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How to figure out whether a force mediator is w+ or w- in feynman diagram?

I always make sure each vertex is zero but sometimes I get my w+ or w- worng.

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  • $\begingroup$ You should just be labelling them as $W$ in any case. $\endgroup$ – jacob1729 May 22 '19 at 12:53
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    $\begingroup$ Did you check for conservation of electric charge? $\endgroup$ – Qmechanic May 22 '19 at 14:46
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You can't tell from a Feynman diagram whether an internal $W$ line represents a $W^+$ going one way, or a $W^-$ going the other way. This is actually one of the key benefits of Feynman diagrams. Older formalisms such as old-fashioned perturbation theory would have represented this as two separate diagrams, but the Feynman propagator accounts for both automatically. For more details, see Schwartz's QFT text.

If you are taking a course where you get marked down for giving the "wrong" assignment, then that course is using some arbitrary convention for which assignment is the "correct" one. In this case the right thing to do is to ask them what convention they want to use. It doesn't actually matter.

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As per this answer it seems to depend on the Feynman diagram itself. In the case where you can say which vertex comes first in time (W is time-like), the charge of the W boson is assigned such that it carries the same charge as the initial and final states. (Time and space in Feynman diagrams are not consistent from source to source, so make sure you know which axis is which!)

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The better idea is to just check the initial charge of ingoing particles and the charge of outgoing particles. For example a neutron in beta decay turns into a proton. Since charge is conserved we know that a negatively charged W boson is emitted.

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