Drawing Feynman diagrams including $\pi^0$-mesons

Question

I am struggling with drawing a Feynman diagram for particle physics processes including $\pi^0$ mesons and hope you can help my clarifying my confusion.

Take, for example, the following process $$D^+ \rightarrow \pi^0 + e^+ + \nu.$$ I know that $D^+ = |c\bar{d}\rangle$ and $\pi^0 = \frac{1}{\sqrt{2}}(|u\bar{u} \rangle - |d\bar{d}\rangle)$.

On the left side of the process, we have a charm-quark but not on the right hand-side. Thus, we deal with weak interaction. My problem is that I do not know how to deal with the "superposition" of $u\bar{u}$ and $d\bar{d}$ in the $\pi^0$-meson.

I've attached my attempt of a Feynman diagram, but there I only describe the conversion of the charm to a down quark, but not how to create the $u\bar{u}$ part (so, in fact, I explain only "half" of the pion). I would imagine that we need another Z-Boson that might create the $u\bar{u}$, but I do not know where I should get this Z-boson from.

I am grateful for any hint/tip/idea that helps to clarify my confusion. Thank you in advance!

Answer 1

The fact that the particle is described by a superposition of states, does not mean that one has to measure both states at once. That's how superposition works, one one makes a measurements and then the wavefunction collapses on one of the possible superimposing states. This means that the neutral pion, when measured, can be found in the $|d\bar{d}\rangle$ state or the $|u\bar{u}\rangle$ state, not both.

This implies that the Feynman diagram you draw is the right tree-level Feynman diagram for the process you are studying $D^+\to\pi^0\,e^+\nu_e$.

More complex Feynman diagrams can be drawn were you find a $|u\bar{u}\rangle$ state. But this surely would comprise of more than one weak decay.