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from Wiki "The W bosons have a positive and negative electric charge of 1 elementary charge respectively and are each other's antiparticle."

Q:If each is the other's antiparticle then which is retrograde in time according to Feynman-Stueckelberg interpretation (FSI)? I mean if by the FSI antiparticles move backwards in time, and if $W^+$ and $W^-$ are each others antiparticle - then which one moves backwards in time?

EDIT: after noting David's and Ron's points I've added a diagram to illustrate the problem.

Suppose out of the vacuum $W^+$ and $W^-$ are created which then decay to electrons and neutrinos W boson creation

It only makes sense if one of the W bosons is retrograde. I have arbitrarily placed the tilt of the W bosons, so they can be interchanged without loss of generality.

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  • $\begingroup$ What you have draw is the independent decay of two weak bosons near each other. There is no six line vertex in the theory. Indeed, I not aware of a three line vertex with weak bosons (it is certainly not among the ones usually draw as characterizing the electroweak interaction), but I don't know enough to say that it is forbidden. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jun 14, 2012 at 0:59
  • $\begingroup$ @dmckee Doesn't the beta decay implicitly have a three vertex with a $\bar{\nu }_e$ , $e^-$ and $W^-$,but yes your right about the whole interaction being incomprehensible - I'll redraw it . $\endgroup$
    – metzgeer
    Commented Jun 14, 2012 at 1:10
  • $\begingroup$ Yes, of course there is the vertex you describe. My poor typing and proofreading skills strike again. That was meant to be I not aware of a three line vertex with two weak bosons. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jun 14, 2012 at 1:12

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If we treat Feynman diagrams containing these particles seriously (and there are a lot of people who will tell about the ways in which that is not physically realistic and why you should not do it), these particle enter in either the time-like direction or the space-like direction.

For time-like lines it is easy enough to follow the charge through the diagram and give a unique answer to the question. In this case the question is exactly analogous to asking "if electrons and positrons are each others antiparticles which one is going forward in time and which backward".

For space-like lines, however, there isn't really a notion of "forward" or "backward" in time at all---tilt the line a little one way and there is a "unique" answer, tilt it the other way and the other answer is obvious and necessary. The perturbation series does not distinguish between the two "tilts", so we just label the line $W^\pm$ or even just $W$ when we're being lazy.

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  • $\begingroup$ Yes I agree and can see how electrons and positrons are each others antiparticles. Where the positron is the negative energy state of the electron taken in the forward direction of time. But note, a $W^+$ decays into a $e^+$ and e-neutrino - doesn't that imply the $W^+$ is like the positron and is the antiparticle to the $W^-$ ? $\endgroup$
    – metzgeer
    Commented Jun 14, 2012 at 0:00
  • $\begingroup$ Lines are labelled with little arrows, not with + or -, for this reason--- the spacelike orientation contains contributions from either spacelike W+'s or spacelike W-'s, you can't tell which. It is conventional which one is going forward in time--- the paremetrization choice is arbitrary. Also, there are lots of people who will say lots of silly things, one should not give them airtime unless one agrees. $\endgroup$
    – Ron Maimon
    Commented Jun 14, 2012 at 0:10
  • $\begingroup$ @RonMaimon Sorry Ron, mea culpa, I meant the positive and negative meant the charges on the W bosons. $\endgroup$
    – metzgeer
    Commented Jun 14, 2012 at 0:51
  • $\begingroup$ @metzgeer: That comment was not directed at you, but at dmckee. $\endgroup$
    – Ron Maimon
    Commented Jun 14, 2012 at 3:19

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