2
$\begingroup$

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

Fully hadronic K^+ meson decay

$\endgroup$
1
$\begingroup$

It is correct , see this table of quark decays:

enter image description here

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.

$\endgroup$
1
  • $\begingroup$ Thank you! This is really helpful, cheers. $\endgroup$
    – ceruwof
    May 31 at 11:37
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.