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If someone would ask me this question I'd say:

"Nothing but their charges. The thing is so because we don't know what exactly happens in the virtual part of the process (say a scattering). Physicists don't know (and, to some extent, they don't really care) if the W being exchanged is going from vertex "a" to vertex "b" or if it's the other way around, so they admit if it goes from a to b, then it is, say, a $W^-$ and if it goes from b to a, then it is a $W^+$. Now, in either case the initial and final states are the same, so it doesn't make any difference wether the boson is positive or negative. Assuming so scientists avoid the problem while it is not essential and can reach very important conclusions."

Though this answer may sound I know what I'm saying I'm not positive about it, so I'd like to know if it's right or not, and how it could be improved.

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    $\begingroup$ Draw the Feynman diagrams for the simple interactions, people love visual aids, you will lose them with words in 5 seconds and end up drawing them anyway. To me at least, seeing how the two diagrams can be combined into one picture reinforces the notion we don't know and we don't care. Apologies if this is too simplistic, but it might improve your answer. $\endgroup$ – user108787 Aug 7 '16 at 0:57
  • $\begingroup$ @count_to_10 That's nice, draws do help a lot. $\endgroup$ – Patrick Aug 7 '16 at 1:12
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The $W^+$ and $W^-$ particles are related to each other by the $CP$, charge parity operations. Since $CPT~=~1$ the they are particle/antiparticles. They are related to each other in much the same way the electron and positron are related to each other.

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