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

Question

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), $\pi^+=u\bar{d}$

Answer 1

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.

Answer 2

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.