Quark decay implies particle decay?

Question

For example, since $$s\rightarrow u+\overline{v_e}+e^{-}$$, then sticking a $\overline{u}$ next to the quarks ($s$ and $u$) we get $$s\overline{u}\rightarrow u\overline{u}+\overline{v_e}+e^{-}$$, which becomes $$K^{-}\rightarrow \pi^0+\overline{v_e}+e^{-}$$ -- a valid decay equation.

So I was wondering: If you take a quark equation (like the $s$ decay one above) and add some other quark to each quark in that equation (like how we added $\overline{u}$ above), do we always get a valid equation (e.g. the $K^{-}$ decay above).

Thanks

Answer 1

In diagrams, you are just adding a disconnected non-interacting piece to the Feynman diagram (just a straight line for the added quark), so sure, just writing one or more additional quarks on both sides always yields an allowed reaction, although you can't be sure the quarks bind into a single state (e.g. if you added just a $u$ to your reaction instead of a $\bar u$, you would get anything, except two $u$-quarks).