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

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! 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.