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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.

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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 \begin{equation} \psi = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} \end{equation} 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 \begin{equation} e_L \mapsto \begin{pmatrix} e_L \\ 0 \end{pmatrix} \end{equation} 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.

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  • $\begingroup$ So I should think of the doublet as $L^T = (\nu_L \quad 0 \quad e_L \quad 0)$? Thank you! $\endgroup$
    – Depenau
    Mar 12, 2023 at 13:25
  • $\begingroup$ 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. $\endgroup$ Mar 12, 2023 at 13:29

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