The leptonic $SU(2)$ left-handed doublet is

$$L_L = \begin{pmatrix}\nu_L\\ e_L\end{pmatrix}.$$

Both $\nu_L$ and $e_L$ are Dirac spinors so both are 4-spinors. The dirac adjoint on 4-spinors is the following:


But this is on a spinor. How would a dirac adjoint on an $SU(2)$ doublet work? I'm having trouble understanding what $\bar{L}_L$ is. I believe the following is correct:


However I want to understand how you would come to this definition of $\bar{L}_L$ or is this just a notation?

  • $\begingroup$ $\gamma_5$ is hermitian, so... What do you mean by "how you would come to that"? $\endgroup$ Jun 17, 2023 at 17:31
  • $\begingroup$ I meant why would the dirac adjoint of a L be a doublet of the dirac adjoints of the two components $\endgroup$ Jun 17, 2023 at 17:32

1 Answer 1


Most good texts cover this. $L_L^i= P_L L^i$, where the i s are SU(2) indices, completely disjoint (direct product) from Dirac (gamma matrix) space.

Since $\gamma_5$ is hermitian, the Dirac adjoint on 4-spinors is
$$\overline { L^i}=L^{i~\dagger}\gamma^0\implies \\ \overline { L^i_L}= L^{i~\dagger} P_L\gamma^0=L^{i~\dagger}\gamma^0 P_R\equiv \overline {L^i_L} P_R.$$ I normally (in a minority) put the L underneath the overbar, to remind readers that at the right of such expressions there is an apparently paradoxical implied chiral R-projector, ultimately turning into an L in the kinetic term, but not in a mass or Yukawa term! It's a stay against confusion.

The weak isospin index i is unaffected by these maneuvers, and if you wanted to represent it as indexing an su(2) column 2-spinor, you might as well transpose the adjoint by laying it on its side as a row spinor, instead, $$\overline{L_L}=(\overline {\nu_L}, \overline{e_L} ) .$$ Some people write it as a column 2-spinor as you do, intending to dot it with the particle column 2-spinor to its side. The above notation doesn't need the dot, and is handier in Yukawa couplings.


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