Getting rid of the theta term in the standard electroweak theory

Question

This has already been asked here more than once, but the existing answers do not tackle my misunderstanding.

A topological $\theta$-term is understood to be physical, in the usual particle model constructions, if it cannot be rotated away by the chiral anomaly simultaneously with a possible $\mathit{CP}$-violating mass phase. That is the case of QCD, in which an $\alpha$ chiral rotation induces a change of $2\alpha$ in $\theta$ and $-2\alpha$ in the mass phase.

Regarding the also non-abelian weak isospin $SU(2)_L$ sector, however, the situation is different. Since, distinctly from QCD, this group is chiral, i.e., only the left-handed fields measure is of consequence, it is said that the $\theta$-term can be rotated away. I do not understand this argument.

My impression is that the partition function can only acquire the $\text{tr}F\wedge F$ increment through axial rotations, i.e., transformations which rotate right and left-handed fields with opposite angle. There is no freedom, in this case, to absorb an arbitrary mass phase in the right-handed fields. What am I missing?

I have also read through standard references which imply that the anomalous symmetry which is used to remove the $\theta$-term is the baryon/lepton number (instead of the axial one), which just further confuses me.

Can someone make these statements precise?

Answer 1

I'm not sure whether this is helpful, but it's in the spirit of Cheng & Li in a schematic toy model of just u and d, without leptons (which only serve to cancel the SM non-abelian anomalies within a family), in the effective SM lagrangian after SSB. Crudely, the following (singlet!) L current has an anomaly, and an explicit breaking term, $$ \partial_\mu (V^\mu-A^\mu)\sim c \theta \operatorname{Tr} F(W)\cdot \tilde{F}(W) + m_u\bar u \gamma_5 u + m_d\bar d \gamma_5 d, $$ where c is an ℏ-dependent quantum anomaly constant and $$ V^\mu \sim \bar u\gamma^\mu u + \bar d\gamma^\mu d,\\ A^\mu \sim \bar u\gamma^\mu \gamma^5 u + \bar d\gamma^\mu \gamma^5 d, $$ the former being the baryon current.

The Ws have no well-defined parity, as it is broken maximally here. The charge of V rotates the quarks with the same phase, but that of A with the opposite. So you transfer θ to the pseudoscalar bilinears through an L-rotation, but you still have an R-rotation available to remove it from there! It's gone.

You see why this logical pathway is not open to parity-preserving QCD!

If you stray from the toy model, in the "real world" (hohoho!), you may observe tweaks that prevent this rotation away of the $θ_{EW}$.