Amplitudes for the Exotic b1π Decay

General Relations

Angular Distribution of Two-Body Decay

Let's begin with a general amplitude for the two-body decay of a state with angular momentum quantum numbers J,m. Specifically, we want to know the amplitude of this state having daughter 1 with trajectory $\Omega =(\phi ,\theta )$. We can also describe the angular momentum between the daughters as being L and spin sum as s. Alternatively, we will label the daughters as having helicities of $\lambda _{1}$ and $\lambda _{2}$ - projections on the direction of decay (specified by daughter 1)

$\langle \Omega \lambda _{1}\lambda _{2}|U|Jm\rangle =\sum _{L,S}\langle \Omega \lambda _{1}\lambda _{2}|Jm\lambda _{1}\lambda _{2}\rangle \langle Jm\lambda _{1}\lambda _{2}|JmLS\rangle \langle JmLS|U|Jm\rangle$	simple insertion of complete sets of states for recoupling
$=\sum _{L,S}\left[{\sqrt {\frac {2J+1}{4\pi }}}D_{m\lambda }^{J*}(\Omega ,0)\right]\left[{\sqrt {\frac {2L+1}{2J+1}}}\left({\begin{array}{cc|c}L&S&J\\0&\lambda &\lambda \end{array}}\right)\left({\begin{array}{cc|c}S_{1}&S_{2}&S\\\lambda _{1}&-\lambda _{2}&\lambda \end{array}}\right)\right]a_{LS}^{J}$	Substitution of each bra-ket with their respective formulae. $\lambda =\lambda _{1}-\lambda _{2}$ Note that in the event of one daughter being spin-less, the second Clebsch-Gordan coefficient is 1

Isospin Projections

One must also take into account the various ways isospin of daughters can add up to the isospin quantum numbers of the parent, requiring a term:

$C^{a,b}=\left({\begin{array}{cc|c}I^{a}&I^{b}&I\\I_{z}^{a}&I_{z}^{b}&I_{z}^{a}+I_{z}^{b}\end{array}}\right)$

where a=1 and b=2, referring to the daughter number. Because an even-symmetric angular wave function (i.e. L=0,2...) imply that 180 degree rotation is equivalent to reversal of daughter identities (a,b becoming b,a) one must write down the symmetrized expression:

$C(L)={\frac {1}{\sqrt {2}}}\left[C^{a,b}+(-1)^{L}C^{b,a}\right]$

Application

$\langle \Omega _{X}0\lambda _{b_{1}}|U_{X}|J_{X}m_{X}\rangle =\sum _{L_{X}}\left[{\sqrt {\frac {2J_{X}+1}{4\pi }}}D_{m_{X}\lambda _{b_{1}}}^{J_{X}*}(\Omega _{X},0)\right]\left[{\sqrt {\frac {2L_{X}+1}{2J_{X}+1}}}\left({\begin{array}{cc|c}L_{X}&1&J_{X}\\0&\lambda _{b_{1}}&\lambda _{b_{1}}\end{array}}\right)\right]a_{L_{X}}^{J_{X}}$

$\langle \Omega _{b_{1}}0\lambda _{\omega }|U_{b_{1}}|1,m_{b_{1}}=\lambda _{b_{1}}\rangle =\sum _{L_{b_{1}}}\left[{\sqrt {\frac {2J_{b_{1}}+1}{4\pi }}}D_{m_{b_{1}}=\lambda _{b_{1}}\lambda _{\omega }}^{1*}(\Omega _{b_{1}},0)\right]\left[{\sqrt {\frac {2L_{b_{1}}+1}{2J_{b_{1}}+1}}}\left({\begin{array}{cc|c}L_{b_{1}}&1&1\\0&\lambda _{\omega }&\lambda _{\omega }\end{array}}\right)\right]b_{L_{b_{1}}}$

$\langle \Omega _{\omega }0\lambda _{\rho }|U_{\omega }|1,m_{\omega }=\lambda _{\omega }\rangle =\sum _{L_{\omega }J_{\rho }}\left[{\sqrt {\frac {2J_{\omega }+1}{4\pi }}}D_{m_{\omega }=\lambda _{\omega }\lambda _{\rho }}^{1*}(\Omega _{\omega },0)\right]\left[{\sqrt {\frac {2L_{\omega }+1}{2J_{\omega }+1}}}\left({\begin{array}{cc|c}L_{\omega }&1&1\\0&\lambda _{\rho }&\lambda _{\rho }\end{array}}\right)\right]c_{L_{\omega }J_{\rho }}$

$\langle \Omega _{\rho }0\lambda _{\rho }|U_{\rho }|J_{\rho },m_{\rho }=\lambda _{\rho }\rangle =\sum _{L_{\rho }}\left[{\sqrt {\frac {2J_{\rho }+1}{4\pi }}}D_{m_{\rho }0}^{J_{\rho }*}(\Omega _{\rho },0)\right]\left[{\sqrt {\frac {2L_{\rho }+1}{2J_{\rho }+1}}}\left({\begin{array}{cc|c}L_{\rho }&0&J_{\rho }\\0&0&0\end{array}}\right)\right]d_{L_{\rho }}=\sum _{L_{\rho }}{\sqrt {\frac {2L_{\rho }+1}{4\pi }}}Y_{m_{\rho }}^{J_{\rho }*}(\Omega _{\rho })d_{L_{\rho }}$

$A^{J_{X}}=\sum _{\lambda _{b_{1}},\lambda _{\omega },\lambda _{\rho }}\langle \Omega _{X}0\lambda _{b_{1}}|U|J_{X}m_{X}\rangle C_{X}(L_{X})k^{L_{X}}\langle \Omega _{b_{1}}0\lambda _{\omega }|U|1,m_{b_{1}}=\lambda _{b_{1}}\rangle C_{b_{1}}(L_{b_{1}})q^{L_{b_{1}}}\langle \Omega _{\omega }0\lambda _{\rho }|U|1,m_{\omega }=\lambda _{\omega }\rangle C_{\omega }(L_{\omega })u^{L_{\omega }}\langle \Omega _{\rho }0\lambda _{\rho }|U|J_{\rho },m_{\rho }=\lambda _{\rho }\rangle C_{\rho }(L_{\rho })v^{L_{\rho }}$