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i know that for an s channel reaction, the quantum numbers of the intermediate particle have to be the same as those of the particles coming in, for example in the reaction $\gamma \pi \rightarrow a_2 \rightarrow \eta \pi$ the in- and outgoing particles have to be in a D-wave in order to produce the quantum numbers of the $a_2$: $I^G(J^{PC})=1^- (2^{++})$.

I was asked to also calculate the amplitude for the t-channel process (with the momentum of the mediating particle being the difference of the Pion momenta) and I do not know how to make a connection between the quantum numbers of the final and initial particles and those of the mediating particle. Imagining the pion to "decay" into the mediating particle and another Pion does not really seem like an option.

I would be grateful for an explanation or a hint where to look this up.

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Look at each vertex independently.

  • The vertex including the photon also involves a hadron, so the exchange is going to be strong mediated.

  • The eta carries certain quark flavors. These can-not have come from the photon, so they came from the exchange particle.

  • You have to conserve angular momentum between the initial and final states, whic may involve a change or orbital state.

  • There is no change in charge at either vertex.

Your exchange particle is therefore strongly interacting, bosonic and uncharged but could have either parity. I believe that at the level of quantum hadrodynamics this is usually envisioned as $$\gamma + \pi \overset{\eta^*}{\to} \eta + \pi \,,$$ where the star implies that the exchange $\eta$ is off-shell.

However, you may find various descriptions at the quark-level to be more helpful. In that view the on-shell photon is often the "exchange" particle. Something like

The photon interacts with one of the valence quarks, hitting it hard enough to pair produce on the flux tube and after hadronization you have the final state particles.

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  • $\begingroup$ I am slightly confused by what exactly you mean by "change in orbital angular momentum". Are you talking about the angular momentum between the photon and the initial pion? You say that at the photon vertex the sum of the exchange particle's spin and the change in angular momentum is $\pm 1$, while it is $0$ at the pion-eta vertex. If your conclusion is, that the exchange particle's spin is 0, how can the change in angular momentum be equal to $0$ and $\pm 1$ at the same time? $\endgroup$
    – Jupith
    Commented Aug 25, 2014 at 14:51
  • $\begingroup$ @Jupith Yeah, that is poorly stated and reflect imprecise thinking. I'll get back to you on it. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 25, 2014 at 19:50
  • $\begingroup$ So I thought about this a little more ;) Did you mean that at each vertex the incoming and outgoing particle's spins couple with their relative angular momentum to give their total angular momentum which is then the spin of the exchange particle? This would give $J^P=L^{(-)^L}$ for the exchange particle though, which still cannot give me $0^-$, as the $\eta$ would have... $\endgroup$
    – Jupith
    Commented Aug 27, 2014 at 7:26
  • $\begingroup$ @Jupith I have not been able to get is squared away in my head, so I'm going to strip the offending paragraphs for now. You have to conserve momentum between the initial and final states, and when the spin math lets you down that requires the addition of a change in orbital state as well. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 28, 2014 at 2:10

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