# $\mathrm{\rho^0}$ meson decay via the weak interaction?

Of course, the $\mathrm{\rho^0}$ meson can decay in $\mathrm{\pi^{+}\ \pi^{-}}$ through the strong interaction. Using Feynman diagrams, I cannot understand why the same decay couldn't happen through the weak interaction. I attach the diagram I've drawn.

Strong decay:

Weak decay:

• "I cannot understand why the same decay couldn't happen through the weak interaction." - Is there a reference saying that it doesn't happen? – ACuriousMind Feb 8 '15 at 0:44
• $\rho^0\to\pi^+ + \pi^-$ has $J^{PC}=1^{--}\to 0^-+0^-$ so expect suppression. – alexchandel Oct 1 '19 at 6:32

• OK, the weak interaction respects fewer symmetries than the strong. We can take advantage of that. Now you're on the right track. This is how the weak interaction form-factors of the proton have been measured: by observing the parity violation rate in $\vec{e}(p,e'p)$ and similar channels. But that is a comparison between a electromagnetic process and weak one and it still requires around $10^{13}$ separate events to extract the results to useful precision. It's going to take more than that to pull a weak signal out from under a strong one. Still impractical, I'm afraid.