Three pions: pulling out the (12)3 orientation

Label the particles $1=\pi^+$, $2=\pi^-$, $3=\pi^-$, and take the input four-vectors in their common rest frame (the GJ frame). The ordered chains are $\tau_{12}=((1,2),3)$ and $\tau_{13}=((1,3),2)$: the positive pion defines the inner angles in both. DPD uses $\tau_{31}=((3,1),2)$ instead. The completed derivation, including reflectivity, is collected in the COMPASS basis note. All statements below concern angular factors, before isobar couplings, lineshapes, or Bose symmetrization.

The convention used by PartialWaveFunctions, and conjugated by CascadeDecays, is

\[D^{j*}_{mn}(\alpha,\beta,\gamma) =e^{im\alpha}d^j_{mn}(\beta)e^{in\gamma},\qquad d^j_{mn}(\beta)=\langle jm|e^{-iJ_y\beta}|jn\rangle.\]

Write $\Omega_{ij}=(\phi_{\rm GJ}^{ij},\theta_{\rm GJ}^{ij},\phi_{\rm TY}^{ij})$ and $O_{ij}=R_z(\phi_{\rm GJ}^{ij})R_y(\theta_{\rm GJ}^{ij})R_z(\phi_{\rm TY}^{ij})$. The $z$ axis points along the pair momentum, and the first particle of the ordered pair has positive $x$ in the resulting decay-plane frame.

Exact identity

For fixed $\lambda$ (with $|\lambda|\leq\min(J,S)$),

\[\begin{aligned} &D^{J*}_{M\lambda}(\phi_{\rm GJ}^{13},\theta_{\rm GJ}^{13},0) D^{S*}_{\lambda0}(\phi_{\rm TY}^{13},\theta_{\rm TY}^{13},0)\\ &\quad=\sum_{\lambda'=-J}^{J} D^{J*}_{M\lambda'}(\Omega_{12}) d^J_{\lambda'\lambda}(\zeta) (-1)^\lambda d^S_{\lambda0}(\theta_{\rm TY}^{13})\\ &\quad=(-1)^S\sum_{\lambda'=-J}^{J} D^{J*}_{M\lambda'}(\Omega_{12}) d^J_{\lambda'\lambda}(\zeta) d^S_{\lambda0}(\theta_{\rm TY}^{31}). \end{aligned}\]

There are three corrections to the proposed expression: the inner angles on the left must be those of 13, the remaining small-$d$ has spin $S$, and using the positive-pion-first angle on the right requires $(-1)^\lambda$. The sum over $\lambda'$ is essential. In a full amplitude, sum these fixed-$\lambda$ identities with the appropriate helicity couplings.

Why the extra phase appears

For a noncollinear three-body event,

\[O_{12}^{-1}O_{31}=R_y(\zeta),\qquad O_{12}^{-1}O_{13}=R_y(\zeta)R_z(\pi).\]

The cyclic (12) and (31) frames have the same oriented plane normal. Choosing particle 1 first in (13) reverses that normal and the transverse $x$ axis. The last $R_z(\pi)$ contributes $e^{i\lambda\pi}=(-1)^\lambda$. Meanwhile the inner azimuth is absorbed into the last Euler angle of the parent matrix:

\[D^{J*}_{M\lambda}(\phi,\theta,0) D^{S*}_{\lambda0}(\varphi,\vartheta,0) =D^{J*}_{M\lambda}(\phi,\theta,\varphi)d^S_{\lambda0}(\vartheta).\]

Reversing the pair gives

\[\theta_{\rm TY}^{31}=\pi-\theta_{\rm TY}^{13},\qquad \phi_{\rm TY}^{31}=\phi_{\rm TY}^{13}+\pi\pmod{2\pi},\]

and, for integer $S,\lambda$,

\[d^S_{\lambda0}(\pi-\theta)=(-1)^{S+\lambda}d^S_{\lambda0}(\theta).\]

Explicitly substituting $\theta_{13}=\pi-\theta_{31}$ gives

\[\begin{aligned} (-1)^\lambda d^S_{\lambda0}(\theta_{13}) &=(-1)^\lambda d^S_{\lambda0}(\pi-\theta_{31})\\ &=(-1)^\lambda(-1)^{S+\lambda}d^S_{\lambda0}(\theta_{31})\\ &=(-1)^{S+2\lambda}d^S_{\lambda0}(\theta_{31})\\ &=(-1)^S d^S_{\lambda0}(\theta_{31}). \end{aligned}\]

Here $\lambda$ is integer, so $(-1)^{2\lambda}=1$. In particular, an odd-spin isobar changes sign when converting its ordered angular factor from (13) to (31). This is an ordering convention; any simultaneous conversion of couplings must use the same convention.

The two-chain amplitude with a common orientation

Let $H^{ij}_\lambda$ denote the parent helicity coupling for an isobar of spin $S$ and a spin-zero spectator, and $h^{ij}=H^{ij\to\pi\pi}_{00}$ the isobar decay coupling. Any scalar lineshape can multiply the corresponding chain throughout. The couplings below retain their original (12) and (13) ordering; the explicit $(-1)^S$ performs the angular conversion to (31). For the two identical negative pions, add the two chains with a plus sign:

\[\begin{aligned} \mathcal A_M={}&\sum_\lambda D^{J*}_{M\lambda}(\phi_{\rm GJ}^{12},\theta_{\rm GJ}^{12},0) H^{12}_\lambda D^{S*}_{\lambda0}(\phi_{\rm TY}^{12},\theta_{\rm TY}^{12},0)h^{12}\\ &+\sum_\lambda D^{J*}_{M\lambda}(\phi_{\rm GJ}^{13},\theta_{\rm GJ}^{13},0) H^{13}_\lambda D^{S*}_{\lambda0}(\phi_{\rm TY}^{13},\theta_{\rm TY}^{13},0)h^{13}. \end{aligned}\]

Absorb each inner azimuth into its parent $D$, pull the (13) orientation into the (12) frame, and perform the explicit angle substitution above:

\[\boxed{ \mathcal A_M=\sum_{\lambda'=-J}^{J} D^{J*}_{M\lambda'}(\phi_{\rm GJ}^{12},\theta_{\rm GJ}^{12},\phi_{\rm TY}^{12}) \left[ H^{12}_{\lambda'}d^S_{\lambda'0}(\theta_{\rm TY}^{12})h^{12} +(-1)^S\sum_\lambda d^J_{\lambda'\lambda}(\zeta) H^{13}_\lambda d^S_{\lambda0}(\theta_{\rm TY}^{31})h^{13} \right]. }\]

The internal sum runs over $|\lambda|\leq\min(J,S)$; the first term is zero for $|\lambda'|>S$. With helicity contractions understood, this is

\[\mathcal A= D^{J*}(\Omega_{12}) \left[ H^{12}d^S(\theta_{12})h^{12} +(-1)^S d^J(\zeta)H^{13}d^S(\theta_{31})h^{13} \right],\qquad \theta_{31}=\pi-\theta_{13}.\]

The common orientation contains all dependence on the GJ axes. The bracket depends on internal kinematics and couplings. For identical-pion symmetrization the two chains use the same coupling functions, evaluated at their respective pair invariants; an overall normalization is conventional.

Which zeta?

This is the parent rotation, wr(2,3,0) in ThreeBodyDecays, with target spectator 2 and reference spectator 3. It is positive in this ordering:

\[\zeta=\arccos\!\left(\frac{\boldsymbol p_2\cdot\boldsymbol p_3} {|\boldsymbol p_2||\boldsymbol p_3|}\right)\in(0,\pi).\]

It is not the final-pion Wigner rotation wr(2,3,1). The final pions have spin zero, so their spin rotations are trivial. In invariant notation, with $s=P^2$, $\sigma_k=(P-p_k)^2$, and the Källén function $\Lambda$,

\[\cos\zeta= \frac{(s+m_2^2-\sigma_2)(s+m_3^2-\sigma_3) -2s(\sigma_1-m_2^2-m_3^2)} {\sqrt{\Lambda(s,m_2^2,\sigma_2)\Lambda(s,m_3^2,\sigma_3)}}.\]

The cyclic-chain convention follows the DPD framework. Here the precise sign is fixed independently by the Cartesian rotation identities above and checked against the installed ThreeBodyDecays implementation.

Executable cross-checks

The COMPASS notebook executes the checks alongside its derivation. It generates the physical events, prints the first event and all three sets of angles, compares the Cartesian and invariant zeta, and tests the pair-ordering identities and reflectivity factorization. Regenerate those results locally with:

quarto render docs/compass-basis.qmd --to gfm