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The three structures are the radiative cascadesIn the quark model, the J/psi meson and $\psi (2S)$ are respectively $1{}^3{S}_1$ and $2{}^3{S}_1$ charmonium states. Thus $L=0$, $S=1$, $J=1$, and meson parity and charge conjugation gives $J^{PC}=1^{--}$. The intermediate chi c mesons are $1{}^3{P}_J$: $L=1$, $S=1$, and addition of angular momentum permits $J=0,1,2$. ConsequentlyThese are electric dipole transitions, changing $L$ by one and preserving $S$ to leading order. A singlet $1{}^1{P}_1$ state has $J^{PC}=1^{+-}$ and is not another member of these spin-triplet cascades.

The primary photon energy is $(M_{2S}^2-M_{{\chi }_J}^2)/(2M_{2S})$, about $261$, $171$ and $128\mathrm{MeV}$ for $J=0,1,2$. The secondary photon remains more energetic even after its boost into the parent rest frame. Hence $m^2(J/\psi {\gamma }_h)=M_{{\chi }_J}^2$: with this quantity on the vertical axis, each resonance gives a horizontal band. A finite decay width and detector measurement resolution broaden each band; photon ordering removes the mirrored duplication of an unordered Dalitz plot.

The faint band beside the negative-slope diagonal boundary is $\psi (2S)\to J/\psi {\pi }^0$, followed by ${\pi }^0\to \gamma \gamma$. The neutral pion is predominantly $(u\bar{u}-d\bar{d})/\sqrt{2}$, a ${}^1{S}_0$ pseudoscalar meson with $J^{PC}=0^{-+}$. This production transition violates isospin, accounting for its small rate. To locate the band, set $X=m^2(J/\psi {\gamma }_l)$ and $Y=m^2(J/\psi {\gamma }_h)$. Four-momentum conservation givesFor the neutral pion, $m_{\gamma \gamma }^2=m_{{\pi }^0}^2$ is small, putting the line just inside the $m_{\gamma \gamma }^2=0$ boundary. The allowed $J/\psi \eta$ channel, with the eta meson also $0^{-+}$, produces a parallel line further inside, at $m_{\gamma \gamma }^2=m_{\eta }^2$. The positive-slope edge $X=Y$ instead comes from ordering the two photon energies; proximity to that edge alone would not identify the two-photon state.