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
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.
No comments:
Post a Comment