I am curious about the allowed Feynman diagrams in Muon decay and whether the $W$ boson can be either the $W^+$ or $W^-$. Muon decay follows the reaction: \begin{equation} \mu^-\rightarrow e^-+\overline{\nu}_{e}+\nu_\mu \end{equation}

I am sure the W$^-$ diagram is valid: enter image description here

But I am not sure if the W$^+$ diagram is allowed since here the W$^+$, $\bar{\nu}_e$, $\nu_\mu$ are created out of the vacuum. And created kind of before the muon decays. enter image description here

My question is whether both Feynman diagrams are valid for muon decay or if only the W$^-$ is allowed?

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    $\begingroup$ You are misreading Feynman diagrams, by taking the x-axis to represent time. If the second diagram were thought to bring momentum and energy Into the muon vertex, it is the same diagram as the first. A W+ into the vertex is the same as a W- out of the vertex. $\endgroup$ Dec 4, 2019 at 20:17
  • $\begingroup$ Thanks Cosmas, this is the correct answer. This forum asked the same question: physicsforums.com/threads/muon-decay-feynman-graph.589383 $\endgroup$
    – Adam
    Dec 4, 2019 at 20:33
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    $\begingroup$ If you thought of charge conservation at each vertex, you would see that the arrow of time for the W should be into the vertex, making it a W- $\endgroup$
    – anna v
    Dec 5, 2019 at 6:00

1 Answer 1


These diagrams represent the same process. A $W^+$ going into a vertex is equivalent to a $W^-$ going out of a vertex


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