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  CP violation from the Electroweak SU(2)$_{weak,flavor}$ by $\int \theta F \wedge F $

+ 4 like - 0 dislike

Question: Why there is NO Charge-Parity (CP) violation from a potential Theta term in the electroweak SU(2)$_{weak,flavor}$ sector by $\theta_{electroweak} \int F \wedge F$?

(ps. an explicit calculation is required.)


We know for a non-Abelian gauge theory, the $F \wedge F $ term is nontrivial and breaks $CP$ symmetry (thus break $T$ symmetry by $CPT$ theorem), which is this term: $$ \int F \wedge F $$ with a field strength $F=dA+A\wedge A$.

$\bullet$ SU(3)$_{strong,color}$ QCD:

To describe strong interactions of gluons (which couple quarks), we use QCD with gauge fields of non-Abelian SU(3)$_{color}$ symmetry. This extra term in the QCD Lagrangian: $$ \theta_{QCD} \int G \wedge G =\theta_{QCD} \int d^4x G_{\mu\nu}^a \wedge \tilde{G}^{\mu\nu,a} $$ which any nonzero $\theta_{QCD}$ breaks $CP$ symmetry. (p.s. and there we have the strong CP problem).

$\bullet$ Compare the strong interactions $\theta_{QCD,strong}$ to U(1)$_{em}$ $\theta_{QED}$: For U(1) electromagnetism, even if we have $\theta_{QED} \int F \wedge F$, we can rotate this term and absorb this into the fermion (which couple to U(1)$_{em}$) masses(?). For SU(3) QCD, unlike U(1) electromagnetism, if the quarks are not massless, this term of $\theta_{QCD}$ cannot be rotated away(?) as a trivial $\theta_{QCD}=0$.

$\bullet$ SU(2)$_{weak,flavor}$ electro-weak:

To describe electroweak interactions, we again have gauge fields of non-Abelian SU(2)$_{weak,flavor}$symmetry. Potentially this extra term in the electroweak Lagrangian can break $CP$ symmetry (thus break $T$ symmetry by $CPT$ theorem): $$ \theta_{electroweak} \int F \wedge F =\theta_{electroweak} \int d^4x F_{\mu\nu}^a \wedge \tilde{F}^{\mu\nu,a} $$ here the three components gauge fields $A$ under SU(2) are: ($W^{1}$,$W^{2}$,$W^{3}$) or ($W^{+}$,$W^{-}$,$Z^{0}$) of W and Z bosons.

Question [again as the beginning]: We have only heard of CKM matrix in the weak SU(2) sector to break $CP$ symmetry. Why there is NO CP violation from a potential Theta term of an electroweak SU(2)$_{weak,flavor}$ sector $\theta_{electroweak} \int F \wedge F$? Hint: In other words, how should we rotate the $\theta_{electroweak}$ to be trivial $\theta_{electroweak}=0$? ps. I foresee a reason already, but I wish an explicit calculation is carried out. Thanks a lot!

This post imported from StackExchange Physics at 2014-06-04 11:35 (UCT), posted by SE-user Idear
asked Dec 27, 2013 in Theoretical Physics by wonderich (1,500 points) [ no revision ]
See section 2 of arxiv.org/abs/hep-ph/9305271

This post imported from StackExchange Physics at 2014-06-04 11:35 (UCT), posted by SE-user Mitchell Porter
@ Mitchell, thanks for the comments, let me take a look.

This post imported from StackExchange Physics at 2014-06-04 11:35 (UCT), posted by SE-user Idear
@ Mitchell, it will also be nice if you can summarize your/their viewpoints.

This post imported from StackExchange Physics at 2014-06-04 11:36 (UCT), posted by SE-user Idear

1 Answer

+ 1 like - 0 dislike

Your question is riddled with ^'s in equations making it hard for me to understand the body of your question. If I understand your question "why is there no weak isospin vacuum angle in analogy with the one in QCD?," then I can answer it easily:

Suppose we write that CP-odd term in the Lagrangian. Then, to remove it, all you need to do is to look for a U(1) transformation of the fermion fields that triggers the anomaly in SU(2) of weak isospin [c.f. Fujikawa], but classically leaves the Lagrangian invariant. In other words, we need to look for classical symmetries that are anomalously violated. In the standard model, we have the vector baryon or lepton transformations. So, just do a U(1) lepton transformation by just the right right amount, and the CP-violating term will go away.

What if the neutrinos are Majorana so that lepton transformations are no longer a classical symmetry? No problem! Just do a baryon transformation instead. You can make the CP-odd term go away like that, too.

This post imported from StackExchange Physics at 2014-06-04 11:36 (UCT), posted by SE-user QuantumDot
answered Mar 4, 2014 by QuantumDot (195 points) [ no revision ]

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