Treating charged pion decay $\pi^{-} \rightarrow \ell^{-} + \bar{\nu}_{\ell}$ (where $\ell$ designates a lepton of first- or second-generation) by the representation of the figure below,

pion decay

Griffiths (Eq.~ 9.68) states that the amplitude "must have the general form

$$\mathcal{M} = \frac{g_w^2}{8(M_W c)^2} \; \left[\bar{u}(3) \gamma_{\mu} (1 - \gamma^5) v(2) \right] \, F^{\mu}$$ where $F^{\mu}$ is a 'form factor' describing the $\pi \rightarrow W$ blob."

What is the basis for this? I understand the appearance of one weak vertex factor

$$\frac{-i g_w}{2 \sqrt{2}} \gamma^{\nu}(1-\gamma^5)$$

(for the normal vertex in the diagram) and the presence of the (approximated) propagator $i g_{\mu\nu}/(M_W c)^2$. However the appearance of the factor $[(g_w/(2\sqrt{2})]^2$ suggests an implicit second vertex proportional to $g_w$, or rather, to $i g_w/(2\sqrt{2})$ (though clearly numbers like $2\sqrt{2}$ could as easily have been placed into the magnitude of the needed vector $F^{\mu}$).

  • $\begingroup$ There are two weak vertices: $\bar{u} d $ coupling to the W and the lepton vertex, no? The form factor describes the $\pi$ converting to the quark-antiquark pair. $\endgroup$ – Cosmas Zachos Aug 1 at 18:55
  • $\begingroup$ Yes, but I thought ostensibly the purpose of the "blob" representation was to back away from the (in some sense privileged) knowledge that the $\pi$ is composed of a quark and anti-quark. $\endgroup$ – JayDee.UU Aug 1 at 20:11
  • $\begingroup$ Possibly; however, the complication is in the strong interactions, and not the weak vertex itself, spectacularly simple. So, even ignoring quarks, you want to separate the hadronic (pion) current $F^\mu$ from the W as much as you can, and keep complete control of the weak parts. Call it a convention. $\endgroup$ – Cosmas Zachos Aug 1 at 22:22
  • $\begingroup$ Thanks for the clarification. $\endgroup$ – JayDee.UU Aug 1 at 22:39

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