# What is a fermion doublet exactly?

I am trying to prove that the Weinberg operator is the only dimension-5 operator that can be constructed out of Standard Model fields. To that end, I've tried to write up all the dimension-5 operators I can think of, and then apply Lorentz, $$U(1)_Y$$, and $$SU(2)_L$$ transformations to see which are invariant. Problem is, I'm very unsure what the properties of the fermion doublets, $$L = \begin{pmatrix} \nu_L \\ e_L \end{pmatrix},$$ are. Schwartz writes that it transforms as a left-handed Weyl spinor under $$SU(2)$$, while the singlet $$e_R$$ transforms as a right-handed Weyl spinor. Does this mean that $$\nu_L$$, $$e_L$$ and $$e_R$$ are Weyl spinors?

I think I have a good understanding of Dirac and Weyl spinors, $$\psi = \begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3 \\ \psi_4 \end{pmatrix} = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}, \quad \psi_L = \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix}, \quad \psi_R = \begin{pmatrix} \psi_3 \\ \psi_4 \end{pmatrix}.$$ I know you can build Dirac spinor bilinears that transform nicely under the Lorentz group using the $$\gamma^\mu$$-matrices, $$\overline{\psi}\psi, \quad \overline{\psi}\gamma^5\psi, \quad \overline{\psi}\gamma^\mu\psi, \quad \overline{\psi}\gamma^5\gamma^\mu\psi, \quad \overline{\psi}[\gamma^\mu,\gamma^\nu]\psi,$$ and I know that for Weyl spinors, the $$\sigma^\mu$$-matrices are used instead. For instance, in the Weyl basis, $$\overline{\psi}\gamma^\mu\psi = \begin{pmatrix} \psi_L^\dagger & \psi_R^\dagger \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^\mu \\ \overline{\sigma}^\mu & 0 \end{pmatrix} \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} = \psi_L^\dagger\overline{\sigma}^\mu\psi_L + \psi_R^\dagger\sigma^\mu\psi_R.$$ But when I write something like $$\overline{L}\gamma^\mu L$$, what exactly does it mean? Is this just confusing notation, or does it really mean $$\overline{L}\gamma^\mu L = \overline{\nu}_L\gamma^\mu\nu_L + \overline{e}_L\gamma^\mu e_L,$$ such that each component in the doublet is a Dirac spinor? I'm very confused.

Everything with a subscript of $$L$$ or $$R$$ is a Weyl spinor and always will be. As you note, a Dirac spinor can be expressed in the Weyl basis as $$$$\psi = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$$$$ meaning the bilinear $$\bar{e}_L \gamma^\mu e_L$$ say, which appears in $$\bar{L} \gamma^\mu L$$, should be evaluated by the rule $$$$e_L \mapsto \begin{pmatrix} e_L \\ 0 \end{pmatrix}$$$$ so that only one block of the $$\gamma$$ matrices act on it. There would be less of an abuse of notation if we wrote $$L^\dagger \bar{\sigma}^\mu L$$ to begin with.
• So I should think of the doublet as $L^T = (\nu_L \quad 0 \quad e_L \quad 0)$? Thank you! Commented Mar 12, 2023 at 13:25
• You can do that as long as you remember that the matrices acting on it will be $\gamma^\mu \otimes I = \mathrm{diag}(\gamma^\mu, \gamma^\mu)$. Another option is keeping indices around. Commented Mar 12, 2023 at 13:29