# How does the charge conjugated lepton doublet Lorentz transform?

According to Schwartz, left- and right-handed Weyl spinors transform, infinitesimally, like $$\delta\psi_R = \frac{1}{2}(i\theta^j\sigma_j + \beta^j\sigma_j)\psi_R, \quad \delta\psi_L = \frac{1}{2}(i\theta^j\sigma_j - \beta^j\sigma_j)\psi_L.$$ He also mentions that lepton doublets transform like left-handed Weyl spinors. But since (to my understanding) charge conjugation, $$C: \quad \psi \longrightarrow \psi^c = -i\gamma^2\psi^{*},$$ takes left-handed Weyl spinors to right-handed Weyl spinors, my expectation is then that $$\delta L^c = \frac{1}{2}(i\theta^j\sigma_j + \beta^j\sigma_j) L^c,$$ i.e., the charge conjugated lepton doublet should transform like a right-handed Weyl spinor. Problem is, I get this: $$\delta L^c = (\delta L)^c = -i\gamma^2(\delta L)^{*} = -i\gamma^2\left(\frac{1}{2}(i\theta^j\sigma_j - \beta^j\sigma_j)L_s\right)^{*} = -i\gamma^2\frac{1}{2}(-i\theta^j\sigma_j^{*} - \beta^j\sigma_j^*)L_s^{*} = \frac{1}{2}(-i\theta^j\sigma_j^{*} - \beta^j\sigma_j^{*})L_s^c.$$ I can't really reconcile this with the Lorentz transformation of a right-handed Weyl spinor. What's going wrong?