snooper007
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The following formula appears in P J Mulders's lecture notes
http://www.nat.vu.nl/~mulders/QFT-0E.pdf
{\cal C}~b(k,\lambda)~{\cal C}^{-1}~=~d(k,{\bar \lambda}) (8.18)
where {\cal C} is charge conjugation operator.
\lambda is helicity.
I don't know why there is {\bar {\lambda}} on the right side,
as is well known that charge conjugation can not change helicity, spin, and momentum.
Thanks
http://www.nat.vu.nl/~mulders/QFT-0E.pdf
{\cal C}~b(k,\lambda)~{\cal C}^{-1}~=~d(k,{\bar \lambda}) (8.18)
where {\cal C} is charge conjugation operator.
\lambda is helicity.
I don't know why there is {\bar {\lambda}} on the right side,
as is well known that charge conjugation can not change helicity, spin, and momentum.
Thanks
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