Higher-order weak interaction decays

[MaximumTaco]

Considering the decay of certain exotic mesons, such as the following:

$$B^{+} --> D^{0} \pi^{+}$$

Apparently the decay proceeds via a weak interaction where multiple W boson exchanges occur.

I was trying to nut out how this actually occurs, and draw up a sensible Feynman diagram representation of the process, but i can't quite nut it out.

Recall that the B+ meson is comprised of a u and anti-b quark, and the D0 is a c and anti-up, and the Pi+ is an up and anti-down, FYR.

Could someone offer me any pointers as to how such decay process usually work? First-order weak process such as beta decay are pretty straightforward, but this idea seems a bit more tricky.

 

[Meir Achuz]

bbar-->cbar + W^+.
W^+--> u + dbar.
That u + cbar--> D^0.
The incoming u + dbar--> pi^+.

 

[MaximumTaco]

But doesn't the D^0 meson contain a c, not a cbar ?

 

[Meir Achuz]

Yes. You fooled me with your first equation. It should be
B^+--D^0bar+pi^+.
If you really meant D^0, then the diagram would be a mess and the decay never seen unless you have some completely new theory.

 

[MaximumTaco]

Could you tell me how the decay could proceed anyway, even though it might never occur in practice?

 

[Meir Achuz]

The final state you want, D_0 + pi^+ has the quarks [c ubar] and
[u dbar]. You could get that from [u bbar] by the steps:
1. bbar -->W^+ + cbar.
2. W^+ -->[c] + [dbar].
3. cbar --> W^- + dbar.
4. W^- + u --> d.
5. dbar +d --> [ubar] + (via strong intgeraction).
This leaves you with the right quarks for D_0 + pi^+.
The decay is doubly weak, so I cannot see how it could be observed.