- #1
KBriggs
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I am trying to work out the Feynman diagram for the decay
[tex]D^0\to K^++\pi^-[/tex]
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 [tex]|\bar{u}c\rangle = |u\bar{c}\rangle[/tex]?
Thanks
[tex]D^0\to K^++\pi^-[/tex]
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 [tex]|\bar{u}c\rangle = |u\bar{c}\rangle[/tex]?
Thanks