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I am slightly confused why the decay $\pi^0 \rightarrow \gamma \rightarrow e^-e^+$ is forbidden. A naive guess would say that the intemediate photon $\gamma$ has a spin 1, the initial pion has a spin $0$ therefore this violates spin conservation. However, on the same reasoning $\pi^+\rightarrow W^+ \rightarrow e^+ \nu_e$ would be forbidden, but it's not. Therefore I cannot see why we cannot have $\pi^0 \rightarrow \gamma \rightarrow e^-e^+$ yet still have $\pi^+\rightarrow W^+ \rightarrow e^+ \nu_e$. Please can someone explain? (p.s. all intermediate particles should be taken as virtual).

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  • $\begingroup$ There's a real issue with your notation. 1-body decays like $\pi^0 \to \gamma$ and $\pi^0 \to W^+$ are kinematically forbidden, since they cannot conserve both energy and momentum. The photon is massless so you can't have $\gamma \to e^+ e^-$ either. $\endgroup$ – dukwon May 28 '17 at 14:29
  • $\begingroup$ @dukwon Sorry I was taking all intermediate particles to be virtual. $\endgroup$ – Quantum spaghettification May 28 '17 at 14:31
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    $\begingroup$ So your question is about Feynman diagrams contributing to decay amplitudes, not the decays themselves. It's important to avoid the trap of thinking about Feynman diagrams too literally. This seems like a duplicate of physics.stackexchange.com/q/233076 $\endgroup$ – dukwon May 28 '17 at 14:36
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The QCD and QED themselves conserve parity. The conclusion of this statement is that all corresponding effective vertices must conserve the parity. The only coupling of $\pi^{0}$ to $\gamma$ conserving the parity is $$ L_{\pi^{0}} \simeq \frac{\pi^{0}}{\Lambda}\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}, $$ which doesn't allow your decay process $\pi^{0} \to \gamma^{*} \to e^{-}e^{+}$. To understand this, note that the pion fields are pseudo-scalars, while the photon field is the vector field, while the Levi-Civita tensor $\epsilon^{\mu\nu\alpha\beta}$ is the pseudo-tensor.

However, it is possible to construct the effective vertex allowing the decay $\pi^{0}\to e^{+}e^{-}$ through $Z$-boson, i.e., through weak interactions. The reason is that they directly violate the parity. Therefore, it is possible to construct phenomenological parity-violating low-dimensional effective interaction vertex $$ L_{\pi^{0}}' \simeq \Lambda'\partial^{\mu}\pi^{0}Z_{\mu}, $$ allowing the decay process $\pi^{0} \to Z^{*} \to e^{+}e^{-}$.

By the same reason, it is easy to construct parity violating vertex $$ L_{\pi^{+}} = \tilde{\Lambda}\partial^{\mu}\pi^{+}W^{-}_{\mu} + \text{ h.c.}, $$ allowing your decay process $\pi^{+}\to W^{+*} \to l^{+}\nu_{l}$.

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  • $\begingroup$ Hi thanks for your answer, in my university lecture notes there is an example of a decay of the $J/\psi$ particle: $J/\psi\to \gamma^* \to q\bar q$ under what you have said here this decay is also forbidden - correct? $\endgroup$ – Quantum spaghettification May 28 '17 at 15:15
  • $\begingroup$ @Quantumspaghettification : the spin of $J/\psi$ meson is one, allowing the effective vertex $\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}^{J/\psi}F_{\alpha\beta}^{\gamma}$. $\endgroup$ – Name YYY May 28 '17 at 15:59
  • $\begingroup$ Ok so its a bit more complicated then saying we need 'angular momentum' to be conserved at each vertex, or we need 'parity' to be conserved - we infact need the correct combination for us to have an allowed effective vertex? Is there any general combinations of spin and parity that will always give you an allowed effective vertex? $\endgroup$ – Quantum spaghettification May 28 '17 at 16:11
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    $\begingroup$ @Quantumspaghettification : yes, combining the Lorentz invariance and parity invariance it is not hard to construct all of the possible effective vertices. $\endgroup$ – Name YYY May 28 '17 at 17:52

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