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Fermion density. Giamarchi claims (eq 2.55), for a non-chiral boson, $ \rho_R(x) + \rho_L(x) = \frac{-1}{\pi}\partial_x \phi $. In addition $\phi_R^{(G))} = K\theta - \phi, \phi_L^{(G))} = K \theta + \phi$. His final action for the chiral bosons looks like (C.12 upon above field redefinition)  $$ -S_G = \frac{1}{4\pi K} \int d\tau dx \left[ i \partial_\tau \phi_R \partial_x \phi_R - i \partial_\tau \phi_L \partial_x \phi_L - ... \right]$$  It looks just like the perfect metal action, but he has (eq 2.22) $\psi_R \propto e^{i k_F x} e^{i \phi_R}, \psi_L \propto e^{-i k_F x} e^{i \phi_L}$ whereas we have $\psi^\dagger $\psi  = e^{-i e^{i  k_F x} \psi^\dagger_R \psi_R  + e^{i e^{-i  k_F x} \psi^\dagger_L$ \psi_L$  where $\psi^\dagger_R $\psi_R  \propto e^{i \phi_I}, \psi^\dagger_L \psi_L  \propto e^{- i \phi_{N+I}}$. That means our chiral fields are off by a minus sign:  $$ \phi_I \equiv \phi^R_I = - \phi_R^{(G))} $$  $$ \phi_{N+I} \equiv \phi^L_I = \phi_L^{(G))} $$