Why is $\pi^0 \rightarrow \gamma \rightarrow e^-e^+$ forbidden but $\pi^+\rightarrow W^+ \rightarrow e^+ \nu_e$ allowed?

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).

• 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. – dukwon May 28 '17 at 14:29
• @dukwon Sorry I was taking all intermediate particles to be virtual. – Quantum spaghettification May 28 '17 at 14:31
• 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 – dukwon May 28 '17 at 14:36

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 $\pi^{+}\to W^{+*} \to l^{+}\nu_{l}$.
• 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? – Quantum spaghettification May 28 '17 at 15:15
• @Quantumspaghettification : the spin of $J/\psi$ meson is one, allowing the effective vertex $\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}^{J/\psi}F_{\alpha\beta}^{\gamma}$. – Name YYY May 28 '17 at 15:59