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\textit{Oh, an empty article!} I want $K = K_R \oplus (minus) K_L$. And$ K_R = K_L = 1$. What does this imply about the signs of the velocities in the fermionic Langrangian?  I think the correct thing is $\psi^\dagger_R i( \partial_t - v \partial_x) \psi_R$. This is what the perfect metal paper has.  Mike, this goes against your equation 1.8.  You can get started by \textbf{double clicking} this text block Next. I agree with your equation 1.7  I want $\psi^\dagger = e^{-i k_F x} \psi^\dagger_R + e^{i k_F x} \psi^\dagger_L$  and begin editing. You can also click in addition $\psi^\dagger_R ~ e^{i \phi_I}, \psi^\dagger_L ~ e^{- i \phi_{N+I}}$  This implies the variations are  $$ \delta \psi_{R,I} = \epsilon(i k_{F,I} +\partial_x \psi_{R,I} )$$  $$ \delta \psi_{L,I} = \epsilon(-i k_{F,I} +\partial_x \psi_{L,I} ) $$  (in particular, we are assuming the Fermi momenta of  the\textbf{Insert} button below to add new block elements. Or you can \textbf{drag and drop an image}  right onto this text. Happy writing! and left movers in channel I coincide)  And similarly, for the bosons, but with no minus sign:  \delta \phi_I = \epsilon( k_{F,I} + \partial_x \phi_I )  \delta \phi_{N+I} = \epsilon( k_{F,I} + \partial_x \phi_{N+I})