# D meson decay

1. Nov 30, 2011

### 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

2. Nov 30, 2011

### Bill_K

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

3. Nov 30, 2011

### 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

Last edited: Nov 30, 2011
4. Nov 30, 2011

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

5. Nov 30, 2011

### 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.