Is it equivalent to talk about a chiral theory in terms of 1. chiral fermions which are differently coupled to the gauge field or 2. fermions coupled to a "vector minus axial" gauge boson?

The question arose noticing that the derivative part of a Lagrangian $$\mathcal{L}_D=\bar\psi_L(i\gamma^\mu D_\mu)\psi_L+\bar\psi_R(i\gamma^\mu\partial_\mu)\psi_R=\bar\psi_L(i\gamma^\mu \partial_\mu+g\gamma^\mu W_\mu)\psi_L+\bar\psi_R(i\gamma^\mu\partial_\mu)\psi_R$$ is the same as $$\mathcal{L}_D=\bar\psi(i\gamma^\mu D_\mu^5)\psi=\bar\psi\left(i\gamma^\mu \partial_\mu+g\gamma^\mu W_\mu\frac{1-\gamma^5}{2}\right)\psi=\bar\psi_L(i\gamma^\mu \partial_\mu+g\gamma^\mu W_\mu)\psi_L+\bar\psi_R(i\gamma^\mu\partial_\mu)\psi_R$$ but I'm not sure that this means there is no difference in the two approaches: in the Standard Model left fermions are in doublets and right fermions in singlets after all.

To be more clear, I'm interested about the chiralily of the setup, the "$\gamma^5$" part of the equation, and about whether to insert such a factor in the fermion or in the boson makes a difference.

  • $\begingroup$ For a singlet $W_\mu$, your two expressions are identical, and there is no difference between you 1. and 2. : The gauge field only couples to L fermions. Your expanding to doublets for non-abelian gauge fields is also a red herring: In the SM the R fermions are also in doublets of $SU(2)_R$, not gauged. They are only singlets under $SU(2)_L$. You may improve your perspective when you learn about the custodial symmetries of the Higgs potential. But Yukawa couplings upend/violate all this!! $\endgroup$ Oct 18, 2021 at 20:18
  • $\begingroup$ @CosmasZachos The equations I wrote come from Peskin&Schroeder p. 735, where they describe "a stripped-down, Abelian version of the coupling of fermions to the weak interaction gauge theory". I don't know if my question is about group theory: if the answer to my question (which I think is pretty clear "are 1. and 2. equivalent") is dependent on the dimension and the commutativity of the gauge group, then yes I guess my question is about group theory. Also, why couldn't $W_\mu=W_\mu^a T^a$, with the fermions in the proper representation? $\endgroup$ Oct 19, 2021 at 14:53

1 Answer 1


For a singlet $W_\mu$, obviously yes: you are really writing the same thing, and it is not relevant whether you consider the chiral projector as part of the fermions or a feature of the chiral gauge field acting on the same fermions, and automatically excluding $R\psi$ from its couplings, as it projects those out. Everybody may understand this in connecting kinetic terms to gauge coupling terms.

The important thing, if you were referring to the Yukawa coupling of the P&S toy model is to also understand the point of their (21.18): the Higgs couples to both $R\psi$ and $L\psi$, even though the W gauge fields don't! (The B field does.) The go-to review for particle physicists is, of course, the PDG.

Because the real-world structure of the Yukawa couplings, the real heart of the SM, despite what many teacher fail to emphasize, is very unfriendly to your second point of view, one almost never uses this "unified" language (2) for the standard model (while keeping it in the back of their minds with the suitable caveats).

Switching off the g' couplings, which have a screwed up chirality, you might still unwisely choose to talk about non-chiral isodoublet fields $$ \Psi\equiv \begin{pmatrix} u\\ d\end{pmatrix}, $$ and (unwisely) organize the terms as per your implied second question.

The Yukawa-free terms you wrote only, may, indeed, be written in terms of doublets, $$ i(\overline{\Psi_L} \gamma^\mu D_\mu\Psi_L+\overline{\Psi_R}\gamma^\mu\partial_\mu\Psi_R) =i(\overline{\Psi_L}(\gamma^\mu \partial_\mu +ig\gamma^\mu {\mathbb W}_\mu) \Psi_L+\overline{\Psi_R}\gamma^\mu\partial_\mu\Psi_R)\\ =i(\overline{\Psi}(\gamma^\mu \partial_\mu +ig\gamma^\mu {\mathbb W}_\mu L) \Psi, $$ where now ${\mathbb W}_\mu=\vec W_\mu\cdot \vec \tau /2 $ are matrices acting on (left) doublets. It is evident you may attach the L projector to them, and skip chiral projectors off the fermions, and appear you have simplified things.

Far from it. You have introduced two symmetries, evident from your separate chiral pieces: a gauged $SU(2)_L$ which acts on the L fermions, and a global $SU(2)_R$, completely independent of it (their generators commute between the two groups), so you may perform rotations in each completely independently, and still have full invariance. So, indeed, you may simplify your structure by attaching a Left projector to the gauge field matrix, and drop the subscripts off the fermion doublets, which might appear simpler.

Far from it. The simplification is restricted to this sector, and hence illusory. Crucially, the $SU(2)_R$ symmetry is absent in the Yukawa sector: I defy you to try... You'll fail. This sector is what makes the SM admirable: it ultimately gives L-gauge-invariant masses to a theory without $SU(2)_R$ invariance!

  • Takeaway: It is the part you did not write, the Yukawa couplings of the SM which makes use of nonchiral isodoublets $\Psi$ pointless and confusing.

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