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I want $K = K_R \oplus - K_L$. 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 $\psi^\dagger_R i( \partial_t - v_F \, \partial_x) \psi_R$. This is what the perfect metal paper has.  Mike, this goes against your equation 1.8.  Next. (opposite velocities to PM paper)  $$ \mathcal{L}_f = \psi^\dagger_R i(\partial_t + v_F \partial_x) \psi_R + \psi^\dagger_L i(\partial_t - v_F \partial_x) \psi_L $$  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 Just like  in addition $\psi^\dagger_R the perfect metal paper:  \begin{align}   \psi^\dagger_R &  \propto e^{i \phi_I}, & e^{-i \phi_I} \\  \psi^\dagger_L &  \propto e^{- i \phi_{N+I}}$ & e^{i \phi_{N+I}}  \end{align}  This implies the variations are  $$ \delta \psi_{R,I} = \epsilon(i k_{F,I} +\partial_x \psi_{R,I} )$$