# Deducing decay processes and Feynman diagrams using Lagrangian and conservation laws

The decay processes of the $W$ bosons are completely governed by the charged current interaction terms of the Standard model:

$$\mathcal{L}_{cc} = ie_{W}\big[W_{\mu}^{+}(\bar{\nu}_{m}\gamma^{\mu}(1-\gamma_{5})e_{m} + V_{mn}\bar{u}_{m}\gamma^{\mu}(1-\gamma_{5})d_{n})\\ + W_{\mu}^{-}(\bar{e}_{m}\gamma^{\mu}(1-\gamma_{5})\nu_{m} + (V^{\dagger})_{mn}\bar{d}_{m}\gamma^{\mu}(1-\gamma_{5})u_{n})\big].$$

I would like to find the possible decay processes of the $W^{-}$ boson.

I would like to find the Feynman diagrams for each of these processes.

I find that I have to use electric charge conservation and lepton charge conservation to deduce that the possible decay processes are

$$W^{-} \to e^{-} \bar{\nu}_{e}$$ $$W^{-} \to \mu^{-}\bar{\nu}_{\mu}$$ $$W^{-} \to \tau^{-}\bar{\nu}_{\tau}$$ $$W^{-} \to d\bar{u}$$ $$W^{-} \to s\bar{c}$$

As the fermion is outgoing, the arrow on its external line in the Feynman diagram must point outwards.

As the anti-fermion is outgoing, the arrow on its external line in the Feynman diagram must point inwards.

1. Is there a way to obtain these possible decay processes without using conservation laws and by simply inspecting the Lagrangian?

2. Is there a way to determine if the arrows on the external fermion and anti-fermion lines must point inwards and outwards without knowing using their particle/anti-particle property and by simply inspecting the Lagrangian?

1. No. The Lagrangian gives you the possible vertices for Feynman diagrams, but it does not straightforwardly encode the possible processes. For a simple example, consider the non-trivial $\gamma\gamma\to\gamma\gamma$ process to which e.g. a box diagram with four external photons and an internal fermion loop contributes, but of which you can see no trace at the level of the Lagrangian of QED itself. However, the distinction between "inspecting the Lagrangian" and "using conservation laws" is weak and artificial, since the conservation laws follow from inspecting the Lagrangian through the lens of Noether's theorem.