- #1
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
[tex]{\cal C}~b(k,\lambda)~{\cal C}^{-1}~=~d(k,{\bar \lambda})[/tex] (8.18)
where [tex]{\cal C}[/tex] is charge conjugation operator.
[tex]\lambda[/tex] is helicity.
I don't know why there is [tex]{\bar {\lambda}}[/tex] 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
[tex]{\cal C}~b(k,\lambda)~{\cal C}^{-1}~=~d(k,{\bar \lambda})[/tex] (8.18)
where [tex]{\cal C}[/tex] is charge conjugation operator.
[tex]\lambda[/tex] is helicity.
I don't know why there is [tex]{\bar {\lambda}}[/tex] on the right side,
as is well known that charge conjugation can not change helicity, spin, and momentum.
Thanks
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