Can Z decay into a neutral pion?

center o bass

I wondered if the following decay was possible:
[tex]\Xi^0 \to \Lambda + \pi^0[/tex]

The only mechanism I can think of is that a strange quark of the Lambda particle emits a Z and becomes a down quark followed by the Z creating a up-antiup pair. But I'm not sure whether the strange to up transition (by emission of Z) is allowed. Neither am I sure about the Z to u-antiu pair is allowed. I know that W can decay into an up and a down quark, but do the Z only interact with neutrinos?

mfb

According to the PDG, this is the dominant decay mode with >99,5% branching fraction.

s->u+W, W->anti-u+d, combine the d with the existing u+s to form the lambda, and use the u anti-u for the pion.

The Z can interact with quarks, but in order to change their flavor with Z emission you need loop processes.

center o bass

Thank you for the reply! So the Z can change flavour anyway? I did some research and found this page on hyperphysics which claimed that a Z boson could not change the flavour of quarks.

http://hyperphysics.phy-astr.gsu.edu...neucur.html#c1

kloptok

It is true that a Z cannot change flavour - there are no so called Flavour Changing Neutral Currents (FCNC) - at tree level (i.e. without loops). A source of confusion might be that FCNC processes are permitted at higher order, that is, when loops are included. See the picture at Wikipedia: http://en.wikipedia.org/wiki/Flavor-...eutral_current

A Z boson will however, never be able to change the flavour of a quark "on its own". So decays of the type s->Z+d are not permitted in the Standard Model.