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

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Discussion Overview

The discussion revolves around the decay process of the D^0 meson into K^+ and π^-. Participants explore the requirements for constructing a Feynman diagram for this decay, particularly whether the D meson is its own antiparticle and the implications of this on the decay process.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the D meson is its own antiparticle, specifically asking if |\bar{u}c\rangle = |u\bar{c}\rangle.
  • Another participant corrects the decay process, stating that the decay is actually D^0 → K^- + π^+, suggesting the consideration of an anti-D^0 instead.
  • A participant proposes the idea of changing D to anti-D by having the constituent quarks exchange a gluon, raising the possibility of the decay proceeding from the antiparticle.
  • Cabibbo suppression is mentioned as a factor affecting the decay, with one participant noting that the decay in question is doubly Cabibbo suppressed.
  • A later reply discusses the possibility of the c quark decaying into a d quark and a W^+ meson, which then decays into u and anti-s, questioning the likelihood of this decay compared to the previously mentioned one.
  • One participant expresses a lack of understanding of Cabibbo suppression and seeks resources for further explanation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the D meson is its own antiparticle, and there are competing views regarding the decay processes and the role of Cabibbo suppression.

Contextual Notes

There are unresolved assumptions regarding the nature of the D meson and the implications of Cabibbo suppression on the decay processes discussed. The discussion also references specific decay modes without fully exploring their implications.

KBriggs
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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
 
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That's because the decay is actually D0 → K- + π+. Maybe you want to start with an anti-D0
 
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:
Do you know what Cabibbo suppression is? That decay is doubly Cabibbo suppressed. That should help you.
 
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.
 

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