If the lepton doublet transforms like a left-handed Weyl spinor under Lorentz transformations, $$ L \longrightarrow exp\left[\frac{1}{2}(i\theta_j\sigma^j - \beta_j\sigma^j)\right]L = \Lambda_{sL}L, $$ then does the Hermitian adjoint transform like $$ L^\dagger \longrightarrow (\Lambda_{sL}L)^\dagger = L^\dagger\Lambda_{sL}^\dagger = L^\dagger exp\left[-\frac{1}{2}(i\theta_j\sigma^j + \beta_j\sigma^j)\right] = L^\dagger\Lambda_{sR}^{-1}? $$ How does the Dirac adjoint lepton then transform? I've tried $$ \overline{L} \longrightarrow (\Lambda_{sL}L)^\dagger\gamma^0 = L^\dagger\Lambda_{sR}^{-1}\gamma^0. $$ $\Lambda_{sR}^{-1}$ and $\gamma^0$ aren't the same dimensions, so do they commute? Do I then get $$ \overline{L} \longrightarrow \overline{L}\Lambda_{sR}^{-1}, $$ such that theres no difference between the transformation properties of Hermitian adjoint and Dirac adjoint lepton doublets?
1 Answer
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$$(\Lambda)^\dagger \neq \Lambda^{-1}$$ instead $$\gamma^0(\Lambda)^\dagger \gamma^0 = \Lambda^{-1}$$
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$\begingroup$ Isn't that only true for the whole $\Lambda_s = exp[i\theta_{\mu\nu}S^{\mu\nu}]$? $\endgroup$– DepenauMar 14 at 15:09
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$\begingroup$ Lorentz transformation is chirality preserving. So the properties I mentioned are true regardless whether you are talking about the chiral Weyl spinor or the whole Dirac spinor. $\endgroup$– MadMaxMar 14 at 15:17