# Dirac adjoint of $SU(2)$ lepton doublet

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:

$$\bar{\psi}=\psi^\dagger\gamma^0.$$

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:

$$\bar{L}_L=\begin{pmatrix}\bar{\nu}_L\\\bar{e}_L\end{pmatrix}=\begin{pmatrix}\nu_L^\dagger\gamma^0\\e_L^\dagger\gamma^0\end{pmatrix}.$$

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

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

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.