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 for having daughter 1 with momentum direction $\Omega =(\phi ,\theta )$ in the center of mass reference frame, and helicity $\lambda _{1}$, while daughter 2 has direction $-\Omega =(\phi +\pi ,\pi -\theta )$ and helicity $\lambda _{2}$.

Let U be the decay operator from the initial state into the given 2-body final state. Intermediate between the at-rest initial state of qn J,m and the final plane-wave state is a basis of outgoing waves describing the outgoing 2-body state in a basis of good J,m and helicities. Insertion of the complete set of intermediate basis vectors, and summing over all intermediate J,m gives

$\langle \Omega \lambda _{1}\lambda _{2}|U|Jm\rangle =\langle \Omega \lambda _{1}\lambda _{2}|Jm\lambda _{1}\lambda _{2}\rangle \langle Jm\lambda _{1}\lambda _{2}|U|Jm\rangle$

This is one way to describe the final state, but it is not the only way. Instead of specifying the final-state particles' spin state via their helicities, we can first couple their spins together independent of their momentum direction, to obtain total spin S, then couple S to their relative orbital angular momentum L to obtain their total angular momentum J. When we do this, we give up our knowledge of the particles' helicities, having replaced those two quantum numbers with the alternative pair L,S. These two bases, the helicity basis and the L,S basis, are each individually complete and orthonormal within themselves. Following on from the above expression, let us insert a sum over the L,S basis.

$\langle \Omega \lambda _{1}\lambda _{2}|U|Jm\rangle =\sum _{L,S}\langle \Omega \lambda _{1}\lambda _{2}|Jm\lambda _{1}\lambda _{2}\rangle \langle Jm\lambda _{1}\lambda _{2}|JmLS\rangle \langle JmLS|U|Jm\rangle$

$=\sum _{L,S}\left[{\sqrt {\frac {2J+1}{4\pi }}}D_{m\lambda }^{J*}(\Omega ,0)\right]\left[{\sqrt {\frac {2L+1}{2J+1}}}\left({\begin{array}{cc|c}L&S&J\\0&\lambda &\lambda \end{array}}\right)\left({\begin{array}{cc|c}S_{1}&S_{2}&S\\\lambda _{1}&-\lambda _{2}&\lambda \end{array}}\right)\right]a_{LS}^{J}$

where $\lambda =\lambda _{1}-\lambda _{2}$,and the double-stacked symbols are Clebsh-Gordon coefficients. The product of CG coefficients in the second brackets on the right-hand side represent the overlap between the basis vectors in the helicity and L,S basis, which turns out to be independent of m, as required by rotational invariance. This overlap integral is somewhat lengthy to calculate, but the result turns out to be fairly simple, as shown above. This expression holds regardless of what axis is used to define the quantization direction for m, but whatever choice is made must server as the z-axis of the reference frame in which the plane wave direction $\Omega$ is defined.

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

Production

Photon-Reggeon-Resonance vertex

Consider the production of the resonance from the photon and reggeon in the reflectivity basis, the eigenstates of the reflectivity operator. (This operator is a combination of parity and $\pi$ rotation about the normal to the production plane (usually y axis.)
$\mathbb {R} |Jm\rangle =P(-1)^{J-m}|J\;-m\rangle$

The eigenstates of the reflectivity operator are formed as follows:
$|Jm\epsilon \rangle =|Jm\rangle +\epsilon P(-1)^{J-m}|J\;-m\rangle$

such that

$\mathbb {R} |Jm\epsilon \rangle =\epsilon (-1)^{2J}|Jm\epsilon \rangle$

The photon linear polarization states turn out to be eigenstates of reflectivity as well:
Let x (y) polarization states be denoted with - (+)

$|\mp \rangle ={\sqrt {\frac {\pm 1}{2}}}\left(|1-1\rangle \mp |1+1\rangle \right)$

$\mathbb {R} |\mp \rangle =\mp 1|\mp \rangle$

Since the production Hamiltonian should commute with reflectivity: $V=\mathbb {R} ^{-1}V\mathbb {R}$

$\langle Jm\epsilon |\mathbb {R} ^{-1}V\mathbb {R} |\mp ;J_{R}\lambda _{R}\epsilon _{R};t,s;\Omega _{0}\rangle =\epsilon (\mp 1)\epsilon _{R}\langle Jm\epsilon |V|\mp ;J_{R}\lambda _{R}\epsilon _{R};t,s;\Omega _{0}\rangle$

Acting with the reflectivity operator on initial and final state brings out the reflectivity eigenvalues of the resonance, photon and reggeon. This result leads to a constraint:
$\epsilon =\mp \epsilon _{R}$

Proton-Reggeon vertex

The amplitude of target proton's emission of an exchange particle, a reggeon, in particular direction and helicity projections can be written as:

$\langle \Omega _{R};J_{R}\lambda _{R}\epsilon _{R};J_{P}\lambda _{p}|W|J_{T}m_{T}\rangle =\langle \Omega _{R};J_{R}\lambda _{R}\;\mp \epsilon ;\textstyle {\frac {1}{2}}\;\lambda _{p}|\textstyle {\frac {1}{2}}\;m_{T}\lambda _{R}\lambda _{p}\rangle \langle \textstyle {\frac {1}{2}}\;m_{T}\lambda _{R}\lambda _{p}|W|\textstyle {\frac {1}{2}}\;m_{T}\rangle$	transition amplitude for $p\rightarrow R+p'$ in the direction $\Omega _{R}$ w.r.t. the coordinate system defined in the resonance RF.
$={\frac {1}{\sqrt {2\pi }}}\left[D_{m_{T}(\lambda _{R}-\lambda _{p})}^{{\frac {1}{2}}*}(\Omega _{R},0)\;w_{\lambda _{R}\;\lambda _{p}}\mp \epsilon P_{R}(-1)^{J_{R}-\lambda _{R}}D_{m_{T}(-\lambda _{R}-\lambda _{p})}^{{\frac {1}{2}}*}(\Omega _{R},0)\;w_{\lambda _{R}\;-\lambda _{p}}\right]$	follows from relations given above

Decay

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

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

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

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

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