# Can the $W^+$ boson couple an antistrange quark decaying into an antiup quark?

I am trying to figure out the Feynman diagram for the fully hadronic $$K^+$$ meson decay $$K^+ \rightarrow \pi^+ + \pi^0$$. I have drawn out my attempt below, but in order for this to work, I would need the antistrange quark to decay into an antiup quark and $$W^+$$ boson. Is this possible, and if so, is this the correct Feynman diagram? I couldn't find anything about antistrange quark decay from my limited google skills.

The quark contents of the mesons are as follows:

$$K^+=u\bar{s}$$, $$\pi^0=u\bar{u}$$ (or $$d\bar{d}$$, depending on state), $$\pi^+=u\bar{d}$$

It is correct , see this table of quark decays:

In general, there exists a particle->antiparticle symmetry in the interactions, no separate tables are given for the decays . The table of elementary particles for example is by default followed with the antiparticle table, without need to write it expliscitly.

• Thank you! This is really helpful, cheers. May 31, 2021 at 11:37

Yes, that Feynman diagram is in fact the tree level diagram for the relevant process $$K^+\rightarrow \pi^+\pi^0$$. The interaction term which governs that transition is given by $$W_\mu^+\bar{u}^i_L \gamma^\mu V^{ij}d^j_L$$ where $$u^i = \{u,c,t\}$$ and $$d^i=\{d,s,b\}$$. Given that the CKM matrix elements are all non zero, you have coupling between all the up quarks to all the down quarks.