Feynman diagram for $\overline{K}\,\!^0$ antimeson production on the quark-level

Question

I've recently stumbled upon a physics problem concerning $\overline{K}\,\!^0$ antimeson production. In this particular example, colliding a $\pi^-$ meson with a stationary proton yields a $K^0$ meson and a $\Lambda^0$ hyperon:

$$\pi^-\,[\overline{u}d] + p\,[uud]\rightarrow K^0\,[d\overline{s}] + \Lambda^0\,[uds]$$

This can be expressed in a Feynman diagram by letting the $u$ and $\overline{u}$ quarks annihilate to a gluon, out of which a pair of $s$-$\overline{s}$-quarks is generated.

However, if a $\overline{K}\,\!^0$ particle would be generated by the same method, in order to conserve the baryon number and the strangeness, more than just a particle must be produced. For example, the following reaction could take place, so that every quantum number is conserved:

$$\pi^-\,[\overline{u}d] + p\,[uud]\rightarrow \overline{K}\,\!^0\,[s\overline{d}] + K^0\,[d\overline{s}] + n\,[udd]$$

However, I can't seem to find a corresponding Feynman diagram for the reaction. I am guessing that the $\Lambda^0$ hyperon decays weakly and somehow yields the antikaon and the neutron, but I can't figure out how... Does anyone have a clue what the Feynman diagram could be?

Answer 1

There are no very simple diagrams. You need at least one pair production and some kind of flavor changing reaction.

This

includes one pair production and a Drell-Yan flavor change.

There will be others but they will presumably all be equally complicated and therefore unlikely. This will be a low rate event in such systems even when the energy available.