# $\mathrm{\rho^0}$ 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? Feb 8, 2015 at 0:44
• $\rho^0\to\pi^+ + \pi^-$ has $J^{PC}=1^{--}\to 0^-+0^-$ so expect suppression. Oct 1, 2019 at 6:32

Assume for the sake of argument that this decay channel happens.

Now, ask yourself how you are going to prove it?

• How about we compute the rate for each channel and see if the real rate is the sum? This works when the added channel is a reasonable fraction of the dominate channel, but you're talking about comparing a strong decay to a weak one. This is hugely impractical.

• 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.

So, long story short: this is an experimentally inaccessible channel.

As far as I understand, the decay of an anti-u quark into an anti-d quark will emmit a $$\mathrm W^+$$ boson. But the anti-u and d quarks into which the boson would decay, have charges $$-2/3$$, $$-1/3$$ respectively. So their total charge will be $$-1$$. A $$\mathrm W^+$$ boson cannot decay into two particles whose sum equals $$-1$$, the sum of the particles should be $$+1$$. So this decay cannot be a weak interaction because the conservation of charge would be violated.

• What makes you disagree with the correct diagram conserving charge at every vertex? A $\bar u$ quark decays to a $W^-$ and a $\bar d$. Where did you freak $W^+$ come from? The future? Sep 17, 2022 at 19:39

Note that rho meson has no strange quark i.e. its strangeness quantum number is zero. Similarly the decay products positive as well as negative pion has no strange quark ie. Strangeness quantum number is zero for decay products too. Thus, strangeness is conserved in this decay process of neutral rho meson. Now note that it is the rule of nature that strangeness is conserved in strong interaction/decay but NOT in weak interaction/decay. Thus this decay of neutral rho meson can proceed only via strong process i.e. via gluon and can never happen through weak process i.e can never happen through W boson, a mediator of weak interaction/decay.