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?

  • $\begingroup$ @CosmasZachos thank you for the references. I am indeed familiar with Schwartz's discussion. There, he makes the claim I stated above -- it's validity is not clear to me and I'm asking for details (preferably some equations) so I can grasp exactly what is happening. In the isospin sector, one can perform a chiral rotation to put $\theta = 0$. This induces a mass phase. The claim is one can rotate $f_R$ to absorb this, but this is equivalent to a chiral+vector rotation. The chiral will reinstate $\theta \neq 0$. I don't know where I'm going wrong. $\endgroup$
    – GaloisFan
    Commented Jul 22, 2023 at 16:24
  • $\begingroup$ @CosmasZachos thank you, but I have also already read that first obvious reference which does not even mention the isospin anomaly. All your comments attempted to do so far, as usual, is make the point that I'm missing obvious stuff I should already know; That the question is poorly formulated; Or make me feel dense for not seeing the obvious. Point taken, as usual. $\endgroup$
    – GaloisFan
    Commented Jul 22, 2023 at 20:47
  • $\begingroup$ Apologies if the links are useless. Will delete forthwith. It is assumed you have computed the anomaly and do not see why the EW F-fdual term is not on the right hand side. $\endgroup$ Commented Jul 22, 2023 at 21:17
  • $\begingroup$ @CosmasZachos Apologies for the rudeness. I have not performed the calculation explicitly for a pure right-handed rotation -- instead, I am trying to understand why the line of reasoning I laid out in the first comment fails. That is: a righthanded rotation of the fermion field, to absorb the mass phase, equals a vector plus an axial one. The vector rotation is a true symmetry of the quantum theory at this stage and is of no consequence, both to the $\theta$-term and to the mass terms, but the chiral one should reintroduce the $\theta$-term. In this train of thought, where is the error? $\endgroup$
    – GaloisFan
    Commented Jul 24, 2023 at 19:30
  • $\begingroup$ I'm not sure which currents you are taking about... Write all currents involved and their conceivable divergences, including the ones involving mass-style terms. There are no R-coupling gauge fields, so a R type rotation eliminates all phases from any mass-with-angle terms on the rhside, as here. $\endgroup$ Commented Jul 24, 2023 at 19:44

1 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 $θ_{EW}$.

  • $\begingroup$ Thank you! I'm close to being convinced of what is happening -- the singlet vector current alone seems to be anomalous for weak isospin as there will be V-A currents in the triangle, coming from the chiral interaction itself. That is, both the axial and the vector singlets are anomalous, with cancelling contributions if $\alpha_A = \alpha_V$, which corresponds to a right-handed rotation. Just a question: you stated that the $V(A)$ rotates with opposite(same) phases. Isn't it the other way around? $\endgroup$
    – GaloisFan
    Commented Jul 24, 2023 at 23:08
  • $\begingroup$ Indeed, the weak currents on two of the triangle vertices select only L fermions in the loop, which also couple to V, as well as A. Yes, you are right, by opposite I meant the antiquark has an opposite phase to the quark, so they'd cancel in the bilinears; I'd better change it for future readers; thanks. $\endgroup$ Commented Jul 25, 2023 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.