Do I Need My Own Antiparticle to Calculate D^0 Decay to K^++π^-?

[KBriggs]

I am trying to work out the Feynman diagram for the decay

$$D^0\to K^++\pi^-$$

But I can't seem to get it unless the D meson is its own antiparticle. Could someone tell me if this is the case? Ie, is $$|\bar{u}c\rangle = |u\bar{c}\rangle$$?

Thanks

 

[Bill_K]

That's because the decay is actually D⁰ → K^- + π⁺. Maybe you want to start with an anti-D⁰

 

[KBriggs]

No, the question gives three decays, one of which is the one you posted and another is the one I posted.

Could I change D to anti-D by having the two constituent quarks exchange a gluon, so that the decay I posted would then proceed from the antiparticle?

See for example decay modes 42 and 213 here: http://pdg.lbl.gov/2008/listings/s032.pdf

 

[Vanadium 50]

Do you know what Cabibbo suppression is? That decay is doubly Cabibbo suppressed. That should help you.

 

[KBriggs]

Update: seem to work if the c decays into d and a W^+ meson, which then decays into u and anti-s

any problems there?

this decay is apparently the much less likely than the one you posted. Why is that?

EDIT: I don't yet know what cabibbo suppression is. Do you know of a resource that explains it? Google turns up a lot of scholarly stuff that assumes I know it already.