This is not possible because lepton number is not conserved. The lepton number of the LHS is 1, while the lepton number on the right is 2.

I would say this is possible. My Feynman diagram is mediated by the [tex]z^0[/tex] boson, i'm pretty sure i've put the arrows in correctly.

I would say this is possible. Once again, my Feynman diagram is mediated using the [tex]z^0[/tex] boson.

This is not possible since baryon number is not conserved.

I would appreciate it if someone could check these. I'm pretty sure they're right, i'm a little sceptical that the mediating boson in every possible case is the [tex]z^0[/tex]!

ok, this is not Atomic, Solid State, Comp. Physics and it is Home Work related question and should not be asked here.

Anyhow, a mentor will move this post.

I will help you with this since you provided attempt to solution.

i) Lepton FAMILIY number is violated. On LHS you have muon lepton number +1, and on the RHS you have muon lepton number 0. Similar for the electron lepton number.

ii) Is not ok, look at the muon lepton number and electron lepton number.

iii) is an ok interaction, but Z-bosons do not participate in fermion decays, you must use W+ boson instead in the lowest order Feynman diagram.

iv) A pion is not a baryon, so think of this decay once more ;-)

You should not that the phrase "draw a Feynman diagram" is different from "draw the lowest order Feynman diagram". You can have loops etc which also contribute to the decay rate.

I also see that I should pay more attention to lepton family number.

I understand what you've written apart from why it's a [tex]w^+[/tex] boson. I thought that the [tex]\tau^{+}[/tex] was going to a [tex]\mu^{+}[/tex] so there was no change in charge.
I know there's something wrong with this logic, what is it?

With regards to the final decay, a pion is a meson so [tex]\tau [/tex] lepton number is conserved (do you say it that way?), charge is conserved (+1 either side) and baryon number is the same both sides (0). Hence the interaction can take place! =p

The Feynman diagram of this would be mediated by a [tex]w^-[/tex] (since a [tex]\tau^-[/tex] is a fermion [tex]z^0[/tex] is out of the question!)

Thanks for your help, I think i'm starting to get it.

I could explain to you WHY Z-bosons DON'T participate in fermion decays, but I don't think you will be able to understand it ;-)

Fermionic decays are mediated by W-bosons, the tau goes into a W-boson and tau-neutrino. The W-boson then goes into a muon and an muon-anti-neutrino.

The rule you can use for the Z-boson is that it always goes into a particle and its antiparticle. e.g. Z-> electron + positron, or Z -> muon-neutrino + muon-anti-neutrino, or Z-> up-quark + anti-up-quark ... (and so on) ..

That makes sense. The [tex]\tau^+[/tex] is losing it's charge when it becomes [tex]\overline{\nu_{\tau}}[/tex]. It's getting rid of it's charge by emitting a [tex]w^+[/tex] boson.

The rule at the start of the quote will prove to be incredibly useful!

well I used particle physics by martin, which is still very very basic (it only requires quantum mechanics). But real particle physics require Quantum Field Theory, which is very advanced physics. :-)

The twisting, turning roller-coaster through particle physics has just begun!! XD

Well i'm only in my first year (i'm actually a maths student which is probably why i'm so shoddy at this) and I should probably know the basics well. It's always good to have a solid foundation to build stuff on next year.

So? I must have misunderstood you. I didn't realise that we were restricting the discussion to the SM. Of course, many of the q-numbers discussed are conserved in the SM, but fermion number is at least more general. Did you mean "leptons" when you said "fermions"? I'm just curious because you made the statement somewhat mysteriously.

Just leptons, right? Now that I think about it, I'm not sure whether the following decay is allowed or disallowed in the SM. Ignoring the extreme CKM suppression, does the SM in principle disallow, e.g.

t -> c,Z

followed by

Z -> c,ubar

I would just like to figure out exactly how the participation of the Z is disallowed.