0}(s)|e^{\pm i\delta_{0}(s)}$, we further derive

$\displaystyle{\rm{disc}}\left[\frac{R_{{\mathbb{B}},0}^{\rm{new}}(s)}{\Omega_{0}(s)}\right]=2i\frac{L_{{\mathbb{B}},0}^{\rm{new}}(s)}{|\Omega_{0}(s)|}\sin\delta_{0}(s)\theta(\sqrt{s}-2M_{\pi}).$

(44)

Therefore, we can derive a DR with $n$ subtractions,

$\displaystyle R_{{\mathbb{B}},0}^{\rm{new}}(s)=\Omega_{0}(s)\left(P_{n-1}(s)+\frac{s^{n}}{\pi}\int_{4M_{\pi}^{2}}^{+\infty}{\rm{d}}z\frac{L_{{\mathbb{B}},0}^{\rm{new}}(z)\sin\delta_{0}(z)}{(z-s)z^{n}|\Omega_{0}(z)|}\right),$

(45)

where $P_{n-1}(s)$ is an arbitrary polynomial of order $n-1$. Finally, we obtain a DR for $T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}^{\rm{new}}(s)$ as

$\displaystyle T_{{\mathbb{B}}\bar{{\mathbb{B}}}\to\pi\pi,0}^{\rm{new}}(s)=L_{{\mathbb{B}},0}^{\rm{new}}(s)+\Omega_{0}(s)\left(P_{n-1}(s)+\frac{s^{n}}{\pi}\int_{4M_{\pi}^{2}}^{+\infty}{\rm{d}}z\frac{L_{{\mathbb{B}},0}^{\rm{new}}(z)\sin\delta_{0}(z)}{(z-s)z^{n}|\Omega_{0}(z)|}\right).$

(46)

For the phase shift $\delta_{0}(s)$, we take the parametrization in Ref. [52]. For the $\Omega_{0}(s)$ Omnès function, we take the $\Omega_{11}(s)$ matrix element of the coupled-channel Omnès matrix for the $\pi\pi$-$K\bar{K}$ $S$-wave interaction obtained in Ref. [53].