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I am trying to figure out why a term like

## L \sim i \bar \psi_L \sigma^{\mu\nu}G_{\mu\nu}^a t^a \psi_R + h.c=##

##= i \bar \psi_L \sigma^{\mu\nu}G_{\mu\nu}^a t^a \psi_R - i \bar \psi_R \sigma^{\mu\nu}G_{\mu\nu}^a t^a \psi_L ##

violates CP by looking at all the terms composing the Lagrangian.

I made a calculation (with the guidance of http://www.physics.princeton.edu/~mcdonald/examples/EP/feinberg_pr_108_878_57.pdf) and I obtained that the fermionic part withouth the ##\gamma^5##, which goes like ## \bar \psi \gamma^\mu \gamma^\nu \psi ##,

transforms as

## [(-1)] [(-1)^\mu (-1)^\nu]##, first square bracket for C and second for P and #(-1)^\mu=1# for #\mu=0# and -1 otherwise

The term with the ##\gamma^5## transforms as

## [(-1)] [(-1) (-1)^\mu (-1)^\nu ]##

and naively, by writing ##G_{\mu\nu}=\partial_\mu G_\nu - \partial_\nu G_\mu## and taking into account the vector properties of ##G_\mu## and that ##C(G)=-1## I assumed that the field strength transforms as

## [(-1)] [(-1)^\mu (-1)^\nu]##

In this way, under CP, the lagrangian goes into

## L \sim i \bar \psi_R \sigma^{\mu\nu}G_{\mu\nu}^a t^a \psi_L + h.c## which implies CP violation, by comparing to the second term of the original lagrangian, since there is a sign difference. In case where there is no ##\gamma^5## CP is preserved.

However I was then thinking a bit more about this and immediately realised that

##G_{\mu\nu}=\partial_\mu G_\nu - \partial_\nu G_\mu + i [G_\mu,G_\nu]##

it seems to me that the derivative terms have the following CP properties

##\mu=0, \nu=i~~ P=-1, C=-1 \to CP=+1##

##\mu=i, \nu=j ~~ P=+1, C=-1 \to CP=-1##

whereas the ##G^2## proportional term seems to go

##\mu=0, \nu=i ~~ P=-1, C=+1 \to CP=-1##

##\mu=i, \nu=j ~~ P=+1, C=+1 \to CP=+1##

where ##C=+1## since I have two fields that have ##C=-1##. This has opposite transformation property of the derivative part. Where I am mistaking?

Thanks a lot!

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# A CP properties of field strength tensor

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