2

I'm reading a paper where they introduce the lepton doublets $L$ and "their $SU(2)_L$ conjugations" $\tilde{L}$, which I'm guessing means $$ \tilde{L} = i\sigma_2L^*. $$ After $\textit{vev}$, they end up with an expression involving charge conjugated weyl spinors $\nu_L^c$ and $e_L^c$, which I think are defined as $$ \nu_L^c = -i\sigma_2\nu_L^* \quad \text{and} \quad e_L^c = i\sigma_2e_L^*, $$ since it works on Dirac spinors like $$ \psi^c = -i\gamma_2e_L^*. $$ But that makes me think that the first equation means $$ \tilde{L} = i\sigma_2L^* = \begin{pmatrix} i\sigma_2\nu_L^* \\ i\sigma_2e_L^* \end{pmatrix}, $$ and not $$ \tilde{L} = i\sigma_2L^* = i \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} \nu_l \\ e_L \end{pmatrix} = i \begin{pmatrix} -e_L^* \\ \nu_L^* \end{pmatrix}, $$ Like I thought. Which is true? And why exactly?

Qmechanic
  • 201,751
Depenau
  • 495
  • 2
    You might have conflated two kinds of Pauli matrices, one for Dirac spinor (as in charge conjugation) and another one for weak doublet (as in $SU(2)$ conjugation). Please double check the definition of the $SU(2)$ conjugation you mentioned. – MadMax Mar 16 '23 at 16:52
  • They don't give the definition in the paper, unfortunately. The definitions I use comes from Schwartz and wikipedia. I don't think they say anything about what kind of Pauli matrix is used. – Depenau Mar 16 '23 at 18:08
  • 1
    Linked, and also. – Cosmas Zachos Mar 16 '23 at 19:04

0 Answers0